heat convection by latif m. jiji - solutions

PROBLEM 1.1

Heat is removed from a rectangular surface by

convection to an ambient fluid at T . The heat transfer

coefficient is h. Surface temperature is given by

sT = 2/1x

A

where A is constant. Determine the steady state heat transfer rate from the plate.

(1) Observations. (i) Heat is removed from the surface by convection. Therefore, Newton's law of cooling is applicable. (ii) Ambient temperature and heat transfer coefficient are uniform. (iii) Surface temperature varies along the rectangle.

(2) Problem Definition. Find the total heat transfer rate by convection from the surface of a plate with a variable surface area and heat transfer coefficient.

(3) Solution Plan. Newton's law of cooling gives the rate of heat transfer by convection. However, in this problem surface temperature is not uniform. This means that the rate of heat transfer varies along the surface. Thus, Newton’s law should be applied to an infinitesimal area dAs and integrated over the entire surface to obtain the total heat transfer.

(4) Plan Execution.

(i) Assumptions. (1) Steady state, (2) negligible radiation, (3) uniform heat transfer coefficient and (4) uniform ambient fluid temperature.

(ii) Analysis. Newton's law of cooling states that

qs = h As (Ts - T ) (a)

where

As = surface area, m2

h = heat transfer coefficient, W/m2-oCqs = rate of surface heat transfer by convection, W

Ts = surface temperature, oC

T = ambient temperature, oC

Applying (a) to an infinitesimal area dAs

d qs = h (Ts - T ) dAs (b)

The next step is to express )(xTs in terms of distance x along the triangle. )(xTs is specified as

sT = 2/1x

A (c)

L

x0 W

x0

L

Wsdq

dx

PROBLEM 1.1 (continued)

The infinitesimal area dAs is given by

dAs = W dx (d) where

x = axial distance, m W = width, m

Substituting (c) and into (b)

d qs = h(2/1x

A - T ) Wdx (e)

Integration of (f) gives qs

qs = sdq =

L

dxTAxhW0

2/1 )( (f)

Evaluating the integral in (f)

LTALhWqs2/12

Rewrite the above

TALhWLqs2/12 (g)

Note that at x = L surface temperature )(LTs is given by (c) as

2/1)( ALLTs (h)

(h) into (g)

TLThWLq ss )(2 (i)

(iii) Checking. Dimensional check: According to (c) units of C are 2/1o C/m . Therefore units

sq in (g) are W.

Limiting checks: If h = 0 then qs = 0. Similarly, if W = 0 or L = 0 then qs = 0. Equation (i)

satisfies these limiting cases.

(5) Comments. Integration is necessary because surface temperature is variable.. The same procedure can be followed if the ambient temperature or heat transfer coefficient is non-uniform.

PROBLEM 1.2

A right angle triangle is at a uniform surface temperature Ts. Heat is removed by convection to

an ambient fluid at T . The heat transfer coefficient h varies along the surface according to

h = C

x1 2/

where C is constant and x is distance along the base measured from the apex. Determine the

total heat transfer rate from the triangle.

(1) Observations. (i) Heat is removed from the surface by convection. Therefore, Newton's law of cooling may be helpful. (ii) Ambient temperature and surface temperature are uniform. (iii) Surface area and heat transfer coefficient vary along the triangle.


(3) Solution Plan. Newton's law of cooling gives the rate of heat transfer by convection. However, in this problem surface area and heat transfer coefficient are not uniform. This means that the rate of heat transfer varies along the surface. Thus, Newton’s law should be applied to an infinitesimal area dAs

and integrated over the entire surface to obtain the total heat transfer.

(4) Plan Execution.

(i) Assumptions. (1) Steady state, (2) negligible radiation and (3) uniform ambient fluid temperature.


qs = h As (Ts - T ) (a)

where





Applying (a) to an infinitesimal area dAs

d qs = h (Ts - T ) dAs (b)

The next step is to express h and dAs in terms of distance x along the triangle. The heat transfer coefficient h is given by

h = C

x1 2/ (c)

The infinitesimal area dAs is given by

W

x

L

dx

dqs


dAs = y(x) dx (d) where

x = distance along base of triangle, m y(x) = height of the element dAs, m

Similarity of triangles give

y(x) = W

Lx (e)

where

L = base of triangle, m W = height of triangle, m

Substituting (c), (d) and (e) into (b)

d qs = C

x1 2/(Ts - T )

W

Lx dx (f)

Integration of (f) gives qs. Keeping in mind that C, L, W, Ts and T are constants, (f) gives

qs = sdq = )( TTL

WCs

L

x

x

02/1

dx (g)

Evaluating the integral in (g)

qs =2

3C W L

1/2 (Ts - T ) (h)

(iii) Checking. Dimensional check: According to (c) units of C are W/m3/2-oC. Therefore units of qs in (h) are

qs = C(W/m3/2-oC) W(m) L1/2(m1/2) (Ts - T )(oC) = W

Limiting checks: If h = 0 (that is C = 0) then qs = 0. Similarly, if W = 0 or L = 0 or Ts = T

then qs = 0. Equation (h) satisfies these limiting cases.

(5) Comments. Integration was necessary because both area and heat transfer coefficient vary with distance along the triangle. The same procedure can be followed if the ambient temperature or surface temperature is non-uniform.

PROBLEM 1.3

A high intensity light bulb with surface heat flux sAq )/( is cooled by a fluid at T . Sketch the

fluid temperature profiles for three values of the heat transfer coefficients: h1, h2, and h3, where

h1 < h2 < h3.

(1) Observations. (i) Heat flux leaving the surface is specified (fixed). (ii) Heat loss from the surface is by convection and radiation. (iii) Convection is described by Newton's law of cooling. (iv). Changing the heat transfer coefficient affects temperature distribution. (v). Surface temperature decreases as the heat transfer coefficient is increased. (vi) Surface temperature gradient is described by Fourier’s law.(vii) Ambient temperature is constant.

(2) Problem Definition. Determine effect of heat transfer coefficient on surface temperature and surface gradient..

(3) Solution Plan. (i) Apply Newton's law of cooling to examine surface temperature. (ii) Apply Fourier’s law to determine temperature gradient at the surface.

(4) Plan Execution.

(i) Assumptions. (1) Steady state, (2) no radiation ,(3) uniform ambient fluid temperature and (4) constant properties.

(ii) Analysis. Newton’s law of cooling

)(/ TThAq ss (a)

Solve for sT

h

AqTT s

s

)/( (b)

This result shows that for constant sAq )/( surface temperature decreases as h is increased.

Apply Fourier’s law

0

/

y

sy

TkAq (c)

where y is the distance normal to the surface. Rewrite (c)

k

Aq

y

T w

y

/

0

(d)

This shows that temperature gradient at the surface remains constant independent of h. Based on (b) and (d) the temperature profiles corresponding to three values of h are shown in the sketch.

(iii) Checking. Dimensional check: (1) Each term in (b) has units of temperature

C)Cw/m(

)w/m()/()C()C( o

o2

2oo

h

AqTT s

s

T

sTT

y

2h

1h

3hsAq )/(


(2) Each term in (d) has units of C/mo

C/mC)W/m-(

)C/m(/C/m)( o

o

2oo

0 k

Aq

y

T w

y

Limiting check: (i) for h = 0 (no heat leaves the surface), surface temperature is infinite. Set h = 0

in (b) gives .sT

(5) Comments. Temperature gradient at the surface is the same for all values of h as long as the thermal conductivity of the fluid is constant and radiation is neglected.

PROBLEM 1.4

Explain why fanning gives a cool sensation.

(1) Observations. (i) Metabolic heat leaves body at the skin by convection and radiation. (ii) Convection is described by Newton's law of cooling. (iii). Fanning increases the heat transfer coefficient and affects temperature distribution, including surface temperature. (iv). Surface temperature decreases as the heat transfer coefficient is increased. (v) Surface temperature is described by Newton’s law of cooling. (vi) Ambient temperature is constant.

(2) Problem Definition. Determine effect of heat transfer coefficient on surface temperature.

(3) Solution Plan. Apply Newton's law of cooling to examine surface temperature.

(4) Plan Execution.

(i) Assumptions. (1) Steady state, (2) no radiation ,(3) uniform ambient fluid temperature, (4) constant surface heat flux and (5) constant properties.


)( TThq ss (a)

where

h = heat transfer coefficient, CW/m o2

sq surface heat flux, 2W/m

sT = surface temperature, Co

T =ambient temperature, Co

Solve (a) for sT

h

qTT s

s (b)

This result shows that for constant sq , surface temperature decreases as h is increased. Since

fanning increases h it follows that it lowers surface temperature and gives a cooling sensation.

(iii) Checking. Dimensional check: Each term in (b) has units of temperature

C)Cw/m(

)w/m()C()C( o

o2

2oo

h

qTT s

s

T

y

TsT

sq

fan

fanno

skin


Limiting check: for h = 0 (no heat leaves the surface), surface temperature is infinite. Set h = 0 in

(b) gives .sT

(5) Comments. (i) The analysis is based on the assumption that surface heat flux remains constant. (ii) Although surface temperature decreases with fanning, temperature gradient at the surface remains constant. This follows from the application of Fourier’s law at the surface

s

sy

Tkq

Solving for syT )/(

k

q

y

T s

s

constant

PROBLEM 1.5

A block of ice is submerged in water at the melting temperature. Explain why stirring the water

accelerates the melting rate.

(1) Observations. (i) Melting rate of ice depends on the rate of heat added at the surface. (ii) Heat is added to the ice from the water by convection. (iii) Newton's law of cooling is applicable. (iv). Stirring increases surface temperature gradient and the heat transfer coefficient. An increase in gradient or h increases the rate of heat transfer. (v) Surface temperature remains constant equal to the melting temperature of ice. (vi) water temperature is constant.

(2) Problem Definition. Determine effect of stirring on surface heat flux.

(3) Solution Plan. Apply Newton's law of cooling to examine surface heat flux.

(4) Plan Execution.

(i) Assumptions. (1) no radiation ,(2) uniform water temperature, (3) constant melting (surface) temperature.


)( TThq ss (a)

where

h = heat transfer coefficient, CW/m o2

sq surface heat flux, 2W/m


T =ambient water temperature, Co

Stirring increases h . Thus, according to (a) surface heat flux increases with stirring. This will accelerate melting.

(iii) Checking. Dimensional check: Each term in (a) has units of heat flux.

Limiting check: For sTT (water and ice are at the same temperature), no heat will be added to

the ice. Set sTT in (a) gives .0sq

(5) Comments. An increase in h is a consequence of an increase in surface temperature gradient. Application of Fourier’s law at the surface gives

s

sy

Tkq (b)

y

Tsq

0sT

stirringstirringno

ice

ice

water


Combining (a) and (b)

TT

y

Tk

hs

s (c)

According to (c), for constant sT and T , increasing surface temperature gradient increases h.

PROBLEM 1.6

Consider steady state, incompressible, axisymmetric parallel flow in a tube of radius or . The

axial velocity distribution for this flow is given by

)(2

2

1or

ruu

where u is the mean or average axial velocity. Determine the three components of the total

acceleration for this flow.

(1) Observations. (i) This problem is described by cylindrical coordinates. (ii) For parallel

streamlines 0vv r . (iii) Axial velocity is independent of axial and angular distance.

(2) Problem Definition. Determine the total acceleration in the r, and z directions.

(3) Solution Plan. Apply total derivative in cylindrical coordinates.

(4) Plan Execution.

(ii) Assumptions. (1) Constant radius tube, (2) constant density and (3) streamlines are parallel to surface.

(ii) Analysis. Total acceleration in cylindrical coordinates is given by

tzrrr

Dd rrz

rr

rr vvv

vvvvv

Dt

v

dt

v2

(1.23a)

tzrrr

Ddz

rr

vvv

vvvvvv

Dt

v

dt

v (1.23b)

tzrr

Dd zzz

zzr

zz vvv

vvvv

Dt

v

dt

v (1.23c)

For streamlines parallel to surface

0vv r (a)

The axial velocity zu v is given by

)(2

2

1o

zr

ruuv (b)

From (b) it follows that

0tz

zz vv (c)

Substituting into (1.23a), (1.23b) and (1.23c)

Radial acceleration: 0Dt

v

dt

v rr Dd

Angular acceleration;Dt

v

dt

v Dd0


Axial acceleration: 0Dt

v

dt

v zz Dd

(5) Comments. All three acceleration components vanish for this flow.

PROBLEM 1.7

Consider transient flow in the neighborhood of a vortex line where the

velocity is in the tangential direction given by

t

r

rtrV o

4exp1

2),(

2

Here r is the radial coordinate, t is time, o is circulation

(constant) is kinematic viscosity. Determine the three components

of total acceleration.

(1) Observations. (i) This problem is described by cylindrical coordinates. (ii) streamlines are

concentric circles. Thus the velocity component in the radial direction vanishes ( 0rv ). (iii)

For one-dimensional flow there is no motion in the z-direction ( 0zv ). (iv) The -velocity

component, v , depends on distance r and time t.

(2) Problem Definition. Determine the total acceleration in the r, and z directions.

(3) Solution Plan. Apply total derivative in cylindrical coordinates.

(4) Plan Execution.

(ii) Assumptions. (1) streamlines are concentric circles (2) no motion in the z-direction.

(ii) Analysis. Total acceleration in cylindrical coordinates is given by

The three components of the total acceleration in the cylindrical coordinates zr ,, are

tzrrr

Dd rrz

rrr

rr vvv

vvvvv

Dt

v

dt

v2

(1.23a)

tzrrr

Ddz

rr

vvv

vvvvvv

Dt

v

dt

v (1.23b)

tzrr

Dd zzz

zzr

zz vvv

vvvv

Dt

v

dt

v (1.23c)

For the flow under consideration the three velocity component, ,rv v and zv are

0rv (a)

t

r

rtrv o

4exp1

2),(

2

(b)

0zv (c)

Radial acceleration: (a) and (c) into (1.23a)

Vr


r

D r2v

Dt

v (d)

(b) into (d) 2

2

32

2

4exp1

4

)

t

r

r

D or

Dt

v(e)

Tangential acceleration: (a) and (c) into (1.23b)

t

D v

Dt

v (f)

(b) into (f)

t

r

tt

r

r

vD o

4exp

1

42

22

Dt (g)

Axial acceleration: (a) and into (1.23c)

0Dt

v zD (h)

(iii) Checking. Dimensional check: Units of acceleration in (e) and (g) are .m/s2 Note that

according to (b), units of o are /sm2 and the exponent of the exponential is dimensionless.

Thus units of (e) are

22

332

242

4exp1

)m(4

)/sm()

t

r

r

vD or

Dt= 2m/s

Units of (g) are

t

r

tt

r

rDt

Dv o

4exp

)s(

1

)s()/sm(4

)m(

)m(2

)/sm( 2

2

222

= 2m/s

Limiting check: (1) For 0o , all acceleration components vanish. Setting 0o in (e) and

(g) gives 0Dt

v

Dt

v DD r

(2) According to (b) at t , the tangential velocity vanishes ( v = 0). Thus all acceleration

components should vanish. Setting t in (e) and (g) gives 0Dt

v

Dt

v DD r .

(5) Comments. The three velocity components must be known to determine the three acceleration components.

PROBLEM 1.8

An infinitely large plate is suddenly moved parallel to its surface with a velocity oU . The

resulting transient velocity distribution of the surrounding fluid is given by

0

2 )exp()/2(1 dUu o

where the variable is defined as

t

ytx

2),(

Here t is time, y is the vertical coordinate and is kinematic viscosity. Note that streamlines

for this flow are parallel to the plate. Determine the three components of total acceleration.

(1) Observations. (i) This problem is described by Cartesian coordinates. (ii) For parallel streamlines the y-velocity component 0v . (iii) For one-dimensional flow there is no motion in

the z-direction (w = 0). The x-velocity component depends on distance y and time t.

(2) Problem Definition. Determine the total acceleration in the x, y and z directions.

(3) Solution Plan. Apply total derivative in Cartesian coordinates.

(4) Plan Execution.

(ii) Assumptions. (1) streamlines are parallel to surface and (2) no motion in the z-direction.

(ii) Analysis. Total acceleration in Cartesian coordinates is given by

t

f

z

fw

y

f

x

fu

Dt

Df

dt

dfv (1.21)

where f represents any of the three velocity components u, v or w. The x-velocity component u is given by

0

2 )exp()/2(1 dUu o (a)

where

t

ytx

2),( (b)

Note that u depends on y and t only. For one-dimensional parallel flow

0wv (c)

Total acceleration in the x-direction, xa . Set f = u in (1.21)

t

u

z

uw

y

u

x

uu

Dt

Du

dt

duax v (d)

oUx

y

plate 0


Since u depends on y and t only, it follows that

0x

u (e)

Substitute (c) and (e) into (d)

t

uax (f)

This derivative is obtained using the chain rule

td

du

t

uax (g)

Using (a)

)exp(2 2oU

d

du(h)

Using (b)

ttt

yt

y

t 4

1

42

2/3 (i)

Substitute (h) and (i) into (g)

t

U

t

ua o

x

)exp(

2

2

(g)

Total acceleration in the y-direction, ya . Set f = v in (1.21)

tzw

yxu

Dt

D

dt

day

vvvv

vvv (h)

Apply (c) to (h)

0ya (i)

Total acceleration in the z-direction, za . Set f = w in (1.21)

t

w

z

ww

y

w

x

wu

Dt

Dw

dt

dwaw v (j)

Apply (c) to (h)

0za (k)

(iii) Checking. Dimensional check: Units of acceleration in (g) are /s.m2 Note that is

dimensionless. Thus units of (g) are

/sm)s(

)exp(

2

)m/s( 22

t

Ua o

x

Limiting check: (1) For 0oU , the acceleration .0xa Setting 0oU in (g) gives .0xa

(2) According to (b) at t , .0),(y Evaluation (a) at 0),(y gives


oUyu ),( (l)

Since u is constant every where it follows that the xa must be zero. Setting 0 and t in

(g) gives .0xa

(5) Comments. The three velocity components must be known to determine the three acceleration components.

PROBLEM 1.9

Consider two parallel plates with the lower plate stationary and the upper plate moving with a

velocity .oU The lower plate is maintained at temperature 1T and the upper plate at .oT The

axial velocity of the fluid for steady state and parallel streamlines is given by

H

yUu o

where H is the distance between the two plates.

Temperature distribution is given by

11

22

)(2

TH

yTT

H

yy

kH

UT o

o

where k is thermal conductivity and is viscosity. Determine the total temperature derivative.

(1) Observations. (i) This problem is described by Cartesian coordinates. (ii) For parallel streamlines the y-velocity component 0v . (iii) For one-dimensional flow there is no motion in

the z-direction (w = 0). The x-velocity component depends on distance y only.

(2) Problem Definition. Determine the total temperature derivative.

(3) Solution Plan. Apply total derivative in Cartesian coordinates.

(4) Plan Execution.

(ii) Assumptions. (1) streamlines are parallel to surface, (2) no motion in the z-direction and (3) temperature distribution s one dimensional, ).(yTT

(ii) Analysis. Total acceleration in Cartesian coordinates is given by

t

f

z

fw

y

f

x

fu

Dt

Df

dt

dfv (1.21)

where f represents temperature. Let f = T in (1.21)

t

T

z

Tw

y

T

x

Tu

Dt

DT

dt

dTv (a)

where

H

yUu o (b)

and

v = w = 0 (c)

Temperature distribution is given by

11

22

)(2

TH

yTT

H

yy

kH

UT o

o (d)

Using (d)

x

y

oU

0

oT

1T


0t

T

x

T (e)

Substituting (b), (c) and (e) into

0Dt

DT

dt

dT (f)

(iii) Checking. Dimensional check: Each term in (d) has units of temperature.

(5) Comments. Velocity and temperature distribution must be know in order to determine the total derivative of temperature.

PROBLEM 1.10

One side of a thin plate is heated electrically such

that surface heat flux is uniform. The opposite side

of the plate is cooled by convection. The upstream

velocity is V and temperature is T . Experiments

were carried out at two upstream velocities, 1V

and 2V where .12 VV All other conditions were unchanged. The heat transfer coefficient

was found to increase as the free stream velocity is increased. Sketch the temperature profile T(y) of the fluid corresponding to the two velocities.

(1) Observations. (i) Heat flux leaving the surface is specified (fixed). (ii) Heat loss from the surface is by convection and radiation. (iii) Convection is described by Newton's law of cooling. (iv). Changing the heat transfer coefficient affects temperature distribution. (v). Surface temperature decreases as the heat transfer coefficient is increased. (vi) Surface temperature gradient is described by Fourier’s law.(vii) Ambient temperature is constant.

(2) Problem Definition. Determine effect of heat transfer coefficient on surface temperature and surface gradient..

(3) Solution Plan. (i) Apply Newton's law of cooling to examine surface temperature. (ii) Apply Fourier’s law to determine temperature gradient at the surface.

(4) Plan Execution.

(i) Assumptions. (1) Steady state, (2) no radiation ,(3) uniform ambient fluid temperature and (4) constant properties.


)( TThq so (a)

Solve for sT

h

qTT o

s (b)

This result shows that for constant oq , surface temperature decreases as h is increased. Apply

Fourier’s law

0y

oy

Tkq (c)

where y is the distance normal to the surface. Rewrite (c)

k

q

y

T o

y 0

(d)

This shows that temperature gradient at the surface remains constant independent of h. Based on (b) and

oq

x

y

T

V

T

T

y

2h

1h

oq2sT


(d) the temperature profiles corresponding to two values of h are shown in the sketch.

(iii) Checking. Dimensional check: (1) Each term in (b) has units of temperature

C)Cw/m(

)w/m()/()C()C( o

o2

2oo

h

AqTT w

s

(2) Each term in (d) has units of C/mo

C/mC)W/m-(

)C/m(/C/m)( o

o

2oo

0 k

Aq

y

T w

y

Limiting check: (i) for h = 0 (no heat leaves the surface), surface temperature is infinite. Set h = 0

in (b) gives .sT

(5) Comments. Temperature gradient at the surface is the same for all values of h as long as the thermal conductivity of the fluid is constant and radiation is neglected.

PROBLEM 1.11

Heat is removed from an L-shaped area by convection. The heat

transfer coefficient is h and the ambient temperature is .T Surface

temperature varies according to

xceos TxT )(

where c and oT are constants. Determine the rate of heat transfer

from the area.

(1) Observations. (i) Heat is removed from the surface by convection. Therefore, Newton's law of cooling is applicable. (ii) Ambient temperature and heat transfer coefficient are uniform. (iii) Surface temperature varies along the area. (iv) The area varies with distance x.


(3) Solution Plan. Newton's law of cooling gives the rate of heat transfer by convection. However, in this problem surface temperature is not uniform. This means that the rate of heat transfer varies along the surface. Thus, Newton’s law should be applied to an infinitesimal area dAs and integrated over the entire surface to obtain the total heat transfer.

(4) Plan Execution.

(i) Assumptions. (1) Steady state, (2) negligible radiation, (3) uniform heat transfer coefficient and (4) uniform ambient fluid temperature.


qs = h As (Ts - T ) (a)

where





Since the L-shaped area varies with distance x, it is divided into two parts, 1 and 2, each having constant width. Applying

(a) to an infinitesimal area 1sdA

d 11 )( sss dATThq (b)

Integration of (b) from x = 0 to x = a gives the total heat from area 1

a2

0

a

a

a

a

a2x

x0

21

a

a

a2

a

sdq

dx dx


a

sss dATThq0

11 )( (c)

Similarly, for area 2

d 2)(2 sss dATThq (d)

Integration form x = a to x = 2a gives the total heat from area 2

a

asss dATThq

2

2)(2 (e)

1sdA and 2sdA are given by

dxadAs1 (f)

dxadAs 22 (g)

Surface temperature )(xTs is specified as

xceos TxT )( (h)

Substitute (f) and (h) into (c)

dxTTahq

axceos

0

1 )(

Evaluate the integral

aTc

Tahq caeo

s )1(1 (i)

Similarly, (g) and (h) into (e)

dxTTahq

a

a

xceos

2

)(22


aTc

Tahq caca eeo

s )(2 22 (j)

The total heat transfer from the L-shaped area is

21 sss qqq aTc

TahaT

c

Tahq cacaca eee oo

s )(2)1( 21

Rearrange

12 2

oos

T

TacTh

c

aq caca ee (k)

(iii) Checking. Dimensional check: According to (a) units of c are1/m . Therefore units sq

each term in the bracket of (k) is dimensionless and the coefficient has units of W.


Limiting checks: (1) If h = 0 then qs = 0. Similarly, if a = 0 the area vanishes and qs = 0.

Equation (i) satisfies these limiting cases.

(2) If 0oT , the entire surface is at uniform temperature .0sT Application of Newton’s law

of cooling (a) gives

hTaqs23 (l)

Setting 0oT in (k) gives same result.

(5) Comments. Integration is necessary because surface temperature is variable. The same procedure can be followed if the ambient temperature or heat transfer coefficient is non-uniform.

PROBLEM 2.1

[a] Consider transient (unsteady), incompressible, three dimensional flow. Write the continuity

equation in Cartesian coordinates for this flow.

[b] Repeat [a] for steady state.

The continuity equation in Cartesian coordinates is

0wzy

uxt

v (2.2a)

[a] For incompressible flow the density is constant. Thus the can be taken out of the

differentiation sign. In addition, for constant density

0t

Equation (2.2a) becomes

0z

w

yx

u v (a)

[b] Equation (a) holds for steady state as well.

PROBLEM 2.2

Far away from the inlet of a tube, entrance

effects diminish and stream lines become

parallel and the flow is referred to as fully

developed. Write the continuity equation in

the fully developed region for incompressible

fluid.

(1) Observations. (i) The fluid is incompressible. (ii) Radial and tangential velocity components are zero. (iii) Streamlines are parallel. (iv) Cylindrical geometry.

(2) Problem Definition. Simplify the continuity equation for this flow.

(3) Solution plan. Apply continuity equation in cylindrical coordinates.

(4) Plan Execution.

(i) Assumptions. (1) Incompressible fluid and (2) radial and tangential velocity components are zero.

(ii) Analysis. The continuity equation in cylindrical coordinates is given by (2.4)

011

zrzr

rrrt

vvv (2.4)

This equation is simplified for:

Incompressible fluid: is constant, 0t

Parallel streamlines (no radial velocity): 0rv

Parallel streamlines (no tangential velocity): 0v

Introducing the above simplifications into (2.4), gives

0z

zv (a)

this result shows that the axial velocity component is invariant with z.

(iii) Checking. Dimensional check: Each term in (2.4) has units of density per unit time.

(5) Comments. (i) The axial velocity varies with radial distance only. (ii) Equation (a) holds for unsteady state as well. The reason is because for incompressible flow steady or unsteady the following applies

0t

(b)

r r

z

ullyf developed

PROBLEM 2.3

Consider incompressible flow between parallel

plates. Far away from the entrance the axial

velocity component does not vary with the axial

distance.

[a] Determine the velocity component in the y-

direction.

[b] Does your result in [a] hold for steady as well as unsteady flow? Explain.

(1) Observations. (i) The fluid is incompressible. (ii) axial velocity is invariant with axial distance. (iii) Plates are parallel. (iv) Cartesian geometry.

(2) Problem Definition. Determine the velocity component v in the y-direction.

(3) Solution plan. Apply continuity equation.

(4) Plan Execution.

(i) Assumptions. (1) Incompressible fluid, (2) axial velocity is invariant with axial distance and (3) two-dimensional flow.

(ii) Analysis. The continuity equation in Cartesian coordinates is

0wzy

uxt

v (2.2a)

For incompressible flow the density is constant. Thus the can be taken out of the

differentiation sign. In addition, for constant density

0t

(a)

Since the axial velocity u is invariant with axial distance x, it follows that

0x

u (b)

For two-dimensional flow

0z

(c)

(a)-(c) into (2.2a)

0y

v (d)

Integrating (d)

),( txfv (e)


At the wall the velocity must vanish. Thus

0v everywhere in the flow field (f)

[b] Equation (f) holds for steady state as well.

PROBLEM 2.4

The radial and tangential velocity components for

incompressible flow through a tube are zero. Show

that the axial velocity does not change in the flow

direction. Is this valid for steady as well as

transient flow?

(1) Observations. (i) The fluid is incompressible. (ii) Radial and tangential velocity components are zero. (iii) Streamlines are parallel. (iv) Cylindrical geometry.

(2) Problem Definition. Show that 0z

v z .

(3) Solution plan. Apply continuity equation in cylindrical coordinates.

(4) Plan Execution.

(i) Assumptions. (1) Incompressible fluid and (2) radial and tangential velocity components are zero.

(ii) Analysis. The continuity equation in cylindrical coordinates is given by (2.4)

011

zrzr

rrrt

vvv (2.4)



No radial velocity: 0rv

No tangential velocity 0v

Introducing the above simplifications into (2.4), gives

0z

zv (a)

this result shows that the axial velocity component is invariant with z.

Equation (a) holds for unsteady state as well. The reason is because for incompressible flow steady or unsteady the following applies

0t

(b)

(iii) Checking. Dimensional check: Each term in (2.4) has units of density per unit time.

(5) Comments. (i) Since the radial and tangential velocity component vanishes everywhere in the flow field, it follows that the streamlines are parallel to the surface. (ii) The axial velocity varies with radial distance only.

PROBLEM 2.5

Show that yxxy

(1) Observations. (i) Shearing stresses are tangential surface forces. (ii) xy and yx are

shearing stresses in a Cartesian coordinate system.(iii) Tangential forces on an element result in angular rotation of the element. (iv) If the net external torque on an element is zero its angular acceleration will vanish.

(2) Problem Definition. Find the relationship between xy and yx acting on an element.

(3) Solution Plan. Apply Newton’s law of angular motion to an element dydx .

(4) Plan Execution.

(i) Assumptions. Continuum,

(ii) Analysis. Consider an element dydx with tangential shearing stresses acting on its four

sides. The depth of the element is unity. Apply Newton’s law of angular motion

I0 (a)

where

I = moment of inertia about 0, 2mKg

= angular acceleration, 2rad/s

0 = torque about center 0, m-N

Note that normal forces (pressure and normal stress, not shown) exert no torque on the element since their resultants pass through the center 0. The moment of inertia of the element dydx about the

center 0 is

dxdydydx

I12

)()( 22

(b)

The sum of all external torques acting on the element due to shearing stresses is

2222][][0

dxdydx

x

dxdy

dydxdy

y

dydx

xyxyxy

yxyxyx (c)

Note that in the above each shearing stress is multiplied by area to obtain force and by the arm to give torque. Equation (c) is simplified by neglecting third order

dxdydx

dydx

dydy

dxdy

dx xyyxxyxyyxyx )(2222

0 (d)

dx

dy

y

x

yx

dyy

yxyx

dxx

xyxy

xy

0


Substituting (b) and (d) into (a)

dxdydydx

dxdyxyyx12

)()()(

22

Simplify

12

)()()(

22 dydxxyyx (e)

The right hand side of (e) is of higher order and thus can be neglected to give

0)( xyyx

Thus

xyyx (f)

(iii) Checking. Dimensional check: Units of (a);

)s

rad()mKg()mN(

22

0 I

This gives

2s

mKgN

which is the correct units for Newton.

(5) Comments. It is incorrect to conclude that xyyx because the element is in static

equilibrium.

PROBLEM 2.6

A fluid flows axially between parallel plates.

Assume: Newtonian fluid, steady state, constant

density, constant viscosity, negligible gravity and

parallel streamlines. Write the three components of

the momentum equations for this flow.

(1) Observations. (i) Properties are constant. (ii) Cartesian coordinates. (iii) Parallel streamlines: no velocity component in the y-direction. (iv) Axial flow: no velocity component in the z-direction. (v) The Navier-Stokes equations give the three momentum equations.

(2) Problem Definition. Determining the three momentum equations for the flow under consideration.

(3) Solution Plan. Apply the Navier-Stokes equations of motion in Cartesian coordinates. Simplify according to the conditions of the problem.

(4) Plan Execution.

(i) Assumptions. (1) Newtonian fluid, (2) constant density and viscosity, (3) steady state, (4) negligible gravity, (5) streamlines are parallel to the plates (no motion in the y-direction) and (6) axial flow (no motion in the z-direction).

(ii) Analysis. The Navier-Stokes for constant properties are

x-direction:2

2

2

2

2

2

z

u

y

u

x

u

x

pg

z

uw

y

u

x

uu

t

uxv (2.10x)

y-direction:2

2

2

2

2

2

zyxy

pg

zw

yxu

ty

vvvvvv

vv (2.10y)

z-direction:2

2

2

2

2

2

z

w

y

w

x

w

z

pg

z

ww

y

w

x

wu

t

wzv (2.10z)

These equations are simplfied as follows:

Steady state: 0t

No gravity: g = 0

Parallel streamlines: 0v

Axial flow: w = 0

Substituting these simplifications inot (2.10)

x-direction:2

2

2

2

y

u

x

u

x

p

x

uu (a)


However, continuity equation gives

0x

w

xx

u

zw

yxu

t

vv (2.2b)

For incompressible flow this simplifies to

0z

w

yx

u v(b)

This simplifies to

0x

u(c)

Substituting (c) into (a)

x-direction:2

2

y

u

x

p (d)

Equations (2.10y) and (2,10z) simplify to

y-direction: 0y

p (e)

z-direction: 0z

p (f)

(iii) Checking. Dimensional check: units of (d)

)sm

m()

ms

kg()

mm

N(

22

2

2 y

u

x

p

2s

m-kgN

Units of (d) are correct.

(5) Comments. The continuity equation provides additional simplification of the Navier-Stokes equations.

PROBLEM 2.7

A fluid flows axially (z-direction) through a tube.

Assume: Newtonian fluid, steady state, constant

density, constant viscosity, negligible gravity and

parallel streamlines. Write the three components of

the momentum equations for this flow.

(1) Observations. (i) Properties are constant. (ii) Cylindrical coordinates. (iii) Parallel streamlines: no velocity component in the r-direction. (iv) Axial flow: no velocity component in the -direction. (v) No variation in the -direction. The Navier-Stokes equations give the three

momentum equations.

(2) Problem Definition. Determining the three momentum equations for the flow under consideration.

(3) Solution Plan. Apply the Navier-Stokes equations of motion in cylindrical coordinates. Simplify according to the conditions of the problem.

(4) Plan Execution.

(i) Assumptions. (1) Newtonian fluid, (2) constant density and viscosity, (3) steady state, (4) negligible gravity, (5) streamlines are parallel to surface (no motion in the r-direction) and (6) axial flow (no motion in the -direction).


r-direction:

2

2

22

2

2

2

21)(

1

zrrr

rrrr

pg

tzrrr

rrr

r

rrz

rrr

vvvv

vvv

vvvvv

(2.11r)

-direction:

2

2

22

2

2

21)(

11

zrrr

rrr

p

rg

tzrrr

r

r

vvvv

vvv

vvvvvv zr

(2.11 )

z-direction:

2

2

2

2

2

11

zrrr

rrz

pg

tzrr

zzzz

zzzz

vvv

vvv

vvvv zr

(2.11z)


Steady state: 0t

Parallel streamlines: 0r

v


Axial flow: v = 0

No gravity: 0zr ggg

No variation in the -direction, 0

Substituting these simplifications into (2.11)

r-direction 0r

p (a)

-direction: 0p

(b)

z-direction:2

21

zrr

rrz

p

z

zzz vvvv z

(c)

Equation (c) is simplified further using the continuity equation in cylindrical coordinates

011

zrzr

rrrt

vvv (2.4)

This equation is simplified to

0z

zv (d)

(d) into (c)

z-direction:r

rrrz

p zv (e)

(iii) Checking. Dimensional check: units of (e)

322223 N/mmkg/s)s/mm())m(

1)mkg/s()N/m(

r

v zrrrz

p

Thus units of (e) are correct.

(5) Comments. The continuity equation provides additional simplification of the Navier-Stokes equations.

PROBLEM 2.8

Consider two-dimensional flow (x,y) between parallel

plates. Assume: Newtonian fluid, constant density and

viscosity. Write the two components of the momentum

equations for this flow. How many unknown do the

equations have? Can they be solved for the unknowns? If not what other equation(s) is needed to obtain a solution?

(1) Observations. (i) Properties are constant. (ii) Cartesian coordinates. (iii) Two dimensional flow (no velocity component in the z-direction. (iv) The Navier-Stokes equations give two momentum equations.

(2) Problem Definition. Determining the two momentum equations for the flow under consideration.

(3) Solution Plan. Apply the Navier-Stokes equations of motion in Cartesian coordinates. Simplify according to the conditions of the problem and count the unknown dependent variables.

(4) Plan Execution.

(i) Assumptions. (1) Newtonian fluid, (2) constant density and viscosity, (3) steady state and (4) two-dimensional flow (no motion in the z-direction).


x-direction:2

2

2

2

2

2

z

u

y

u

x

u

x

pg

z

uw

y

u

x

uu

t

uxv (2.10x)

y-direction:2

2

2

2

2

2

zyxy

pg

zw

yxu

ty

vvvvvv

vv (2.10y)


Steady state: 0t

Two dimensional flow: z

w = 0

Substituting these simplifications inot (2.10)

x-direction:2

2

2

2

y

u

x

u

x

pg

y

u

x

uu xv

(a)

y-direction:2

2

2

2

yxy

pg

yxu y

vvvv

v(b)

These two equations contains three unknowns: ,u v and p. A third equation is needed to obtain a

solution. This equation is continuity


0x

w

xx

u

zw

yxu

t

vv (2.2b)

For incompressible two-dimensional flow this simplifies to

0xx

u v(c)

(iii) Checking. Dimensional check: Each term in (a) and (b) has units of 22 ms

kg.

(5) Comments. It is not surprising that continuity is needed to obtain a solution to the flow field. Conservation of mass (continuity) and momentum (Navier-Stokes equations) must be satisfied.

PROBLEM 2.9

Consider Two-dimensional (r,z) flow through a tube. Assume: Newtonian, constant density and viscosity.

Write the two components of the momentum equations for this flow. How many unknowns do the equations have?Can the equations be solved for the unknowns? If not what other equation(s) is needed to obtain a solution?

(1) Observations. (i) Properties are constant. (ii) Cylindrical coordinates. (iii) Two dimensional flow (no velocity component in the -direction. (iv) The Navier-Stokes equations give two

momentum equations.

(2) Problem Definition. Determining the two momentum equations for the flow under consideration.

(3) Solution Plan. Apply the Navier-Stokes equations of motion in cylindrical coordinates. Simplify according to the conditions of the problem and count the unknown dependent variables.

(4) Plan Execution.

(i) Assumptions. (1) Newtonian fluid, (2) constant density and viscosity, (3) steady state and (4) two-dimensional flow (no motion in the -direction).

(ii) Analysis. The Navier-Stokes in the r and z directions for constant properties are

r-direction:

2

2

22

2

2

2

21)(

1

zrrr

rrrr

pg

tzrrr

rrr

r

rrz

rrr

vvvv

vvv

vvvvv

(2.11r)

z-direction:

2

2

2

2

2

11

zrrr

rrz

pg

tzrr

zzzz

zzzz

vvv

vvv

vvvv zr

(2.11z)


Steady state: 0t

Two dimensional flow: v = 0


r r

z


r-direction:2

2

)(1

zr

rrrr

pg

zrr

rr

rz

rr

vv

vv

vv (a)

z-direction:2

21

zrr

rrz

pg

zr

zzz

zz vvvv

vv zr (b)

These two equations contains three unknowns: ,rv zv and p. A third equation is needed to obtain

a solution. This equation is continuity in cylindrical coordinates

The continuity equation in cylindrical coordinates is given by (2.4)

011

zrzr

rrrt

vvv (2.4)



Equation (2.4) becomes

01

zr

rr

zr

vv (c)

(iii) Checking. Dimensional check: Each term in (a) and (b) has units of 22 ms

kg.

(5) Comments. It is not surprising that continuity is needed to obtain a solution to the flow field. Conservation of mass (continuity) and momentum (Navier-Stokes equations) must be satisfied.

PROBLEM 2.10

In Chapter 1 it is stated that fluid motion and fluid nature play a role in convection heat transfer. Does the energy equation substantiate this observation? Explain.

(1) Observations. (i) Motion in energy consideration is represented by velocity components. (ii) Fluid nature is represented by fluid properties.

(2) Problem Definition. Determine if the energy equation depends on velocity and fluid properties.

(3) Solution Plan. Write the energy equation and determine if it depends on velocity and properties.

(4) Plan Execution.

(i) Assumptions. (1) Continuum, (2) Newtonian fluid and (3) negligible nuclear, radiation and electromagnetic energy transfer.

(ii) Analysis. The energy equation in given by

Dt

DpTTk

Dt

DTc p (2.15)

where

pc specific heat at constant pressure

k thermal conductivity

p pressure

coefficient of thermal expansion or compressibility

= dissipation function

The coefficient of thermal expansion is a property of material defined as

pT

1 (2.16)

The dissipation function is associated with energy dissipation due to friction. It is important in high speed flow and for very viscous fluids. In Cartesian coordinates is given by

2

3

2

2222222

z

w

yx

u

z

u

x

w

y

w

zxy

u

z

w

yx

u

v

vvv

(2.17)

Note that the total temperature derivative in (2.15) is defined as

t

T

z

Tw

y

T

x

Tu

Dt

DTv (a)


Examination of the above equations shows that energy equation (2.15) depends on the velocity

components u, v and w. In addition, (2.15) depends on , ,pc , k and . These are properties

of fluid.

(5) Comments. To determine temperature distribution it is necessary to know the velocity distribution.

PROBLEM 2.11

A fluid flows axially (x-direction) between parallel plates. Assume: Newtonian fluid, steady state, constant density,constant viscosity, constant conductivity, negligible

gravity and parallel streamlines. Write the energy

equation for this flow.

(1) Observations. (i) Properties are constant. (ii) Cartesian coordinates. (iii) Parallel streamlines: no velocity component in the y-direction. (iv) Axial flow: no velocity component in the z-direction.

(2) Problem Definition. Determining the energy equations for the flow under consideration.

(3) Solution Plan. Apply the energy equations in Cartesian coordinates. Simplify according to the conditions of the problem.

(4) Plan Execution.

(i) Assumptions. (1) Newtonian fluid, (2) constant density, viscosity and conductivity, (3) steady state, (4) negligible gravity, (5) streamlines are parallel to the plates (no motion in the y-direction), (6) axial flow (no motion in the z-direction) and (7) negligible nuclear, radiation and electromagnetic energy transfer.

((4) Plan Execution.

(i) Assumptions. (1) Continuum, (2) Newtonian fluid, (3) no energy generation and (4) constant properties.

(ii) Analysis. The energy equation for this case is given by

2

2

2

2

2

2

z

T

y

T

x

Tk

z

Tw

y

T

x

Tu

t

Tc p v (2.19b)

where



p pressure

T temperature density


The dissipation function in Cartesian coordinates is given by

2

3

2

2222222

z

w

yx

u

z

u

x

w

y

w

zxy

u

z

w

yx

u

v

vvv

(2.17)



Steady state: 0t

No gravity: g = 0


Axial flow: 0z

w

Substituting these simplifications into (2.19b)

2

2

2

2

y

T

x

Tk

x

Tuc p (a)

Similarly (2.17) simplifies to

2

3

222

2x

u

y

u

x

u (b)

Further simplifications are obtained using continuity equation (2.2b)

0x

w

xx

u

zw

yxu

t

vv (2.2b)

For incompressible parallel flow this becomes

0x

u(c)

(c) into (b)

2

y

u (d)

Substitute (d) into (a) gives the energy equation for this flow

2

2

2

2

2

y

u

y

T

x

Tk

x

Tuc p (e)

(iii) Checking. Dimensional check: Each term in (e) has units of 3W/m .

(5) Comments. The continuity equation provides additional simplification of the dissipation function.

PROBLEM 2.12

An ideal gas flows axially (x-direction) between parallel

plates. Assume: Newtonian fluid, steady state, constant

viscosity, constant conductivity, negligible gravity and

parallel streamlines. Write the energy equation for this

flow.

(1) Observations. (i) Properties are constant. (ii) Cartesian coordinates. (iii) Parallel streamlines: no velocity component in the y-direction. (iv) Axial flow: no velocity component in the z-direction. (v) The fluid is an ideal gas.

(2) Problem Definition. Determining the energy equations for the flow under consideration.

(3) Solution Plan. Apply the energy equations in Cartesian coordinates. Simplify according to the conditions of the problem.

(4) Plan Execution.

(i) Assumptions. (1) Newtonian fluid, (2) constant density, viscosity and conductivity, (3) steady state, (4) negligible gravity, (5) streamlines are parallel to the plates (no motion in the y-direction), (6) axial flow (no motion in the z-direction), (7) negligible nuclear, radiation and electromagnetic energy transfer and (8) Ideal gas.

(ii) Analysis. The energy equation for this case in given by

VpTkDt

DTcv (2.23)

Rewriting and noting that k is constant

Vpz

T

y

T

x

Tk

t

T

z

Tw

y

T

x

Tuc

2

2

2

2

2

2

vv (a)

where

vc specific heat at constant pressure


p pressure

T temperature wu ,,v velocity components in x, y and z directions

density


The dissipation function in Cartesian coordinates is given by

2

3

2

2222222

z

w

yx

u

z

u

x

w

y

w

zxy

u

z

w

yx

u

v

vvv

(2.17)



Constant k

Steady state: 0t


Axial flow: 0z

w

Incompressible fluid: 0V

Substituting these simplifications into (a)

2

2

2

2

y

T

x

Tk

x

Tucv (b)

Similarly (2.17) simplifies to

2

3

222

2x

u

y

u

x

u (b)


0x

w

xx

u

zw

yxu

t

vv (2.2b)

For incompressible parallel flow this becomes

0x

u(c)

(c) into (b)

2

y

u (d)

Substitute (d) into (a) gives the energy equation for this flow

2

2

2

2

2

y

u

y

T

x

Tk

x

Tucv (e)

(iii) Checking. Dimensional check: Each term in (e) has units of 3W/m .

(5) Comments. The continuity equation provides additional simplification of the dissipation function.

PROBLEM 2.13

Consider two-dimensional free convection over a vertical plate. Assume:Newtonian fluid, steady state, constant viscosity, Boussinesq approximation and

negligible dissipation. Write the governing equations for this case. Can the flow field be determined independently of the temperature field?

(1) Observations. (i) This is a two-dimensional free convection problem. (ii) The flow is due to gravity. (iii) The flow is governed by the momentum and energy equations. Thus the governing equations are the Navier-Stokes equations of motion and the energy equation. (iv) The geometry is Cartesian.

(2) Problem Definition. Determine: the x and y components of the Navier-Stokes equations of motion, and the energy equation for the flow under consideration .

(3) Solution Plan. Start with the Cartesian coordinates Navier-Stokes equations of motion and energy equation for constant properties. Simplify them for this special case.

(4) Plan Execution.

(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3) constant properties, (4) Boussinesq approximations and (5) negligible nuclear, electromagnetic and radiation energy transfer.

(ii) Analysis. Momentum equations. The Navier Stokes equations of motion for free convection are given in (2.29)

VvppTTgDt

VD 21 (2.29)

This vector equation gives the x and y components

2

2

2

2

2

21)(

z

u

y

u

x

u

x

pTTg

z

uw

y

u

x

uu

t

uv (a)

2

2

2

2

2

2

zyxy

p

zw

yxu

t

vvvvvv

vv (b)

Gravity is assumed to point in the negative x-direction. The Cartesian coordinates energy equation for incompressible constant conductivity fluid is given by equation (2.19b)

2

2

2

2

2

2

z

T

y

T

x

Tk

z

Tw

y

T

x

Tu

t

Tc v (2.19b)

where is the dissipation function. These equations are simplified based on the following assumptions

Steady state: 0t

Axial flow: 0z

w

y

x

g

u


No dissipation: 0

(a), (b) and (2.19b) become

2

2

2

21)(

y

u

x

u

x

pTTg

y

u

x

uu v (c)

2

2

2

21

yxy

p

yxu

vvvv

v (d)

2

2

2

2

y

T

x

Tk

y

T

x

Tuc p v (e)

Equations (c), (d) and (e) are the governing equations for this flow. Examination of momentum equations (c) and (d) shows that they contain the unknown temperature variable T. Thus these equations can not be solved for the flow field without invoking the energy equation. Note that the three equations contain four unknowns: ,u ,v p and T. Continuity provides the fourth

equation.

(iii) Checking. Dimensional check: Each term of momentum equations (c) and (d) has units of 2m/s . Each term in (e) has units of 3W/m .

Limiting check: If the fluid is not moving, the energy equation should reduce to pure conduction. Setting 0u in (e) gives

02

2

2

2

y

T

x

T

This is the correct equation for this limiting case.

(5) Comments. (i) Governing equations (c), (d) and (e) are coupled. Thus they must be solved, together with continuity, for the flow field and temperature field. (ii) In energy equation (e),

properties kc p , and represent fluid nature. Velocity components u and v represent fluid

motion. This confirms the observation made in Chapter 1 that fluid motion and nature play a role in convection heat transfer (temperature distribution).

PROBLEM 2.14

Discuss the condition(s) under which the Navier-Stokes equations of motion can be solved

independently of the energy equation.

Solution

Examination of the smallest rectangle in Table 2.1 shows that for constant properties (density and viscosity), continuity and momentum (4 equations) contain the four flow field unknowns u,

v, w and p. Thus for constant properties the Navier-Stokes equations and continuity can be

solved for the flow field independently of the energy equation.

PROBLEM 2.15

Consider a thin film of liquid condensate which is falling over a flat surface by

virtue of gravity. Neglecting variations in the z-direction and assuming Newtonian fluid, steady state, constant properties and parallel streamlines.

[a] Write the momentum equation(s) for this flow.

[b] Write the energy equation including dissipation effect

(1) Observations. (i) The flow is due to gravity. (ii) For parallel streamlines the

velocity component v = 0 in the y-direction. (iii) Pressure at the free surface is

uniform (atmospheric). (iv) Properties are constant. (v) The geometry is Cartesian.

(2) Problem Definition. Determine: [a] the x and y components of the Navier-Stokes equations of motion, and [b] the energy equation for the flow under consideration .


(4) Plan Execution.

(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3) flow is in the x-direction only, (4) constant properties, (5) uniform ambient pressure, (6) parallel streamlines. (7) negligible nuclear, electromagnetic and radiation energy transfer.

(ii) Analysis. [a] Momentum equations. The Navier Stokes equations of motion in Cartesian coordinates for constant properties are given in equations (2.10x ) and (2.10y)

2

2

2

2

2

2

z

u

y

u

x

u

x

pg

z

uw

y

u

x

uu

t

uxv (2.10x)

2

2

2

2

2

2

zyxy

pg

zw

yxu

ty

vvvvvv

vv (2.10y)

The gravitational components are

,gg x 0yg (a)

Based on the above assumptions, equations (2.10) are simplified as follows:

Steady state: 0tt

u v (b)

Axial flow (x-direction only): 0z

w (c)

Parallel flow: 0v (d)

Substituting (a)-(d) into (2.10x) and (2.10y), gives

gx

uu

2

2

2

2

y

u

x

u

x

p (e)

and

y

x

g

Problem 2.15 (continued)

0y

p (f)

The x-component (e) can be simplified further using the continuity equation for incompressible flow, equation (2.3)

0zyx

uV

wv (g)

Substituting (c) and (d) into (g), gives

0x

u (h)

Using (h) into (e) gives the x-component

g2

2

y

u

x

p= 0 (i)

Integrating (f) with respect to y)(xfp (j)

where f(x) is the constant of integration. At the free surface, ,Hy the pressure is uniform

equal to .p Therefore, setting Hy in (j) gives

pxf )( (k)

Substituting (k) into (j) gives the pressure solution

pp (l)

Differentiating (k) with respect to x givesx

p = 0 (m)

Substituting (m) into (i) gives the x-component of the Navier-Stokes equations

02

2

yd

udg (n)

[b] Energy equation. The Cartesian coordinates energy equation for incompressible constant conductivity fluid is given by equation (2.19b)

2

2

2

2

2

2

z

T

y

T

x

Tk

z

Tw

y

T

x

Tu

t

Tc v (2.19b)

where the dissipation function in Cartesian coordinates is given by equation (2.17)

2

3

2

2222222

z

w

yx

u

z

u

x

w

y

w

zxy

u

z

w

yx

u

v

vvv

(2.17)

Problem 2.15 (continued)

Based on the above assumptions, these equations are simplified as follows:

Steady state: 0t

T (o)

Substituting (c), (d) and (o) into (2.19b), gives

2

2

2

2

y

T

x

Tk

x

Tuc p (p)

The dissipation function (2.17) is simplified using (c), (d) and (h)

2

y

u (q)

Substituting (q) into (p) gives the energy equation

2

2

2

2

2

y

u

y

T

x

Tk

x

Tuc (r)

(iii) Checking. Dimensional check: Each term of the x-component equation (n) must have the same units

g = kg/m2-s2

2

2

yd

ud = (kg/m-s)

m s

m

/2

= kg/m2-s2

Each term in (r) has the same units of 3W/m .

Limiting check: If the fluid is not moving, the energy equation should reduce to pure conduction. Setting 0u in (i) gives

02

2

2

2

y

T

x

T

This is the correct equation for this limiting case.

(5) Comments. (i) For two-dimensional incompressible parallel flow, the momentum equations

are considerably simplified because the vertical velocity component v = 0.

(ii) The flow is in fact one-dimensional since u does not change with x and is a function of yonly.

(iii) In energy equation (r), properties ,,kc p and represent fluid nature. The velocity u

represents fluid motion. This confirms the observation made in Chapter 1 that fluid motion and nature play a role in convection heat transfer (temperature distribution).

(ii) The last term in energy equation (r) represents dissipation.

PROBLEM 2.16

A wedge is maintained at T1 along one side and T2 along

the opposite side. A solution for the flow field is obtained

based on Newtonian fluid and constant properties. The

fluid approaches the wedge with uniform velocity and

temperature. Examination of the solution shows that the

velocity distribution is not symmetrical with respect to the

x-axis. You are asked to support the argument that the

solution is incorrect.

(1) Observations. (i) This is a forced convection problem. (ii) Flow properties (density and viscosity) are constant. (iii) Upstream conditions are uniform (symmetrical) (iv) The velocity vanishes at both wedge surfaces (symmetrical). (v) Surface temperature is asymmetric. (vi) Flow field for constant property fluids is governed by the Navier-Stokes and continuity equations. (vii) If the governing equations are independent of temperature, the velocity distribution over the wedge should be symmetrical with respect to x. (viii) The geometry is Cartesian.

(2) Problem Definition. Determine if the governing equations for the velocity distribution is independent of temperature.

(3) Solution Plan. Examine the Navier-Stokes and continuity equations in Cartesian coordinates for dependency on temperature.

(4) Plan Execution.

(i) Assumptions. (1) Newtonian fluid, (2) two-dimensional (x and y), (3) constant properties and (4) uniform upstream conditions.

(ii) Analysis. The Navier Stokes equations of motion in Cartesian coordinates for constant properties are given in equations (2.10x ) and (2.10y)

2

2

2

2

2

2

z

u

y

u

x

u

x

pg

z

uw

y

u

x

uu

t

uxv (2.10x)

2

2

2

2

2

2

zyxy

pg

zw

yxu

ty

vvvvvv

vv (2.10y)

For two-dimensional conditions (w = 0) these equations become

2

2

2

2

y

u

x

u

x

pg

y

u

x

uu

t

uxv (a)

2

2

2

2

yxy

pg

yxu

ty

vvvv

vv (b)

These equations contain three unknowns: u, v and p. Continuity provides the fourth equation

0wzy

uxt

v (2.2a)

For two-dimensional constant density (2.2a) simplifies to

1T

2TT

V

y

x


0yx

u v (b)

Since properties are constant, and are constant. Thus (a), (b) and (c) are independent of

temperature. It follows that the solution to these equations for u, v and p is independent of the

asymmetry of the boundary temperature. A solution based on the assumption of constant property that give asymmetrical velocity distribution must be incorrect.

(iii) Checking. Dimensional check: Each term in (a) and (b) has units of kg/m2-s2

(5) Comments. (i) Although the flow was assumed two-dimensional, the same conclusion applies to three-dimensional flow as long as the geometry is symmetrical about the x-axis and upstream conditions are uniform. (ii) Examination of the smallest rectangle in Table 2.1 shows that for constant properties (density and viscosity), continuity and momentum (4 equations)

contain the four flow field unknowns u, v, w and p. Thus for constant properties the Navier-

Stokes equations and continuity can be solved for the flow field independently of the energy equation. This is valid for steady as well as transient flow.

PROBLEM 2.18

Consider two-dimensional (x and y), steady, constant

properties, parallel flow between two plates

separated by a distance H. The lower plate is

stationary while the upper plate moves axially with a

velocity oU . The upper plate is maintained at

uniform temperature oT and the lower plate is cooled

with a flux oq . Taking into consideration dissipation, write the Navier-Stokes equations of

motion, energy equation and boundary conditions at the two plates.

(1) Observations. (i) The geometry is Cartesian. (ii) Properties are constant. (ii) Axial flow (no

motion in the z-direction), (iv) Parallel streamlines means that the normal velocity component is

zero. (v) Specified flux at the lower plate and specified temperature at the upper plate.

(2) Problem Definition. Determine: [a] the x and y components of the Navier-Stokes equations of motion, [b] the energy equation for the flow under consideration, and [c] velocity and temperature boundary conditions at the lower and upper plates.


(4) Plan Execution.

(i) Assumptions. (1) Newtonian fluid, (2) flow is in the x -direction, (3) constant properties

and (4) negligible nuclear, electromagnetic and radiation energy transfer.

(ii) Analysis. The Navier Stokes equations of motion in Cartesian coordinates for constant properties are given in equations (2.10x ) and (2.10y)

2

2

2

2

2

2

z

u

y

u

x

u

x

pg

z

uw

y

u

x

uu

t

uxv (2.10x)

2

2

2

2

2

2

zyxy

pg

zw

yxu

ty

vvvvvv

vv (2.10y

The energy equation given by (2.19b)

2

2

2

2

2

2

z

T

y

T

x

Tk

z

Tw

y

T

x

Tu

t

Tc p v (2.19b)

The dissipation function in Cartesian coordinates is given by (2.17)

2

3

2222222

2z

w

yx

u

z

u

x

w

y

w

zxy

u

z

w

yx

u vvvv(2.17)




Axial flow: 0z

w


0z

w

yx

u

zw

yxu

t

vv (2.2b)

For incompressible parallel flow this reduces to

0x

u (a)

Substituting these simplifications into (2.10x), (2.10y), (2.19b) and (2.17)

2

2

y

u

x

pg

t

ux

(b)

0y

pg y (c)

2

2

2

2

y

T

x

Tk

x

Tu

t

Tc p (d)

2

y

u(e)

The boundary conditions on velocity components are

(1) 0)0,(xu

(2) 0)0,(xv

(3) oUHxu ),(

(4) 0),( Hxv

The boundary conditions on temperature are

(1) oqy

xTk

)0,(

(2) oTHxT ),(

(iii) Checking. Each term in momentum equations (b) and (c) has units of 22 skg/m . Each

term in energy equation (d) has units of .W/m3

Limiting check: If the upper plate is stationary and there is no axial pressure gradient, the problem reduces to one-dimensional transient conduction. Since 0vu , (d) becomes

2

2

2

2

y

T

x

Tk

t

Tc p (f)

However, since the boundary conditions on temperature are independent of x, axial temperature


gradient vanishes and (f) simplifies further to

2

2

y

Tk

t

Tc p (g)

This is the one-dimensional transient conduction equation.

(5) Comments. (i) The continuity equation provides important simplifications in the momentum and energy equations. (ii) For steady state set tT / is set equal to zero.

PROBLEM 2.19

A shaft or radius 1r rotates concentrically inside a sleeve of inner

radius 2r . Lubrication oil fills the clearance between the shaft and

the sleeve. The sleeve is maintained at uniform temperature oT .

Neglecting axial variation and taking into consideration dissipation,

write the Navier-Stokes equations of motion, energy equation and

boundary conditions for this flow. Assume constant properties.

(1) Observations. (i) The geometry is cylindrical. (ii) No variation in

the axial and angular directions. (iii) Properties are constant.

(2) Problem Definition. Determine: [a] the r and components of the Navier-Stokes equations

of motion, [b] the energy equation and [c] boundary conditions for the flow under consideration.

(3) Solution Plan. Start with the cylindrical coordinates Navier-Stokes equations of motion and energy equation for constant properties. Simplify them for this special case.

(4) Plan Execution.

(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3) flow is in the -direction, (4)

constant properties, (5) negligible nuclear, electromagnetic and radiation energy transfer and (6) negligible gravity.

(ii) Analysis. [a]. The Navier Stokes equations of motion in cylindrical coordinates for constant properties are given in equations (2.11r) and (2.11 )

r-direction:

2

2

22

2

2

2

21)(

1

zrrr

rrrr

pg

tzrrr

rrr

r

rrz

rrr

vvvv

vvv

vvvvv

(2.11r)

-direction:

2

2

22

2

2

21)(

11

zrrr

rrr

p

rg

tzrrr

r

r

vvvv

vvv

vvvvvv zr

(2.11 )


Steady state: 0t

No gravity: 0ggr

No axial flow:z

0z

v

r

1r2r

shaft

oT


Symmetry: 0

Substituting these simplifications inot (2.11r)

)(1

2

rr

r vvv

v rrrrr

p

rr (a)

Similarly, (2.11 ) become

)(1

vvvv

v r rrrrrr

r (b)


011

zrzr

rrrt

vvv (2.4)

For incompressible fluid and 0z

, this simplifies to

0rrr

v (c)

Integrating

Cr rv

where C is constant of integration. Since 0)( 1rrv it follows that C = 0. Therefore

0rv (d)

Introducing (d) into (a) and (b)

r-direction:r

p

r

2v

(e)

-direction 0)(1

vrrrr

(f)

[b] Energy equation. For constant properties the energy equation is given by

2

2

2

2

2 0

11

z

TT

rr

Tr

rrk

z

TT

rr

T

t

Tc zrp v

vv (2.24)

where the dissipation function is


22

2222

0

1

0

12

122

rzzr

rrrzrrr

zrz

rzrr

vvvv

vvvvvvv

(2.25)

Equations (2.24) and (2.25) are simplified for the conditions of this problem

r

Tr

rrk

10 (g)

2

rr

0 vv (h)

[c] Boundary conditions. The boundary conditions for Navier-Stokes equations (e) and (f) are

(1) 11)( rrv

(2) 0)( 2rv

The boundary conditions on energy equation (g) are

(1) Insulated surface at :1r 0)( 1

r

rT

(2) Specified temperature at :2r oTrT )( 2

(iii) Checking. Dimensional check: Units of (e)

)mm

N()

ms

m()

m

kg(

22

22

3 r

p

r

v

22222 sm

kg

mm

N

sm

kg

Units of each term in (g)

ms

kg

m

W)C/m()m(

)m()m(

1)

Cm

W(

33

o

o r

Tr

rrk

ms

kg)s/1()

ms

kg(

3

2

(5) Comments. The continuity equation provides additional simplification of the Navier-Stokes and energy equations.

PROBLEM 2.20

A rod of radius ir moves axially with velocity

oU inside a concentric tube of radius or . A

fluid having constant properties fills the space

between the shaft and tube. The tube surface is

maintained at uniform temperature .oT Write

the Navier-Stokes equations of motion, energy

equation and surface boundary conditions

Taking into consideration dissipation. Assume that the streamlines are parallel to the surface.

(1) Observations. (i) The geometry is cylindrical. (ii) No variation in the angular direction. (iii)

Properties are constant. (iv) Parallel streamlines means that the radial velocity component is

zero.

(2) Problem Definition. Determine: [a] the r and z components of the Navier-Stokes equations of motion, [b] the energy equation for the flow under consideration and [c] boundary conditions.


(4) Plan Execution.

(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3) flow is in the z -direction, (4) constant properties, (5) negligible nuclear, electromagnetic and radiation energy transfer and (6) negligible gravity.

(ii) Analysis. [a]. The Navier Stokes equations of motion in cylindrical coordinates for constant properties are given in equations (2.11r) and (2.11 z )

r-direction:

2

2

22

2

2

2

21)(

1

zrrr

rrrr

pg

tzrrr

rrr

r

rrz

rrr

vvvv

vvv

vvvvv

(2.11r)

z-direction:

2

2

2

2

2

11

zrrr

rrz

pg

tzrr

zzzz

zzzz

vvv

vvv

vvvv zr

(2.11z)


Steady state: 0t

No gravity: 0zr gg

No tangential flow: 0v


Symmetry: 0

Parallel streamlines: 0r

v


r-directionr

p0 (a)

Similarly, (2.11 z ) become

:2

21

zrr

rrz

p

z

zzz vvvv z (b)


011

zrzr

rrrt

vvv (2.4)

For incompressible fluid and 0rv , this simplifies to

0z

zv (c)

Introducing (d) into (b)

z-directionr

rrrz

p zv10 (d)


2

2

2

2

2

11

z

TT

rr

Tr

rrk

z

TT

rr

T

t

Tc zrp v

vv (2.24)


22

2222

1

12

122

rzzr

rrrzrrr

zrz

rzrr

vvvv

vvvvvvv

(2.25)

Equations (2.24) and (2.25) are simplified for the conditions of this problems

2

21

z

T

r

Tr

rrk

z

Tc zpv (g)

2

r

zv (h)



(1) oiz Ur )(v

(2) 0)( oz rv


(1) Equality of temperature at :ir ),()( , zrTzrT iri

(2) Equality of flux at :irr

zrTk

r

zrTk ir

ri ),(),(

(3) Specified temperature at :or oo TzrT )),(

where the subscript r refers to the rod.

(iii) Checking. Dimensional check: United of each term in (d)

32 m

N)

mm

N(

z

p

322 m

N

ms

kg

m)(

m/s)(m)(

m)(m)(

1)

ms

kg(

rr

rr

zv

Units of each term in (h)

3

o

o3 m

W

m

C

s

m

Ckg

J

m

kg

z

Tc zpv

32

o

o2

2

m

W

m

C

Cm

W

z

Tk

ms

kg)s/1()

ms

kg(

3

2

(5) Comments. (i) The continuity equation provides additional simplification of the Navier-Stokes and energy equations. (ii) The temperature distribution is two-dimensional. (iii) To solve for the temperature distribution it is necessary to write a heat equation for the rod as well as thermal boundary conditions at two axial locations.

PROBLEM 2.21

A rod or radius ir rotates concentrically inside a tube of inner radius or .

Lubrication oil fills the clearance between the shaft and the tube. Tube

surface is maintained at uniform temperature oT . The rod generates heat

volumetrically at uniform rate q . Neglecting axial variation and taking

into consideration dissipation, write the Navier-Stokes equations of

motion, energy equation and boundary conditions for this flow. Assume

constant properties.

(1) Observations. (i) The geometry is cylindrical. (ii) No variation in

the axial and angular directions. (iii) Properties are constant.

(2) Problem Definition. Determine: [a] the r and components of the Navier-Stokes equations

of motion, [b] the energy equation and [c] boundary conditions for the flow under consideration.


(4) Plan Execution.

(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3) flow is in the -direction, (4)

constant properties, (5) negligible nuclear, electromagnetic and radiation energy transfer and (6) negligible gravity.

(ii) Analysis. [a]. The Navier Stokes equations of motion in cylindrical coordinates for constant properties are given in equations (2.11r) and (2.11 )

r-direction:

2

2

22

2

2

2

21)(

1

zrrr

rrrr

pg

tzrrr

rrr

r

rrz

rrr

vvvv

vvv

vvvvv

(2.11r)

-direction:

2

2

22

2

2

21)(

11

zrrr

rrr

p

rg

tzrrr

r

r

vvvv

vvv

vvvvvv zr

(2.11 )


Steady state: 0t

No gravity: 0ggr

No axial flow:z

0z

v

oT

0

oq

ir

or


Symmetry: 0


)(1

2

rr

r vvv

v rrrrr

p

rr (a)

Similarly, (2.11 ) become

)(1

vvvv

v r rrrrrr

r (b)


011

zrzr

rrrt

vvv (2.4)

For incompressible fluid and 0z

, this simplifies to

0rrr

v (c)

Integrating

Cr rv

where C is constant of integration. Since 0)( or rv it follows that C = 0. Therefore

0rv (d)

Introducing (d) into to (a) and (b)

r-direction:r

p

r

2v

(e)

-direction 0)(1

vrrrr

(f)


2

2

2

2

2 0

11

z

TT

rr

Tr

rrk

z

TT

rr

T

t

Tc zrp v

vv (2.24)


22

2222

0

1

0

12

122

rzzr

rrrzrrr

zrz

rzrr

vvvv

vvvvvvv

(2.25)

Equations (2.24) and (2.25) are simplified for the conditions of this problems


r

Tr

rrk

10 (g)

2

rr

0 vv (h)


(1) ii rr )(v

(2) 0)( orv


(1) Specified flux at :ir ii q

r

rTk

)(

(2) Specified temperature at :or oo TrT )(

Conservation of energy for the rod gives the flux iq :

Energy generated in rod = energy leaving surface at ir

iii qrrq 22

2

ii

rqq (i)

(iii) Checking. Dimensional check: United of (e)

)mm

N()

ms

m()

m

kg(

22

22

3 r

p

r

v

22222 sm

kg

mm

N

sm

kg

Units of each term in (g)

ms

kg

m

W)C/m()m(

)m()m(

1)

Cm

W(

33

o

o r

Tr

rrk

ms

kg)s/1()

ms

kg(

3

2

(5) Comments. The continuity equation provides additional simplification of the Navier-Stokes and energy equations.

PROBLEM 2.22

Air flows over the two spheres. The radius of sphere 2 is double

that of sphere 1. However, the free stream velocity for sphere 1 is

double that for sphere 2. Determine the ratio of the average heat

transfer coefficients 21 / hh for the two spheres.

(1) Observations. (i) This is a forced convection problem. (ii) The same fluid flows over both spheres. (iii) Sphere diameter and free stream velocity affect the Reynolds number which in turn affect the heat transfer coefficient.

(2) Problem Definition. Since the average heat transfer coefficient h is correlated in terms

of the Nusselt number, the problem becomes one of determining the Nusselt number for each sphere and taking their ratio.

(3) Solution Plan. Use the results of dimensional analysis to obtain a relationship between the Nusselt number and the significant parameters in forced convection.

(4) Plan Execution.

(i) Assumptions. (1) Newtonian fluid, (2) steady state and (3) constant properties.

(ii) Analysis. Non-dimensional form of the governing equations for convection gives

DNu =h D

k = f( DRe , Pr, DrG , E) (a)

where

D = diameter of sphere, m E = Eckert number

DrG = Grashof number

h = average heat transfer coefficient, W/m2-oCk = thermal conductivity of fluid, W/m-oC

DNu = average Nusselt number

Pr = Prandtl number

DRe = Reynolds number

Assume that free convection is negligible compared to forced convection. This eliminates the Grashof number. Furthermore, neglect dissipation effects. This eliminates the Eckert number. Thus (a) is simplified to

DNu =k

Dh = f( DRe , Pr) (b)

Since the same fluid flows over both spheres, it follows that the Prandtl number is the same for both. Thus (b) becomes

DNu =k

Dh = f( DRe ) (c)

The Reynolds number is defined as

2

1

1V

2V


DRe =DV

(d)

where

V free stream velocity, m/s

= kinematic viscosity, m2/s

Solving equation (c) for h

h =k

Df( DRe ) (e)

According to (e), to calculate h for each sphere it is necessary to determine: (1) the exact form

of the function f( DRe ), (2) the diameter, (3) the thermal conductivity and (4) the Reynolds

number. However, of interest is determining the ratio of the heat transfer coefficients for two spheres. Applying (e) to the two spheres and taking the ratio of the resulting equation

h

h

1

2

=21

12

D

D

RefD

RefD (f)

where the subscripts 1 and 2 refer to sphere 1 and 2, respectively. Using (d) to determine the two Reynolds numbers

111

DVReD (g)

and

222

DVReD (h)

HoweverD2 = 2 D1 (i)

and

2/12 VV (j)

Substituting (i) and (j) into (h)

2

2 112

DVReD =

111

DReDV

(k)

Thus the Reynolds number is the same for both spheres. It follows that

f(2DRe ) = f(

1DRe ) (l)

Substituting this result into (f) gives

h

h

1

2

=D

D

2

1

= 2 (m)

(iii) Checking. Qualitative check: From (e) one concludes that for the same fluid (same k)and Reynolds number, the heat transfer coefficient is inversely proportional to the diameter. This confirms the result in (m).

2DRe2DRe


(5) Comments. (i) For constant Reynolds and Prandtl numbers the heat transfer coefficient increases as the diameter decreases. (ii) The ratio of the total heat transfer rate from the two spheres is obtained from Newton's law of cooling

2

1

))((

))((

2

1

221

212

222

211

2

1

D

D

DD

DD

TTDh

TTDh

q

q

s

s

Thus, although the heat transfer coefficient for the small sphere is greater than that of the large sphere, its total heat transfer rate is smaller by a factor of two.

PROBLEM 2.23

The average Nusselt number for laminar free convection over an isothermal vertical plate is

determined analytically and is given by

)(43

44/1

PrfGr

k

LhNu L

L

where LGr is the Grashof number based on the length of the plate L and f(Pr) is a function of the

Prandtl number. Determine the percent change in the average heat transfer coefficient if the

length of the plate is doubled.

(1) Observations. (i) This is a free convection problem. (ii) The average heat transfer

coefficient h depends on the vertical length L of the plate. (iii) L appears in the Nusselt number as well as the Grashof number.

(2) Problem Definition. Derive a relationship between the average heat transfer coefficient h

and the length of a vertical plate L.

(3) Solution Plan. Solve the given Nusselt number correlation equation for h in terms of length L.

(4) Plan Execution.

(i) Assumptions. (1) Laminar flow and (2) given correlation equation for Nusselt number applies to both plates.

(ii) Analysis. The percent change in h is given by

% change in h = 100 11001

2

1

12

h

h

h

hh (a)

where the subscripts 1 and 2 refer to plates of length L and 2L, respectively and h is the average

heat transfer coefficient. The average Nusselt number LNu is given by

k

LhNu L

4/1

43

4 LGrf(Pr) (b)

where

f(Pr) = function of Prandtl numberGrL = Grashof number

h = average heat transfer coefficient, W/m2-oCk = thermal conductivity, W/m-oCL = plate length, m

NuL = average Nusselt number

Pr = Prandtl number

The Grashof number is defined as

GrL =2

3)( LTTg s (c)

where


g = gravitational acceleration, m/s2

sT = surface temperature, oC


= coefficient of thermal expansion, 1/K (or 1/oC)


Substituting (c) into (b) and solving for h

h =

4/1

24

)(

3

4 TTgk s

4/1

1)(

LPrf (d)

Applying (d) to the two plates

h1 =

4/1

24

)(

3

4 TTgk s

4/11

1)(L

Prf (e)

and

h2 =

4/1

24

)(

3

4 TTgk s f

4/12

1Pr

L (f)

Taking the ratio of (e) and (f) 4/1

2

1

1

2

L

L

h

h (g)

Substituting (g) into (a)

% change in h = 100 1)( 4121

/L/L (h)

(iii) Computations. For the case where the length L2 = 2L1, equation (h) gives

% change in h = 100 [(1/2)1/4 1] = 9.15 %

(iv) Checking. Dimensional check: Units of h in (d) should be W/m2-oC:

h =

4/1

222

o2oo

)/sm(4

)C)()(m/s()C/1()CW/m(

3

4 TTgk s f(Pr)

4/14/1 )m(

1

L= W/m2-oC

Qualitative check: According to (d) the average heat transfer coefficient is inversely proportional to L

1/4. Thus increasing L, decreases h . This is consistent with the negative result obtained ( 9.15 %) which indicates a decrease in h .

(5) Comments. Although h decreases as the length of the plate is increased, the total heat transfer rate increases. Newton's law of cooling gives

4/3

1

2

1

2

4/1

2

1

11

22

11

22

1

2

L

L

L

L

L

L

Lh

Lh

TTLh

TTLh

q

q

s

s = (2)3/4 = 1.68

PROBLEM 2.24

An experiment was performed to determine the average heat transfer coefficient for forced convection over spheres. In the experiment a sphere of diameter 3.2 cm is maintained at uniform

surface temperature. The free stream velocity of the fluid is 23.4 m/s. Measurements showed that

the average heat transfer coefficient is 62 CW/m o2 .

[a] Predict the average heat transfer coefficient for the same fluid which is at the same free

stream temperature flowing over a sphere of diameter 6.4 cm which is maintained at the same

surface temperature. The free stream velocity is 11.7 m/s.

[b] Which sphere transfers more heat?

(1) Observations. (i) This is a forced convection problem. (ii) The same fluid flows over both spheres. (iii) Sphere diameter and free stream velocity affect the Reynolds number which in turn affect the heat transfer coefficient. (iv) Newton’s law of cooling gives the heat transfer rate.

(2) Problem Definition. Since the average heat transfer coefficient h is expressed in terms

of the Nusselt number, the problem becomes one of determining the Nusselt number for each sphere and taking their ratio.

(3) Solution Plan. Use the results of dimensional analysis to obtain a relationship between the Nusselt number and the significant parameters in forced convection. Apply Newton’s law of cooling to determine heat transfer rate.

(4) Plan Execution.

(i) Assumptions. (1) Newtonian fluid, (2) steady state and (3) constant properties.

(ii) Analysis and computations. Non-dimensional form of the governing equations for forced convection gives

DNu =h D

k = f( DRe , Pr, E) (a)

where

D = diameter of sphere, m E = Eckert number

h = average heat transfer coefficient, W/m2-oCk = thermal conductivity of fluid, W/m-oC


Pr = Prandtl number


Assume that dissipation is negligible, equation (a) is simplified to

DNu =h D

k = f( DRe , Pr) (b)

Since the same fluid flows over both spheres, it follows that the Prandtl number is the same for both. Thus (b) becomes


DNu =h D

k = f( DRe ) (c)


DVReD (d)

where



Solving equation (c) for h

h =k

Df( DRe ) (e)

Applying (2) to the two spheres and taking the ratio to eliminate k

h

h

1

2

=)(

)(

21

12

D

D

RefD

RefD (f)

where the subscripts 1 and 2 refer to sphere 1 and 2, respectively. Using (d) to determine the two Reynolds numbers

1DRe = 11 DV (g)

and

222

DVReD (h)

The total heat transfer rate is determined using Newton’s law of cooling

)( TTAhq sT (i)

where A is surface area of sphere given by

2DA (j)

(j) into (i)

)( TTDhq s (k)

Applying (k) to the two spheres and taking their ratio

22

11

22

11

2

1

)(

)(

Dh

Dh

TTDh

TTDh

q

q

s

s (l)

(iii) Computations. Substituting numerical values in (g) and (h)

)/sm(

)/sm(88.74

)/sm(

)m(2.3)m/s(4.232

2

21DRe

and


)/sm(

)/sm(88.74

)/sm(

)m(4.6)m/s(7.112

2

22DRe

Thus the two Reynolds numbers are identical. It follows that

)()(21 DD RefRef (i)

Substituting (i) into (f)

h

h

1

2

=D

D

2

1

(j)

Solving (j) for 2h

CW/m31m)(4.6

m)(2.3)CW/m(62 o2o2

2

112

D

Dhh

Substitute (j) into (l)

121

12

2

1

DD

DD

q

q

Checking. Dimensional check: computations showed that the Reynolds number is dimensionless and that units of h are correct.

Qualitative check: From (e) one concludes that for the same fluid (same k) and Reynolds number, the heat transfer coefficient is inversely proportional to the diameter. Results show that increasing the diameter by a factor of 2 reduces the heat transfer coefficient by the same factor.

(5) Comments. For constant Reynolds and Prandtl numbers the heat transfer coefficient increases as the diameter decreases.

PROBLEM 2.25

Atmospheric air flows between parallel plates with a mean velocity of m/s10 . One plate is

maintained at C25 o while the other at C.115 o

[a] Calculate the Eckert number. Can dissipation be neglected?

[b] Use scale analysis to compare the magnitude of normal conduction, ,/ 22 yTk with

dissipation, .)/( 2yu Is dissipation negligible compared to conduction?

(1) Observations. (i) Dissipation is important when the Eckert number is high compared to unity. (ii) If the ratio of dissipation to conduction is small compared to unity, it can be neglected.

(2) Problem Definition. [a] Compute the Eckert number. [b] Estimate normal conduction and dissipation using scaling.

(3) Solution Plan. [a] Using the definition of the Eckert number, compute its value for the given data. [b] Use scaling to estimate the ratio of dissipation to normal conduction.

(4) Plan Execution.

(i) Assumption. (1) Newtonian fluid and (2) continuum.

(ii) Analysis. The Eckert number is defined in equation (2.43) as

)( 21

2

ss TTc

u

p

(a)

where

pc specific heat, CJ/kg o

1sT temperature of plate 1 = 25 Co

2sT temperature of plate 2 = 115 Co

u mean axial velocity = 10 m/s

Dissipation is given by

dissipation =

2

y

u(b)

where

u axial velocity, m/s

y normal coordinate, m

viscosity, kg/s-m

Normal conduction is given by

conduction = 2

2

y

Tk (c)

To scale dissipation and conduction, the following scales are introduced

0

2sT

1sT

uH y


Scale for :T )( 21 ss TTT

Scale for :u uu

Scale for :y Hy

where H is spacing between plates. Rewrite (b) and (c)

dissipation =

2

y

u(d)

conduction = 2)(

)(

y

Tk (e)

Use the above scales to estimate (d) and (e)

dissipation =2

2

H

u(f)

conduction = 2

21 )(

H

TTk ss (g)

Taking the ratio of (f) and (g)

)(conduction

ndissipatio

21

2

ss TTk

u (h)

(iii) Computations. Properties of air are determined at the average temperatureT

C702

)C)(11525(

2

oo

21 ss TTT

7.1008pc CJ/kg o

02922.0k CW/m o

61047.20 mkg/s

Substitute into (a)

0011.0sJ

mkg0011.0

C)25)(-C)(115J/kg(7.1008

)/sm()10(2

2

oo

222

E

Substitute into (h)

conduction

ndissipatio00078.0

W

W00078.0

s-W

m-kg00078.0

C)25)(-C)(115m-0.02922(W/

m/s)(m)(10)kg/s(1047.203

2

oo

226

Computations show that the Eckert number is small compared to unity. Thus dissipation can be neglected. Computations also show that dissipation is small compared to normal conduction. Thus it can be neglected.

(iv) Checking. Dimensional check: Computations show that units for dissipation and conduction are correct.


(5) Comments. (i) The spacing between the two plates, H, need not be specified to compare dissipation with conduction. Their ratio in (h) is independent of H. (ii) The Eckert number is a measure of the importance of dissipation.

PROBLEM 2.26

An infinitely large plate is immersed in an infinite fluid. The plate is suddenly moved along its

plane with velocity .oU Neglect gravity and assume constant properties.

[a] Show that the axial Navier-Stokes equation is given by

2

2

y

u

t

u

[b] Due to viscous forces, the effect of plate motion

penetrates into the fluid. The penetration depth )(t

increases with time. Use scaling to derive an expression

for )(t .

(1) Observations. (i) The plate is infinite. (ii) No changes take place in the axial direction (infinite plate). (iii) This is a transient problem. (iv) Constant properties. (v) Cartesian coordinates.

(2) Problem Definition. [a] Determine the equation of motion for resulting from a suddenly accelerated plate. [b] Use scaling to estimate ).(t

(3) Solution Plan. [a] Apply the Navier-Stokes equations of motion. Introduce continuity to identify simplifying conditions. [b] Assign scales to each variable in the governing equation to estimate ).(t

(4) Plan Execution.

(i) Assumptions. (1) Newtonian fluid, (2) constant properties, (3) no motion in the z-direction and (4) negligible gravity.

(ii) Analysis. [a] The Navier-Stokes for two-dimensional constant properties are

x-direction:2

2

2

2

2

2

z

u

y

u

x

u

x

pg

z

uw

y

u

x

uu

t

uxv (2.10x)

y-direction:2

2

2

2

2

2

zyxy

pg

zw

yxu

ty

vvvvvv

vv (2.10y)


No gravity: g = 0

No axial variation: 0x

No motion in the z-direction: 0z

w



x-direction:2

2

y

u

y

u

t

uv (a)

y-direction:2

2

yy

p

yt

vvv

v (b)


0z

w

yx

u

zw

yxu

t

vv (2.2b)

For two-dimensional incompressible flow this simplifies to

0yx

u v (c)

This simplifies to

0y

v (d)

Integration of (d) gives

)(tfv (e)

where f(t) is “constant” of integration. This time function is determined from the no slip boundary condition on v

0)0,(xv (f)

Applying (f) to (e) gives

0)(tf (g)

(g) into (e)

0v (h)

Substitute (h) into (a) and (b)

x-direction:2

2

y

u

t

u (i)

y-directiony

p0 (j)

[b] To obtain scaling estimate of )(t rewrite (i)

y

u

yt

u(k)

Scale for :u )0(Uu

Scale for :u )0(tt

Scale for :y )0(y


Substitute into (k)

2

U

t

Ul

Introduce the definition of kinematic viscosity

(m)

Substitute (m) into (l) and solve for

tt)( n

(iii) Checking. Dimensional check: (i) Each term in (i) has units of 22 mkg/s .

)m/sm()mkg/s()m/s()(kg/m 2

2

223

y

u

t

u= 22 mkg/s

(ii) Units of (n) should be length:

m(s)/s)m()( 2 tt

Limiting check: Initially the penetration thickness is zero. Setting t = 0 in (n) gives 0)0( ,

which is the correct result.

(5) Comments. (i) A major simplification of this problem is due to the assumption of infinite plate. Due to this assumption all derivatives with respect to x vanish. (ii) The same governing equation (i) applies to an oscillating plate moving in a plane normal to y. (iii) Scaling estimate of the penetration thickness )(t is independent of plate velocity U.

PROBLEM 2.27

An infinitely large plate is immersed in an infinite fluid at

uniform temperature iT . The plate is suddenly maintained at

temperature .oT Assume constant properties and neglect gravity.

[a] Show that the energy equation is given by

2

2

y

T

t

T

[b] Due to conduction, the effect of plate temperature propagates into the fluid. The penetration

depth )(t increases with time. Use scaling to derive an expression for )(t .

(1) Observations. (i) The plate is infinite. (ii) No changes take place in the axial direction (infinite plate). (iii) This is a transient problem. (iv) Constant properties. (v) Cartesian coordinates. (vi) Gravity is neglected. Thus there is no free convection. (vii) The fluid is stationary.

(2) Problem Definition. [a] Determine the energy equation resulting from a step change in surface temperature. [b] Use scaling to estimate the thermal penetration thickness ).(t

(3) Solution Plan. [a] Apply energy equation and simplify it for the conditions of the problem. [b] Assign scales to each variable in the governing equation for temperature distribution.

(4) Plan Execution.

(i) Assumptions. (1) Newtonian fluid, (2) constant properties, (3) stationary fluid, and (4) negligible gravity.

(ii) Analysis. (i) Assumptions. (1) Continuum, (2) Newtonian fluid, (3) constant properties and (4) negligible nuclear, radiation and electromagnetic energy transfer.

(ii) Analysis. The energy equation for this case is given by

2

2

2

2

2

2

z

T

y

T

x

Tk

z

Tw

y

T

x

Tu

t

Tc p v (2.19b)

where



p pressure

T temperature density


Stationary fluid: 0wu v


No axial variation: 0x

No variation in the z-direction: 0z

Substituting these simplifications into (2.19.b)

2

2

y

Tk

t

Tc p

(a)

Introduce the definition of thermal diffusivity

k

c p (b)

(b) into (a)

2

2

y

T

t

T (c)

[b] To obtain scaling estimate of )(t rewrite (i)

y

T

yt

T(d)

Scale for :T )( io TTT

Scale for :u )0(tt

Scale for :y )0(y

Substitute into (d)

2

ioio TT

t

TTe

Solve for

tt)( f

(iii) Checking. Dimensional check: (i) Each term in (i) has units of. C/so :

C/s)C/m(/s)m()C/s( o2o

2

22o

y

T

t

T

(ii) Units of (f) should be length:

m(s)/s)m()( 2 tt

Limiting check: Initially the penetration thickness is zero. Setting t = 0 in (f) gives 0)0( ,

which is the correct result.

(5) Comments. (i) A major simplification of this problem is due to the assumption of infinite plate. Consequently, all derivatives with respect to x vanish. (ii) Since the fluid is stationary, the problem is one of pure conduction.

PROBLEM 3.1

A large plate moves with constant velocity oU parallel to a stationary plate separated by a

distance H. An incompressible fluid fills the channel formed by the plates. The stationary plate

is at temperature 1T and the moving plate is at temperature oT . Taking into consideration

dissipation, determine the maximum temperature and

the heat flux at the moving plate. Assume laminar

flow and neglect gravity effect and pressure

variation in the channel.

(1) Observations. (i) Moving plate sets fluid in motion in the x-direction. (ii) Since plates are infinite the flow field does not vary in the axial direction x. (iii) The effect of pressure gradient is negligible. (iv) The fluid is incompressible (constant density). (v) Use Cartesian coordinates.

(2) Problem Definition. Determine the velocity and temperature distribution.

(3) Solution Plan. Apply continuity and Navier-Stokes equations to determine the flow field. Apply the energy equation to determine the temperature distribution.

(4) Plan Execution.

(i) Assumptions. (i) Steady state, (ii) laminar flow, (iii) constant properties (density, viscosity and conductivity), (iv) infinite plates, (v) no end effects, (vi) uniform pressure (vi) negligible gravitational effect.

(ii) Analysis. Since the objective is the determination of temperature distribution and heat transfer rate, it is logical to begin the analysis with the energy equation. The energy equation for constant properties is given by (2.19b)

2

2

2

2

2

2

z

T

y

T

x

Tk

z

Tw

y

T

x

Tu

t

Tc v (2.19b)

where the dissipation function is given by (2.17)

2

3

2

2222222

z

w

yx

u

z

u

x

w

y

w

zxy

u

z

w

yx

u

v

vvv

(2.17)

Thus it is clear from (2.19b) and (2.17) that the determination of temperature distribution requires the determination of the velocity components u , v and w. This is accomplished by

applying continuity and the Navier-Stokes equations. We begin with the continuity equation in Cartesian coordinates

0z

w

yx

u

zw

yxu

t

vv (2.2b)

For constant density


0zyxt

(a)

Since plates are infinite

0wzx

(b)

Substituting (a) and (b) into (2.2b), gives

0y

v (c)

Integrating (c) )(xfv (d)

To determine the “constant” of integration )(xf we apply the no-slip boundary condition at the

lower plate 0)0,(xv (e)

Equations (d) and (e) give 0)(xf

Substituting into (d) 0v (f)

Since the vertical component v vanishes everywhere, it follows that the streamlines are parallel. To determine the horizontal component u we apply the Navier-Stokes equation in the x-direction,(2.10x)

2

2

2

2

2

2

z

u

y

u

x

u

x

pg

z

uw

y

u

x

uu

t

uxv (2.10x)

This equation is simplified as follows: Steady state

0t

u (g)

Negligible gravity effect

0xg (h)

Negligible axial pressure variation

0x

p (i)

Substituting (b) and (f)-(i) into (2.10x) gives

02

2

dy

ud (j)

The solution to (j) is

21 CyCu (k)

where 1C and 2C are constants of integration. The two boundary conditions on u are:

0)0(u and oUHu )( (l)

These conditions give


H

UC o

1 and 02C (m)

Substituting (m) into (k)

H

y

U

u

o

(3.8)

With the velocity distribution determined, we return to the dissipation function and energy equation. Substituting (b) and (f) into (2.17) gives

2

y

u (n)

Using solution (3.8) into (n) gives

2

2

H

U o (o)

Noting that for steady state 0/ tT and using (b), (f) and (o), the energy equation (2.10b)

simplifies to

02

2

2

2

H

U

dy

Tdk o (p)

In arriving at (p) axial temperature variation was neglected. This is valid for infinite plates at uniform surface temperature. Equation (p) is solved by direct integration

432

2

2

2CyCy

kH

UT o (q)

where 3C and 4C are constants of integration. The two boundary conditions on (q) are

1)0( TT and oTHT )( (r)

These boundary conditions and solution (q) give

Hk

U

H

TTC oo

2

21

3 and 14 TC (s)

Substituting (s) into (q) and rearranging the result in dimensionless form, give

H

y

H

y

TTk

U

H

y

TT

TT

o

o

o

1)(2 1

2

1

1 (t)

This can be written in terms of the Eckert and Prandtl numbers as

H

y

H

y

k

c

TTc

U

H

y

TT

TT p

op

o

o

1)(2 1

2

1

1

H

y

H

yPrE

H

y

TT

TT

o

121

1 (u)

where


)( 1

2

TTc

UE

op

o andk

cPr

p (v)

The maximum temperature occurs where the temperature gradient is zero. Differentiating (u),

setting the result equal to zero and solving for position of maximum temperature my

PrEH

ym 1

2

1 (w)

Substituting (w) into (u) gives the maximum temperature mT

8

PrE

PrETT

TT

o

m

2

1

2

1

1

1 (x)

The heat flux at the moving surface is determined by applying Fourier’s law at Hy

dy

HdTkHq

)()(

Using (u) into the above

12

)()( 1 PrE

H

TTkHq o (y)

(iii) Checking. Dimensional check: Each term in (3.8), (t), (u), (w) and (x) is dimensionless.

Units of (y) should be 2W/m :

2

o1

o

m

W

)m(

)C)()(CW/m()(

H

TTkHq o

Differential equation check: Velocity solution (3.8) satisfies equation (j) and temperature solution (t) satisfies (p).

Boundary conditions check: Velocity solution (3.8) satisfies boundary conditions (l) and temperature solution (t) satisfies boundary conditions (r).

Limiting check: (i) If the upper plate is stationary the fluid will also be stationary. Setting 0oU

in (3.8) gives .0)(yu

(ii) If the upper plate is stationary, dissipation will vanish, temperature distribution will be linear and surface flux at the upper plate will be due to conduction between the two surfaces. Setting

0oU in (v) gives E = 0. When this is substituted into (u) and (y) gives the anticipated linear

temperature distribution and a surface flux of

H

TTkHq o )(

)( 1

(iii) If the fluid is inviscid, dissipation will vanish and temperature should be linear. Setting 0 in (v ) gives 0Pr . When this is substituted into (u) gives a linear temperature

distribution.


(5) Comments. (i) Treating the plate as infinite is one of the key simplifying assumptions. This eliminates the x-coordinate as a variable and results in governing equations that are ordinary. Alternatively, one could state that the streamline are parallel. This means that .0/ yvv

substituting this into the continuity equation for two-dimensional incompressible flow gives .0/ xu This is identical equation (b) which is based on assuming infinite plate.

(ii) According to (w), maximum temperature occurs in the upper half of the channel.

k

UTT o

o2

)0(2

PROBLEM 3.2

A large plate moves with constant velocity oU

parallel to a stationary plate separated by a

distance H. An incompressible fluid fills the

channel formed by the plates. The upper plate

is maintained at uniform temperature oT and

the stationary plate is insulated. A pressure gradient dxdp / is applied to the fluid. Taking into

consideration dissipation, determine the temperature of the insulated plate and the heat flux at

the upper plate. Assume laminar flow and neglect gravity effect.

(1) Observations. (i) Moving plate sets fluid in motion in the x-direction. (ii) Since plates are infinite the flow field does not vary in the axial direction x. (iii) The effect of pressure gradient must be included. (iv) The fluid is incompressible. (v) Using Fourier’s law, Temperature distribution gives surface heat flux of the moving plate. (vi) Use Cartesian coordinates.



(4) Plan Execution.

(i) Assumptions. (i) Steady state, (ii) laminar flow, (iii) constant properties (density, viscosity and conductivity), (iv) infinite plates, (v) no end effects and (v) negligible gravitational effect.


2

2

2

2

2

2

z

T

y

T

x

Tk

z

Tw

y

T

x

Tu

t

Tc v (2.19b)


2

3

2

2222222

z

w

yx

u

z

u

x

w

y

w

zxy

u

z

w

yx

u

v

vvv

(2.17)



0z

w

yx

u

zw

yxu

t

vv (2.2b)


0zyxt

(a)



0wzx

(b)


0y

v (c)


To determine the “constant” of integration )(xf , we apply the no-slip boundary condition at the




Since the vertical component v vanishes everywhere, it follows that the streamlines are parallel. To determine the horizontal component u we apply the Navier-Stokes equations (2.10)

2

2

2

2

2

2

z

u

y

u

x

u

x

pg

z

uw

y

u

x

uu

t

uxv (2.10x)

2

2

2

2

2

2

zyxy

pg

zw

yxu

ty

vvvvvv

vv (2.10y)

These equations are simplified as follows: Steady state

0t

u (g)


0yx gg (h)

Substituting (b) and (f)-(h) into (2.10x) and (2.10y) gives

2

2

dy

ud

x

p (i)

and

0y

p (j)

Equation (j) shows that pressure does not vary in the y-direction and thus it can either be a function of x or constant. Integrating (i) twice

212

2

1CyCy

dx

dpu (k)



(1) 0)0(u

(2) oUHu )(


dx

dpH

H

UC o

21 and 02C (l)

Substituting (l) into (k)

H

y

dx

dp

U

H

H

y

U

u

oo

12

12

(m)


2

y

u (n)

Using solution (m) into (n) gives 2

1

2y

dx

dp

dx

dpH

H

Uo (o)


simplifies to

01

2

2

2

2

ydx

dp

dx

dpH

H

U

dy

Tdk o (p)


214

2

32

21

12

1

2322ByBy

dx

dp

ky

dx

dp

dx

dpH

H

U

ky

dx

dpH

H

U

kT oo (q)

where 1B and 2B are constants of integration. The two boundary conditions on (q) are

(1) 0)0(

dy

dT

(2) oTHT )(


01B (r)

4

2

32

2

2

1

12

1

2322H

dx

dp

kH

dx

dp

dx

dpH

H

U

kH

dx

dpH

H

U

kTB oo

o (s)

Substituting (r) into (q) and rearranging the result in dimensionless form, give


4

4

3

3

22

2

2

224

112

111

2

8

111

2

8

1

H

y

H

y

dx

dpH

U

H

y

dx

dpH

U

dx

dp

k

H

TT ooo (t)

Surface temperature of the insulated plate, T(0), is obtained by setting y = 0 in (t)

12

11

2

8

11

2

8

1)0(

2

2

224

dx

dpH

U

dx

dpH

U

dx

dp

k

H

TT ooo (u)

Surface heat flux at the upper plate is obtained by applying Fourier’s law at y = H

dy

HdTkHq

)()(

Using (t) into the above

3

11

2

2

11

2

4

1)(

2

2

223

dx

dpH

U

dx

dpH

U

dx

dpH

Hq oo (v)

(iii) Checking. Dimensional check: Each term in (m), (t), (u) and (v) is dimensionless. Each term in (q) has units of temperature.

Differential equation check: Velocity solution (m) satisfies equation (i) and temperature solution (t) satisfies (p).

Boundary conditions check: Velocity solution (m) and temperature solution (t) satisfy their respective boundary conditions.

Limiting check: (i) If the upper plate is stationary and there is no axial pressure gradient the fluid

will also be stationary. Setting 0/ dxdpUo in (m) gives .0)(yu

(ii) If the upper plate is stationary and there is no axial pressure gradient, there will be no fluid

motion and dissipation will vanish. The temperature distribution will be uniform equal to .oT

Setting 0/ dxdpUo in (t) gives .)( oTyT Similarly, surface heat flux will vanish. Setting

0/ dxdpUo in (v) gives .0)(Hq

(5) Comments. (i) Treating the plate as infinite is one of the key simplifying assumptions. This eliminates the x-coordinate as a variable and results in governing equations that are ordinary. (ii) According to (t), maximum temperature occurs at the insulated plate. (iii) According to the dimensionless form of solutions (u) and (v), the problems is characterized by the single dimensionless parameter

dx

dpH

Uo

2

2

PROBLEM 3.3

Incompressible fluid is set in motion between two

large parallel plates by moving the upper plate with

constant velocity oU and holding the lower plate

stationary. The clearance between the plates is H.

The lower plate is insulated while the upper plate

exchanges heat with the ambient by convection. The heat transfer coefficient is h and the

ambient temperature is .T Taking into consideration dissipation determine the temperature of

the insulated plate and the heat flux at the moving plate. Assume laminar flow and neglect

gravity effect.

(1) Observations. (i) Moving plate sets fluid in motion in the x-direction. (ii) Since plates are infinite the flow field does not vary in the axial direction x. (iii) The fluid is incompressible (constant density). (iv) Use Cartesian coordinates.



(4) Plan Execution.



2

2

2

2

2

2

z

T

y

T

x

Tk

z

Tw

y

T

x

Tu

t

Tc v (2.19b)


2

3

2

2222222

z

w

yx

u

z

u

x

w

y

w

zxy

u

z

w

yx

u

v

vvv

(2.17)



0z

w

yx

u

zw

yxu

t

vv (2.2b)



0zyxt

(a)


0wzx

(b)


0y

v (c)







2

2

2

2

2

2

z

u

y

u

x

u

x

pg

z

uw

y

u

x

uu

t

uxv (2.10x)


0t

u (g)


0xg (h)


0x

p (i)


02

2

dy

ud (j)


21 CyCu (k)





H

UC o

1 and 02C (m)


H

y

U

u

o

(3.8)


2

y

u (n)


2

2

H

U o (o)


simplifies to

02

2

2

2

H

U

dy

Tdk o (p)


432

2

2

2CyCy

kH

UT o (q)


(1) 0)0(

dy

dT

(2) TyThdy

HdTk )(

)(


03C (r)

k

U

khH

UTC oo

2

22

4 (s)

Substituting (r) and (s) into (q) and rearranging the result in dimensionless form, give

2

2

2 2

1

2

1

H

y

hH

k

k

U

TT

o

(t)

The dimensionless parameter khH / is known as the Biot number, Bi . It is associated with

convection boundary conditions.


The temperature of the insulated surface, T(0), is obtained by evaluating (t) at y = 0.

hH

k

k

U

TT

o2

1)0(2

(u)


dy

HdTkHq

)()(

Substituting (t) into the above

H

UHq o

2

)( (v)

(iii) Checking. Dimensional check: Each term in (3.8), (t) and (u) is dimensionless. Each term

in solution (v) has units of 2W/m .

Differential equation check: Velocity solution (3.8) satisfies equation (j) and temperature solution (t) satisfies (p). Boundary conditions check: Velocity solution (3.8) satisfies boundary conditions (l) and temperature solution (t) satisfies boundary conditions (r).

Limiting check: (i) If the upper plate is stationary the fluid will also be stationary. Setting 0oU

in (3.8) gives .0)(yu

(ii) If the upper plate is stationary, dissipation will vanish, temperature distribution will be

uniform equal to the ambient temperature .T Setting 0oU in (u) gives .)( TyT Similarly

the heat flux )(Hq vanishes. Substituting 0oU (v) gives .0)(Hq

(iii) If the fluid is inviscid, dissipation will vanish and temperature should be uniform equal to

.T Setting 0 in (u) gives .)( TyT Similarly the heat flux )(Hq vanishes. Substituting

0 (v) gives .0)(Hq

Global energy balance: Energy leaving the channel must equal to the work done to move the plate. Consider the work done by the plate on the fluid

ooUW (w)

where

W work done per unit surface area by the plate on the fluid

o shearing stress at the moving plate

However, shearing stress is given by

y

Huo

)( (x)

(3.8) into (x)

H

Uoo (y)

(y) into (w)


H

UW o

2

(z)

This is identical to the heat removed from the upper plate given in equation (v).

(5) Comments. (i) Treating the plate as infinite is one of the key simplifying assumptions. This eliminates the x-coordinate as a variable and results in governing equations that are ordinary. Alternatively, one could state that the streamline are parallel. This means that .0/ yvv (ii)

The solutions is characterized by a single dimensionless parameter ,/ khH which is the Biot

number. (iii) The Nusselt number at moving plate, ),(HNu is defined as

k

hHHNu )( (w)

The heat transfer coefficient h is based on the overall temperature drop, defined as

)()0(

)(

HTT

Hqh (x)

Centerline temperature and moving plate temperature are obtained by evaluating (t) at y = 0 and y = H

hH

k

k

UTT o

2

1)0(

2

and

Hh

UTHT o

2

2

1)(

The above two equations give

k

UHTT o

2

2

1)()0( (y)

(v) and (y) into (x)

H

kh 2

Substituting into (w) gives the Nusselt number

2)(HNu (z)

PROBLEM 3.4

Two parallel plates are separated by a

distance 2H. The plates are moved in

opposite direction with constant velocity oU .

Each plate is maintained at uniform

temperature oT . Taking into consideration

dissipation determine the heat flux at the

plates. Assume laminar flow and neglect

gravity effect

(1) Observations. (i) Moving plates set fluid in motion in the positive and negative x-direction.(ii) Since plates are infinite the flow field does not vary in the axial direction x. (iii) The fluid is incompressible (constant density). (iv) The fluid is stationary at the center plane y = 0. (v) Symmetry dictates that no heat is conducted through the center plane. (vi) Use Cartesian coordinates.



(4) Plan Execution.


(ii) Analysis. Taking advantage of symmetry only the upper half of the channel is analyzed. Since the objective is the determination of temperature distribution and heat transfer rate, it is logical to begin the analysis with the energy equation. The energy equation for constant properties is given by (2.19b)

2

2

2

2

2

2

z

T

y

T

x

Tk

z

Tw

y

T

x

Tu

t

Tc v (2.19b)


2

3

2

2222222

z

w

yx

u

z

u

x

w

y

w

zxy

u

z

w

yx

u

v

vvv

(2.17)



0z

w

yx

u

zw

yxu

t

vv (2.2b)



0zyxt

(a)


0wzx

(b)


0y

v (c)







2

2

2

2

2

2

z

u

y

u

x

u

x

pg

z

uw

y

u

x

uu

t

uxv (2.10x)


0t

u (g)


0xg (h)


0x

p (i)


02

2

dy

ud (j)


21 CyCu (k)





H

UC o

1 and 02C (m)


H

y

U

u

o

(3.8)


2

y

u (n)


2

2

H

U o (o)


simplifies to

02

2

2

2

H

U

dy

Tdk o (p)


432

2

2

2CyCy

kH

UT o (q)


(1) 0)0(

dy

dT

(2) oTHT )(


03C (r)

k

UTC o

o2

2

4 (s)

Substituting (r) and (s) into (q) and rearranging the result in dimensionless form, give

2

2

21

2

1

H

y

k

U

TT

o

o (t)

The temperature of the insulated surface, T(0), is obtained by evaluating (t) at y = 0.



dy

HdTkHq

)()(

Substituting (t) into the above

H

UHq o

2

)( (u)

Similarly, the heat flux at the lower plate is

H

UHq o

2

)( (v)

(iii) Checking. Dimensional check: Each term in (3.8) and (t) is dimensionless. Each term in

solutions (u) and (v) has units of 2W/m .

Differential equation check: Velocity solution (3.8) satisfies equation (j) and temperature solution (t) satisfies (p).

Boundary conditions check: Velocity solution (3.8) and temperature solution (t) satisfy their respective boundary conditions.

Limiting check: (i) If the two plates are stationary, the fluid will also be stationary. Setting

0oU in (3.8) gives .0)(yu

(ii) If the upper and lower plates are stationary, dissipation will vanish, temperature distribution

will be uniform equal to the ambient temperature .oT Setting 0oU in (t) gives .)( oTyT

Similarly the heat flux )(Hq vanishes. Substituting 0oU (u) gives .0)(Hq

(iii) If the fluid is inviscid, dissipation will vanish and temperature should be uniform equal to

.oT Setting 0 in (t) gives .)( oTyT Similarly the heat flux )(Hq vanishes. Substituting

0 (u) gives .0)(Hq

(5) Comments. (i) Treating the plate as infinite is one of the key simplifying assumptions. This eliminates the x-coordinate as a variable and results in governing equations that are ordinary. Alternatively, one could state that the streamline are parallel. This means that .0/ yvv (ii)

Symmetry provides additional simplification. (iii) The Nusselt number at the upper moving plate, ),(HNu is defined as

k

hHHNu )( (w)

Defining the heat transfer coefficient is defined as

oTT

Hqh

)0(

)((x)

Centerline temperature is obtained by evaluating (t) at y = 0

k

UTT o

o

2

2

1)0( (y)


(u) and (y) into (x) gives h

H

kh 2

Substituting into (w) gives the Nusselt number

2)(HNu (z)

A more appropriate definition of the heat transfer coefficient is based on the mean temperature,

,mT rather than the temperature at the center. That is

om TT

Hqh

)(

PROBLEM 3.5

Incompressible fluid flows in a long tube of radius

or . Fluid motion is driven by an axial pressure

gradient ./ zp The tube exchanges heat by

convection with an ambient fluid. The heat transfer

coefficient is h and the ambient temperature is .T

Taking into consideration dissipation, assuming

laminar incompressible axisymmetric flow, and neglecting gravity, axial temperature variation

and end effects, determine:

[a] Surface temperature.

[b] Surface heat flux.

[c] Nusselt number based on [ )()0( orTT ].

(1) Observations. (i) Fluid motion is driven by axial pressure drop. (ii) For a very long tube the flow field does not vary in the axial direction z. (iii) The fluid is incompressible (constant density). (iv) Heat is generated due to viscous dissipation. It is removed from the fluid by convection at the surface. (v) The Nusselt number is a dimensionless heat transfer coefficient. (vi) To determine surface heat flux and heat transfer coefficient requires the determination of temperature distribution. (vii) Temperature distribution depends on the velocity distribution. (viii) Use cylindrical coordinates.


(3) Solution Plan. Apply continuity and Navier-Stokes equations in cylindrical coordinates to determine the flow field. Apply the energy equation to determine temperature distribution. Fourier’s law gives surface heat flux. Equation (1.10) gives the heat transfer coefficient.

(4) Plan Execution.

(i) Assumptions. (i) Steady state, (ii) laminar flow, (iii) axisymmetric flow, (iv) constant properties (density, viscosity and conductivity), (v) no end effects, (vi) uniform ambient temperature, (vii) uniform heat transfer coefficient and (viii) negligible gravitational effect.

(ii) Analysis. [a] Since temperature distribution is obtained by solving the energy equation, we begin the analysis with the energy equation. The energy equation in cylindrical coordinates for constant properties is given by (2.24)

2

2

2

2

2 0

11

z

TT

rr

Tr

rrk

z

TT

rr

T

t

Tc zrP v

vv (2.24)


22

0

2

0

222

0

1

0

12

122

rzzr

rrrzrrr

zrz

rzrr

vvvv

vvvvvvv

(2.25)

Equations (2.24) and (2.25) show that the determination of temperature distribution requires the

determination of the velocity components rv , v and .zv The flow field is determined by


solving the continuity and the Navier-Stokes equations. We begin with the continuity equation in cylindrical coordinates

011

zrzr

rrrt

vvv (2.4)


0zrt

(a)

For axisymmetric flow

0v (b)

For a long tube with no end effects axial changes in velocity are negligible

0z

(c)

Substituting (a)-(c) into (2.4)

0rrdr

dv (d)

Integrating (d)

)(zfr rv (e)

To determine the “constant” of integration )(zf we apply the no-slip boundary condition at the

surface

0),( zrov (f)

Equations (e) and (f) give 0)(zf

Substituting into (e)

0rv (g)

Since the radial component rv vanishes everywhere, it follows that the streamlines are parallel

to the surface. To determine the axial component zv we apply the Navier-Stokes equation in the

z-direction, (2.11z)

2

2

2

2

2

11

zrrr

rrz

pg

tzrr

zzzz

zzzz

vvv

vvv

vvvv zr

(2.11z)

This equation is simplified as follows:

Steady state

0t

(h)


0zr gg (i)

Substituting (b), (c) and (g)-(i) into (2.11z) gives


01

dr

dr

dr

d

rz

p zv (3.11)

Since zv depends on r only, equation (3.11) can be written as

)(1

rgdr

dr

dr

d

rz

p zv (j)

Integrating (j) with respect to z

oCzrgp )( (k)

where oC is constant of integration. We turn our attention now to the radial component of

Navier-Stokes equation, (2.11r)

2

2

22

2

2

2

21)(

1

zrrr

rrrr

pg

tzrrr

rrr

r

rrz

rrr

vvvv

vvv

vvvvv

(2.11r)

Substituting (b), (g) and (i) into (2.11r), gives

0r

p (l)

Integrating (l) )(zfp (m)

where )(zf is “constant” of integration. We now have two solutions for the pressure p: (k) and

(m). Equating the two, gives

)()( zfCzrgp o (n)

One side of (n) shows that the pressure depends on z only while the other side shows that it depends on r and z. This, of course, is a contradiction. The only possibility for reconciling this is by requiring that

g(r) = C (o)

where C is a constant. Substituting (o) into (j)

Cdr

dr

dr

d

rz

p zv1 (p)

Thus the axial pressure gradient in the tube is constant. Equation (p) can now be integrated to give the axial velocity distribution. Integrating once

12

2

1Cr

zd

pd

dr

dr zv

Separating variables and integrating again


212 ln

4

1CrCr

zd

pdzv (q)

where 1C and 2C are constants of integration. The two boundary conditions on zv are

,0)0(

dr

d zv 0)( oz rv (r)

Equations (q) and (r) give 1C and 2C

,01C 22

4

1or

zd

pdC

Substituting into (q)

)(4

1 22oz rr

zd

pdv (3.12)

With the velocity distribution determined we return to the energy equation (2.24) and the dissipation function (2.25). We note that for a long tube at uniform surface temperature with no end effects, axial temperature variation can be neglected. Thus

02

2

z

T

z

T (s)

Substituting (b), (c), (g), (h) and (s) into (2.24)

01

dr

dTr

dr

d

rk (t)

Using (b), (c) and (g) into (2.25) gives the dissipation function for this flow

2

dr

d zv

Substituting the velocity solution (3.11) into the above, gives

2

2

2

1r

zd

pd (u)

Using (u) to eliminate in (t) and rearranging, we obtain

dr

dTr

dr

d 3

2

4

1r

zd

pd

k

Integrating the above twice

434

2

ln64

1CrCr

zd

pd

kT (v)

Two boundary conditions are needed to evaluate the constants of integration 3C and 4C . They

are:

(1) 0)0(

dr

dT


(2) ])([)(

TrThdr

rdTk o

o

Equations (v) and the two boundary conditions give the two constants

,03C ]14[64

24

4o

o

hr

k

zd

pd

k

rTC

Substituting the above into (v)

2424

64]14[

64 zd

pd

k

r

hr

k

zd

pd

k

rTT

o

o (w)

This solution can be expressed in dimensionless form as

4

4

2414

64

ooo

r

r

hr

k

zd

pd

k

r

TT (x)

The dimensionless parameter k

hro in (x) is known as the Biot number.

Surface temperature is obtained by setting orr in (w)

hzd

pdrTrT o

o

1

16)(

23

(y)

[b] Surface heat flux )( orq is obtained by applying Fourier’s law

dr

rdTkrq o

o

)()(

Using (w) into the above 23

16)(

zd

pdrrq o

o (z)

[c] The Nusselt number is defined as

k

hr

k

hDNu o2

(z-1)

where D is tube diameter. The heat transfer coefficient h is determined using equation (1.10)

dr

rdT

rTT

kh o

o

)(

)]()0([ (z-2)

Substituting (w) into the above

or

kh

4 (z-3)

Introducing (z-3) into (z-1) 8Nu (z-4)


(iii) Checking. Dimensional check: Each term in (3.12) has units of velocity. Each term in (w) has units of temperature. Each term in (x) is dimensionless.

Differential equation check: Velocity solution (3.12) satisfies equation (p) and temperature solution (w) satisfies (t).

Boundary conditions check: Velocity solution (3.12) satisfies boundary conditions (r) and temperature solution (w) satisfies boundary conditions (1) and (2).

Limiting check: (i) If pressure is uniform ( 0/ dzdp ) the fluid will be stationary. Setting

0/ dzdp in (3.12) gives .0zv

(ii) If pressure is uniform ( 0/ dzdp ) the fluid will be stationary and no dissipation takes place

and thus surface heat transfer should vanish Setting 0/ dzdp in (z) gives .0)( orq

(iii) Global conservation of energy. Heat transfer rate leaving the tube must be equal to the work required to pump the fluid. Pump work for a tube section of length L is

QppW )( 21 (z-1)

Where

1p = upstream pressure

2p = downstream pressure

Q = volumetric flow rate, given by

or

v0

2 rdrQ z

Substituting (3.12) into the above and integrating

4

8or

dz

dpQ (z-2)

Combining (z-1) and (z-2))

)(8

21

4

ppdz

dprW o (z-3)

Work per unit area W is

Lr

WW

o2

Substituting (z-3) into the above

L

pp

dz

dprW o )(

16

213

(z-4)

However

dz

dp

L

pp )( 21

Combining this result with (z-4) gives 23

16 dz

dprW o


This result is identical to surface heat transfer rate given in (z)

(5) Comments. (i) The assumption of a long tube with negligible end effects is a key factor in simplifying the problem. This assumption eliminates the z-coordinate as a variable and results in governing equations that are ordinary. Assuming parallel streamlines is equivalent to neglecting end effects since it leads to the same mathematical simplifications. (ii) Solution (w) shows that maximum temperature occurs at the center .0r

(iii) The Nusselt number is constant independent of Reynolds and Prandtl numbers.

PROBLEM 3.6

Fluid flows axially in the annular space

between a cylinder and a concentric rod . The

radius of the rod is ir and that of the cylinder

is .or Fluid motion in the annular space is

driven by an axial pressure gradient ./ zp

The cylinder is maintained at uniform

temperature .oT Assume incompressible laminar axisymmetric flow and neglect gravity and end

effects. Show that the axial velocity is given by

1)/ln()/ln(

)/(1)/(

4

22

2

o

io

oio

oz rr

rr

rrrr

dz

dprv

(1) Observations. (i) Fluid motion is driven by axial pressure drop. (ii) For a very long tube the flow field does not vary in the axial direction z. (iii) The fluid is incompressible (constant density). (iv) Use cylindrical coordinates.

(2) Problem Definition. Determine the velocity distribution.

(3) Solution Plan. Apply continuity and Navier-Stokes equations in cylindrical coordinates to determine the flow field.

(4) Plan Execution.

(i) Assumptions. (i) Steady state, (ii) laminar flow, (iii) axisymmetric flow, (iv) constant properties (density and viscosity), (v) no end effects and (vi) negligible gravitational effect.

(ii) Analysis. The flow field is determined by solving the continuity and the Navier-Stokes equations. We begin with the continuity equation in cylindrical coordinates

011

zrzr

rrrt

vvv (2.4)


0zrt

(a)


0v (b)


0z

(c)


0rrdr

dv (d)

Integrating (d)


)(zfr rv (e)


surface

0),( zrov (f)



0rv (g)




2

2

2

2

2

11

zrrr

rrz

pg

tzrr

zzzz

zzzz

vvv

vvv

vvvv zr

(2.11z)


Steady state

0t

(h)


0zr gg (i)


01

dr

dr

dr

d

rz

p zv (3.11)


)(1

rgdr

dr

dr

d

rz

p zv (j)


oCzrgp )( (k)



2

2

22

2

2

2

21)(

1

zrrr

rrrr

pg

tzrrr

rrr

r

rrz

rrr

vvvv

vvv

vvvvv

(2.11r)



0r

p (l)




)()( zfCzrgp o (n)


g(r) = C (o)


Cdr

dr

dr

d

rz

p zv1 (p)


12

2

1Cr

zd

pd

dr

dr zv


212 ln

4

1CrCr

zd

pdzv (q)


,0)( iz rv 0)( oz rv (r)

Equations (q) and (r) give 1C and 2C1

221 ln)(

4

1

i

ooi

r

rrr

zd

pdC

2

1

222

4

1lnln)(

4

1ii

i

ooi r

zd

pdr

r

rrr

zd

pdC

Substituting into (q) and rearranging

1)/ln()/ln(

)/(1)/(

4

22

2

o

io

oio

oz rr

rr

rrrr

dz

dprv (3.12)

(iii) Checking. Dimensional check: Each term in (3.12) has units of velocity.

Differential equation check: Velocity solution (3.12) satisfies equation (p).


Boundary conditions check: Velocity solution (3.12) satisfies boundary conditions (r).



(5) Comments. The assumption of a long tube with negligible end effects is a key factor in simplifying the problem. This assumption eliminates the z-coordinate as a variable and results in governing equations that are ordinary. Assuming parallel streamlines is equivalent to neglecting end effects since it leads to the same mathematical simplifications.

PROBLEM 3.7

A rod of radius ir is placed concentrically inside a cylinder of radius .or The rod moves axially

with constant velocity oU and sets the

fluid in the annular space in motion. The

cylinder is maintained at uniform

temperature .oT Neglect gravity and end

effects, and assume incompressible

laminar axisymmetric flow

[a] Show that the axial velocity is given by

)/ln()/ln(

ooi

oz rr

rr

Uv

[b] Taking into consideration dissipation, determine the heat flux at the outer surface and the

Nusselt number based on [ oi TrT )( ]. Neglect axial temperature variation.

(1) Observations. (i) Fluid motion is driven by axial motion of the rod. Thus motion is not due to pressure gradient. (ii) For a very long tube the flow field does not vary in the axial direction z.(iii) The fluid is incompressible (constant density). (iv) Heat is generated due to viscous dissipation. It is removed from the fluid by conduction at the surface. (v) The Nusselt number is a dimensionless heat transfer coefficient. (vi) To determine the heat transfer coefficient require the determination of temperature distribution. (vii) Temperature distribution depends on the velocity distribution. (viii) Use cylindrical coordinates.


(3) Solution Plan. Apply continuity and Navier-Stokes equations in cylindrical coordinates to determine the flow field. Apply the energy equation to determine temperature distribution. Fourier’s law gives surface heat flux. Equation (1.10) gives the heat transfer coefficient.

(4) Plan Execution.

(i) Assumptions. (i) Steady state, (ii) laminar flow, (iii) axisymmetric flow, (iv) constant properties (density, viscosity and conductivity), (v) no end effects, (vi) uniform surface temperature and (vii) constant rod velocity, (viii) negligible axial pressure gradient and (ix) negligible gravitational effect.

(ii) Analysis. [a] Velocity distribution is governed by the continuity equation and Navier-Stokes equations of motion. The flow field is determined by solving the continuity and the Navier-Stokes equations. We begin with the continuity equation in cylindrical coordinates

011

zrzr

rrrt

vvv (2.4)


0zrt

(a)



0v (b)


0z

(c)


0rrdr

dv (d)

Integrating (d)

)(zfr rv (e)


surface

0),( zrov (f)



0rv (g)




2

2

2

2

2

11

zrrr

rrz

pg

tzrr

zzzz

zzzz

vvv

vvv

vvvv zr

(2.11z)


Steady state

0t

(h)


0zr gg (i)


01

dr

dr

dr

d

rz

p zv (3.11)


)(1

rgdr

dr

dr

d

rz

p zv (j)



oCzrgp )( (k)



2

2

22

2

2

2

21)(

1

zrrr

rrrr

pg

tzrrr

rrr

r

rrz

rrr

vvvv

vvv

vvvvv

(2.11r)


0r

p (l)




)()( zfCzrgp o (n)


g(r) = C (o)


Cdr

dr

dr

d

rz

p zv1 (p)


12

2

1Cr

zd

pd

dr

dr zv


212 ln

4

1CrCr

zd

pdzv (q)


,0)( iz rv 0)( oz rv (r)

Equations (q) and (r) give 1C and 2C

,01C 22

4

1or

zd

pdC

Substituting into (q)


)(4

1 22oz rr

zd

pdv (3.12)

Since temperature distribution is obtained by solving the energy equation, we begin the analysis with the energy equation. The energy equation in cylindrical coordinates for constant properties is given by (2.24)

2

2

2

2

2 0

11

z

TT

rr

Tr

rrk

z

TT

rr

T

t

Tc zrP v

vv (2.24)


22

0

2

0

222

0

1

0

12

122

rzzr

rrrzrrr

zrz

rzrr

vvvv

vvvvvvv

(2.25)

Equations (2.24) and (2.25) show that the determination of temperature distribution requires the

determination of the velocity components rv , v and .zv

With the velocity distribution determined we return to the energy equation (2.24) and the dissipation function (2.25). We note that for a long tube at uniform surface temperature with no end effects, axial temperature variation can be neglected. Thus

02

2

z

T

z

T (s)

Substituting (b), (c), (g), (h) and (s) into (2.24)

01

dr

dTr

dr

d

rk (t)

Using (b), (c) and (g) into (2.25) gives the dissipation function for this flow

2

dr

d zv

Substituting the velocity solution (3.11) into the above, gives 2

2

2

1r

zd

pd

(u)

Using (u) to eliminate in (t) and rearranging, we obtain

dr

dTr

dr

d 3

2

4

1r

zd

pd

k (3.13)

Integrating the above twice

434

2

ln64

1CrCr

zd

pd

kT (v)


Two boundary conditions are needed to evaluate the constants of integration 3C and 4C . They

are:

0)0(

dr

dT and oo TrT )( (w)

Equations (v) and (w) give the two constants

,03C 4

2

464

1oo r

zd

pd

kTC

Substituting the above into (v)

4

424

164

o

oo

r

r

zd

pd

k

rTT (3.14a)


4

4

241

64

oo

o

r

r

zd

pd

k

r

TT (3.14b)

[b] Surface heat flux )( orq is obtained by applying Fourier’s law

dr

rdTkrq o

o

)()(

Using (3.14) into the above 23

16)(

zd

pdrrq o

o (3.15)

[c] The Nusselt number is defined as

k

hr

k

hDNu o2

(x)

where D is tube diameter. The heat transfer coefficient h is determined using equation (1.10)

dr

rdT

TT

kh o

o

)(

])0([ (y)

Substituting (3.14a) into (y) or

kh

4 (z)

Substituting (z) into (x) 8Nu (3.16)

(iii) Checking. Dimensional check: Each term in (3.12) has units of velocity. Each term in

)14.3( a as units of temperature. Each term in (3.15) has units of .W/m2

Differential equation check: Velocity solution (3.12) satisfies equation (p) and temperature solution (3.14) satisfies (3.13).


Boundary conditions check: Velocity solution (3.12) satisfies boundary conditions (r) and temperature solution (3.14) satisfies boundary conditions (w).



(ii) If pressure is uniform ( 0/ dzdp ) the fluid will be stationary and no dissipation takes place

and thus surface heat transfer should vanish Setting 0/ dzdp in (3.15) gives .0)( orq

(iii) Global conservation of energy. Heat transfer rate leaving the tube must be equal to the rate of work required to pump the fluid. Pump work for a tube section of length L is

QppW )( 21 (z-1)

Where

1p = upstream pressure

2p = downstream pressure

Q = volumetric flow rate, given by

or

v0

2 rdrQ z

Substituting (3.12) into the above and integrating

4

8or

dz

dpQ (z-2)

Combining (z-1) and (z-2))

)(8

21

4

ppdz

dprW o (z-3)

Work per unit area W is

Lr

WW

o2

Substituting (z-3) into the aboveL

pp

dz

dprW o )(

16

213

(z-4)

However

dz

dp

L

pp )( 21

Combining this result with (z-4) gives 23

16 dz

dprW o

This result is identical to surface heat transfer rate given in (3.15)


(5) Comments. (i) The assumption of a long tube with negligible end effects is a key factor in simplifying the problem. This assumption eliminates the z-coordinate as a variable and results in governing equations that are ordinary. Assuming parallel streamlines is equivalent to neglecting end effects since it leads to the same mathematical simplifications.

(ii) Solution (3.14) shows that maximum temperature occurs at the center .0r

(iii) The Nusselt number is constant independent of Reynolds and Prandtl numbers.

PROBLEM 3.8

A liquid film of thickness H flows down an inclined plane due to

gravity. The plane is maintained at uniform temperature oT and

the free film surface is insulated. Assume incompressible laminar

flow and neglect axial variation of velocity and temperature and

end effects.


2

22

2

1sin

H

y

H

ygHu

[b] Taking into consideration dissipation, determine the heat flux at the inclined plane.

(1) Observations. (i) Fluid motion is driven by gravity. (ii) No velocity and temperature variation in the axial direction. (iii) The fluid is incompressible (constant density). (iv) Heat is generated due to viscous dissipation. (v) Temperature distribution depends on the velocity distribution. (vi) Use Cartesian coordinates.


(3) Solution Plan. Apply Navier-Stokes and continuity equations in Cartesian coordinates to determine the flow field. Apply the energy equation to determine temperature distribution.

(4) Plan Execution.

(i) Assumptions. (i) Steady state, (ii) laminar flow, (iii) constant properties (density, viscosity and conductivity), (iv) no end effects, (v) no motion in the z-direction and (vi) velocity and temperature do not vary in the axial direction. .

(ii) Analysis. [a] The Navier-Stokes equation for constant properties are

x-direction:2

2

2

2

2

2

z

u

y

u

x

u

x

pg

z

uw

y

u

x

uu

t

uxv (2.10x)

y-direction:2

2

2

2

2

2

zyxy

pg

zw

yxu

ty

vvvvvv

vv (2.10y)

Note that the sign of the gravity force in (2.10y) is negative since this component of gravity points in the negative y-direction. That is

singg x and cosgg y

These equations are simplified as follows:

0zx

u

tt

u v


2

2

y

u

x

pg

y

uxv (a)

2

2

2

2

yxy

pg

yxu y

vvvv

v (b)

The continuity equation introduces additional simplifications. For two-dimensional constant properties, the continuity equations gives

0yx

u v

Since 0x

u, it follows that

0y

v (c)

IntegratingCv (d)

However, the no slip condition at the wall gives

0(0)v

Applying this condition to (d) gives

C = 0 Thus

0v (d) Substituting (d) into (a) and (b)

2

2

0y

u

x

pg x (e)

y

pg y0 (f)

Integrating (f)

1Cygp y (g)

The pressure at the free surface is atmospheric. Thus

apHp )(

(g) gives

1CHgp ya

HgpC ya1

Substituting into (g)

ay pyHgp )( (h)

This result shows that pressure is independent of x. thus


0x

p (i)


xgy

u2

2

(j)

Integrating twice

32

2

2CyC

ygu x (k)

The two boundary conditions are

(1) 0)0(u

(2) 0)(

dy

Hdu

These boundary conditions give

Hg

C x2 and 03C

Substituting into (k)

2

22

2

1

H

y

H

yHgu x (l)

However

singg x (m)

Introducing (m) into (l)

2

22

2

1sin

H

y

H

ygHu (n)

[b] The energy equation is the starting point for determining the temperature distribution. The energy equation for constant properties is given by (2.19b)

2

2

2

2

2

2

z

T

y

T

x

Tk

z

Tw

y

T

x

Tu

t

Tc p v (2.19b)


2

3

2

2222222

z

w

yx

u

z

u

x

w

y

w

zxy

u

z

w

yx

u

v

vvv

(2.17)

Equation (2.19b) simplifies to

02

2

y

Tk (o)


The dissipation function simplifies to

2

y

u (p)

Using (n) into (p) 2

2

2

1)sin(

H

ygH (q)

Introducing (q) into (o)

01)sin(

22

2

2

H

ygH

dy

Tdk (r)

Integrating (r) twice

212

4322

1232

)sin(AyA

H

y

H

yy

k

gHT (s)


(1) oTT )0(

(2) 0)(

dy

HdT

The two boundary conditions give the constants 1A and 2A

23

1 )sin(3

gk

HA

oTA2

Substituting into (s) and rearranging the result in dimensionless form

4

4

3

3

2

2

24 12

1

3

1

2

1

3

1

)sin( H

y

H

y

H

y

H

y

k

gH

TT o (t)

Surface heat flux is determined using Fourier’s law

dy

dTkq

)0()0(

3

)sin()0(

23 gHq (u)

(iii) Checking. Dimensional check: Equation (n) has units of velocity. Each term in (t) is dimensionless. Each term in (u) has units of heat flux.

Differential equation check: Velocity solution (n) satisfies equation (j) and temperature solution (t) satisfies (r).

Boundary conditions check: Velocity solution (n) satisfies the two boundary conditions following equation (k) and temperature solution (t) satisfies the boundary conditions following equation (s).


Limiting check: (i) If gravity or inclination angle vanishes the fluid will be stationary. Setting 0g or 0 in (n) gives .0u

(ii) If gravity or inclination angle vanishes the fluid will be stationary, dissipation will also

vanish and the temperature will be uniform throughout equal to .oT Setting 0g or 0 in

(t) gives .oTT

(5) Comments. (i) Neglecting axial variation of u and T are key simplifying assumptions in this problem. (ii) Surface heat flux is negative since all energy generated due to friction must leave through the inclined plane.

PROBLEM 3.9

A liquid film of thickness H flows down an inclined plane due to gravity. The plane exchanges

heat by convection with an ambient fluid. The heat transfer coefficient is h and the ambient

temperature is .T The inclined surface is insulated. Assume incompressible laminar flow and

neglect axial variation of velocity and temperature and end effects.


2

22

2

1sin

H

y

H

ygHu

[b] Taking into consideration dissipation, determine the heat flux

at the free surface.

(1) Observations. (i) Fluid motion is driven by gravity. (ii) No velocity and temperature variation in the axial direction. (iii) The fluid is incompressible (constant density). (iv) Heat is generated due to viscous dissipation. (v) Temperature distribution depends on the velocity distribution. (vi) the inclined surface is at specified temperature and the free surface exchanges heat by convection with the ambient. (vii) Use Cartesian coordinates.


(3) Solution Plan. Apply Navier-Stokes and continuity equations in Cartesian coordinates to determine the flow field. Apply the energy equation to determine temperature distribution.

(4) Plan Execution.

(i) Assumptions. (i) Steady state, (ii) laminar flow, (iii) constant properties (density, viscosity and conductivity), (iv) no end effects, (v) no motion in the z-direction, (vi) uniform heat transfer coefficient and ambient temperature and (vii) velocity and temperature do not vary in the axial direction.

(ii) Analysis. [a] The Navier-Stokes equati on for constant properties are

x-direction:2

2

2

2

2

2

z

u

y

u

x

u

x

pg

z

uw

y

u

x

uu

t

uxv (2.10x)

y-direction:2

2

2

2

2

2

zyxy

pg

zw

yxu

ty

vvvvvv

vv (2.10y)

Note that the sign of the gravity force in (2.10y) is negative since this component of gravity points in the negative y-direction. That is

singg x and cosgg y

These equations are simplified as follows:

0zx

u

tt

u v


2

2

y

u

x

pg

y

uxv (a)

2

2

2

2

yxy

pg

yxu y

vvvv

v (b)

The continuity equation introduces additional simplifications. For two-dimensional constant properties, the continuity equations gives

0yx

u v

Since 0x

u, it follows that

0y

v (c)

Integrating

Cv

However, the no slip condition at the wall gives

0(0)v

Applying this condition gives

C = 0 Thus

0v (d) Substituting (d) into (a) and (b)

2

2

0y

u

x

pg x (e)

y

pg y0 (f)

Integrating (f)

1Cygp y (g)

The pressure at the free surface is atmospheric. Thus

apHp )(

(g) gives

1CHgp ya

HgpC ya1


ay pyHgp )( (h)

This result shows that pressure is independent of x. thus


0x

p (i)


xgy

u2

2

(j)

Integrating twice

32

2

2CyC

ygu x (k)


(1) 0)0(u

(2) 0)(

dy

Hdu

These boundary conditions give

Hg

C x2 and 03C


2

22

2

1

H

y

H

yHgu x (l)

However

singg x (m)

Introducing (m) into (l)

2

22

2

1sin

H

y

H

ygHu (n)

[b] The energy equation is the starting point for determining the temperature distribution. The energy equation for constant properties is given by (2.19b)

2

2

2

2

2

2

z

T

y

T

x

Tk

z

Tw

y

T

x

Tu

t

Tc p v (2.19b)


2

3

2

2222222

z

w

yx

u

z

u

x

w

y

w

zxy

u

z

w

yx

u

v

vvv

(2.17)


Equation (2.19b) simplifies to

02

2

y

Tk (o)

The dissipation function simplifies to

2

y

u (p)

Using (n) into (p) 2

2

2

1)sin(

H

ygH (q)

Introducing (q) into (o)

01)sin(

22

2

2

H

ygH

dy

Tdk (r)

Integrating (r) twice

232

4322

1232

)sin(CyC

H

y

H

yy

k

gHT (s)


(1) 0)0(

dy

dT

(2) THThdy

HdTk )(

)(

The two boundary conditions give the constants 3C and 4C

03C

43

)sin( 32

4

H

h

k

k

HgTC

Substituting into (s) and rearranging

4

4

3

3

2

224

12

1

3

1

2

1

4

1

3

1)sin(

H

y

H

y

H

y

hH

k

k

gHTT (t)

Rewriting (t) in dimensionless form

4

4

3

3

2

2

24 12

1

3

1

2

1

4

1

3

1

)sin( H

y

H

y

H

y

hH

k

k

gH

TT (v)

Surface heat flux is determined using Fourier’s law

dy

HdTkHq

)()(

(t) into the above


3

)sin()(

23 gHHq (w)

(iii) Checking. Dimensional check: Equation (n) has units of velocity. Each term in (t) has units of temperature. Each term in (w) has units of flux. Each term in (v ) and (y) is dimensionless.

Differential equation check: Velocity solution (n) satisfies e quation (j) and temperature solution (t) satisfies (r).

Boundary conditions check: Velocity solution (n) and temperature solution (t) satisfy their respective boundary conditions.

Limiting check: (i) If gravity or inclination angle vanishes the fluid will be stationary. Setting 0g or 0 in (n) gives .0u

(ii) If gravity or inclination angle vanishes the fluid will be stationary, dissipation will also

vanish and the temperature will be uniform throughout equal to .T Setting 0g or 0 in

(t) gives .TT

Qualitative check: All dissipation heat must leave the free surface. Equation (w) shows that surface heat flux is positive (leaving the fluid).

(5) Comments. (i) Neglecting axial variation in u and T are key simplifying assumptions in this problem. (ii) Distinction should be made between the ambient heat transfer coefficient and the liquid film heat transfer coefficient. They are not identical.

PROBLEM 3.10

Lubricating oil fills the clearance space of between a rotating

shaft and its housing. The shaft radius is cm6ir and housing

radius is cm.1.6ir The angular velocity of the shaft is

RPM3000 and the housing temperature is C.40ooT Taking

into consideration dissipation, determine the maximum oil

temperature and the heat flux at the housing. Neglect end effects

and assume incompressible laminar flow. Properties of lubricating

oil are: CW/m138.0 ok and skg/m0356.0 .

Solution

This problem is identical to Example 3.3. The maximum temperature occurs at the shaft surface

irr . This temperature is given in equation (3.21)

1)/ln(2)/()/(1

2

4)( 2

2

2 iooi

oi

ioi rrrr

rr

r

kTrT (3.21)

where

k thermal conductivity = CW/m138.0 o

ir shaft radius = 0.06 m

or housing radius = 0.061 m

oT housing temperature = 40 Co

viscosity = skg/m0356.0

angular velocity = 3000 RPM = 100 rad/s

The heat flux at the housing surface per unit length is given in (3.22)

2

2

)/(1

)(4)(

oi

io

rr

rrq (3.22)

Computation.

Substituting into (3.21)

1m)(06.0

m)(061.0ln2(m/m))061.0/06.0(

(m/m))061.0/06.0(1

)(0.06(m)s/rad)(100(2

C)W/m138.0(4

skg/m0356.0C)(40)( 22

2

22oo

max irTT

C8.86 omaxT


22

2

(m/m)1)(0.06/0.061

(0.06)(m))s/rad)(100(m)/s0.0356)(kg(4)( orq = 4,888

3s

mkg = 4,888

m

W

Comments. (i) The key simplifying assumption is axisymmetry. This resulted in concentric streamlines with vanishing normal velocity and angular changes.


(ii) Temperature rise of the lubricating oil and energy dissipation increase as the clearance between the shaft and the housing is decreased. This is evident from equations (3.22) which

show that in the limit as 1)/( oi rr , .q

(iii) The energy dissipated due to friction is considerable. The heat dissipated for a housing radius of 1.0 cm is 48.9 W.

PROBLEM 3.11

Consider lubrication oil in the clearance between a shaft and its

housing. The radius of the shaft is ir and that of the housing is or .

The shaft rotates with an angular velocity and its housing

exchanges heat by convection with the ambient fluid. The heat

transfer coefficient is h and the ambient temperature is .T Taking

into consideration dissipation, determine the maximum

temperature of the oil and surface heat flux at the housing. Assume

incompressible laminar flow and neglect end effects.

(1) Observations. (i) Fluid motion is driven by shaft rotation (ii) The housing is stationary. (iii) Axial variation in velocity and temperature are negligible for a very long shaft. (iv) Velocity and temperature do not vary with angular position. (v) The fluid is incompressible (constant density). (vi) Heat ge nerated by viscous dissipation is removed from the oil at the housing. (vii) No heat is conducted through the shaft. (viii) The maximum temperature occurs at the shaft. (ix) Heat fl ux at the housing is determined from temperature distribution and Fourier’s law of conduction. (x) Use cylindrical coordinates.

(2) Problem Definition. Determine the velocity and temperature distribution in the oil.

(3) Solution Plan. This problem is similar to Example 3.3. The flow field is given by (3.18). The energy equation is given in (l) of Example (3.18). The only difference between this problem and Example 3.3 is the boundary condition at the housing. Housing heat flux can be determined using the solution to temperature distribution and Fourier’s law.

(4) Plan Execution

(i) Assumptions. (I) Steady state, (ii) laminar flow, (iii) axisymmetric flow, (iv) constant properties (density, viscosity and conductivity), (v) no end effects, (vi) uniform ambient temperature and (vii) negligible gravitational effect.

(ii) Analysis. The velocity distribution is given by

1)/(

)/()/()/()(2

2

io

iiio

i rr

rrrrrr

r

rv (3.18)

Following Example 3.3, the energy equation is

01

dr

dTr

dr

d

rk (a)

and the solution is

432

2

2

2

ln1

)/(1

2

4)( CrC

rrr

r

krT

oi

i (b)

where 3C and 4C are the integration constants. Two boundary conditions are needed to

determine 3C and 4C . They are:


(1) 0)(

dr

rdT i

(2) TrThdr

rdTk o

o )()(

These boundary conditions give the two constants

2

2

2

2

3

1

)/(1

2

2ioi

i

rrr

r

kC

and

o

o

i

o

ooi

i

ohr

kr

r

r

hr

k

rr

r

krTC ln212

)/(1

2

4 2

22

2

2

24

Substituting the above into (b) and rearranging

1)/(12)/ln()/(2)/()/(1

2

4)( 222

2

22

2

io

o

oioo

oi

i

o

i rrhr

krrrrrr

rr

r

r

r

kTrT (c)


112ln24

1

)/(1

2

)(2

2

2

2

2

2

2

2

2

2

i

o

o

oo

i

o

o

i

oi

ir

r

hr

k

r

r

r

r

r

r

r

r

rr

r

k

TrT (d)

The maximum temperature is at the shaft’s surface. Setting irr in (c) gives

22

2

2max )/(1)/(2)/ln(21)/(1

2

4)( oioi

o

oi

oi

ii rrrr

hr

krr

rr

r

kTrTT (e)

Energy generated due to dissipation per unit shaft length, ),( orq is determined by applying

Fourier’s law at the housing. Thus

dr

rdTkrrq o

oo

)(2)(

Using (c), the above gives

2

2

)/(1

)(4)(

oi

io

rr

rrq (f)

(iii) Checking. Dimensional check: each term in solutions (c) has units of temperature. (f) has the correct units of W/m. and (3.20b) is dimensionless. Equation (3.22) has the correct units of W/m.

Boundary conditions check: Temperature solution (c) satisfies the two boundary conditions on temperature.

Limiting check: (i) If the shaft does not rotate no dissipation takes place and thus surface heat

transfer should vanish. Setting 0 in (f) gives .0)( orq


(ii) If the fluid is inviscid no dissipation takes place and thus surface heat transfer should vanish.

Setting 0 in (f) gives .0)( orq

Global conservation of energy. Heat transfer rate from the hous ing must equal to work required to overcome friction at the shaft’s surface. The rate of shaft work per unit length is given by

iii rrrW )(2 (p)

where

W = work done on the fluid per unit shaft length

)( ir = shearing stress at the shaft’s surface, given by

irrrdr

dri

vv 0)( (q)

Substituting (3.18) into the above

2)/(12)(

oi

irr

r (r)

Combining (p) and (r) and rearranging, gives

2

2

)/(1

)(4

oi

i

rr

rW (s)

This result is identical to surface heat transfer rate given in (f)

(5) Comments. (i) The key simplifying assumption is axisymmetry. This resulted in concentric streamlines with vanishing normal velocity and angular changes.

(ii) Temperature rise of the lubricatig oil and energy dissipation increase as the clearance between the shaft and the housing is decreased.

(iii) Velocity distributions are governed by a single parameter )./( oi rr Temperature distribution

is governed by two parameters: )/( oi rr and the Biot number ./ khro

(iv) Heat transfer rate at the housing, equation (f), is identical to that of Example 3.3 given in equation (3.22). This is not surprising since dissipation energy for constant property fluids is a function of flow field. Thus, dissipation energy is the same for problems with identical flow fields even if they have different temperature boundary conditions.

PROBLEM 3.12

A rod of radius ir is placed concentrically inside a sleeve of radius

or . Incompressible fluid fills the clearance between the rod and the

sleeve. The sleeve is maintained at uniform temperature oT while

rotating with constant angular velocity . Taking into consideration

dissipation, determine the maximum fluid temperature and surface

heat flux at the sleeve. Assume incompressible laminar flow and

neglect end effects.

(1) Observations. (i) Fluid motion is driven by sleeve rotation (ii) The shaft is stationary. (iii) Axial variation in velocity and temperature are negligible for a very long shaft. (iv) Velocity and temperature do not vary with angular position. (v) The fluid is incompressible (constant density). (vi) Heat ge nerated by viscous dissipation is removed from the oil at the housing. (vii) No heat is conducted through the shaft. (viii) The maximum temperature occurs at the shaft. (ix) Use cylindrical coordinates.

(2) Problem Definition. Determine the velocity and temperature distribution in the oil.

(3) Solution Plan. Apply continuity and Navier-Stokes equations in cylindrical coordinates to determine the flow field. Use the energy equation to determine temperature distribution. Apply Fourier’s law at the housing to determine the rate of energy generated by dissipation.

(4) Plan Execution

(i) Assumptions. (I) Steady state, (ii) laminar flow, (iii) axisymmetric flow, (iv) constant properties (density, viscosity and conductivity), (v) no end effects, (vi) uniform surface temperature and (vii) negligible gravitational effect.

(ii) Analysis. Temperature distribution is obtained by solving the energy equation. Thus we begin the analysis with the energy equation. The energy equation in cylindrical coordinates for constant properties is given by (2.24)

2

2

2

2

2 0

11

z

TT

rr

Tr

rrk

z

TT

rr

T

t

Tc zrP v

vv (2.24)


22

0

2

0

222

0

1

0

12

122

rzzr

rrrzrrr

zrz

rzrr

vvvv

vvvvvvv

(2.25)

The solution to (2.24) requires the determination of the velocity components rv , v and .zv

These are determined by solving the continuity and the Navier-Stokes equations in cylindrical coordinates. The continuity equation is given by equation (2.4)

011

zrzr

rrrt

vvv (2.4)



0zrt

(a)


0 (b)

For a long shaft with no end effects axial changes are negligible

0z

zv (c)


0rrdr

dv (d)

Integrating (d)

Cr rv (e)

To determine the constant of integration C we apply the no-slip boundary condition at the housing surface

0)( orrv (f)

Equations (e) and (f) give 0C


0rv (g)

Since the radial component rv vanishes everywhere, it follows that the streamlines are

concentric circles. To determine the tangential velocity v we apply the Navier-Stokes equation

in the -directions, equation (2.11 )

2

2

22

2

2

21)(

11

zrrr

rrr

p

rg

tzrrr

r

r

vvvv

vvv

vvvvvv zr

(2.11 )

For steady state

0t

(h)

Neglecting gravity and applying (b),(c), (g) and (h), equation (2.11 ) simplifies to

0)(1

vrdr

d

rdr

d (i)

Integrating (i) twice

r

Cr

C 21

2v (j)

where 1C and 2C are constants of integration. The two boundary conditions on v are


0)( irv , oo rr )(v (j)

Boundary conditions (j) give 1C and 2C

22

2

1

2

io

o

rr

rC ,

22

22

2

io

oi

rr

rrC (k)

Substituting (k) into (j) and rearranging in dimensionless form, gives

r

r

r

r

rr

rr

r

r i

ioi

oi

o2)/(1

)/()(v (l)

We now return to the energy equation (2.24) and the dissipation function (2.25). Using (b), (c), (g) and (h), equation (2.24) simplifies to

01

dr

dTr

dr

d

rk (m)

The dissipation function (2.25) is simplified using (b), (c) and (g)

2

0

rdr

d vv

Substituting the velocity solution (l) into the above, gives

4

2

2

21

)/(1

2

rrr

r

oi

i (n)

Combining (n) and (m) and rearranging, we obtain

dr

dTr

dr

d3

2

2

21

)/(1

2

rrr

r

koi

i (o)

Integrating (o) twice

432

2

2

2

ln1

)/(1

2

4)( CrC

rrr

r

krT

oi

i (n)



0)(

dr

rdT i and oo TrT )( (o)

Equations (n) and (o) give the two constants

2

2

2

2

3

1

)/(1

2

2ioi

i

rrr

r

kC

and

o

iooi

io r

rrrr

r

kTC ln

21

)/(1

2

4 22

2

2

2

4


Substituting the above into (o)

)/ln(2)/()/()/(1

2

4)( 22

2

2rrrrrr

rr

r

kTrT oioi

oi

io

(p)


)/ln(2)/()/(

)/(1

2

4

)( 22

2

2

rrrrrr

rr

r

k

TrToioi

oi

i

o (q)

The maximum temperature is at the shaft’s surface. Setting irr in (p) gives

)/ln(2)/(1)/(1

2

4)( 2

2

2 iooi

oi

ioi rrrr

rr

r

kTrT (r)

Energy generated due to dissipation per unit shaft length, ),( orq is determined by applying

Fourier’s law at the housing. Thus

dr

rdTkrrq o

oo

)(2)(

Using (q), the above gives

2

2

)/(1

)(4)(

oi

io

rr

rrq (s)

(iii) Checking. Dimensional check: each term in solutions (l) and (q) is dimensionless. Equation (s) has the correct units of W/m.

Differential equation check: Velocity solution (l) satisfies equation (i) and temperature solution (p) satisfies (o).

Boundary conditions check: Velocity solution l) satisfies boundary conditions (j) and temperature solution (p) satisfies boundary conditions (o).

Limiting check: (i) If sleeve does not rotate the fluid will be stationary. Setting 0 in (l)

gives .0v

(ii) If the sleeve does not rotate no dissipation takes place and thus surface heat transfer should

vanish. Setting 0 in (s) gives .0)( orq

(5) Comments. (i) The key simplifying assumption is axisymmetry. This resulted in concentric streamlines with vanishing normal velocity and angular changes.

(ii) Temperature rise of the lubricating oil and energy dissipation increase as the clearance between the shaft and the housing is decreased. This is evident from equation (s) which show

that in the limit as 1)/( oi rr , .q

(iii) Velocity and temperature distribu tion are governed by a single parameter )./( oi rr

PROBLEM 3.13

A hollow shaft of outer radius or rotates with constant angular

velocity while immersed in an infinite fluid at uniform

temperature .T Taking into consideration dissipation, determine

surface temperature and heat flux. Assume incompressible laminar

flow and neglect end effects.

(1) Observations. (i) Fluid motion is driven by shaft rotation (ii) Axial variation in velocity and temperature are negligible for a very long shaft. (iii) Velocity, pressure and temperature do not vary with angular position. (iv) The fluid is incompressible (constant density). (v) Heat ge nerated by viscous dissipation is conducted radially. (vi) The determination of surface temperature and heat flux requires the determination of temperature distribution in the rotating fluid. (vii) Use cylindrical coordinates.

(2) Problem Definition. Determine the velocity and temperature distribution in the rotating fluid.

(3) Solution Plan. Apply continuity and Navier-Stokes equations in cylindrical coordinates to determine the flow field. Use the energy equation to determine temperature distribution.

(4) Plan Execution.

(i) Assumptions. (I) Steady state, (ii) laminar flow, (iii) axisymmetric flow, (iv) constant properties (density, viscosity and conductivity), (v) no end effects, (vi) no angular and axial variation of velocity, pressure and temperature and (vii) negligible gravitational effect.

(ii) Analysis. Temperature distribution is obtained by solving the energy equation. Thus we begin the analysis with the energy equation. The energy equation in cylindrical coordinates for constant properties is given by (2.24)

2

2

2

2

2 0

11

z

TT

rr

Tr

rrk

z

TT

rr

T

t

Tc zrP v

vv (2.24)


22

0

2

0

222

0

1

0

12

122

rzzr

rrrzrrr

zrz

rzrr

vvvv

vvvvvvv

(2.25)

The solution to (2.24) requires the determination of the velocity components rv , v and .zv

These are determined by solving the continuity and the Navier-Stokes equations in cylindrical coordinates. The continuity equation is given by equation (2.4)

011

zrzr

rrrt

vvv (2.4)



0zrt

(a)


0 (b)

For a long shaft with no end effects axial changes are negligible

0z

zv (c)


0rrdr

dv (d)

Integrating (d)

Cr rv (e)

To determine the constant of integration C we apply the no-slip boundary condition at the housing surface

0)( irrv (f)

Equations (e) and (f) give 0C


0rv (g)

Since the radial component rv vanishes everywhere, it follows that the streamlines are

concentric circles. To determine the tangential velocity v we apply the Navier-Stokes equation

in the -direction, equation (2.11 )

2

2

22

2

2

21)(

11

zrrr

rrr

p

rg

tzrrr

r

r

vvvv

vvv

vvvvvv zr

(2.11 )

For steady state

0t

(h)

Neglecting gravity and applying (b),(c), (g) and (h), equation (2.11 ) simplifies to

0)(1

vrdr

d

rdr

d (i)

Integrating (3.17) twice

r

Cr

C 21

2v (j)

where 1C and 2C are constants of integration. The two boundary conditions on v are


oo rr )(v , 0)(v

These boundary conditions give 1C and 2C

01C , 22 orC (k)

Substituting (k) into (j) and rearranging in dimensionless form, gives

r

r

r

r o

o

)(v (l)

We now return to the energy equation (2.24) and the dissipation function (2.25). Using (b), (c), (g) and (h), equation (2.24) simplifies to

01

dr

dTr

dr

d

rk (m)

The dissipation function (2.25) is simplified using (b), (c) and (g)

2

0

rdr

d vv

Substituting the velocity solution (l) into the above, gives

4

22 1

2r

ro (n)

Combining (m) and (n) and rearranging

dr

dTr

dr

d3

22 1

2r

rk

o (o)

Integrating (o) twice

432

22 ln

12

4)( CrC

rr

krT o (p)



(1) )(T finite

(2) TT )(

Boundary condition (1) gives

03C

Boundary condition (2) gives

TC4

Substituting the above into (n)


2

22 1

24

)(r

rk

TrT o (q)


2

2

2)(

)(

r

r

rk

TrT o

o

(r)

Surface temperature obtained by setting orr in (q)

2)()( oo rk

TrT (s)

Surface heat flux per unit shaft length, ),( orq is determined by applying Fourier’s law at orr

dr

rdTkrrq o

oo

)(2)(

Using (q) the above gives 2)(4)( oo rrq (t)

(iii) Checking. Dimensional check: each term in solutions (l) and (r) is dimensionless.

Equation (q) has the correct units of Co and equation (t) has units of W/m.

Differential equation check: Velocity solution (l) satisfies equation (i) and temperature solution (q) satisfies (o).

Boundary conditions check: Velocity solution (l) and temperature solution (q) satisfy their respective boundary conditions.

Limiting check: (i) If shaft does not rotate the fluid will be stationary. Setting 0 in (l) gives

.0v

(ii) If the shaft does not rotate no dissipation takes place and fluid temperature should be uniform

equal to T and surface heat transfer should vanish. Setting 0 in (q) gives TrT )( .

Setting 0 in (t) gives .0)( orq

Global conservation of energy: Surface heat transfer rate must equal to work required to overcome friction at the shaft’s surface. The rate of shaft work per unit length is given by

ooo rrrW )(2 (u)

where

W = work done on the fluid per unit shaft length

)( or = shearing stress at the shaft’s surface, given by

orrrdr

dro

vv0)( (v)

Substituting (l) into the above

2)( ir (w)


Combining (u) and (w) 2)(4 orW (y)

This result is identical to surface heat transfer rate given in (t).

(5) Comments. (i) The key simplifying assumption is axisymmetry. This resulted in concentric streamlines with vanishing normal velocity and angular changes. (ii) Surface temperature is lowest in the entire region. (iii) Heat flow direct ion is negative. (iii) This problem was solved by specifying two conditions at infinity. If surface temperature is specified instead of fluid temperature at infinity, the solution determines ).(T

PROBLEM 3.14

Two large porous plates are separated by a distance H. An incompressible fluid fills the channel

formed by the plates. The lower plate is maintained at temperature 1T and the upper plate at 2T .

An axial pressure gradient dxdp / is applied to the

fluid to set it in motion. A fluid at temperature 1T is

injected through the lower plate with a normal

velocity .ov Fluid is removed along the upper plate

at velocity .ov The injected fluid is identical to the

channel fluid. Neglect gravity, dissipation and

axial variation of temperature.


H

y

Hv

yv

dx

dp

v

Hu

o

o

o )/exp(1

)/exp(11

[b] Determine surface heat flux at each plate.

(1) Observations. (i) Axial pressure gradient sets fluid in motion. (ii) The fluid is incompressible. (iii) The flow field is determined by solving the continuity and Navier-Stokes equations. (iv) Energy equation gives the temperature distribution. (v) Fourier’s law and temperature distribution give surface heat flux. (vi) Axial variation of temperature is neglected. (viii) Use Cartesian coordinates.



(4) Plan Execution.

(i) Assumptions. (i) Steady state, (ii) laminar flow, (iii) constant properties (density, viscosity and conductivity), (iv) infinite plates, (v) no axial variation of temperature, (vi) negligible gravitational and (vii) negligible dissipation.

(ii) Analysis. [a] Velocity distributi on. Applying continuity and the Navier-Stokes equations. We begin with the continuity equation in Cartesian coordinates

0z

w

yx

u

zw

yxu

t

vv (2.2b)


0zyxt

(a)


0wzx

(b)



0y

v (c)

Integrating (c) Cv (d)

where C is constant of integration. The boundary condition on v is

ovv )0( (e)

Equations (d) and (e) give

ovC

Substituting into (d)

ovv (f)

To determine the horizontal component u we apply the Navier-Stokes equations (2.10)

2

2

2

2

2

2

z

u

y

u

x

u

x

pg

z

uw

y

u

x

uu

t

uxv (2.10x)

2

2

2

2

2

2

zyxy

pg

zw

yxu

ty

vvvvvv

vv (2.10y)

These equations are simplified as follows: Steady state

0t

u (g)


0yx gg (h)

Substituting (b) and (f)-(h) into (2.10x) and (2.10y) gives

2

21

dy

ud

x

p

dy

duvo (i)

and

0y

p (j)

where ./ Equation (j) shows that pressure does not vary in the y-direction and thus it can

be a function of x or constant. Integrating (i) once

dy

duy

xd

pdCu

vo 11 (k)

To solve this equation it is rewritten first as

)( yQuPdy

du (l)

where

ovP , 1

1)( Cy

dx

dpyQ (m)


The solution to (l) is

2)( CdyyQeeuPdyPdy

(n)

substituting (m) into (n) and evaluating the integrals

yo

o

o

o

veCC

vvy

vdx

dpu

)/(21)/(

1 (o)

The constants 1C and 2C are determined form the boundary conditions on u

(1) 0)0(u

(2) 0)(Hu


Hvedx

dpHC

oHov

1

1/1 (p)

Substituting (l) into (k)

1

1/2 Ho

ov

edx

dpH

vC (q)

(p) and (q) into (o)

H

y

e

e

dx

dp

v

Hu

Ho

yo

ov

v

1

11/

/

(r)

[b] Temperature distribution and Nusselt number. With the velocity distribution determined, the energy equation is applied to determine temperature distribution. The energy equation for constant properties is given by (2.19b)

2

2

2

2

2

2

z

T

y

T

x

Tk

z

Tw

y

T

x

Tu

t

Tc v (2.19b)

Neglecting dissipation and using (a) and (b) this equation is simplified

2

2

dy

Td

dy

dTov (s)

where pck / is thermal diffusivity. To Integrate (s) it is rewritten as

dy

dy

dT

dy

dTd

ov

Integrating

3lnln Cydy

dT ov

Rewriting


ydy

dT

C

ov

3

1ln

/3

yoeCdy

dT v

Integrating again

4/

3 CeCTy

o

ov

v (t)


(1) 1)0( TT

(2) 2)( THT

These boundary conditions and solution (t) give

1/12

3 H

o

oe

TTC

v

v (u)

1

)(/

/12

24 H

H

o

o

e

eTTTC

v

v

(w)

Substituting (u) into (w) into (t) and rearranging the result in dimensionless form, give

1]/exp[

)]/)(/exp[(]/exp[)( 212

H

HyHHTTTT

o

oo

v

vv (x)

This result in now expressed in terms of the Prandtl number. Note that

Pr

Substituting into (x)

1])/(exp[

)]/()/(exp[])/(exp[)( 212

PrHv

HyPrHvPrHvTTTT

o

oo (y)

This result can be rearranged in dimensionless form as

1])/(exp[

)]/()/(exp[])/(exp[

21

2

PrHv

HyPrHPrHv

TT

TT

o

oo v (z)

Surface heat flux is determined by applying Fourier’s law at each plate

dy

dTkq

)0()0( (z-1)

dy

HdTkHq

)()( (z-2)

Substituting (y) into (z-1) and (z-2)


1])/(exp[)0( 21

PrHv

TTkPr

vq

o

o (z-3)

1])/(exp[

])/(exp[)()( 21

PrHv

PrHvTTkPr

vHq

o

oo (z-4)

Expressed in dimensionless form, (K) and (L) become

1])/(exp[

1

)(

)0(

21

PrHvTTPrk

v

q

oo

(z-5)

1])/(exp[

])/(exp[

)(

)(

21

PrHv

PrHv

TTPrkv

Hq

o

o

o

(z-6)

(iii) Checking. Dimensional check: Each term in (z), (z-5), and (z-6) is dimensionless. The exponents of all exponentials are dimensionless.

Differential equation check: Velocity solution (r) satisfies e quation (i) and temperature solution (x) satisfies (s).

Boundary conditions check: Velocity solution (r) and temperature solution (x) satisfy their respective boundary conditions.

Limiting check: (i) If there is no axial pressure gradient, the fluid will be stationary. Set 0/ dxdp in (r) gives .0)(yu

(ii) If 21 TT , surface heat flux will vanish. Set 21 TT in (z-3) and (z-4) gives

.0)()0( Hqq

(5) Comments. (i) Treating the plate as infinite is one of the key simplifying assumptions. This eliminates the x-coordinate as a variable and results in governing equations that are ordinary. (ii) Contrary to expectation, the axial velocity plays no role in the temperature distribution This is evident from energy equation (s) and temperature solutions (y).

(iii) According to the dimensionless form of solutions (z), (z -5) and (z-6), the problems is characterized by the following single dimensionless parameter

PrHvo

Note that this parameter is a combination of Prandtl number, geometry, injection velocity, and kinematic viscosity , a property.

(iv) Taking the ratio of (N) to (M) provides a co mparison of surface heat flux at the two plates

])/(exp[)0(

)(PrHv

q

Hqo

This result indicates that heat flux at the upper plate is higher than that at the lower plate.

PROBLEM 4.1

Put a check mark in the appropriate column for each of the following statements.

Statement true false may be

(a) 0)/()/( yxu v is valid for transient flow. x

(b) The y-momentum equation is neglected in boundary layer flow.

x

(c) Boundary layer equatio ns are valid for all Reynolds numbers.

x

(d) Pressure gradient is zero outside the boundary layer.

x

(e)

2

2

2

2

y

u

x

u for a streamlined body. x

(f) In boundary layer flow fluid velocity upstream of an object is undisturbed.

x

(g) Axial pressure gradient is neglected in boundary layer flow. x

(i) Axial conduction is neglected in boundary layer flow.

x

PROBLEM 4.2

Examine the three governing equations, (2.2), (4.13) and (4.18) for two-dimensional, constant

properties, laminar boundary layer flow.

[a] How many dependent variables do these equations have?

[b] How is the pressure p determined?

[c] If streamlines are parallel in the boundary layer what terms will vanish?

[d] Can (2.2) and (4.13) be solved for the velocity field u and v independently of the energy

equation (4.18)?

Solution: The three equations are:

0yx

u v (2.2)

:2

21

y

u

dx

dp

y

u

x

uu v (4.13)

2

2

y

T

y

T

x

Tu v (4.18)

[a] There are four dependent variables: ,u ,v ,p and T.

[b] The pressure p is determined form the inviscid external flow. The solution to the Navier-

Stokes equations with 0 (Euler’s equations of motion) for the flow over the same object

gives p .

[c] The following terms will vanish if streamlines are parallel:

If 0v then 0y

v. When this is substituted into (2.2) gives 0

x

u. Equations (4.13) and

(4.18) become:

2

210

y

u

dx

dp

and

2

2

y

T

x

Tu

[d] Equations (2.2) and (4.13) can be solved for u and v independently of energy equation (4.18)

since temperature does not enter in (2.2) and (4.13).

PROBLEM 4.3

Air flows over a semi-infinite plate with a free stream velocity V = 0.4 m/s and a free stream

temperature C.20oT The plate is maintained at C.60osT Can boundary layer

approximations for the flow and temperature fields be applied at:

[a ] location x = 1.5 mm?[b] location x = 15 mm?

Note: Evaluate air properties at the average film temperature .2/)( TTT sf

(1) Observations. (i) This is forced convection flow over a streamlined body. (ii) Viscous (velocity) boundary layer approximations can be made if the Reynolds number Rex > 100. (iii) Thermal (temperature) boundary layer approximations can be made if the Peclet number Pex = Rex Pr > 100. (iv) The Reynolds number decreases as the distance along the plate is decreased.

(2) Problem Definition. Determine the local Reynolds and Peclet numbers at the locations of interest.

(3) Solution Plan. Write the definitions of Rex and Pex and calculate their values at x = 1.5 mm and x = 15 mm.

(4) Plan Execution.

(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3) constant properties, (4) two-dimensional flow and (5) streamlined body.

(ii) Analysis. The local Reynolds and Peclet numbers are defined as number

Rex =V x

(a)

andPex = Rex Pr (b)

where

Pex = local Peclet number Pr = Prandtl number Rex = local Reynolds number

V = upstream velocity = 0.4 m/s x = distance from the leading edge of the plate, m

= kinematic viscosity, m2 /s

Properties are evaluated at the film temperature Tf defined as

Tf = T Ts

2 (c)

where

Tf = film temperature, oCTs = surface temperature = 60oC

T = free stream temperature = 20oC

t

y

Ts

T V,

x


(iii) Computations. Use (c) to calculate Tf

Tf = (60 + 20)( oC)/2 = 40oC

At this temperature Appendix C gives

Pr = 0.71

= 16.96 10-6 m2/s

[a] At x = 1.5 mm = 0.0015 m, equation (a) gives

Rex =)s/m(1096.16

)m(0015.0)s/m(4.026

= 35.4

Since this value is less than 100, it follows that velocity boundary layer approximations can not be made. Using (b) to calculate Pex

Pex = 35.4 x 0.71 = 25.1

Since this is smaller than 100, it follows that temperature boundary layer approximations can not be made.

[b] At x = 15 mm = 0.015 m the Reynolds number and Peclet number of part [a] will increase by a factor of 10. Thus

Rex = 354

and

Pex = 251

Since both Rex and Pex are larger than 100, it follows that both velocity and temperature boundary layer approximations can be made at this location.

(iv) Checking. Dimensional check: Computations showed that equation (a) is dimensionally consistent.

(5) Comments. The Reynolds and Peclet numbers should be calculated to establish if boundary layer approximations can be made.

PROBLEM 4.4

Water at C25o flows with uniform velocity V = 2 m/s over a streamlined object. The object is 8

cm long and its surface is maintained at C.85osT Use scaling to: [a] show that 1/ L , [b]

evaluate the inertia termsx

uu and

y

uv , and [c] evaluate the viscous terms

2

2

x

uand .

2

2

y

u

(1) Observations. (i) The surface is streamlined. (ii) The fluid is water. (iii) Inertia and viscous effects can be estimated using scaling. (iv) If a viscous term is small compared to inertia, it can

be neglected. (v) Properties should be evaluated at the film temperature .2/)( TTT sf

(2) Problem Definition. Estimate the magnitudes ,/ Lx

uu ,

y

uv ,

2

2

x

uand .

2

2

y

u

(3) Solution Plan. Use the scaling to estimate the magnitudes of the above terms.

(4) Plan Execution.

(i) Assumptions. (1) Continuum and (2) streamlined surface.

(ii) Analysis. Scales:

u V (a)

Lx (b)

y (c)

Scaling of continuity equation gives a scale for v

LVv (4.7d)

[a] A balance between inertia term x

uu and viscous term

2

2

y

u gives

LReL

1 (4.14b)

where LRe is the Reynolds number defined as

LVReL (d)

where

08.0L m

2V m/s

kinematic viscosity, /sm2

[b] Scales for inertia terms are:

L

VV

x

uu (e)

and


V

y

uvv

Using (4.7d) to eliminate v in the above, gives

L

VV

y

uv (f)

Thus the two inertia terms are of the same magnitude.

[c] The two viscous terms are scaled as: First term:

2

2

x

u2L

V (g)

Second term:

2

2

y

u2

V (h)

For 1/ L , comparing (g) with (h) shows that

2

2

x

u<<

2

2

y

u (4.2)

(iii) Computation. Properties are evaluated at the film temperature fT

C552

)C)(2585(

2

ooTT

T sf

s

m105116.0

26

[a] Substituting into (d)

5

2610127.3

/s)m(105116.0

)m(08.0)m/s(2Re

Equation (4.14b) gives

00179.010127.3

1

5L

m000143.0)m(08.000179.0

Thus 1/ L .

[b] The two inertia terms are of the same order of magnitude, given by (e) or (f)

)m/s(50)m(08.0

)m/s(2)m/s(2 2

x

uu

[c] The first viscous term is given by (g)


2

2

x

u 2m/s00016.0)m()08.0(

)m/s(2)/sm(105116.0

22

26

The second viscous term is given by (h)

2

2

y

u 2m/s50)m()000143.0(

)m/s(2)/sm(105116.0

22

26

Thus the first viscous term can be neglected since it is much smaller than the second term.

(iv) Checking: Dimensional check: Inertia and viscous terms have the same units.

(5) Comments. Computation showed that the second viscous term is identical to the inertia term. This is consequence of equating the two terms to derive (4.14b).

PROBLEM 4.5

Water at C25o flows with uniform velocity V = 2 m/s over a streamlined object. The object is 8

cm long and its surface is maintained at C.85osT Use scaling to: [a] show that 1/ Lt , [b]

evaluate the convection terms x

Tu and

y

Tv , and [c] evaluate the conduction terms

2

2

x

T

and .2

2

y

T

(1) Observations. (i) The surface is streamlined. (ii) The fluid is water. (iii) Convection and conduction effects can be estimated using scaling. (iv) If a conduction term is small compared to

convection, it can be neglected. (v) The scale for Lt / depends on whether t or .t

(vi) Properties should be evaluated at the film temperature .2/)( TTT sf

(2) Problem Definition. Estimate the magnitudes ,/ Ltx

Tu ,

y

Tv ,

2

2

x

Tand .

2

2

y

T

(3) Solution Plan. Use the scaling to estimate the magnitudes of the above terms.

(4) Plan Execution.


(ii) Analysis. Scales: Two scales are used for u depending on whether t or .t

Case (1): t .

u V (a)

Lx (b)

ty (c)

TTT s (d)

Scaling of continuity equation gives a scale for v

vL

V t (4.23)

Case (2): t .

u tV (4.29)

Scaling of the continuity equation gives

vL

V t2

(4.30)

[a] Scaling of Lt / depends on whether t or .t The two cases are considered.

Case (1): t . A balance between convection x

Tu and normal conduction

2

2

y

T gives


LePrRL

t 1 (4.24)

where Pr is the Prandtl number and LRe is the Reynolds number defined as

LVReL (f)

where

08.0L m

2V m/s

kinematic viscosity, /sm2

Case (2): t . A balance between convection x

Tu and normal conduction

2

2

y

T gives

LeRPrL

t1/3

1 (4.31)

[b] Scales for convection terms.

Case (1): t .

L

TTV

x

Tu s (g)

and

t

s TT

y

Tvv

Using (4.23) to eliminate v in the above, gives

L

TTV

y

T sv (h)

Thus the two convection terms are of the same magnitude.

Case (2): t . Using (d) and (4.29)

L

TTV

x

Tu st (i)

where is scaled as

LRe

L (4.14b)

Substituting (4.14b) into (i)

2L

TTReV

x

Tu s

Lt (j)


Similarly, using (c), (d), (4.14b) and (4.30) give the same result of (j).

[c] Scale for conduction terms are: Axial conduction is scaled as

2

2

x

T2L

TTs (i)

normal conduction is scaled as

2

2

y

T2t

s TT (j)

For 1/ Lt , comparing (i) with (j) shows that

2

2

x

T<<

2

2

y

T (4.2)


C552

)C)(2585(

2

ooTT

T sf

Pr = 3.27

s

m10566.1

27

s

m105116.0

26

[a] Substituting into (f)

5

2610127.3

/s)m(105116.0

)m(08.0)m/s(2Re

Case (1): t . Equation (4.24) gives

4

51089.9

10127.3)(27.3(

1

L

t

m1091.7)m(08.01089.9 54t

Thus 1/1 L for this case.

Case (2): t . Equation (4.31) gives

3

53/110205.1

10127.3)27.3(

1

L

t

m1064.9)m(08.010205.1 53t

[b] The two convection terms are of the same order of magnitude for both cases.


Case (1): t . Using (g)

s

C1500

)m(08.0

)C)(2585()m/s(2

oo

x

Tu

Case (2): t . Noting that m1064.9 5t , and using (j)

1011)m()08.0(

)C)(2085(10127.3)m(1064.9)m/s(2

22

o55

x

Tu

[c] Axial conduction is given by (i)

2

2

x

T

s

C10175.1

)m()08.0(

)C)(2585()/sm(10566.1

o4

22

o27

Normal conduction is given by(j). Two cases are considered:

Case (1): t .

m1091.7 5t

2

2

y

T2s

m1502

)m()1091.7(

)C)(2585()/sm(10566.1

227

o27

Case (2): t .

m1064.9 5t

2

2

y

T2s

m1011

)m()1064.9(

)C)(2585()/sm(10566.1

227

o27

Thus axial conduction can be neglected since it is much smaller than normal conduction.

(iv) Checking: Dimensional check: convection and conduction terms have the same units.

(5) Comments. (i) 1/ Lt . (ii) Axial conduction is small compared to normal conduction.

(iii) Computation showed that the normal conduction is identical to the convection term for both cases. This is a consequence of equating the two terms to derive (4.24) and (4.31).

PROBLEM 4.6

Atmospheric air at 25 Co flows over a surface at 115 Co . The free stream velocity is 10 m/s.

[a] Calculate the Eckert number.

[b] Use scale analysis to show that the dissipation term 2)/( yu is small compared to the

conduction term )./( 22 yTk

(1) Observations. (i) The fluid is air. (ii) Dissipation and conduction can be estimated using scaling. (iii) Dissipation is negligible if the Eckert number is small compared to unity.

(2) Problem Definition. Compute the Eckert number. Estimate the magnitudes of dissipation and conduction terms.

(3) Solution Plan. Use the definition of Eckert number to compute its value. Apply scaling to estimate the dissipation and conduction terms.

(4) Plan Execution.

(i) Assumptions. Continuum.

(ii) Analysis. [a] The Eckert number is defined as

)(

2

TTc

VE

sp

(a)

where

pc specific heat, Ckg-

Jo

E Eckert number

sT = surface temperature = 115 Co

T = free stream temperature = 25 Co

[b] The ratio of dissipation to condu ction is estimated using scaling.

Dissipation =

2

y

u

Conduction =2

2

y

Tk

where

k thermal conductivity, Cm-

Wo

viscosity,s-m

kg

Scales:


u V (a)

Lx (b)

y (c)

TTT s (d)

The dissipation and conduction terms are estimated using scales (a)-(d).

Dissipation

2V

(e)

Conduction2

TTk s (f)

where

sT surface temperature = 115 Co

T free stream temperature = 25 Co

Taking the ratio of (e) to (f)

Conduction

nDissipatio

)(

2

TTk

V

s

(g)


C702

)C)(25115(

2

ooTT

T sf

pc 1008.7Ckg-

Jo

02922.0kCm-

Wo

61047.20s-m

kg

[a] Substituting into (a)

0011.0-sJ

-mkg0011.0

C))(25115(C)J/kg(7.1008

)m/s()10(2

2

oo

22

E

[b]Conduction

nDissipatio00078.0

)C)(25115)(CW/m(02922.0

)s/m()10()skg/m(1047.20oo

2226

(iv) Checking: Dimensional check: The ratio of dissipation to conduction in (g) must be dimensionless:


1s-W

m-kg

)C)()(CW/m(

)s/m()skg/m(3

2

oo

222

TTk

V

s

Limiting check: If 0V there is no dissipation. Setting 0V in (e) gives the correct result.

(5) Comments. (i) Since the Eckert number is small compared to unity dissipation is negligible. (ii) Dissipation is negligible compared to conduction.

PROBLEM 4.7

Air at C20o flows over a streamlined surface with a free stream velocity of m/s10 . Use scale

analysis to determine the boundary layer thickness at a distance of 80 cm from the leading edge.

(1) Observations. (i) The surface is streamlined. (ii) The fluid is air.

(2) Problem Definition. Estimate the magnitude of boundary layer thickness at a specified

distance from the leading edge.

(3) Solution Plan. Use the scaling to estimate the magnitudes of .

(4) Plan Execution.


(ii) Analysis. Scale analysis gives as

LReL

1 (4.14b)

where LRe is the Reynolds number defined as

LVReL (a)

where

8.0L m

10V m/s

61009.15 /sm2

(iii) Computation. Substituting into (a)

5

2610302.5

/s)m(1009.15

)m(8.0)m/s(10Re

Equation (4.14b) gives

00137.010302.5

1

5L

m0011.0)m(8.000137.0

(iv) Checking: Dimensional check: The Reynolds number is dimesionles.

(5) Comments. .1/ L

PROBLEM 4.8

In boundary layer flow, pressure gradient normal to the

flow direction is assumed zero. That is .0/ yp If this is

correct, how do you explain lift on the wing of an airplane

in flight?

Solution

Although 0/ yp in boundary layer flow, .0/ xp Thus, pressure distribution around the

surface is accounted for. Lift is th e net force acting on a surface in the y-direction.

PROBLEM 4.9

Derive an equation describing the vertical velocity component v at the edge of the boundary

layer for two-dimensional incompressible flow over a semi-infinite flat plate. Assume laminar

flow. Compare your result with scaling estimate.

(1) Observations. (i) This is a forced convection problem over a flat plate. (ii) At the edge of the

thermal boundary layer, the axial velocity is Vu . (iii) Blasius solution gives the distribution

of the velocity components u(x,y) and v(x,y). (iv) Scaling gives an estimate of v(x,y).

(2) Problem Definition. Determine the vertical velocity at the edge of the viscous boundary layer, ).,(xv

(3) Solution Plan. Use Blasius solution for ),( yxv . Evaluate ),( yxv at .y

(4) Plan Execution.

(i) Assumptions. All assumptions leading to Blasius so lution are applicable. These are: (1) Newtonian fluid, (2) steady state, (3) constant properties, (4) two-dimensional, (5) laminar flow

(Rex < 5 105), (6) uniform upstream velocity and temperature, (7) flat plate and (8) negligible changes in kinetic and potential energy.

(ii) Analysis. Blasius solution for ),( yxv is given by

fd

df

xVV 2

1v (4.43)

where V free steam velocity and is kinematic viscosity. The variable and f are defined as

x

Vyyx ),( (4.41)

d

df

V

u (4.42)

At the edge of the thermal boundary layer, y

1d

df

V

u (a)

According to Blasius solution, at 1d

df

V

u, Table 4.1,

8),(x (b)

and27923.6)(f (c)

Substituting(b) and (c) into (4.43)


)27923.68(2

1

xVV

)v(x,

xReV

8604.0)v(x, (d)

where the Reynolds number is defined as xV

Rex .

Scaling estimate of the velocity component v in the boundary layer is given by

vx

V (4.7d)

where

xRex

1 (4.16)

substituting (4.16) into (4.7d) and rearranging

V

v

xRe

1 (e)

(iii) Checking. Dimensional check: Since the Reynolds number is dimensionless it follows that each term in (d) is dimensionless.

Qualitative check: The layer thickness increases with distance x. Solution (c) confirms this behavior.

(5) Comments. Recalling that Blasius solution gives

xRex

2.5 (4.46)

Comparing (d) with (4.46) shows that t for 8.9Pr . Examination of Fig. 4.6 shows that

t for all fluids with 0.1Pr and that t for 0.1Pr .

PROBLEM 4.10

Sketch the streamlines in boundary layer flow over a semi-infinite flat plate.

Since the flow within the boundary layer is two-dimensional the vertical velocity component does not vanish. Thus stream lines are not parallel.

PROBLEM 4.11

Define the thickness of the velocity boundary layer in Blasius solution as the distance y where

the velocity u = 0.988 V . Derive an expression for /x.

(1) Observations. (i) This is a laminar boundary layer flow problem. (ii) Blasius solution gives the velocity distribution for the flow over a semi-infinite flat plate. (iii) A solution for the boundary layer thickness depends on how the thickness is defined.

(2) Problem Definition. Determine the distance y from the surface of the plate to the location

where the velocity ratio u/V = 0.988.

(3) Solution Plan. Use Blasius solution, Table 4.1, to dete rmine the location of the edge of the

velocity boundary layer where u/V = 0.988.

(4) Plan Execution.

(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3) two-dimensional flow, (4)

laminar flow (Rex < 5 105), (5) constant properties, (6) uniform upstream velocity, (7) flat plate and (8) boundary layer flow (Rex > 100).

(ii) Analysis. Blasius solution, Table 4.1, gives the axial velocity ratio u/V as a

function of the variable which is defined as

= yx

V (a)

where

V = free stream velocity, m/sx = axial coordinate, m y = normal coordinate, m

= dimensionless variable


The viscous boundary layer thickness is the value of y where the velocity ratio u/V reaches

an arbitrarily selected value. At f ( ) = u/V = 0.988, Table 47.1 gives = 4.8. This value of

corresponds to the edge of the boundary layer where y = . Thus (a) gives

4.8 = = yx

V =

x

V(b)

Solving for

V

x8.4 (c)

Dividing both sided of (c) by x, rearranging and using the definition of Reynolds number gives

xxVx Re

8.48.4 (d)

(iii) Checking. Dimensional check: in equation (c) should have units of length


)sm/(

)m()s/m( 2

V

x = m

(5) Comments. (i) The thickness of the viscous boundary layer depends on how it is defined.

Thus, it is not uniquely determined. (ii) Regardless of how is defined, the solution takes the

form of equation (d). Only the cons tant in (d) changes according to how is defined.

PROBLEM 4.12

Water flows over a semi-infinite plate with an upstream velocity of 0.2

m/s. Blasius solution is used to calculate at three locations along the

plate. Results are tabulated. Are these results valid? Explain.

(1) Observations. (i) Blasius solution is valid for laminar boundary layer flow over a semi-infinite plate. (ii) The transition Reynolds number from laminar to

turbulent flow is 5105 . (iii) Boundary layer approximations are valid if the Reynolds number is

greater than 100.

(2) Problem Definition. Determine the Reynolds number at each location.

(3) Solution Plan. Determine the Reynolds number at each location to establish the applicability of Blasius solution. Where applicable, use Bl asius result to determine the boundary layer thickness.

(4) Plan Execution.

(i) Assumptions. All assumptions leading to Blasius solution are applicable. These are:(1) Newtonian fluid, (2) steady state, (3) constant properties, (4) two-dimensional, (5) laminar flow

(Rex < 5 105), (6) viscous boundary layer flow (Rex > 100), (7) uniform upstream velocity, (8)

flat plate, (9) negligible changes in kinetic and potential energy and (10) no buoyancy ( = 0 or g= 0).

(ii) Analysis. The Reynolds number is computed to establish if the flow is laminar and if boundary layer approximations can be made. The Reynolds number is defined as

Rex =xV

(a)

where

Rex = Reynolds number


= kinematic viscosity = 0.5116 610 m2 /s

To determine if the flow is laminar or turbulent, compare the Reynolds number with the

transition Reynolds number. For flow over a flat plate the transition Reynolds number txRe is

5105txRe (b)

The flow is laminar if Rex < txRe . Viscous boundary layer approximations are valid for

100xRe (c)

Blasius solution for the boundary layer thickness is

xRex

2.5 (d)

x(cm) (cm)

300 1.441

40 0.526

0.01 0.0083

V

y

x0


(iii) Computations. Evaluating the Reynolds number at x = 300 cm, equation (a) gives

6

2610173.1

)/sm(105116.0

)m(3)m/s(2.0xRe

Since this value is larger than the transition Reynolds number given in (b), it follows that the flow is turbulent and thus Blasius solu tion (d) does not apply. Using (d) gives

cm441.1m01441.0)m(310173.1

2.5

6

Although this is the reported value for , it is incorrect.

Evaluating the Reynolds number at x = 40 cm, equation (a) gives

5

26105637.1

)/sm(105116.0

)m(4.0)m/s(2.0xRe

Since this value is smaller than the transition Reynolds number given in (b), and since it is larger than 100, it follows that the flow is laminar and thus Blasius solution (d) is applicable. Using (d) gives

cm526.0m00526.0)m(4.010173.1

2.5

6

Thus the reported value for is correct.

Evaluating the Reynolds number at x = 0.01 cm, equation (a) gives

39)/sm(105116.0

)m(0001.0)m/s(2.026xRe

Since this value is smaller than 100, it follows that boundary layer approximations are not valid and thus Blasius solution (d) doe s not apply. Using (d) gives

cm0083.0m000083.0)m(0001.039

2.5

Although this is the reported value for , it is incorrect

(iv) Checking. Dimensional check: Computations showed that equations (a) and (d) are dimensionally correct.

(5) Comments. In applying Blasius results it is important to verify that the conditions leading to Blasius solution are satisfied.

PROBLEM 4.13

Consider laminar boundary layer flow over a semi-infinite flat plate. Evaluate the wall shearing

stress at the leading edge. Comment on your answer. Is it valid? If not explain why.

(1) Observations. (i) This is an external flow problem over a flat plate. (ii) Blasius’s solution for the velocity distribution and wall shearing stress is assumed to be applicable. (iii) Of interest is the value of the local stress at the leading edge of the plate.

(2) Problem Definition. Determine the local wall shearing stress for laminar boundary layer flow over a flat plate.

(3) Solution Plan. Use Blasius solution for the local wall stress. Evaluate wall stress at the leading edge.

(4) Plan Execution.

(i) Assumptions. All assumptions leading to Blasius so lution are applicable. These are:(1) Newtonian fluid, (2) steady state, (3) constant properties, (4) two-dimensional, (5) laminar flow

(Rex < 5 105), (6) viscous boundary layer flow (Rex > 100, (7) uniform upstream velocity and (8) flat plate.

(ii) Analysis. Blasius solution gives

o =x

VV33206.0 (4.47)

where

V = free stream velocity, m/s x = axial distance measured from the leading edge, m

viscosity, mkg/s


o wall shearing stress, 2N/m

Noting that all quantities in equation (4.47) are constant except the variable x, (4.47) can be rewritten as

x

o

constant (a)

(iii) Computations. To determine the shearing stress at the leading edge, set x = 0 in (a)

0

constant)0(o =

(iv) Checking. Dimensional check: Units of in (4.47) should be N/m2:

222 m

N

ms

kg

m)(/s)m(

m/s)(m/s)()mkg/s(

x

VVo


(5) Comments. The shearing stress cannot be infinite. This suggests that Blasius solution is not valid at the leading edge. One of the assumptions leading to Blasius solution is Rex > 100. However, at the leading edge x = 0, the Reynolds number is given by

Rex =V x

= 0

Therefore, Blasius solution ca nnot be used to determine o at the leading edge. In fact the

solution breaks down at small values of x where the corresponding local Reynolds number and Peclet number are smaller than 100.

PROBLEM 4.14

Water at 20 Co flows over a m2m2 plate with a free stream velocity of 0.18 m/s. Determine

the force needed to hold the plate in place. Assume laminar boundary layer flow.

(1) Observations. (i) This is an external flow problem over a flat plate. (ii) The force needed to hold the plate in place is equal to the total shearing force by the fluid on the plate, (iii) Integration of wall shear over the surface gives the total shearing force. (iv) Blasius’s solution for the velocity distribution and wall shearing stress is assumed to be applicable.

(2) Problem Definition. Determine the local wall shearing stress for laminar boundary layer flow over a flat plate.

(3) Solution Plan. Use Blasius solution for the local wall stress. Integrate shearing force over the total surface area of plate.

(4) Plan Execution.

(i) Assumptions. All assumptions leading to Blasius so lution are applicable. These are:(1) Newtonian fluid, (2) steady state, (3) constant properties, (4) two-dimensional, (5) laminar flow

(Rex < 5 105), (6) viscous boundary layer flow (Rex > 100, (7) uniform upstream velocity and (8) flat plate.

(ii) Analysis. Since shearing stress is not uniform, shearing force must be integrated over the area to obtain the total restraining force. Thus

L

o WdxxF0

)( (a)

where

L plate length = 2 m W plate width = 2 m

x distance along surface, m

o wall shearing stress, 2N/m

The local shearing stress is given by Blasius solution

o =x

VV33206.0 (4.47)

where

V = free stream velocity = 0.18 m/s

viscosity = 310003.1 mkg/s

= kinematic viscosity = 610004.1 m2/s

Substituting (4.47) into (a)

L

dxxV

WVF0

2/133206.0

Evaluating the integral

x

dxF

L

W

V


LVWVF 66412.0 (b)

(iii) Computations. Equation (b) gives

s

mkg144.0

/s)m(10004.1

(m/s)2(m)18.0)2(m)m)0.18(m/skg/s)(10003.1(66412.0

26

3F

F = 0.144 newton

(iv) Checking. Dimensional check: Computations showed that units of force are correct.

(5) Comments. According to (4.47) shearing stress decreases with distance from the leading

edge. Thus doubling the length of the plate increases the total force by a factor of 2 . This is

evident in equation (b) which shows that LF .

PROBLEM 4.15

Consider Blasius solution for uniform flow over a semi-infinite plate. Put a check mark in the

appropriate column for each of the following statements.

Solution.

Statement true False Undetermined

(a)0

dx

dpbecause the flow is laminar. x (1)

(b) Wall shearing stress increases with distance from the leading edge of plate.

x (2)

(c) Solution is not valid for 100xRe . x

(d) Solution is not valid for 5105xRe . x

(e) The solution is valid for 100xRe . x(3)

(f) Boundary layer thickness is uniquely defined. x

(g) Solution is not valid for a curved plate. x(4)

(h) The solution for the wall shear at the leading edge (x = 0) is not valid.

x(5)

(i) The plate does not disturb upstream flow. x

(j) Solution is not valid for 5105xRe . x(3)

(1) In boundary layer flow 0dx

dp for flat plate only.

(2) See equation (7.12).

(3) Blasius solution is valid for 5105100 xRe .

(4) For a curved surface 0dx

dp.

(5) 0xRe at x = 0. Boundary layer approximation is not valid at 100xRe .

PROBLEM 4.16

Imagine a cold fluid flowing over a thin hot plate. Using your intuition, would you expect the

fluid just upstream of the plate to experience a temperature rise due to conduction from the hot plate? How do you explain the assumption in Pohlhausen's solution that fluid temperature is

unaffected by the plate and therefore T(0, y) = T ?

(1) Observations. (i) This is an external forced convection problem for flow over a flat plate. (ii) Of interest is the region where the upstream fluid reaches the leading edge of the plate. (iii) The fluid is heated by the plate. (iv) Heat fr om the plate is conducted through the fluid in all directions. (v) Pohlhausen’s solution assumes that heat is not conducted upstream from the plate and therefore fluid temperature at the leading edge is the same as upstream temperature.

(2) Problem Definition. Determine the conditions under which axial conduction in force convection flow can be neglected.

(3) Solution. In reality heat from the plate is conducted in all directions and thus one would expect the temperature of the incoming fluid to be affected by the presence of the plate. This becomes more obvious if one imagines that fluid velocity is decreased and in the limit the fluid becomes stationary. Clearly, for a stationary fluid heat is conducted from the plate through the fluid in all directions. Howeve r, as the velocity of the fluid increases, energy conducted upstream is carried downstream by convection. This tends to minimize the effect of axial conduction. and eventually may be ignored when compared with normal conduction. The condition for this approximation is

Pe = Rex Pr > 100 (a)where

Pe = Peclet number Pr = Prandtl number Rex = local Reynolds number

Pohlhausen’s solution is based on the above boundary layer approximation. By neglecting axial conduction the temperature of the incoming fluid at the leading edge is assumed to be the same as the free stream temperature. That is

T(0,y) = Twhere

T = temperature distribution in the fluid, oC

T = free stream temperature, oCy = coordinate normal to plate, m

x

y

00 Ts

T

V T(0,y) = ?

t

PROBLEM 4.17

Consider laminar boundary layer flow over a semi-infinite flat plat. The plate is maintained at

uniform temperature .sT Assume constant properties and take into consideration dissipation.

[a] Does Blasius solution apply to this case? Explain.

[b] Does Pohlhausen’s solution apply to this case? Explain.

Solution

[a] Blasius solution is based on the assumption that properties are constant independent of temperature. Thus, Blasius solu tion applies to this problem.

[b] Pohlhausen’s solution neglects dissipati on. Thus it does not apply to this problem.

PROBLEM 4.18

A fluid with Prandtl number 9.8 flows over a semi-infinite flat plat. The plate is maintained at

uniform surface temperature. Derive an expression for the variation of the thermal boundary

layer thickness with distance along the plate. Assume steady state laminar boundary layer flow

with constant properties and neglect dissipation. Express your result in dimensionless form.

(1) Observations. (i) This is a forced convection problem over a flat plate. (ii) At the edge of the thermal boundary layer, fluid

temperature is TT . (iii) Pohlhausen’s

solution gives the temperature distribution in the boundary layer. (iv) The thermal boundary

layer thickness t increases with distance from the leading edge. (v) t depends on the Prandtl

number.

(2) Problem Definition. Determine the variation of the thermal boundary layer thickness with distance for a fluid with Prandtl number of 9.8.

(3) Solution Plan. Use Pohlhausen’s graphical solution for the temperature distribution of laminar flow over a flat plate to determine the thermal boundary layer thickness.

(4) Plan Execution.

(i) Assumptions. All assumptions leading to Pohlhausen's solution are applicable. These are: (1) Newtonian fluid, (2) steady state, (3) constant properties, (4) two-dimensional, (5) laminar

flow (Rex < 5 105), (6) viscous and thermal boundary layer flow (Rex > 100 and Pex > 100), (7) uniform upstream velocity and temperature, (8) uniform surface temperature, (9) flat plate, (10) negligible changes in kinetic and potential energy, (11) negligible dissipation, (12) no buoyancy

( = 0 or g = 0) and (13) no energy generation ( 0q ). In addition, (14) neglect radiation.

(ii) Analysis. Pohlhausen’s solution is shown in graphical form in fig. 4.6

Fig. 4.6

0.1

6.0

4.0

2.0

2 4 6 8 12 1410

01.0Pr

1.0

)(7.0 air110100

0

8.0

xVy

s

s

TT

TT


At the edge of the thermal boundary layer, ty the dimensionless temperature in Fig. 4.6 is

equal to unity

1s

s

TT

TT (a)

For a fluid with 8.9Pr the corresponding value of ),( tx is obtained from Fig. 4.6 as

8.2x

Vt (b)

Solving (b) for t

V

xt 8.2 (c)

Expressing this result in dimensionless form

x

t

RexVx

8.28.2 (d)

(iii) Checking. Dimensional check: Since the Reynolds number is dimensionless it follows that each term in (d) is dimensionless.

Qualitative check: The thermal boundary layer thickness increases with distance x. Solution (c) confirms this behavior.

(5) Comments. Recalling that Blasius solution gives

xRex

2.5 (4.46)

Comparing (d) with (4.46) shows that t for 8.9Pr . Examination of Fig. 4.6 shows that


PROBLEM 4.19

Use Pohlhausen’s solution to determine the heat flux at the leading edge of a plate. Comment on your answer. Is it valid? If not explain why.

(1) Observations. (i) This is an external forced convection problem for flow over a flat plate. (ii) Pohlhausen’s solution for the temperature distribution and heat transfer coefficient is assumed to be applicable. (iii) Of interest is the va lue of the local heat flux at the leading edge of the plate. (iv) Knowing the local transfer coefficient and using Newton’s law, gives the heat flux

(2) Problem Definition. Determine the local heat transfer coefficient for laminar boundary layer flow over a flat plate.

(3) Solution Plan. Apply Newton’s law of cooling to determine the local heat flux. Use Pohlhausen’s solution for the local heat transfer coefficient. Apply the solution at the leading edge.

(4) Plan Execution.

(i) Assumptions. All assumptions leading to Pohlhausen's solution are applicable. These are:(1) Newtonian fluid, (2) steady state, (3) constant properties, (4) two-dimensional, (5)

laminar flow (Rex < 5 105), (6) viscous and thermal boundary layer flow (Rex > 100 and Pex > 100), (7) uniform upstream velocity and temperature, (8) uniform surface temperature, (9) flat plate, (10) negligible changes in kinetic and potential energy, (11) negligible dissipation, (12) no

buoyancy ( = 0 or g = 0) and (13) no energy generation ( 0q ).

(ii) Analysis. Application of Newton’s law gives

][)( TThxq ss (a)

where

h = local heat transfer coefficient, W/m2-oC

sq = surface heat flux, W/m2


T = free stream temperature, oC

Based on the above assumptions, Pohlhausen’s soluti on for the local heat transfer coefficient is

d

d

x

Vkh

)0( (b)

where

k = thermal conductivity, W/m- oC

V = free stream velocity, m/s T = temperature variable, oC

)/()( ss TTTT , dimensionless temperature

x = axial distance measured from the leading edge, m


00

sTx

y

t

V

T


= yV

x, dimensionless variable

y = vertical coordinate, m

Noting that all quantities in equation (a) are constant except the variable x, (a) can be rewritten as

h = x

constant (c)

(iii) Computations. To determine the heat transfer coefficient at the leading edge, set x = 0 in (c)

h(0) = cons ttan

0 =

Substituting into (a)

][)( TTxq ss

(iv) Checking. Dimensional check: Units of h in (a) should be W/m2-oC:

h = k (W/m-oC)V

x

m s

m s m

/

/2d

d )0((1/1) = W/m2-oC

(5) Comments. Physically, the heat transfer coefficient cannot be infinite. This suggests that Pohlhausen’s solution is not valid at the leading edge. One of the assumptions leading to Pohlhausen’s solution is Rex > 100. However, at the leading edge x = 0, the Reynolds number is given by

Rex =V x

= 0

Therefore, Pohlhausen’s solution cannot be used to determine h at the leading edge. In fact the solution breaks down at small values of x where the corresponding local Reynolds number and Peclet number are smaller than 100.

PROBLEM 4.20

Consider laminar boundary layer flow over a semi-infinite flat plate at uniform surface

temperature .sT The free stream velocity is V and the Prandtl number is 0.1. Determine the

temperature gradient at the surface ./)0,( dyxdT

(1) Observations. (i) This is an external forced convection problem for flow over a flat plate. (ii) Pohlhausen’s solution for the temperature distribution is assumed to be applicable. (iii) Of interest is the value of the normal temperature gradient at the surface.

(2) Problem Definition. Determine the normal temperature gradient at the surface, ,/)0,( dyxdT for laminar boundary layer flow over a flat plate.

(3) Solution Plan. Apply Pohlhausen’s solution for the temperature distribution for laminar flow over a flat plate.

(4) Plan Execution.




(ii) Analysis. Surface temperature gradient is given by

dy

d

d

d

d

dT

y

yxT ),( (a)

where

x

Vyyx ),( (4.41)

s

s

TT

TT (4.58)

Substituting (4.41) and (4.58) into (a)

d

d

x

VTT

y

yxTs )(

),( (b)

Evaluating (b) at 0y gives the temperature gradient at the surface

d

d

x

VTT

y

xTs

)0()(

)0,( (c)

The problem reduces to determining ./)0( dd This is given by Pohlhausen’s solution as


dd

fdd

d

Pr

Pr

2

2

332.0)0( (4.64)

where )(f is given by Blasius solution an d is listed in Table 4.1. The integral in (6.64) must be

evaluated numerically. An alternate approximate method for determining dd /)0( is using

Fig. 4.6 to determine the slope of the 098.0Pr curve.

(iii) Computations.

Numerical integration of (4.64) gives

1568.0)0(

d

d (d)

Substituting into (c)

x

VTT

y

xTs )(1568.0

)0,( (e)

Approximating the Prandtl number of 0.098 by 0.1 and using Fig. 4.6, gives the slope at the wall as

125.0)0(

d

d (f)

(iv) Checking. Dimensional check: Surface temperature gradient in (e) should have units ofC/m:

m

C

)(m/s)m(

)m/s()C/m)((

)0,( o

2

o

x

VTT

y

xTs

Limiting check: Surface temperature gradient should vanish for TTs . Setting TTs in (e)

gives .0/)0,( yxT

Qualitative check: Surface temperature gradient should increase as the free stream velocity is increased. This is confirmed by ©.

Comments. (1) Using numerical integration to evaluate the integral in (4.64) is necessary since this integral cannot be evaluated analytically. (2) As the thermal boundary layer thickness increases surface heat flux, and thus surface temperature gradient, should decrease. This follows from the observation that the thermal boundary layer acts as an insulation layer. Equation (e) show that surface temperature gradient decreases with distance x. (3) Fig. 4.6 provides a reasonable estimate surface temperature gradient.

PROBLEM 4.21

Fluid flows between two parallel plates. It enters with uniform velocity V and temperature .T

The plates are maintained at uniform surface temperature .sT Assume laminar boundary layer

flow at the entrance. Can Pohlhausen solution be applied to determine the heat transfer

coefficient? Explain.

Solution

The velocity distribution in Pohlhausen’s solution is based on Blasius solution. Blasius solution is limited to the flow over a single plate. For a single plate axial pressure gradient is set equal to zero. That is, Blasius solution is based on

0x

p

For flow between parallel plates axial pressure gradient does not vanish. That is

0x

p

In fact, pressure decreases in the flow direction. Thus, Blasius solution does not apply to the flow between parallel plates. It follows that Pohlhausen’s also does not apply.

PROBLEM 4.22

Two identical rectangles, A and B, of dimensions

L1 L2 are drawn on the surface of a semi-infinite

flat plate as shown. Rectangle A is oriented with

side L1 along the leading edge while rectangle B is

oriented with side L2 along the edge. The plate is

maintained at uniform surface temperature.

[a] If the flow over rectangle A is laminar, what is

it for B ?

[b] If the heat transfer rate from plate A is

435 W, what is the rate from plate B ?

(1) Observations. (i) This is an external forced convection problem over two flat plates. (ii) Both plates have the same surf ace area. (iii) For flow over a flat plate, the heat transfer coefficient h decreases with distance from the leading edge. (iv) Since the length in the flow direction is not the same for the two plates, the average heat transfer coefficient is not the same. It follows that the total heat transfer rate is not the same. (v) The flow over a flat plate

is laminar if the Reynolds number is less than 5 105.

(2) Problem Definition. Determine the Reynolds number at the trailing end of plate B.Obtain a solution for the average heat transf er coefficient for laminar forced convection over a flat plate.

(3) Solution Plan. Examine the Reynolds number at the trailing end of plate B to establish if the flow is laminar or turbulent. Use Newton’s law of cooling to determine the heat transfer from each plate. Use Pohlhausen’s solution to obtain a solution for the average heat transfer coefficient.

(4) Plan Execution.

(i) Assumptions. All assumptions leading to Pohlhausen's solution are applicable. These are: (1) Newtonian fluid, (2) steady state, (3) constant properties, (4) two-

dimensional, (5) laminar flow (Rex < 5 105), (6) viscous and thermal boundary layer flow (Rex > 100 and Pex > 100), (7) uniform upstream veloc ity and temperature, (8) uniform surface temperature, (9) flat plate, (10) negligible changes in kinetic and potential energy,

(11) negligible dissipation, (12) no buoyancy ( = 0 or g = 0) and (13) no energy generation ( 0q ). In addition, (14) neglect radiation.

(ii) Analysis. [a] To establish if the flow is laminar or turbulent, compare the Reynolds number with the transition Reynolds number. For the flow over a flat plate the transition Reynolds number is

txRe = 5 105 (a)

The flow is considered laminar if Rex < txRe . The Reynolds number for plate B is

1LRe = 1LV (b)

T

sTV B

A2L

1L

1L

2L

viewtop

PROBLEM 4.22 (continued)where

1LRe = Reynolds number at trailing end of plate B

L1 = length of plate B in the flow direction, m

V = upstream velocity, m/s


Similarly, for plate A

2LRe = 2LV (c)

where

2LRe = Reynolds number at trailing end of plate A

L2 = length of plate A in the flow direction, m

Taking the ratio of (b) and (c) and rearranging

1LRe =2

1

L

L2LRe (d)

Since L1 < L2, equation (d) gives

1LRe < 2LRe (e)

Thus,1LRe <

2LRe <txRe . Since the flow is laminar for plate A, it follows that it is also

laminar for plate B.

[b] Application of Newton' s law of cooling gives

q = h A (Ts - T ) (f) where

A = surface area, m2

h = average heat transfer coefficient, W/m2-oCq = total surface heat transfer rate, W



Applying (f) to the two plates and taking the ratio of the resulting equations

q

q

h

h

B

A

B

A

(g)

where the subscripts refer to plates A and B. The average heat transfer coefficient for laminar flow over a flat plate is given by Pohlhausen's solution, equations (4.67) and (4.71b)

3/1)(664.0 PrL

Vkh (h)

where


k = thermal conductivity, W/m-oC Pr = Prandtl number

Applying (h) to plates A and B, noting that L = L2 for A and L = L1 for B, and taking the ratio of the results

h

h

L

L

B

A

2

1

(i)

Substituting into (g) and solving for qB

q qB A

L

L

2

1

(j)

(iii) Computations. With qA = 435 W, equation (j) gives

qB = 435(W)L

L

2

1

Since L2 > L1, it follows that the heat transfer rate from B is greater than that from A.

(iv) Checking. Dimensional check: Units of h in equation (h) should be W/m2-oC

h = k(W/m-oC)V

L )m)(s/m(

s/m2

Pr1/3

= W/m2-oC

Limiting check: For the special case of L1 = L2 (square plate), the heat transfer rate from the two plates should be the same. Setting L1 = L2 in equation (j) gives q qB A .

(5) Learning and Generalizing. To maximize the rate of heat transfer from a flat rectangular plate under laminar flow conditions, the long side of the plate should face flow direction.

PROBLEM 4.23

A semi-infinite plate is divided into four equal sections of one centimeter long each. Free

stream temperature and velocity are uniform and the flow is laminar. The surface is maintained

at uniform temperature. Determine the ratio of the heat transfer rate from the third section to

that from the second section.

(1) Observations. (i) This is an external forced convection problem for flow over a flat plate. (ii) Pohlhausen’s solution for the temperature distribution and heat transfer coefficient is assumed to be applicable. (iii) Of interest is the value of the heat transfer rate from a section of the plate at a specified location and of a given width. (iv) Newton’s law of cooling gives the heat transfer rate.

(2) Problem Definition. Determine the local heat transfer coefficient for laminar boundary layer flow over a flat plate.

(3) Solution Plan. Apply Newton’s law of cooling to determine the local heat flux. Use Pohlhausen’s solution for the local heat transfer coefficient.

(4) Plan Execution.




(ii) Analysis. Application of Newton’s law to the surface element Wdx gives

WdxTTxhdq s ][)( (a)

where

h = local heat transfer coefficient, W/m2-oC q = surface heat flux, W Ts = surface temperature, oC

T = free stream temperature, oC W = plate width, m


The local heat transfer coefficient is given by

d

d

x

Vkxh

)0()( (4.66)

where


V = free stream velocity, m/s

T

V

sT

4321

x

1x 2x 3x4x

dx

W




= yV




x

dx

d

dVWkTTdq s

)0(][ (b)

Integration of (b) between two values of x gives the heat transfer rate for that section. Application of (b) to sections 3 and 4, give

)()0(

][2)0(

][ 232

3

2

xxd

dVWkTT

x

dx

d

dVWkTTq ss

x

x

(c)

)()0(

][)0(

][ 343

4

3

xxd

dVWkTT

x

dx

d

dVWkTTq ss

x

x

(d)

where

01.02x m

02.03x m

03.04x m

Taking the ratio of (c) and (d)

23

34

2

3

xx

xx

q

q (e)

(iii) Computations. Equation (e) gives

7673.0)m(01.0)m(02.0

)m(02.0)m(03.0

2

3

q

q

(iv) Checking. Dimensional check: Units of q in (c) should be W:

W)m(])[1/1()0(

/sm

(m/s)C)m/W()m()C]([ 342

oo xxd

dVkWTTq s

Limiting check: The heat transfer rate should vanish for TTs . Setting TTs in (c) and (d)

gives .032 qq

Qualitative check: The heat transfer rate should increase as the free stream velocity, plate width, or thermal conductivity are increased. This is confirmed by (c) and (d).

Comments. Although each section is rectangular in shape, the same procedure can be followed to determine the heat transfer rate from any configuration drawn on the plate.

PROBLEM 4.24

A fluid at a uniform velocity and temperature flows over a semi-infinite flat plate. The

surface temperature is uniform. Assume laminar boundary layer flow.

[a] What will be the percent change in the local heat transfer coefficient if the free

stream velocity is reduced by a factor of two?

[b] What will be the percent change in the local heat transfer coefficient if the distance from

the leading edge is reduced by a factor of two?

(1) Observations. (i) This is an external forced convection problem for flow over a flat plate. (ii) Of interest is the variation of the local heat transfer coefficient with free stream velocity and distance from the leading edge. (iii) Pohlhausen's solution applies to this problem.

(2) Problem Definition. Obtain a solution for the local heat transfer coefficient and examine its dependency on the free stream velocity and distance from the leading edge.

(3) Solution Plan. Use Pohlhausen's solution for the loca l heat transfer coefficient over a semi-infinite flat plate.

(4) Plan Execution.

(i) Assumptions. All assumptions leading to Pohlhausen's solution are applicable. These are:(1) Newtonian fluid, (2) steady state, (3) constant properties, (4) two-dimensional,

(5) laminar flow (Rex < 5 105), (6) viscous and thermal boundary layer flow (Rex > 100 and Pex > 100), (7) uniform upstream velocity and te mperature, (8) uniform surface temperature, (9) flat plate, (10) negligible changes in kinetic and potential energy, (11) negligible

dissipation, (12) no buoyancy ( = 0 or g = 0) and (13) no energy generation ( 0q ).

(ii) Analysis. Pohlhausen's solution for the local heat transfer coefficient gives

d

d

x

Vkh

)0( (a)

where

h = local heat transfer coefficient, W/m2-oCk = thermal conductivity, W/m-oC

V = free stream velocity, m/s T = temperature variable, oCTs = surface temperature, oC



x = axial coordinate, measured from the leading edge, m y = vertical coordinate, measured from the surface, m



= yV


However, k, and d

)0( are constants. Thus, equation (a) can be rewritten as

x

VCh (b)

where C is a constant. The percent change in h is given by

Percent change = 100 ( 12 hh )/h1

= 100 (h2/h11) (c)

where subscripts 1 and 2 refer to the initial and changed conditions, respectively. Applying (b) to the two locations and substituting into (c)

Percent change = 100 121

12

xV

xV (d)

(iii) Computations.

[a] Percent change in h if the free stream velocity V is reduced by a factor of two:

Applying (d) for 2/12 VV and x1 = x2, gives

Percent change = 100 11

2

V

V100 )12/1( 3.29 %

[b] Percent change in h if the distance x from the leading edge is reduced by a factor of two:

Applying (d) for x2 = x1/2 and 21 VV , gives

Percent change = 100 %4.41)12(10012

1

x

x

(iv) Checking. Dimensional check: Units of h in (a) should be W/m2-oC:

h = k (W/m-oC)V

x

m s

m s m

/

/2d

dT )0(*

(1/1) = W/m2-oC

Qualitative check: As the free stream velocity decreases, the local heat transfer coefficient should decrease. Computations confirm this as indicated by the negative sign in the percent change when free stream velocity is reduced by a factor of two.

(5) Comments. (i) The local heat transfer coefficient increases as the distance from the leading edge is decreased and as the free stream velocity is increased. (ii) Since percent change involves taking ratios, the problem is solved without knowing the nature of the fluid

and the magnitudes of V and x.

PROBLEM 4.25

Use Pohlhausen's solution to derive an expression for the ratio of the thermal boundary layer

thickness for two fluids. The Prandtl number of one fluid is 1.0 and its kinematic viscosity is

012 10 6. m / s2 . The Prandtl number of the second fluid is 100 and its kinematic viscosity is

68 10 6. m / s2 .

(1) Observations. (i) This is an external flow problem. (ii) At the edge of the thermal

boundary layer, ty , fluid temperature

approaches free stream temperature. That is,

TT and 1)/()(*ss TTTTT . (iii)

According to Pohlhausen's solution, Fig. 4.6, the thermal boundary layer thickness depends

on the Prandtl number, free stream velocity V ,

kinematic viscosity and location x.

(2) Problem Definition. Derive an expression for the thermal boundary layer thickness for laminar flow over a semi-infinite flat plate.

(3) Solution Plan. Use Pohlhausen's solution, Fig. 4.6, to determine t at Pr = 1 and Pr = 100.

(4) Plan Execution.




(ii) Analysis and Computations. At the edge of the thermal boundary layer, ty , fluid

temperature is

TxT t),(

and therefore the dimensionless temperature *T is

1)/()()/()(*ssss TTTTTTTTT (a)

where

T = temperature variable, oCTs = surface temperature, oC

T = free stream temperature, oC*T = dimensionless temperature, oC

y = coordinate normal to the plate, m

t = thermal boundary layer thickness = value of y where *T = 1, m

Let the subscripts 1 and 2 de note conditions corresponding to Pr = 1 and Pr = 100, respectively.

0

0

2 4 6 8 10 12 14

0.2

0.4

0.6

0.8

1.0

T T

T T

s

s

y V x/

Pr = 0,01

0.1

1.01

00

0.71 (air)

Fig. 7.2


Using Fig. 4.6 at Pr = 1, the value of corresponding to *T = 1 is

2.51

11x

Vt (b)

Similarly, for Pr = 100

5.12

22x

Vt (c)

where

V = free stream velocity, m/s x = coordinate along the plate, m

= dimensionless variable = xVy /

1 = kinematic viscosity (of fluid with Pr = 1) = 0.12 610 m2/s

2 = kinematic viscosity (of fluid with Pr = 100) = 6.8 610 m2/s

Taking the ratio of (b) and (c) gives

46.0)s/m(108.6

)s/m(1012.0

5.1

2.5

5.1

2.526

26

2

1

2

1

t

t

(iii) Checking. Dimensional check: The right hand side of equation (b) should be dimensionless:

)m()/sm(

)(m/s)m(

21

11x

Vt = dimensionless

(5) Learning and Generalizing. (i) In calculating the ratio 21/ tt , the location x and the free

stream velocity V are assumed to be the same for both fluids. (ii) The ratio 21/ tt is constant

independent of location.

PROBLEM 4.26

Water at 25oC flows over a flat plate with a uniform velocity of 2 m/s. The plate is maintained at

85oC. Determine the following:

[a] The thermal boundary layer thickness at a distance of 8 cm from the leading edge.[b] The heat flux at this location.[c] The total heat transfer from the first 8 cm of the plate.

[d] Whether Pohlhausen's solution can be used to find the heat flux at a distance of 80 cm from

the leading edge.

(1) Observations. (i) This is an external forced convection problem for flow over a flat plate. (ii) The Reynolds number and Peclet number should be checked to determine if the flow is laminar and if boundary layer approximations are valid. (iii) Pohlhausen's solution is applicable

if 100 < Rex < 100 105 and Pex = Rex Pr > 100. (iv) Thermal boundary layer thickness and heat transfer coefficient vary along the plate. (v) Newton’s law of cooling gives local heat flux. (vi) The fluid is water.

(2) Problem Definition. The problem is determining the temperature distribution in the fluid. Knowing the temperature distribution for laminar flow (Pohlhausen's solu tion), the thermal boundary layer thickness, local and average heat transfer coefficients, local heat flux and total heat transfer rate can be determined.

(3) Solution Plan. Check the Reynolds and Peclet numbers to determine if boundary layer approximations can be made, the flow is laminar and if Pohlhausen’s solution is applicable. Use Pohlhausen’s solution to determine the local heat transfer coefficient and Newton’s law of cooling to determine heat flux and heat transfer rate.

(4) Plan Execution.

(i) Assumptions. Assume and verify that Pohlhausen’s solution is applicable. Assumptions leading to Pohlhausen's solution are: (1) Ne wtonian fluid, (2) steady state, (3) constant

properties, (4) two-dimensional, (5) laminar flow (Rex < 5 105), (6) viscous and thermal boundary layer flow (Rex > 100 and Pex > 100), (7) uniform upstream velocity and temperature, (8) uniform surface temperature, (9) flat plate, (10) negligible changes in kinetic and potential

energy, (11) negligible dissipation, (12) no buoyancy ( = 0 or g = 0) and (13) no energy generation ( 0q ). In addition, (14) no radiation.

(ii) Analysis. The Reynolds and Peclet numbers are defined as

Rex =V x

(a)

andPex = Rex Pr (b)

where

Pex = Peclet number Pr = Prandtl number Rex = Reynolds number

V = upstream velocity = 2 m/s x = distance from the leading edge of the plate, m

00

sT x

yt

V

T

)(xqs

)(xh




Tf = T Ts

2 (c)

where

Tf = film temperature, oCTs = surface temperature = 85oC


To determine if the flow is laminar or turbulent, compare the Reynolds number with the

transition Reynolds number. For the flow over a flat plate, transition Reynolds number tx

Re is

txRe = 5 105 (d)

The flow is laminar if Rex < txRe . Viscous boundary layer approximations are valid for

100xRe

Thermal boundary layer approximations are valid for

Pex > 100Substituting into (c)

Tf = (85 + 25)( oC)/2 = 55oC

Properties of water at this temperature are

k = 0.6458 W/m-oCPr = 3.27

= 0.5116 10-6 m2/s

The Reynolds and Peclet numbers are determined at x = 0.08 m

Rex = 2 0 08

0 5116 10312 744

6 2

( / ) . ( )

. /,

m s m

m s

and

Pex = 312,744 3.27 = 1.0227 106

Therefore, boundary layer approximations can be made and the flow is laminar. Pohlhausen’s solution is applicable.

[a] Boundary layer thickness t . Pohlhausen’s solution, Fig. 4.6, is used to determine the

thermal boundary layer thickness. At the edge of the thermal boundary layer, y = t , fluid

temperature is approximately the same as ambient temperature. That is

T(x, t ) T (e)

Or, in terms of the dimensionless temperature *T

*T =s

st

TT

TxT ),( 1 (f)


T = temperature variable, oC*T = dimensionless temperature

y = distance from the plate, m

t = thermal boundary layer thickness, m

At T = 1 and Pr = 3.27, Fig. 4.6 gives the thermal boundary layer thickness in terms of the

dimensionless variable

3 = yx

V=

x

Vt (g)

Solving (g) for t

t = 3V

x (h)

[b] Local surface heat flux qs . This is the local heat transfer per unit area. Applying Newton's

law of cooling at location x gives

qs = h (Ts - T ) (i)

where

h = local heat transfer coefficient, W/m2-oCqs = local surface heat flux, W/m2

The local heat transfer coefficient is given by Pohlhausen’s solution. For 0.5 < Pr < 50, equation (7.24b) gives

h = 0.332 kV

xPr

1/3 (j)

Substituting (j) into (i) gives

qs = 0.332 k (Ts - T )V

xPr

1/3 (k)

[c] Total heat transfer rate qT. Applying Newton’s law of cooling and using the average heat transfer coefficient gives the total heat transfer from the plate

qT = )( TTLWh s (l)

where

h = average heat transfer coefficient, W/m2-oCL = length of plate = 0.08 m W = width of plate, m

The average heat transfer coefficient is given by equations (4.67) and (4.71b)

3/1664.0 Pr

vL

Vkh (m)

Substituting (m) into (l) and rearranging

1/3)()(664.0 PrReTTkW

qLs

T (n)


where ReL is the Reynolds number at x = L.

(iii) Computations.

[a] Boundary layer thickness t . Substituting into equation (h)

t = 3)s/m(2

)m(08.0)s/m(105116.0 26

= 0.00043 m = 0.43 mm

[b] Surface heat flux qs . Equation (k) gives

qs = 0.332 (0.6458)(W/m-oC) (85 -25)(oC)2

05116 10 0 086 2

( / )

. ( / ) . ( )

m s

m s m (3.27)1/3

=133,477 W/m2

[c] Total heat transfer rate qT . Equation (n) gives

q

WT = 0.664 (0.6458)(W/m-oC) (85 - 25)(oC) (312,744)1/2 (3.27)1/3 = 21,356 W/m

[d] Validity of Pohlhausen’s solution at x = 0.8 m. The Reynolds number at x = 0.8 m is

Rex = 2 0 8

0 5116 103127 10

6 2

6( / ) . ( )

. /.

m s m

m s

Since the Reynolds number is greater than the transition Reynolds number the flow is turbulent and therefore Pohlhausen’s solution does not apply.

(iv) Checking. Dimensional check: Computations showed that units for equations (a), (h), (k) and (n) are dimensionally consistent.

Quantitative check: Equations (4.66) and (4.67) show that for a plate of length L, the average heat transfer coefficient is twice the local coefficient at x = L. This means that the average heat flux is twice the flux at x = L. Thus, the average flux for this problem is 2(133,477)(W/m2) = 266,954(W/m2). Therefore

sT qLWq (o)

where sq is the average heat flux. Substituting into (o)

W

qT = 0.08(m)266,954(W/m2) = 21,356 W/m

Limiting check: If Ts = T , the local heat flux and the total heat transfer rate should vanish.

Setting Ts = T in equations(k) and (n) gives the expected results.

(5) Comments. (i) It is important to check the conditions for boundary layer flow and for laminar flow before proceeding with the solution (i.e. check the Reynolds and Peclet numbers). (ii) Typically, the thickness of the thermal boundary layer is relatively small. In this example, at a distance of 8 cm from the leading edge is 0.43 mm.

PROBLEM 4.27

The cap of an electronic package is cooled by forced convection. The free stream temperature is

25oC. The Reynolds number at the downstream end of the cap is 110,000. Surface temperature

was found to be 145oC. However, reliability requires that

surface temperature does not exceed 83oC. One possible

solution to this design problem is to increase the free stream

velocity by a factor of 3. You are asked to determine if

surface temperature under this plan will meet design

specification.

(1) Observations. (i) This is an external forced convection problem over a flat plate. (ii) Increasing the free stream velocity, increases the average heat transfer coefficient. This in turn causes surface temperature to drop. (iii) Based on this observation, it is possible that the proposed plan will meet design specification. (iv) Since the Reynolds number at the downstream end of the package is less than 500,000, it follows that the flow is laminar throughout. (v) Increasing the free stream velocity by a factor of 3, increases the Reynolds number by a factor of 3 to 330,000. At this Reynolds number the flow is still laminar. (vi) The power supplied to the package is dissipated into heat and transferred to the surroundings from the surface. (vii) Pohlhausen's solution can be applied to this problem. (viii) The ambient fluid is unknown.

(2) Problem Definition. Find the relationship between surface temperature, free stream velocity and heat transfer coefficient.

(3) Solution Plan. Surface temperature is related to heat transfer coefficient by Newton's law of cooling. Heat transfer coefficient is related to free stream velocity in Pohlhausen’s solution. Apply Newton’s law of cooling and Pohlhausen’s solution to the cap.

(4) Plan Execution.



( = 0 or g = 0) and (13) no energy generation ( 0q ). In addition, (14) neglect radiation and

(15) all power supplied to package leaves the cap as heat.

(ii) Analysis. Application of Newton's law of cooling gives

Tq = h A (Ts - T ) (a)

where


h = average heat transfer coefficient, W/m2-oC

Tq = total surface heat transfer rate, W



V T,cap at Ts


The heat transfer rate Tq is determined by the power supplied to the package which is assumed

constant. Changing the free stream velocity results in a change in the average heat transfer

coefficient and surface temperature. However, Tq , A and T remain the same. Let subscripts 1

and 2 refer to conditions corresponding to the two velocities V 1 and V 2, where

V 2 = 3 V 1 (b)

Applying (a) to the two cases gives

Tq = h1A (Ts1 - T ) (c)

and

Tq = h2 A (Ts2 - T ) (d)

Taking the ratio of (c) and (d)

11 1

2 2

h T T

h T T

s

s

( )

( )

Solving the above for Ts2

Ts2 = T + )( 1

2

1 TTh

hs (e)

The average heat transfer coefficient for laminar flow over a flat plate is given by Pohlhausen's solution (7.25)

3/1)(664.0 PrL

Vkh (f)

where

k = thermal conductivity, W/m-oC L = length of cap, m

Pr = Prandtl number



Properties in Pohlhausen’s solution are evaluated at the film temperature, 2/)( TTT sf .

Since surface temperature is expected to change, it follows that properties will also change. However, this effect cannot be considered in th e solution since the cooling fluid is not known. As a first approximation, changes in properties due to changes in surface temperature will be

neglected. Thus, k, Pr and are assumed to remain constant and equation (f) simplifies to

VCh (g)

where C is a constant. Applying (g) to the two cases and taking the ratio of the resulting equations

2

1

2

1

V

V

h

h (h)

Substituting (h) into (e) gives


Ts2 = T +2

1

V

V (Ts1 - T ) (i)

(iii) Computations. Setting V 2 = 3 V 1 and Ts1 = 145oC in (i) gives

Ts2 = 25 (oC) + )C)(25145(3/1 o = 94.3oC

Since this temperature is above the design limit of 83oC, the proposed plan will not work.

(iv) Checking. Dimensional check: Since the ratio of velocities is dimensionless, it follows that equation (i) is dimensionally consistent.

Qualitative check: As V is increased, surface temperature Ts should decrease. This is confirmed by equation (i).

Limiting check: There should be no change in surface temperature for the special case of V 2 =

V 1. Setting V

V

1

2

= 1 in (i) gives the expected result of Ts2 = Ts1.

(5) Comments. (i) This problem is solved without knowing the size of the cap, the magnitude of the free stream velocity and the nature of the fluid. (ii) Neglecting radiation is a conservative assumption since it overestimates surface temperature. Taking radiation into consideration results in a lower temperature than 94.3oC. Thus, the proposed design modification should not be rejected without first examining the effect of radiation. (iii) If the cooling fluid is known, changes in properties due to changes in surface temperature can be accounted for using a trial

and error procedure. A value for Ts2 is assumed, properties at Tf2 are determined and 21 / hh is

calculated and used in equation (e) to calculate Ts2. If the calculated Ts2 is not close to the assumed value, the procedure is repeated until a satisfactory agreement is obtained between

assumed and calculated values. (iv) For V 2 = 4.3V 1, the resulting surface temperature will be 83oC.This meets design specification.

PROBLEM 4.28

The back of the dinosaur Stegosaurus has two rows

of fins. Each row is made up of several fins arranged

in line and separated by a space. One theory

suggests that providing a space between neighboring

fins reduces the weight on the back of the dinosaur

when compared with a single long fin along the back.

On the other hand, having a space between

neighboring fins reduces the total surface area. This

may result in a reduction in the total heat loss.

Model the fins as rectangular plates positioned in line as shown. The length of each plate is L

and its height is H. Consider two fins separated by a distance L. Compare the heat loss from the

two fins with that of a single fin of length 3L and height H. Does your result support the

argument that spaced fins lead to a reduction in heat loss? To simplify the analysis assume

laminar flow.

(1) Observations. (i) This is an external forced convection problem of flow over a flat plate. (ii) Convection heat transfer from a surface can be determined using Newton’s law of cooling. (iii) The local heat transfer coefficient changes along the plate. The total heat transfer rate can be determined using the average heat transfer coefficient. (iv) For laminar flow, Pohlhausen's solution gives the heat transfer coefficient. (v) For two in-line fins heat transfer from the down stream fin is influenced by the upstream fin. The further the two fins are apart the less the interference will be.

(2) Problem Definition. Determine the average heat transfer coefficient for plates of length Land 3L.

(3) Solution Plan. Use Pohlhausen’s solution to determine the average heat transfer coefficient. Apply Newton's law of cooling to determine the h eat transfer rate from two plates of length L each and compare with the heat transfer from a single plate of length 3L.

(4) Plan Execution.

(i) Assumptions. All assumptions leading to Pohlhausen's solution are applicable. These are: (1) Newtonian fluid, (2) steady state, (3) constant properties, (4) two-dimensional, (5)


buoyancy ( = 0 or g = 0) and (13) no energy generation ( 0q ). In addition, (14) neglect

radiation and (15) neglect interference between in-line plates.

(ii) Analysis. Of interest is the ratio of the total heat transfer rate from two single plates of length L each and a single plate of length 3L. Newton’s law of cooling applied to a single plate of length L gives

)(1 TTAhq s (a)


where

A plate area, 2m

h average heat transfer coefficient, CW/m o2

1q heat transfer rate from a single plate of length L, W

sT surface temperature, Co

T free stream temperature, Co

The area of a single plate of length L and height H is

A = HL (b)

The average Nusselt number is obtained from Pohlhausen's solution

LL RePrk

LhNu 3/1664.0 (c)

where

k = thermal conductivity, W/m-oC

L = plate length, m Pr Prandtl number

LRe Reynolds number


LVRe (d)

where



Substituting (b)-(d) into (a)

LTTV

kHPrq s )(664.0 3/11 (e)

The total heat transfer rate from two in-line fins, 2q , is

LTTV

kHPrqq s )(664.022 3/112 (f)

The heat transfer rate 3q from a single plate of length 3L is obtained from (e) by replacing L

with 3L

LTTV

kHPrq s 3)(664.0 3/13 (g)

Taking the ratio of (f) to (g)

15.13

2

3

2

q

q (h)


(iii) Checking. Dimensional check: Units of q1 in equation (e) should be W

m)()C)(()s/m(

s)/m()m()CW/m(664.0 o

2

o3/11 LTT

VHkPrq s = W

Limiting check: If TTs the heat transfer rate should vanish regardless of plate arrangement.

Setting TTs in (f) and (g) gives

032 qq

(5) Comments. (i) Based on the assumption of no inte rference between neighboring in-line plates, the heat transfer rate from two in-line plates of length L each separated by a distance Lexceeds that of a single plate of length 3L by 15%. The weight of tw o in-line plates of length Leach is 2/3rd of that of a single plate of length 3L. Thus the two in-line plate arrangement has advantages in heat transfer rate and weight when compared to a single plate of length 3L. (iii) The analysis can be generalized to n in-line plates each of length L separated by spacing L as compared with a single plate of length (2n-1)L. The ratio of the heat transfer rates for this case is given by

1212 n

n

q

q

n

n (i)

For example, for n = 10

29.21102

10

19

10

q

q

Thus the heat transfer rate from 10 in-line plates of length L each separated by a distance L

exceeds that of a single plate of length 19L by 129%. The weight is reduced by approximately ½.

PROBLEM 4.29

A fluid with Prandtl number 0.098 flows over a semi-

infinite flat plate. The free stream temperature is T and

the free stream velocity is V . The surface of the plate is

maintained at uniform temperature .sT Assume laminar

flow.

[a] Derive an equation for the local Nusselt number.

[b] Determine the heat transfer rate from a section of the plate between 1x and 2x . The width of

the plate is W.

[c] Derive an equation for the thermal boundary layer thickness ).(xt

(1) Observations. (i) This is an external forced convection problem for flow over a flat plate. (ii) Pohlhausen’s solution for the temperature distribution and heat transfer coefficient is assumed to be applicable. (iii) Knowing the heat transfer coefficient, the local Nusselt number can be determined. (iv) the Newton’s law of cooling gives the heat transfer rate. (iv) Pohlhausen’s solution gives the thermal boundary layer thickness.

(2) Problem Definition. Determine the local heat transfer coefficient and thermal boundary layer thickness for laminar boundary layer flow over a flat plate.

(3) Solution Plan. Use Pohlhausen’s solution to determine the local heat transfer coefficient. Apply Newton’s law of cooling to determine the heat transfer rate. Use Fig. 4.6 to determine the thermal boundary layer thickness.

(4) Plan Execution.




(ii) Analysis. [a]: The local heat tran sfer coefficient is given by

d

d

x

Vkxh

)0()( (4.66)

where





= yV



T

V

x

1x 2x

W0

dx


The local Nusselt number is defined as

k

hxNux (a)


d

dReNu xx

)0( (b)

where xRe is the local Reynolds number defined as

xVRex (c)

The constant dd /)0( is given in Table 4.2

138.0)0(

d

d (d)

Substituting (d) into (b)

xx ReNu 138.0 (e)

[b] Application of Newton’s law to the surface element Wdx gives

WdxTTxhdq s ][)( (f)

where

h = local heat transfer coefficient, W/m2-oC q = surface heat flux, W Ts = surface temperature, oC

T = free stream temperature, oC W = plate width, m


Substituting (4.66) into (f)

x

dx

d

dVWkTTdq s

)0(][ (g)

Integration of (g) from 1x to 2x

)()0(

][2)0(

][ 12

2

1

xxd

dVWkTT

x

dx

d

dVWkTTq ss

x

x

Introducing the definition of the local Reynolds number (c) and using (d), the above gives

)(][2276.012

xx ReReWkTTq s (h)

[c] At the edge of the thermal boundary layer, ty the dimensionless temperature in Fig. 4.6

is equal to unity

1s

s

TT

TT (i)

For a fluid with 098.0Pr the corresponding value of ),( tx is obtained from Fig. 4.6 as


12x

Vtt (j)

Solving (b) for t

V

xt 12 (k)

Expressing this result in dimensionless form

x

t

RexVx

1212 (l)

(iii) Checking. Dimensional check: (1) Each term in (e) and (l) is dimensionless.

(2) Units q in equation (h) are

WC)m/W()m()C]([ oo kWTTq s

Limiting check: The heat transfer rate should vanish for TTs . Setting TTs in (h) gives

.0q

Qualitative check: The thermal boundary layer thickness increases with distance x. Solution (k) confirms this behavior.

(5) Comments. (i) The value of dd /)0( was obtained by interpolation in Table 4.2. A more

accurate value can be obtained using equation (6.46)

dd

fdd

d

Pr

Pr

2

2

332.0)0( (4.64)

However, this requires numerical ev aluation of the integral in (6.46).

(ii)Recalling that Blasius solution gives

xRex

2.5 (4.46)

Comparing (l) with (4.46) shows that t for 098.0Pr . Examination of Fig. 4.6 shows that


PROBLEM 4.30

Two identical triangles are drawn on the surface of a flat

plate as shown. The plate, which is maintained at uniform

surface temperature, is cooled by laminar forced

convection. Determine the ratio of the heat transfer rate

from the two triangles, q1/q2.

(1) Observations. (i) This is an external forced convection problem of flow over a flat plate. (ii) Convection heat transfer from a surface can be determined using Newton’s law of cooling. (iii) The local heat transfer coefficient changes along the plate. (iv) For each triangle the area changes with distance along the plate. (v) The total heat transfer rate can be determined by integration along the length of each triangle. (vi) Pohlhausen's solution may be a pplicable to this problem.

(2) Problem Definition. Determine the local heat transfer coefficient along each triangle.

(3) Solution Plan. Apply Newton's law of cooling to an element of each triangle, ydx, determine the local heat transfer coefficient along the plate and integrate over the area.

(4) Plan Execution.




radiation.

(ii) Analysis. Of interest is the ratio of the total heat transfer rate from triangle 1 to triangle 2. Since both the heat transfer coefficient and area vary along each triangle, it follows that Newton's law of cooling s hould be applied to an element dA at a distance x from the leading edge:

dq = h(x) (Ts - T )dA (a) where

dA = area of element, m2

dq = rate of heat transfer from element, W h = local heat transfer coefficient, W/m2-oC


T = free stream temperature, oCx = distance along plate, m

The local heat transfer coefficient h(x) is obtained from Pohlhausen's solution (4.66)

d

d

x

Vkh

)0( (b)

where

TsT

V

viewtop

1 2H

Lx

1y

2y






= x

Vy , dimensionless variable


Noting that all quantities in equation (b) are constant except the variable x, (b) is rewritten as

h = x

constant (c)

The next step is to determine the infinitesimal area dA for each triangle. Using the subscripts 1 and 2 to refer to triangles 1 and 2, respectively,

dxxydA )(11 (d)

and

dxxydA )(22 (e)

where

)(1 xy = side of element in triangle 1, m

)(2 xy = side of element in triangle 2, m

Similarity of triangles gives

)()(1 xLL

Hxy (f)

xL

Hxy )(2 (g)

Substituting (f) into (d) and (g) into (e) gives

dA1 = (H/L)(L x )dx (h)

dA2 = (H/L) x dx (i)where

H = base of triangle, m L = length of triangle, m

Substituting (c) and (h) into (a) and integrating from x = 0 to x = L, gives

q1 =

L

s

L

s dxx

xL

L

HTTCdx

x

xL

L

HTTCdq

02/1

02/11 )()(

Carrying out the integration yields

2/11 )()3/4( HLTTCq s (j)

Similarly, substituting (c) and (i) into (a) and integration from x = 0 to x = L gives


q2 =

L

s

L

s dxxL

HTTCdx

x

x

L

HTTCdq

0

2/1

02/12 )()(


q2 = (2/3)C(Ts - T )H L1/2 (k)

Taking the ratio of (j) and (k)

22

1

q

q (l)

(iii) Checking. Dimensional check: Units of q1 in equation (j) should be W. First, units of C are determined

C = k(W/m-oC)[V (m/s)/ (m2/s)]1/2

d

d )0((1/1) = W/m3/2-oC

Units of 1q are

1q = C(W/m3/2-oC)(Ts- T )(oC)H(m)L1/2 (m1/2) = W

Since 2q has the same form as 1q , it follows that the units of 2q in equation (k) are also correct.

Qualitative check: The result shows that the rate of heat transfer from triangle 1 is greater than that from triangle 2. This is expected since the heat transfer coefficient increases as the distance from the leading edge is decreased and triangle 1 has its base at x = 0 where h is maximum.

(5) Comments. (i) Although the two triangles have the same area, the rate of heat transfer from triangle 1 is double that from triangle 2. Thus, orientation and proximity to the leading edge of a flat plate play an important role in determining the rate of heat transfer. (ii) The same approach can be used to determine heat transfer for configurations other than rectangles, such as circles and ellipses.

PROBLEM 4.31

An isosceles triangle is drawn on a semi-infinite flat

plate at a uniform surface temperature .sT Consider

laminar uniform flow of constant properties fluid

over the plate. Determine the rate of heat transfer

between the triangular area and the fluid.

(1) Observations. (i) This is an external forced convection problem of flow over a flat plate. (ii) Heat transfer rate can be determined using Newton’s law of cooling. (iii) The local heat transfer coefficient changes along the plate. (iv) The area changes with distance along the plate. (v) The total heat transfer rate can be determined by integration along the length of the triangle. (vi) Pohlhausen's solu tion may be applicable to this problem.

(2) Problem Definition. Determine the local heat transfer coefficient along the triangle.

(3) Solution Plan. Apply Newton's law of cooling to an element ydx, determine the local heat transfer coefficient along the plate and integrate over the area.

(4) Plan Execution.




radiation.

(ii) Analysis. Taking advantage of symmetry, we consider the right angle triangle representing half the isosceles triangle. Newton's law of cooling applied to an element ydx at a distance x from the leading edge gives

dq = 2h(x) (Ts - T )ydx (a) where h = local heat transfer coefficient, W/m2-oC dq = rate of heat transfer from 2 elements, W Ts = surface temperature, oC


y = y(x) height of element

Note that the factor 2 is introduced to account for heat transfer from the two right angle triangles representing the isosceles triangles. The local heat transfer coefficient h(x) is obtained from Pohlhausen's solution

d

d

x

Vkh

)0( (4.66)

TsT

V

viewtop

H

L

dx

y


where





= x


The variable y(x) is given by

xL

Hxy )( (b)

where

H = base of triangle, m L = length of triangle, m

Substituting (4.66) and (b) into (a) and integrating from x = 0 to x = L, gives

L

s dxxd

dVk

L

HTTq

0

2/1)0()(2


d

dLVHTTkq s

)0()(

3

4 (c)

This result can be expressed in terms of the Reynolds number as

d

dReHTTkq Ls

)0()(

3

4 (d)

where

LVReL (e)

(iii) Checking. Dimensional check: Units of q in equation (d) should be W

q = k(W/m-oC) )C)(( oTTs H (m)d

d )0((1/1) = W

Limiting check: The heat transfer rate should vanish for TTs . Setting TTs in (d) gives

.0q

(5) Comments. Examination of (4.66) shows that the heat transfer coefficient decreases as x is increased. Thus, the same triangle rotated 180 degrees, with its base at x = 0, will have a higher heat transfer rate.

PROBLEM 4.32

Determine the total heat transfer rate from a half circle

drawn on a semi-infinite plate as shown. Assume laminar

two-dimensional boundary layer flow over the plate.

(1) Observations. (i) This is an external forced convection problem of flow over a flat plate. (ii) Heat transfer rate can be determined using Newton’s law of cooling. (iii) The local heat transfer coefficient changes along the plate. (iv) The area changes with distance along the plate. (v) The total heat transfer rate can be determined by integration along over the area of the semi-circle. (vi) Pohlhausen's solution gives the heat transfer coefficient.

(2) Problem Definition. Determine the local heat transfer coefficient along the semi-circle.


(4) Plan Execution.




radiation.

(ii) Analysis. Taking advantage of symmetry, we consider the upper half the semi-circle. Newton's law of cooling applied to an element ydx at a distance x from the leading edge gives

dq = 2h(x) (Ts - T )ydx (a) where

h = local heat transfer coefficient, W/m2-oC dq = rate of heat transfer from upper and lower elements, W Ts = surface temperature, oC



Note that the factor 2 is introduced to account for heat transfer from the two halves of the semi-circle. The local heat transfer coefficient h(x) is obtained from Pohlhausen's solution

d

d

x

Vkh

)0( (4.66)

where






= x


The variable y(x) is given by 22 )()( oo rxrxy

This can be simplified to 22)( xxrxy o (b)

where

or radius of semi-circle, m

Substituting (4.66) and (b into (a) and integrating from orx to orx 2 , gives

o

o

r

rdxxr

d

dVkTTq os

2

)2()0(

)(2


d

dVrTTkq os

)0()(

3

4 2/3 (c)


d

dRerTTkq

oros

)0()(

3

4 (d)

where

orVRe

or (e)


q = k(W/m-oC) )C)(( oTTs or (m)d

d )0((1/1) = W


.0q

(5) Comments. Examination of (4.66) shows that the heat transfer coefficient decreases as x is

increased. Thus, the same semi-circle rotated 180 degrees and remaining at distance or from the

leading edge of the plate, will have a lower heat transfer rate.

PROBLEM 4.33

Consider steady, two-dimensional, laminar boundary

layer flow over a semi-infinite plate. The surface is

maintained at uniform temperature .sT Determine the

total heat transfer rate from the surface area described

by LxHxy /)( as shown.

(1) Observations. (i) This is an external forced convection problem of flow over a flat plate. (ii) Heat transfer rate can be determined using Newton’s law of cooling. (iii) The local heat transfer coefficient changes along the plate. (iv) The area changes with distance along the plate. (v) The total heat transfer rate can be determined by integration along the length of the triangle. (vi) Pohlhausen's solution may be applicable to this problem.



(4) Plan Execution.




radiation.

(ii) Analysis. Newton's law of cooling applied to an element ydx at a distance x from the leading edge gives

dq = h(x) (Ts - T )ydx (a) where h = local heat transfer coefficient, W/m2-oC dq = rate of heat transfer from 2 elements, W Ts = surface temperature, oC



The local heat transfer coefficient h(x) is obtained from Pohlhausen's solution

d

d

x

Vkh

)0( (4.66)

where


T

V

L

xHy

y

x L

H

sT viewtop





= x


The variable y(x) is given by

L

xHxy )( (b)

where

H = height at x = L, m L = length, m

Substituting (4.66) and (b) into (a) and integrating from x = 0 to x = L, gives

L

s dxd

dVk

L

HTTq

0

)0()(


d

dLVHTTkq s

)0()( (c)


d

dReHTTkq Ls

)0()( (d)

where

LVReL (e)


q = k(W/m-oC) )C)(( oTTs H (m)d

d )0((1/1) = W


.0q

(5) Comments. Examination of (4.66) shows that the heat transfer coefficient decreases as x is increased. Thus, the same shape rotated 180 degrees, with its base H at x = 0, will have a higher heat transfer rate.

PROBLEM 4.34

Fluid flows over a semi-infinite flat plat which is maintained

at uniform surface temperature. It is desired to double the

rate of heat transfer from a circular area of radius 1R by

increasing its radius to 2R . Determine the percent increase

in radius need to accomplish this change. In both cases the

circle is tangent to the leading edge. Assume laminar

boundary layer flow with constant properties.

(1) Observations. (i) This is an external forced convection problem of flow over a flat plate. (ii) this problem involves determining the heat transfer rate from a circle tangent to the leading edge of a plate. (iii) Heat transfer rate can be determined using Newton’s law of cooling. (iv) The local heat transfer coefficient changes along the plate. (v) The area changes with distance along the plate. (vi) The total heat transfer rate can be determined by integration along the length of the triangle. (vii) Pohlhausen's so lution may be applicable to this problem.


(3) Solution Plan. Determine the heat transfer rate from a circle which is tangent to the leading edge of a plate. Apply Newton's law of cooling to an element ydx, determine the local heat transfer coefficient along the plate and integrate over the area.

(4) Plan Execution.




radiation.

(ii) Analysis. Consider a circle of radius R

which is tangent to the leading edge of the plate. Taking advantage of symmetry, we consider the upper half of the circle. Newton's la w of cooling applied to an element ydx at a distance x from the leading edge gives

dq = 2h(x) (Ts - T )ydx (a) where

h = local heat transfer coefficient, W/m2-oC dq = rate of heat transfer from 2 elements, W Ts = surface temperature, oC



T sT

V

viewtop

1R 2R

T

Vx0

R

xR

y


Note that the factor 2 is introduced to account for the two haves of the circle. The local heat transfer coefficient h(x) is obtained from Pohlhausen's solution

d

d

x

Vkh

)0( (4.66)

where





= x


The variable y(x) is given by 222 2)()( xRxxRRxy (b)

Substituting (4.66) and (b) into (a) and integrating from x = 0 to x = 2R, gives

R

s dxxRd

dVkTTq

2

0

)2()0(

)(2


2/32/3 )0(

)(3

)2(2R

d

dVTTkq s

(c)

Applying (c) to two circles of radii 1R and 2R and taking the ratio of the two results, gives

2/31

2/32

1

2

R

R

q

q (d)

Doubling the heat transfer rate gives

21

2

q

q (e)

Thus the radius of the circle needed to double the heat transfer rate is obtained by substituting (e)

into (d) and solving for 2R

113/2

2 587.1)2( RRR (f)

Thus percent increase in R is

% increase = 7.58587.1

1001

11

1

12

R

RR

R

RR

(iii) Checking. Dimensional check: Units of q in equation (c) should be W


q = k(W/m-oC) )m()/sm(

)m/s()C)(( 3/22/3

2

o RV

TTs (m)d

d )0((1/1) = W

Limiting check: (1) The heat transfer rate should vanish for TTs . Setting TTs in (c) gives

.0q

(2) The heat transfer rate must vanish for a circle of radius R = 0. Setting R = 0 in (c) gives .0q

(5) Comments. Examination of (4.66) shows that the heat transfer coefficient decreases as x is increased. Thus, moving a circle in the x-direction decreases the rate of heat transfer.

PROBLEM 4.35

Liquid potassium (Pr << 1) flows over a semi-

infinite plate. Assume laminar boundary layer

flow, suggest a simplified velocity profile for

solving the energy equation

Solution

For Pr << 1 the viscous boundary layer

thickness is much thinner than the thermal boundary layer thickness t . Thus, throughout

much of the thermal boundary layer the axial velocity is .Vu Therefore, the term convective

term is approximate as

x

TV

x

Tu .

PROBLEM 4.36

For very low Prandtl numbers the thermal boundary layer is much thinner than the viscous

boundary layer. Thus little error is introduced if the velocity everywhere in the thermal boundary

layer is assumed to be the free stream velocity .V Show that for laminar boundary layer flow

over a flat plate at low Prandtl numbers the local Nusselt number is given by

2/12/1564.0 RePrNux

How does this result compare with scaling prediction?

(1) Observations. (i) The flow field for this boundary layer problem is simplified by assuming that the axial velocity is uniform throughout the thermal boundary layer. (ii) Since velocity distribution affects temperature distribution, the solution for the local Nusselt number can be expected to differ from Pohlhausen’s solution. (iii) The Nusselt number depends on the temperature gradient at the surface.

(2) Problem Definition. Determine the temperature gradient at the surface for laminar flow over a flat plate using a uniform velocity throughout the thermal boundary layer.

(3) Solution Plan. Follow Pohlhausen’s solution replacing Bl asius solution to the flow field with uniform axial velocity equal to the free stream velocity.

(4) Plan Execution.




(15) assume uniform axial velocity equal to the free stream velocity.

(ii) Analysis. The local Nusselt number for laminar flow over a flat plate is given by

xx Red

dNu

)0( (4.68)

where

xVRex


x

Vyyx ),(

s

s

TT

TT




The temperature gradient at the surface d

d )0( is given in equation (b) of Appendix B

1

0 0

)(2

exp)0(

ddfPr

d

d (b)

where )(f is the solution to the velocity field. In Pohlhausen’s solution )(f is obtained from

Blasius solution to the flow fiel d for laminar boundary layer flow over a semi-infinite flat plate. Rather than use Blasius solution we now use a simp lified flow field of uniform velocity given by

1d

df

V

u (c)

Integrating (c) f (d)

Substituting (d) into (b) 1

0 02exp

)0(dd

Pr

d

d

Evaluating the integral 1

0

2

4exp

)0(d

Pr

d

d (e)

The definite integral in (e) is recognized as the error function. Thus

PrPr

dPr

d

d564.0

4exp

)0(1

0

2 (f)

Introducing (f) into (4.68)

xx RePrNu 564.0 (g)

Scaling results for the case of 1Pr is given by

xx eRPrNu , for Pr <<1 (4.55)

This compares favorably with the exact solution (g).

(iii) Checking. Dimensional check; Each term in (g) is dimensionless.

(5) Comments. Scaling prediction of the local Nusselt number gives the same dependency on the Prandtl and Reynolds numbers as the exact solution. The constant 0.564 in the exact solution is replaced by unity in the scaling prediction.

PROBLEM 4.37

Consider laminar boundary layer flow over a flat plate at a uniform temperature .sT When the

Prandtl number is very high the viscous boundary layer is much thicker than the thermal

boundary layer. Assume that the thermal boundary layer is entirely within the part of the

velocity boundary layer in which the velocity profile is approximately linear. Show that for such

approximation the Nusselt number is given by

2/13/1339.0 RePrNux

Note:0

3/13 )3/1(

3)exp(

cdxcx , where is the Gamma function.

(1) Observations. (i) The flow field for this boundary layer problem is simplified by assuming that the axial velocity varies linearly in the y-direction. (ii) Since velocity distribution affects temperature distribution, the solution for the local Nusselt number can be expected to differ from Pohlhausen’s solution. (iii) The Nusselt number depends on the temperature gradient at the surface.

(2) Problem Definition. Determine the temperature gradient at the surface for laminar flow over a flat plate using a uniform velocity throughout the thermal boundary layer.

(3) Solution Plan. Follow Pohlhausen’s solution replacing Bl asius solution to the flow field with linear axial velocity distribution.

(4) Plan Execution.




(15) assume uniform axial velocity equal to the free stream velocity.

(ii) Analysis. The local Nusselt number for laminar flow over a flat plate is given by

xx Red

dNu

)0( (4.68)

where

xVRex


x

Vyyx ),(


s

s

TT

TT



The temperature gradient at the surface d

d )0( is given in equation (b) of Appendix B

1

0 0

)(2

exp)0(

ddfPr

d

d (b)

where )(f is the solution to the velocity field. In Pohlhausen’s solution )(f is obtained from

Blasius solution to the flow fiel d for laminar boundary layer flow over a semi-infinite flat plate. Rather than use Blasius solution we now use a si mplified flow field of linear velocity given by

Ad

df

V

u (c)

where A is constant. Integrating (c)

BAf2

)(2

(d)

The constants A and B are determined from Blasius bound ary condition (4.45b) and solution. Boundary condition (4.45b) gives

0)0(f (4.45b)

Table 7.2 of Blasius solution gives

33206.0)0(

2

2

d

fd (e)

These two conditions give A and B

16603.0A , 0B (f)

Equation (d) becomes

216603.0)(f (g)

Substituting (g) into (b) 1

0 0

2

216603.0exp

)0(dd

Pr

d

d

Evaluating the indefinite integral1

0

3

616603.0exp

)0(d

Pr

d

d (h)

The definite integral in (h) is evaluated next. Let


3

6

16603.0Prz

It follows that

3/13/1

16603.0

6z

Pr

Differentiating

dzzPr

d 3/23/1

16603.0

6

3

1

Substituting into (h)

1

3/2

0

3/1

16603.0

6

3

1)0(dzze

Prd

d z (i)

The definite integral in (i) is recognized as the Gamma function given by

)(0

1 ndzze nz 1n (j)

Comparing the integral in (i) with (j) gives

3

1n (k)

Using (j) and (k), equation (i) becomes

13/1

)3/1(16603.0

6

3

1)0(

Prd

d (l)

The value of )3/1( is obtained from tables of Gamma function

679.2)3/1( (m)

1/3339.0)0(

Prd

d (n)

Substituting (n) into (4. 2/13/1339.0 xx RePrNu (o)

(iii) Checking. Dimensional check: (1) The exponent of the exponential in(h) is dimensionless. (2) Each term in (o) is dimensionless.

(5) Comments. Scaling prediction of the local Nusselt number gives the same dependency on the Prandtl and Reynolds numbers as the exact solution (o). The constant 0.339 in the exact solution is replaced by unity in the scaling prediction. Scaling results for the case of 1Pr isgiven by

2/11/3xx eRPrNu , for Pr >>1 (4.57)

PROBLEM 4.38

Consider steady, two-dimensional, laminar boundary

layer flow over a porous flat plate at uniform surface

temperature. The plate is subject to a uniform suction

ovv )0,(x . Far away downstream both the axial

velocity and the temperature may be assumed to be

functions of y only. Free stream velocity is V and

free stream temperature is .T Determine the heat transfer coefficient and Nusselt number in

this region.

(1) Observations. (i) The flow and temperature fields for this boundary layer problem are simplified by assuming that the axial velocity and temperature do not vary in the x-direction. (ii) The heat transfer coefficient depends on the temperature gradient at the surface. (iii) Temperature distribution depends on the flow field. (iv) The effect of wall suction must be taken into consideration.

(2) Problem Definition. Determine the temperature gradient at the surface for laminar flow over a flat plate.

(3) Solution Plan. Introduce simplifying assumptions in the energy equation for boundary layer flow and solve for the temperature distribution for laminar flow over a plate with suction.

(4) Plan Execution.

(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3) constant properties, (4) two-

dimensional, (5) laminar flow (Rex < 5 105), (6) viscous and thermal boundary layer flow (Rex > 100 and Pex > 100), (7) uniform upstr eam velocity and temperature, (8) uniform surface temperature, (9) flat plate, (10) negligible changes in kinetic and potential energy, (11) negligible

dissipation, (12) no buoyancy ( = 0 or g = 0) and (13) no energy generation ( 0q ). In

addition, (14) neglect radiation, (15) uniform suction velocity and (16) negligible axial variation of velocity and temperature.

(ii) Analysis. The local heat transfer coefficient h is given by

TT

y

xT

khs

)0,(

(1.10)

where

k thermal conductivity, CW/m o

T fluid temperature, Co



y normal coordinate, m

The local Nusselt number is defined as

ov

T

Vx

y


k

hxNux (a)

Temperature distribution in (1.10) is obtained from the solution to boundary layer energy equation

2

2

y

T

y

T

x

Tu v (4.18)

where


v normal velocity, m/s

thermal diffusivity, /sm2

Since temperature variation in the axial direction is neglected, it follows that

0x

T (b)

(b) into (a)

2

2

y

T

y

Tv (c)

To solve (c) for the temperature distribution, velocity component v must be determined. The continuity equation for two-dimensional incompressible flow is given by

0yx

u v (2.2)

Since velocity variation in the axial direction is neglected, it follows that

0x

u (d)

(d) into (2.2)

0y

v (e)

Integration of (e) gives cv (f)

where c is constant of integration. Surface boundary condition on v is

ovv (g)

where

ov surface suction velocity, m/s

Substituting (g) into (c)

2

2

y

T

y

Tov-

Since T is independent of x, it follows that the above can be written as


2

2

dy

Td

dy

dTov- (h)

The solution to (h) requires two boundary conditions. They are:

sTxT )0,( (i)

TxT ),( (j)

To solve (h), it is rewritten as

dy

dy

dT

y

Td

ov (k)

Integrate (j)

1lnln Cydy

dT ov (l)

where 1C is constant of integration. To integrate (l) again, it is first rewritten as

ydy

dT

C

ov

1

1ln

or

)exp(1 yCdy

dT ov (m)

Integrate (m)

21 )exp( CyCT o

o

v

v (n)

where 2C is constant of integration. Application of boundary conditions (i) and (j) gives 1C and

2C

)(1 TTC sov

(o)

TC2 (p)

(o) and (p) into (n)

)exp()( yTTTT sov

(q)

Substituting (q) into (1.10) gives the heat transfer coefficient h

ovkh (r)

However, is defined as

pc

k (s)


where

pc specific heat, CJ/kg o

density, 3kg/m

(s) into (r)

ovpch (t)

Substituting (t) into (a) gives the local Nusselt number

k

xcNu

px

ov

The above can be written in a more revealing way as

x

k

cNu

px

ov

This is recognized as

xx RePrNu (u)

where the local Reynolds number is defined as

xRex

ov (v)

(iii) Checking. Dimensional check: (1) The exponent of the exponential in(q) is dimensionless. (2) Each term in (q) has units of temperature. (3) Equation (t) give the correct units for h.

Boundary conditions check: Solution (q) satisfies conditions (i) and (j).

Limiting check: For the special case of TTs , fluid temperature should be uniform equal to

.T . Setting TTs in (q) gives .TT

(5) Comments. (i) Neglecting axial variation of velocity and temperature are the key

simplifying assumptions in this problem. (ii) Free stream velocity V does not enter into the

solution for the temperature distribution and heat transfer coefficient. (iii) The Reynolds number in solution (u) depends on the suction velocity and not free stream velocity.

PROBLEM 4.39

A semi infinite plate is heated with uniform flux q along its length. The free stream temperature

is T and free stream velocity is .V Since the heat transfer coefficient varies with distance

along the plate, Newton’s law of cooling requires that surface temperature must also vary to

maintain uniform heat flux. Consider the case of laminar boundary layer flow over a plate whose

surface temperature varies according to

nCxTxTs )(

Working with the solution to this case, show that 2/1n corresponds to a plate with uniform

surface flux.

(1) Observations. (i) This is a forced convection flow over a plate with variable surface temperature. (ii) The local heat flux is determined by Newton’s law of cooling. (iii) The local heat transfer coefficient and surface temperature vary with distance along the plate. The variation of surface temperature and heat transfer coefficient must be such that Newton’s law gives uniform heat flux. (iv) The local heat transfer coefficient is obtained from the local Nusselt number.

(2) Problem Definition. Determine the value of the exponent n that results in uniform heat flux.

(3) Solution Plan. Apply Newton’s law of cooling to determine the local surface heat flux. Compute the Reynolds number to establish if the flow is laminar or turbulent. If the flow is laminar use the result of numerical solution to forced convection over a plate with variable surface temperature to determine the local Nusselt number and heat transfer coefficient.

(4) Plan Execution.


dimensional, (5) laminar flow (Rex < 5 105, to be verified), (6) viscous and thermal boundary layer flow (Rex > 100 and Pex > 100, to be verified), (7) uniform upstream velocity and temperature, (8) flat plate, (9) negligible changes in kinetic and potential energy, (10) negligible

dissipation, (11) no buoyancy ( = 0 or g = 0), (12) no energy generation ( 0q ) and (14)

negligible radiation

(ii) Analysis. Newton’s law of cooling gives

))(( TTxhq s (a)

where

)(xh local heat transfer coefficient, CW/.m o2

q surface heat transfer rate, 2W/m

)(xTs surface temperature, Co


Surface temperature variation is given by

nCxTxTs )( (b)

Substituting (b) into (a)


)(xhCxq n (c)

According to (c), surface flux is constant if

nxxh /1)( (d)

The local heat transfer coefficient is obtained from the local Nusselt number defined as

k

hxNux (e)

Solving for h(x)

x

kNuxh x)( (f)

The solution to the Nusselt number for a plate with surface temperature variation described in equation (b) is

xx Red

dNu

)0( (4.80)

where dd /)0( is the dimensionless surface temperature gradient. It is a constant which

depends on the Prandtl number and the exponent n. Substituting (4.80) into (f)

x

kRe

d

dxh x

)0()(

Using the definition of Reynolds number into the above

2/1)0()0()( x

V

d

dk

x

kxV

d

dxh (g)

Using (g) into (c)

2/1)0(x

V

d

dkCxq n (h)

Examination of (h) shows that for q to be constant independent of x, the exponent n must be

2

1n (i)

(iii) Checking. Dimensional check: (1) Each term in (4.80) is dimensionless. (2) Equations(g) and (h) are dimensionally correct.

Limiting check: (1) For the special case of C = 0, surface temperature according to (b) will be the

same as ambient temperature. That is .TTs The corresponding heat flux for this case should

vanish. Setting C = 0 in (h) gives .0q

(5) Comments. (i) The key to the solution to this problem is the determination of the variation of h with distance x.

PROBLEM 4.40

Water flows over a semi-infinite flat plate which is maintained at a variable surface

temperature sT given by

75.0)( CxTxTs

where

C = 54.27 oC / (m)0.75

T = free stream temperature = C.3o

x = distance from the leading edge, m

Determine the average heat transfer coefficient for a plate if length L = 0.3 m. Free stream

velocity is 1.2 m/s.

(1) Observations. (i) This is a forced convection flow over a plate with variable surface temperature. (ii) The Reynolds number should be computed to determine if the flow is laminar or turbulent. (iii) The local heat transfer coefficient and surface temperature vary with distance along the plate. (iv) The local heat transfer coefficient is obtained from the solution to the local Nusselt number. (v) The determination of the Nusselt number requires determining the temperature gradient at the surface.

(2) Problem Definition. Determine surface temperature gradient for a plate at variable surface temperature.

(3) Solution Plan. Compute the Reynolds number to establish if the flow is laminar or turbulent. If the flow is laminar use the result of numerical solution to forced convection over a plate with variable surface temperature to determine surface temperature gradient.

(4) Plan Execution.



dissipation, (11) no buoyancy ( = 0 or g = 0), (12) no energy generation( 0q ) and (13)


(ii) Analysis.

The maximum Reynolds number is given by

LVReL (a)

where

L = plate length = 0.3 m

V free stream velocity = 1.2 m/s

kinematic

Properties are evaluated at the film temperature defined as

T

V

L

x

75.0xCTTs

W


2

sf

TTT (b)

where


sT average surface temperature, Co

The average surface temperature is defined as

2

)()0( LTTT ss

s (c)

Surface temperature is given by 75.0)( CxTxTs (d)

where

0.75

o

m

C27.54C

Evaluating sT at x = 0 and x = 0.3 m and substituting into (d) gives

C142

(m))(0.3)C/m(27.45C3C3

2

o0.750.750.75ooo75.0CLTT

Ts

Substituting into (b)

C5.82

C))(143( oo

fT


Cm

W5791.0

ok

942.9Pr

s

m103716.1

26


)/sm(103716.1

)m(3.0)m/s(2.126LRe = 262,467

Thus this is laminar boundary layer flow. The local Nusselt number for laminar flow over a flat plate with variable surface temperature is given by

xx Red

dNu

)0( (4.80)

where


k

hxNux (e)

)(xh local heat transfer coefficient, CW.m o2


Substitute (e) into (4.80) and solve for h

xRex

k

d

dh

)0( (f)

The dimensionless surface temperature gradient, ,/)0( dd is obtained from the

numerical solution for laminar flow over a flat plate at a surface temperature of the form

ns CxTxT )( (g)

The solution is presented in Fig. 4.8. The exponent n in (g) characterizes surface temperature variation.

(iii) Computations. For the problem under consideration 75.0n and 942.9Pr , Fig. 4.8

gives

1.1.

)0(

d

d (h)

Substituting into (f)

xRex

kh 1.1 (i)

The corresponding Nusselt number is

xx Rek

hxNu 1.1 (j)

(iv) Checking. Dimensional check: (1) Equation (f) has the correct units for heat transfer coefficient. (2) The Reynolds number in (a) and Nusselt number in (j) are dimensionless.

(5) Comments. (i) The key to the solution to this problem is the determination of the constant ./)0( dd This constant is determined through the use of Fig. 4.8. (ii) Fig. 4.8 is applicable to a

class of variable surface temperature described by equation (g) only

ns CxTxT )( (g)

(iii) According to (d) surface temperature varies from C3o at the leading edge to C25o . Clearly,

surface temperature is not uniform. For uniform surface temperature the Nusselt number is given by (4.72c)

10for,339.0 3/1 PrRePrNu xx (4.72c)

0 1.00.5 1.5

1.0

d

d )0(

0.7Pr

n

10

30

2.0

re, temperatusurfaceing with varrplatefor(0)

d

d4.8Fig.

ns CxT-T [4]


In this problem 942.9Pr . Using this value in (4.72c) gives

xx ReNu 729.0 (k)

This is 33% smaller than the variable surface temperature solution given in (j).

PROBLEM 4.41

Air flows over a plate which is heated non-uniformly such that its surface temperature increases linearly as the distance from the leading edge is increased according to

CxTxTs )(

where

C = 24 oC /cm

T = free stream temperature = C20o


Determine the total heat transfer rate from a square

plate cm.10cm10 Free stream velocity is 3.2 m/s.

(1) Observations. (i) This is a forced convection flow over a plate with variable surface temperature. (ii) The Reynolds number should be computed to determine if the flow is laminar or turbulent. (iii) Newton’s law of cooling gives the heat transfer rate from the plate. (iv) The local heat transfer coefficient and surface temperature vary with distance along the plate. Thus determining the total heat transfer rate requires integration of Newton’s law along the plate. (v) The local heat transfer coefficient is obtained from the local Nusselt number.

(2) Problem Definition. Determine the average Nusselt number for forced convection over a flat plate with variable surface temperature.

(3) Solution Plan. Apply Newton’s law of cooling to determine the total heat transfer rate form the plate. Compute the Reynolds number to establish if the flow is laminar or turbulent. If the flow is laminar use the result of numerical solution to forced convection over a plate with variable surface temperature to determine the average Nusselt number.

(4) Plan Execution.



dissipation, (11) no buoyancy ( = 0 or g = 0), (12) no energy generation ( q ) and (13)


(ii) Analysis. The total heat transfer from the plate is given by Newton’s law of cooling

L

s WdxTTxhq0

))(( (a)

where

)(xh local heat transfer coefficient, CW.m o2

L plate length = 0.1 m q heat transfer rate, W

)(xTs surface temperature, Co

CxTTs

T

V

L

x



W plate width = 0.1 m

x distance along plate, m

The local heat transfer coefficient is obtained from the local Nusselt number defined as

k

hxNux (b)

The maximum Reynolds number is given by

LVReL (c)

where

V free stream velocity = 3.2 m/s

kinematic

Properties are evaluated at the film temperature defined as

2

sf

TTT (d)

where

sT average surface temperature, Co

The average surface temperature is defined as

2

)()0( LTTT ss

s (e)

Surface temperature is given by

CxTxTs )( (f)

where C = 24 oC /cm. Evaluating sT at x = 0 and x = 0,1 m and substituting into (e) gives

C1402

)C/cm)10(cm24(C20C20

2

ooooCLTT

Ts


C802

C))(14020( oo

fT

Properties of air at this temperature are

Cm

W02991.0

ok

706.0Pr

s

m1092.20

26 0 1.00.5 1.5

1.0

d

d )0(

0.7Pr

n

10

30

2.0

re, temperatusurfaceing with varrplatefor(0)

d

d4.8Fig.

ns CxT-T [4]

PROBLEM 4.41 (continued)Substituting into (c)

)/sm(1092.20

)m(1.0)m/s(2.326LRe =15,296

Thus this is laminar boundary layer flow. The local Nusselt number for laminar flow over a flat plate is given by

xx Red

dNu

)0( (4.80)

where ./)0( dd is the dimensionless surface temperature gradient. Numerical solution to

dd /)0( for laminar flow over a flat plate at a surface temperature of the form

ns CxTxT )( (g)

is given in Fig. 4.8. The exponent n in (g) characterizes surface te mperature variation.. Using the definition of local Nusselt number and Reynolds number, (4.80) is solved for the local heat transfer coefficient

x

V

d

dk

xV

x

k

d

dh

1)0()0( (h)

Substituting (f) and (h) into (a)

L

dxxV

d

dkWCq

0

)0(


2/3)0(

3

2L

V

d

dkWCq

Expressing this result in terms of the Reynolds number

LV

d

dkWLCq

)0(

3

2

LRed

dkWLCq

)0(

3

2 (i)

Equation (i) is rewritten in dimensionless form as

LRed

d

kWLC

q )0(

3

2 (j)

(iii) Computations. For the problem under consideration 1n and 706.0Pr , Fig. 4.8 gives

48.0.

)0(

d

d (h)

Substituting into (i)


W4.28296,15.480m/m)C/cm)100(c4(2m)0.1(m))(1.0)(CW/m)(02991.0(3

2 ooq

(iv) Checking. Dimensional check: (1) Computations showed that (i) has the correct units for heat transfer rate. (2) Each term in (j) is dimensionless.

Limiting check: (1) If the width of plant is zero the heat transfer rate will vanish. Setting W = 0 in (i) gives q = 0. (2) For the special case of C = 0, surface temperature according to (f) will be the

same as ambient temperature. That is .TTs The corresponding heat transfer rate for this case

is q = 0. Setting C = 0 in (i) gives q = 0.

(5) Comments. (i) The key to the solution to this problem is the determination of the constant ./)0( dd This constant is determined through the use of Fig. 4.8. (ii) Fig. 4.8 is applicable to a

class of variable surface temperature described by equation (g) only

ns CxTxT )( (g)

PROBLEM 4.42

The surface temperature of a plate varies with distance

from the leading edge according to

8.0)( CxTxTs

Two identical triangles are drawn on the surface as

shown. Fluid at uniform upstream temperature T and

uniform upstream velocity V flows over the plate. Assume laminar boundary layer flow.

Determine the ratio of the heat transfer rate from the two triangles, q1/q2.

(1) Observations. (i) This is an external forced convection problem of flow over a flat plate. (ii) Convection heat transfer from a surface can be determined using Newton’s law of cooling. (iii) The local heat transfer coefficient and surface temperature vary along the plate. (iv) For each triangle the area varies with distance along the plate. (v) The total heat transfer rate can be determined by integration along the length of each triangle.

(2) Problem Definition. Determine the local heat transfer coefficient along each for laminar boundary layer flow over a plate with variable surface temperature.

(3) Solution Plan. Apply Newton's law of cooling to an element of each triangle, ydx, determine the local heat transfer coefficient along the plate and integrate over the area.

(4) Plan Execution.


dimensional, (5) laminar flow (Rex < 5 105), (6) viscous and thermal boundary layer flow (Rex > 100 and Pex > 100), (7) uniform upstr eam velocity and temperature, (8) uniform surface temperature, (9) flat plate, (10) negligible changes in kinetic and potential energy, (11) negligible

dissipation, (12) no buoyancy ( = 0 or g = 0) and (13) no energy generation ( 0q ). In

addition, (14) neglect radiation.

(ii) Analysis. Of interest is the ratio of the total heat transfer rate from triangle 1 to triangle 2. Since both the heat transfer coefficient and area vary along each triangle, it follows that Newton's law of cooling s hould be applied to an element dA = ydx at a distance x from the leading edge:

dq = h(x) (Ts - T ) y(x)dx (a) where

dq = rate of heat transfer from element, W h = local heat transfer coefficient, W/m2-oC



y = y(x) = width of element

T

V

viewtop

1 2 H

L

)(xTs

TsT

V

viewtop

1 2H

Lx

1y

2y


Integration of (a) gives the total heat transfer rate

dxxyTTxhq

L

s )())((0

(b)

To evaluate the integral we must specify ( TTs ), y(x) and h(x). Surface temperature variation

is given as8.0)( CxTxTs (c)

Similarity of triangles gives

)()(1 xLL

Hxy (d)

xL

Hxy )(2 (e)

The local heat transfer coefficient h(x) for a plate with variable surface temperature described by (c) is given by (4.78)

d

d

x

Vkh

)0( (4.78)

where





= x


y = coordinate normal to plate, m

The dimensionless surface temperature gradient, ,/)0( dd is obtained from the numerical

solution for laminar flow over a flat plate at a surface temperature of the form

ns CxTxT )( (f)

Fig. 4.8 gives dd /)0( . As shown in Fig. 4.8, dd /)0( depends on the Prandtl number and

the exponent n in (f). Since both triangles have the same surface temperature distribution and the same fluid, it follows that dd /)0( is the same for both triangles.

To determine the heat transfer rate, ,1q from triangle 1, equations (c), (d) and (4.78) are

substituted into (b)

dxxx

xL

d

dV

L

HkCq

L

0

8.01

)()0(



99.2

)0( 3.1

1

L

d

dVkCHq (g)

Similarly, for triangle 2, equations (c), (e) and (4.78) are substituted into (b)

dxxx

x

d

dV

L

HkCq

L

0

8.01

)0(


3.2

)0( 3.1

2

L

d

dVkCHq (h)

Taking the ratio of (g) and (h)

769.099.2

3.2

2

1

q

q (i)

(iii) Checking. Dimensional check: (1) Each term in (d) and (e) has units of length. (2)

Noting that units of C are 8.0o C/m , equations (g) and (h) have the correct units for heat.

Limiting check: For the limiting case of a plate which is maintained at the free stream

temperature, that is TTs , the corresponding heat transfer rate should vanish for both

triangles. According to (f), TTs when C = 0. Setting C = 0 in (g) and (h) gives .021 qq

(5) Comments. (i) According to (i) the heat transfer ratio .1/ 21 qq By contrast, in Problem

4.30 in which surface temperature is uniform, 1/ 21 qq . The reason for this reversal is the

increase in surface temperature with x favors triangle 2 where the area also increases with x.

Although the two triangles have the same area, the rate of heat transfer from triangle 1 is double that from triangle 2. Thus, orientation and proximity to the leading edge of a flat plate play an important role in determining the rate of heat transfer. (ii) The same approach can be used to determine heat transfer for configurations other than rectangles, such as circles and ellipses.

PROBLEM 4.43

Construct a plot showing the variation of xx ReNu / with wedge angle. Where xNu is the local

Nusselt number and xRe is the local Reynolds number. Assume laminar boundary layer flow of

air.

(1) Observations. (i) This is a forced convection boundary layer flow over a wedge. (ii) Wedge surface is maintained at uniform temperature. (iii) The flow is laminar. (iv) The fluid is air. (v) Similarity solution for the local Nusselt number is presented in Section 4.4.3. (vi) The Nusselt number depends on the Reynolds number and the dimensionless temperature gradient at the surface ./)0( dd (vii) Surface temperature

gradient depends on wedge angle.

(2) Problem Definition. Determine the variation of dd /)0( with wedge angle.

(3) Solution Plan. Use the wedge solution of Section 4.4.3 to determine the variation of local Nusselt number with wedge angle.

(4) Plan Execution.


dimensional, (5) laminar flow (Rex < 5 105), (6) viscous and thermal boundary layer flow (Rex > 100 and Pex > 100), (7) uniform upstream velocity and temperature, (8) uniform surface temperature, (9) symmetrical flow, (10) negligible changes in kinetic and potential energy, (11)

negligible dissipation, (12) no buoyancy ( = 0 or g = 0) and (13) no energy generation ( 0q ).

(ii) Analysis. The local Nusselt number for laminar boundary layer flow over a wedge is given by

xx Red

dNu

)0( (4.96)

where

xVRex , local Reynolds number

)(xV external flow velocity over the wedge, m/s

x

xVyyx

)(),( , similarity variable

s

s

TT

TT, dimensionless temperature


Rewrite (496)

d

d

Re

Nu

x

x )0( (a)

Surface temperature gradient depends on wedge angle and Prandtl number. Table 4.3 lists dd /)0( corresponding to four wedge angles and five Prandtl numbers. For air with Prandtl

number 7.0Pr , Table 4.3 gives the values of dd /)0( used to construct a pot of

xx ReNu / vs. wedge angle .

(4) Comments. (i) The local heat transfer coefficient decreases with distance along the surface. (ii) Local Nusselt number and heat transfer coeffi cient increase with wedge angle. The increase is approximately linear.

Pr = 0.7

Wedgeangle

d

d )0(

0 0.292

5/ 0.331

2/ 0.384

0.496

x

x

Re

Nu

wedge angle

0.5

0.4

0.3

0 60 120 180

deg,

PROBLEM 4.44

Consider laminar boundary layer flow over a wedge. Show that the average Nusselt number Nu

for a wedge of length L is given by

LRed

d

mNu

)0(

1

2

where the Reynolds number is defined as .)(LLV

ReL

(1) Observations. (i) This is a forced convection boundary layer flow over a wedge. (ii) Wedge surface is maintained at uniform temperature. (iii) The flow is laminar. (iv) The average Nusselt number depends on the average heat transfer coefficient.. (v) Similarity solution for the local heat transfer coefficient is presented in Section 4.4.3.

(2) Problem Definition. Determine the average heat transfer coefficient for laminar flow over a wedge at uniform surface temperature.

(3) Solution Plan. Start with the definitions of the average Nusselt number. Use the wedge solution of Section 4.4.3 to determine the local heat transfer coefficient.

(4) Plan Execution.


dimensional, (5) laminar flow (Rex < 5 105), (6) viscous and thermal boundary layer flow (Rex > 100 and Pex > 100), (7) uniform upstr eam velocity and temperature, (8) uniform surface temperature, (9) symmetrical flow, (10) negligible changes in kinetic and potential energy, (11)

negligible dissipation, (12) no buoyancy ( = 0 or g = 0) and (13) no energy generation ( 0q ).

(ii) Analysis. The average Nusselt number for a wedge of length L is defined as

k

LhNu L (a)

where

h = average heat transfer coefficient, CW/m o2

k = thermal conductivity, CW/m o

L = wedge length, m

The average heat transfer coefficient is defined as

L

dxxhL

h0

)(1

(b)

where the local heat transfer coefficient, h(x), is given by (4.95)

d

d

x

xVkxh

)0()()( (4.95)


where


x = distance along wedge surface form the leading end, m

x

xVyyx


s

s

TT


= kinematic viscosity, sm2

(4.95) into (b)

dxx

xV

d

dkxh

L

0

)()0()( (c)

The external flow velocity, ),(xV is given by

mCxxV )( (4.82)

where C is constant and m is a measure of wedge angle , defined as

2m (4.83)

(4.82) into (c) L

m dxxd

dC

L

kh

0

2/)1()0(


2/)1()0(

1

2 mLd

dC

L

k

mh (e)

(e) into (a)

2/)1()0(

1

2 mL L

d

dC

mNu

To express in terms of the Reynolds number, rewrite the above

LCL

d

d

mNu

m

L)()0(

1

2

(4.82) into the above

LLV

d

d

mNu L

)()0(

1

2 (f)

Introduce the definition of Reynolds number

LL Red

d

mNu

)0(

1

2 (g)


(iii) Checking: Dimensional check: (1) Units of the constant C is determined from (4.82) as

mms

mC . Using this shows that (e) has the correct units. (2) Equation (f) is dimensionless,

Limiting check: For a wedge with zero angle ( m = 0), the solutions should reduce to the flat

plate solution of Pohlhausen. Setting m = 0 in (g) gives

LL Red

dNu

)0(2 (h)

This agrees with Pohlhausen’s result of equation (4.69).

(4) Comments. In determining the average heat transfer coefficient for a wedge, the variation of the velocity outside the boundary layer with distance x must be taken into consideration. That is,

unlike Pohlhausen’s solution for the flat plate, for the wedge ).(xVV

PROBLEM 4.45

Compare the total heat transfer rate from a o90 wedge, ,wq with that from a flat plate, ,pq of

the same length. Construct a plot of pw qq / as a function of Prandtl number.

(1) Observations. (i) Newton’s law of cooling gives the heat transfer rate from a surface. (ii)

Total heat transfer from a surface depends on the average heat transfer coefficient h . (iii) Both

flat plate and wedge are maintained at uniform surface temperature. (iv) Pohlhausen’s solution

gives h for a flat plate. (v) Similarity solution for the local heat transfer coefficient for a wedge is presented in Section 4.4.3.

(2) Problem Definition. Determine the average heat transfer coefficient for laminar flow over a flat plate and a wedge.

(3) Solution Plan. Apply Newton’s law of cooling to flat plate and

wedge. Use Pohlhausen’s and wedge solutions to h .

(4) Plan Execution.

(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3)

constant properties, (4) two-dimensional, (5) laminar flow (Rex < 5 105), (6) viscous and thermal boundary layer flow (Rex > 100 and Pex > 100), (7) uniform upstream velocity and temperature, (8) uniform surface temperature, (9) symmetrical flow over wedge, (10) negligible changes in

kinetic and potential energy, (11) negligible dissipation, (12) no buoyancy ( = 0 or g = 0) and (13) no energy generation ( 0q ).


)( TTAhq s (a)

where

A = surface area, 2m

h average heat transfer coefficient, CW/m o2

q heat transfer rate, W



Let the subscript p denote plate and w denote wedge. Apply (a) to the plate and wedge and take their ratio

w

p

w

p

h

h

q

q(b)

The problem reduces to determining ph and wh .

The average heat transfer coefficient for boundary layer laminar flow over a flat plate at uniform surface temperature is given by Pohlhausen’s solution (4.67)


p

pLp

d

dRe

L

kh

)0(2 (4.67)

where


L = wedge length, m

LReLV


x

Vyyxp ),( , dimensionless similarity variable

x = axial coordinates, m y = normal coordinates, m

s

spp

TT


),( yxTT pp , temperature distribution


Using the definition of Reynolds number, (4.67) us rewritten as

p

pp

d

dLV

L

kh

)0(2 (c)

The average heat transfer coefficient foe a wedge is defined as

L

ww dxxhL

h0

)(1

(d)

The local heat transfer coefficient for a wedge, )(xhw , is given by (4.95)

w

ww

d

d

x

xVkxh

)0()()( (4.95)

where


x

xVyyxw


s

sww

TT


),( yxTT ww , temperature distribution



(4.95) into (d)

dxx

xV

d

dkxh

L

ww

0

)()0()( (e)

The external flow velocity, ),(xV is given by

mCxxV )( (4.82)

where C is constant and m is a measure of wedge angle , defined as

2m (4.83)

(4.82) into (e) L

mww dxx

d

dC

L

kh

0

2/)1()0(


2/)1()0(

1

2 mww L

d

dC

L

k

mh (f)

Substitute (c) and (f) into (a)

p

p

wm

p

w

d

d

d

d

V

CL

mq

q

)0(

)0(

1

1 (g)

(iii) Checking: Dimensional check: (1) Equation (4.82) shows that mCL has units of velocity. It

follows that (g) is dimensionless. Similarly, units of (f) are correct.

Limiting check: For a wedge with zero angle ( m = 0), the wedge solution should reduce to

the flat plate solution and the heat ratio should be unity. For this case, according to (4.82),

.VC Setting m = 0 in (g) gives .1/ pw qq

(4) Comments. (i) The local heat transfer coefficient decreases with distance along the surface. (ii) Local Nusselt number and heat transfer coeffi cient increase with wedge angle. The increase is approximately linear.

PROBLEM 4.46

For very low Prandtl numbers the thermal boundary layer is much thinner than the viscous boundary layer. Thus little error is introduced if the velocity everywhere in the thermal boundary

layer is assumed to be the free stream velocity .V Show that for laminar boundary flow layer

flow over a wedge at low Prandtl numbers the local Nusselt number is given by

xx RePrm

Nu)1(

(1) Observations. (i) The flow field for this boundary layer problem is simplified by assuming that axial velocity within the thermal boundary layer is the same as that of the external flow. (ii) Since velocity distribution affects temperature distribution, the solution for the local Nusselt number differs from the exact case of Section 4.4.3. (iii) The local Nusselt number depends the local heat transfer coefficient which depends on the temperature gradient at the surface.

(2) Problem Definition. Determine the temperature gradient at the surface for laminar flow over a wedge for the simplified velocity field described above.

(3) Solution Plan. Follow analysis of section 4.4.3 for determining the local Nusselt number using a simplified flow field solution.

(4) Plan Execution.



negligible dissipation, (12) no buoyancy ( = 0 or g = 0), (13) no energy generation ( 0q ) and

(14) Pr << 1. Thus assume that axial velocity with in the thermal boundary layer is the same as that of the external flow.

(ii) Analysis. The local Nusselt number, xNu , for a wedge is given by equation (4.96) of

Section 4.4.3

xx Red

dNu

)0( (4.68)

where

xxVRex

)(


x

xVyyx



s

s

TT



The problem reduces to determining the dimensionless temperature gradient at the surface dd /)0( given by equation (4.94)

1

0 0

)(2

1)(mexp

)0(ddF

Pr

d

d (4.94)

where m is measure of wedge angle , defined as

2m (4.83)

The function )(F is obtained from the solution to the flow field over the wedge. It is defined in

(4.87) as

)(

),(

xV

yxu

d

dF (4.87)

where u(x,y) is the axial velocity within the thermal boundary layer. However, for Pr << 1 we assume that

)(),( xVyxu (a)

(a) into (4.87)

1d

dF (b)

Integration of (b) gives F (c)

(b) into (4.94) 1

0 02

1)(exp

)0(dd

Prm

d

d (d)

Evaluating the integral in the integrand

1

0

2

2

1)(exp

)0(d

Prm

d

d (e)

The definite integral in (e) is recognize d as the error function. To proceed, let

4

1)( Prmz (f)

Thus

dzPrm

d1)(

4 (g)


Substitute (f) and (g) into (e)

1

0

2 )exp(1)(

4)0(dzz

Prmd

d (h)

Howeverz

z dzz e0

22erf (i)

and

12

erf

0

2

dzze (j)

(i) into (h)

Prm

Prmd

d 1)(

1)(

)0(1

(k)

(k) into (4.68)

xx RePrm

Nu)1(

(l)

(iii) Checking. Dimensional check; All equations are dimensionless.


plate solution with Pr << 1 and .Vu . Setting m = 0 in (l) gives

xx PrReNu 564.0 (m)

This agrees with the solution to Problem 4.36.

(5) Comments. The assumption that axial velocity within the thermal boundary layer is the same as that of the external flow provided a major simplification in the solution. It made it possible to obtain a solution for the Nusselt number without the need for numerical integration.

PROBLEM 4.47

Consider laminar boundary layer flow over a wedge at a uniform temperature .sT When the

Prandtl number is very high the viscous boundary layer is much thicker than the thermal

boundary layer. Assume that the velocity profile within the thermal boundary layer is

approximately linear. Show that for such approximation the local Nusselt number is given by

2/11/3)0()1(489.0 RePrFmNux

Note: 0

3/13 )3/1(

3)exp(

cdxcx , where is the Gamma function.

(1) Observations. (i) The flow field for this boundary layer problem is simplified by assuming that axial velocity within the thermal boundary layer varies linearly with the normal distance. (ii) Since velocity distribution affects temperature distribution, the solution for the local Nusselt number differs from the exact case of Section 4.4.3. (iii) The local Nusselt number depends on the local heat transfer coefficient which depends on the temperature gradient at the surface.

(2) Problem Definition. Determine the temperature gradient at the surface for laminar flow over a wedge for the simplified velocity field described above.

(3) Solution Plan. Follow the analysis of section 4.4.3 for determining the local Nusselt number using a simplified flow field solution.

(4) Plan Execution.



negligible dissipation, (12) no buoyancy ( = 0 or g = 0), (13) no energy generation ( 0q ) and

(14) Pr >> 1. Thus assume that axia l velocity within the thermal boundary layer varies linearly with distance normal to wedge surface.

(ii) Analysis. The local Nusselt number, xNu , for a wedge is given by equation (4.96) of

Section 4.4.3

xx Red

dNu

)0( (4.68)

where

xxVRex

)(



x

xVyyx


s

s

TT



The problem reduces to determining the dimensionless temperature gradient at the surface dd /)0( given by equation (4.94)

1

0 0

)(2

1)(mexp

)0(ddF

Pr

d

d (4.94)

where m is a measure of wedge angle , defined as

2m (4.83)

The function )(F is obtained from the solution to the flow field over the wedge. It is defined in

(4.87) as

)(

),(

xV

yxu

d

dF (4.87)

where u(x,y) is the axial velocity within the thermal boundary layer. However, for Pr >>1 we use a simplified flow field of linear velocity given by

Ad

dF

V

u (c)

where A is constant. Integrating (c)

BAF2

)(2

(d)

The constant B is determined from wedge flow boundary condition. (4.89b) and solution. Boundary condition (4.89b) gives

0)0(F (4.89b)

Apply (4.89b) to (d)B = 0 (e)

Differentiate (d) twice and apply at the surface

)0(2

2

Fd

FdA (f)

The constant )0(F depends on wedge angle. It is obtained from the solution to flow field over

the wedge. Values of )0(F for four angles are listed in Table 4.3. Thus (d) can be written as


2

2

)0()(

FF (g)

Substitute (g) into (4.94) 1

0 0

2

4

(0)1)(exp

)0(dd

PrFm

d

d \ (h)

Evaluating the integral in the integrand

1

0

3

12

(0)1)(exp

)0(d

PrFm

d

d (i)

The definite integral in (i) is evaluated next. Let

3

12

(0)1)( PrFmz

It follows that

3/13/1

(0)1)(

12z

PrFm

Differentiating

dzzPrFm

d 3/23/1

(0)1)(

12

3

1


1

3/2

0

3/1

(0)1)(

12

3

1)0(dzze

PrFmd

d z (j)

The definite integral in (j) is recognized as the Gamma function given by

)(0

1 ndzze nz 1n (k)

Comparing the integral in (j) with (k) gives

3

1n (l)

Using (k) and (l), equation (j) becomes

13/1

)3/1((0)1)(

12

3

1)0(

PrFmd

d (m)


The value of )3/1( is obtained from tables of Gamma function

679.2)3/1( (n)

(n) into (m)

1/33/1(0)1)(489.0)0(

PrFmd

d (o)

Substitute (n) into (o)

2/11/33/1(0)1)(489.0 xx RePrFmNu (o)

(iii) Checking. Dimensional check: (1) The exponent of the exponential in (j) is dimensionless. (2) All equations are dimensionless.


plate solution with Pr >> 1 and linear axial velocity. Setting m = 0 in (o) gives

2/11/33/1(0)489.0 xx RePrFNu (p)

However, for m = 0 the flow field reduces to Blasius solution. Thus, using Table 4.1, gives

33206.0)0()0( fF (q)

(q) into (p) 2/11/3339.0 xx RePrNu

This agrees with the solution to Problem 4.37.

(5) Comments. The assumption that axial velocity within the thermal boundary layer varies linearly in the direction normal to the surface provided a major simplification in the solution. It made it possible to obtain a solution for the Nusselt number without the need for numerical integration.

PROBLEM 5.1

For fluids with 1Pr the thermal boundary

layer thickness is much larger than the viscous

boundary layer. That is .1/t It is

reasonable for such cases to assume that fluid

velocity within the thermal layer is uniform

equal to the free stream velocity. That is

Vu

Consider uniform laminar boundary layer flow over a flat plate. The surface is maintained at

uniform temperature sT and has an insulated leading section of length .ox Assume a third degree

polynomial temperature profile, show that the local Nusselt number is given by

2/12/1

2/1

153.0 xo

x eRPrx

xNu

where the local Reynolds number is ./Re xVx

(1) Observations. (i) In general, to determine the Nusselt number it is necessary to determine the velocity and temperature distribution. (ii) Fluid velocity for 1Pr is assumed to be uniform,

Vu . This represents a significant simplification.

(2) Problem Definition. Determine the temperature distribution for boundary layer flow over a flat plate with insulated leading section.

(3) Solution Plan. Start with equating Newton’s law with Fourier’s law to obtain an equation for the heat transfer coefficient h. Apply the integral form of the energy equation using a third degree polynomial temperature profile.

(4) Plan Execution. (i) Assumptions. (1) Steady state, (2) constant properties, (3) two-

dimensional, (4) laminar flow (Rex < 5 105), (5) thermal boundary layer (Pe > 100), (6) uniform upstream velocity and temperature , (7) flat plate, (8) uniform surface temperature, (9) negligible changes in kinetic and potential energy, (10) negligible axial conduction, (11) negligible

dissipation, (12) no buoyancy ( = 0 or g = 0) and (13) uniform velocity within the thermal boundary layer ( 1Pr ).

(ii) Analysis. The local Nusselt number is defined as

k

hxNux (a)

where the heat transfer coefficient h is given by equation (1.10)

TT

y

xTk

hs

)0,(

(1.10)

Thus h depends on the temperature distribution ).,( yxT The integral form of the energy equation

is used to determine the temperature distribution


)(

0

)(0,

xt

dyTTudx

d

y

xT (5.7)

However, For 1Pr the velocity boundary layer thickness is much smaller than the thermal boundary layer thickness. Thus we assume

Vu (b)

For the temperature profile we assume a third degree polynomial

33

2210, yxbyxbyxbxbyxT (c)

The boundary conditions on the temperature are

(1) sTxT 0,

(2) TxT t,

(3) 0,

y

xT t

(4) 00,

2

2

y

xT

Equation (c) and the four boundary conditions give the coefficients )(xbn

,0 sTbt

sTTb1

)(2

31 , ,02b

33

1)(

2

1

t

sTTb

Substituting the above into (c)

3

3

2

1

2

3)(),(

ttss

yyTTTyxT (d)

Substituting (d) into (1.10)

t

kh

2

3 (e)

Combining (a) and (e)

tx

xNu

2

3 (f)

The problem reduces to finding t which is obtained using the energy equation. Substituting (b)

and (d) into (5.7) )(

0

3

3

22

31)(

2

)(3xt

tts

t

s dyyy

dx

dTTV

TT

Evaluating the integral in the above

dx

dV

dx

dV t

tttt 8

3)8/1()4/3(

2

3

Separating variables and rearranging


dxV

d tt

4 (f)

Integrating

CxV

t

82 (g)

where C is constant of integration. The boundary condition on t is

0)( ot x (h)

Applying (h) to (g) gives

oxV

C8

Substituting into (g) and solving for t

x

x

xVx

ot 18

(i)

Substituting (i) into (f)

)/(1

1

24

3

xx

xVNu

ox (j)

Noting that

p

p

ckk

cPr

/

/(k)

Using (k), equation (j) is expressed in terms of Prandtl and Reynolds numbers

)/(1530.0

xx

PrReNu

o

xx (l)

(iii) Checking. Dimensional check: All terms in equations (i)-(l) are dimensionless.

Boundary conditions check: Assumed temperature profile satisfies the four boundary conditions on temperature.

Comparison with scaling results: For the special case of 0ox , scaling estimate of the Nusselt

number for Pr << 1 is given in equation (4.55)

xx ePrRNu , for Pr <<1 (4.55)

Thus, the dependency on the Prandtl and Reynolds number is the same in (l) and (4.55).

(5) Comments. (i) For the special case of no leading insulated section, equation (l) reduces to

xx PrReNu 530.0 (m

The exact solution to this case gives

xx PrReNu 564.0 (n)

Thus the error in the integral solution is 6%.

PROBLEM 5.2

For fluids with 1Pr the thermal boundary layer

thickness is much smaller than the viscous boundary

layer. That is .1/t It is reasonable for such

cases to assume that fluid velocity within the thermal

layer is linear given by

yVu

Consider uniform laminar boundary layer flow over a flat plate with an insulated leading section

of length .ox The plate is maintained at uniform surface temperature .sT Assume a third degree

polynomial temperature profile, show that the local Nusselt number is given by

2/13/13/1

4/3/1319.0 eRPrxxNu ox

(1) Observations. (i) In general, to determine the Nusselt number it is necessary to determine the velocity and temperature distribution. (ii) Fluid velocity for 1Pr is assumed to be linear,

)/(yVu .

(2) Problem Definition. Determine the temperature distribution for boundary layer flow over a flat plate with insulated leading section.

(3) Solution Plan. Start with equating Newton’s law with Fourier’s law to obtain an equation for the heat transfer coefficient h. Apply the integral form of the energy equation using a linear velocity profile and a third degree polynomial temperature profile.

(4) Plan Execution. (i) Assumptions. (1) Steady state, (2) two-dimensional, (3) constant

properties, (4) laminar flow (Rex < 5 105), (5) thermal boundary layer (Pe > 100), (6) uniform upstream velocity and temperature , (7) flat plate, (8) uniform surface temperature, (9) negligible changes in kinetic and potential energy, (10) negligible axial conduction, (11) negligible

dissipation, (12) no buoyancy ( = 0 or g = 0) and (13) linear velocity within the thermal boundary layer ( 1Pr ).


k

hxNux (a)


TT

y

xTk

hs

)0,(

(1.10)



)(

0

)(0,

xt

dyTTudx

d

y

xT (5.7)


However, For 1Pr the velocity boundar (5. (5. (5. (5. (5. (5. (5. (5.r than the thermal boundary layer thickness. Thus we assume

yVu (b)

where is the thickness of the velocity boundary layer. Application of the integral form of the

momentum equation, (5.5) gives ).(x The solution for )(x for this case is detailed in Example

5.1 and is given by

xRex

12 (5.26)

This gives

xV

12 (c)


33

2210 )()()()(),( yxbyxbyxbxbyxT (d)


(1) sTxT 0,

(2) TxT t,

(3) 0,

y

xT t

(4) 00,

2

2

y

xT

Equation (d) and the four boundary conditions give the coefficients )(xbn

,0 sTbt

sTTb1

)(2

31 , ,02b

33

1)(

2

1

t

sTTb

Substituting the above into (d)

3

3

2

1

2

3)(),(

ttss

yyTTTyxT (e)

Substituting (e) into (1.10)

t

kh

2

3 (f)

Combining (a) and (f)

tx

xNu

2

3 (g)

The problem reduces to finding t which is obtained using the energy equation. Substituting (b)

and (e) into (5.7)


)(

0

3

3

22

31)(

2

)(3xt

tts

t

s dyyyy

dx

dTTV

TT

Evaluating the integral in the above2

102

3 t

t dx

dV

Multiplying and dividing by and rearranging the above

2

2115 tt

dx

d

V (h)

Define

tr (i)

substituting (i) into (h)

2115r

dx

dr

V (j)

However, )(x is given in (c). Substituting (c) into the above and rearranging

21

12

15rx

dx

dr

x (k)

Differentiating the right hand side and noting that /Pr

32

2

12

125.1r

xdx

drrx

xPr (l)

Rearranging and separating variables

3

2

-)/5.2(

4

rPr

drr

x

dx (m)

Integrating and using the boundary condition 0)( ot x

rx

x rPr

drr

x

dx

o 03

2

-)/5.2(

4 (n)

Evaluating the integrals

3/43)/5.2(lnln rPrx

x

o

The above can be written as

3/43)/5.2( rPrx

x

o

(o)


Using the definition of r in (i) and solving (o) for r

3/14/3]/[1

5.2xx

Prr o

t (p)

Substituting (c) for

3/14/3]/[1

5.212xx

Prx

Vot (q)

Using the definition of Reynolds number

3/14/3

1/21/3

3/1 )/(11

)5.2(12 xxRePrx

ot (r)

Substituting (r) into (g)

2/13/13/1

4/3/1319.0 eRPrxxNu ox (s)

(iii) Checking. Dimensional check: All terms in equations (g)-(j) and (m)-(p) are dimensionless.


Comparison with scaling results: For the special case of 0ox , scaling estimate of the Nusselt

number for Pr >> 1 is given in equation (4.57)

xx eRPrNu 1/3)( , for Pr >>1 (4.57)

Thus, the dependency on the Prandtl and Reynolds number is the same in (s) and (4.57).

(5) Comments. (i) For the special case of no leading insulated section, equation (s) reduces to

xx RePrNu3/1

)(319.0 (t)

The exact solution to this case gives

xx RePrNu 3/1)(339.0 , for Pr >> 1 (n)

Thus the error in the integral solution is 5.9%.

PROBLEM 5.3

A square array of chips is mounted flush on a flat plate. The array measures cmcm LL . The

forward edge of the array is at a distance ox from the leading edge of the plate. The chips

dissipate uniform surface flux .sq The plate is cooled by forced convection with uniform

upstream velocity V and temperature T . Assume laminar boundary layer flow. Assume

further that the axial velocity within the thermal

boundary layer is equal to the free stream velocity.

Use a third degree polynomial temperature profile.

[a] Show that the local Nusselt number is given by

)/(175.0

xx

eRPrNu

o

xx

[b] Determine the maximum surface temperature.

(1) Observations. (i) The velocity is assumed to be uniform, ,Vu throughout the thermal

boundary layer. (ii) A leading section of length ox is unheated. (iii) at oxx , surface heat flux

is uniform. (iv) The determination of the Nusselt number requires the determination of the temperature distribution. (v) Surface temperature is unknown. (vi) The maximum surface temperature for a uniformly heated plate occurs at the trailing end.

(2) Problem Definition. Determine the temperature distribution for boundary layer flow over a flat plate which heated with uniform surface flux. This reduces to determining the thermal boundary layer thickness.

(3) Solution Plan. Start with the definition of local Nusselt number. Equate Newton’s law with Fourier’s law to obtain an equation for the heat transfer coefficient h. Apply the integral form of the energy equation using a third degree polynomial temperature profile to determine the temperature distribution.

(4) Plan Execution.

(i) Assumptions. (1) Steady state, (2) two-dimensional, (3) constant properties, (4) laminar

flow (Rex < 5 105), (5) thermal boundary layer (Pe > 100), (6) flat surfa ce, (7) uniform surface heat flux, (8) negligible changes in kinetic and potential energy, (9) negligible axial conduction,

(10) negligible dissipation and (11) no buoyancy ( = 0).


k

hxNux (a)


TxT

y

xTk

hs )(

)0,(

(1.10)

Substitute (1.10) into (a)


xTxT

y

xT

Nus

x)(

)0,(

(b)

Thus the Nusselt number depends on the temperature distribution ).,( yxT Note that surface

temperature varies with location x and is unknown. The integral form of the energy equation is used to determine the temperature distribution

)(

0

)(0,

xt

dyTTudx

d

y

xT (5.7)

The axial velocity u is assumed to be uniform equal to the free stream velocity. Thus

Vu (c)

(c) into (5.7) )(

0

)(0,

xt

dyTTdx

dV

y

xT (d)

Once temperature distribution is determine d, surface temperature and maximum surface temperature will be known.

We assume a third degree polynomial temperature profile

33

2210 )()()()(),( yxbyxbyxbxbyxT (e)


(1) sqy

xTk

)0,(

(2) TxT t ),(

(3) 0,

y

Tx t

(4) 0)0,(

2

2

y

xT

Equation (e) and the four boundary conditions give the coefficients )(xbn

ts

k

qTb

3

20

k

qb s

1 , ,02b23

1

3 t

s

k

qb

Substitute the above into (e)

2

3

33

2),(

t

ts y

yk

qTyxT (f)


Surface temperature is obtained by setting y = 0 in (f)

ts

sk

qTxTxT

3

2)0,()( (g)

The Nusselt number is obtained by substituting (f) and (g) into (b)

tx

xNu

2

3 (h)

The problem reduces to determining .t Substitute (f) into (d) and simplify

)(

033

22

3xt

t

t dyy

ydx

d

V

Evaluate the integral in the above

222

12

1

2

1

3

2ttt

dx

d

V

dx

d

V

t2

4

1 (i)

Separate variables and integrate

2

4

1tddx

V

tx

Evaluate the integrals

CxV

t24 (j)

where C is constant of integration determined from the boundary condition on t

0t at oxx (k)

Apply (k) to (j)

oxV

C 4

Substitute into (j) and solve for t

)(4 ot xxV

(l)

The Nusselt number is obtained by substituting (l) into (h)

)(42

3

o

x

xxV

xNu


This simplifies to

)/1(4

3

xx

xVNu

ox (m)

Noting that pck / , (m) is rewritten as

)/1(

1

4

3

)/1(4

3

xx

xV

k

c

xxk

xVcNu

o

p

o

px

Introduce the definitions of Prandtl and Reynolds numbers, the above gives

)/(175.0

xx

eRPrNu

o

xx (n)

Surface temperature is obtained by substituting (l) in (g)

)(3

4)( o

ss xx

Vk

qTxT (o)

According to (o), the maximum surface temperature occurs at the trailing end of the plate.

For the special case where heating begins at the leading edge, ,0ox equations (n) and (o) give

xx eRPrNu 75.0 (p)

Vk

qTxT s

s3

4)( (q)

(iii) Checking. Dimensional check: (1) All term in (b), (h), (m) and (n) are dimensionless. (2) Each term in (f) and (o) has units of temperature.


Limiting check: If 0sq surface temperature will be the same as free stream temperature.

Setting 0sq in (0) gives .)( TxTs

(5) Comments. For the special case of 0ox , scaling estimate of the Nusselt number for

1Pr is given in equation (4.55)

xx eRPrNu , for Pr <<1 (4.55)

Thus, the dependency on the Prandtl and Reynolds number is the same in (p) and (4.55).

PROBLEM 5.4

A liquid film of thickness H flows by gravity down

an inclined surface. The axial velocity u is given by

2

2

2H

y

H

yuu o

where ou is the free surface velocity. At 0x the

surface is maintained at uniform temperature .sT

The fluid temperature upstream of this section is

.T Assume laminar boundary layer flow and that .1/ Ht Determine the local Nusselt

number and the total surface heat transfer from a section of width W and length L. Neglect heat

loss from the free surface. Use a third degree polynomial temperature profile.

(1) Observations. (i) The velocity distribution is known. (ii) Surface temperature is uniform. (iii) The determination of the Nusselt number requires the determination of the temperature distribution. (iv) Newton’s law of cooling gives the heat transfer rate. This requires knowing the local heat transfer coefficient.

(2) Problem Definition. Determine the temperature distribution for boundary layer flow over a flat plate at uniform surface temperature. This reduces to determining the thermal boundary layer thickness.

(3) Solution Plan. Start with the definition of local Nusselt number. Equate Newton’s law with Fourier’s law to obtain an equation for the heat transfer coefficient h. Apply the integral form of the energy equation using a third degree polynomial temperature profile to determine the temperature distribution. Apply Newton’s law of cooling to determine the heat transfer rate.

(4) Plan Execution.


flow (Rex < 5 105), (5) thermal boundary layer (Pe > 100), (6) flat surfa ce, (7) uniform surface temperature, (8) negligible changes in kinetic and potential energy, (9) negligible axial

conduction, (10) negligible dissipation and (11) no buoyancy ( = 0).

(ii) Analysis The local Nusselt number is defined as

k

hxNux (a)


TT

y

xTk

hs

)0,(

(1.10)

Substitute (1.10) into (a)


xTT

y

xT

Nus

x

)0,(

(b)

Thus, the Nusselt number depends on the temperature distribution ).,( yxT Integration of

Newton’s law of cooling gives the total heat transfer rate

L

s dxxhWTTq0

)()( (c)

The integral form of the energy equation is used to determine the temperature distribution

)(

0

)(0,

xt

dyTTudx

d

y

xT (5.7)

The velocity distribution is given by

2

2

2H

y

H

yuu o (d)


33



(1) sTxT )0,(

(2) TxT t ),(

(3) 0,

y

Tx t

(4) 0)0,(

2

2

y

xT

Equation (e) and the four boundary conditions give the coefficients )(xbn

,0 sTbt

sTTb1

)(2

31 , ,02b

33

1)(

2

1

t

sTTb

Substitute the above into (e)

3

3

2

1

2

3)(),(

ttss

yyTTTyxT (f)

The Nusselt number is obtained by substituting (f) into (b)

tx

xNu

2

3 (g)


The problem reduces to determining .t Substitute (d) and (f) into (5.7)

)(

0

12

1

2

32

1

2

33

3

2

2xt

tto

t

dyyy

H

y

H

y

dx

du

Expand the integrand

)(

0

5

2

13

2

324131

2

32

2

22

2

33

xt

tttto

t

dyH

yy

H

y

Hy

H

y

H

y

Hdx

du


23

2 5

1

24

11

2

3tto

t HHdx

du (h)

To solve this differential equation for t it is first simplified by noting that

23

2 5

1

24

1tt

HH

Thus (h) is approximated by

2

5

11

2

3to

t Hdx

du (i)

Rewrite (i)

dx

dH

u

tt

to

22

15


t

tto

ddxu

Hx

2

4

15


Cxu

Ht

o

3

4

45 (j)

where C is constant of integration determined from the boundary condition on t

0t at 0x (k)

Apply (k) to (j) gives 0C

Substitute into (j) and solve for t

3/1

4

45x

u

H

ot (l)

The Nusselt number is obtained by substituting (l) into (g)


xxH

uNu o

x

3/11

45

4

2

3

Noting that pck / , (m) is rewritten as

3/1

2

2

3/1

3/1

2

23/1

10

1

45

4

4

3

H

x

k

c

k

Hu

H

x

k

HucNu

poopx

Introduce the definitions of Prandtl and Reynolds numbers, the above gives

3/1

2

2

3/110

1

H

xeRPrNu Hx (m)

where the Reynolds number is defined as

HueR o

x (n)

With the Nusselt number determined, the local heat transfer coefficient can be formulated and the total heat transfer rate computed. Equate (a) and (m)

3/1

2

2

3/110

1

H

xeRPr

k

hxH

Solve for h

3/13/1

3/23/1)()(

10xRePr

H

kh H (o)

Substitute (o) into (c)

L

Hs dxxPrReH

kWTTq

0

3/13/1

3/23/1)(

10)( (p)


3/23/1

3/1)/()()(

10

1

2

3HLRePrWkTTq Hs (q)

Rewrite (q) in dimensionless form

3/23/1

3/1)/()(

10

1

2

3

)(HLRePr

WkTT

qH

s

(r)

(iii) Checking. Dimensional check: (1) Each term in (f) has units of temperature. (2) Each term is (g) has units of heat. (3) each term in (b), (g), (m) and (r) is dimensionless.


Limiting check: If surface temperature is the same as free stream temperature, ,TTs the heat

transfer rate will vanish. Setting TTs in (q) gives q = 0.


Qualitative check: The total heat transfer rate is expected to increase with increasing length L.This is in agreement with result (q).

(5) Comments. (i) The Nusselt number in (m) depends on a single parameter PrRe . This

product is known as the Peclet number Pe. (ii) The dimensionless heat transfer rate depends on two parameters: the Peclet number, Pe, and the geometric parameter L/H. (iii) The solution is

not valid for .Ht Thus there is a maximum length for which the solution is valid. This

maximum length, ,maxL is determined from (l) by setting Ht and letting .maxLx The

result is 2

max45

4 HuL o (s)

Expressed in dimensionless form, (s) becomes

HRePrH

L

45

4max (t)

PROBLEM 5.5

A thin liquid film flows under gravity down an

inclined surface of width W. The film thickness

is H and the angle of inclination is . The

solution to the equations of motion gives the

axial velocity u of the film as

2

22

22

sin

H

y

H

ygHu

Heat is added to the film along the surface

beginning at 0x at uniform flux .sq Determine the total heat added from 0x to the section

where the thermal boundary layer penetrates half the film thickness. Assume laminar boundary

layer flow and that .1/ Ht Neglect heat loss from the free surface. Use a third degree

polynomial temperature profile.

(1) Observations. (i) The velocity distribution is known. (ii) Total heat transfer is equal to heat flux times surface area. (iii) Heat fl ux is given. However, the distance x = L at which

2/Ht is unknown.

(2) Problem Definition. Determine the thickness of the thermal boundary layer ).(xt

(3) Solution Plan. (i) Express total heat in terms of heat flux and surface area. (ii) Use the

integral form of the energy equation to determine ).(xt

(4) Plan Execution.


flow (Rex < 5 105), (5) thermal boundary layer (Pe > 100), (6) flat surfa ce, (7) uniform surface heat flux, (8) negligible changes in kinetic and potential energy, (9) negligible axial conduction,

(10) negligible dissipation and (11) no buoyancy ( = 0).

(ii) Analysis. The total heat transfer rate from the surface is

LWqAqq ssT (a)

where

A surface area, 2m

L = distance along surface where 2/Ht , m

Tq total heat transfer rate, W

W = width of surface, m

To determine )(xt , we apply the integral form of the energy equation

)(

0

)(0,

xt

dyTTudx

d

y

xT (5.7)

where the axial velocity u is given by


2

22

22

sin

H

y

H

ygHu (b)


33

2210 )()()()(),( yxbyxbyxbxbyxT (c)


(1) sqy

xTk

)0,(

(2) TxT t ),(

(3) 0,

y

Tx t

(4) 0)0,(

2

2

y

xT


ts

k

qTb

3

20

k

qb s

1 , ,02b23

1

3 t

s

k

qb


2

3

33

2),(

t

ts y

yk

qTyxT (d)

Substituting (b) and (d) into (5.7) and recalling that pck /

)(

033

22

2

sin2

3

2

22xt

t

tss dy

yy

k

q

H

y

H

ygH

dx

d

c

q

p

)(

036

1

2

1

333

2sin 5

2

324

2

2

xt

t

t

t

t dyyyyyH

HyyHdx

dg

c

k

p


43

72

1

15

sintt

H

dx

dg

c

k

p


43

72

1

15

sintt

Hd

gdx


Integrating and noting that 0)0(t

43

72

1

15

sintt

Hgx (e)

Evaluating the above at Lx , where L is the distance form the leading end to the location

where 2/Ht , gives

sin

5760

43)2/(

72

1)2/(

15

sin 443 gH

HHHg

L (f)

Substituting (f) into (a)

sT qWgH

qsin

5760

43 4

(g)

(iii) Checking. Dimensional check: (i) Each term in (d) has units of temperature. (ii) Each term is (g) has units of watts.


Qualitative check: The total heat transfer rate is expected to increase with increasing angle .

This is in agreement with result (g).

(5) Comments. (i) Equation (g) shows that Tq is proportional to .4H Thus film thickness has

significant effect on the total heat transfer rate.

PROBLEM 5.6

A plate is cooled by a fluid with Prandtl number 1Pr . Surface temperature varies with

distance from the leading edge according to

xCTxTs )0,(

where C is constant. For such a fluid it is

reasonable to assume that .Vu Use a

third degree polynomial temperature

profile to show that the local Nusselt

number is given by

2/12/175.0 eRPrNux

and that surface heat flux is uniform. Assume laminar boundary layer flow.

(1) Observations. (i) The determination of the Nusselt number requires the determination of the velocity and temperature distributions. (ii). Velocity is assumed uniform. (iii) Surface temperature is variable. (iv) Newton’s law of cooling gives surface heat flux. This requires knowing the local heat transfer coefficient.

(2) Problem Definition. Determine the temperature distribution for boundary layer flow over a flat plate at variable surface temperature. This reduces to determining the thermal boundary layer thickness.

(3) Solution Plan. Start with the definition of local Nusselt number. Equate Newton’s law with Fourier’s law to obtain an equation for the heat transfer coefficient h. Apply the integral form of the energy equation using a third degree polynomial temperature profile to determine the temperature distribution. Apply Newton’s law of cooling to determine surface heat flux.

(4) Plan Execution.

(i) Assumptions. (1) Steady state, (2) constant properties, (3) two-dimensional, (4)

laminar flow (Rex < 5 105), (5) viscous boundary layer flow (Rex > 100), (6) thermal boundary layer (Pe > 100), (7) uniform upstream velocity and temperature , (8) flat plate, (9) (10) negligible changes in kinetic and potential energy, (11) negligible axial conduction,

(12) negligible dissipation, (13) no buoyancy ( = 0 or g = 0) and (14) 1Pr .


k

hxNux (a)

The heat transfer coefficient h given by equation (1.10)

TxT

y

xTk

hs )(

)0,(

(1.10)


Thus the temperature distribution ),( yxT must be determined. Surface heat flux is obtained

using Newton’s law of cooling

)( TThq ss (b)

The integral form of the energy equation is used to determine temperature distribution

)(

0

)(0,

xt

dyTTudx

d

y

xT (5.7)

Axial velocity distribution u(x,y) for 1Pr is assume to be the same as free stream velocity. Thus

Vu (c)

We assume a third degree polynomial

33

2210, yxbyxbyxbxbyxT (d)

The temperature boundary conditions are:

(1) )(0, xTxT s

(2) TxT t,

(3) 0,

y

xT t

(4) 00,

2

2

y

xT

The four boundary conditions are used to determine the coefficient in (d). The assumed profile becomes

3

3

2

1

2

3)()(),(

ttss

yyxTTxTyxT (e)

Substitute (e) into (1.10)

t

kxh

2

3)( (f)

Introducing (f) into (a)

tx

xNu

2

3 (g)

Thus the problem reduces to determining the thermal boundary layer thickness t . This is

accomplished using the integral form of the energy equation (5.7). Substituting (c) and (e) into (5.7)


)(

0

3

3

2

1

2

31)(

)(

2

3xt

tts

t

s dyyy

TxTdx

dTxT

V


tst

s TxTdx

dTxT

V)(

8

3)(

2

3 (h)

However, surface temperature is given by

xCTTs (i)

Substitute into (h)

tt

xdx

dx

V4 (j)

To solve (j) for )(xt we let

2zx t (k)

Solve (k) for t

x

zt

2

(l)

Substitute (k) and (l) into (j)

dx

dzz

dx

dzzx

V

3242

2

Separate variables

dzzdxxV

32

Integrate

oCzxV

4

4

12 (m)

where oC is a constant determined from the boundary condition on )(xt :

0)0(t (n)

Apply (n) to (l) gives z(0) = 0 (o)

Apply (o) to (m) gives oC = 0. Equation (m) becomes

424 zxV

(p)

Use (l) to eliminate z in (p)


224 txxV

Solve the above for t

xV

t 2 (q)

Substitute (q) into (g) gives the local Nusselt number

xVNux

4

3

Noting that pck / , the above becomes

xV

k

c

k

xVcNu

ppx

4

3

4

3 (r)

Expressing this result in terms of the Prandtl and local Reynolds number, gives

2/12/175.0 RePrNux (s)

The heat transfer coefficient is determined to examine surface heat flux. Substitute (q) into (f), gives h(x)

x

Vkxh

4

3)( (t)

Substitute (i) and (t) into (b), simplify

VkCqs

4

3 (u)

This result shows that surface heat flux is uniform.

(5) Checking. Dimensional check: (i) Equations (g), (r) and (s) are dimensionless. (ii) Equations (e), (f), (h), (q), (t) and (u) are dimensionally correct.

Boundary conditions check: Assumed temperature profile (e) satisfies the four boundary conditions.

Limiting check: If surface temperature is the same as free stream temperature, the heat flux

will be zero. According to (i) TxTs )( when C = 0. Setting C = 0 in (u) gives 0sq .

(6) Comments. The solution to the case of a specified uniform surface flux is given in Section 5.7.3. The corresponding surface temperature is given in (5.35)

1/21/3396.2)(

x

ss

RerP

x

k

qTxT (5.35)

Note that the above can be rewritten as

xCTxTs )(


where C is constant. This is identical to equation (i) which gives the specified surface

temperature in this problem.

PROBLEM 5.7

A plate is cooled by a fluid with Prandtl number 1Pr . Surface temperature varies with

distance form the leading edge according to

xCTxTs )0,(

where C is constant. For such a fluid it

is reasonable to assume that axial

velocity within the thermal boundary

layer is linear given by

yVu

Determine the local Nusselt number and show that surface heat flux is uniform. Use a third

degree polynomial temperature profile and assume laminar boundary layer flow.

(1)Observations. (i) The determination of the Nusselt number requires the determination of the velocity and temperature distributions. (ii). Velocity is assumed linear. (iii) Surface temperature is variable. (iv) Newton’s law of cooling gives surface heat flux. This requires knowing the local heat transfer coefficient.

(2) Problem Definition. Determine the velocity and temperature distribution for boundary layer flow over a flat plate at variable surface temperature. This reduces to determining the viscous and thermal boundary layer thickness.

(3) Solution Plan. Start with the definition of local Nusselt number. Equate Newton’s law with Fourier’s law to obtain an equation for the heat transfer coefficient h. Apply the integral form of the momentum and energy equations to determine the velocity and temperature distribution. Apply Newton’s law of cooling to determine surface heat flux. Since

(4) Plan Execution.





k

hxNux (a)


TxT

y

xTk

hs )(

)0,(

(1.10)




)( TThq ss (b)


)(

0

)(0,

xt

dyTTudx

d

y

xT (5.7)

Axial velocity distribution u(x,y) for 1Pr is assume to vary linearly with normal distance y from the plate. Thus

yVu (c)

where is the viscous boundary layer thickness. Before proceeding with the determination

of temperature distribution, must be determined. We apply the integral form of the

momentum equation (5.5)

dyudx

dudy

dx

dV

y

xuv

xx

0

2

0

0, (5.5)

Substitute (c) into (5.5)

dyydx

dVydy

dx

dV

Vv

xx

0

2

2

2

0

2 11

Evaluate the integrals and simplify

dx

dVv

6

1


1

2

26 Cdx

V (d)

The constant of integration 1C is obtained from the boundary condition on

0)0(

This condition gives 1C = 0. Substituting into (d) and solving for

xV

12 (e)

Turning now to the temperature distribution, we assume a third degree polynomial

33

2210, yxbyxbyxbxbyxT (f)



(1) )(0, xTxT s

(2) TxT t,

(3) 0,

y

xT t

(4) 00,

2

2

y

xT

The four boundary conditions are used to determine the coefficient in (f). The assumed profile becomes

3

3

2

1

2

3)()(),(

ttss

yyxTTxTyxT (g)

Substitute (g) into (1.10)

t

kxh

2

3)( (h)

Introducing (h) into (a)

tx

xNu

2

3 (i)


accomplished using the integral form of the energy equation (5.7). Substituting (c) and (g) into (5.7)

)(

0

3

3

2

1

2

31)(

)(

2

3

xt

tts

t

s dyyy

TxTy

dx

dTxT

V


2)(

10

1)(

2

3t

s

t

s TxT

dx

dTxT

V (j)


xCTTs (k)

Substitute into (j)

215 tt

x

dx

dx

V (l)

To solve (l) for )(xt we first rewrite (l)


2

15 t

t

xdx

dx

V (m)

Use (e) to eliminate 1/ on the left side of (m) and on the right side

2

4

5 t

t

xdx

d (n)

To solve (n) for /t , let

2zx t (o)

substitute into (n)

dx

dzzx 2

8

5


232/3

3

1

24

10Czx (p)

where 2C is a constant determined from the boundary condition on )(xt

0)0(t (q)

Apply (q) to (o) z(0) = 0 (r)

Apply (r) to (p) gives 1C = 0. Equation (p) becomes

32/3

4

5zx (s)

Use (o) to eliminate z in (s) 3

4

5 t

Use (e) to eliminate and solve for x

t

xVx

t3/1

2

35 (t)

Note that

Pr (u)

and


xVRex (v)

Substitute (u) and (v) into (t)

2/13/1

2

35x

t RePrx

(w)

(w) into (i) gives the local Nusselt number

2/13/1

5

3xx RePrNu (x)

To examine surface heat flux we determine the heat transfer coefficient. Substitute (a) into (w), and solve for h(x)

2/13/1

5

3xRePr

x

kh

Use (v) to eliminate the Reynolds number in the above

xV

Prkh

2/13/1

5

3 (y)

Substitute (k) and (y) into (b), simplify 2/1

3/1

5

3 VPrkCqs (z)

This result shows that surface heat flux is uniform.

(5) Checking. Dimensional check: (i) Equations (i), (t), (w) and (x) are dimensionless. (ii) Equations (e), (g), (h), (y), and (z) are dimensionally correct.

Boundary conditions check: Assumed temperature profile (g) satisfies the four boundary conditions.

Limiting check: If surface temperature is the same as free stream temperature, the heat flux

will be zero. According to (k) TxTs )( when C = 0. Setting C = 0 in (z) gives 0sq .

(6) Comments. The solution to the case of a specified uniform surface flux is given in Section 5.7.3. The corresponding surface temperature is given in (5.35)

1/21/3396.2)(

x

ss

RerP

x

k

qTxT (5.35)

Note that the above can be rewritten as

xCTxTs )(

where C is constant. This is identical to equation (k) which gives the specified surface

temperature in this problem.

PROBLEM 5.8

Surface temperature of a plate increases exponentially with distance from the leading edge

according to

)exp()0,( xCTxTs

where C and are constants. The plate is cooled

with a low Prandtl number fluid ( 1Pr ). Since

for such fluids t , it is reasonable to assume

uniform axial velocity within the thermal boundary

layer. That is

Vu

Assume laminar boundary layer flow and use a third degree polynomial temperature

profile.


2/12/12/1)exp(175.0 xx eRPrxxNu

[b] Determine surface flux distribution.

(1) Observations. (i) The determination of the Nusselt number requires the determination of the velocity and temperature distributions. (ii). Velocity is assumed uniform. (iii) Surface temperature is variable. (iv) Newton’s law of cooling gives surface heat flux. This requires knowing the local heat transfer coefficient.

(2) Problem Definition. Determine the temperature distribution for boundary layer flow over a flat plate at variable surface temperature. This reduces to determining the thermal boundary layer thickness.

(3) Solution Plan. Start with the definition of local Nusselt number. Equate Newton’s law with Fourier’s law to obtain an equation for the heat transfer coefficient h. Apply the integral form of the energy equation using a third degree polynomial temperature profile to determine the temperature distribution. Apply Newton’s law of cooling to determine surface heat flux.

(4) Plan Execution.





k

hxNux (a)



TxT

y

xTk

hs )(

)0,(

(1.10)



)( TThq ss (b)


)(

0

)(0,

xt

dyTTudx

d

y

xT (5.7)

Axial velocity distribution u(x,y) for 1Pr is assume to be the same as free stream velocity. Thus

Vu (c)

We assume a third degree polynomial

33



(1) )(0, xTxT s

(2) TxT t,

(3) 0,

y

xT t

(4) 00,

2

2

y

xT

The four boundary conditions are used to determine the coefficient in (d). The assumed profile becomes

3

3

2

1

2

3)()(),(

ttss

yyxTTxTyxT (e)

Substitute (e) into (1.10)

t

kxh

2

3)( (f)

Introducing (f) into (a)

tx

xNu

2

3 (g)



accomplished using the integral form of the energy equation (5.7). Substituting (c) and (e) into (5.7)

)(

0

3

3

2

1

2

31)(

)(

2

3

xt

tts

t

s dyyy

TxTdx

dTxT

V


tst

s TxTdx

dTxT

V)(

8

3)(

2

3 (h)


)exp( xCTTs (i)

Substitute into (h)

)exp()exp(

4 xdx

dx

Vt

t

(j)

Differentiate the right side of (j), the above becomes

dx

dxx

x

V

tt

t

)exp()exp()exp(

4

The above is simplified to

dx

d

V

ttt

24

Rewrite as

dx

d

V

ttt

24

Separate variables

24 t

tt

V

ddx

Integrate

ot CV

x 24ln2

1 (k)

where oC is a constant determined from the boundary condition on )(xt

0)0(t (l)

Apply (l) to (k) gives

VCo 4ln

2

1 (m)

(m) into (k)


V

Vx

t

4

4

ln2

2

(n)

Rewrite

V

Vx

t

4

4

)2exp(

2

Solve the above for t

)2exp(14 xV

t (o)

Substitute (o) into (g) gives the local Nusselt number

)2exp(1

1

42

32

x

xVNux (p)

To express (p) in terms of the Prandtl and local Reynolds numbers, rewrite the above and

note that pck /

)2exp(1

)(

4

3

x

xxV

k

cNu

px

This is written as

xx PrRex

xNu

)2exp(1

)(75.0 (r)

The heat transfer coefficient is determined to examine surface heat flux. Substitute (o) into (f), gives h(x)

)2exp(1

1

4

3)(

x

Vkxh (s)

Substitute (i) and (s) into (b), simplify

)2exp(1

)2exp(

4

3

x

xVkCqs (t)

This result shows that surface heat flux is varies with distance along the surface.

Expressing the above in dimensionless form, gives

)2exp(1

)2exp(

4

3

x

x

VkC

q (u)


(5) Checking. Dimensional check: (i) Equations (g), (p), (r) and (u) are dimensionless. (ii) Equations (e), (f), (k), (o), (s) and (t) are dimensionally correct.

Boundary conditions check: Assumed temperature profile (e) satisfies the four boundary conditions.

Limiting check: (i) If surface temperature is the same as free stream temperature, the heat

flux will be zero. According to equation (i), TxTs )( when C = 0. Setting C = 0 in (u)

gives 0sq .

(ii) According to equation (i), surface temperature is the same as free stream temperature for .0 this case should give zero heat flux. Setting 0 in (t) and applying L’Hospital’s

rule, give 0sq .

(6) Comments. Solution (t) shows that surface heat flux increases with x. Recall that for uniform surface temperature Pohlhausen’s solution shows that the heat transfer coefficient and surface flux decrease with x. However, in this problem surface temperature and the heat transfer coefficient increase with x. This results in a flux that increases with x.

PROBLEM 5.9

A square array of chips of side L is mounted flush on a flat

plate. The chips dissipate non-uniform surface flux according to

x

Cqx

The plate is cooled by forced convection with uniform upstream

velocity V and temperature T . Assume laminar boundary

layer flow with .1/t Use third degree polynomials for the

axial velocity and temperature.


2/13/1331.0 eRPrNux

[b] Show that surface temperature is uniform.

(1) Observations. (i) In general, to determine the Nusselt number it is necessary to determine the velocity and temperature distribution. (ii) Surface heat flux is variable. It decreases with distance x. (iii) Surface temperature is unknown. (iv) Newton’s law of cooling gives surface temperature.

This requires knowing the local heat transfer coefficient. (v) .1/t

(2) Problem Definition. Determine the velocity and temperature distribution for boundary layer flow over a flat plate at variable surface temperature. This reduces to determining the viscous and thermal boundary layer thickness.

(3) Solution Plan. Start with the definition of local Nusselt number. Equate Newton’s law with Fourier’s law to obtain an equation for the heat transfer coefficient h. Apply the integral form of the momentum and energy equations using a third degree polynomial profiles to determine the velocity and temperature distribution. Apply Newton’s law of cooling to determine surface temperature.

(4) Plan Execution. (i) Assumptions. (1) Steady state, (2) constant properties, (3) laminar flow

(Rex < 5 105), (4) two-dimensional, (5) viscous boundary layer flow (Rex > 100), (6) thermal boundary layer (Pe > 100), (7) uniform upstrea m velocity and temperature , (8) flat plate, (9) negligible changes in kinetic and potential energy, (10) negligible axial conduction, (11)

negligible dissipation and (12) no buoyancy ( = 0 or g = 0).

(ii) Analysis. [a] The local Nusselt number is defined as

k

hxNux (a)


TxT

y

xTk

hs )(

)0,(

(1.10)




)(

0

)(0,

xt

dyTTudx

d

y

xT (5.7)

The velocity solution, ),( yxu , for an assumed third degree polynomial is solved in Section 5.7.1

and given by equation (5.9) 3

2

1

2

3 yy

V

u (5.9)

where the integral solution to )(x is

xRex

64.4 (5.10)

The local Reynolds number is defined as

xVRex (b)


33



(1) x

C

y

xTk

0,

(2) TxT t,

(3) 0,

y

xT t

(4) 00,

2

2

y

xT


,3

20 t

x

C

kTb

x

C

kb

11 , ,02b

x

C

kb

t23

1

3

1


3

23

1

3

2),( yy

xk

CTyxT

t

t (d)

Surface temperature is obtained by setting 0y in (d)


tsxk

CTxTxT

3

2)0,()( (e)

Substituting (d) and (e) into (1.10)

t

kh

2

3 (f)


tx

xNu

2

3 (g)

Thus to determine surface temperature and Nusselt number requires the determination of t .

Application of the energy equation gives .t Substitute (5.9) and (d) into (5.7)

)(

0

3

2

3

3

1

3

2

2

1

2

311

xt

t

t dyyyyy

xdx

d

xV

Expanding the integrand and evaluating the integral in the above

532

/140

1/

10

11tt

xdx

d

xV (h)

Since 1/t , it follows that

35/

10

1/

140

1tt

Dropping the last term in (h), gives 3

1110 t

xdx

d

xV (i)


tt

xdx

ddx

xV

x 311

10


ot C

xx

V

31

20 (j)

where oC is constant of integration. The boundary condition on t is

0)0(t (k)

Apply (k) to (j) gives 0oC . Substitute into (j) and solve for t

3/1

20 xV

t (l)

Use (5.10) to eliminate in (l)


3/12

)64.4)(20(x

tRe

x

V (m)

Express (m) in dimensionless form

3/1

1)64.4)(20(

x

t

RexVx

Note that pck / , the above is rewritten as

3/1

15274.4

xp

t

RexVc

k

x

Introduce the definition of the Prandtl number, the above gives

2/13/1

5274.4

x

t

RePrx (n)

Substitute (n) into (g) gives the local Nusselt number

2/13/1331.0 xx RePrNu (o)

[b] Surface temperature is obtained by apply Newton’ s law of cooling or substituting (e) into (e). Newton’s law gives

h

qTT s

s (p)

Substitute (a) into (o) gives h

2/13/1331.0 xRePrx

kh (q)

Surface flux is given by

x

Cqs (r)

(b), (q) and (r) into (p)

xxV

Prk

xCTTs

3/1331.0

Note that in the above variable x cancels out to give

VPrk

CTTs

3/1331.0

(s)


This result can be expressed in dimensionless form as

3/1331.0

1

Pr

Vk

C

TTs (t)

This result shows that surface temperature is uniform.

(iii) Checking. Dimensional check: (i) Equations (5.9), (5.10), (b) and (n) are dimensionless. (ii) Equations (d), (f), (l), (m) and (s) are dimensionally correct.

Boundary conditions check: Assumed temperature profile (d) satisfies the four boundary conditions.

Limiting check: If surface heat flux vanishes, surface temperature will be the same as free stream

temperature. Zero flux corresponds to C = 0. Setting C = 0 in (s) gives .)( TxTs

(5) Comments. (i)The solution is valid for .1/t This corresponds to .1Pr

(ii) Pohlhausen’s solution for a plate at uniform surface temperature gives h(x) as

d

d

x

Vkxh

)0()( (4.66)

The corresponding surface heat flux is

d

d

x

VTTkTTxhq sss

)0()())((

This result can be expressed as

x

Cqs (u)

In this problem a surface flux of the form (u) results in uniform surface temperature. This is in agreement with Pohlhausen’s solution. Thus this problem is identical to Pohlhausen’s problem of flow over a plate at uniform surface temperature.

(iii) Pohlhausen’s solution for the local Nusselt number is

2/13/1331.0 xx RePrNu 10Pr (4.72c)

This is in good agreement with the integral solution(o).

PROBLEM 5.10

A square array of chips of side L is mounted flush on a flat plate. The forward edge of the array

is at a distance ox from the leading edge of the plate. The heat dissipated in each row increases

with successive rows as the distance from the forward edge increases. The distribution of surface

heat flux for this arrangement may be approximated by

2Cxqs

where C is constant. The plate is cooled by forced

convection with uniform upstream velocity V and

temperature T . Assume laminar boundary layer

flow. Assume further that the axial velocity within

the thermal boundary layer is equal to the free

stream velocity, Vu . Use a third degree

polynomial temperature profile.


2/12/12/1

3)/(13.1 eRPrxxNu ox

[b] Determine the maximum surface temperature

[c] How should the rows be rearranged to reduce the maximum surface temperature?


Vu . This represents a significant simplification. (iii) Surface heat flux is variable. It

increases with distance x. (iv) Surface temperature is unknown. Since flux increases with x and heat transfer coefficient decreases with x, surface temperature is expected to increase with x.Thus maximum surface temperature is at the trailing end x = L. (v) Newton’s law of cooling gives surface temperature. This requires knowing the local heat transfer coefficient.

(2) Problem Definition. Determine the temperature distribution for boundary layer flow over a non-uniformly heated flat plate with insulated leading section.



(Rex < 5 105), (4) two-dimensional, (5) thermal boundary layer (Pe > 100), (6) uniform upstream velocity and temperature , (7) flat plate, (8) negligible changes in kinetic and potential energy,

(9) negligible axial conduction, (10) negligible dissipation, (11) no buoyancy ( = 0 or g = 0) and (12) uniform velocity within the thermal boundary layer ( 1Pr ).


k

hxNux (a)



TT

y

xTk

hs

)0,(

(1.10)



)(

0

)(0,

xt

dyTTudx

d

y

xT (5.7)


Vu (b)


33



(1) 20,Cx

y

xTk

(2) TxT t,

(3) 0,

y

xT t

(4) 00,

2

2

y

xT


,3

2 20 tx

k

CTb 2

1 xk

Cb , ,02b

2

2

33

t

x

k

Cb


2

32

3

1

3

2),(

tt

yyx

k

CTyxT (d)


ts xk

CTxTxT 2

3

2)0,()( (e)


t

kh

2

3 (f)



tx

xNu

2

3 (g)


Application of the energy equation gives .t Substituting (b) and (d) into (5.7)

)(

0

2

322

3

1

3

2

xt

tt dy

yyx

dx

dVx


2222222

412

1

2

1

3

2tttt x

dx

dVx

dx

dVx


dxxV

xd t222 4

Integrating

dxxV

xd t222 4

Performing the integration

ot CxV

x 322

3

4 (h)


0)( ot x (i)

Applying (i) to (h) gives

3

3

4oo x

VC

Substituting into (h) and solving for t

xxxV

ot3)/(1

3

4 (j)

Substituting (j) into (g)

3)/(1

1

4

33

xx

xVNu

o

x (k)

Noting that

p

p

ckk

cPr

/

/(l)

[a] Using (l), equation (k) is expressed in terms of Prandtl and Reynolds numbers

2/12/12/1

3)/(13.1 eRPrxxNu ox (m)


[b] Surface temperature is obtained by substituting (j) into (e)

32/5 )/(133

4xxx

Vk

CTT os (n)

This result shows that surface temperature increases with x. Thus maximum temperature is at the trailing end :Lx

32/5max )/(1

33

4)( LxL

Vk

CTT os (o)

[c] Since heat transfer coeffici ent decreases with distance from the leading end, rows of high power density chips should be placed near the leading edge and low density rows towards the trailing end.

(iii) Checking. Dimensional check: Dimensional check: (1) Equations (g), (k) and (l) are dimensionless. (2) Equations (d), (f), (j) and (n) are dimensionally correct.

Boundary conditions check: Assumed temperature profile (d) satisfies the four boundary conditions on temperature.


temperature. Zero flux corresponds to C = 0. Setting C = 0 in (n) gives .)( TxTs

(5) Comments. (i) For the special case of no leading insulated section, equations (m) and (n) reduce to

xx PrReNu 3.1 (p)

2/5

33

4x

Vk

CTTs (q)

(ii) Application of (m) at oxx gives infinite Nusselt number. This anomaly is due to the fact

that boundary layer approximations (neglecting axial conduction) is not valid near the leading edge of the thermal boundary layer.

PROBLEM 5.11

Repeat Problem 5.10 using a linear surface flux distribution .Cxqs


2/12/12/1

3)/(106.1 eRPrxxNu ox

[b] Determine the maximum surface temperature

[c] How should the rows be rearranged to reduce the

maximum surface temperature?


Vu . This represents a significant simplification. (iii) Surface heat flux is variable. It

increases with distance x. (iv) Surface temperature is unknown. Since flux increases with x and heat transfer coefficient decreases with x, surface temperature is expected to increase with x.Thus maximum surface temperature is at the trailing end x = L. (v) Newton’s law of cooling gives surface temperature. This requires knowing the local heat transfer coefficient.

(2) Problem Definition. Determine the temperature distribution for boundary layer flow over a non-uniformly heated flat plate with insulated leading section.



(Rex < 5 105), (4) two-dimensional, (5) thermal boundary layer (Pe > 100), (6) uniform upstream velocity and temperature , (7) flat plate, (8) negligible changes in kinetic and potential energy,

(9) negligible axial conduction, (10) negligible dissipation, (11) no buoyancy ( = 0 or g = 0) and (12) uniform velocity within the thermal boundary layer ( 1Pr ).

(ii) Analysis. [a] The local Nusselt number is defined as

k

hxNux (a)


TT

y

xTk

hs

)0,(

(1.10)


is used to determine the temperature distribution )(

0

)(0,

xt

dyTTudx

d

y

xT (5.7)



Vu (b)


33



(1) xCy

xTk

0,

(2) TxT t,

(3) 0,

y

xT t

(4) 00,

2

2

y

xT


,3

20 tx

k

CTb x

k

Cb1 , ,02b

233

t

x

k

Cb


3

2

1

3

1

3

2),( yy

k

xCTyxT

tt (d)


ts xk

CTxTxT

3

2)0,()( (e)


t

kh

2

3 (f)


tx

xNu

2

3 (g)


Application of the energy equation gives .t Substituting (b) and (d) into (5.7)

)(

0

2

3

3

1

3

2

xt

tt dy

yyx

dx

dVx



2222

4

1

12

1

2

1

3

2tttt x

dx

dVxxx

dx

dVx


xdxV

xd t

42

Integrating

xdxV

xd t

42


ot CxV

x 22 2 (h)


0)( ot x (i)

Applying (i) to (h) gives

22oo x

VC

Substituting into (h) and solving for t

xxxV

ot2)/(1

2 (j)

Substituting (j) into (g)

2)/(1

1

22

3

xx

xVNu

o

x (k)

Noting that pck / , the above is expressed as

2

2/12/1

)/(106.1

xx

eRPrNu

o

x (l)

[b] Surface temperature is obtaine d by substituting (j) into (e)

32)/(13

22xxx

Vk

CTT os (m)

This result shows that surface temperature increases with x. Thus maximum temperature is at the trailing end :Lx

32)/(13

22max, LLx

Vk

CTT os (n)

[c] According to (l), the heat transfer coefficien t decreases with distance from the leading end. Thus rows of high power density chips should be placed near the leading edge and low density rows towards the trailing end.


(iii) Checking. Dimensional check: Dimensional check: (i) Equations (g), (k) and (l) are dimensionless. (ii) Equations (d), (f), (j) and (m) are dimensionally correct.

Boundary conditions check: Assumed temperature profile (d) satisfies the four boundary conditions.


temperature. Zero flux corresponds to C = 0. Setting C = 0 in (m) gives .)( TxTs

(5) Comments. (i) For the special case of no leading insulated section, equations (l) and (m) reduce to

xx PrReNu 06.1 (o)

3

3

22x

Vk

CTTs (p)

(ii) Application of (l) at oxx gives infinite Nusselt number. This anomaly is due to the fact

that boundary layer approximations (neglecting axial conduction) is not valid near the leading edge of the thermal boundary layer.

PROBLEM 5.12

A fluid at temperature oT and flow rate om is injected radially between parallel plates. The

spacing between the plates is H. The upper plate is insulated and the lower plate is maintained at

uniform temperature sT along oRr and is insulated along .0 oRr Consider laminar

boundary layer flow and assume that the radial velocity u does not vary in the direction normal

to the plates (slug flow).

[a] Show that for a cylindrical element rdrt 2 the external mass flow edm to the thermal

boundary layer is

to

e dH

mdrru

dr

ddm

t

0

2

[b] Show that the integral form of conservation of energy is

dyTTdr

d

H

cm

r

rTkr o

pot

)(2

)0,(

0

[c] Assume a linear temperature profile, show that the local Nusselt number is

2/12/12/1

2/1

2

1ror eRPrrRNu

where

H

murRe o

r2

(1) Observations. (i) This problem is described by cylindrical coordinates. (ii) Velocity variation with y is negligible. (iii) Conservation of mass requires that radial velocity decrease with radial distance r. (iv) Surface temperature is uniform.

(2) Problem Definition. Determine the integral formulation of conservation of mass and energy. Determine the temperature distribution within the thermal boundary layer. This requires the

determination of ).(xt

(3) Solution Plan. (i) Apply conservation of mass and energy to a cylindrical element

.2 rdrt (ii) Use the integral form of the energy equation to determine ).(xt


(4) Plan Execution.

(i) Assumptions. (1) Steady state, (2) two-dimensional temperature field, (3) slug flow (velocity varies radially only), (4) constant properties, (5) laminar flow, (6) thermal boundary layer flow, (7) flat surface, (8) uniform surface temperature, (9) negligible changes in kinetic and potential energy, (10) negligible axial conduction, (11) negligible dissipation and (12) no

buoyancy ( = 0).

(ii) Analysis. [a] External mass flow rate. Consider a

volumetric cylindrical element rdrt 2 . The diagram

shows the cross section of the element and the external and radial mass flow rates. Conservation of mass gives

drdr

dmmmdm r

rre

or

re dmdm (a)

where

edm external mass flow element into the boundary layer

rm radial mass flow rate entering element

Integral formulation of rm is

rdyum

t

r

0

2

where u is radial velocity and is density. For constant density and slug flow the above

simplifies to

dyurm

t

r

0

2 (b)


tr urm 2 (c)

Application of conservation of mass between the inlet and channel section location r, gives

omrHu2

Solving for u

rH

mu o

2 (d)

(d) into (c)

to

rH

mm (e)

Differentiating (e)

to

r dH

mdm (f)

drdr

dmm r

rrm

edm

dr

t


(f) into (a)

to

e dH

mdm (g)

[b] Conservation of energy fo r the volumetric cylindrical

element rdrt 2 shown gives

drdr

dEEdEEdq r

rers

This simplifies to

drdr

dEdEdq r

es (h)

where

edE = energy supplied by the external mass

rE = energy convected radially through the boundary layer

sdq = energy conducted to the element through the surface

We now formulate the three energy components:

eope dmTcdE

(d) into the above

to

ope dH

mTcdE (i)

ruTdycE pr 2

(f) into the above

TdyH

mcE

top

r

0

Differentiating the above

Tdydr

d

H

mc

dr

dE topr

0

(j)

Fourier’s law gives

dry

rTrkdqs

)0,(2 (k)

Substituting (i)-(k) into (h)

drTdydr

d

H

mcd

H

mTcdr

y

rTrk

top

to

op

0

)0,(2

Dividing through by dr2 and rearrange the above

dr

t

drdr

dEE r

rrE

edE

sdq


dr

dTTdy

dr

d

H

mc

y

rTrk t

oop

t

0

)0,( (l)

However, the last term in (l) can be written as

t

dyTdr

d

dr

dT o

to

0

(m)

(m) into (l)

dyTTdr

d

H

cm

r

rTkr o

pot

)(2

)0,(

0

(n)

[c} Nusselt number. The local Nusselt number is defined as

k

hrNur (o)


os TT

y

rTk

h

)0,(

(1.10)

Thus h depends on the temperature distribution ).,( yrT The integral form of the energy equation

is used to determine the temperature distribution. We assume a linear temperature profile

yrbrbyrT )()(),( 10 (p)


(1) sTrT )0,(

(2) oTrT t ),(

Equation (p) and the two boundary conditions give

tss

yTTTyrT o )(),( (q)

Substituting (q) into (1.10)

t

kh

When this is substituted into(o), we obtain

tr

rNu (s)

Thus the problem becomes one of determining the thermal boundary layer thickness .t The

integral form of conservation of energy is used to determine .t Substituting (q) into (n)


dyy

dr

dTT

H

cmTTkr

t

tso

po

t

so

0

1)(2

Evaluating the integral and simplifying

dr

d

H

cmkr tpo

t 4

Separating variables

rdrcm

kHd

pott

4

Integrating

opo

t Crcm

kH 22

2

2 (t)


0)( ot R (u)

(t) and (u) give

22o

poo R

cm

kHC

Substituting into (t) and solving for t

224o

pot Rr

cm

kH (v)

(v) into (s)

2/122

4o

por Rr

kH

cmrNu

Substituting (d) into the above and rearranging

2/122

2o

pr Rr

k

crurNu

This result can be expressed in terms of the Reynolds and Prandtl numbers as

2/12/12/1

2/1

2

1ror eRPrrRNu (w)

urRer


(iii) Checking. Dimensional check: (1) Units of (d), (e), (i), (k), (n) and (v) are correct. (2) Equations (s) and (w) are dimensionless.

Boundary conditions check: Assumed temperature profile (q) satisfies the two boundary conditions on temperature.

(5) Comments. (i) The assumption of bulk flow provided a significant simplification. (ii)

Application of (w) at oRr gives infinite Nusselt number. This anomaly is due to the fact that

boundary layer approximations (neglecting axial conduction) is not valid near the leading edge of the thermal boundary layer.

PROBLEM 5.13

The lower plate in Problem 5.12 is heated with uniform flux sq along oRr and insulated

along .0 oRr

[a] Show that for a cylindrical element drrt 2 the external mass flow edm to the thermal

boundary layer is

to

e dH

mdrru

dr

ddm

t

0

2

[b] Show that the integral form of conservation of energy is

dyTTdr

d

H

cmq o

pos

t

)(2 0

[c] Assume a linear temperature profile show that the local Nusselt number is

2/12/12/1

2/1 ror eRPrrRNu

where

H

murRe o

r2

(1) Observations. (i) This problem is described by cylindrical coordinates. (ii) Velocity variation with y is negligible. (iii) Conservation of mass requires that radial velocity decrease with radial distance r. (iv) Surface heat flux is uniform. (v) Surface temperature is unknown.

(2) Problem Definition. Determine the integral formulation of conservation of mass and energy. Determine the temperature distribution within the thermal boundary layer. This requires the

determination of ).(xt

(3) Solution Plan. (i) Apply conservation of mass and energy to a cylindrical element

.2 rdrt (ii) Use the integral form of the energy equation to determine ).(xt

(4) Plan Execution.

(i) Assumptions. (1) Steady state, (2) two-dimensional temperature field, (3) slug flow (velocity varies radially only), (4) constant properties, (5) laminar flow, (6) thermal boundary layer, (7) flat surface, (8) uniform surface heat flux, (9) negligible changes in kinetic and potential energy, (10) negligible axial conduction, (11) negligible dissipation and (12) no

buoyancy ( = 0).


(ii) Analysis. [a] External mass flow rate. Consider a

volumetric cylindrical element rdrt 2 . The diagram

shows the cross section of the element and the external and radial mass flow rates. Conservation of mass gives

drdr

dmmmdm r

rre

or

re dmdm (a)

where

edm external mass flow element into the boundary layer

rm radial mass flow rate entering element

Integral formulation of rm is

rdyum

t

r

0

2

where u is radial velocity and is density. For constant density and slug flow the above

simplifies to

dyurm

t

r

0

2 (b)


tr urm 2 (c)

Application of conservation of mass between the inlet and channel section location r, gives

omrHu2

Solving for u

rH

mu o

2 (d)

(d) into (c)

to

rH

mm (e)

differentiating (e)

to

r dH

mdm (f)

(f) into (a)

to

e dH

mdm (g)

[b] Conservation of energy for th e volumetric cylindrical element rdrt 2 shown gives

drdr

dmm r

rrm

edm

dr

t


drdr

dEEdEEdq r

rers

This simplifies to

drdr

dEdEdq r

es (h)

where

edE = energy supplied by the external mass

rE = energy convected radially through the boundary

layer

sdq = energy conducted to the element through the surface

We now formulate the three energy components:

eope dmTcdE

(d) into the above

to

ope dH

mTcdE (i)

ruTdycE pr 2

(f) into the above

TdyH

mcE

top

r

0

Differentiating the above

Tdydr

d

H

mc

dr

dE topr

0

(j)

Fourier’s law gives

rdrqdq ss 2 (k)

where

sq surface heat flux

Substituting (i)-(k) into (h)

drTdydr

d

H

mcd

H

mTcdrqr

t

op

to

ops

0

2

Dividing through by dr2 and rearrange the above

dr

dTTdy

dr

d

H

mcrq t

o

op

s

t

0

(l)

However, the last term in (l) can be written as

dr

t

drdr

dEE r

rrE

edE

sdq


t

dyTdr

d

dr

dT o

to

0

(m)

(m) into (l)

dyTTdr

d

H

cmrq o

po

s

t

)(2 0

(n)

[c} Nusselt number. The local Nusselt number is defined as

k

hrNur (o)


oTxT

y

rTk

hs )(

)0,(

(1.10)

Thus h depends on the temperature distribution ).,( yrT The integral form of the energy equation

is used to determine the temperature distribution. We assume a linear temperature profile

yrbrbyrT )()(),( 10 (p)


(1) sqy

rTk

)0,(

(2) oTrT t ),(

Equation (p) and the two boundary conditions give

)(),( yk

qTyrT t

so (q)

Surface temperature is obtained by setting 0y in (q)

tk

qTrTxT s

os )0,()( (r)

Substituting (q) and (r) into (1.10)

t

kh

When this is substituted into(o), we obtain

tr

rNu (s)

Thus the problem becomes one of determining the thermal boundary layer thickness .t The

integral form of conservation of energy is used to determine .t Substituting (q) into (n)


dyydr

d

Hk

cmr

t

t

po

0

)(2


dr

d

kH

cmr tpo

2

4

Separating variables

rdrcm

kHd

po

t

42

Integrating

o

po

t Crcm

kH 22 2 (t)


0)( ot R (u)

(t) and (u) give

22o

poo R

cm

kHC

Substituting into (t) and solving for t

222o

po

t Rrcm

kH (v)

(v) into (s)

)(

1

2 22o

po

rRrkH

cmrNu

Substituting (d) into the above and rearranging

2/122o

p

r Rrk

crurNu

This result can be expressed in terms of the Reynolds and Prandtl numbers as

2/12/12/1

2/1 ror eRPrrRNu (w)

urRer

Substituting (v) in (r) gives surface temperature distribution

tk

qTrTxT s

os )0,()(

This result can be written in dimensionless form as


22)0,()( o

p

sos Rr

ruc

k

k

qTrTxT

1)/(1)( 2

oos

os RrPrRe

k

Rq

TxT

r

(iii) Checking. Dimensional check: (1) Units of (d), (e), (i), (k), (n) and (v) are correct. (2) Equations (s) and (w) are dimensionless.

Boundary conditions check: Assumed temperature profile (q) satisfies the two boundary conditions on temperature.

Limiting check: If surface heat flux vanishes, surface temperature will be the same as inlet fluid

temperature .oT Setting 0sq in (r) gives .)( os TxT

(5) Comments. (i) The assumption of bulk flow provided a significant simplification. (ii) The

solution does not apply to the limiting case of 0oR since the flow is three-dimensional and

thus cannot be approximated by bulk conditions. (ii) Application of (w) at oRr gives infinite

Nusselt number. This anomaly is due to the fact that boundary layer approximations (neglecting axial conduction) is not valid near the leading edge of the thermal boundary layer.

PROBLEM 5.14

A porous plate with an impermeable and insulated leading section of length ox is maintained at

uniform temperature sT along .oxx The plate is cooled by forced convection with a free

stream velocity V and temperature .T

Fluid at temperature oT is injected through

the porous surface with uniform velocity .ov

The injected and free stream fluids are

identical. Assume laminar boundary layer

flow, introduce axial velocity simplification

based on 1Pr and use a linear

temperature profile to determine the local

Nusselt number.


Vu . This represents a significant simplification. (iii) The plate is porous. (iv) Fluid is

injected through the plate with uniform velocity. (v) The plate is maintained at uniform surface temperature. (vi) A leading section of the plate is insulated.

(2) Problem Definition. Determine the temperature distribution for boundary layer flow over a porous flat plate with surface injection and insulated leading section.


(4) Plan Execution.

(i) Assumptions. (1) Steady state, (2) constant properties, (3) two-dimensional, (4) laminar

flow (Rex < 5 105), (5) thermal boundary layer (Pe > 100), (6) uniform upstream velocity and temperature , (7) flat plate, (8) uniform surface temperature, (9) negligible changes in kinetic and potential energy, (10) negligible axial conduction, (11) negligible dissipation, (12) no buoyancy

( = 0 or g = 0), (13) uniform axial velocity within the thermal boundary layer ( 1Pr ), (14) uniform porosity and (15) injected fluid is at uniform velocity and temperature and is the same as the external fluid.


k

hxNux (a)


TT

y

xTk

hs

)0,(

(1.10)




TTPcudydx

dTcuTdyc

dx

d

y

xTPk oop

t

p

t

p

xx

v

)()(

00

0,1 (5.6)

This equation is simplified for constant properties

TTPdyTTudx

d

y

xTP oo

tx

v

)(

0

)(0,

1 (b)


Vu (b)

Assume a linear temperature profile

yxbxbyxT 10, (c)


(1) sTxT 0,

(2) TxT t,

Equation (d) and the two boundary conditions give the coefficients )(xbn

,0 sTbt

sTTb1

)(1


tss

yTTTyxT )(),( (d)

Substituting (e) into (1.10)

t

kh (e)

Combining (a) and (e)

t

x

xNu (f)

The problem reduces to finding t which is obtained using the energy equation. Substituting (c)

and (e) into (b)

TTPdyy

dx

dTTV

TTP oo

t

ts

t

s

x

v

)(

0

1)()(

1

Evaluating the integral and rearranging


TT

TT

V

P

dx

d

V

P

s

oot

t

v

2

111 (h)

Rewriting (h)

dx

d

V

P t

t 2

111 (i)

where is constant, defined as

TT

TT

V

Pv

s

oo (j)

Equation (i) is solved for t by separating variables

t

tt

V

P

ddx

212

(k)

Integrating (k) and noting that 0)( ot x

t

t

tt

o

V

P

ddx

x

x 0 212

(l)

Evaluating the integrals

t

to

s

oo

P

VV

P

TT

TT

P

Vxx

)1(1

1ln

2

1

)(

)(

2 2v (m)

This result gives an implicitly solution for t . The procedure for determining the Nusselt number

at a given x is to select a value for x, use (m) to determine the corresponding t and substitute into

(g).

(iii) Checking. Dimensional check: Equations (h), (i) and (j) are dimensionally correct.

Boundary conditions check: Assumed temperature profile (d) satisfies the two boundary conditions on temperature.

(5) Comments. Limiting checks are on solution (m) do not yield useful results. For example, for a solid plate, 0P , the first term on the right side of (m) becomes infinite. For the limiting

case of TTs corresponds to . When this is substituted into (m) gives ./ These

difficulties arise because solution (m) is not valid for 0P or . The type of

differential equation (i) changes for these limiting values and consequently solutions different from (m) must be obtained.

PROBLEM 5.15

A porous plate with an impermeable and

insulated leading section of length ox is

heated with uniform surface flux sq along

.oxx The plate is cooled by forced

convection with a free stream velocity V

and temperature .T Fluid at temperature

oT is injected through the porous surface

with uniform velocity .ov The injected and free stream fluids are identical. Assume laminar

boundary layer flow and introduce axial velocity simplification based on 1Pr . Use a third

degree polynomial temperature profile to determine the local Nusselt number.


Vu . This represents a significant simplification. (iii) The plate is porous. (iv) Fluid is

injected through the plate with uniform velocity. (v) The plate is heated with uniform surface flux. (vi) Surface temperature is unknown, (vii) A leading section of the plate is insulated.

(2) Problem Definition. Determine the temperature distribution for boundary layer flow over a uniformly heated porous flat plate with surface injection and insulated leading section.


(4) Plan Execution.


flow (Rex < 5 105), (5) thermal boundary layer (Pe > 100), (6) uniform upstream velocity and temperature , (7) flat plate, (8) uniform surface heat flux, (9) negligible changes in kinetic and potential energy, (10) negligible axial conduction, (11) negligible dissipation, (12) no buoyancy

( = 0 or g = 0), (13) uniform axial velocity within the thermal boundary layer ( 1Pr ), (14) uniform porosity and (15) injected fluid is at uniform velocity and temperature and is the same as the external fluid.


k

hxNux (a)


TT

y

xTk

hs

)0,(

(1.10)




TTPcudydx

dTcuTdyc

dx

d

y

xTPk oop

t

p

t

p

xx

v

)()(

00

0,1 (5.6)

This equation is simplified for constant properties

TTPdyTTudx

d

y

xTP oo

tx

v

)(

0

)(0,

1 (b)


Vu (c)


33



(1) sqy

xTk

0,

(2) TxT t,

(3) 0,

y

xT t

(4) 00,

2

2

y

xT

Equation (d) and the four boundary conditions give the coefficients )(xbn

,3

20 t

s

k

qTb

k

qb s

1 , ,02b2

1

33

tk

qb s


2

3

3

1

3

2),(

t

ts y

yk

qTyxT (e)


ts

sk

qTxTxT

3

2)0,()( (f)

Substituting (e) and (f) into (1.10)

t

kh

2

3 (g)


Combining (a) and (g)

tx

xNu

2

3 (h)


Substituting (c) and (e) into (b)

TTqV

Pkdy

yy

dx

d

V

Po

s

o

t

t

t

x

v

)(

0

2

3

3

1

3

21

Evaluating the integral and rearranging

)(4

112

TTqV

P

dx

d

V

Po

s

ot vk (h)

Rewriting (h)

)(414

2

TTqV

Pk

V

P

dx

do

s

ot v (i)

Note that the right side of (i) is constant. The boundary condition on (i) is

0)( ot x (j)

Integrating (i) and using (j)

)()(4142

oos

ot xxTT

qV

Pk

V

P v

Solving for t

2/1

2/1

)()(414

oos

ot xxTT

qV

Pk

V

P v (k)

(k) into (h) gives the local Nusselt number

2/1

2/1

)()(

414

2

3

o

os

ox

xx

xTT

qV

Pk

V

PNu

v (l)

This result can be expressed in terms of the Prandtl and local Reynolds number as

2/1

2/1

2/12/1

)()()1(

Re

4

3

oos

op

xx

xxTTq

PcP

PrNu

v

(m)

Surface temperature is obtained by substituting (k) into (f)

2/1

2/1

)()(414

3

2)( oo

s

oss xxTT

qV

Pk

V

P

k

qTxT

v (n)


(iii) Checking. Dimensional check: (1) Equations (e), (h), (k) and (n) are dimensionally correct. (2) Equations (l) and (m) are dimensionless.

Boundary conditions check: Assumed temperature profile (e) satisfies the two boundary conditions on temperature.

Limiting check: For the special case of solid plate, P = 0, (m) and (n) reduce to

2/12/12/1 )(Re4

3oxx xxPrNu (o)

2/1)(3

2)( o

ss xx

k

qTxT (p)

Equation (o) is the correct result for this case (see Problem 5.3). For 0ox , equation (p) shows

that surface temperature varies with x . This result is correct (see Problem 5.6).

(5) Comments. The effect of wall injection on surface temperature is can be evaluated using

solution (n) for ).(xTs If the temperature of the injected fluid is greater than the free stream

temperature, TTo , injection increases surface temperature. On the other hand, if TTo ,

injection will lower surface temperature.

PROBLEM 5.16

Consider steady two-dimensional laminar flow in the inlet region of two parallel plates. The

plates are separated by a distance H. The lower plate is maintained at uniform temperature oT

while heat is removed from the upper plate at uniform flux .oq The inlet temperature is .iT

Determine the distance from the inlet where the lower and upper thermal boundary layers meet.

Use a linear temperature profile and assume that velocity is uniform equal to .iV Express your

result in terms of dimensionless quantities.

(1) Observations. (i) There are two thermal boundary layers in this problem. (ii) The upper and lower plates have different boundary conditions. Thus, temperature distribution is not symmetrical. (iii) The lower plate is at uniform temperature while heat is removed at uniform flux along the upper plate. (iv) Fluid velocity is assumed uniform throughout the channel.

(2) Problem Definition. Determine the temperature distribution for boundary layer flow over the lower and upper plates.

(3) Solution Plan. Apply the integral form of the energy equation using linear temperature profiles for both plates.

(4) Plan Execution.


flow (Rex < 5 105), (5) thermal boundary layer (Pe > 100), (6) uniform velocity throughout, (7) uniform upstream temperature , (7) flat plates, (8) uniform surface temperature at the lower plate, (9) uniform surface flux at the upper plate, (10) negligible changes in kinetic and potential

energy, (11) negligible axial conduction, (12) negligible dissipation and (13) no buoyancy ( = 0 or g = 0).

(ii) Analysis. At the location where the two thermal boundary layers meets, we have

HLL tt )()( 21 (a)

where

H spacing between the two plates L distance from inlet to location where the two thermal layers meet

1t thermal boundary layer for the lower plate

2t thermal boundary layer for the upper plate

The integral form of conservation of energy is given in equation (5.7)

)(

0

)(0,

xt

dyTTudx

d

y

xT (5.7)

However, we assume

iViT

0 x

L

H

oT

oq

1t

2t


iVu (b)

where

iV = fluid axial velocity

Substitute (b) into (a) )(

0

)(0,

xt

ii

dyTTdx

d

y

xT

V (c)

where

iT fluid temperature outside the thermal boundary layers

(1) Lower plate. Assume a linear temperature profile

yxbxbyxT 10, (d)


(1) oTxT )0,(

(2) it TxT ),( 1


,0 oTb1

1)(1

t

oi TTb

Substitute the above into (d)

1)(),(

t

oio

yTTTyxT (e)

Substitute (e) into (c) 1

11

0

1)()(

t

t

oi

t

oi

i

dyy

dx

dTT

TT

V (f)

Evaluate the integral and simplify

)(2

11 1

1 t

ti dx

d

V

Separate variables

dx

ddx

V

tt

i

112

Integrate and use boundary condition 0)0(1t

2)(4 1t

i

xV

Solve for 1t


xVi

t 21 (g)

(1) Upper plate. Assume a linear temperature profile

yxbxbyxT 10, (d)


(1) oqy

xTk

)0,(

(2) it TxT ),( 2


20 t

oi

k

qTb

k

qb o

1

Substitute the above into (d)

)(),( 2 yk

qTyxT t

oi (h)

Substitute (h) into (c) 2

2

0

)(

t

too

i

dyyk

q

dx

d

k

q

V

Evaluate the integral and simplify

)(2

1 2t

i dx

d

V

Separate variables

)(2 2t

i

ddxV

Integrate and use boundary condition 0)0(2t

2)(2 2t

i

xV

Solve for 2t

xVi

t 22 (i)

Let Lx in (g) and (i), substitute into (a)

LV

LV

Hii

22

Solve for L


VHL

2

2

)22( (j)

Rewrite (j) in dimensionless form and use the pck /

k

HVc

H

L p

2)22(

1

This can be written in terms of the Prandtl number and local Reynolds number

HPrReH

L2)22(

1 (k)

where

HVReH (l)

(iii) Checking. Dimensional check: (1) Equations (e), (g), (h), (i) and (j) are dimensionally correct. (2) Equations (k) and (l) are dimensionless.

Boundary conditions check: Assumed temperature profiles (e) and (h) satisfy their respective boundary conditions.

Qualitative check: Increasing the free stream velocity decreases the thermal boundary layer,

resulting in an increase in L. Solution (j) shows that L is directly proportional to V .

(5) Comments. (1) To increase L the Reynolds number should be increased. (2) Taking the ratio of (g) and (i)

22

1

t

t (m)

Thus the thermal boundary layer for constant wall temperature is thicker than that of uniform

surface flux by a factor of .2

PROBLEM 6.1

Use scaling to determine the ratio ./ ht LL Compare scaling estimates with exact solutions.

(1) Observations. (i) This is an internal forced convection problem. (ii) Scaling gives estimates

of hL and .tL (iii) Exact solutions for hL and tL are available for laminar flow through

channels. (iv) Exact solutions for tL depend on channel geometry and surface boundary

conditions.

(2) Problem Definition. Determine the ratio ht LL / using scaling and using exact solutions.

(3) Solution Plan. Apply scaling results (6.2) and (6.3) to estimate ./ ht LL Apply (6.5) and (6.6)

to obtain an exact solution for ./ ht LL

(4) Plan Execution.

(i) Assumptions. (1) Steady state, (2) laminar flow, (3) constant properties, (4) uniform surface heat flux or uniform surface temperature, (5) negligible axial conduction (6) negligible changes in kinetic and potential energy and (7) negligible dissipation

(ii) Analysis.

Scaling estimates of hL and tL are given by equations (6.2) and (6.3)

1~/

2/1

D

h

Re

DL (6.2)

1~/

2/1

rPRe

DL

D

t (6.3)

From (6.2) and (6.3) we obtain

1~/

Pr

LL ht (a)

Exact solutions for hL for laminar flow is given by equation (6.5)

eDhh ReC

D

L

e

(6.5)

where eD is the equivalent diameter, defined as

P

AD

fe

4

where fA is channel flow area and P is channel perimeter. The coefficient hC depends on

channel geometry and is given in Table 6.1. Similarly, exact solutions for tL for laminar flow is

given by equation (6.6)


eDtt PrReC

D

L

e

(6.6)

where tC is a constant which depends on channel geometry as well as boundary conditions and

is given in Table 6.1. Taking the ratio of (6.6) to (6.5) and rearranging

h

tht

C

C

Pr

LL / (b)

(iii) Computations. Using Table 6.1,

the ratio ht CC / is computed for the six

geometries listed in the table for both uniform surface flux and uniform surface temperature. Comparisons between scaling estimate and exact solutions are tabulated. Tabulation results show that scaling estimate is close to exact solutions for the six geometries examined.

(5) Comments. (i) Scaling estimate of hL

and tL does not take into consideration

channel geometry. In addition, scaling does not distinguish between laminar and turbulent flow. (ii) Examination of the tabulated results show that scaling

provides reasonable estimates of ht LL /

for all Prandtl numbers.

Scaling Estimate

1~/

Pr

LL ht

Exact Solution

h

tht

C

C

Pr

LL /

Geometry uniform surface flux

uniform surface temperature

0.77 0.60

ab

a

a/b =10.73 0.46

a

b a/b = 2 0.67 0.57

a

b a/b = 4 0.56 0.72

1.09 0.73

PROBLEM 6.2

Use scaling to estimate the hydrodynamic and thermal entrance lengths for the flow of air in a

cm3cm3 square duct . The mean velocity is 0.8 m/s. Compare scaling estimates with exact

solutions. Evaluate properties at C.50o

(1) Observations. (i) This is an internal forced convection problem. (ii) Scaling gives estimates

of hL and .tL (iii) Exact solutions for hL and tL are available for laminar flow through

channels. (iv) Exact solutions for tL depend on channel geometry and surface boundary

conditions.

(2) Problem Definition. Determine the ratio ht LL / using scaling and using exact solutions.

(3) Solution Plan. Apply scaling results (6.2) and (6.3) to estimate ./ ht LL Apply (6.5) and (6.6)

to obtain an exact solution for ./ ht LL

(4) Plan Execution.

(i) Assumptions. (1) Steady state, (2) laminar flow, (3) constant properties, (4) uniform surface heat flux or uniform surface temperature, (5) negligible axial conduction (6) negligible changes in kinetic and potential energy and (7) negligible dissipation

(ii) Analysis.

Scaling estimates of hL and tL are given by equations (6.2) and (6.3)

1~/

2/1

D

h

Re

DL (6.2)

1~/

2/1

rPRe

DL

D

t (6.3)

Exact solutions for hL for laminar flow is given by equation (6.5)

eDhh ReC

D

L

e

(6.5)


eD

DuRe

e (a)

where eD is the equivalent diameter, defined as

P

AD

fe

4 (b)

fA = channel flow area, 2m

P = channel perimeter, m u mean flow velocity = 0.8 m/s

kinematic viscosity, 2m /s


The coefficient hC depends on channel geometry and is given in Table 6.1. Similarly, exact

solutions for tL for laminar flow is given by equation (6.6)

eDtt PrReC

D

L

e

(6.6)

where tC is a constant which depends on channel geometry as well as boundary conditions and

is given in Table 6.1.

(iii) Computations. For a square duct of side 0.03 m,

m03.0m))(03.0(4

)m()03.0(4

22

eD

Properties of air at C50o are

Pr 0.709

s

m1093.17

26

The Reynolds number is

3.1339s)/(m1093.17

)m)(03.0)(s/m(08.026eDRe

Scaling estimates: Substituting into (6.2)

1~3.1339

)m(03.0/2/1

hL

2.40~hL m

Equation (6.3) gives

1~)3.1339)(709.0(

)m(03.0/2/1

tL

28.5~tL m

Exact solution: For a square channel, Table 6.1 gives:

09.0hC

066.0tC , for uniform surface heat flux

041.0tC , for uniform surface temperature

Equation (6.5) gives hL

3.1339)m)(03.0(09.0hL = 3.61 m


Equation (6.5) gives tL

88.1)3.1339)(709.0)(m)(03.0(066.0tL m, for uniform surface heat flux.

17.1)3.1339)(709.0)(m)(03.0(041.0tL m, for uniform surface temperature.

(5) Comments. (i) Scaling estimate of hL and tL does not take into consideration channel

geometry or surface thermal condition.(ii) Scaling overestimates hL and tL by and order of

magnitude.

PROBLEM 6.3

Far away from the entrance of a channel the velocity and temperature become fully developed. It

can be shown that under such conditions the Nusselt number becomes constant. Consider air

flowing with a mean velocity of 2 m/s through a long tube of diameter 1.0 cm. The mean

temperature at a section in the fully developed

region is 35oC. The surface of the tube is

maintained at a uniform temperature of 130oC.

What is the length of the tube section needed for

the mean temperature to reach 105oC? The Nusselt

number for this case is given by

657.3DNu

(1) Observations. (i) This is an internal force convection problem. (ii) The channel is a long tube. (iii) The surface is maintained at a uniform temperature. (iv) Since the tube section is far away from the entrance, the velocity and temperature can be assumed fully developed. (v) Tube diameter, mean velocity and inlet, outlet and surface temperatures are known. The length is unknown. (vi) The fluid is air.

(2) Problem Definition. Determine the tube length needed to raise the mean temperature to a specified level.

(3) Solution Plan. Use the analysis of flow in tubes at uniform surface temperature to determine the required tube length.

(4) Plan Execution.

(i) Assumptions. (1) Steady state, (2) fully developed flow, (3) axisymmetric flow, (4) constant properties, (5) uniform surface temperature, (6) negligible changes in kinetic and potential energy, (7) negligible axial conduction, (8) negligible dissipation and (9) no energy generation.

(ii) Analysis. For flow in a tube at uniform surface temperature, conservation of energy and Newton's law of cooling lead to equation (6.13)

][)()( exp xmc

hPTTTxT

psmism (a)

cp = specific heat, J/kg-oC

h = average heat transfer coefficient for a tube of length L, W/m2-oCm = mass flow rate, kg/s P = tube perimeter, m Tm(x) = mean temperature at x, oCTmi = mean inlet temperature = 35oCTs = surface temperature = 130oCx = distance from inlet of heated section, m

Applying (a) at the outlet of the heated section (x = L) and solving for L

mos

misp

TT

TT

hP

cmL ln (b)

usT

L


Tmo = mean outlet temperature = 105oC

To compute L using (b), it is necessary to determine cp, P, m , and h . Air properties are determined

at the mean temperature mT , defined as

mT = T Tmi mo

2 (c)

The perimeter P and flow rate m are given by

P = D (d) And

uD

m4

2

(e)

where

D = inside tube diameter = 1 cm = 0.01 m u = mean flow velocity = 2 m/s

= density, kg/m3

The heat transfer coefficient for this case is determined from the Nusselt number, given by

657.3k

DhNuD (f)

(iii) Computations. Properties are determined at the mean temperature mT . Using (c)

mT = C702

)C)(10535( oo

Properties of air at this temperature are:

cp = 1008.7 J/kg-oC

k = 0.02922 W/m-oCPr = 0.707

= 19.9 10-6 m2/s

= 1.0287 kg/m3

Substituting into (d), (e) and (f)

P = 0.01(m) = 0.03142 m

kg/s0.0001616)2(m/s)m1.0287(kg/4

)(m(0.01) 322

m

h = 3.657 )m(01.0

)Cm/W(02922.0 o

= 10.69 W/m2-oC


)C)(105130(

)C)(35130(ln

)Cm/W(69.10)m(03142.0

)Ckg/J(7.1008)s/kg(0001616.0o

o

o2

o

L = 0.65 m


(iv) Checking. Dimensional check: Computations showed that equations (b), (d), (e) and (f) are dimensionally consistent.

Limiting checks: (1) For the special case of Tmo = Tmi , the required length should vanish. Setting Tmo = Tmi in (b) gives L = 0.

(2) The required length for the outlet temperature to reach surface temperature is infinite. Setting

Tmo = Ts in (b) gives L = .

Quantitative checks: (1) An approximate check can be made using conservation of energy and Newton’s law of cooling. Conservation of energy is applied to the air between inlet and outlet

Energy added at the surface = Energy gained by air (g)

Assuming that air temperature in the tube is uniform equal to mT , Newton’s law of cooling gives

Energy added at surface = h D L (Ts mT ) (h)

Neglecting axial conduction and changes in kinetic and potential energy, energy gained by air is

Energy gained by air = m cp(Tmo miT ) (i)

Substituting (h) and (i) into (g) and solving for the resulting equation for L

)(

)(

ms

mimop

TTDh

TTmcL (j)

Equation (j) gives

)C)(70130)(m)(01.0()Cm/W(69.10

)C)(35105)(Ckg/J(7.1008)s/kg(0001616.0oo2

oo

L = 0.57 m

This is in reasonable agreement with the more exact answer obtained above.

(2) The value of h appears to be low compared with typical values listed in Table 1.1 for forced convection of gases. However, it shoul d be kept in mind that values of h in Table 1.1 are for typical applications. Exceptions should be expected.

(5) Comments. Equation (f) gives the Nusselt number and heat transfer coefficient for this case. This equation is valid under certain conditions. Key among the restrictions are: fully developed laminar flow in tubes at uniform surface temperature.

PROBLEM 6.4

A fluid is heated in a long tube with uniform surface flux. The resulting surface temperature

distribution is found to be higher than design specification. Two suggestions are made for

lowering surface temperature without changing surface flux or flow rate: (1) increasing the

diameter, (2) decreasing the diameter. You are asked to determine which suggestion to follow. The

flow is laminar and fully developed. Under such conditions the Nusselt number is given by

364.4DNu

(1) Observations. (i) This is an internal force convection in a tube. (ii) The surface is heated at uniform flux. (iii) Surface temperature increases along the tube and is unknown. (iv) The flow is assumed laminar and fully developed. (v) The heat transfer coefficient for fully developed flow through channels is constant. (vi) According to Newton’s law of cooling, surface temperature is related to mean fluid temperature, surface heat flux and heat transfer coefficient.

(2) Problem Definition. Derive an equation for surface temperature variation in terms of tube diameter.

(3) Solution Plan. Apply surface temperature solution for fully developed laminar flow through a tube with constant surface flux.

(4)Plan Execution.

(i) Assumptions. (1) Steady state, (2) laminar, (3) fully developed flow, (4) axisymmetric flow, (5) constant properties, (6) uniform surface heat flux, (7) negligible changes in kinetic and potential energy, (8) negligible axial conduction, (9) negligible dissipation and (10) no energy generation.

(ii) Analysis. Application of conservation of energy and Newton’s law of cooling give the variation of surface temperature with distance. The solution is given by equation (7.61)

hcm

PxqTxT

psmis

1)( (a)

where

cp = specific heat, J/kg-oC h = heat transfer coefficient, W/m2-oC

m = mass flow rate, kg/s x = distance from inlet, m

P = tube perimeter, m

sq = surface heat flux, W/m2

Ts (x) = local surface temperature, oC T

mi = mean inlet temperature, oC

The perimeter is given by

P = D (b)


For fully developed laminar flow through tubes at constant surface flux the Nusselt number is given by

k

hDNuD = 4.364 (c)

where

D = tube diameter, m k = thermal conductivity, W/m-oCNuD = Nusselt number

Solving (c) for h

h = 4.364k

D (d)

Substituting (b) and (d) into (a)

Ts (x) = Tmi + qD x

mc

D

ks

p .4 364 (e)

Examination of (e) shows that decreasing the diameter will decrease surface temperature.

(iii) Checking. Dimensional check: The right hand side of (e) should have units of oC.

qs (W/m2)D x

m c

D

kp

m m

kg / s J / kg C

m

W / m Co o

( )

( ) ( )

( )

. ( )4 364= W

o oC

J / s

C

W= oC

Qualitative check: Increasing surface flux, increases Ts. Decreasing the mass flow rate, increases Ts. This behavior is confirmed by equation (e).

Limiting check: If surface flux qs = 0, fluid outlet temperature remains constant equal to the inlet

temperature. Setting qs = 0 in (e) gives Ts = Tmi.

(5) Comments. The effect of diameter, surface flux, mass flow rate, distance along the tube and fluid properties on surface temperature can be evaluated using the result obtained in (e). However, attention should be given to the assumptions leading to this result.

PROBLEM 6.5

Two identical tubes are heated with the same uniform flux at their surfaces. Air flows through

one tube while water flows at the same rate through the other. The mean inlet temperature for

both tubes is the same. Which tube will have a higher surface temperature distribution? Assume

laminar flow and neglect entrance effects. For this case the Nusselt number is given by

364.4DNu

(1) Observations. (i) This is an internal force convection in a tube. (ii) The surface is heated at uniform flux. (iii) Surface temperature increases along the tube and is unknown. (iv) The flow is assumed laminar and fully developed. (v) The heat transfer coefficient for fully developed flow through channels is constant. (vi) According to Newton’s law of cooling, surface temperature is related to mean fluid temperature, surface heat flux and heat transfer coefficient.

(2) Problem Definition. Derive an equation for surface temperature variation in terms of tube diameter.

(3) Solution Plan. Apply surface temperature solution for fully developed laminar flow through a tube with constant surface flux.

(4)Plan Execution.

(i) Assumptions. (1) Steady state, (2) laminar, (3) fully developed flow, (4) axisymmetric flow, (5) constant properties, (6) uniform surface heat flux, (7) negligible changes in kinetic and potential energy, (8) negligible axial conduction, (9) negligible dissipation and (10) no energy generation.

(ii) Analysis. Application of conservation of energy and Newton’s law of cooling give the variation of surface temperature with distance. The solution is given by equation (6.10)

Ts (x) = Tmi + hcm

Pxq

ps

1 (6.10)

where

cp = specific heat, J/kg-oCh = heat transfer coefficient, W/m2-oCm = mass flow rate, kg/s

x = distance from inlet, m

moT

tL

hL

L

miT

m

u D

sq

sq

sq

x


P = tube perimeter, m qs = surface heat flux, W/m2

Ts (x) = local surface temperature, oCT

mi = mean inlet temperature, oC

The perimeter is given by

P = D (a)

For fully developed laminar flow through tubes at constant surface flux the Nusselt number is given by

NuD = hD

k = 4.364 (b)

where

D = tube diameter, m k = thermal conductivity, W/m-oCNuD = Nusselt number

Solving (b) for h

h = 4.364k

D (c)

Substituting (a) and (c) into (6.10)

Ts (x) = Tmi + qD x

mc

D

ks

p .4 364 (d)

Examination of this result shows that decreasing the diameter will decrease surface temperature.

(iii) Checking. Dimensional check: The right hand side of (d) should have units of oC.

qs (W/m2))CW/m(364.4

)m(

)CJ/kg()kg/s(m

)m(moo k

D

c

xD

p

= W

o oC

J / s

C

W= oC

Qualitative check: Increasing surface flux, increases Ts. Decreasing the mass flow rate, increases Ts. This behavior is confirmed by equation (d).

Limiting check: If surface flux qs = 0, fluid outlet temperature remains constant equal to the

inlet temperature. Setting qs = 0 in (d) gives Ts = Tmi.

(5) Comments. The effect of diameter, surface flux, mass flow rate, distance along the tube and fluid properties on surface temperature can be evaluated using the result obtained in (d). However, attention should be given to the assumptions leading to this result.

PROBLEM 6.6

Water flows through a tube with a mean velocity of 0.2 m/s. The mean inlet temperature is 20oC

and the inside diameter of the tube is 0.5 cm. The water is heated to 80oC with uniform surface

heat flux of 0.6 W/cm2. Determine surface temperature at the outlet. If entrance effects can be

neglected the Nusselt number for fully developed flow is constant given by

364.4DNu

Is it justifiable to neglect entrance effects?

(1) Observations. (i) This is an internal forced convection problem in a tube. (ii) The surface is heated at uniform flux. (iii) Surface temperature changes along the tube and is unknown. (iv) The Reynolds number should be checked to determine if the flow is laminar or turbulent. (v) If hydrodynamic and thermal entrance lengths are small compared to tube length, the flow can be assumed fully developed throughout. (vi) For fully developed flow, the heat transfer coefficient is uniform. (vii) The length of the tube is unknown. (viii) The fluid is water.

(2) Problem Definition. (i) Find the required length to heat the water to a given temperature and (ii) determine the surface temperature at the outlet.

(3) Solution Plan. (i) Since surface flux, mean velocity, diameter, inlet and outlet temperatures are known, apply conservation of energy between the inlet and outlet to determine the required tube length. (ii) Check the Reynolds number to determine if the flow is laminar or turbulent. (iii) Calculate the hydrodynamic and thermal entrance lengths and compare with the tube length. (iv) Apply surface temperature solution for flow through a tube with constant surface flux.

(4) Plan Execution.

(i) Assumptions. (1) Steady state, (2) constant properties, (3) axisymmetric flow, (4) uniform surface heat flux, (5) negligible changes in kinetic and potential energy, (6) negligible axial conduction, (7) negligible dissipation and (8) no energy generation.

(ii) Analysis. Determination of tube length. Application of conservation of energy between the inlet and outlet of the tube gives

sqLD = mcp(Tmo - Tmi) (a)

where

pc = specific heat, J/kg-oC

D = tube diameter = 0.5 cm = 0.005 m L = tube length, m m = mass flow rate, kg/s

miT = mean temperature at the inlet = 20oC

moT = mean temperature at the outlet = 80oC

sq = surface heat flux = 0.6 W/cm2 = 6000 W/m2

Solving (a) for L gives

L =s

mimop

qD

TT mc )( (b)


The mass flow rate m is given by

uDm 2)4/( (c)

where

u = mean flow velocity = 0.2 m/s

= density, kg/m3

To determine surface temperature at the outlet, use the solution for surface temperature distribution for flow through a tube with uniform surface flux, given by equation (6.10)

Ts (x) = Tmi + )(

1

xhcm

Pxq

ps (d)

where

h = local heat transfer coefficient, W/m2-oCP = tube perimeter, m Ts (x) = local surface temperature, oCx = distance from inlet of heated section, m

The perimeter P is given by

P = D (e)

Surface temperature at the outlet Ts(L) is obtained by setting x = L in (d). Substituting (e) into (d) and letting x = L gives

Ts (L) = Tmi +)(

1

Lhcm

LDq

ps (f)

The determination of h(L) requires establishing if the flow is laminar or turbulent and if it is fully developed at the outlet. Thus, the Reynolds number should be determined. It is defined as

ReD

u D (g)

where

ReD = Reynolds number


Properties of water are determined at the mean temperature T defined as

T = T Tmi mo

2 (h)


T = C502

)C)(8020( oo

Properties of water at this temperature are given in Appendix D

cp = 4182 J/kg-oC k = 0.6405 W/m-oC Pr = 3.57


= 0.5537 10-6

m2/s

= 988 kg/m3


ReD

0 2 0 005

05537 1018066 2

. ( / ) . ( )

. ( / )

m s m

m s

Since the Reynolds number is less than 2300, the flow is laminar. The next step is calculating the hydrodynamic and thermal entrance lengths Lh and Lt to see if the flow is fully developed at the outlet. For laminar flow in a tube the hydrodynamic and thermal lengths are given by (7.43)

Lh = Ch D ReD (i)

Lt = Ct D ReD Pr (j)

where

Ch = hydrodynamic entrance length constant (Table 6.1) = 0.056 Ct = thermal entrance length constant (Table 6.1) = 0.043 Lh = hydrodynamic entrance length, m Lt = thermal entrance length, m

Substituting numerical values into (i) and (j)

Lh = 0.056 0.005 (m) 1806 = 0.506 m

and

Lt = 0.043 0.005 (m) 1806 3.57 = 1.386 m

If tube length L is larger than Lh and Lt, the flow is fully developed. Thus, it is necessary to compute L using equation (b). The mass flow rate in equation (b) is given by (c)

m = 988(kg/m3) 0.2(m/s) (0.005)2(m2)/4 = 0.00388kg/s


L =)m/cm(10)cm/W(6.0)m(005.0

)C)(2080)(Ckg/J(4182)s/kg(00388.02242

oo

= 10.33 m

Since L > Lt> Lh, the flow is fully developed at the outlet. The heat transfer coefficient for fully developed laminar flow through a tube with uniform surface flux is given by

NuD = hD

k = 4.364 (k)

where

k = thermal conductivity = 0.6405 W/m-oCNuD = Nusselt number

Solving (k) for hDkh /364.4 (l)


(iii) Computations. To determine surface temperature at the outlet we first use (l) to compute h(L)

h(L) = 4.364 0 6405

0 005

. ( / )

. ( )

W m C

m

o

= 559 W/m2-oC

With L, m and h(L) determined, equation (f) gives the surface temperature at the outlet

Ts (L) = 20oC + )Cm/W(559

1

)Ckg/J(4182)s/kg(00388.0

)m(43.10)m(005.0)m/W(6000

o2o

2 = 91.3oC

(iv) Checking. Dimensional check: Computations showed that equations (b), (c), (e), (f), (g), (i), (j) and (l) are dimensionally correct.

Quantitative checks: (1) Alternate approach to determining Ts(L): Application of Newton’s law of cooling at the outlet gives

sq = h [Ts(L) - Tmo ] (m)

solving for Ts(L)

Ts(L) = Tmo + q

h

s= 80 (oC) +

0 6 10

559

2 4 2 2

2

. ( / ) ( / )

( / )

W cm cm m

W m Co = 90.7oC

(2) The value of h is within the range reported in Table 1.1 for forced convection of liquids.

Limiting check: If Tmi = Tmo, the required length should vanish. Setting Tmi = Tmo into (b) gives L= 0.

(5) Comments. (i) As long as the outlet is in the fully developed region, surface temperature at the outlet is determined entirely by the local heat transfer coefficient. Therefore, it is not necessary to justify neglecting entrance length to solve the problem. (ii) In solving internal forced convection problems, it is important to establish if the flow is laminar or turbulent and if it is developing or fully developed.

PROBLEM 6.7

Fluid flows with a mean axial velocity u in a tube of

diameter D. The mean inlet temperature is miT . The

surface is maintained at uniform temperature .sT

Show that the average Nusselt number for a tube of

length L is given by

sm

smiDL

TLT

TTPrReNu

)(ln

4

wherek

LhNu L

L ,Du

ReD and Lh is the average

heat transfer coefficient over the length L.

(1) Observations. (i) This is an internal force convection problem. (ii) The channel is a tube. (iii) The surface is maintained at a uniform temperature. (iv) Entrance effect is important in this problem. (v) The average Nusselt number for a tube of length L depends on the average heat transfer coefficient over the length.

(2) Problem Definition. Determine the average heat transfer coefficient for a tube of length L

which is maintained at uniform surface temperature.

(3) Solution Plan. Start with the definition of the average Nusselt number. Use the analysis of flow in tubes at uniform surface temperature to determine the average heat transfer coefficient.

(4) Plan Execution.

(i) Assumptions. (1) Steady state, (2) axisymmetric flow, (3) constant properties, (4) uniform surface temperature, (5) negligible changes in kinetic and potential energy, (6) negligible axial conduction, (7) negligible dissipation and (8) no energy generation.

(ii) Analysis. The average Nusselt number for a tube of length L is given by

k

LhNu L

L (a)

where Lh is the average heat transfer coefficient over the length L, defined as

L

L dxxhL

h

0

)(1

(6.12)

For flow in a tube at uniform surface temperature, conservation of energy and Newton's law of cooling lead to equation (6.13)

][)()( exp xmc

hPTTTxT

psmism (6.13)

pc = specific heat, J/kg-oC

Lhh = average heat transfer coefficient for a tube of length x, W/m2-oC

m = mass flow rate, kg/s

usT

L

miT


P = tube perimeter, m Tm(x) = mean temperature at x, oCTmi = mean inlet temperature, oCTs = surface temperature, oCx = distance from inlet of heated section, m

Applying (a) at the outlet of the heated section (x = L) and solving for Lh

)(ln

LTT

TT

LP

cmh

ms

mispL (b)

where

)(LTm = mean outlet temperature

Substitute (b) into (a)

)(ln

LTT

TT

kP

cmNu

ms

mispL (c)

However, the mass flow rate m is given by

uD

m4

2

(d)

where

D = inside tube diameter, m u = mean flow velocity, m/s

= density, kg/m3

The perimeter of a tube is DP (e)

Substitute (d) and (e) into (c)

)(ln

4 LTT

TT

k

cuDNu

ms

mispL (f)

The coefficient in (f) can be expressed in terms of Prandtl and Reynolds number as

Dpp

PrReuD

k

c

k

cuD (g)

(g) into(f)

)(ln

4 LTT

TTPrReNu

ms

misDL (h)

(iii) Checking. Dimensional check: (1) Equations (6.12), (6.13), (b) and (d) are dimensionally consistent. (2) Equations (c) and (f) are dimensionless.

(5) Comments. Equation (f) or (h) can be used to experimentally determine the average heat transfer coefficient and average Nusselt number.

PROBLEM 6.8

Water flows through a cm75.0cm75.0 square duct with

a mean velocity of 0.12 m/s. The duct is heated with a

uniform surface flux of 0.25 W/cm2. The mean inlet

temperature is 25oC. The maximum allowable surface

temperature is 95oC. Justify neglecting entrance effects.

And determine maximum outlet mean temperature.

(1) Observations. (i) This is an internal forced convection problem. (ii) The fluid is heated at uniform wall flux. (iii) Surface temperature changes with distance along the channel. It reaches a maximum value at the outlet. (iv) The Reynolds and Peclet numbers should be checked to establish if the flow is laminar or turbulent and if this is an entrance or fully developed problem. (v) The channel has a square cross-section. (vi) Application of Newton’s law of cooling at the outlet relates outlet temperature to surface temperature, surface flux and heat transfer coefficient. (vii) Application of conservation of energy gives a relationship between heat added, inlet temperature, outlet temperature, specific heat and mass flow rate.

(2) Problem Definition. [a] Determine the outlet temper ature corresponding to a specified surface temperature and flux. [b] Determine the require d channel length to heat the water to outlet temperature and compare with entrance lengths.

(3) Solution Plan. Apply Newton’s law of cooling at the outlet to determine the mean outlet temperature. This requires determining the heat transfer coefficient. Check the Reynolds and Peclet numbers to establish if the flow is laminar or turbulent and if this is an entrance or fully developed problem. Obtain a solution to the heat transfer coefficient (Nusselt number).

(4) Plan Execution.

(i) Assumptions. (1) Steady state, (2) constant properties, (3) uniform surface heat flux, (4) negligible changes in kinetic and potential energy, (5) negligible axial conduction, (6) negligible dissipation and (7) no energy generation.

(ii) Analysis. [a] Determination of Tmo. Applying Newton’s law of cooling at the outlet

qs = h(L) [Ts (L) moT ] (a)

Solving (a) for Tmo

Tmo = Ts(L)Lh

qs (b)

moT

tL

hL

L

miT

m

u

sqsq

sq

)(LTs

S

S

moT

miT

u

L

sT


where

)(Lh = heat transfer coefficient at the outlet, W/m2-oC

L = channel length, m qs = surface heat flux = 0.25 W/cm2 = 2500 W/m2

moT = mean outlet temperature, oC

)(LTs = surface temperature at the outlet = 95oC

Equation (b) gives Tmo in terms of the heat transfer coefficient at the outlet, h(L). The value of h(L) depends on whether the flow is laminar or turbulent and if the flow is developing or fully developed at the outlet. To establish these conditions, the Reynolds and Prandtl numbers are determined. The Reynolds number for flow through a square channel is defined as

ReDeeuD

(c)

where

De = equivalent diameter, m ReDe = Reynolds numberu = mean velocity = 0.12 m/s


The equivalent diameter for a square channel is defined as

De = 4A

P = 4

4

2S

S = S (d)

where

A = channel flow area = S2, m2

P = channel perimeter in contact with the fluid = 4S, m S = side dimension of the square channel = 0.75 cm = 0.0075 m

Water properties are evaluated at the mean temperature, Tm , defined as

mT = (Tmi + Tmo)/2 (e)

where

mT = mean fluid temperature in channel, oC

miT = mean inlet temperature = 25oC

However, since Tmo is unknown, a solution is obtained using a trial and error procedure. A value

for Tmo is assumed, (e) is used to calculate mT and (b) is used to calculate Tmo. The calculated Tmo

is compared with the assumed value. The procedure is repeated until a satisfactory agreement is obtained between assumed and calculated values of Tmo.

Let Tmo = 85oC

Equation (e) gives

Tm = (25 + 85)( oC)/2 = 55oC


cp = specific heat = 4184 J/kg-oCk = thermal conductivity = 0.6458 W/m-oC

PROBLEM 6.8 (continued)Pr = Prandtl number = 3.27

= kinematic viscosity = 0.5116 10-6 m2/s

= density =985.7 kg/m3

Using (d) to calculate eD

De = 0.0075 m


DeRe = )s/m(105116.0

)m(0075.0)s/m(12.026

= 1759

Since the Reynolds number is smaller than 2300, the flow is laminar. To establish if the flow is developing or fully developed at the outlet, the hydrodynamic length, thermal entrance length and tube length must be determined. Equations (7.43a) and (7.43b) give

Lh = Ch Dee ReD (f)

and

Deett ReDCL Pr (g)

where

Ch = velocity entrance length constant (Table 7.2) = 0.09 Ct = temperature entrance length constant (Table 7.2) = 0.066 Lh = hydrodynamic entrance length, m Lt = thermal entrance length, m

Substituting numerical values into (f) and (g)

Lh = 0.09 (0.0075)(m)(1759) = 1.19 m and

Lt = 0.066(0.0075)(m)(1759) (3.27) = 2.85 m

These two lengths should be compared with the tube length L. To determine L, conservation of energy is applied to the fluid between inlet and outlet

Energy added at the surface = Energy gained by the fluid (h)

Neglecting axial conduction and changes in kinetic and potential energy, (h) gives

qs (4 S L) = mcp(Tmo miT )

or

L =m c T T

S q

p mo mi

s4 (i)

where the mass flow rate m is given by

m = u A = u S2 (j)

Since Tmo is unknown, L cannot be computed. To proceed, assume that the flow is fully developed at the outlet, determine h(L), use (b) to compute Tmo and (i) to compute L. If the computed length is larger than Lh and Lt, the assumption of fully developed flow is verified. The Nusselt number for fully developed laminar flow in a square channel with uniform surface heat flux is given by equation (7.58) and Table 7.3


k

hDNu e

De = 3.608 (k)

or

h = 3.608 eDk / (l)

(iii) Computations. [a] Determination of Tmo. Equation (l) gives h

h = 3.608(0.6458)(W/m-oC)/0.0075(m) = 310.7 W/m2-oC


Tmo = 95(oC))Cm/W(7.310

)m/W(2500o2

2

= 87oC

This is close to the assumed value of 85oC used to obtain approximate water properties.

[b] Determination of channel length L. Equation (j) gives m

m = 985.7(kg/m3)0.12(m/s)(0.0075)2(m2) = 0.00665 kg/s


L =)m/W(2500)m)(0075.0(4

)C(2587)Ckg/J(4184)s/kg(00665.02

oo

= 23 m

Since L > Lt > Lh, the flow is fully developed at the outlet.

(iv) Checking. Dimensional check: Computations showed that equations (b), (c), (f), (g), (i), (j) and (l) are dimensionally consistent.

Quantitative check. The value of h is within the range listed in Table 1.1.

Qualitative check: As surface heat flux is decreased, channel length should increase. Equation (i) shows that L is inversely proportional to qs .

Limiting check. If Tmo = Tmi, channel length should be zero. Setting Tmo = Tmi in (i) gives L = 0.

(5) Comments. (i) This problem illustrates how analysis cannot always be completed without carrying out some computations. This can occur if it is necessary to establish if the flow is laminar or turbulent or if entrance effects can be neglected or not. (ii) A solution is obtained without the need to neglect entrance effects. As long as the outlet is in the fully developed region, water outlet temperature is determined entirely by the local heat transfer coefficient.

PROBLEM 6.9

Two experiments were conducted on fully developed laminar flow through a tube. In both

experiments surface temperature is C180o and the mean inlet temperature is C20o . The mean

outlet temperature for the first experiment is found to be C120o . In the second experiment the

flow rate is reduced by a factor of 2. All other conditions remained the same. Determine:

[a] The outlet temperature of the second experiment.

[b] The ratio of heat transfer rate for the two experiments.

(1) Observations. (i) This is an internal forced convection problem in tubes. (ii) The flow is laminar and fully developed. (iii) The surface is maintained at uniform temperature. (iv) All conditions are identical for two experiments except the flow rate through one is half that of the other. (v) The total heat transfer rate depends on the outlet temperature.

(2) Problem Definition. Determine the outlet temperature for fully developed laminar flow through a tube at uniform surface temperature.

(3) Solution Plan. Use the analysis of fully developed laminar flow in tubes at uniform surface temperature to determine the outlet temperature. Apply conservation of energy to obtain an equation for the heat transfer.

(4) Plan Execution.

(i) Assumptions. (1) Steady state, (2) fully developed laminar flow, (3) axisymmetric flow, (4) constant properties, (5) uniform surface temperature, (6) negligible changes in kinetic and potential energy, (7) no axial conduction, (8) no dissipation and (9) no energy generation.

(ii) Analysis. For flow through tubes with uniform surface temperature, conservation of energy and Newton’s law of cooling lead to equation (6.13)

][)()( exp xcm

hPTTTxT

psmism (a)

pc = specific heat, CJ/kg o

h = average heat transfer coefficient for a tube of length x, CW/m o2



)(xTm = mean temperature at x, Co

miT = mean inlet temperature = 20 Co


x = distance from inlet of heated section, m

Applying (a) at the outlet ( Lx )

][)()( exp Lcm

hPTTTxT

psmismo (b)

where

L = tube length, m

moT = mean outlet temperature, Co


The quantities L, P and pc are the same for both experiments. Furthermore, the flow remains

laminar and fully developed when the flow rate is reduced in the second experiment. Thus the heat transfer coefficient is the same for both experiments. Equation (b) is rewritten in dimensionless form as

)/exp( mTT

TTC

smi

som (c)

where

pc

hPLC (d)

Rewriting (c)

mTT

TT C

som

smiln (e)

Let the subscripts 1 and 2 refer to the first an d second experiments. Applying (e) to the two experiments gives

11

lnmTT

TT C

som

smi (f)

22

lnmTT

TT C

som

smi (g)

Taking the ratio of (g) to (f) and rearranging

som

smi

som

smi

TT

TT

mTT

TT m

12

1

2

lnln

Or

2

1

12

m

m

som

smi

som

smi

TT

TT

TT

TT

Solving for 2moT

2

1

12 )(

m

m

smi

somsmisom

TT

TTTTTT (h)

This result gives the outlet temperature when the flow rate is reduced. Application of conservation of energy to the fluid between the inlet and outlet gives the heat transfer rate q

mimop TTcmq (i)

Applying (i) to the flow in the two tubes and taking the ratio of the two results

mimo

mimo

TTm

TTm

q

q

11

22

1

2 (j)

(iii) Computations. Substituting numerical values into (h) and noting that 2/ 21 mm


2

o

ooo

2)C)(18020(

)C)(180120()C)(18020()C(180

omT = 157.5 Co

Equation (j) gives

6875.020120

205.157

2

1

1

2

q

q

(iv) Checking. Dimensional check: Each term in (h) has units of temperature.

Limiting check: If 21 mm , the two outlet temperatures must be the same. Setting 21 mm in

(h) gives 12 momo TT .

(5) Comments. (i) Although tube size, fluid nature a nd flow rate are not known, it was possible to obtain a solution to the problem. Taking the ratio of two operating conditions results in the cancellation of the unknown factors. (ii) Although the outlet temperature increases as the flow rate is decreased, the rate of heat transfer decreases. (iii) Decreasing the flow rate without changing inlet and surface temperatures and heat transfer coefficient is expected to increase the

outlet temperature. In the limit as the flow rate approaches zero ),0( 2m the corresponding

outlet temperature becomes equal to surface temperature. This follows from (h).

PROBLEM 6.10

A long rectangular duct with a cm8cm4 cross section is

used to heat air from –19.6oC to 339.6

oC. The mean

velocity in the duct is 0.2 m/s and surface temperature is

340oC. Determine the required duct length. Is neglecting

entrance effects justified?

(1) Observations. (i) This is an internal forced convection problem. (ii) The channel has a rectangular cross section. (iii) Surface temperature is uniform. (iv) The Reynolds and Peclet numbers should be checked to establish if the flow is laminar or turbulent and if entrance effects can be neglected. (v) Channel length is unknown. (vi) The fluid is air.

(2) Problem Definition. Determine the required channel length to heat air to a specified outlet temperature. Determine the hydrodynamic and thermal entrance lengths and compare with channel length.

(3) Solution Plan. Use the analysis of flow through tubes with uniform surface temperature to determine channel length. Check the Reynolds and Peclet numbers to establish if the flow is laminar or turbulent. Compute the hydrodynamic and thermal entrance lengths and compare with channel length to determine if entrance effects can be neglected. Obtain a solution to the heat transfer coefficient (Nusselt number).

(4) Plan Execution.

(i) Assumptions. (1) Steady state, (2) constant properties, (3) uniform surface temperature, (4) negligible changes in kinetic and potential energy, (5) negligible axial conduction, (6) negligible dissipation and (7) no energy generation.

(ii) Analysis. For flow through tubes with uniform surface temperature, conservation of energy and Newton’s law of cooling lead to equation (6.13)

][)()( exp xcm

hPTTTxT

psmism (a)

where


h = average heat transfer coefficient for a channel of length x, W/m2-oCm = mass flow rate, kg/s P = channel perimeter, m Tm(x) = mean temperature at x, oCTmi = mean inlet temperature, -19.6oC

Ts = surface temperature = 340oCx = distance from inlet of heated section, m

u

moT

miT

sTL

tL

hL

m

L

moT

miT

b

a

sT

uu


Applying (a) at the outlet, x = L, and solving for L

mos

misp

TT

TT

hP

cmL ln (b)

where

L = channel length, m

Tmo = mean outlet temperature = 339.6oC

Equation (b) gives L in terms of cp, P, m, and h . The average heat transfer coefficient depends on whether the flow is laminar or turbulent and if the flow is developing or fully developed. To establish these conditions, the Reynolds and Prandtl numbers are determined. The Reynolds number for flow through a rectangular channel is defined as

ReDeeuD

(c)

where

eD = equivalent diameter, m

ReDe = Reynolds number u = mean velocity = 0.2 m/s


The equivalent diameter for a rectangular channel is defined as

eD = 4A

P =

)(2

ba

ab (d)

where

A = channel flow area = ab, m2

P = channel perimeter in contact with the fluid = 2(a + b), m a = width of rectangular channel = 8cm = 0.08 m b = height of rectangular channel = 4 cm = 0.04 m


P = 2(a + b) (e) and

uabuAm )( (f)

where

= density, kg/m3

Air properties are evaluated at the mean temperature, Tm , defined as

mT = (Tmi + Tmo)/2 (g)

where

mT = mean fluid temperature in channel, oC

The mean temperature is calculated in order that properties are determined. Substituting into (g)

mT = C1602

)C)(6.3396.19( oo


Properties of air at this temperature are given in Appendix C

cp = 1018.5 J/kg-oCk = 0.3525 W/m-oCPr = 0.701

= 29.75 10-6, m2/s

= 0.8342 kg/m3


eD =)m)(08.004.0(

)m(08.0)m(04.02 = 0.0533 m


ReDe 3.358)/sm(1075.29

)m(0533.0)/(2.026

sm

Since the Reynolds number is smaller than 2300, the flow is laminar. To determine if entrance effects can be neglected, the hydrodynamic and thermal entrance length must be compared with channel length. For laminar flow, equations (6.5) and (6.6) give

Lh = Ch Dee ReD (h)

and

Deett ReDCL Pr (i)

where

Ch = hydrodynamic entrance length constant (Table 6.1, for a/b = 0.08m/0.04m = 2) = 0.085 Ct = thermal entrance length constant (Table 6.1, for a/b = 0.08m/0.04m = 2) = 0.049 Lh = hydrodynamic entrance length, m Lt = thermal entrance length, m

Substituting numerical values into (h) and (i)

Lh = 0.085 (0.0533)(m)(358.3) = 1.623 m and

Lt = 0.049(0.0533)(m) (358.3) (0.706) = 0.656 m

These two lengths should be compared with channel length L. However, L can be determined

only after h is computed. To compute h , L must be known! To proceed, assume that entrance

length effects are negligible (fully developed flow throughout channel), determine h , use (b) to compute L and compare it with Lh and Lt. For fully developed laminar flow through a rectangular channel at constant surface temperature, the Nusselt number is given in Table 6.2

k

DhNu e

De = 3.391 (j)

or

h = 3.391 eDk / (k)

(iii) Computations. Equations (e), (f) and (k) give

P = 2(0.08 + 0.04)(m) = 0.24 m


m = 0.0.8342(kg/m3)0.08(m)0.04(m)0.2(m/s) = 0.0005339 kg/s

h = 3.391(0.03525)(W/m-oC)/0.0533(m) = 2.242 W/m2-oC


C)339.6)((340

C)19.6)((340ln

)C42(W/m0.24(m)2.2

C)5(J/kgkg/s)1018.0.0005339(o

o

o2

o

L = 6.873 m

Comparing Lh and Lt with L

Lh/L = 1.623(m)/6.873(m) = 0.236

and

Lt/L = 0.656(m)/6.87(m) = 0.095

(iv) Checking. Dimensional check: Computations showed that equations (b) )f( , (h), (i)

and (l) are dimensionally consistent.

Limiting checks: (1) For the special case of Tmo = Tmi , the required length should vanish. Setting Tmo = Tmi in (b) gives L = 0. (2) The required length for the outlet temperature to reach

surface temperature is infinite. Setting Tmo = Ts in (b) gives L = .

Quantitative check: The value of h appears to be low compared with typical values listed in Table 1.1 for forced convection of gases. Howe ver, equation (l) shows that for laminar flow

through channels, h is inversely proportional to D. A large D can result in a small h . In

addition, values of h in Table 1.1 are for typical applications. Exceptions should be expected.

(5) Comments. (i) This problem illustrates the importance of establishing if entrance effects can

be neglected or not. (ii) Neglecting thermal entrance length is justified since tL is less than 10%

of L. However, neglecting the viscous entrance length requires careful judgment. This

assumption underestimates h and consequently, according to (b), it overestimates L. Thus it is a conservative assumption.

PROBLEM 6.11

A rectangular duct with inside dimensions of

cm4cm2 is used to heat water from 25 Co to

115 Co . The mean water velocity is 0.018 m/s. The

surface of the duct is maintained at 145 Co . Determine

the required duct length. Assume fully developed flow

conditions throughout.

(1) Observations. (i) This is an internal force convection problem. (ii) The channel is a

rectangular duct. (iii) The surface is maintained at a uniform temperature. (iv) The velocity and

temperature are fully developed. (v) The Reynolds number should be checked to determine if the

flow is laminar or turbulent. (vi) Duct size, mean velocity and inlet, outlet and surface

temperatures are known. The length is unknown. (vii) Duct length depends on the heat transfer

coefficient. (vii) The fluid is water.

(2) Problem Definition. Determine the duct length needed to raise the mean temperature to a specified level. This requires determining the heat transfer coefficient.

(3) Solution Plan. Use the analysis of flow in channels at uniform surface temperature to determine the required duct length.

(4) Plan Execution.

(i) Assumptions. (1) Steady state, (2) fully developed flow, (3) constant properties, (4) uniform surface temperature, (5) negligible changes in kinetic and potential energy, (6) negligible axial conduction, (7) negligible dissipation and (8) no energy generation.

(ii) Analysis. For flow in a channel at uniform surface temperature, conservation of energy and Newton's law of cooling lead to equation (6.13)

][)()( exp xcm

hPTTTxT

psmism (a)

where


h = average heat transfer coefficient for a channel of length L, CW/m o2


P = cross section perimeter, m

Tm(x) = mean temperature at x, Co

Tmi = mean inlet temperature = 25 Co

Ts = surface temperature = 145 Co


Applying (a) at the outlet (x = L) and solving for L

mos

misp

TT

TT

hP

mcL ln (b)

u

moT

miT

sTL


where

Tmo = mean outlet temperature = 115 Co

To compute L using (b), it is necessary to determine cp, P, m , and h . Water properties are

determined at the mean temperature mT , defined as

mT = T Tmi mo

2 (c)

The perimeter P is given by P = 2(a + b) (d)

Where

a = duct width = 4 cm = 0.04 m b = duct height = 2 cm = 0.02 m

The mass flow rate m is given by uabm )( (e)

where

= density, 3kg/m

The heat transfer coefficient for fully developed flow is uniform along a channel. Its value depends on whether the flow is laminar or turbulent. To proceed, it is necessary to calculate the Reynolds number to determine if the flow is laminar or turbulent. For flow in a rectangular duct the Reynolds number is defined as

eDe

DuRe (f)

where


DeRe = Reynolds number

u = mean velocity = 0.018 m/s

= kinematic viscosity, /sm2

The equivalent diameter for a rectangular channel is defined as

eD =P

A4 =

)(2

ba

ab (g)

where

A = duct flow area = ab, m2

The mean temperature is calculated to determine water properties. Substituting into (c)

mT = C702

)C)(11525( oo

Properties of water at this temperature are:

pc = 4191 CJ/kg o


k = 0.6594 CW/m o

Pr = 2.57

= 6104137.0 /sm2

= 977.7 3kg/m


eD = m02667.0m))(02.004.0(

m)(02.0)m(04.02

Equation (h) gives

1160/s)m(104137.0

m)(02667.0)m/s(018.026DeRe

Since the Reynolds number is smaller than 2300, the flow is laminar. The Nusselt number for fully developed laminar flow through rectangular channels at uniform surface temperature is given by equation (7.57) and Table 7.3. Thus

k

DhNu e

De = 3.391 (h)

Solving for h

eD

kh 391.3 (i)

(iii) Computations. Substituting into (d), (e) and (i)

)m)(02.004.0(2P = 0.12 m

kg/s01408.0)m/s(018.00.02(m)m)(04.0)kg/m(7.977 3m

h = )m(02667.0

)CW/m(6594.0391.3

o

= 83.84 CW/m o2


)C)(115145(

)C)(25145(ln

)CW/m(84.83m)(12.0

)s/kg(01408.0)CJ/kg(4191o

o

o2

o

L 8.13 m

(iv) Checking. Dimensional check: Computations showed that equations (b), (d), (e), (f), (g) and (i) are dimensionally consistent.

Limiting checks: (1) For the special case of Tmo = Tmi , the required length should be zero. Setting Tmo = Tmi in (b) gives L = 0.

(2) The required length for the outlet temperature to reach surface temperature is infinite. Setting

Tmo = Ts in (b) gives L = .

Quantitative checks: (1) An approximate check can be made using conservation of energy and Newton’s law of cooling. Conservation of energy is applied to the water between inlet and outlet

Energy added at the surface = Energy gained by water (j)


Assuming that water temperature in the tube is uniform equal to mT , Newton’s law of cooling

gives

Energy added at surface = h P L (Ts mT ) (k)

Neglecting axial conduction and changes in kinetic and potential energy, energy gained by the water is

Energy gained by air = mcp(Tmo miT ) (l)

Substituting (k) and (l) into (j) and solving for the resulting equation for L

)(

)(

ms

mimop

TTPh

TTcmL (m)

Equation (m) gives

)C)(70145)(m)(12.0)CW/m(84.83

)C)(25115)(CJ/kg(4191)kg/s(01408.0oo2

oo

L = 7.04 m

This is in reasonable agreement with the more exact answer obtained above.

(2) The value of h within the range listed in Table 1.1 for forced convection of liquids.

(5) Comments. This problem is simplified by two conditions: fully developed and laminar flow.

PROBLEM 6.12

Air is heated in a cm4cm4 square duct at

uniform surface flux of 590 .W/m2 The mean air

velocity is 0.32 m/s. At a section far away from

the inlet the mean temperature is 40 Co . The

mean temperature is 120 Co . Determine the

maximum surface temperature.

(1) Observations. (i) This is an internal forced

convection problem in a channel. (ii) The surface is heated at uniform flux. (iii) Surface

temperature changes along the channel. It reaches a maximum value at the outlet. (iv) The

Reynolds number should be checked to determine if the flow is laminar or turbulent. (v) Velocity

and temperature profiles become fully developed far away from the inlet. (vi) The heat transfer

coefficient is uniform for fully developed flow. (vii) The channel has a square cross section.

(viii) tube length is unknown. (ix) The fluid is air.

(2) Problem Definition. (i) Find the required length to heat the air to a given temperature and (ii) determine surface temperature at the outlet.

(3) Solution Plan. (i) Since surface flux, mean velocity, duct size, inlet and outlet temperatures are known, application of conservation of energy between the inlet and outlet gives the required duct length. (ii) Check the Reynolds number to determine if the flow is laminar or turbulent. (iii) Apply surface temperature solution for flow through a channel with constant surface flux.

(4) Plan Execution.

(i) Assumptions. (1) Steady state, (2) constant properties, (3) uniform surface heat flux, (4) negligible changes in kinetic and potential energy, (5) negligible axial conduction, (6) negligible dissipation and (7) no energy generation.

(ii) Analysis. Application of conservation of energy between the inlet and outlet gives the required channel length

)( mimops TTcmqLP (a)

where


L = channel length, m m = mass flow rate, kg/s P = perimeter, m

sq = surface heat flux = 590 2W/m

miT 40 Co

moT 120 Co

Solving (a) for L

qP

TTcmL

mimop )( (b)

moT

miT

L

u

sq


The mass flow rate and perimeter are given by

uSm 2 (c)

SP 4 (d)

where

S = duct side = 0.04 m

u = mean flow velocity = 0.32 m/s

= density, 3kg/m

Substituting (c) and (d) into (b)

q

TTcuSL

mimop

4

)( (e)

To determine surface temperature at the outlet, use the solution for surface temperature distribution for channel flow with uniform surface flux, given by equation (6.10)

Ts (x) = Tmi +)(

1

xhcm

Pxq

ps (f)

where

)(xh = local heat transfer coefficient, CW/m o2

)(xTs = local surface temperature, Co


Surface temperature at the outlet, Ts(L), is obtained by setting x = L in (f). Substituting (c) and (d) into (f)

Ts (L) = Tmi + )(

14

LhcuS

Lq

p

s (g)

Finally, it remains to determine the heat transfer coefficient at the outlet, h(L). This requires establishing whether the flow is laminar or turbulent. Thus, the Reynolds number should be determined. The Reynolds number for flow through a square channel is defined as

eDe

DuRe (h)

where



The equivalent diameter for a square channel is defined as

De = P

A4 =

S

S

44

2

= S (i)

Substituting (i) into (h)


SuReDe (j)

Properties of air are determined at the mean temperature mT defined as

mT = T Tmi mo

2 (k)


mT = C802

)C)(12040( oo

Properties of air at this temperature are:

pc = 1009.5 CJ/kg o

k = 0.02991 CW/m o

Pr = 0.706

= 20.92 610 m2/s

= 0.9996 kg/m3

Substituting into (j)

9.611/s)m(1092.20

m)(04.0)m/s(32.026DeRe

Since the Reynolds number is smaller than 2300, the flow is laminar. The heat transfer coefficient for fully developed laminar flow through a square channel with uniform surface flux is constant. It is given by equation (6.55) and Table 6.2

k

DhNu e

De = 3.608 (l)

where hh . Solving (k) for h

eD

kh 608.3 (m)

(iii) Computations. Substituting numerical values in (e) gives required channel length

4378.0)W/m(590)4(

C))(40120(C)J/kg-(5.1009m/s)(32.0)m(04.0)kg/m(9996.02

oo3

L m

To determine surface temperature at the outlet, the heat transfer coefficient is computed using (m)

h(L) = h =)m(04.0

)CW/m(02991.0608.3

o

= 2.7 CW/m o2

Equation (g) gives the surface temperature at the outlet


C)W/m(7.2

1

C)k/J(5.1009m/s)(32.0)m(04.0)kg/m(9996.0

)m)(4378.0(4)W/m(590)C(40

o2o3

2o)(g

LTs

)(LTs = 338.5 Co

(iv) Checking. Dimensional check: Computations showed that equations (e), (g), (j), and (m) are dimensionally correct.

Quantitative checks: (1) Alternate approach to determining Ts(L): Application of Newton’s law of cooling at the outlet gives

qs = h [Ts(L) - Tmo ] (n)

solving for Ts(L)

Ts(L) = Tmo + h

qs = C5.338)CW/m(7.2

)W/m(590)C(120 o

o2

2o

(2) The value of h is within the range reported in Table 1.1 for forced convection of liquids.

Limiting check: If Tmi = Tmo, the required length should be zero. Setting Tmi = Tmo into (e) gives L = 0.

(5) Comments. (i) As long as the outlet is in the fully developed region, surface temperature at the outlet is determined entirely by the local heat transfer coefficient. (ii) In solving internal forced convection problems, it is important to establish if the flow is laminar or turbulent and if it is developing or fully developed.

PROBLEM 6.13

Consider fully developed laminar flow in two tubes having the same length. The flow rate, fluid,

inlet temperature and surface temperature are the same for both tubes. However, the diameter

of one tube is double that of the other. Determine the ratio of the heat transfer rate from the two

tubes.

(1) Observations. (i) This is an internal forced convection problem in tubes. (ii) The flow is laminar and fully developed. (iii) The surface is maintained at uniform temperature. (iv) All conditions are identical for two tubes except the diameter of one is twice that of the other. (v) The total heat transfer in each tube depends on the outlet temperature.

(2) Problem Definition. Compare the outlet temperatures of the two tubes.

(3) Solution Plan. Apply conservation of energy to obtain an equation for the heat transfer in each tube. Use the analysis of fully developed laminar flow in tubes at uniform surface temperature to determine the outlet temperatures.

(4) Plan Execution.

(i) Assumptions. (1) Steady state, (2) fully developed laminar flow, (3) axisymmetric flow, (4) constant properties, (5) uniform surface temperature, (6) negligible changes in kinetic and potential energy, (7) no axial conduction, (8) no dissipation and (9) no energy generation.

(ii) Analysis. Application of conservation of energy to the fluid between the inlet and outlet of tube, gives

q = mimop TTcm (a)

where

cp = specific heat, J/kg-oCm = mass flow rate, kg/s q = rate of heat transfer, W Tmi = inlet mean temperature, oCTmo = outlet mean temperature, oC

Applying (a) to the flow in the two tubes, noting that cp, m and Tmi are the same for both tubes, and taking the ratio of the two results

mimo

mimo

TT

TT

q

q

1

2

1

2 (b)

where the subscripts 1 and 2 refer to the small tube and large tube, respectively. Thus, the problem becomes one of determining the outlet temperatures. For flow through tubes with uniform surface temperature, conservation of energy and Newton’s law of cooling lead to equation (7.13)

][)()( exp xcm

hPTTTxT

psmism (c)

h = average heat transfer coefficient for a tube of length x, W/m2-oCm = mass flow rate, kg/s P = tube perimeter, m Tm(x) = mean temperature at x, oCTmi = mean inlet temperature, 35oC


Ts = surface temperature, oCx = distance from inlet of heated section, m

Applying (c) at the outlet where x = L and Tm(x) = Tmo and rearranging

)]/1)[( exp( pmismiom cmLhPTTTT (d)

where

L = tube length, m Tmo = mean outlet temperature, oC

Examination of (d) shows that all quantities are identical for both tubes except P and h .Applying (d) to the two tubes and taking the ratio of the resulting equations

)/exp(1

)/exp(1

11

22

1

2

p

p

mimo

mimo

cmLhP

cmLhP

TT

TT (e)

Thus, the two outlet temperatures will differ according to how the product of P h differs for the two tubes. The perimeter P is given by

P = D (f)

where D is tube diameter. For fully developed laminar flow with uniform surface temperature, the heat transfer coefficient is uniform along the tube, given by (7.57)

NuD = hD

k= 3.66 (g)

where

NuD = Nusselt number k = thermal conductivity of fluid, W/m-oC

The product P h can now be constructed from (f ) and (g)

P h = 3.66 Dk

D= 3.66 k (h)

Thus, the product P h is independent of tube size. It follows from (e) that the two outlet temperatures are identical

1)/66.3exp(1

)/66.3exp(1

1

2

p

p

mimo

mimo

cmkL

cmkL

TT

TT (i)

Substituting (i) into (b) gives

1/ 12 qq (j)

(iii) Checking. Dimensional check: The exponent in equation (d) should be dimensionless.

pcm

LhP ( )( / )( )

( / )( / ) /

m W m C m

kg s J kg C

W

J s

W

W

o

o

2

= 1

Limiting check: In the limit as L , the outlet temperature becomes equal to the surface

temperature regardless of tube size. Setting L = in (d) gives Tmo = Ts.

(5) Comments. The result is somewhat surprising. One would expect that increasing the diameter, increases the heat transfer rate. However, according to (g), the heat transfer coefficient is inversely proportional to diameter. On the other hand, the perimeter is


(6) directly proportional to diameter. These two effects cancel each other resulting in identical outlet temperatures regardless of tube size. This is true for the assumptions listed above and as long as all conditions are the same for both tubes.

PROBLEM 6.14

To evaluate the accuracy of scaling prediction of the thermal entrance length and Nusselt number, compare scaling estimates with the exact results of Graetz solution for flow through

tubes.

(1) Observations. (i) This is an internal forced convection problem. (ii) Equation (6.3) gives scaling estimate of the thermal entrance length. (iii) Equation (6.20b) gives scaling estimate of the local Nusselt number. (iv) The Graetz problem d eals with laminar flow in the entrance of a tube at uniform surface temperature. (v) Grae tz solutions gives the thermal entrance length (distance to reach fully developed temperature) and local Nusselt number.

(2) Problem Definition. Determine the thermal entrance length and Nusselt number using scaling and compare with Graetz results.

(3) Solution Plan. (i) Use Graetz solution (Table 6.4 or Fig. 6.9) to determine the distance from the entrance to the section where the Nusselt number is constant (fully developed temperature). Compare with scaling estimate, equation (6.3) (ii) Use Graetz so lution (Table 6.4) to determine the Nusselt number at various distances from the entrance. Compare with scaling estimate, equation (6.20b).

(4) Plan Execution.

(i) Assumptions. (1) Steady state, (2) constant properties, (3) uniform surface temperature, (4) negligible axial conduction (5) negligible changes in kinetic and potential energy and (6) negligible dissipation

(ii) Analysis. Equation (6.3) gives scaling estimate of the thermal entrance length tL

1~/

2/1

rPRe

DL

D

t (a)

where

D = diameter

Pr = Prandtl number


Scaling estimate of the Nusselt number is given by equation (6.20b)

1~2/1

x/D

RerP

uN D (b)

Graetz solution for the variation of the local Nusse lt number with distance from the entrance is presented in Table 6.4. Since the Nusselt is constant in the fully developed region, Graetz solution can also be used to determine the entrance length. The fully developed Nusselt number is

66.3DNu

Table 6.4 gives the dimensionless distance corresponding to 66.3DNu as


1.0//

PrRe

DL

PrRe

Dx

D

t

D

(c)

To compare with scaling result (a), equation (c) is rewritten in the same form as (a)

316.01.0/

2/1

rPRe

DL

D

t (d)

Thus the Graetz solution constant 0.316 is replaced by unity in scaling.

Graetz solution, Table 6.4, shows that the Nusse lt number depends on the dimensionless axial distance , defined in (c). Rewriting scaling result (b) in terms of , gives

1~DuN (e)

To facilitate comparison of scaling estimate (e) with Graetz

solution, Table 6.4 is modified to include .DuN

Examination the result shown in Table 6.4a shows that

exact values of .DuN range from 0.286 to 1.157. Scaling

predicts these constants to be unity, as shown in (e).

(iv) Checking. Dimensional check: All equations are dimensionless.

(5) Comments. (i) To compare scaling estimate with exact solution for the Nusselt number, it is necessary to cast both results in the same form. (ii) Scaling estimate of the Nusselt number is surprisingly good.

Table 6.4a Local Nusselt number for tube at

uniform surface temperature

=PrRe

Dx

D

/ )(Nu )(Nu

0.0005 12.8 0.286

0.002 8.03 0.359

0.005 6.00 0.424

0.02 4.17 0.590

0.04 3.77 0.754

0.05 3.71 0.830

0.1 3.66 1.157

PROBLEM 6.15

Use scaling to estimate the heat transfer coefficient for plasma at a distance of 9 cm from the

entrance of a vessel. The mean plasma velocity is 0.042 m/s and the diameter is 2.2 mm.

Properties of plasma are:

CJ/kg3900 opc , CW/m5.0 ok , /sm1094.0 26 , 3kg/m1040

(1) Observations. (i) This is an internal forced convection problem. (ii) Equation (6.20b) gives scaling estimate of the local Nusselt number. (iii) The Graetz problem deals with laminar flow in the entrance of a tube at uniform surface temperature.

(2) Problem Definition. Determine the Nusselt number using scaling.

(3) Solution Plan. Use (6.20b) to estimate the Nusselt number in the entrance region of a tube.

(4) Plan Execution.

(i) Assumptions. (1) Steady state, (2) constant properties and (3) uniform surface temperature.

(ii) Analysis. Scaling estimate of the Nusselt number is given by equation (6.20b)

2/1

~x/D

RerPuN D

D (a)

where

D = diameter = 2 mm = 0.002 m

Pr = Prandtl number



DuReD (b)

where

u mean velocity = 0.042 m/s

The Prandtl number is given by

k

c

k

cPr

pp (c)

where

pc = specific heat = CJ/kg3900 o

k thermal conductivity = CW/m5.0 o

kinematic viscosity = /sm1094.0 26

= density = 3kg/m1040

The Nusselt number is defined as


k

hDNuD (d)

h heat transfer coefficient

Substitute (d) into (a) and solve for h

2/1

~x/D

RerP

D

kh D (e)

(iii) Computations. Use (b) and (c) to compute the Reynolds and Prandtl numbers

3.98/sm1094.0

)m(0022.0)m/s(024.026DRe

CW/m5.0

/s)m(1094.0)kg/m(1040C)J/kg(3900o

o 263

Pr = 7.63

Substitute into (e)

C-W/m9730022(m)0.09(m)/0.

98.37.63

0.0022(m)

C)0.5(W/m-~ o2

2/1o

h

The exact solution to this problem is given in Table 6.4 and Fig. 6.9. The local Nusselt is given in terms of the dimensionless distance , defined as

PrRe

Dx

D

/ (f)

Computing

98.37.63

0022(m)0.09(m)/0.= 0.05454

At this value of , Table 6.4 gives

7.3DNu


C-W/m8410.0022(m)

C)0.5(W/m-3.7 o2

o

h

(iv) Checking. Dimensional check: (1) Each term in (e) has units of heat transfer coefficient. (2) Equations (a)-(d) are dimensionless.

(5) Comments. Scaling estimate of the heat transfer coefficient is surprisingly good.

PROBLEM 6.16

Air flows with fully developed velocity through a tube of inside diameter 2.0 cm. The flow is fully

developed with a mean velocity of 1.2 m/s. The surface is maintained at a uniform temperature

of 90oC. Inlet temperature is uniform equal 30

oC. Determine the length of tube needed to

increase the mean temperature to 70oC.

(1) Observations. (i) This is an internal forced convection problem in a tube. (ii) The velocity is fully developed. (iii) The temperature is developing. (iv) Surface is maintained at uniform temperature. (v) The Reynolds number should be computed to establish if flow is laminar or turbulent. (vi) Tube length is unknown. (vii) The determination of tube length requires determining the heat transfer coefficient.

(2) Problem Definition. Find the required tube length to increase the air temperature to a specified level. This reduces to determining the heat transfer coefficient.

(3) Solution Plan. Compute the Reynolds number to establish that the flow is laminar. Use the results of Section 6.5 on flow through tubes at uniform surface temperature. Use Graetz solution for fully developed laminar flow and developing temperature in tubes to determine the heat transfer coefficient.

(4) Plan Execution.

(i) Assumptions. (1) Steady state, (2) fully developed velocity, (3) axisymmetric flow, (4) constant properties, (5) uniform surface temperature, (6) negligible changes in kinetic and potential energy, (7) negligible axial conduction, (8) negligible dissipation and (9) no energy generation.

(ii) Analysis.. For flow in a tube at uniform surface temperature, conservation of energy and Newton's law of cooling lead to equation (6.13)

][)()( exp xcm

hPTTTxT

psmism (6.13)


h = average heat transfer coefficient for a tube of length L, W/m2-oCL = length of tube, m m = mass flow rate, kg/s

P = tube perimeter, m Tm(x) = mean temperature at x, oC


sT = surface temperature = 90oC


Applying (6.13) at the outlet (x = L) and solving for L

tL

hL

m

L

moT

miT

sT

uu


mos

misp

TT

TT

hP

cmL ln (a)

where

moT = mean outlet temperature = 70oC

To compute L using (b), it is necessary to determine cp, P, m , and h . All properties are

determined at the mean temperature mT defined as

mT = T Tmi mo

2 (b)


P = D (c) and

uD

m4

2

(d)

where

D = inside tube diameter = 2 cm = 0.02 m u = mean flow velocity = 1.2 m/s

= density, kg/m3

The average heat transfer coefficient, h , for fully developed velocity and developing temperature is given in Graetz solution, Section 6. 8 (Table 6.4 and Fig. 6.9). To proceed, it is necessary to calculate the Reynolds number to determine if the flow is laminar or turbulent. For flow in a tube the Reynolds number is defined as

ReD

u D (e)

where



The mean temperature is calculated in order that properties are determined. Substituting into (b)

mT = C502

)C)(7030( oo


cp = 1007.4 J/kg-oCk = 0.02781 W/m-oCPr = 0.709

= 17.92 10-6, m2/s

= 1.0924 kg/m3


ReD

12 0 02

17 92 101339 36 2

. ( / ) . ( )

. ( / ).

m s m

m s


Since the Reynolds number is smaller than 2300, the flow is laminar. Attention is focus on the

determination of h using either Table 6.4 or Fig. 6.9. However, h depends on the length L

which is unknown. Thus the problem is solved by trial and error. Assume L, use Table 6.4 or Fig.

6.9 to determine h and substitute into (a) to calculate L. If the calculated L is not the same as the

assumed value, the procedure is repeated until a satisfactory agreement between assume and calculated L is obtained.

Table 6.4 gives the average Nusselt number, )(Nu , as a function of dimensionless axial distance

. These are defied as

k

xhNu )( (f)

The variable is defined in (6.21) as

DPrRe

Dx / (g)

Solving (f) for h

D

kNuh )( (h)

(iii) Computations. Substituting into (d) and (e)

P = 0.02(m) = 0.06283 m

kg/s0.00041181.2(m/s))m1.0924(kg/4

)(m(0.02) 322

m

The result of the trial and error procedure described above is:

Assume: x = L = 1.15 m. Substitute into (g)

0606.03.1339709.0

)m)/0.02(m(14.1

At this value of Table 6.4 gives

536.4)(Nu

Substitute into (h)

C)W/m307.6536.40.02(m)

C)m0.02781(W/ o2o

h

Substitute into (a)

m15.1C)70)((90

C)30)((90ln

C)(W/m307.60.06283(m)

)C4(J/kgkg/s)1007.0.0004118(o

o

o2

o

L

Thus the calculated value of L is the same as the assumed value.

(iv) Checking. Dimensional check: Computations showed that equations (a), (b), (c), (e), (g) and (h) are dimensionally consistent.


Limiting check: For the special case of mimo TT , the required length should vanish. Setting

mimo TT in (a) gives x = L = 0.

Quantitative checks: The value of h appears to be low compared with typical values listed in Table 1.1 for forced convection of gases. Howe ver, it should be remembered that values of h in Table 1.1 are for typical applications. Exceptions should be expected.

(5) Comments. (i) The thermal entrance length tL is determined using equation (6.6)

Dtt PrReC

D

L (6.6)

Table 6.1 gives 033.0tC . Substituting into (6.6) gives 627.0tL m. Since this is not small

compared to L , entrance length must be taken into consideration in solving this problem.

(ii) If entrance effects are neglected and temperature is assumed full developed, the corresponding Nusselt number will be 3.66. Substituting this value in (a) gives L = 1.425 m. This is 24% larger than the more accu rate result of entrance length analysis.

PROBLEM 6.17

Air flows with a mean velocity of 2 m/s through a tube of diameter 1.0 cm and length 14 cm. The

velocity is fully developed throughout. The mean temperature at the inlet is 35oC. The surface of

the tube is maintained at a uniform temperature of 130oC. Determine the outlet temperature.

(1) Observations. (i) This is an internal forced convection problem in a tube. (ii) The velocity is fully developed. (iii) The temperature is developing. (iv) Surface is maintained at uniform temperature. (v) The Reynolds number should be computed to establish if flow is laminar or turbulent. (vi) Outlet mean temperature is unknown. (vii) The determination of outlet temperature requires determining the heat transfer coefficient. (viii) Since outlet temperature is unknown, air properties can not be determined. Thus a trial and error procedure is needed to solve the problem.

(2) Problem Definition. Find the outlet temperature of air heated in tube at uniform surface temperature. This reduces to determining the heat transfer coefficient.

(3) Solution Plan. Compute the Reynolds number to establish that the flow is laminar. Use the results of Section 6.5 on flow through tubes at uniform surface temperature. Use Graetz solution for fully developed laminar flow and developing temperature in tubes to determine the heat transfer coefficient.

(4) Plan Execution.

(i) Assumptions. (1) Steady state, (2) fully developed velocity, (3) axisymmetric flow, (4) constant properties, (5) uniform surface temperature, (6) negligible changes in kinetic and potential energy, (7) negligible axial conduction, (8) negligible dissipation and (9) no energy generation.


][)()( exp xcm

hPTTTxT

psmism (6.13)


h = average heat transfer coefficient for a tube of length L, W/m2-oCm = mass flow rate, kg/s

P = tube perimeter, m Tm(x) = mean temperature at x, oC




Applying (6.13) at the outlet (x = L) and solving for moT

tL

hL

m

L

moT

miT

sT

uu


p

missmocm

hLPTTTT exp)( (a)

where

L = length of tube = 0.14 m

moT mean outlet temperature, oC

To compute moT using (b), it is necessary to determine cp, P, m , and h . All properties are


mT = T Tmi mo

2 (b)


P = D (c) and

uD

m4

2

(d)

where

D = inside tube diameter = 1 cm = 0.01 m u = mean flow velocity = 2 m/s

= density, kg/m3

The average heat transfer coefficient, h , for fully developed velocity and developing

temperature is given in Graetz solution, Section 6.8 (Table 6.4 and Fig. 6.9). However, this solution is valid for laminar flow. Thus, to proceed with the analysis, it is necessary to calculate the Reynolds number to determine if the flow is laminar or turbulent. For flow in a tube the Reynolds number is defined as

ReD

u D (e)

where



Since mean outlet temperature moT is unknown, properties can not be determined using (b). A

trial an error procedure is required in which a value for moT is assumed, properties determined

using the assumed value and (a) is used to calculate moT . If the calculated moT is equal to the

assumed value, the procedure is repeated until a satisfactory agreement is obtained.

Assume C65omoT . (b) gives

mT = C502

)C)(6535( oo


cp = 1007.4 J/kg-oCk = 0.02781 W/m-oC


Pr = 0.709

= 17.92 10-6, m2/s

= 1.0924 kg/m3


ReD 1.1116/s)(m1017.92

(m)2(m/s)0.0126

Since the Reynolds number is smaller than 2300, the flow is laminar. Attention is now focus on

the determination of h using either Table 6.4 or Fig. 6.9. Table 6.4 gives the average Nusselt

number, )(Nu , as a function of dimensionless axial distance . These are defied as

k

xhNu )( (f)

The variable is defined in (6.21) as

DPrRe

Dx / (g)

Solving (f) for h

D

kNuh )( (h)


P = 0.01(m) = 0.03141 m

kg/s0001716.02(m/s))m1.0924(kg/4

)(m(0.01) 322

m

To determine h , use (g) and Table 6.4

01769.007.1116709.0

)m)/0.01(m(14.0

At this value of Table 6.4 gives

95.5)(Nu

Substitute into (h)

C)W/m55.1695.50.01(m)

C)m0.02781(W/ o2o

h

Substitute into (a)

C6.67)CJ/kg(4.1007)kg/s(0001716.0

)CW/m(55.16)m(03141.0)m(14.0exp)35130()C(130 o

o

o2o

moT

This is close to the assume value. Thus C6.67 omoT .


(iv) Checking. Dimensional check: Computations showed that equations (a), (b), (c), (e), (g) and (h) are dimensionally consistent.

Limiting check: For the special case of mis TT , the outlet temperature should be the same as

the inlet. Setting mis TT in (a) gives mio TT .

Quantitative checks: The value of h appears to be low compared with typical values listed in Table 1.1 for forced convection of gases. Howe ver, it should be remembered that values of h in Table 1.1 are for typical applications. Exceptions should be expected.

(5) Comments. (i) The thermal entrance length tL is determined using equation (6.6)

Dtt PrReC

D

L (6.6)

Table 6.1 gives 033.0tC . Substituting into (6.6) gives 261.0tL m. Since this is not small

compared to L , entrance length must be taken into consideration in solving this problem. In fact the thermal boundary layer is developing throughout the tube.

(ii) Because outlet temperature is unknown propertie s can not be determined a priori. Thus trial and error procedure is needed to solve the problem.

PROBLEM 6.18

A research apparatus for a pharmaceutical laboratory requires heating plasma in a tube 0.5 cm

in diameter. The tube is heated by uniformly wrapping an electric element over its surface. This

arrangement provides uniform surface heat flux. The plasma is monitored in a 15 cm long

section. The mean inlet temperature to this

section is C18o and the mean velocity is 0.025

m/s. The maximum allowable temperature is

C.42o You are asked to provide the designer of

the apparatus with the outlet temperature and

required power corresponding to the maximum

temperature. Properties of plasma are:

CJ/kg3900 opc , CW/m5.0 ok , /sm1094.0 26 , 3kg/m1040

(1) Observations. (i) This is an internal forced convection problem in a tube. (ii) The velocity is fully developed and the temperature is developing. (iii) The surface is heated with uniform flux. (iv) The Reynolds number should be computed to establish if the flow is laminar or turbulent. (v) Compute thermal entrance length to determine if it can be neglected. (vi) Surface temperature varies with distance from entrance. It is maximum at the outlet. Thus surface temperature at the outlet is known. (vii) Analysis of uniformly heated channels gives a relationship between local surface temperature, heat flux and heat transfer coefficient. (viii) The local heat transfer coefficient varies with distance form the inlet. (ix) Knowing surface heat flux, the required power can be determined. (x) Newton’s law of cooling applied at the outlet gives outlet temperature.

(2) Problem Definition. Determine the local heat transfer coefficient at the outlet ).(Lh

(3) Solution Plan. Apply channel flow heat transfer analysis for uniform surface flux to determine surface heat flux. Compute the Reynolds number to establish if the flow is laminar or turbulent. Compute entrance length to determine if it can be neglected. Select an applicable equation for determining the Nusselt number at the outlet. Apply Newton’s law of cooling at the outlet to determine outlet temperature.

(4) Plan Execution.

(i) Assumptions. (1) Steady state, (2) constant properties, (e) uniform surface heat flux, (4) negligible changes in kinetic and potential energy, (5) negligible axial conduction and (6) negligible dissipation.

(ii) Analysis. Equation (6.10) gives the result of heat transfer analysis for channel flow with uniform surface heat

)(

1)(

xhcm

PxqTxT

psmis (6.10)

where

pc = specific heat, CJ/kg3900 o


+

-L

miT

moT

uDsectiontest




miT mean inlet temperature = C18o

sq = surface heat flux, 2W/m

)(xTs = local surface temperature, Co


Apply (6.10) at the outlet, x = L, and solve for sq

1

)(

1)(

Lhcm

PLTLTq

pmiss (a)

where

L = 0.15 m

42)(LTs Co

The perimeter P is DP (b)

where

D = tube diameter = 0.005 cm

The flow rate is given by

uD

m4

2

(c)

where

u mean velocity = 0.025m/s

density = 3kg/m1040

The heat transfer coefficient at the outlet, h(L). This requires establishing whether the flow is laminar or turbulent. Thus, the Reynolds number should be computed

DuReD (d)

= kinematic viscosity /sm1094.0 26

The Peclet number, Pe, is computed to determine if axial conduction can be neglected

DPrRePe (e)

The Prandtl number, Pr, is given by

k

c

k

cPr

pp (f)

where is viscosity.

The thermal entrance length tL is determined using equation (6.6)

Dtt PrReC

D

L (6.6)


where tC is a constant given in Table 6.1. For tubes at uniform surface heat flux, 033.0tC .

To determine the required power, the surface flux is multiplied by the total surface are

sqDLPower (g)

To determine outlet temperature, apply Newton’s law of cooling at the outlet

moss TTLhq )(

Solve for moT

)(Lh

qTT s

smo (h)

Substitute into (c)

DRe 133/s)(m1094.0

0.005(m)0.025(m/s)26

Thus the flow is laminar. Substitute into (f)

rP 625.7C)-(W/m5.0

/s)(m1094.0)mC)1040(kg/3900(J/kg-o2

263o

Substitute into (e)

1014133625.7Pe

Thus axial conduction can be neglected.

Equation (6.6) is used to compute tL

m166.0132625.7)m(005.0033.0tL

Therefore, the thermal boundary layer is still developing at the outlet. It follows that entrance effects are important and that the heat transfer coefficient at the outlet should be obtained from Fig. 6.11. For laminar flow in the entrance region of a tube at fully developed velocity profile and uniform heat flux Fig. 6.11 gives the local Nusselt numbers Nu(x) as a function of dimensionless axial distance , defined as

rPRe

Dx

D

/ (i)

The local heat transfer coefficient is given by

)()( xNuD

kxh (j)

(iii) Computation. Compute at x = L

0296.0625.7133

)m(005.0/)m(15.0

At 0296.0 , Fig. 6.11 gives 5)(LNuD . Equation (j) gives


CW/m50050.005(m)

C)0.5(W/m)( o

o2Lh

Equation (b) is used to compute P

m01571.0)m)(005.0(P

Equation (c) gives the flow rate

kg/s0.0005105m/s)(025.0)kg/m(10404

)m()005.0( 32

m

Substitute into (a)

2

2W/m7540

CW/m500

1

)CJ/kg-(3900)kg/s(0005105.0

)m(15.0)m(0157.0)C(18)C(42

1

oooo

sq

Substitute into (g)

W77.17)W/m(7540)m(15.0)m(005.0 2Power

Substitute into (h)

C9.26)CW/m(500

)W/m(7540)C(42 o

o2

2o

moT

(iv) Checking. Dimensional check: (i) Computations showed that equations (a), (b), (c), (e), (g), (h) and (j) are dimensionally consistent. (ii) Equations (d), (f) and (i) are dimensionless.

Limiting checks: For the special case of mis TT , the required surface flux should vanish and

.mimo TT Setting mis TT in (a) gives .0sq When this result is substituted into (g) gives

.mimo TT

Quantitative checks: (i) The value of h within the range listed in Table 1.1 for forced convection of liquids.

Global energy balance: energy added at the surface (power) should be equal to energy change of mass flow rate

)( mimops TTmcqDL

or

1)( mimop

s

TTmc

qDL

The above gives

0026.1C))(189.26)(CJ/kg-(3900)kg/s(0005105.0

)W/m(7540)m(15.0)m(005.0oo

2

(5) Comments. (i) Using Fig. 6.1 to determine h introduces a small error. (ii) If entrance effects are neglected and the temperature is assumed fully developed at the outlet, the corresponding


Nusselt number is 4.364. Using this value gives C,W/m4.436 o2h ,W/m6978 2sq power =

16.45 W and C.28 omoT

(iii) In solving internal forced convection problems, it is important to establish if the flow is laminar or turbulent and if it is developing or fully developed.

PROBLEM 6.19

An experiment is designed to investigate heat transfer in rectangular ducts at uniform surface

temperature. One method for providing heating at uniform surface temperature is based on

wrapping a set of electric elements around the surface. Power supply to each element is

individually adjusted to provide uniform surface temperature. This experiment uses air flowing

in a cm8cm4 rectangular duct 32 cm

long. The air is to be heated from C22o

to C.98o The velocity is fully developed

with a mean value of 0.15 m/s. Your task

is to provide the designer of the

experiment with the heat flux distribution

along the surface. This data is needed to

determine the power supplied to the

individual elements.

(1) Observations. (i) This is an internal forced convection problem in a rectangular channel. (ii) The velocity is fully developed and the temperature is developing. (iii) The surface is maintained at uniform temperature. (iv) The Reynolds number should be computed to establish if the flow is laminar or turbulent. (v) Compute entrance lengths to determine if they can be neglected. (vi) Surface flux varies with distance from entrance. It is minimum at outlet. (vii) Newton’s law gives surface flux in terms of the local heat transfer coefficient )(xh and the local mean

temperature )(xTm . (viii) The local and average heat transfer coefficient decrease with distance

form the inlet. (ix) The local mean temperature depends on the local average heat transfer

coefficient ).(xh (x) Surface temperature is unknown.

(2) Problem Definition. Determine the local and average heat transfer coefficient temperature to

a specified level. This requires determining the local heat transfer coefficients )(xh and )(xh and

surface temperature.

(3) Solution Plan. Apply Newton’s law of cooling. Use the analysis of flow in channels at

uniform surface temperature to determine the local mean temperature ).(xTm Compute the

Reynolds number to establish if the flow is laminar or turbulent. Compute entrance lengths to determine if entrance or fully developed analysis is required. If the thermal entrance can be neglected, use fully developed Nusselt number results. If entrance region is significant, use Graetz solution to determine the local a nd average heat transfer coefficients.

(4) Plan Execution.

(i) Assumptions. (1) Steady state, (2) constant properties, (e) uniform surface temperature, (4) negligible changes in kinetic and potential energy, (5) negligible axial conduction and (6) negligible dissipation.


)()()( xTTxhxq mss (a)

where


+

L

miT mo

T

u

+ + + +

---- - -a

b


)(xqs = local surface heat flux, 2W/m

)(xTm = local mean temperature, Co


For flow through a channels at uniform surface temperature, conservation of energy and Newton's law of cooling lead to equation (6.13)

][)()( exp xcm

hPTTTxT

psmism (6.13)

where


h = average heat transfer coefficient for a channel of length x, CW/m o2


P = duct perimeter, m

miT mean inlet temperature C22o



])(

[))(()( exp xcm

xhPTTxhxq

pmiss (b)

Thus we need to determine: ,pc P, m, ,sT )(xh and ).(xh Properties are determined at the mean

temperature mT , defined as

mT = 2

momi TT (c)

Surface temperature is determined by applying (6.13) at the outlet (x = L) where mom TLT )( ,

and solving for sT

)/exp()/exp(1

1pmomi

ps mcLhPTT

mcLhPT (d)

where

L = channel length = 32 cm = 0.32 m

moT = outlet temperature = C98o

The perimeter P is )(2 baP (e)

where

a = channel height = 8 cm = 0.08 m b = channel width = 4 cm = 0.04 m

x

sT

0umiT

t

a

b

sq


Mass flow rate is given by uabm )( (f)

where

= density, 3kg/m


The determination of )(xh and )(xh requires computing the Reynolds number to establish if the

flow is laminar or turbulent and computing the thermal entrance lengths to determine if it is important. The Reynolds number is

eeD

DuRe (g)

where



The equivalent diameter is defined as

)(

2

)(244

ba

ab

ba

ab

P

AD

fe (h)

where fA is flow area. To proceed with the analysis the Reynolds number must be computed

first. Properties are determined at mT

mT = C602

)C)(9822( oo


pc = 1008 J/kg- oC

k = 0.02852 W/m-oC

Pr = 0.708

= 18.9 10-6

m2/s

= 1.0596 kg/m3

Equation (h) gives

2m05333.0)m)(04.008.0(

)m)(04.0)m)(08.0(2eD


2DRe 28.423/s)(m109.81

.053333(m)0.15(m/s)026

Since the Reynolds number is less than 2300, the flow is laminar. The next step is to compute

the thermal entrance length tL . For laminar flow through channels equation (6.6) gives

eDett PrReDCL (6.6)


tC = thermal entrance length coefficient, uniform surface temperature (Table 6.1) = 0.049

Substituting numerical values into (6.6)

tL = 0.049 0.053333 (m) 423.28 0.708 = 0.783 m

Since tL is larger than channel length L, it follows that entrance effects must be taken into

consideration in determining )(xh and ).(xh For laminar flow in the entrance region of a tube at

fully developed velocity profile and uniform surface temperature, Graetz solution gives )(xh and

).(xh Fig. 6.9 and Table 6.4 give the average the local and average Nusselt numbers Nu(x) and

Nu as a function of dimensionless axial distance , defined as

rPRe

Dx

eD

e/ (i)

The average heat transfer coefficient )(xh is given by

)()( xNuD

kxh (j)

Similarly, the local heat transfer coefficient is given by

)()( xNuD

kxh (k)

(iii) Computation. Surface temperature is determined using (d). This requires computing

).(Lh Thus we compute at x = L

02002.0708.028.423

)m(053333.0/)m(32.0

At 02.0 , Table 6.4 gives 81.5Nu


CW/m11.381.5)0.053333(m

C)m0.02852(W/)( o

o2Lh

Equation (c) is used to compute P

m24.0)m)(04.0()m)(08.0(2P

Equation (d) gives the flow rate

kg/s0.0005086m/s)(15.0m))(04.0(m))(08.0()kg/m(0596.1 3m

Before substituting into (d), the exponent of the exponential is calculated

4659.0C)J/kgkg/s)1008(0.0005086(

C)0.32(m)(W/m)11.3(m)(24.0o

o2

pmc

LhP



C07.2264659.0exp()C(98)C(22)4659.0exp(1

1 ooosT

Equation (b) is used to determine the heat flux variation with x. The procedure is to select a value of x, use (i) to compute the corresponding , use Table 6.4 or Fig. 6.9 to determine )(xh and

).(xh Substituting in (b) gives the local heat flux. For example, to determine the heat flux at

,Lx we apply (b) at Lx

])(

[))(()( exp Lcm

LhPTTLhLq

pmiss (l)

Table 6.4 gives the local Nusselt number Nu(L) at x = L ( 02.0 ) as 17.4Nu . Thus

CW/m23.217.4)0.053333(m

C)m0.02852(W/)( o

o2Lh

Substituting into (l)

22 W/m6.285)4659.0)C)(2207.226)(CW/m(23.2)( (expooLqs

The flux at other locations x along the channel is given in the following table.

(iv) Checking. Dimensional check: (i) Computations showed that equations (a)-(k) are dimensionally consistent. (ii) The Reynolds number and the exponent of the exponential are dimensionless.

Limiting checks: For the special case of smmi TxTT )( , the required surface flux should

vanish. Setting sm TxT )( in (a) gives .0sq

Qualitative check: As anticipated, the local and average heat transfer coefficients and surface heat flux decrease with distance from the inlet.

Quantitative checks: (i) The value of h is outside the range listed in Table 1.1 for forced convection of gases. Examination of equation (j) of (k) shows that the heat transfer coefficient is inversely proportional to the diameter. Thus, as diameter increases the heat transfer coefficient decreases.

x(m)

Nu(x) h(x)

( CW/m o2 )

)(xNu )(xh

( CW/m o2 )

)(xqs

2W/m 0 0

0.008 0.0005 12.8 6.85 19.29 10.32 1344.9

0.016 0.001 10 5.35 16 8.56 1024

0.032 0.002 8.03 4.29 12.09 6.47 794.6

0.08 0.005 6.0 3.21 8.92 4.77 547.9

0.12 0.0075 5.5 2.94 8 4.28 471.9

0.16 0.01 5 2.67 7 3.74 411.8

0.32 0.02 4.17 2.23 5.81 3.11 285.6


(5) Comments. (i) Using Fig. 6.9 to determine h and h introduces a small error. (ii) If entrance effects are neglected and the flow is assumed fully developed throughout, the corresponding

Nusselt number is 3.66. Using this value gives C.W/m96.1 o2h

PROBLEM 7.1

Explain why

(a) t can not be larger than .

(b) can be larger than .t

Solution

[a] The driving force in free convection is buoyancy. Thus, wherever the local temperature is different from the ambient temperature, the

fluid will move. This means that t can not be

larger than .

[b] Fluid inside the thermal boundary layer moves due to buoyancy force. Because of viscous forces this moving fluid drags fluid layers outside the thermal boundary layer and cause it to

move. That is, fluid motion outside t is due to viscous force and not buoyancy. Consequently,

can be larger than t .

PROBLEM 7.2

A vertical plate 6.5 cm high and 30 cm wide is maintained at

82 Co . The plate is immersed in water at 18 Co . Determine:

(a) The viscous boundary layer thickness.(b) The thermal boundary layer thickness at the trailing end of

the plate.(c) The average heat transfer coefficient.

(d) Total heat added to water.

(1) Observations. (i) This is an external free convection problem over a vertical plate. (ii) The Rayleigh number should be computed to determine if the flow is laminar or turbulent. (iii) The solution for laminar flow is given in Section 7.4. (iv) For laminar flow, Fig.7.2 gives the viscous

boundary layer thickness and Fig. 7.3 gives the thermal boundary layer thickness t . (v)

Newton’s law of cooling gives the heat transfer rate. (vi) Equation (7.23) gives the average heat

transfer coefficient h . (vii) The fluid is water.

(2) Problem Definition. Determine flow and heat transfer characteristics for free convection over a vertical plate at uniform surface temperature.

(3) Solution Plan. Compute the Rayleigh number to determine if the flow is laminar. For

laminar flow use the result of Section 7.4 to determine , t and .h Apply Newton's law of

cooling to determine the total heat transfer .Tq

(4) Plan Execution.

(i) Assumptions. (1) Continuum, (2) Newtonian, (3) steady state, (4) two-dimensional, (5) constant properties (except in buoyancy), (6) boundary layer flow, (7) laminar flow, (8) uniform surface temperature, (9) negligible radiation and (10) quiescent fluid.

(ii) Analysis. The Rayleigh number RaL is calculated first to determine the appropriate

correlation equation for h . The Rayleigh number is defined as

RaL= PrLTTg s

2

3

(a)

where

g = gravitational acceleration = 9.81 m/s2

L = plate dimension in the direction of gravity = 6.5 cm = 0.065 m Pr = Prandtl number RaL = Rayleigh number

= coefficient of thermal expansion, 1/K

v = kinematic viscosity, m2/s

Properties of air are determined at the film temperature Tf defined as

Tf=

T Ts

2 (b)

g

T

W

L

sT


Ts

= surface temperature = 82oC

T = ambient air temperature = 18oC

Viscous boundary layer thickness is determined using Fig. 7.2. This requires computing the

following:

x

yGrx4/1

4 (c)

and

2

3xTTgGr s

x (d)

Thermal boundary layer thickness t is determined using Fig. 7.3. The average heat transfer

coefficient is given by equation (7.27)

d

dGr

L

kh L )0(

43

44/1

(7.27)



Temperature gradient dd /)0( is given in Table 7.1. Newton’s law of cooling gives the total

heat transfer rate

Tq = h A (Ts - T ) (e)

where

A = surface area of the two vertical sides, m2

Tq = heat transfer from the surface to the ambient air, W

Surface area of the two vertical sides is given by

A = 2 LW (f)W = plate width = 30 cm = 0.3 m

(iii) Computations. Properties of air are determined at the film temperature Tf defined as

Tf=

T Ts

2 =

2

)C)(1882( o

= 50oC

Water properties at this temperature

k = 0.64056 W/m-oC Pr = 3.57

310462.0 1/K6105537.0 m2/s


RaL = 57.3)/s(m)10(0.5537

)(mC)(0.065)18)()(82m/s(1/K)9.81(100.4622426

33o23

= 0.927577 910


At Pr = 3.57, Fig. 7.2 gives

54

4/1

L

GrL (g)

Use (d) to compute LGr at the trailing end x = L

9

2426

33o23

10259826.0)/s(m)10(0.5537

)(mC)(0.065)18)()(82m/s(1/K)9.81(100.462LGr


)m(065.04

10259826.05

4/19

Solving the above for

mm62.3m1062.3 3

Fig. 7.3 gives the thermal boundary layer thickness t . At Pr = 3.57, Fig. 7.3 gives

)m(065.04

10259826.05.2

4/19

tt

Solving for mm1.81m1081.1 3t

At Pr = 3.57, Table 7.1 gives the gradient dd /)0(

d

d )0(0.86


Cm

W10140.86

4

100.25982

0.065(m)

C)0.6405(W/m

3

4o2

1/49o

h


Tq = W1265C)18)(0.3(m)(82C)0.065(m)14(W/m10 oo2

(iv) Checking. Dimensional check: Computations showed that equations (a)-(g) are dimensionally consistent.

Quantitative check: The magnitude of h is within the range given in Table 1.1 for free convection of gases.

Qualitative check: Increasing surface temperature Ts should increase the heat transfer rate.

According to equation (e), Tq is directly proportional to h and Ts. According to (7.27), an

increase in sT brings about an increase in .h

(5) Comments. The computed values of and t are approximate since the corresponding

values of cannot be read accurately from Fig. 7.2 and Fig. 7.3.

PROBLEM 7.3

Use Fig. 7.3 to determine dd /)0( for Pr = 0.01 and 100. Compare your result with the value

given in table 7.1.

(1) Observations. (i) This is an external free convection problem for flow over a vertical plate. (ii) Laminar flow solution for temperature distribution for a plate at uniform surface temperature is given in Fig. 7.3 . (iii) The dimensionless temperature gradient at the surface is given in Table 7.1. (iv) The solution depends on the Prandtl number.

(2) Problem Definition. Determine the normal temperature gradient at the surface, ,/)0( dd

for laminar boundary layer flow over a vertical plate.

(3) Solution Plan. Use the temperature solution presented graphically in Fig. 7.3 and compare with the exact value listed in Table 7.1.

(4) Plan Execution.

(i) Assumptions (a)

(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3) Boussinesq approximations, (4)

two-dimensional, (5) laminar flow ( 910xRa ),

(6) vertical flat plate (7) uniform surface temperature, (8) no dissipation and (9) no radiation.

(ii) Analysis. Fig. 7.3 is a plot of the dimensionless temperature vs. the similarity

variable . Temperature gradient dd /)0(

can be determined from this figure by graphically evaluating the slope is the slope at

.0 This slope is expressed as

0

)0(

d

d (a)

(iii) Computations. For Pr = 0.01, equation (a) and Fig. 7.3 give

076.06

545.01)0(

0d

d

Table 7.1 gives 0806.0)0(

d

d.

For Pr = 100, equation (a) and Fig. 7.3 give


04.249.0

1)0(

0d

d

Table 7.1 gives 191.2)0(

d

d.

(iv) Checking. Quantitative check: Numerical results obtained using Fig. 7.3 are in good agreement with the exact solutions of Table 7.1.

Comments. Using Fig. 7.3 to determine dd /)0( has an inherent error associated with reading

its scale. Nevertheless, for Prandtl numbers 0.01 and 100 the error is less than 7%.

PROBLEM 7.4

In designing an air conditioning system for a pizza restaurant an estimate of the heat added to

the kitchen from the door of the pizza oven is needed. The rectangular door is

cm201cm50 with its short side along the vertical direction. Door surface temperature is

110oC. Estimate the heat loss from the door if the ambient air temperature is 20oC.

(1) Observations. (i) This is a free convection problem. (ii) Heat is lost from the door to the surroundings by free convection and radiation. (iii) To determine the rate of heat loss, the door can by modeled as a vertical plate losing heat by free convection to an ambient air. (iv) As a first approximation, radiation can be neglected. (v) Newton’s law of cooling gives the rate of heat transfer. (vi) The Rayleigh number should be computed to determine if the flow is laminar or turbulent. (vii) For laminar flow the solution of Section 7.4 is applicable.

(2) Problem Definition. Determine the average heat transfer coefficient for free convection from a vertical plate.

(3) Solution Plan. Apply Newton's law of cooling to the door. For laminar flow use results of Section 7.4.

(4) Plan Execution.

(i) Assumptions. (1) steady state, (2) two-dimensional, (3) constant properties (except in buoyancy), (4) uniform surface temperature, (5) negligible radiation, (6) quiescent ambient fluid and (7) door is in the closed position at all times.

(ii) Analysis. Application of Newton's law of cooling to the surface gives

q = h A ( sT - T ) (a)

where


h = average heat transfer coefficient, W/m2-oC q = heat transfer rate, W



Surface area is given by

A = LW (b) where

L = door height = 50 cm = 0.5 m W = door width = 120 cm = 1.2 m

To determine the average heat transfer coefficient h , the Rayleigh number RaL is calculated first to determine if the flow is laminar or turbulent.

L

W

T

gsT


RaL = PrLTTg s

2

3

(c)

where


L = door side in the direction of gravity = 50 cm = 0.5 m Pr = Prandtl number of air RaL = Rayleigh number



For ideal gases, the coefficient of thermal expansion is given by

= 003411.015.273)C(20

1

)K(

1oT

(1/K)

Properties are determined at the film temperature Tf defined as

Tf=

2

TTs = 2

)C)(20110( o

= 65oC


k = 0.02887 W/m-oCPr = 0.7075

= 19.4 10-6 m2/s

Substituting into (c) gives

RaL = 7075.0)s/m()104.19(

)m()5.0)(C)(20110)(s/m(81.9)C/1(003411.02426

33o2o

= 0.70766 109

Since RaL

< 109, the flow is laminar and h is given by (7.23)

d

dGr

L

kh L )0(

43

44/1

(7.23)

where LGr is the Grashof number given by

LGr =Pr

RaLTTg Ls

2

3

(d)

and dd /)0( is a dimensionless temperature gradient which depends on the Prandtl number. It

is listed in Table 7.1.

(iii) Computations. Substitution into (d) gives

99

1000226.17075.0

1070766.0LGr

PROBLEM 7.4 (continued)At Pr = 0.7075, Table 7.1 gives

501.0)0(

d

d

Substitute into (7.23)

CW/m853.4501.04

1000226.1

0.5(m)

C)m0.02887(W/

3

4 o2

4/19o

h

Substitute into (a) and use (b) give the heat transfer rate from door

q = 4.853 (W/m2-oC) (0.5)(m)(1.2)(m) (110 )20 (oC) = 261.9 W

(iii) Checking. Dimensional check: Computations showed that equations (a), (c) and (7.23) are dimensionally consistent.

Quantitative check: The magnitude of h is within the range given in Table 1.1.

(5) Comments. (i) The model used to solve this problem is not conservative due to neglecting radiation. If the door is treated as a small object enclosed by a much larger surface at the ambient temperature, an estimate can be made of the radiation heat loss. Assuming that the door is made of stainless steel with an emissivity of 0.25, radiation heat loss will be 120 W. This is significant when compared with free convection heat loss. (ii) Opening and closing the door results in transient effects not accounted for in the above model. In addition, when the door is open radiation from the interior of oven may be significant.

PROBLEM 7.5

To compare the rate of heat transfer by radiation with that by free convection, consider the following test case. A vertical plate measuring 12 cm 12 cm is maintained at a uniform

surface temperature of 125oC. The ambient air and the surroundings are at 25oC. Compare the

two modes of heat transfer for surface emissivities of 0.2 and 0.9. A simplified model for heat

loss by radiation rq is given by

)( 44sursr TTAq

where A is surface area, is emissivity and 428 KW/m1067.5 . Surface and

Surroundings temperatures are measured in degrees kelvin

(1) Observations. (i) This is a free convection and radiation problem. (ii) The geometry is a vertical plate. (iii) Surface temperature is uniform. (iv) Newton’s law of cooling gives convection heat transfer rate while Stefan-Boltzmann law gives radiation heat transfer rate. (v) The Rayleigh number should be computed to determine if the flow is laminar or turbulent. (vi) For laminar flow the solution of Section 7.4 is applicable. (vii) Since radiation heat transfer is considered in this problem, all temperatures should be expressed in degrees kelvin. (viii) The fluid is air.

(2) Problem Definition. Determine heat transfer rate by free convection and by radiation from a vertical plate in air. Convection requires the determination of the heat transfer coefficient.

(3) Solution Plan. Apply Newton's law of cooling to determine the rate of heat loss by convection. Apply Stefan-Boltzmann radiation law to determine the rate of heat loss by radiation. Compute the Rayleigh number and select an appropriate correlation equations to obtain the average heat transfer coefficient. For laminar flow use results of Section 7.4.

(4) Plan Execution.

(i) Assumptions. (1) Steady state, (2) two-dimensional, (3) constant properties (except in buoyancy), (4) uniform surface temperature, (5) the plate is a small surface enclosed by a much larger surface at a uniform temperature and (6) quiescent ambient.

(ii) Analysis. Application of Newton's law of cooling to the vertical plate gives

)( TTAhq sc (a)

where

A = surface area of vertical side, m2

h = average heat transfer coefficient, W/m2-oC or W/m2-K

cq = convection heat transfer rate, W

sT = surface temperature = 125(oC) + 273.15 = 398.15 K

T = ambient temperature = 25(oC) + 273.13 = 298.15 K

Surface area is

L

T

g sT

gssurroundin

surT

L


A = L2 (b)where

L = side of square plate = 12 cm = 0.12 m


RaL = PrLTTg s

2

3

(c)

where


Pr = Prandtl number of air RaL = Rayleigh number




Tf=

T Ts

2 =

( . . )( )29815 39815

2

K = 348.15K


k = 0.02957 W/m-oCPr = 0.7065

= 20.41 10-6 m2/s

For ideal gases the coefficient of thermal expansion is given by

= 002872.015.348

1

)K(

1

fT(1/K)


RaL = 7065.0)s/m()1041.20(

)m()12.0)(C)(25125)(s/m(81.9)C/1(002872.02426

33o2o

= 8.257 610

Since RaL


d

dGr

L

kh L )0(

43

44/1

(7.23)


LGr =Pr

RaLTTg Ls

2

3

(d)




Radiation heat loss rq is given by the Stefan-Boltzmann law. Assuming that the plate is a small

surface which is surrounded by a much larger surface, rq is given by

rq = A ( 44surs TT ) (e)

where

rq = radiation heat loss, W

Tsur = surroundings temperature = 25(oC) + 273.13 = 298.15 K

= emissivity

= Stefan-Boltzmann constant = 5.67 10-8 W/m2-K4

(iii) Computations. Convection heat loss: substitution into (d) gives

66

10687.117065.0

10257.8LGr

At Pr = 0.7065, Table 7.1 gives

5009.0)0(

d

d


CW/m804.65009.04

10687.11

0.12(m)

C)m0.02957(W/

3

4 o2

4/16o

h

Substitute into (a) and use (b) give the heat transfer rate from door

cq = 6.804 (W/m2-oC) (0.12)(m)(0.12)(m) (125 -25)(oC) = 9.798 W

Radiation heat loss: equation (e) for = 0.2, equation (e) gives

rq = 0.2 5.67 10-8 (W/m2-K4) 0.12(m) 0.12(m) ( 44 15.29815.398 )(K4) = 2.81 W

For = 0.9:

rq = 0.9 5.67 10-8 (W/m2-K4) 0.12(m) 0.12(m) ( 44 15.29815.398 )(K4) = 12.66 W

(iv) Checking. Dimensional check: Computations showed that equations (a), (c), (7.23) and (e) are dimensionally consistent.


Limiting check: For surs TTT , heat transfer by convection and radiation vanish. Setting

surs TTT in (a) and (e) gives .0rc qq

(5) Comments. (i) When compared with free convection, radiation heat loss can be significant


and in general should not be neglected. (ii) The magnitude of is the same whether it is

expressed in units of degree Celsius or kelvin. The reason is that is measured in terms of degree change. One degree change on the Celsius scale is equal to one degree change on the kelvin scale. This is also true of units of heat transfer coefficient and specific heat. (iii) Because temperature in the Stefan-Boltzmann radiation law must be expressed in degrees kelvin, care should be exercised in using the correct units when carrying radiation computations.

PROBLEM 7.6

A sealed electronic package is designed to be cooled by free convection. The package consists of components which are mounted on the inside surfaces of

two cover plates measuring cm5.7cm5.7 cm each. Because the plates are

made of high conductivity material, surface temperature may be assumed uniform. The maximum allowable surface temperature is 70oC. Determine

the maximum power that can be dissipated in the package without violating design constraints. Ambient air temperature is 20oC.

(1) Observations. (i) This is a free convection problem. (ii) The power dissipated in the electronic package is transferred to the ambient fluid by free convection. (iii) As the power is increased, surface temperature increases. (iv) The maximum power dissipated corresponds to the maximum allowable surface temperature. (v) Surface temperature is related to surface heat transfer by Newton’s law of cooling. (vi) The problem can be modeled as free convection over a vertical plate. (vii) The Rayleigh number should be computed to determine if the flow is laminar or turbulent. (viii) For laminar flow the solution of Section 7.4 is applicable. (ix) The fluid is air.

(2) Problem Definition. Since dissipated power is related to heat loss from the surface to the surroundings, Newton's law of cooling should be applied. Thus, the problem becomes finding the average heat transfer coefficient.

(3) Solution Plan. Apply Newton's law of cooling to the surface of the package. Check the Rayleigh number to determine if the flow is laminar or turbulent. Use appropriate correlation equations to determine the average heat transfer coefficient.

(4) Plan Execution.

(i) Assumptions. (1) Steady state, (2) two-dimensional, (3) constant properties (except in buoyancy), (4) uniform surface temperature, (5) negligible heat loss from the top and bottom surfaces, (6) the sides are vertical flat plates, (7) negligible radiation and (8) quiescent fluid.


P = q = h A ( TTs ) (a)

where


h = average heat transfer coefficient, W/m2-oC P = power dissipated in package, W

q = heat transfer from the surface to the ambient air, W




A = 2 LW (b)

gT

components

air

g

T

W

L

sT


where

L = package height = 7.5 cm = 0.75 m W = package width = 7.5 cm = 0.75 m


RaL = PrLTTg s

2

3

(c)

where


Pr = Prandtl number of air RaL = Rayleigh number




Tf=

T Ts

2 =

( )( )70 20

2

oC = 45oC

Air properties at this temperature are

k = 0.02746 W/m-oCPr = 0.7095

= 17.44 10-6 m2/s


= 003143.015.273)C(45

1

)K(

1o

fT(1/K)


RaL = 7095.0)/s(m)10(17.44

)(mC)(0.075)20)()(70C)9.81(m/s/0.003143(12426

33o2o

= 1.5171 610

Since RaL


d

dGr

L

kh L )0(

43

44/1

(7.23)


LGr =Pr

RaLTTg Ls

2

3

(d)





66

105171.17065.0

10257.8LGr


5017.0)0(

d

d


CW/m08.65017.04

105171.1

0.075(m)

C)m0.02746(W/

3

4 o2

4/16o

h

Equations (a) and (b) give the maximum power dissipated from the two sides of the package

P = q = 2 6.08 (W/m2-oC) (0.075)(m)(0.075)(m) (70-20)(oC) = 3.42 W

(iv) Checking. Dimensional check: Computations showed that equations (a), (c) and (7.23) dimensionally consistent.


Qualitative check: Increasing the allowable surface temperature sT should increase the maximum

power P. According to equation (a), q is directly proportional to sT . Furthermore, h increases

when sT is increased.

Limiting check: The dissipated power should vanish (P = 0) if heat cannot be removed by free

convection (q = 0). Setting h = 0 in (a) gives P = q = 0.

(5) Comments. (i) The model used to solve this problem is conservative due to both neglecting radiation and heat loss from the side surfaces. (ii) The maximum power dissipated is relatively small, indicating the limitation of free convection in air as a cooling mode for such applications.

The maximum dissipated power in water is 494.7 W (Problem 7.7). (iii) The magnitude of is

the same whether it is expressed in units of degree Celsius or kelvin. The reason is that is measured in terms of degree change. One degree change on the Celsius scale is equal to one degree change on the kelvin scale. This is also true of units of heat transfer coefficient and specific heat.

PROBLEM 7.7

Assume that the electronic package of Problem 7.6 is to be used in an underwater application.

Determine the maximum power that can be dissipated if the ambient water temperature is 20oC.

(1) Observations. (i) This is a free convection problem. (ii) The power dissipated in the electronic package is transferred to the ambient fluid by free convection. (iii) As the power is increased, surface temperature increases. (iv) The maximum power dissipated corresponds to the maximum allowable surface temperature. (v) Surface temperature is related to surface heat transfer by Newton’s law of cooling. (vi) The problem can be modeled as free convection over a vertical plate. (vii) The Rayleigh number should be computed to determine if the flow is laminar or turbulent. (viii) For laminar flow the solution of Section 7.4 is applicable. (ix) The fluid is water.



(4) Plan Execution.



P = q = h A ( TTs ) (a)

where


h = average heat transfer coefficient, W/m2-oC P = power dissipated in package, W

q = heat transfer from the surface to the ambient air, W




A = 2 LW (b)where

L = package height = 7.5 cm = 0.75 m W = package width = 7.5 cm = 0.75 m


g

T

W

L

sT


RaL = PrLTTg s

2

3

(c)

where


Pr = Prandtl number of water RaL = Rayleigh number




Tf=

T Ts

2 =

( )( )70 20

2

oC = 45oC

Water properties at this temperature are

k = 0.6286 W/m-oCPr = 4.34

= 0.000389 1/K

= 0.6582 10-6 m2/s


RaL = 34.4)/s(m)10(0.6582

)(mC)(0.075)10)()(70C)9.81(m/s/0.000389(12426

33o2o

= 0.96778 910

Since RaL


d

dGr

L

kh L )0(

43

44/1

(7.23)


LGr =Pr

RaLTTg Ls

2

3

(d)




99

1022299.034.4

1096778.0LGr


9108.0)0(

d

d



CW/m5.8799108.04

1022299.0

0.075(m)

C)0.6286(W/m

3

4 o2

4/19o

h


P = q = 2 879.5 (W/m2-oC) (0.075)(m)(0.075)(m) (70-20)(oC) = 494.7 W

(iv) Checking. Dimensional check: Computations showed that equations (a), (c) and (7.23) dimensionally consistent.

Quantitative check: The magnitude of h is within the range given in Table 1.1 for free convection of liquids

Qualitative check: Increasing the allowable surface temperature sT should increase the maximum

power P. According to equation (a), q is directly proportional to sT . Furthermore, h increases

when sT is increased.



(5) Comments. (i) The model used to solve this problem is conservative due to neglecting heat loss from the side surfaces. (ii) The maximum power dissipated is relatively large, indicating the effectiveness of water as a free convection medium. The maximum power in air (Problem 7.6) is

3.42 W. (iii) The magnitude of is the same whether it is expressed in units of degree Celsius or

kelvin. The reason is that is measured in terms of degree change. One degree change on the Celsius scale is equal to one degree change on the kelvin scale. This is also true of units of heat transfer coefficient and specific heat.

PROBLEM 7.8

Consider laminar free convection from a vertical plate at uniform

surface temperature. Two 45 triangles are drawn on the plate as

shown. Determine the ratio of the heat transfer rates from two

triangles.

(1) Observations. (i) This is a free convection problem. (ii) The surface is maintained at uniform temperature. (iii) Newton’s law of cooling determines the heat transfer rate. (iv) Heat transfer rate depends on the heat transfer coefficient. (v) The heat transfer coefficient decreases with distance from the leading edge of the plate. (vi) The width of each triangle changes with distance from the leading edge. (vii) For laminar flow the solution of Section 7.4 is applicable.

(2) Problem Definition. Examine the variation of local heat transfer coefficient with distance and determine the heat transfer rate from each triangle.

(3) Solution Plan.

Apply Newton’s law of cooling to an element of each triangle. Formulate an equation for element area and heat transfer coefficient )(xh for laminar free convection over a flat plate.

(4) Plan Execution.

(i) Assumptions. (1) Laminar flow, (2) steady state, (3) two-dimensional, (4) constant properties (except in buoyancy), (5) uniform surface temperature, (6) quiescent fluid and (7) no radiation.

(ii) Analysis. Consider triangle 1 first. Application of Newton’s law of cooling to the element

dxxb )(1

dxxbTTxhdq s )())(( 11

IntegrateH

s dxxbxhTTq0

11 )()()( (a)

where

)(1 xb width of element 1, m

)(xh local heat transfer coefficient, CW/m o2

1q = heat transfer rate, W

sT surface temperature, oC


x = distance along plate, m

Similarly, for the second triangle H

s dxxbxhTTq0

22 )()()( (b)

where

x

dx

dx

b1(x)

b2(x)

g

T

12

H

B

gT

1

2


)(2 xb width of element 2, m

To evaluate the integrals in (a) and (b) it is necessary to determine the variation with x of h(x),

)(1 xb and )(2 xb . The local heat transfer coefficient for free convection laminar flow over a

vertical plate is given by (7.21)

d

dGr

x

kh x )0(

4

4/1

(7.21)

where

k thermal conductivity, W/m-oC

dxd /)0( dimensionless temperature gradient at the surface

xGr = Grashof number, defined as

2

3)( xTTgGr s

x (c)




(c) into (7.21)

d

dxTTg

x

kh s )0(

4

)(4/1

2

3

(d)

According to (d), )(xh decreases varies with x as

4/1)( xCxh (e)

The geometric functions )(1 xb and )(2 xb are determined using similarity of triangles. Thus

b1(x) = B(1 - x

H) (f)

and

b2(x) = Bx

H (g)

where

B = base of triangle, m H = height of triangle, m

Substitute (e) and (f) into (a)

dxxHxTTBCqH

s0

4/11 /1


4/31 ))(21/16( BCHTTq s (h)

Similarly, substitute (d) and (g) into (b)


dxxHxTTBCqH

s0

4/12 )/(

Evaluate the integral 4/3

2 ))(7/4( BCHTTq s (i)

Taking the ratio of (i) and (i) q

q

1

2

= 4

3 (j)

(iii) Checking. Dimensional check: Units of C are determined using (d):

C = W/m7/4-oC

Thus units of h(x) in (d) are

h(x) = C (W/m7/4-oC) x-1/4(m)-1/4 = W/m2-oC

Examining units of q1 in (h)

q1 = B(m) C(W/m7/4-oC) (Ts -T ) (oC) 4/3H (m3/4) = W

Limiting check: If = 0 or g = 0 or Ts = T , no free convection takes place and consequently q1

= q2 = 0. Any of these limiting cases give C = 0. Thus, according to (h) and (i), q1 = q2 = 0.

(5) Comments. (i) More heat is transferred from triangle 1 than triangle 2. This follows from the fact that h decrease with distance x. This favors triangle 1 since its large base is at x = 0 where his maximum. (ii) The result applies to any right angle triangles and is not limited to 45o

triangles. (iii) Heat transfer from a surface of fixed area depends on its orientation relative to the leading edge. (iv) This problem illustrates how integration is used to account for variations in element area and heat transfer coefficient. The same approach can be applied if surface temperature and/or ambient temperature vary over a surface area.

PROBLEM 7.9

A vertical plate measuring 21 cm 21 cm is at a uniform surface temperature of 80oC. The

ambient air temperature is 25oC. Determine the heat flux at 1 cm, 10 cm and 20 cm from the

lower edge.

(1) Observations. (i) This is a free convection problem over a vertical plate. (ii) The surface is maintained at uniform temperature. (iii) Local heat flux is determined by Newton’s law of cooling. (iv) Heat flux depends on the local heat transfer coefficient. (v) Free convection heat transfer coefficient for a vertical plate decreases with distance from the leading edge. Thus, the flux also decreases. (vi) The Rayleigh number should be computed to determine if the flow is laminar or turbulent. For (vii) Laminar flow the solution of Section 7.4 is applicable. (viii) The fluid is air.

(2) Problem Definition. Determine the local heat transfer coefficient for free convection over a vertical plate at uniform surface temperature.

(3) Solution Plan. Apply Newton’s law of cooling, compute the Rayleigh number and select an appropriate Nusselt number correlation equation.

(4) Plan Execution.

(i) Assumptions. (1) Steady state, (2) two-dimensional, (3) constant properties (except in buoyancy), (4) uniform surface temperature, (5) quiescent fluid and (6) negligible radiation.

(ii) Analysis. Application of Newton's law gives

q = h(x) (Ts - T ) (a)

where

h(x) = local heat transfer coefficient, W/m2-oCq = local heat flux, W

Ts


T = ambient temperature = 25oC

The Rayleigh is computed to determine if the flow is laminar or turbulent.

Rax = PrLTTg s

2

3

(b)

where


Pr = Prandtl number

LRa = Rayleigh number at LL = plate height = 0.21 m



Properties of air are determined at the film temperature fT defined as

Tf= 2/)( TTs = 2/)C)(2580( o = 52.5oC

L

T

g sT

L



k = 0.02799 W/m-oCPr = 0.709

= 18.165 10-6 m2/s


= 003071.015.273)C(5.52

1

)(

1oKT

(1/K)


RaL = 709.0)/s(m)10(18.165

)(mC)(0.2)25)()(80/sC)9.81(m/0.003071(12426

33o2o

= 2.8482 107

Since RaL

< 109, the flow is laminar over the region of interest. The local heat transfer

coefficient for free convection laminar flow over a vertical plate is given by (7.21)

d

dGr

x

kh x )0(

4

4/1

(7.21)

where

dxd /)0( dimensionless temperature gradient at the surface

xGr = Grashof number, defined as

2

3)( xTTgGr s

x (c)

Note that dxd /)0( depends on the Prandtl number and is listed in Table T.1.

(iii) Computations. At Pr = 0.709, Table 7.1 gives

5015.0)0(

d

d=

At x = 0.01 m, (c) gives

)/s(m)10(18.165

)(mC)(0.01)25)()(80/sC)9.81(m/0.003071(12426

33o2o

xGr = 0.5022 410


)5015.0(4

5022

m)(01.0

)CW/m(02799.04/1o

h = 8.356 W/m2-oC

Substitute into (a)

2oo2 W/m459.6C))(2580)(CW/m(355.8q


The same procedure is followed to determine the flux at x = 10 and x = 20 cm. Results are tabulated below.

x (cm) Grx h(x)(W/m2-oC) q (W/m2)

1 4105022.0 8.355 459.6

10 7105022.0 4.699 258.4

20 8104018.0 3.951 217.3

(iv) Checking. Dimensional check: Computations showed that equations (a), (b) and (7.21) are dimensionally consistent.

Quantitative check: The values of h are approximately within the range given in Table 1.1 for free convection of gases.

Limiting check: The flux should vanish for Ts = T . Setting Ts = T in (a) gives 0q .

(5) Comments. According to (7.21) and (a), surface heat flux decreases with distance from the leading edge as

4/1x

Cq

Thus, high heat flux components should be place close to the leading edge.

PROBLEM 7.10

200 square chips measuring cm1cm1 each are mounted on both

sides of a thin vertical board .cm01cm10 The chips dissipate 0.035

W each. Assume uniform surface heat flux. Determine the maximum surface temperature in air at 22oC.

(1) Observations. (i) This is a free convection problem over a vertical plate. (ii) The power dissipated in the chips is transferred to the air by free convection. (iii) This problem can be modeled as free convection over a vertical plate with constant surface heat flux. (iv) Surface temperature increases as the distance from the leading edge is increased. Thus, the maximum surface temperature occurs at the top end of the plate (trailing end). (v) The Rayleigh number should be computed to determine if the flow is laminar or turbulent. (vii) For laminar flow the analysis of Section 7.5 gives surface temperature

distribution. (vii) The fluid is air. (viii) Properties depend on the average surface temperature sT .

Since sT is unknown, the problem must be solved by trail and error.

(2) Problem Definition. Determine surface temperature distribution for a vertical plate with uniform surface heat flux under free convection conditions.

(3) Solution Plan. Apply the analysis of Section 7.5 for surface temperature distribution of a vertical plate with uniform surface heat flux in laminar free convection. Compute the Rayleigh number to confirm that the flow is laminar.

(4) Plan Execution.

(i) Assumptions. (1) Steady state, (2) two-dimensional, (3) constant properties (except in buoyancy), (4) uniform surface heat flux, (5) all dissipated power leaves surface as heat, (6) negligible radiation and (7) quiescent ambient fluid.

(ii) Analysis. Surface temperature distribution for a vertical plate with uniform surface flux is given by equation (7.27)

)0()(

5)(

5/1

4

42

xkg

qTxT s

s (7.27)

where


k = thermal conductivity, W/m-oCPr = Prandtl number

sq = surface flux, W/m2

RaL = Rayleigh number at x = L


T = ambient temperature = 22oCx = distance from leading edge, m

= coefficient of thermal expansion, 1/K )0( = dimensionless surface temperature


The dimensionless surface temperature, )0( , depends the Prandtl number. Values corresponding

to four Prandtl numbers are listed in Table 7.2.

T g


For laminar flow

RaL = PrLTTg s

2

3

< 910 (a)

where

L = vertical side of plate = 10 cm = 0.1 m

If all dissipated power in the chip leaves the surface, conservation of energy gives

A

Pqs (b)

where

A = chip surface area = 1 cm2 = 0.0001 m2

P = power dissipated in chip = 0.035 W

Properties are evaluate at the film temperature defined as

2

)2/( TLTT s

f (c)

where Ts(L/2) is surface temperature at the mid-point. However, since Ts(x) is unknown, an iterative procedure is required to obtain a solution. An assumed value for Ts(L/2) is used to calculate the film temperature at which properties are determined. Equation (a) is then used to calculate Ts(L/2). If the calculated value does not agree with the assumed temperature, the procedure is repeated until a satisfactory agreement is obtained.

(iii) Computations. Equation (b) gives surface flux

sq = 0.035(W)/0.0001(m2) = 350 W/m2

Assume Ts (L/2) = 58oC. Equation (c) gives

Tf = (58 + 22)(oC)/2 = 40oC


cp = 1006.8 J/kg- oCk = 0.0271 W/m-oCPr = 0.71

= 16.96 10-6 m2/s

Coefficient of thermal expansion for an ideal gas is given by

= 003193.015.273)C(40

1

15.273

1o

fT1/K

At Pr = 0.71, Table 7.2 gives .806.1)0(

Substituting into (7.27) and letting x = L/2 = 0.1(m)/2 = 0.05 m

806.1)m(05.0C)0.0271(W/m

)350(W/m

)C)9.81(m/s/0.003193(1

(m/s))10(16.965)C(22/2

5/14

o

2

2o

226oLTs = 87.7 Co


Ts(L/2) = 87.7oC

Since this is higher than the assumed value of 58oC, the procedure is repeated with a new assumed temperature at mid-point. Assume Ts(L/2) = 78oC. The following results are obtained

Tf = 50oCcp = 1007.4 J/kg- oCk = 0.02781 W/m-oCPr = 0.709

= 25.27 10-6 m2/s

= 0.0030945 1/K

= 17.92 10-6 m2/s

= 1.0924 kg/m3

Substituting into (a) gives Ts(L/2) = 76.8oC. This is close to the assumed value of 78oC.

Surface temperature at the trailing end is now computed by evaluating (a) at x = L = 0.1 m

5/14

o

2

2o

26262/1o )m(1.0

)Cm/W(02781.0

)m/W(350

)s/m(81.9)C/1(0030945.0

)s/m(1092.17)s/m(1027.25

709.0

)709.0(10)709.0(94)C(22/2LTs

Ts(L) = 84.9oC

The condition on the Rayleigh number in equation (b) is verified next. Substituting into (c)

RaL = 709.0)s/m()1092.17(

)m()1.0)(C)(229.84)(s/m(81.9)C/1(0030945.02426

33o2o

= 4.22 106

This satisfies the condition on RaL given in equation (b).

(iv) Checking. Dimensional check: Equations (a), (c) and (d) are dimensionally consistent..

Quantitative check: The heat transfer coefficient at the mid-point, h(L/2), can be calculated using Newton's law of cooling:

h(L/2) = ])2/(/[ TLTq ss = 350(W/m2)/(76.8 - 22)(oC) = 6.39 W/m2-oC

This is within the range given in Table 1.1 for free convection of gases.

Validity of correlation equation (a): The conditions listed in (b) are met.

(5) Comments. (i) Surface temperature is determined without calculating the heat transfer coefficient. This is possible because equation (a) combines the correlation equation for the heat transfer coefficient and Newton's law of cooling to eliminate h and obtain an equation for surface

temperature in terms of surface heat flux. (ii) The magnitude of is the same whether it is

expressed in units of degree Celsius or kelvin. The reason is that is measured in terms of degree change. One degree change on the Celsius scale is equal to one degree change on the kelvin scale. This is also true of units of heat transfer coefficient and specific heat.

PROBLEM 7.11

cm21cm12 power board dissipates 15 watts uniformly. Assume that all energy leaves the

board from one side. The maximum allowable surface temperature is .C82o The ambient fluid is

air at .C24o Would you recommend cooling the board by free convection?

(1) Observations. (i) This is a free convection problem over a vertical plate. (ii) The power dissipated in the chips is transferred to the air by free convection. (iii) This problem can be modeled as free convection over a vertical plate with constant surface heat flux. (iv) Surface temperature increases as the distance from the leading edge is increased. Thus, the maximum surface temperature occurs at the top end of the plate (trailing end). (v) The Rayleigh number should be computed to determine if the flow is laminar or turbulent. (vii) For laminar flow the analysis of Section 7.5 gives surface temperature distribution. (vii) The fluid is air. (viii)

Properties depend on the average surface temperature sT . Since sT is unknown, the problem

must be solved by trail and error.


(3) Solution Plan. Apply the analysis of Section 7.5 for surface temperature distribution of a vertical plate with uniform surface heat flux in laminar free convection. Compute the Rayleigh number to confirm that the flow is laminar.

(4) Plan Execution.


(ii) Analysis. Surface temperature distribution for a vertical plate with uniform surface flux is given by equation (7.27)

)0()(

5)(

5/1

4

42

xkg

qTxT s

s (7.27)

where






= coefficient of thermal expansion, 1/K )0( = dimensionless surface temperature


The dimensionless surface temperature, )0( , depends the Prandtl number. It can be determined

from Table 7.2 or using correlation equation (7.29):

L

T

g

L

0

x

sq


5/12/1 1094

)0(25Pr

PrPr, 1000001.0 Pr (7.29)

The heat flux is defined as

A

Pqs (a)

where

A = surface area, 2m

P = dissipated power = 15 W

Surface area is2LA (b)

where

L = side of power board = 0.12 m

The Rayleigh number is used to determine if the flow is laminar. The criterion is

RaL = PrLTTg s

2

3

< 910 (c)

Equation (7.27) is used to determine maximum surface temperature, )(LTs , corresponding to the

specified surface flux. Properties are evaluate at the film temperature defined as

2

)2/( TLTT s

f (d)

where Ts(L/2) is surface temperature at the mid-point. However, since Ts(x) is unknown, an iterative procedure is required to obtain a solution. An assumed value for Ts(L/2) is used to calculate the film temperature at which properties are determined. Equation (7.27) is then used to calculate Ts(L/2). If the calculated value does not agree with the assumed temperature, the procedure is repeated until a satisfactory agreement is obtained.

(iii) Computations. Assume Ts (L/2) = 76oC. Equation (d) gives

Tf = (76 + 24)(oC)/2 = 50oC


k = 0.02781 W/m-oCPr = 0.709

= 17.92 10-6 m2/s


= 0030945.015.273)C(50

1

15.273

1o

fT1/K

At Pr = 0.709, equation (7.29) gives )0(


49337.1)709.0(5

)709.0(10709.094)0(

5/1

2

Equations (a) and (b) give surface heat flux

22 m

W67.1041

)0.12(m0.12

15(W)sq

Substituting into (7.27) and letting x = L/2 = 0.12(m)/2 = 0.06 m

C9.15949337.1)m(06.0C)m0.02781(W/

)m1041.67(W/

)C)9.81(m/s/0.00309451

(m/s))10(17.925)C(24/2 o

5/14

o

2

2o

226oLTs


Tf = 90oCk = 0.03059 W/m-oCPr = 0.705

= 0.0027537 1/K

= 21.35 10-6 m2/s4958.1)0(

Substituting into (7.27) gives Ts(L/2) = 162.4oC. Further iteration will bring )2/(LTs between

156 oC and 162 oC. Surface temperature at the trailing end will be even higher. Therefore, board

temperature will exceed the maximum allowable of .C82o It follows that cooling by free

convection is not recommended.

Surface temperature at the trailing end is now computed by evaluating (7.27) at x = L = 0.12 m

4958.10.12(m)C)m0.03059(W/

)m1041.67(W/

)C)9.81(m/s1/0.0027537(

/s)(m)10(21.355C)(24/2

5/14

o

2

2o

2226oLTs

)(LTs 183.1oC

The Rayleigh number is computed to confirm that the flow is laminar. Substitute into (a)

RaL = 7

2426

33o2o

101149.050.70)/s(m)10(21.35

)(mC)(0.12)24)()(183.1C)9.81(m/s1/0.0027537(

Since this is less than 910 , the flow is laminar.

(iv) Checking. Dimensional check: Equations (7.27) and (a)-(d) are dimensionally consistent..



h(L/2) = ])2/(/[ TLTq ss = 1041.67(W/m2)/(159.0 - 24)(oC) = 7.66 W/m2-oC


(5) Comments. (i) High surface temperature is due to high surface heat flux. Forced convection cooling is required to meet design conditions on maximum temperature. (ii) A trial and error procedure was required to solve this problem because properties depend on surface temperature which is unknown a priori.

PROBLEM 7.12

Use the integral method to obtain a solution to the local Nusselt number for laminar flow

over a vertical plate at uniform surface temperature .sT Assume t and a velocity and

temperature profiles given by

33

2210 )()()()(, yxayxayxaxayxu

and3

32

210 )()()()(),( yxbyxbyxbxbyxT

Since there is a single unknown ),(x either the momentum or energy equation may be

used. Select the energy equation to determine ).(xt

(1) Observations. (i) This is a free convection problem over a vertical plate at uniform surface temperature. (ii) In general, to determine the Nusselt number it is necessary to determine the velocity and temperature distribution. (iii) The integral method can be used to determine the velocity and temperature distribution. (iv) Application of the integral method reduces to determining the velocity and temperature boundary layer thickness.

(2) Problem Definition. Determine the temperature distribution for boundary layer flow over a flat plate at uniform surface temperature.

(3) Solution Plan. Start with the definition of local Nusselt number. Equate Newton’s law with Fourier’s law to obtain an equation for the heat transfer coefficient h. Apply the integral form of the energy equation using a third degree polynomial temperature profile.

(4) Plan Execution. (i) Assumptions. (1) Steady state, (2) constant properties except for buoyancy, (3) Boussinesq approximations are valid, (4) two-dimensional, (5) laminar flow

( 910LRa ), (6) flat plate, (7) uniform surface temperature, (8) negligible changes in kinetic

and potential energy, (9) negligible axial conduction, (10) negligible dissipation, and (11)

t .


k

hxNux (a)

where

h local heat transfer coefficient, CW/m o2


xNu local Nusselt number

x distance along plate measured from the leading edge, m


TT

y

xTk

hs

)0,(

(1.10)

where



T ambient temperature, Co

y coordinate normal to plate, m

Thus h depends on the temperature distribution ).,( yxT The integral form of the energy

equation, (7.36), is used to determine the temperature distribution

t

dyTTudx

d

y

xT

0

)(0,

(b)

where



Assume a third degree polynomial for the axial velocity u(x,y)

33

2210 )()()()(, yxayxayxaxayxu (c)

The coefficients )(xan are determined using the following known exact and approximate

boundary conditions on the velocity

(1) 0)0,(xu

(2) 0),( txu

(3) 0),(

y

xu t

(4) )()0,(

2

2

TTg

y

xus

Condition (4) is obtained by setting 0y in the x-component of the equations of motion,

(7.5). Equation (c) and the four boundary conditions give the coefficients )(xan

,00a ,4

)(1 t

s TTga ,

2

)(2

TTga s

t

s TTga

1

4

)(3

Substituting the above into (c) and rearranging

2

2

214

)(

ttt

s yyy

TTgu (d)

Note that velocity profile (d) is based on the assumption that .t



33



(1) sTxT 0,

(2) TxT t,

(3) 0,

y

xT t

(4) 00,

2

2

y

xT


,0 sTbt

sTTb1

)(2

31 , ,02b

33

1)(

2

1

t

sTTb

Substituting the above into (e)

3

3

2

1

2

3)(),(

ttss

yyTTTyxT (f)

Substituting (f) into (1.10)

t

kh

2

3 (g)


tx

xNu

2

3 (h)

The problem reduces to finding t which is obtained using the energy equation. Substituting

(d) and (f) into (b))(

0

3

3

2

2

2

1

2

31)(1

4

)(1

2

)(3

xt

tts

ttt

s

t

s dyyy

TTyy

ydx

dTTgTT

Expand the integrand and simplify

t

tttttt

s

t

dyyyyyy

ydx

dTTg

0

5

6

4

5

2

3

3

42

2

14

2

7

4

)(1

2

3


333333

14

1

6

1

5

1

6

7

2

1

4

)(1

2

3tttttt

s

t dx

dTTg

Simplify and rearrange


)(2

3153

TTgdx

d

s

tt

Rewrite the above

)(2

1053

TTgdx

d

s

tt

Separating variables and integrating

t

stt

t

dxTTg

d

0

3

0)(2

105

Evaluate the integrals and rearrange

xTTg s

t

)(2

105

4

4

Solve for xt /

4/1

3)(

210

xTTgxs

t (i)

This result can be expressed in terms of the Rayleigh number as

4/1

806.3

x

t

Rax (j)

Substitute (j) into (h) 4/1394.0 xx RaNu (k)

An alternate form is 4/1

394.0 PrGrNu xx (l)

(iii) Checking. Dimensional check: Equations (a), (h), (i), (j) and (k) are dimensionless. Units of (b), (d) and (f) are correct.

Boundary conditions check: Velocity profile (d) and temperature profile (f) satisfy their respective boundary conditions.

(5) Comments. (i) The integral form of the momentum equation was not used in the solution. Therefore the result does not satisfy momentum and thus it is not expected to be accurate. (ii) To examine the accuracy of this model, equation (l) is rewritten as

4/14/1

(394.04

Pr)NuGr

xx (m)

This result is compared with similarity solution (7.49) and integral solution (7.50) which satisfies both momentum and energy. As expected, the accuracy of the integral method


deteriorates when the momentum equation is neglected. Only at Prandtl numbers of order unity good accuracy is obtained.

xx Nu

Gr4/1

4

Exact Integral

Pr Eq. (7.49) Momentum & energy, Eq (7.50)

EnergyEq. (m)

0.01 0.0806 0.176 0.0725

0.09 0.219 0.305 0.2166

0.5 0.442 0.469. 0.4627

0.72 0.5045 0.513 0.5361

1.0 0.5671 0.557 0.6078

2.0 0.7165 0.663 0.7751

10 1.1649 0.991 1.2488

100 2.191 1.762 2.2665

PROBLEM 7.13

Consider laminar free convection over a vertical plate at uniform surface flux sq . Assume

t and a third degree polynomial velocity profile given by

2

1)(,yy

xuyxu o

Show that:

[a] An assumed second degree polynomial for the temperature profile gives

2

22

1),(

yy

k

qTyxT s

[b] The local Nusselt number is given by

5/1

4

2

2

4536

)(4x

k

qg

Pr

PrNu s

x

(1) Observations. (i) This is a free convection problem over a vertical plate at uniform surface heat flux. (ii) In general, to determine the Nusselt number it is necessary to determine the velocity and temperature distribution. (iii) The integral method can be used to determine the velocity and temperature distribution. (iv) Application of the integral method reduces to determining the velocity and temperature boundary layer thickness.

(2) Problem Definition. Determine the temperature distribution for boundary layer flow over a flat plate which is heated with uniform surface flux.

(3) Solution Plan. Start with the definition of local Nusselt number. Equate Newton’s law with Fourier’s law to obtain an equation for the heat transfer coefficient h. Apply the integral form of the energy equation using a second degree polynomial temperature profile.

(4) Plan Execution. (i) Assumptions. (1) Steady state, (2) constant properties except for buoyancy, (3) Boussinesq approximations are valid, (4) two-dimensional, (5) laminar flow

( 910LRa ), (6) flat plate, (7) uniform surface temperature, (8) negligible changes in kinetic

and potential energy, (9) negligible axial conduction, (10) negligible dissipation, and (11)

t .


k

hxNux (a)

where

h local heat transfer coefficient, CW/m o2



xNu local Nusselt number

x distance along plate measured from the leading edge, m


TT

y

xTk

hs

)0,(

(1.10)

where


T ambient temperature, Co

y coordinate normal to plate, m

Thus h depends on the temperature distribution ).,( yxT The integral form of the energy

equation, (7.36), is used to determine the temperature distribution

t

dyTTudx

d

y

xT

0

)(0,

(b)

where



A third degree polynomial for the axial velocity u(x,y) gives

2

1)(,yy

xuyxu o (c)

Assume a second degree polynomial temperature profile

2210 )()()(),( yxbyxbxbyxT (d)


(1) sqy

xTk

)0,(

(2) TxT t,

(3) 0,

y

xT t

Equation (d) and the three boundary conditions give the coefficients )(xbn

,2

0k

qTb s

k

qb s

1 ,1

22

k

qb s



2

22

),(y

yk

qTyxT s (e)

Surface temperature is determined by setting y = 0 in (e)

k

qTyxT s

2),( (f)

Substituting (f) into (1.10)

kh 2 (g)


xNux 2 (h)

The problem reduces to finding . The two unknown functions )(xuo and )(x are

determined using momentum equation (7.35) and energy equation (7.36):

dyudx

ddyTTg

y

xu

0

2

0

)()0,(

(7.35)

)(

0

)(0,

x

dyTTudx

d

y

xT (7.36)

Substitute (c) and (e) into (7.35)

dyy

yu

dx

ddy

yy

k

qg

u oso

0

42

2

2

0

2

122

(i)

Evaluate the integrals in (i)

22

105

1

6o

so udx

d

k

qgu (j)

Similarly, substitute (c) and (e) into (7.36)

)(

0

22

212

x

dyy

yy

yu

dx

d

k

q

k

q oss (k)

Evaluate the integrals and rearrange

260 oudx

d (l)

Equations (j) and (l) are two simultaneous firs t order ordinary differential equations. The

two dependent variables are )(x and ).(xuo We assume a solution of the form


mo Axxu )( (m)

nBxx)( (n)

where A, B, m and n are constants. To determine these constants we substitute (m) and (n) into (j) and (l) to obtain

12222

105

2

6

nmnsnm BxAnm

xBk

qgvx

B

A (o)

and122)2(60 nmxABnm (p)

To satisfy (o) and (p) at all values of x, the exponents of x in each term must be identical. Thus, (o) requires that

122 nmnnm (q)

Similarly, (p) requires that

012nm (r)

Solving (q) and (r) for m and n gives

5

3m ,

5

1n (s)

Introducing (s) into (o) and (p) gives two simultaneous algebraic equations for A and B

BAnm

Bk

qg

B

A s 22

105

2

6 (t)

and2)2(60 ABnm (u)

Solving equations (t) and (u) for A and B, gives

5/2

5

436060

sqg

kA (v-1)

and5/1

2

)6/(

60)75/3600(

kqgB

s

(w-1)

Note that

Pr (x)

Substitute into (v-1) and (w-1) 5/2

2

5

436060 Pr

qg

kA

s

(v-2)


5/12 )5/4(360

sqg

PrkaB (w-2)

Substitute (s) and (w-2) into (n), rearranging and introducing the definition of Rayleigh number, gives the solution to xx /)(

1)5/1(

5/12 )5/4(360

xqg

Prka

x s

(y)

Introduce (y) into (h), use (x) and rearrange gives the local Nusselt number

5/1

4

24536

4x

k

qg

Pr

PrNu s

2

x (z)

(iii) Checking. Dimensional check: Equations (c), (e), (j), (l), (t), (u), (v-1), (v-2), (w-1) and are dimensionally correct. Equations (a), (h), (y) and (z) are dimensionless.

Boundary conditions check: Temperature profile (e) satisfy the three listed boundary conditions.

Limiting check: If 0sq , the Nusselt number should vanish. Setting 0sq in (z) gives

.0xNu

(5) Comments. (i) The same approach can be used to solve the corresponding problem of

variable surface flux, ).(xqs

(ii) The accuracy of the integral solution can be evaluated by comparing (z) with the exact solution. Equation (7.32) gives the exact solution to the local Nusselt number for free convection over a vertical plate at uniform surface flux

)0(

1

5

5/1

4

2x

k

qgNu s

x (7.32)

where the parameter )0( is determined using correlation equation (7.33)

5/1

2

2/1

5

1094)0(

Pr

PrPr, 1000001.0 Pr (7.33)

To facilitate the comparison, the two solutions are rearranged. Integral solution (z) is rewritten as

5/15/1

4

2 4536

4

Pr

Prx

k

qgNu

2s

x (A-1)

Similarly, exact solution (7.32) is rewritten using (7.33) to eliminate )0(


5/1

2/1

215

4

2 1094 PrPr

Prx

k

qgNu s

x (A-2)

Thus the accuracy of the integral solution can be evaluated by comparing (A-1) and (A-2). The following table compares the two solutions.

The agreement between the two solution is excellent. At Pr = 0.01 the error is 8.3%. At all other Prandtl numbers from 0.1 to 100 the error ranges from 1% to 3.3%.

15

4

2x

k

qgNu s

x

Pr Integral (A-1) Exact (A-2

0.01 0.1019 0.1111

0.1 0.2505 0.2637

0.5 0.4432 0.4388

1.0 0.5479 0.5340

5 0.8254 0.8046

10 0.9618 0.9453

100 1.5455 1.5567

PROBLEM 8.1

Water at 120oC boils inside a channel with a flat

surface measuring 45 cm 45 cm. Air at 62 m/s

and 20oC flows over the channel parallel to the

surface. Determine the heat transfer rate to the air.

Neglect wall resistance.

(1) Observations. (i) This is an external forced convection problem. (ii) The geometry can be modeled as a flat plate. (iii) Surface temperature is uniform. (iv) Newton’s law of cooling gives heat transfer rate from the surface to the air. (v) The average heat transfer coefficient must be determined. (vi) The Reynolds number should be evaluated to establish if the flow is laminar, turbulent or mixed. (vii) Analytic or correlation equations give the heat transfer coefficient.

(2) Problem Definition. Determine the average heat transfer coefficients for forced convection over a flat plate.

(3) Solution Plan. Compute the Reynolds number at the trailing edge to establish if it is laminar or turbulent. Apply Newton’s law of cooling. Use analytic or correlation equations to determine the local heat transfer coefficient.

(4) Plan Execution.

(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3) two-dimensional flow, (4) uniform upstream velocity and temperature, (5) flat plate, (6) uniform surface temperature, (7)

constant properties, (8) no buoyancy ( = 0 or g = 0) and (9) no radiation

(ii) Analysis. Newton’s law of cooling gives the total heat transfer rate

)(2 TTLhq sT (a)


L = length and width of the flat surface = 0.45 m

Tq = total heat transfer rate, W


T = free stream temperature = 20 Co

To determine the average heat transfer coefficient, the Reynolds number is computed to establish if the flow is laminar or turbulent. The Reynolds number is defined as

xRe =xV

(b)

where

xRe = Reynolds number

V = upstream velocity = 62 m/s

x = distance from the leading edge of the plate, m


Properties are evaluated at the film temperature fT defined as

T

V

air

water water


2

TTT s

f (c)

For the flow over a flat plate, transition Reynolds number txRe is

txRe = 5 105 (d)

The flow is laminar if txx ReRe . Substituting into (c)

fT = (120 + 20)( Co )/2 = 70oC

Properties of air at this temperature are given in Appendix D

k = 0.02922 CW/m o

Pr = 0.707

= 19.9 10-6 /sm2

Evaluating the Reynolds number in (b) at Lx

LRe = 6

2610402.1

)/sm(109.19

)m(45.0)m/s(62

Therefore, the flow is mixed over the plate. The average Nusselt, LNu , number for a plate with

laminar and turbulent flow is given by equation (8.7b)

3/15/45/42/1 )()(037.0)(664.0 PrReReRek

LhNu

tLtL xx (e)

Equation (e) is subject to the limitations on th e Pohlhausen’s solution and following conditions

flat plate, constant sT

5 510 < xRe < 710

0.6 < Pr < 60

properties at fT

All conditions are satisfied.

(iii) Computations. Solving (e) for h

h = 3/15/455/462/15o

)707.0(])105()104.1[(037.0)105(664.0)m(45.0

)CW/m(02922.0

h = 126.6 CW/m o2


Tq = 126.6( CW/m o2 ) )m()45.0( 22 (120 - 20)( Co ) = 2,562.8 W

(iv) Checking. Dimensional check: Computations showed that equations (a), (b) and (e) are dimensionally consistent.

Quantitative check: The magnitudes of the heat transfer coefficient is within the range given in Table 1.1 for force convection of gases.


Limiting check: If the flow is laminar over the entire plate, ReL =txRe , equation (e) reduces to

Pohlhausen’s solution.

(5) Comments. (i) It is important to compute the Reynolds number to determine if the flow is laminar or turbulent. (ii) For laminar forced convection over a flat plate, the heat transfer coefficient decreases as the distance from the leading edge is increased. However, if transition takes place, the heat transfer coefficient increases at the transition location and drops as the distance from the leading edge is increased.

PROBLEM 8.2

Steam at 105oC flows inside a specially designed narrow channel.

Water at 25oC flows over the channel with a velocity of 0.52 m/s.

Assume uniform outside surface temperature Ts = 105oC.

[a] Determine surface heat flux at 20 cm and 70 cm downstream

from the leading edge of the channel.

[b] Determine the total heat removed by the water if the length is L

= 80 cm and the width is W = 100 cm.

(1) Observations. (i) This is an external forced convection problem. (ii) The geometry can be modeled as a flat plate. (iii) Surface temperature is uniform. (iv) To determine the heat flux at a given location, the local heat transfer coefficient must be determined. (v) The average heat transfer coefficient is needed to determine the total heat transfer rate. (vi) Newton’s law of cooling gives surface flux and total heat transfer rate. (vii) The Reynolds number should be checked to establish if the flow is laminar, turbulent or mixed. (viii) Analytic or correlation equations give the heat transfer coefficient.

(2) Problem Definition. Determine the local and average heat transfer coefficients for forced convection over a flat plate.

(3) Solution Plan.

[a] Apply Newton's law of cooling to determine the local heat flux. Check the Reynolds number at 0.2 m and 0.7 m from the leading edge to see if it is laminar or turbulent. Use analytic or correlation equations for the local heat transfer coefficient.

[b] Apply Newton's law of cooling to the entire plate to determine the total heat loss. Check the Reynolds number at 0.8 m from the leading edge. Select an appropriate equation to determine the average heat transfer coefficient for the plate.

(4) Plan Execution.


constant properties, (8) no buoyancy ( = 0 or g = 0) and (9) no radiation.

(ii) Analysis.

[a] Newton’s law of cooling gives the local heat flux

qs = h (Ts - T ) (a)

where

h = local heat transfer coefficient, W/m2-oCqs = local surface heat flux, W/m2

Ts = surface temperature = 105oC


To determine the local heat transfer coefficient, the Reynolds number is computed to establish if the flow is laminar or turbulent. The Reynolds number is defined as

Rex =V x

(b)

Where

steam

water

x

sT

0

L

WVT

sq

steam

water

W

L






Tf = 2

TTs (c)

For the flow over a flat plate, transition Reynolds number tx

Re is

txRe = 5 105 (d)

The flow is laminar if Rex < txRe .Substituting into (c)

Tf = (105 + 25)(oC)/2 = 65oC


k = 0.6553 W/m-oCPr = 2.77

= 0.4424 10-6 m2/s

The Reynolds number at x = 20 cm = 0.2 m is

Rex = 081,235)/sm(104424.0

)m(2.0)m/s(52.026

Comparing this with the transition Reynolds number in (d) shows that the flow is laminar at x = 20 cm. Similarly, the Reynolds number at x = 70 cm = 0.7 m is

Rex = 785,822)/(104424.0

)(7.0)/(52.026 sm

msm

Thus, the flow is turbulent at x = 70 cm. The local heat transfer coefficient for laminar flow is given by Pohlhausen’s solution. For 0.5 < Pr < 50, equation (7.24b) gives

h = 0.332 kV

xPr

1/3 (e)

The local Nusselt number for turbulent flow over a flat plate is given by equation (8.4a)

3/15/4 )()(0296.0 PrRek

hxNu xx (f)

This correlation equation applies toflat plate, constant Ts

5 105 < Rex < 107

0.6 < Pr < 60 properties at Tf (g)

Since all conditions in (g) are satisfied, equation (f) is applicable to this case.

[b] Newton’s law of cooling gives the total heat transfer rate

qT = )( TTLWh s (h)


where

h = average heat transfer coefficient, W/m2-oCL = length of plate = 80 cm = 0.8 m


W = width of plate = 100 cm = 1.0 m

The Reynolds number at x = L = 0.8 m is

ReL = 325,940)s/m(104424.0

)m(8.0)s/m(52.026

Therefore, the flow is mixed over the plate. The average Nusselt number for a plate with laminar and turbulent flow is given by equation (8.7b)

3/15/45/42/1 )()(037.0)(664.0 PrReReRek

LhNu

tt xLxL (i)

where


Equation (i) is subject to the condi tions listed in (g) and the limitations on Pohlhausen’s solution. Both are satisfied by this case.

(iii) Computations.

[a] Heat flux at x = 20 cm. The local heat transfer coefficient for laminar flow at x = 20 cm is determined using (e). Solving (e) for h

h = 3/12/1o

)77.2()081,235(332.0)m(2.0

)Cm/W(6553.0= 740.7 W/m2-oC

Substituting this result in (a)

sq = 740.7(W/m2-oC)(105 - 25)(oC) = 59,256 W/m2

Heat flux at x = 70 cm. The local heat transfer coefficient is given by (f). Solving (f) for h

h = 0.0296)m(7.0

)Cm/W(6553.0 o

(822,785)4/5 (2.77)1/3 = 2,101 W/m2-oC


sq = 2101(W/m2-oC)(105 - 25)(oC) = 168,080 W/m2

[b] Total heat transfer from plate. Solving (i) for h

3/15/45/42/1 )()(037.0)(664.0 PrReReReL

kh

tt xLx

h = 3/15/45/42/1o

)77.2(])000,500()325,940[(037.0)000,500(664.0)m(8.0

)Cm/W(6553.0= 1554.3 W/m2-oC


qT

= 1554.3(W/m2-oC) 0.8(m)1.0(m) (105 - 25)(oC) = 99,475 W


(iv) Checking. Dimensional check: Computations showed that units of equations (a), (b), (e), (f), (h) and (i) are dimensionally consistent.

Quantitative check: The magnitudes of heat transfer coefficients are within the range given in Table 1.1 for force convection of liquids.

Limiting check: If the flow is laminar over the entire plate, ReL =txRe , equation (i) reduces to

Pohlhausen’s solution.

(5) Comments. (i) It is important to check the Reynolds number to determine if the flow is laminar or turbulent. (ii) For laminar forced convection over a flat plate, the heat transfer coefficient decreases as the distance from the leading edge is increased. However, if transition takes place, the heat transfer coefficient increases at the transition location and drops as the distance from the leading edge is increased. (iii) The fact that the average heat transfer coefficient for the entire plate (L = 0.8 m) is larger than the local heat transfer coefficient at x = 0.2 m is due to transition from laminar to turbulent flow.

PROBLEM 8.3

Electronic components are mounted on one side of a circuit board. The board is cooled on the

other side by air at 23oC flowing with a velocity of 10 m/s. The length of the board is L = 20 cm

and its width is W = 25 cm. Assume uniform board temperature.

[a] Determine the maximum power that can be dissipated in the package if surface temperature

is not to exceed 77oC. Assume that all dissipated power is conducted through the plate to the air.

[b] To increase the maximum power without increasing surface temperature, it is recommended

that the boundary layer be tripped to turbulent flow very close to the leading edge. Is this a valid

recommendation? Substantiate your view.

(1) Observations. (i) This is an external forced convection problem of flow over a flat plate. (ii) Surface temperature is assumed uniform. (iii) The heat transfer coefficient in turbulent flow is greater than that in laminar flow. Thus higher heat transfer rates can be sustained in turbulent flow than laminar flow. (iv) The Reynolds number should be checked to establish if the flow is laminar, turbulent or mixed. (v) Heat loss from the surface is approximately equal to the power dissipated in the package. (vi) Newton’s law of cooling gives a relationship between heat transfer rate, surface area, heat transfer coefficient, surface temperature and ambient temperature. (vii) The fluid is air.

(2) Problem Definition. Determine the average heat transfer coefficient.

(3) Solution Plan. Apply Newton’s law of cooling to determine the maximum heat removed, check the Reynolds number to establish if the flow is laminar, turbulent or mixed and use appropriate average Nusselt number solutions or correlation equations to determine the average heat transfer coefficient.

(4) Plan Execution.


constant properties, (8) energy leaves the top surface only, (9) no buoyancy ( = 0 or g = 0) and (9) negligible radiation.

(ii) Analysis.

[a] Applying Conservation of energy to the package gives

P = q (a) where

P = power dissipated in package, W q = heat removed from top surface, W

Application of Newton’s law of cooling to the top surface

q = h A (Ts - T ) (b) where

A = plate area, m2

L

W

components

V

T


h = average heat transfer coefficient, W/m2-oCTs = surface temperature = 77oC


Surface area is A = LW (c)

where

L = plate length = 20 cm = 0.2 m W = plate width = 25 cm = 0.25m

To determine h it is necessary to first establish if the flow is laminar or turbulent. If the Reynolds number at the trailing end is smaller than the transition number, the flow is laminar throughout. Define

ReL =LV

(d)

and

txRe = 5 105 (e)

where

LRe = Reynolds number at the trailing end of plate

txRe = transition Reynolds number for flow over a flat plate

V = free stream velocity = 10 m/s


Properties are evaluated at the film temperature Tf

Tf = (Ts + T )/2 = (77 + 23)(oC)/2 = 50oC


k = thermal conductivity = 0.02781 W/m-oCPr = Prandtl number = 0.709



LRe =10 0 2

17 92 10 6 2

( / ) . ( )

. ( / )

m s m

m s = 111,607

Since this is smaller than txRe , it follows that the flow is laminar throughout. Therefore,

Pohlhausen's solution (7.26)for the average Nusselt number is applicable.

3/12/1664.0 PrRe

k

LhNu LL (f)

where LNu is the average Nusselt number. Solving (f) for h

h3/12/1

664.0 PrReL

kL (g)

[b] The average Nusselt number for mixed flow over a flat plate of length L is given by (8.7b)


3/15/45/42/1 )()(037.0)(664.0 PrReReRek

LhNu

tt xLxL (h)

If the boundary layer is tripped at the leading edge, the flow will be turbulent throughout. Since transition is assumed to take place at x = 0, it follows that

txRe = 0 (i)

Substituting (i) into (h)

037.0k

LhNu L ( LRe )4/5 (Pr)1/3 (j)

Solving (j) for h

h = 0.037 k

L ( LRe )4/5 (Pr)1/3 (k)

(iii) Computations.

[a] Laminar flow. Equation (c) gives surface area

A = 0.2(m) 0.25(m) = 0.05 m2

Use (g) to calculate h

h 0 6640 02781

0 2111 607 0 709

1 2 1 3.

. ( / )

. ( ), .

/ /W m C

m

o

= 27.5 W/m2-oC

Substituting into (b) and using (a) gives

P = q = 27.5 (W/m2-oC) 0.05(m2) (77 - 23)(oC) = 74.25 W

[b] Turbulent flow. Use (k) to obtain h

h = 0.037 0 02781

0 2

. ( / )

. ( )

W m C

m

o

(111,607)4/5 (0.709)1/3 = 50.09 W/m2-oC

Substituting into (b) and using (a)

P = q = 50.09 (W/m2-oC) 0.05(m2) (77 - 23)(oC) = 135.2 W

(iv) Checking. Dimensional check: Computations showed that equations (b), (c), (d), (g) and (k) are dimensionally consistent.

Quantitative check: The magnitude of the heat transfer coefficient for both laminar and turbulent flow is within the range of values listed in Table 1.1. Furthermore, as expected, the average heat transfer coefficient for turbulent flow is higher than laminar flow.

(5) Comments. (i) Tripping the boundary layer increases the maximum power by 82 % without increasing surface temperature. (ii) The disadvantage of tripping the boundary layer is the corresponding increase in pressure drop.

PROBLEM 8.4

Water at 15oC flows with a velocity of 0.18 m/s over a plate of length L = 20 cm and width W =

25 cm. Surface temperature is 95oC. Determine the heat transfer rate from the leading and

trailing halves of the plate.

(1) Observations. (i) This is an external forced convection problem. (ii) The geometry is a flat plate. (iii) Surface temperature is uniform. (iv) Newton’s law of cooling gives the heat transfer rate. (v) The Reynolds number should be checked to establish if the flow is laminar, turbulent or mixed. (vi) Analytic or correlation equations give the heat transfer coefficient. (vii) If the flow is laminar throughout, heat transfer from the first half should be greater than that from the second half. (viii) Second half heat transfer can be obtained by subtracting first half heat rate from the heat transfer from the entire plate. (ix) The fluid is water.

(2) Problem Definition. Determine the average heat transfer coefficient for the first half and for the entire plate.

(3) Solution Plan. Apply Newton’s law of cooling to the first half and to the entire plate. Check the Reynolds number at end of the first half and second half to establish if the flow is laminar, turbulent or mixed. Use analytic or correlation equations for the average heat transfer coefficient.

(4) Plan Execution.


constant properties, (8) no buoyancy ( = 0 or g = 0 ) and (9) no radiation.

(ii) Analysis. Application of Newton’s law of cooling to the first half gives

1q = 1h A1(Ts - T ) (a)

where

1A = surface area of half plate, m2

1h = average heat transfer coefficient for the first half, W/m2-oC

1q = heat transfer rate from the first half, W



Heat transfer from the second half is given by

12 qqq T (b)

where

2q = heat transfer rate from the second half, W

Tq = heat transfer rate from entire plate, W

Heat transfer from the entire plate is given by

Tq = Th AT(Ts - T ) (c)

where

TA = surface area of entire plate of length L, m2

Th = average heat transfer coefficient for entire plate of length L, W/m2-oC

2q1q

x

2/L 2/L

T

V

0


The areas A1 and AT areA1 = WL/2 (d)

andAT = WL (e)

where

L = plate length = 20 cm = 0.2 m W = plate width = 25 cm = 0.25 m

To determine the average heat transfer coefficient, the Reynolds number is computed to establish if the flow is laminar or turbulent. The Reynolds number is defined as

Rex =xV

(f)

where




Properties of water are evaluated at the film temperature Tf defined as

Tf = T Ts

2 (g)

For the flow over a flat plate, transition Reynolds number txRe is

txRe = 5 105 (h)

The flow is laminar if Rex < txRe .Substituting into (g)

Tf = (95 + 15)(oC)/2 = 55oC


k = 0.6458 W/m-oCPr = 3.27

= 0.5116 10-6 m2/s

The Reynolds number for the first half is evaluated at x = L/2 = 20 cm/2 = 10 cm = 0.1 m

ReL/2 = 184,35

s/m105116.0

)m(1.0)s/m(18.026

Therefore, the flow is laminar over the first half. The average Nusselt number for laminar flow is given by Pohlhausen’s solution. For 0.5 < Pr < 50, equation (7.26) gives

2/12/

1/312/ )(664.0

)2/(LL RePr

k

LhNu (i)

where

2/LNu = average Nusselt number for the first half

The Reynolds number for the entire plate is evaluated at x = L = 20 cm = 0.2 m

ReL = 367,70

s/m105116.0

)m(2.0)s/m(18.026


Therefore, the flow is laminar over the entire plate. The average Nusselt number for laminar is

2/11/3664.0 LT

L RePrk

LhNu (j)

(iii) Computations. Substituting into equations (d) and (e)

A1 = 0.2(m)0.25(m)/2 = 0.025 m2

andAT = 0.2(m)0.25(m) = 0.05 m2

Solving (i) for 1h

2/12/

1/31 664.0

2LRePr

L

kh = 2/13/1

o

)184,35()27.3(664.0)m(2.0

)Cm/W)(6458.0(2= 1193.9 W/m2-oC

Similarly, equation (j) gives

2/11/3 )(664.0 LT RePrL

kh = 2/13/1

o

)367,70()27.3(664.0)m(2.0

)Cm/W)(6458.0(= 844 W/m2-oC


1q = 0.025(m2)1193.9(W/m2-oC)(95 15 )(oC) = 2388 W

Equation (c) gives

Tq = 0.05(m2)844(W/m2-oC)(95 15 )(oC) = 3376 W

Substituting into (b) gives the heat transfer rate from the second half

2q = 3376 (W) 2388(W) = 988 W

(iv) Checking. Dimensional check: Computations showed that units of equations (a), (c)-(f), (i) and (j) are dimens ionally consistent.

Quantitative check: The magnitudes of heat transfer coefficients are within the range given in Table 1.1 for forced convection of liquids.

Qualitative check: As anticipated, heat transfer from the first half is greater than that from the second half.

(5) Comments (i) For laminar flow, two half plates oriented in parallel (side by side) transfer

more heat than two placed in series. (ii) An alternate method for determining 2q is to determine

the average heat transfer coefficient for the second half. This requires integration of the local heat transfer coefficient from x = 0.1 m to x = 0.2 m.

PROBLEM 8.5

A chip measuring 5 mm 5 mm is placed flush on a flat plate 18 cm from the leading edge. The

chip is cooled by air at 17oC flowing with a velocity of 56 m/s. Determine the maximum power

that can be dissipated in the chip if its surface temperature is not to exceed 63oC. Assume no

heat loss from the back side of the chip.

(1) Observations. (i) The chip is cooled by forced convection. (ii) This problem can be modeled as a flat plate with an unheated leading section. (iii) Newton's law of cooling can be applied to determine the rate of heat transfer between the chip and the air. (iv) Check the Reynolds number to establish if the flow is laminar or turbulent.

(2) Problem Definition. Find the average heat transfer coefficient over the chip.

(3) Solution Plan. Apply Newton’s law of cooling to determine the maximum heat removed from the chip, check the Reynolds number to establish if the flow is laminar or turbulent, model the chip as flat heated surface with an insulated leading section and use appropriate Nusselt number solutions or correlation equations to determine the average heat transfer coefficient.

(4) Plan Execution.

(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3) two-dimensional flow, (4) uniform upstream velocity and temperature, (5) flat plate, (6) chip surface temperature is uniform, (7) constant properties, (8) all power dissipated in the chip is transferred to the air from

its surface, (9) the chip is mounted on an insulated plate, (10) no buoyancy ( = 0 or g = 0 ) and (10) no radiation.

(ii) Analysis. Applying Conservation of energy to the chip

P = q (a) where

P = power dissipated in chip, W q = heat removed from top surface, W

Application of Newton’s law of cooling to the top surface


A = surface area of chip, m2



The surface area is A = LW (c)

where

L = chip length = 5 mm = 0.005 m W = chip width = 5 mm = 0.005 m

To determine h it is necessary to first establish if the flow over the chip is laminar or turbulent.

0x

S

xo

T

V chip


If the Reynolds number at the leading end of the chip is larger than the transition number, the flow is turbulent over the chip. Define

Rex =V x

(d)

and

txRe = 5 105 (e)

where

xRe = local Reynolds number


V = free stream velocity = 56 m/s x = variable, measured from the leading edge of plate, m



Tf = (Ts + T )/2 = (63 + 17)(oC)/2 = 40oC



= kinematic viscosity = 16.96 10-6 (m2/s)

Evaluating the Reynolds number in (d) at the leading edge of the chip x = xo = 0.18 m

oxRe =56 018

16 96 10 6 2

( / ) . ( )

. ( / )

m s m

m s = 594,340

Since this is larger than txRe , it follows that the flow is turbulent over the chip. Equation (8.8)

gives the local Nusselt number for flow over a plate with an insulated leading section

Nuhx

k x xx

x0 0296

1

4 5

9 10 1 9

. ( ) (

( / )

/

/ /

Re Pr)1/3

o

(f)

where

Nux = local Nusselt number xo = distance from the leading edge of the plate to the chip = 0.18 m

However, what is needed in equation (b) is the average heat transfer coefficient. Since the chip is small compared to the distance xo, it is reasonable to assume that the average Nusselt number is approximately equal to the local value at the center of the chip, s

9/110/9o

1/35/4

)/(1

)()(0296.0

sx

PrRe

k

shNu s

s (g)

where

Nus = average Nusselt number over the chip

s = distance from leading edge of plate to center of chip, m


Reynolds number Res and distance s are given by

Res = sV

(h)

ands = xo + L/2 (i)

(iii) Computations. Substituting into (i) and (h)

s = 0.18(m) + 0.005(m)/2 = 0.1825 m

Res =56 01825

16 96 10 6 2

( / ) . ( )

. ( / )

m s m

m s = 602,594

Solving (g) for h and substituting numerical values

9/110/9o

1/35/4

)/(1

)()(0296.0

sx

PrRe

s

kh s

9/110/9

3/15/4o

])m1825.0/m18.0(1[

)71.0()594,602(0296.0

)m(1825.0

)Cm/W(0271.0

h = 268.8 W/m2-oC

Substituting into (b) and using (a) and (c)

P = q = 268.8 (W/m2-oC)0.005(m)0.005(m) ( 1763 )(oC) = 0.309 W

(iv) Checking. Dimensional check: Computations showed that equations (b)-(d) and (g)-(i) are dimensionally consistent.

Quantitative check: The magnitude of the heat transfer coefficient is within the range of values listed in Table 1.1 for forced convection of gases.

(5) Comments. The assumption that the average heat transfer coefficient over the chip is approximately equal to the local value at the center of the chip was made to avoid the need to integrate the local value, equation (f), over the surface of the chip. This approximation becomes less reasonable as the dimension of the chip in the x direction becomes large.

PROBLEM 8.6

A 1.2 m 1.2 m solar collector is mounted flush on

the roof of a house. The leading edge of the collector

is located 5 m from the leading edge of the roof.

Estimate the heat loss to the ambient air on a typical

winter day when wind speed parallel to the roof is 12

m/s and air temperature is 5oC. Outside collector

surface temperature is estimated to be 35oC.

(1) Observations. (i) Heat transfer from the collector to the air is by forced convection. (ii) This problem can be modeled as a flat plate with an unheated leading section. (iii) Newton's law of cooling can be applied to determine the rate of heat transfer between the collector and air. (iv) The heat transfer coefficient varies along the collector. (v) The Reynolds number should be computed to establish if the flow is laminar or turbulent.

(2) Problem Definition. Determine the heat transfer coefficient over the collector.

(3) Solution Plan. Apply Newton’s law of cooling to the collector, compute the Reynolds number to establish if the flow is laminar or turbulent, model the collector surface as flat heated surface with an insulated leading section and use appropriate Nusselt number solutions or correlation equations to determine the heat transfer coefficient.

(4) Plan Execution.

(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3) two-dimensional flow, (4) uniform upstream velocity and temperature, (5) flat surface, (6) collector surface temperature is

uniform, (7) constant properties, (8) roof is insulated, (9) no buoyancy ( = 0 or g = 0 ) and (9) no radiation.

(ii) Analysis. Since the heat transfer coefficient varies with distance along the collector, integration of Newton’s law of cooling gives the total heat transfer to the air

L

sT dxxhwTTq0

)()( (a)

where


L = length of collector = 1.2 m



T = ambient temperature = 5 Co

w = width of collector = 1.2 m

To determine the local heat transfer coefficient, the Reynolds number is computed to establish if the flow is laminar or turbulent. If the Reynolds number at the leading end of the collector is larger than the transition number, the flow is turbulent over the collector. Define

Rex =xV

(b)

and

txRe = 5 105 (c)

V

T

m6.5

solarcollector

PROBLEM 8.6 (continued) where

xRe = local Reynolds number


V = wind speed = 12 m/s

x = distance from the leading edge of the roof, m


Properties are evaluated at the film temperature fT , defined as

2

TTT s

f (d)


fT = (35 + 5)( Co )/2 = 20oC


k = 0.02564 CW/m o

Pr = 0.713

= 15.09 10-6 /sm2

Evaluating the Reynolds number in (b) at 5oxx m

LRe = 6

26109761.3

)/sm(1009.15

)m(5)m/s(12

Since the Reynolds number is greater than the transition number it follows that the flow is turbulent over the collector. Equation (8.8) gives the local Nusselt number for flow over a plate with an insulated leading section

Nuhx

k x xx

x0 0296

1

4 5

9 10 1 9

. ( ) (

( / )

/

/ /

Re Pr)1/3

o

(e)

where

Nux = local Nusselt number xo = distance from the leading edge of the roof to the collector = 5 m

Solving (e) for h

9/110/9o

1/35/4

)/(1

)()(0296.0

xx

PrRe

x

kh x (f)

However, when (f) is substituted in (a), the resulting integral can not be evaluated analytically. Thus, numerical integration is required. An approximate approach is to assume that the heat transfer coefficient over the collector is uniform equal to the local value at the center of the collector. Thus

)()( cxhxh (g)

where

cx = distance from the leading edge of the roof to the center of collector = 5.6 m

Substituting (g) into (a) and evaluating the integral


wLhTTq csT )( (h)

Evaluating (f) at cxx

9/110/9o

1/35/4

)/(1

)()(0296.0

c

cx

c

c

xx

PrRe

x

kh (i)

(iii) Computations. The Reynolds number at collector center 6.5cxx m is

xcRe = 6

26104533.4

)/sm(1009.15

)m(6.5)m/s(12


CW/m7.32])m6.5/m5(1[

)713.0()104533.4(0296.0

)m(6.5

)CW/m(02564.0 o2

9/110/9

3/15/46o

ch

Substituting (g)

W6.1412)m(2.1)m(2.1)CW/m(7.32)C)(535( o2oTq

(iv) Checking. Dimensional check: Computations showed that equations (b), (h) and (i) are dimensionally consistent.

Quantitative check: The magnitude of the heat transfer coefficient is within the range of values listed in Table 1.1 for forced convection of gases.

(5) Comments. The assumption that the average heat transfer coefficient over the collector is approximately equal to the local value at the center of the collector was made to avoid the need to numerically evaluate the integral in equation (a). This approximation becomes less reasonable as the dimension of the collector in the x direction becomes large.

PROBLEM 8.7

Water at 20oC flows over a rectangular plate of length L =

1.8 m and width W = 0.3 m. The upstream velocity is 0.8

m/s and surface temperature is 80oC. Two orientations are

considered. In the first orientation the width W faces the

flow and in the second the length L faces the flow. Which

orientation should be selected to minimize heat loss from the

plate? Determine the heat loss ratio of the two orientations.

(1) Observations. (i) This is an external forced convection problem. (ii) The flow is over a flat plate. (iii) Surface temperature is uniform. (iv) Plate orientation is important. (v) Variation of the heat transfer coefficient along the plate affects the total heat transfer. (vi) The heat transfer coefficient for laminar flow decreases as the distance from the leading edge is increased. However, at the transition point it increases and then decreases again. (vii) Higher rate of heat transfer may be obtained if the wide side of a plate faces the flow. On the other hand, higher rate may be obtained if the long side of the plate is in line with the flow direction when transition takes place. (viii) The fluid is water.

(2) Problem Definition. Determine the average heat transfer coefficient for the flow over a rectangular plate for two orientations: [a] wide side facing the flow and [b] short side facing the flow.

(3) Solution Plan. Apply Newton's law of cooling for flow over a flat plate. Check the Reynolds number for the two orientations and select appropriate equations for the average Nusselt number to obtain the average heat transfer coefficient.

(4) Plan Execution.


constant properties, (8) no buoyancy ( = 0 or g = 0 ) and (9) negligible radiation.

(ii) Analysis. Newton's law of cooling gives the total heat transfer rate

q = h A(Ts - T ) (a) where

A = plate area, m2

h = average heat transfer coefficient, W/m2-oCq = heat transfer rate, W



Surface area is A = LW (b)

where

L = plate length = 1.8 m W = plate width = 0.3 m

To determine h it is necessary to first establish if the flow is laminar or turbulent. If the Reynolds

0

V

Tx

x

2

1

2W

1W

2L

1L

0


number at the trailing end is smaller than the transition number, the flow is laminar throughout. Define

ReL =V L

(c)

and

txRe = 5 105 (d)

where

ReL = Reynolds number at the trailing end of plate

txRe = transition Reynolds number for flow over plate




Tf = (Ts + T )/2 = (80 + 20)(oC)/2 = 50oC




Consider two orientations, 1 and 2 as shown. In orientation 1, the short side faces the flow. In orientation 2, the long side faces the flow. The Reynolds numbers corresponding to the two orientations are computed using (c)

1LRe =)s/m(105537.0

)m(8.1)s/m(8.026

= 2,600,686, turbulent

Similarly, for orientation 2

2LRe =)s/m(105537.0

)m(3.0)s/m(8.026

= 433,448, laminar

where

L1 = L = 1.8 m L2 = W = 0.3 m

Comparing1LRe and

2LRe withtxRe shows that the flow is mixed (laminar and turbulent) for

orientation 1, and laminar for orientation 2. The average Nusselt number for mixed flow over a flat plate of length L1 is given by (8.7b)

3/15/45/4

1

2/111

1037.0664.0 PrReReRe

k

LhNu

tt xLxL (e)

Valid for flat plate, constant Ts

5 105 < Rex < 107

0.6 < Pr < 60 properties at Tf (f)

Since all conditions in (f) are satisfied, equation (e) is applicable to this case. For orientation 2, Pohlhausen's solution (7.26) gives the average Nusselt number


3/12/1222 )(664.0

2PrRe

k

LhNu LL (g)

(iii) Computations. Solving (E) for 1h and substituting numerical values

3/15/45/42/1

1

1 )()(037.0)(664.01

PrReReReL

kh

tt xLx

1h 3/15/45/42/1o

)57.3()000,500()686,600,2(037.0)000,500(664.0)m(8.1

)Cm/W(6405.0

1h = 2253.5 W/m2-oC

Similarly, solving (g) for 2h and substituting numerical values

2h = 3/12/1

2

)(664.02

PrReL

kL

2h = 0.664 3/12/1o

)57.3()448,433()m(3.0

)Cm/W(6405.0= 1426.5W/m2-oC

Substituting into equation (a) and using (b) gives the heat transfer rate for each orientation

1q = 2253.5(W/m2-oC)1.8(m)0.3(m)( 2080 )(oC) = 73,013 W

and

2q = 1426.5(W/m2-oC)1.8(m)0.3(m)( 2080 )(oC) = 46,219 W

The ratio of the two heat transfer rates is

)W(219,46

)W(013,73

2

1

q

q= 1.58

(iv) Checking. Dimensional check: Computations showed that equations (a), (b), (c), (e) and (g) are dimensionally consistent.

Quantitative check: Values of the heat transfer coefficients are within the range listed in Table 1.1 for forced convection of liquids.

Limiting check: For the special case of a square plate L1 = L2, the ratio of the two heat transfer

rates should be unity. Either equation (e) or (g) gives 1h = 2h . Equation (a) gives 1q = 2q .

(5) Comments. If the flow is laminar for both orientations, orientation 1 will have a lower heat transfer rate than orientation 2. However, if transition takes place in one or both orientations, numerical calculations must be carried out to determine which orientation has the higher heat transfer rate.

PROBLEM 8.8

100 flat chips are placed on a 10 cm 10 cm circuit board and cooled by forced convection of

air at 27oC. Each chip measures 1 cm 1 cm and dissipates 0.13 W. The maximum allowable

chip temperature is 83oC. Free stream air velocity is 5 m/s. Tests

showed that several chips near the trailing end of the board

exceeded the allowable temperature. Would you recommend

tripping the boundary layer to turbulent flow at the leading edge to solve the overheating problem? Substantiate your recommendation.

(1) Observations. (i) This is an external forced convection problem. (ii) The flow is over a flat plate. (iii) The problem can be modeled as flow over a flat plate with uniform surface heat flux. (iv) Surface temperature varies with distance along plate. The highest surface temperature is at the trailing end. (v) Tripping the boundary layer at the leading edge changes the flow from laminar to turbulent. This increases the heat transfer coefficient and lowers surface temperature. (vi) Newton’s law of cooling gives surface temperature.

(2) Problem Definition. Determine the local heat transfer coefficient at the trailing end for turbulent flow.

(3) Solution Plan. Apply Newton's law of cooling at the trailing end and use a correlation equation for turbulent flow over a flat plate at constant surface heat flux to determine the local heat transfer coefficient. Solve Newton's law for the surface temperature.

(4) Plan Execution.

(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3) two-dimensional flow, (4) uniform upstream velocity and temperature, (5) flat plate, (6) uniform surface heat flux, (7)

constant properties, (8) no buoyancy ( = 0 or g = 0 ), (9) negligible radiation and (10) turbulent flow.

(ii) Analysis. Apply Newton's law of cooling

qs = h(x) [Ts(x) - T ] (a)

where

h(x) = local heat transfer coefficient, W/m2-oCqs = surface heat flux = 0.13 W/cm2 = 1300 W/m2

Ts(x) = local surface temperature, oC



Solving equation (a) for Ts(x)

)()(

xh

qTxT s

s (b)

The local Nusselt number for turbulent flow over a flat plate at constant surface heat flux is given by equation (8.9)

Nuhx

kx x0 030 4 5 1 3. / /Re Pr (c)

V

T

x0

L

V

T

sq

)(xTs


where

k = thermal conductivity, W/m-oCNux = local Nusselt number

Rex = local Reynolds number = /xV

Pr = Prandtl number



Equation (c) gives the local heat transfer coefficient. Properties are evaluated at the film temperature defined as

2/)( TTT sf (d)

where

fT = film temperature, oC

sT = average surface temperature, oC

Since surface temperature varies along the plate, an average value, sT , is used to determine film

temperature. sT is approximated by

2/)]([ LTTT ss (e)

(iii) Computations. To determine air properties, surface temperature at the trailing end, Ts(L), is needed to compute Tf. However, Ts(L) is unknown. In fact, the objective of the problem is determining Ts(L). To proceed, a solution is obtained by trial and error procedure. A value for Ts(L)is assumed, (d) and (e) are used to determine Tf, (c) is used to calculate h(L) and (b) is used to calculated Ts(L). If the calculated Ts(L) is not equal to the assumed value, the process is repeated until a satisfactory agreement is obtained between assumed and calculated values.

Assume Ts(L)= 79oC. Equations (e) and (d) give

sT = (27 + 79)(oC)/2 = 53oC

andTf = (53 + 27)(oC)/2 = 40oC


k = 0.0271 W/m-oCPr = 0.71

= 16.96 10-6 m2/s

Thus, the Reynolds number at x = L = 0.1 m is

ReL =LV

481,29)s/m(1096.16

)m(1.0)s/m(526

Solving (c) for h and evaluating the resulting equation at x = L = 0.1 m

3/15/4 )()(03.0)( PrReL

kLh L

3/15/4o

)71.0()481,29()m(1.0

)Cm/W(0271.003.0 = 27.3 W/m2-oC



Ts(L) = 27oC + )Cm/W(3.27

)m/W(1300o2

2

= 74.6oC

This is reasonably close to the assumed value of 79oC. Repeating the calculation with a new assumed value of 75oC will result in a minor change in the result.

(iv) Checking. Dimensional check: Computations showed that units for equations (b) and (c) are consistent.

Quantitative check: the value of h(L) is within the range given in Table 1.1 for forced convection of gases.

(5) Comments. (i) By tripping the boundary layer to cause transition to turbulent flow, surface temperature at the trailing end will not exceed the maximum allowable level. (ii) Since the

Reynolds number at the trailing end is less than the transition value of 5 105, the flow will be laminar if it is not tripped. To determine surface temperature under laminar flow conditions, the corresponding heat transfer coefficient must be computed. The Nusselt number for laminar flow over a flat plate with uniform surface flux is given by equation (7.31)

2/13/1453.0 xx RePrk

hxNu (f)

Solving this equation for h(L)

2/13/1 )()(453.0)( LRePrL

kLh 2/13/1

o

)481,29()71.0()m(1.0

)Cm/W(0271.0453.0 = 18.8 W/m2-oC


Ts(L) = 27oC + )Cm/W(8.18

)m/W(1300o2

2

= 96.1oC

Since this is considerably higher than the assumed value of 79oC, the procedure is repeated with a new assumed value of Ts(L) = 99oC. This yields a calculated value of Ts(L) = 96.2oC. Thus for laminar flow, surface temperature exceeds the allowable maximum level of 83oC.

PROBLEM 8.9

Water at 27oC flows normally over a tube with a velocity of 4.5 m/s. The outside diameter of the

tube is 2 cm. Condensation of steam inside the tube results in a uniform outside surface

temperature of 98oC. Determine the length of tube needed to transfer 250,000 W of energy to the

water.

(1) Observations. (i) This is an external forced convection problem. (ii) The flow is normal to a tube. (iii) Surface temperature is uniform. (iv) Tube length is unknown. (v) Newton’s law of cooling can be used to determine surface area. Tube length is related to surface area. (vi) The fluid is water.

(2) Problem Definition. The required tube length can be determined from Newton's law of cooling. Thus, the problem is finding the average heat transfer coefficient for flow normal to a cylinder.

(3) Solution Plan. Apply Newton's law of cooling to the flow over tube. Use forced convection correlation equation to determine the average Nusselt number.

(4) Plan Execution.

(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3) two-dimensional flow, (4) uniform upstream velocity and temperature, (5) uniform surface heat flux, (6) constant

properties, (7) no buoyancy ( = 0 or g = 0 ) and (8) no radiation.

(ii) Analysis. Newton's law of cooling gives

q = h A(Ts T ) (a)

where

A = plate area, m2


q = heat transfer rate = 250,000 W



Surface area is A = DL (b)

where

D = outside tube diameter = 2 cm = 0.02 mL = tube length, m

Substituting (b) into (a) and solving the resulting equation for L

)( TThD

qL

s

(c)

Equation (8.10a) gives a correlation for the average Nusselt number for forced convection normal to a cylinder

5/48/5

4/13/2

3/12/1

000,282/1

/41

62.03.0 D

DD Re

Pr

PrRe

k

DhuN (d)

sTD

L

V T TV


Valid for: flow normal to cylinder Pe = Re PrD > 0.2

properties at Tf (e)where



Pe = Peclet number = DPrRe

Pr = Prandtl number



Reynolds number is defined as

DVReD (f )

where



Properties are determined at the film temperature fT defined as

Tf = 2/)( TTs (g)

The Reynolds number is computed to establish if (e) is satisfied. Substituting into (g)

Tf = (98 + 27)(oC)/2 = 62.5oC




Thus

DRe)s/m(104586.0

)m(02.0)s/m(5.426

= 196,249

and

Pe = PrReD = 196,249 (2.885) = 5.66 105

Therefore, (e) is satisfied and correlation equation (d) is applicable.

(iii) Computations. Equation (d) gives the average heat transfer coefficient

5/48/5

4/13/2

1/32/1

000,282/249,1961

885.2/4.01

(2.885))249,196(62.03.0

k

DhuN D = 589

Solving for h


h =D

k589 =

)m(02.0

)Cm/W(653.0589

o

= 19,231 W/m2-oC

Substituting into (c) gives the required length

L =)C)(2798)(CW/m(231,19)m)(02.0(

)W(000,250oo2

= 2.91 m

(iv) Checking. Dimensional check: Computations showed that units for equations (c), (d) and (f) are dimensionally consistent.

Quantitative check: The value of h is within the range given in Table 1.1 for forced convection of liquids.

Limiting check: An infinitely long tube is needed if TTs . Setting TTs in equation (c)

gives L = .

(5) Comments. (i) The required length to transfer 250 kw of heat is only 2.43 m. This appears unreasonably short. A review of the analysis and calculations uncovered no errors. The relatively short length needed is due to the high heat transfer coefficient associated with forced convection of water. Note that the calculated heat transfer coefficient is at the high end of values given in Table 1.1. (ii) It was not necessary to consider the thermal interaction between the surface and the fluid inside the tube because outside surface temperature was specified.

PROBLEM 8.10

A proposed steam condenser design for marine applications is based on the concept of rejecting

heat to the surrounding water while a boat is in motion. The idea is to submerge a steam-

carrying tube in the water such that its axis is normal to boat velocity. Estimate the rate of

steam condensation for a 75 cm long tube with an outside diameter of 2.5 cm. Assume a

condensation temperature of 90oC and a uniform surface temperature of 88

oC. Ambient water

temperature is 15oC and boat speed is 8 m/s.

(1) Observations. (i) Heat is removed by the water from the steam causing it to condense. (ii) The rate at which steam condenses inside the tube depends on the rate at which heat is removed from the outside surface. (iii) Heat is removed from the outside surface by forced convection. (iv) This is an external forced convection problem of flow normal to a tube. (v) Newton’s law of cooling gives the rate of heat loss from the surface.

(2) Problem Definition. Determine the rate of heat transfer from the outside surface to the ambient water.

(3) Solution Plan. Apply conservation of energy to the condensing steam inside the tube. Use Newton’s law of cooling to determine the heat removed from the tube. Use correlation equation for forced convection flow normal to a tube to determine the average heat transfer coefficient (Nusselt number).

(4) Plan Execution.

(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3) two-dimensional flow, (4) uniform upstream velocity and temperature, (5) uniform surface temperature, (6) constant properties, (7) no radiation, (8) negligible changes in kinetic and potential energy of steam, (9)

negligible axial conduction in tube and steam, constant properties, (10) no buoyancy ( = 0 or g= 0 ) and (11) no radiation.

. (ii) Analysis. Applying conservation of energy to the steam between the inlet and outlet of the tube gives

q = m h hi o = fghm ˆ (a)

or, solving (a) for m

m =

fgh

q

ˆ (b)

where

hi = steam enthalpy at inlet of tube, J/kg

ho = steam enthalpy at outlet of tube, J/kg

fgh = latent heat of condensation = 2283.2 kJ/kg (at 90 oC)

m = rate of steam condensation, kg/s q = rate of heat removed from steam, W

The latent heat, fgh , is defined by the temperature of the condensing steam. Thus to determine

steam condensation rate from (b), the rate of heat removal from steam must be obtained. Applying conservation of energy to the tube gives


q = Energy removed from steam and added to tube

= Energy removed from tube and added to water (c)

Newton’s law of cooling gives the energy removed from tube surface by convection and added to water

q = h DL (Ts - T ) (d) where

D = diameter of tube = 2.5 cm = 0.025 m

h = average heat transfer coefficient at the outer surface of tube, W/m2-oCL = length of tube = 75 cm = 0.75 m Ts = surface temperature = 88oC



5/48/5

4/13/2

3/12/1

000,282/1

/4.01

62.03.0 D

DD Re

Pr

PrRe

k

DhuN (e)

Valid for: flow normal to cylinder

Pe = Re PrD > 0.2

properties at Tf (f)where




Pr = Prandtl number




DVReD (g)

where




Tf = 2/)( TTs (h)

The Reynolds number is computed to establish if (f) is satisfied. Substituting into (h)

Tf = (88 + 15)(oC)/2 = 51.5oC


k = thermal conductivity = 0.6421 W/m-oC

T

VsT


Pr = Prandtl number = 3.48


Thus

DRe)s/m(105411.0

)m(025.0)s/m(826

= 369,617

and

Pe = PrReD = 369,617(3.48) = 1.286 106

Therefore, (f) is satisfied and correlation equation (e) is applicable.

(iii) Computations. Equation (e) gives the average Nusselt number and average heat transfer coefficient

5/48/5

4/13/2

1/32/1

000,282/617,3691

48.3/4.01

(3.48))617,369(62.03.0

k

DhuN D = 1012.3

Solving for h

h =D

k3.1012 =

)m(025.0

)Cm/W(6421.03.1012

o

= 26,000 W/m2-oC


q = 26,000(W/m2-oC) 0.025(m) 0.75(m) (88 - 15)(oC) = 111,801 W = 111.8 kW

Latent heat of condensation at 90oC is

fgh = 2283.2 kJ/kg

Substituting into (b) gives the condensation rate m

m = )kg/kJ(2.2283

)kW(8.111 = 0.04897

)kg/kJ(

)s/kJ(= 0.04897 kg/s

(iv) Checking. Dimensional check: Computations showed that equations (b), (d), (e) and (g) are dimensionally consistent.

Quantitative check: The value of h is outside the range given in Table 1.1 for forced convection of liquids. A review of the analysis and calculations uncovered no errors. It should be kept in mind that Table 1.1 gives rough estimates of h for typical applications. Exceptions are expected.

Qualitative check: Increasing the tube’s length and/or boat speed, increases condensation rate. Equation (b) shows that condensation rate is directly proportional to q. According to equation

(d), q increases as L is increased. Similarly, according to (e), an increase in V results in an

increase in h which in turn increases q.

(5) Comments. Although condensation rate may be adequate when the boat is in motion, it decreases when the boat is stationary.

PROBLEM 8.11

An inventive student wanted to verify the speed of a boat

using heat transfer analysis. She used a 10 cm long

electrically heated tube with inside and outside diameters

of 1.1 cm and 1.2 cm, respectively. She immersed the tube

in the water such that its axis is normal to boat velocity.

She recorded the following measurements:

Water temperature = 16.5oC

Outside surface temperature of tube = 23.5oC

Electric energy dissipated in tube = 480 W

Determine the speed of the boat.

(1) Observations. (i) Electric power is dissipated into heat and is removed by the water. (ii) This velocity measuring concept is based on the fact that forced convection heat transfer is affected by fluid velocity. (iii) velocity affects the heat transfer coefficient which in term affects surface temperature. (iv) Newton’s law of cooling relates surface heat loss to the heat transfer coefficient, surface area and surface temperature. (v) This problem can be modeled as external flow normal to a cylinder. (vi) The fluid is water.

(2) Problem Definition. Formulate a relationship between fluid velocity, heat transfer rate and surface temperature for flow normal to a cylinder.

(3) Solution Plan. Apply conservation of energy and Newton's law of cooling to the tube. Use a correlation equation to relate heat transfer coefficient to fluid velocity.

(4) Plan Execution.

(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3) constant properties, (4) no axial conduction through tube, (5) constant boat velocity, (6) uniform surface flux, (7) uniform surface

temperature (8) uniform water temperature, (9) no buoyancy ( = 0 or g = 0) and (10) negligible radiation.

(ii) Analysis. Conservation of energy applied to the tube gives

qP (a)

where

P = electric power supplied to tube = 480 W q = heat transfer rate from tube surface to water, W

Application of Newton’s law of cooling to the tube and gives

)( TTLDhq so (b)

where

oD = outside tube diameter = 1.2 cm = 0.012 m


L = tube length = 10 cm = 0.1 m

sT = surface temperature = 23.5 Co

T = water temperature = 16.5 Co

+

-T

sT

V


Substituting (a) into (b)

)( TTLDhP so

Since h is expected to depend on velocity, equation (a) is solved for h

)( TTLD

Ph

so

(c)


5/48/5

4/13/2

3/12/1

000,282/1/4.01

62.03.0 D

DoD Re

Pr

PrRe

k

DhuN (d)


properties at Tf

where




Pr = Prandtl number




oD

DVRe (e)

where

V = boat speed, m/s


Substituting (e) into (d) and solving for h

5/48/5

4/13/2

3/12/1

000,282/)(1/4.01

)(62.03.0 /DV

Pr

Pr/DV

D

kh o

o

o

(f)

Properties of water are determined at the film temperature fT defined as

2

TTT s

f (g)

Equation (f) gives a relationship between h and the boat speed V . Substituting (f) into (c) gives

the desired relationship between P , V and Ts. However, the resulting equation cannot be solved

explicitly for V . The solution is obtained by trial and error. Equation (c) is used to calculate h ,

a value for V is selected and substituted into (f) to calculate h . If the calculated h using (f) is


not the same as that obtained from (c), the procedure is repeated until a satisfactory agreement is obtained between the two values.

(iii) Computations. Equation (c) is used to calculate h

CW/m189,18)C)(5.165.23()m(1.0)m(012.0

)W(480 o2

ch

Equation (f) is used to calculate fT

Tf= (23.5 + 16.5)( Co )/2 = 20 Co


k = 0.5996 CW/m o

Pr = 6.99

= 1.004 10-6 /sm2

Assume V = 10 m/s. Substituting into (f)

h)m(012.0

)CW/m(5996.03.0

o

)m(012.0

)CW/m(5996.0

000,282)/sm(10004.1

)m(012.0)m/s(101

99.6/4.01

)99.6()/sm(10004.1

)m(012.0)m/s(1062.0

o5/4

8/5

264/13/2

3/1

1/2

26

h = 28,606 CW/m o2

Since this is larger than h = 18,189 CW/m o2 obtained from (c), the procedure is repeated

using a lower value for V . Assume 5V m/s and substituting into (f) gives h = 18,089

CW/m o2 . This agrees within 0.5% of the value obtained from (c). Thus the speed of the boat is

5 m/s.

(iv) Checking. Dimensional check: Computations showed that equations (c) and (f) are dimensionally consistent.

Quantitative check: The calculated value of h = 18,189 CW/m o2 is within the range given in

Table 1.1 for forced convection in liquids.

(5) Comments. (i) The velocity measuring method suggested by the student is indeed sound. It is based on the observation that fluid velocity affects heat transfer coefficient. Thus heat transfer coefficient may be used as a measure of velocity. (ii) Since correlation equations for heat transfer coefficients are not exact, a velocity measuring instrument which is based on this concept must be calibrated to obtain accurate velocity measurements.

PROBLEM 8.12

A thin electric heater is wrapped around a rod of diameter 3 cm. The heater dissipates energy

uniformly at a rate of 1300 W/m. Air at 20oC flows normal to the rod with a velocity of 15.6 m/s.

Determine the steady state surface temperature of the heater.

(1) Observations. (i) This is an external forced convection problem. (ii) The flow is normal to a rod. (iii) Surface heat transfer rate per unit length is known. However, surface temperature is unknown. (iv) In general, surface temperature varies along the circumference. However, the rod can be assumed to have a uniform surface temperature. (v) This problem can be modeled as forced convection normal to a rod with uniform surface flux or temperature. (vi) Newton’s law of cooling gives surface temperature. (vii) The fluid is air.

(2) Problem Definition. Surface temperature can be determined from Newton's law of cooling if the heat transfer coefficient is known. Thus, the problem is finding the average heat transfer coefficient for flow normal to a cylinder.

(3) Solution Plan. Apply Newton's law of cooling to the flow over a rod. Use forced convection correlation equation to determine the average Nusselt number.

(4) Plan Execution.

(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3) two-dimensional flow, (4) uniform upstream velocity and temperature, (5) uniform surface flux, (6) constant properties, (7)

all energy dissipated in electric heater leaves surface (no axial conduction), (8) no buoyancy ( = 0 or g = 0) and (9) negligible radiation.


q = h DL ( sT T ) (a)

where

D = diameter of rod = 3 cm = 0.03 m


L = tube length, m q = heat transfer rate, W

sT = average surface temperature, oC


Solving (a) for sT

hD

LqTTs

/ (b)

where

q/L = energy dissipated per unit length = 1300 W/m

Equation (8.10a) gives a correlation for the average Nusselt number for forced convection normal to a cylinder at uniform surface temperature or surface flux

D

L

V T TV

sT_


5/48/5

4/13/2

3/12/1

000,282/1/4.01

62.03.0 D

DD Re

Pr

PrRe

k

DhuN (c)


properties at Tf (d)where




Pr = Prandtl number




DVReD (e)

where




Tf = 2/)( TTs (f)

Since surface temperature is unknown, a trial and error procedure is needed to solve the problem.

A value for sT is assumed, (f) is used to determine Tf and (b) is used to calculate sT . If the

calculated sT is not equal to the assumed value, the process is repeated until a satisfactory

agreement is obtained between assumed and calculated sT .

Assume sT = 100oC. Equation (f) gives

Tf = (100 + 20)(oC)/2 = 60oC

At this temperature air properties are



The Reynolds number is computed to establish if (d) is satisfied. Substituting into (e)

DRe)s/m(109.18

)m(03.0)s/m(6.1526

= 24,762

and


Pe = PrReD = 24,762(0.708) = 1.753 104

Therefore, (d) is satisfied and correlation equation (c) is applicable.

(iii) Computations. Equation (c) gives the average Nusselt number and average heat transfer coefficient

5/48/5

4/13/2

1/32/1

000,282/762,241708.0/4.01

(0.708))762,24(62.03.0

k

DhuN D = 89.42

Solving for h

h =D

k42.89 =

)m(03.0

)Cm/W(02852.042.89

o

= 85 W/m2-oC

Substituting into (b) gives surface temperature

sT = 20oC +)Cm/W(85)m)(03.0(

)m/W(300,1o2

= 182.3oC

This value is considerably higher than the assumed one of 100oC. The procedure is repeated with

another assumed temperature. Assume sT = 180oC. This gives Tf = 100oC, h = 83.2 W/m2- oC

and a calculated surface temperature sT = 185.8oC. This is close to the assumed value. Thus the

resulting surface temperature is 185.8oC.

(iv) Checking. Dimensional check: Computations showed that units for equations (b), (c) and (e) are dimensionally consistent.

Quantitative check: The value of h is within the range given in Table 1.1 for forced convection of gases.

Limiting check: If the heater is turned off, surface temperature should be the same as free stream

temperature. Setting q = 0 in (b) gives sT T .

(5) Comments. (i) The trial and error procedure converges rapidly towards a satisfactory solution. This is because heat transfer coefficient is not very sensitive to the temperature at which properties are determined. Changing film temperature from 60oC to 100oC in the above example

resulted in a 2% change in h . (ii) The assumption that radiation loss is negligible should be examined in view of the high surface temperature. Since surface emissivity is unknown one can only make a rough approximation of radiation. Using Stefan-Boltzmann law, assuming that the rod is a small surface enclosed by a much larger surface and assuming an emissivity of 1.0, radiation loss is found to be 296 W/m. However, if the surface emissivity is 0.1, then the radiation loss is 2.2%, which is negligible.

PROBLEM 8.13

A fluid velocity measuring instrument consists of a wire which is heated electrically. By

positioning the axis of the wire normal to flow direction and measuring surface temperature and

dissipated electric power, fluid velocity can be estimated. Determine the velocity of air at 25oC

for a wire diameter of 0.5 mm, dissipated power 35 W/m and surface temperature 40oC.

(1) Observations. (i) Electric power is dissipated into heat and is removed by the fluid. (ii) This velocity measuring instrument is based on the fact that forced convection heat transfer is affected by fluid velocity. (iii) velocity affects the heat transfer coefficient which in term affects surface temperature and heat flux. (iv) Newton’s law of cooling relates surface heat loss to the heat transfer coefficient, surface area and surface temperature. (v) This problem can be modeled as external flow normal to a cylinder. (vi) The fluid is air.

(2) Problem Definition. Formulate a relationship between fluid velocity, heat transfer and surface temperature for flow normal to a cylinder.

(3) Solution Plan. Apply Newton's law of cooling to the wire. Use a correlation equation to relate heat transfer coefficient to the fluid velocity.

(4) Plan Execution.

(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3) constant properties, (4) no axial conduction through wire, (5) uniform upstream velocity and temperature, (6) constant properties,

(7) uniform surface flux and surface temperature, (8) no buoyancy ( = 0 or g = 0) and (9) no radiation.

(ii) Analysis. Application of Newton’s law of cooling to the wire gives

q = h DL (Ts - T ) (a) where

D = wire diameter = 0.5 mm = 0.0005 m

h = average heat transfer coefficient, W/m2-oCL = wire length, m Ts = surface temperature = 40oC


Since h is expected to depend on velocity, equation (a) is

solved for h

h =)(

/

TTD

Lq

s

(b)

whereq

L = power or heat dissipated in wire per unit length = 35 W/m

Equation (8.10a) gives a correlation for the average Nusselt number for forced convection normal to a cylinder at uniform surface temperature or surface flux

_

V

TLq /

sT


5/48/5

4/13/2

3/12/1

000,282/1

/4.01

62.03.0 D

DD Re

Pr

PrRe

k

DhuN (c)




NuD = average Nusselt number


Pr = Prandtl number




ReD = V D

(e)

where


= kinematic viscosity = m2/s

Properties of air are evaluated at the film temperature Tf defined as

Tf = (Ts + T )/2 (f)

The objective is to express h in terms of free stream velocity V , substitute into (b) and obtain

an equation relating q/L, V and Ts. Substituting (e) into (c) and solving the resulting equation

for h gives

D

kDV

Pr

PrDV

D

kh

5/48/5

4/13/2

3/11/2

000,282

/1

/4.01

/62.03.0 (g)

Equation (g) gives a relationship between h and the free stream velocity V . Substituting this

result into (b) gives the desired relationship between q/L, V and Ts. However, the resulting

equation cannot be solved explicitly for V . The solution is obtained by trial and error. Equation

(b) is used to calculate h , a value for V is selected and substituted into (g) to calculate h . If the

calculated h using (g) is not the same as that obtained from (b), the procedure is repeated until a satisfactory agreement is obtained between the two values.

(iii) Computations. Equation (f) is used to calculate Tf

Tf= (40 + 25)(oC)/2 = 32.5oC


k = 0.02656 W/m-oCPr = 0.7115


= 16.2475 10-6 m2/s


h =)C)(2540)(m(0005.0

)m/W(35o

= 1485.4 W/m2-oC

Assume V = 50 m/s. Substituting into (g)

h)m(0005.0

)Cm/W(02656.03.0

o

)(0005.0

)/(02656.0

000,282)/(1024.16

)(0005.0)/(501

7115./4.01

)7115.0()/(1024.16

)(0005.0)/(5062.0 5/4

8/5

264/13/2

3/1

1/2

26

m

CmW

sm

msm

0

sm

msm

o

h = 1060.1 W/m2-oC

Since this is less than h = 1485.4 W/m2-oC, the procedure is repeated using a higher value for

V . The results of five trials are tabulated below.

Assumed V Calculated h m/s W/m2-oC

50 1059.9

70 1259.7

90 1434.9

100 1515.6

96 1484.2

The result shows that the free stream velocity is V = 96 m/s. With V determined, it remains to verify that condition (d) is satisfied by calculating the Peclet number.

Pe = ReD Pr = PrDV )/( = 7115.0)]s/m(102475.16/)m(0005.0)s/m(96[ 26 = 2102

Therefore, condition (d) is satisfied.

(iv) Checking. Dimensional check: Computations showed that equations (b), (e) and (g) are dimensionally consistent.

Qualitative check: If q/L is held constant and V is increased, surface temperature should

decrease. According to (g), increasing V increases h . An increase in h results in a decrease in Ts, as indicated by equation (a).

Quantitative check: The calculated value of h = 1484.2 W/m2-oC is outside the range suggested in Table 1.1. This is due to the fact that the diameter of the wire is very small (0.0005 m). This is not among the typical application considered in Table 1.1.

(5) Comments. This velocity measuring instrument is based on the observation that fluid velocity affects heat transfer coefficient. Since correlation equations are not exact, it is necessary to calibrate such an instrument to obtain accurate velocity measurements.

PROBLEM 8.14

Students were asked to devise unusual methods for determining the height of a building. One

student designed and tested the following system. A thin walled copper balloon was heated to

133oC and parachuted from the roof of the building. Based on aerodynamic consideration, the

student reasoned that the balloon dropped at approximately constant speed. The following

measurements were made:

D = balloon diameter = 13 cm

M = mass of balloon = 150 grams

oT = balloon temperature at landing = 47oC


U = balloon velocity = 4.8 m/s

Determine the height of the building.

(1) Observations. (i) The sphere cools off as it drops. Heat loss from the sphere is by forced convection. (ii) The height of the building can be determined if the time it takes the sphere to land is known. (iii) Time to land is the same as cooling time. (iv) Transient conduction determines cooling time. (v) If the Biot number is less than 0.1, lumped capacity method can be used to determine transient temperature. (vi) Cooling rate depends on the heat transfer coefficient.

(2) Problem Definition. Determine transient temperature of sphere. This requires determining the drop time and the heat transfer coefficient.

(3) Solution Plan. Apply Newton’s law of motion to the falling sphere and use lumped capacity method to determine drop time. Use correlation equation to determine the heat transfer coefficient.

(4) Plan Execution.

(i) Assumptions. (1) Newtonian fluid, (2) Constant heat transfer coefficient, (3) Biot number < 0.1 (to be verified ), (4) constant properties, (5) constant sphere velocity, (6) uniform ambient conditions, (7) negligible wind speed, (8) no radiation and (9) no heat loss to the sphere interior.

(ii) Analysis. Application of Newton’s law of motion to the falling sphere gives

otUH (a)

where

H = building height, m

ot = drop time, s

U = balloon velocity, m/s

The dorp time is determined from transient temperature analysis. For Bi << 1, the lumped capacity model gives the transient temperature solution for the sphere. Equation (5.7) gives

)()( TTTtT i ])/(exp[ tVcAh ps (b)

where

sA = surface area, m2


pc = specific heat of copper = 385 CJ/kg o


t = time, s

)(tT = temperature variable, Co


iT = initial temperature of sphere = 133 Co

V = volume, m3

= density of copper = 8933 kg/m3

The product of V in equation (b) is equal to the mass of the sphere. That is

V = M (c) where

M = mass of sphere = 150 g = 0.15 kg

Surface area of sphere is 2DAs (d)

where

D = diameter = 13 cm = 0.13 m

Substituting (c) and (d) into (b) and solving the resulting equation for t

TtT

TT

hD

Mct ip

)(ln

2 (e)

Applying this result at the drop time, ott

TT

TT

hD

Mct

o

ip

o ln2

(f)

where

oT = sphere temperature at landing = 47 Co

Thus the only unknown in (f) is the average heat transfer coefficient. Equation (8.12a) gives a correlation for the average Nusselt number for forced convection over a sphere

Nuh D

kD D D

s2 0 4 0 061 2 2 3 0 4

1 4

. ./ / ./

Re Re Pr (g)

subject to the following limitations:

3.5 < ReD < 7.6 104

0.71 < Pr < 380

1.0 < ( / s) <3.2

properties at T , s at Ts where

k = thermal conductivity of air = 0.02564 W/m-oC


Pr = Prandtl number of air = 0.713



= viscosity of air at the free stream temperature = 18.17 10-6 kg/m-s

s = viscosity of air at surface temperature, kg/m-s


UDReD (h)

where

= kinematic viscosity of air = 15.09 10-6 m2/s

Since surface temperature changes with time, s should be evaluated at the average surface

temperature Ts defined as

2

ois

TTT (i)

(iii) Computations. The average heat transfer coefficient h is computed using (g). Substituting numerical values into (i)

C902

)C)(47133( oo

sT

The viscosity at this temperature is

61035.21s skg/m


352,41)/sm(1009.15

m)(13.0)m/s(8.426DRe


4/1

6

64.03/22/1

)skg/m(1035.21

)skg/m(1017.18)713.0()352,41(06.0)352,41(4.02

k

DhNu D

k

DhNu D 130.4

Solving for h

m)(13.0

)CW/m(025644.1304.130

o

D

kh 25.7 CW/m o2


6.60)2047(

)20133(ln

)C-m/W(7.25)(m)13.0(

kg)(15.0)CJ/kg(385o222

o

ot s

Substituting into (a) gives the building height

9.290)s(6.60)m/s(8.4H m

The Biot number can now be computed to establish the validity of the lumped capacity method. It is defined as


ck

hBi (j)

where

ck = thermal conductivity of copper = 397 CW/m o (at 90 Co )

= wall thickness of sphere, m

c = density of copper = 8933 3kg/m

Sphere thickness is determined from its mass, density and volume

m10162.3)m()13.0()kg/m(8933

)kg(15.0 4

2232D

M

c


5

o

4o2

10047.2)CW/m(397

m)(10162.3)CW/m(7.25Bi

Since the Biot number is much smaller than unity it follows that the lumped capacity method is applicable.

(iv) Checking. Dimensional check: Computations showed that equations (a), (f), (g), (h) and (j) are dimensionally consistent.

Limiting check: If building height is infinite, the final temperature should be the same as ambient

temperature. Setting TTo in (f) gives ot . When this is substituted into (a) gives H .

Quantitative check: The value of h is within the range listed in Table 1.1 for forced convection of gases.

(5) Comments. Neglecting wind speed is a key assumption in the method used. Wind speed can introduce significant error in estimating the height of the building.

PROBLEM 8.15

A 6 cm diameter sphere is used to study skin friction characteristics at elevated temperatures.

The sphere is heated internally with an electric heater and placed in a wind tunnel. To obtain a nearly uniform surface temperature the sphere is made of copper. Specify the required heater capacity to maintain surface temperature at 140oC. Air velocity in the wind tunnel is 18 m/s and its temperature is 20oC.

(1) Observations. (i) The electric energy dissipated inside the sphere is removed from the surface as heat by forced convection. (ii) This problem can be modeled as external flow over a sphere. (iii) Newton’s law of cooling relates heat loss from the surface to heat transfer coefficient, surface area and surface temperature. (iv) The fluid is air.

(2) Problem Definition. Determine the rate of heat transfer from the surface of a sphere by forced convection.

(3) Solution Plan. Apply conservation of energy to the sphere. Apply Newton's law of cooling to obtain a relationship between heat removed from the sphere and its surface temperature. Use a correlation equation to determine the average heat transfer coefficient for the flow over a sphere.

(4) Plan Execution.

(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3) constant properties, (4) uniform surface flux, (5) uniform surface temperature, (6) uniform wind tunnel conditions, (7) constant

properties, (8) no buoyancy ( = 0 or g = 0) and (9) no radiation.

(ii) Analysis. Applying Conservation of energy to the sphere

P = q (a)where

P = electric power dissipated in sphere, W q = heat removed from surface, W

Application of Newton’s law of cooling to the surface of sphere





Surface area of sphere is

A = D

2 (c) where

D = diameter = 0.06 m

The only remaining unknown in (b) is the heat transfer coefficient h . Equation (8.12a) gives a correlation for the average Nusselt number for forced convection over a sphere

4/14.03/22/1 )(06.04.02

sPrReRe

k

DhuN DDD (d)

T

V

sT

P

q

+ -


Valid for:

3.5 < ReD < 7.6 104

0.71 < Pr < 380

1.0 < ( / s) <3.2

properties at T , s at Ts (e) where



Pr = Prandtl number of air = 0.713 ReD = Reynolds number


s = viscosity of air at surface temperature = 23.44 10-6 kg/m-s

The Reynolds number is given by

ReD = V D

(f)

where



(iii) Computations. Appendix C gives air properties at T = 20oC. Substituting into (f)

ReD = 18 0 06

1509 10 6 2

( / ) . ( )

. ( / )

m s m

m s = 71,571

Thus, the Reynolds and Prandtl numbers are within the limitations in (e). Next compute / s

/ s = 18.17 10-6 (kg/m-s)/ 23.44 10-6 (kg/m-s) = 0.775

Although this is outside the range given in (e), it represents a small deviation particularly since this ratio is raised to the 1/4 power. Substituting into (d)

NuhD

kD =

4/1

6

64.03/22/1

)sm/kg(1044.23

)sm/kg(1017.18)713.0()571,71(06.0)571,71(4.02 = 174.5

Solving the above for h

h = 174.5 k

D = 174.5 (0.02564) (W/m-oC)/0.06(m) = 74.57 W/m2-oC

Substituting into (b) and using (c) gives the required heater capacity P

P = q = 74.57(W/m2-oC) (0.06)2(m2) (140 -20)(oC) = 101.2 W

(iv) Checking. Dimensional check: Computations showed that equations (b), (d) and (f) are dimensionally consistent.

Quantitative check: The value of heat transfer coefficient is within the range shown in Table 1.1.


Limiting check: For the special case of Ts = T the required power should be zero. This is confirmed by equations (a) and (b).

(5) Comments. (i) Since not all conditions listed in (e) on correlation equation (d) have been met, the result should be viewed as an approximation. An alternate approach is to search the literature for another correlation equation that will meet the conditions of this problem. (ii) If radiation is included in the analysis, the required heater capacity will be greater than that determined by the above model.

PROBLEM 8.16

A hollow aluminum sphere weighing 0.2 kg is initially at 200oC. The sphere is parachuted from

a building window m100 above street level. You are challenged to catch the sphere with your

bare hands as it reaches the street. The sphere drops with an average velocity of 4.1 m/s. Its

diameter is 40 cm and the ambient air temperature is 20oC. Will you accept the challenge?

Support your decision.

(1) Observations. (i) The sphere cools off as it drops. Heat loss from the sphere is by forced convection. (ii) This is an external flow problem with a free stream velocity that changes with time. (iii) This is a transient conduction problem. The cooling time is equal to the time it takes the sphere to drop to street level. (iv) If the Biot number is less than 0.1, lumped capacity method can be used to determine transient temperature. (v) Cooling rate depends on the heat transfer coefficient.

(2) Problem Definition. Determine transient temperature of sphere. This requires determining the drop time and the heat transfer coefficient.

(3) Solution Plan. Use the lumped capacity method to determine transient temperature. Use correlation equation to determine the heat transfer coefficient.

(4) Plan Execution.

(i) Assumptions. (1) Newtonian fluid, (2) constant heat transfer coefficient, (3) Biot number < 0.1 (to be verified ), (4) constant properties, (5) constant sphere velocity, (6) uniform ambient conditions, (7) negligible wind speed, (8) no heat loss to the air inside the sphere, (9) no

buoyancy ( = 0 or g = 0) and. (10) no radiation and

(ii) Analysis. Equation (5.7) gives the transient temperature for the lumped capacity model

]exp[)()( tVc

AhTTTtT

p

si (a)

where

sA = surface area, m2

pc = specific heat of aluminum = 902 J/kg- oC

h = average heat transfer coefficient, W/m2-oCt = time, s

)(tT = temperature variable, oC


iT = initial temperature of sphere = 200oC

V = volume, m3

= density of aluminum = 2702 kg/m3

The product of V in equation (a) is equal to the mass of the sphere. That is

V = M (b) where

M = mass of sphere = 0.2 kg

Surface area of sphere is

sA = D2 (c)


D = diameter = 0.4 m

There are two unknowns in (a): drop time to and the average heat transfer coefficient h . For constant sphere velocity, drop time is

V

sto (d)

where

s = drop distance = 100 m

V = sphere velocity = 4.1 m/s

Equation (8.12a) gives a correlation for the average Nusselt number for forced convection over a sphere

Nuh D

kD D D

s2 0 4 0 061 2 2 3 0 4

1 4

. ./ / ./

Re Re Pr (e)

Valid for:

3.5 < ReD < 7.6 104

0.71 < Pr < 380

1.0 < ( / s) <3.2

properties at T , s at Ts (f) where



Pr = Prandtl number of air = 0.713 ReD = Reynolds number


s = viscosity of air at surface temperature, kg/m-s


ReD = V D

(g)

where


Since surface temperature changes with time, s should be evaluated at the average surface

temperature Ts defined as

sT = (Ti + To)/2 (h)

where To is the final sphere temperature obtained from (a). Since this temperature is unknown, the procedure becomes one of trial and error. A value for To is assumed and air viscosity

s determined at sT . Equation (a) is used to calculate To. The calculated To is compared with the

assumed value and the process is repeated until a satisfactory agreement is obtained between assumed and calculated values.

(iii) Computations. To calculate the final sphere temperature from (a), the drop time to and

the average heat transfer coefficient h need to be determined. Equation (d) gives to

to =100(m)/4.1(m/s) = 24.39 s


To determine s at the mean surface temperature sT , assume a final sphere temperature To =

80oC. Equation (h) gives

sT = (200 + 80)(oC)/2 = 140oC

Thus

s = 23.44 10-6 kg/m-s


ReD = )s/m(1009.15

)m(4.0)s/m(1.426

= 108,681

This is somewhat outside the range of applicability of correlation equation (e). Therefore, results based on using this equation are approximate. Substituting into (e)

k

DhNu D =

4/1

6

64.03/22/1

)sm/kg(1044.23

)sm/kg(1017.18)713.0()108681(06.0)108681(4.02 = 222.1

h = 222.1(k/D) = 222.1 (0.02564) (W/m-oC)/0.4(m) = 14.2 W/m2-oC

Substituting into (a) and using (c)

To = 20(oC) + [(200 20 )(oC)])Ckg/J(902)kg(2.0

)s(39.24)m()4.0()Cm/W(2.14exp

o

22o2

= 88.6oC

Although this is slightly different from the assumed value of To = 80oC, repeating the procedure with a new value of To = 88.6oC will have a minor effect on the result.

The Biot number can now be checked to establish the validity of the lumped capacity method. The Biot number is defined as

Bi = h /kal (i)where

Bi = Biot number kal = thermal conductivity of aluminum = 236 W/m-oC

= wall thickness of sphere, m

The thickness is determined from its mass and volume

= M/ al D2 = 0.2(kg)/2702(kg/m3) (0.4) 2(m2) = 0.000147 m

When this is substituted into (i) gives Bi = 8.8 10-6. Thus, the use of the lumped capacity method is justified.

(iv) Checking. Dimensional check: Computations showed that equations (a), (d), (e) and (g) are dimensionally consistent.

Limiting check: If the sphere is dropped from an infinite height, its temperature at landing should

be equal to the ambient temperature. Setting s = in (d) gives to = . When this is substituted

into (a) gives To = T .

Quantitative check: The value of h is slightly outside the range listed in Table 1.1 for forced convection of gases.


(5) Comments. (i) The analysis shows that it is not safe to catch the sphere since its temperature at landing is 88.6oC. However, assumptions (8) and (10) are conservative since they result in an overestimate of surface temperature at landing. (ii) Both the Reynolds number and the viscosity correction factor in (e) are outside the range of applicability of correlation equation (e). However, the effect on the accuracy of the result should be minor.

PROBLEM 8.17

Steam condenses on the outside surface of a 1.6 cm diameter tube. Water enters the tube at

12.5oC and leaves at 27.5

oC. The mean water velocity is 0.405 m/s. Outside surface temperature

is 34 oC. Neglecting wall thickness, determine tube length.

(1) Observations. (i) This is an internal forced convection problem. (ii) The channel is a tube. (iii) The outside surface is maintained at a uniform temperature. (iv) Neglecting tube thickness resistance means that the inside and outside surface temperatures are identical. (v) Fluid temperature is developing. (vi) Inlet and outlet temperatures are known. (vii) The Reynolds number should be determined to establish if the flow is laminar or turbulent. (viii) The required tube length depends on the heat transfer coefficient. (ix) The fluid is water.

(2) Problem Definition. Determine the required tube length to heat water at a specified flow rate to a specified temperature. This requires the determination of the average heat transfer coefficient in a tube.

(3) Solution Plan. Apply the analysis of internal flow through a tube at uniform surface temperature. Compute the Reynolds number to establish if the flow is laminar or turbulent. Use analytic solutions or correlation equations to determine the average heat transfer coefficient.

(4) Plan Execution.

(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3) axisymmetric flow, (4) constant properties, (5) uniform surface temperature, (6) negligible changes in kinetic and potential energy, (7) negligible axial conduction, (8) negligible dissipation, (9) negligible wall thickness, (10) no energy generation and (11) smooth tube.

(ii) Analysis. For flow through a tube at uniform surface temperature, conservation of energy and Newton's law of cooling lead to equation (7.64)

])()( [exp xTTTxT

p

smismcm

hP (a)

where


h = average heat transfer coefficient for a tube of length L, CW/m o2



)(xTm = mean temperature at x, Co

miT = mean inlet temperature = 12.5 Co



The perimeter P is given by DP (b)

where

L

miTm

u

sT

D

moTx0


D = inside tube diameter = 1.6 cm = 0.016 m

Substituting (b) into (a) and applying the resulting equation at the outlet, x = L

])( [exp LTTTT

p

smismocm

hPD (c)

where

moT = outlet temperature = 27.5 Co

Solving (c) for the length L

smo

smip

TT

TT

hD

cmL ln (d)

Conservation of mass gives the flow rate m

4

2DuAum (e)

where


= density, 3kg/m

To compute L using (d), it is necessary to determine h . The Reynolds number is computed to establish if the flow is laminar or turbulent. Reynolds number is defined as

DuReD (f)

where



Water properties are determined at the mean temperature mT , defined as

mT = T Tmi mo

2 (g)

Substituting numerical values into (g)

mT = C202

)C)(5.275.12( oo


pc = 4182 CJ/kg o

k = 0.5996 CW/m o

Pr = 6.99

= 6101004 /sm2

= 998.3 3kg/m



4

)(m)016.0(m/s)(405.0)kg/m(34.998 223

m = 0.08129 kg/s

Finally, the heat transfer coefficient is needed to determine L from equation (d). The Reynolds number is computed to determine if the flow is laminar or turbulent. Equation (f) gives

6454)/sm(10004.1

m)(016.0)m/s(405.026DRe

Since this is greater than the transition number of 2300, it follows that the flow is turbulent. The appropriate correlation for the Nusselt number is given by the Gnielin ski equation (8.17a)

1)8/(7.121

1000)8/(

3/22/1 Prf

PrRef

k

DhuN D

D 1 2 3( / ) /D L (h)

Valid for valid for 0 < D/L <1 developing or fully developed turbulent flow through tubes

2300 < ReD < 5 106

0.5<Pr < 2000 properties at Tm

where

DuN = average Nusselt number in turbulent flow

f = friction factor

For a smooth pipe f is given by 2)64.1ln79.0( DRef (i)

(iii) Computations. The mass flow rate is calculated using (e)

kg/s08129.04

)m()016.0()m/s(405.0)kg/m(3.998 223

m

The Nusselt number is calculated using (h). The friction factor f is determined using (i)

0357.0)64.15.3746ln79.0( 2f

To determine h from equation (h) the length L must be known. Since L is unknown, a trial and

error procedure is used. A value for L is assumed and equation (h) is used to obtain a first

approximation for h . This approximate value is used in (d) to compute L. An improved value for

h is then obtained by substituting the calculated L into (h). Equation (d) is used again to obtain

an new value for L. This procedure is repeated until a satisfactory agreement is obtained between assumed and calculated L. As a first approximation, let L = 0. Equation (f) gives

k

DhuN D 3.52

1)99.6()8/0401.0(7.121

99.610006454)8/0357.0(

3/22/1

Solving for h

DNuD

kh = 52.3

)m(016.0

)CW/m(5996.0 o

= 1960 CW/m o2



m13.4)C)(345.27(

)C)(345.12(ln

)C-W/m(1960)m()016.0(

)CJ/kg-(4182)kg/s(08129.0o

o

o2

o

L

Substituting this value into (h) gives 2.1984h CW/m o2 . With this value of h equation (d)

gives 077.4L m. Further iteration gives 1985h CW/m o2 and 075.4L m.

(iv) Checking. Dimensional check: Computations showed that equations (d)-(i) are dimensionally consistent.

Quantitative check: (1) The value of h is within the range given in Table 1.1 for forced

convection of liquids.

(2) An approximate value for L can be obtained based on the assumption that the fluid is at an

average temperature C20omT . Conservation of energy gives

)()( msmomip TThDLTTcm

Solving for L

)(

)(

ms

momip

TThD

TTcmL

Substituting numerical values into the above

m65.3)C)(2034(

)C)(5.125.27(

)C-W/m(1985)m()016.0(

)CJ/kg-(4182)kg/s(08129.0o

o

o2

o

L

This result differs from the exact solution by 10%.

(5) Comments. The fact that assuming L = 0 in calculating the heat transfer coefficient from

equation (h) introduces a small error in h implies that entrance effects are negligible.

PROBLEM 8.18

A 150 cm long tube with 8 mm inside diameter passes through a laboratory chamber. Air enters

the tube at 12oC with fully developed velocity and a flow rate 0.0005 kg/s. Assume uniform

surface temperature of 25oC, determine outlet air temperature.

(1) Observations. (i) This is an internal force convection problem. (ii) The channel is a tube. (iii) The surface is maintained at a uniform temperature. (iv) The velocity is fully developed. (v) The temperature is developing. (vi) The outlet temperature is unknown..(vii) The Reynolds number should be checked to establish if the flow is laminar or turbulent. (viii) The fluid is air.

(2) Problem Definition. Determine the outlet air temperature. This requires the determination of the average heat transfer coefficient.

(3) Solution Plan. Apply the analysis of internal flow through a tube at uniform surface temperature. Check the Reynolds number to determine if the flow is laminar or turbulent. Use analytic solutions or correlation equations to determine the average heat transfer coefficient.

(4) Plan Execution.

(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3) fully developed velocity, (4) axisymmetric flow, (5) constant properties, (6) uniform surface temperature, (7) negligible changes in kinetic and potential energy, (8) negligible axial conduction, (9) no dissipation, (10) no energy generation and (11) smooth tube.


])()( [exp xTTTxT

p

smismcm

hP (a)

where

cp = specific heat, J/kg- oC

h = average heat transfer coefficient for a tube of length L, W/m2-oCm = mass flow rate = 0.0005 kg/s P = tube perimeter, m Tm(x) = mean temperature at x, oCTmi = mean inlet temperature = 12oCTs = surface temperature = 25oCx = distance from inlet of heated section, m


P = D (b) where

D = inside tube diameter = 8 mm = 0.008 m

Substituting (b) into (a) and setting x = L gives the outlet temperature

])()( [exp LTTTxT

p

smismocm

hD (c)

L

miTm

u

sT

D

moTx0


where

L = tube length = 150 cm = 1.5 m Tmo = mean outlet temperature, oC

To compute Tmo using (c), it is necessary to determine h . The Reynolds number is calculated to establish if the flow is laminar or turbulent. Reynolds number is defined as

ReD /Du (d)

where

ReD = Reynolds number u = mean flow velocity, m/s


The mean velocity is determined from mass flow rate

4/2DuAum

or2/4 Dmu (e)

where

A = cross section area, m2

= density, kg/m3

Air properties are determined at the mean temperature mT , defined as

mT = T Tmi mo

2 (f)

Since outlet temperature Tmo is unknown, the solution is obtained by a trial and error procedure.

A value for Tmo is assumed, (f) is used to calculate mT , a first approximation of properties is

determined at this temperature and (c) is used to calculate Tmo. If the calculated value is not the same as the assumed value, the procedure is repeated until a satisfactory agreement is obtained.

Assume Tmo = 18oC. Equation (f) gives

mT = C152

)C)(1812( oo


cp = 1005.95 J/kg- oCk = 0.02526 W/m-oCPr = 0.7145

= 14.64 10-6 m2/s

= 1.22545 kg/m3

The mean velocity is obtained from (e)

u =)m()008.0)(m/kg)(22545.1(

)s/kg)(0005.0(4223

= 8.117 m/s

The Reynolds number is obtained from (d)

ReD 5.4435)s/m(1064.14

)m(008.0)s/m(117.826


Since this is greater than the transition number of 2300, it follows that the flow is turbulent. The

appropriate correlation for the Nusselt number is given by the Gnielin ski equation (8.17a)

1)8/(7.121

1000)8/(

3/22/1 Prf

PrRef

k

DhuN D

D 1 2 3( / ) /D L (g)

Valid for valid for 0< D/L <1 developing or fully developed turbulent flow through tubes

2300 < ReD < 5 106

0.5<Pr < 2000 properties at Tm (h)where


f = friction factor

For a smooth pipe f is given by

f = (0.79lnReD 1.64)–2 (i)

(iii) Computations. The Nusselt number is calculated using properties at the assumed outlet temperature Tmo = 18oC. The friction factor f is determined first using (i)

f = (0.79 ln 4435.5 - 1.64)-2 = 0.0401


k

DhuN D

3/2

3/22/1)]m(5.1/)m(008.0[1

1)7145.0()8/0401.0(7.121

7145.010005.4435)8/0401.0(= 15.47

Solving for h

DNuD

kh = 15.47

)m(008.0

)Cm/W(02526.0 o

= 48.8 W/m2-oC

Substituting into (c) gives the outlet temperature

C7.24)m(5.1)CJ/kg(95.1005)kg/s(0005.0

)CW/m(8.48)m)(008.0([exp)C)(2512()C(25 o

o

o2oo

moT

Since this is greater than the assumed value of 18oC, the procedure is repeated with a new assumed value of 23oC. Based on this value, the calculated outlet temperature is found to be 24.7oC. Thus, the outlet temperature is 24.7oC.

(iv) Checking. Dimensional check: Computations showed that equations (c), (d), (e) and (g) are dimensionally consistent.

Quantitative check: The value of h is within the range given in Table 1.1 for forced convection

of gases.

(5) Comments. (i) The outlet temperature is within 0.3oC of the maximum value that it can reach. (ii) If one incorrectly assumes that the flow is laminar, the corresponding Nusselt number is 3.66. This is considerably smaller than the turbulent Nusselt number calculated above.

PROBLEM 8.19

Water enters a tube with a fully developed velocity and uniform temperature Tmi = 18oC. The

inside diameter of the tube is 1.5 cm and its surface temperature is uniform at Ts = 125oC.

Neglecting wall thickness, determine the length of the tube needed to heat the water to 82oC at a

flow rate of 0.002 kg/s.

(1) Observations. (i) This is an internal forced convection problem. (ii) Tube surface is maintained at uniform temperature. (iii) The velocity is fully developed. (iv) The length of tube is unknown. (v) The temperature is developing. However, depending on tube length relative to the thermal entrance length, temperature may be considered fully developed throughout. (vi) The Reynolds number should be checked to determine if the flow is laminar or turbulent. (vii) The fluid is water.

(2) Problem Definition. Determine the length of tube needed to increase the temperature of water to a specified level at a given flow rate.

(3) Solution Plan. Apply the analysis of internal flow through a tube at uniform surface temperature. Use analytic solutions or correlation equations to determine the average heat transfer coefficient.

(4) Plan Execution.

(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3) constant properties, (4) axisymmetric flow, (5) uniform surface temperature, (6) fully developed velocity, (7) no axial conduction, (8) negligible changes in potential and kinetic energy, (8) no dissipation, (9) no energy generation and (10) smooth tube.


])()( [exp xTTTxT

pcm

hPsmism (a)

where



m= mass flow rate = 0.002 kg/s P = perimeter, m Ts = surface temperature = 125oCTmi = inlet temperature = 18oCTm (x) = mean temperature at any location x along the tube, oCx = distance along the tube measured from the inlet, m

To determine the required tube length, set x = L and Tm (L) = Tmo and solve (a) for L

mos

misp

TT

TT

Ph

cmL ln (b)

where

LTs

Ts

Du

Tmi

xTmo




P = D (c)where

D = inside diameter = 1.5 cm = 0.015 m

Thus, all quantities in (b) are known except h . To determine h , the Reynolds number is calculated to establish if the flow is laminar or turbulent. The Reynolds number is defined as

ReD = u D

(d)

where

u = mean velocity, m/s


The mean velocity is determined from mass flow rate equation

m uD2

4 (e)

where

= density, kg/m3

Solving (e) for u

u = 4

2

m

D (f)

Properties are evaluated at the mean temperature Tm , the average of inlet and outlet temperatures

C502

)C)(8218(

2

oo

momim

TTT


cp = specific heat = 4182 J/kg- oCk = thermal conductivity = 0.6405 W/m-oCPr = Prandtl number = 3.57


= density = 988 kg/m3

Equations (f) and ( d) give

u = 4 0 002

988 0 0153 2 2

( . )( / )

( / ) ( . ) ( )

kg s

kg m m= 0.01146 m/s

and

ReD =0 01146 0 015

05537 10 6 2

. ( / ) . ( )

. ( / )

m s m

m s= 310

Since ReD < 2300, the flow is laminar. The next step is establishing if thermal entrance length Lt

is small compared to tube length L. If this is the case, the flow can be considered thermally fully developed throughout. The thermal entrance length for laminar flow in a constant temperature tube is given by equation (7.43b)

Lt = 0.033 D ReD Pr (g)



Lt = 0.033 (0.015)(m) (310) (3.57) = 0.548 m

Since L is unknown, comparison can not be made with Lt. Assuming that Lt is not small compared to L, the flow must be treated as thermally developing and correlation equation (8.14a) should be used

3/1))/(0.041

)/(0668.066.3

PrReLD

PrReLD

k

DhuN

D

DD (h)

Valid for entrance region of tube uniform surface temperature Ts

fully developed laminar flow (ReD < 2300) developing temperature

properties at 2/)( momim TTT (i)

To calculate h from this equation, the length L must be known. Thus, a trial and error procedure

is required to solve the problem. A value for L is assumed, equation (h) is used to calculate h

and the result substituted into (b) to determine L. The procedure is repeated until a satisfactory agreement is obtained between assumed and calculated values.

(iii) Computations. Assume L = 0.5 m. Substituting into (h)

k

DhuN D 229.5

310(3.57).5(m)0.015(m)/004.01

)57.3(310])m(5.0/)m(015.0[0668.066.3

3/2

Solving for h

DNuD

kh = Cm/W29.223229.5

)m(015.0

)Cm/W(6405.0 o2o


m725.0)C)(82125(

)C)(18125(ln

)m)(015.0()Cm/W(29.223

)Ckg/J(4182)s/kg(002.0o

o

o2

o

L

Since calculated L is not equal to the assumed value of 0.5 m, the process is repeated with another assumed value of L = 0.73 m. Results of four trials are tabulated below.

Assumed L Calculated h Calculated L

m W/m2-oC m

0.5 223.29 0.725

0.73 205.38 0.788

0.79 202.23 0.8

0.80 201.74 0.802

Therefore, L = 0.802 m

(iv) Checking. Dimensional check: Computations showed that equations (b), (d), (f), (g) (h)


and (i) are dimensionally consistent.

Quantitative check: The value of the heat transfer coefficient is within the range shown in Table 1.1 for forced convection of liquids.

Qualitative check: The value of h for developing temperature should be larger than h for fully developed temperature. Equation (7.57) gives the Nusselt number for the fully developed case

66.3k

DhuN D (j)

Cm/W3.156)m(015.0

)Cm/W(6405.066.366.3 o2

o

D

kh

This is smaller than h = 201.74 W/m2-oC for the developing case.

Limiting check: If Tmo = Tmi, the required length should be zero. Setting Tmo = Tmi in (b) gives the L = 0.

(5) Comments. (i) Since Lh is not small compared to L, temperature entrance effects can not be neglected. (ii) Equation (h) converges to the limiting case of fully developed temperature when L

. Setting L = in (h) gives the fully developed solution (j).

PROBLEM 8.20

Cold air is supplied to a research apparatus at a rate of 0.14 g/s. The air enters a 20 cm long

tube with uniform velocity and uniform temperature of 20oC. The inside diameter of the tube is

5 mm. The inside surface is maintained at 30oC. Determine the outlet air temperature.

(1) Observations. (i) This is an internal forced convection problem. (ii) Tube surface is maintained at a uniform temperature. (iii) The velocity and temperature are developing. Thus, entrance effects may be important. (iv) The outlet temperature is unknown. (v) The fluid is air.

(2) Problem Definition. Determine air outlet temperature. This requires determining the average heat transfer coefficient.

(3) Solution Plan. Apply the analysis of internal flow through a tube at uniform surface temperature. Check the Reynolds number to determine if the flow is laminar or turbulent. Compute the hydrodynamic and thermal entrance lengths to establish if entrance effects can be neglected. Use an appropriate correlation equation to compute the average heat transfer coefficient.

(4) Plan Execution.

(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3) constant properties, (4) axisymmetric flow, (5) uniform surface temperature, (6) uniform inlet velocity and temperature, (7) no axial conduction, (8) negligible changes in potential and kinetic energy, (9) no dissipation and (10) no energy generation.

. (ii) Analysis. For flow in a tube at uniform surface temperature, conservation of energy and Newton's law of cooling lead to equation (7.64)

])()( exp[ xTTTxT

p

smismcm

hP (a)

where


h = average heat transfer coefficient for a tube of length L, W/m2-oCm = mass flow rate = 0.14 g/s = 0.00014 kg/s P = tube perimeter, m Tm(x) = mean temperature at x, oCTmi = mean inlet temperature = 20 oC

Ts = surface temperature = 30oCx = distance from inlet of heated section, m


P = D (b)where

D = inside tube diameter

= 5 mm = 0.005 m

Substituting (b) into (a) and setting x = L gives the outlet temperature

L

Ts

Du

Tmi

xTmo


])()( exp[ LTTTxT

p

smismocm

hD (c)

where

L = tube length = 20 cm = 0.2 m Tmo = mean outlet temperature, oC

To compute Tmo using (c), it is necessary to determine h . The Reynolds number is calculated to establish if the flow is laminar or turbulent. Reynolds number is defined as

ReD

u D (d)

where

ReD = Reynolds number u = mean flow velocity, m/s


The mean velocity is determined from mass flow rate

4/2DuAum

or2/4 Dmu (e)

where

A = cross section area, m2

= density, kg/m3

Air properties are determined at the mean temperature mT , defined as

mT = T Tmi mo

2 (f)

Since outlet temperature Tmo is unknown, the solution is obtained by a trial and error procedure.

A value for Tmo is assumed, (f) is used to calculate mT , a first approximation of properties is

determined at this temperature and (c) is used to calculate Tmo. If the calculated value of Tmo is not the same as the assumed value, the procedure is repeated until a satisfactory agreement is obtained.

Assume Tmo = 0oC. Equation (f) gives

mT = C102

)C)(020( oo


cp = 1005.6 J/kg- oC

k = 0.02329 W/m-oCPr = 0.721

= 12.46 0-6 m2/s

= 1.3414 kg/m3

The mean velocity is obtained from (e)

u =)m()005.0)(m/kg)(3414.1(

)s/kg)(00014.0(4223

= 5.315 m/s


The Reynolds number is obtained from (d)

ReD 2133)s/m(1046.12

)m(005.0)s/m(315.526

Since this is less than the transition number of 2300, it follows that the flow is laminar. The next step is to determine if entrance effects can be neglected. For laminar flow, the hydrodynamic entrance length Lh for a constant surface temperature tube is given by equation (7.43a)

Lh/D = 0.056 ReD (g)

The thermal entrance length Lt is given by equation (7.43b)

Lt/D = 0.033 ReD Pr (h)

Equations (g) and (h) give

Lh = 0.056(2,133)0.005(m) = 0.597m

andLt = 0.033(2,133)(0.721) 0.005(m) = 0.254 m

Thus, both velocity and temperature are developing. The appropriate correlation for the average Nusselt number is given by equation (8.15a)

14.03/1

)/(86.1s

DD PrReLDk

DhNu (i)

Valid for: entrance region of tube uniform surface temperature Ts

laminar flow (ReD < 2300) developing velocity and temperature 0.48 < Pr < 16700

0.0044 < s < 9.75

properties at Tm , s at Ts (j)

where

= viscosity at mean temperature = 16.71 10-6 kg/s-m

s = viscosity at surface temperature = 18.65 10-6 kg/s-m

Conditions (j) are satisfied.

(iii) Computations. Equation (i) is used to calculate h

k

DhNu D

14.0

6

63/1

)ms/kg(1065.18

)ms/kg(1071.16)721.0(1332

0.2(m)

)m(005.086.1 = 6.18

or

DNuD

kh 18.6

)m(005.0

)Cm/W(02329.0 o

= 28.8 W/m2-oC



moT = 30(oC) )2030( (oC) 7.3)Ckg/J(6.1005)s/kg(00014.0

)m(2.0)Cm/W(8.28)m)(005.0(exp

o

o2oC

Since this is higher than the assumed temperature, the above procedure is repeated with assumed

Tmo = 4oC. The corresponding heat transfer coefficient and calculated outlet temperature are h =28.94 W/m2-oC and Tmo = 3.8oC.

(iv) Checking. Dimensional check: Computations showed that units of equations (b)-(i) are dimensionally consistent.

Qualitative check: Since entrance effects are important, it follows that the heat transfer coefficient is greater than that of fully developed flow. For laminar fully developed flow, the Nusselt number is given by Equation (7.57)

66.3k

DhuN D (k)

Using this equation to compute h

Cm/W17)m(005.0

)Cm/W(02329.066.366.3 o2

o

D

kh

This is smaller than h = 28.8 W/m2-oC for the developing case.

Limiting check: In the limit as L 0, the outlet temperature approaches inlet temperature. Setting L = 0 in equation (c) gives Tmo = Tmi.

(5) Comments. (i) If entrance effects are neglected, h will be underestimated ( h = 17 W/m2-oC) and the corresponding outlet temperature will be Tmo = 4 oC. (ii) The Reynolds number is very close to the transition Reynolds number of 2300. Since this value of transition Reynolds number is not exact, depending on surface roughness and other factors, it is uncertain if the flow is laminar or turbulent for this case.

PROBLEM 8.21

Water flows through a tube of inside diameter 2.5 cm. The inside surface temperature is 230oC

and the mean velocity is 3 cm/s. At a section far away from the inlet the mean temperature is

70oC.

[a] Calculate the heat flux at this section

[b] What will the flux be if the mean velocity is increased by a factor of ten?

(1) Observations. (i) This is an internal forced convection problem. (ii) Tube surface is maintained at uniform temperature. (iii) The section of interest is far away from the inlet. This means that flow and temperature can be assumed fully developed and the heat transfer coefficient uniform. (iv) It is desired to determine the surface flux at this section. Newton’s law of cooling gives a relationship between local flux, surface temperature and heat transfer coefficient. (v) The Reynolds number should be checked to determine if the flow is laminar or turbulent. (vi) The fluid is water.

(2) Problem Definition. Determine surface heat flux corresponding to two mean flow velocities. Since the flux can be obtained from Newton's law of cooling, the problem is one of finding the heat transfer coefficient or fully developed flow corresponding to the two velocities.

(3) Solution Plan. Apply Newton's law of cooling at the specified section of the tube. Use analytic solutions or correlation equations to determine the average heat transfer coefficient.

(4) Plan Execution.

(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3) constant properties, (4) axisymmetric flow, (4) uniform surface temperature, (5) fully developed velocity and temperature, (6) no axial conduction, (7) negligible changes in potential and kinetic energy, (8) no dissipation, (9) no energy generation and (10) smooth tube.

(ii) Analysis. Apply Newton’s law of cooling

q = )( TTh s (a)

where

h = local heat transfer coefficient, W/m2-oC

q = surface flux, W/m2

Ts = surface temperature = 230oCTm = mean temperature of water = 70oC

Since flow and temperature are fully developed, the heat transfer coefficient is uniform (local and average coefficients are identical). To determine h, the Reynolds number is calculated to establish if the flow is laminar or turbulent. The Reynolds number is defined as

ReD = u D

(b)

where

D = inside tube diameter = 2.5 cm = 0.025 m ReD = Reynolds number u = mean velocity = 3 cm/s = 0.03 m/s


Dx mT

sT

u

sT

fully developed


Properties are evaluated at the mean temperature Tm = 70oC. Appendix D gives

k = thermal conductivity = 0.6594 W/m-oCPr = Prandtl number =2.57

Equation (b) gives

ReD =0 03 0 025

0 4137 10 6 2

. ( / ) . ( )

. ( / )

m s m

m s= 1,813

Since ReD < 2300 it follows that the flow is laminar. The Nusselt number for fully developed laminar flow in tubes at uniform surface temperature is given by equation (7.57)

DNu = NuD =k

hD= 3.66 (c)

When the velocity is increased by a factor of 10 the Reynolds number increases to

ReD = 18,130

Thus, the flow becomes turbulent. For this case the average Nusselt number is given by the Gnielinski equation (8.17a)

Nuh D

kPr

D

D

f

f

8

8

1000

1 12 7 11 2

2 3

Re Pr

./

/

1 2 3( / ) /D L (d)


2300 < ReD < 5 106

0.5 <Pr < 2000 properties at Tm (e)where

NuD = average Nusselt number in turbulent flow

f = friction factor


f = (0.79lnReD 1.64)–2 (f)

For fully developed flow the factor D/L in equation (d) is set to zero.

(iii) Computations.

[a] For the case where u = 0.03 m/s and the flow is laminar. Equation (c) gives

h = 3.66k

D = 3.66

0 6594

0 025

. ( / )

. ( )

W m C

m

o

= 96.5 W/m2-oC


q = 96.5(W/m2-oC)(230 - 70)(oC) = 15,440 W/m2


[b] For the case where the flow is turbulent, equations (d) and (f) give h . The Reynolds number for this case is 18,130. Equation (f) gives f

f = [0.79ln (18,130) 1.64]–2 = 0.02682

Substituting into (d) and setting D/L = 0

Nuh D

kD

0 02682 8 18 130 1000 2 57

1 12 7 0 02682 8 2 57 11 2 2 3

. / , .

. . / ./ /

= 89.76


h = 89.76k

D = 89.76

0 6594

0 025

. ( / )

. ( )

W m C

m

o

= 2367.5 W/m2-oC

Equation (a) gives the flux

q = 2367.5 (W/m2-oC)(230 - 70)(oC) = 378,800 W/m2

(iv) Checking. Dimensional check: Computations showed that equations (a)-(d) are dimensionally consistent.

Qualitative check: As expected turbulent heat flux is greater than that of laminar flow.

Limitations on correlation equation (d): The conditions listed in (e) are met.

(5) Comments. Although the velocity is increased by a factor of 10, the flux is increased by a factor of 25. This large increase is a result of transition from laminar to turbulent. If the flow remains laminar, then according to (c), there will be no change in h even though the Reynolds number is increased.

PROBLEM 8.22

Air flows through a tube of inside diameter 5 cm. At a section far away from the inlet the mean

temperature is 30oC. At another section further downstream the mean temperature is 70

oC.

Inside surface temperature is 90oC and the mean velocity is 4.2 m/s. Determine the length of this

section.

(1) Observations. (i) This is an internal forced convection problem in a tube. (ii) Both velocity and temperature are fully developed. (iii) Tube surface is maintained at uniform temperature. (iv) The Reynolds number should be computed to establish if flow is laminar or turbulent. (v) Mean velocity, mean inlet and outlet temperatures and tube diameter are known. (vi) The fluid is air.

(2) Problem Definition. Find the required tube length to increase the air temperature by a given amount.

(3) Solution Plan. Use the analysis of fully developed laminar flow in tubes at uniform surface temperature to determine the required tube length.

(4) Plan Execution.

(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3) fully developed flow, (4) axisymmetric flow, (5) constant properties, (6) uniform surface temperature, (7) negligible changes in kinetic and potential energy, (8) negligible axial conduction, (9) no dissipation, (10) no energy generation and (11) smooth tube.


])()( exp[ xTTTxT

p

smismcm

hP (a)


h = average heat transfer coefficient for a tube of length L, W/m2-oCL = length of tube, m m = mass flow rate, kg/s P = tube perimeter, m Tm(x) = mean temperature at x, oCTmi = mean inlet temperature = 30oCTs = surface temperature = 90oCx = distance from inlet of heated section, m

Applying (a) at the outlet of the heated section (x = L) and solving for L

tL

hL

iT

iV

t

L

miT m

u

sT

D

moT

x0


Lmc

P h

T T

T T

p s mi

s mo

ln (b)

where


To compute L using (b), it is necessary to determine cp, P, m , and h . All properties are


mT = T Tmi mo

2 (c)


P = D (d) and

4/2 uDm (e)

where

D = inside tube diameter = 5 cm = 0.05 m u = mean flow velocity = 4.2 m/s

= density, kg/m3

The heat transfer coefficient for fully developed flow is uniform along a channel. Its value depends on whether the flow is laminar or turbulent. To proceed, it is necessary to calculate the Reynolds number to determine if the flow is laminar or turbulent. For flow in a tube the Reynolds number is defined as

ReD

u D (f)

where



The mean temperature is calculated in order that properties are determined. Substituting into (c)

mT = C502

)C)(7030( oo


cp = 1007.4 J/kg- oCk = 0.02781 W/m-oCPr = 0.709

= 17.92 10-6, m2/s

= 1.0924 kg/m3


ReD 719,11)s/m(1092.17

)m(05.0)s/m(2.426

Since the Reynolds number is greater than 2300, the flow is turbulent. The Nusselt number for turbulent flow through a tube is given by the Gnielinski equation (8.17a)


17.121

1000

3/22/1

8

8

Pr

PrRe

k

DhuN

f

fD

D 1 2 3( / ) /D L (g)


2300 < ReD < 5 106

0.5<Pr < 2000 properties at Tm (h)where


f = friction factor


f = (0.79lnReD 1.64)–2 (i)

For fully developed flow, set D/L = 0 in (g)

1)8/(7.121

1000)8/(3/22/1 Prf

PrRef

k

DhuN D

D (j)

where the average and local heat transfer coefficients, h and h are identical in the fully

developed region.


P = 0.05(m) = 0.1571 m

s/kg009009.0)s/m(2.4)m/kg(0924.14

)m()05.0( 322

m

The Nusselt number is calculated using equation (j). The friction factor f is determined using (i)

f = (0.79 ln11,719 64.1 )-2 = 0.0301


k

DhuN D

1)709.0()8/0301.0(7.121

709.01000719,11)8/0301.0(

3/22/1= 34.2

Solving for h

DNuD

kh = 34.2

)m(05.0

)Cm/W(02781.0 o

= 19.02 W/m2-oC

Substituting into (b) gives the required tube length

)C)(7090(

)C)(3090(ln

)Cm/W(3.17)m(1571.0

)Ckg/J(4.1007)s/kg(009009.0o

o

o2

o

L = 3.34 m

(iv) Checking. Dimensional check: Computations showed that equations (b), (d), (e), (f), and (j) are dimensionally consistent.


Limiting check: For the special case of Tmo = Tmi , the required length should vanish. Setting Tmo

= Tmi in (b) gives L = 0.

Quantitative check: The value of h is on the low end of values listed in Table 1.1 for forced convection of gases. It should be kept in mind that values of h in Table 1.1 are for typical applications. Exceptions should be expected.

(5) Comments. For fully developed laminar flow in tubes at uniform surface temperature, the Nusselt number is given by equation (7.57)

k

hDNuD = 3.66 (k)

If one incorrectly uses this equation, the heat transfer coefficient and required length become

h = 1.85 W/m2-oC

and

L = 34.3 m

Thus, the error in using (k) is significant.

PROBLEM 8.23

Two identical tubes have inside diameters of 6 mm. Air flows through one tube at a rate of 0.03

kg/hr and through the other at a rate of 0.4 kg/hr. Far away from the inlets of the tubes the

mean temperature is 120oC for both tubes. The air is heated at a uniform surface temperature

which is identical for both tubes. Determine the ratio of the heat flux of the two tubes at this

section.

(1) Observations. (i) This is an internal forced convection problem. (ii) The surface of each tube is maintained at uniform temperature which is the same for both. (iii) The velocity and temperature are fully developed. Thus, the heat transfer coefficient is uniform. (iv) Air flows through each tube at different rates. (v) The Reynolds number should be computed to establish if the flow is laminar or turbulent. (vi) Surface heat flux depends on the heat transfer coefficient.

(2) Problem Definition. Determine the ratio of surface heat flux for the two tubes at sections here the mean temperature is the same for both.

(3) Solution Plan. Apply Newton’s law of cooling to each tube. Compute the Reynolds number to establish if the flow is laminar or turbulent. Use analytic solutions or correlation equations to determine the heat transfer coefficient.

(4) Plan Execution.

(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3) constant properties, (4) axisymmetric flow, (5) uniform surface temperature, (6) no axial conduction, (7) fully developed velocity and temperature, (8) no changes in potential and kinetic energy and (0) no dissipation.

(ii) Analysis. Local heat flux is given by Newton’s law of cooling

q = h (Ts - Tm) (a)

where

h = heat transfer coefficient, W/m2-oCq = surface heat flux, W/m2

Tm = mean temperature = 120oCTs = surface temperature, oC

Applying (a) to the two tubes, noting that Ts and Tm are the same for both tubes, and taking the ratio of the two equations

q

q

h

h

1

2

1

2

(b)

where the subscripts 1 and 2 refer to tubes 1 and 2. To determine the ratio of the heat transfer coefficients, the Reynolds number for each tube should be computed to establish if the flow is laminar or turbulent. The Reynolds number is defined as

ReD = uD

(c)

where

D = tube diameter = 0.006 m

u = mean velocity, m/s ReD = Reynolds number



Properties of air are determined at the mean temperature Tm = 120oC. The mean velocity is determined from the flow rate

/m u D2 4 (d)

or

um

D

42 (e)

where


= density = 0.8979 kg/m3

The mass flow rates for the two tubes are

m1 = 0.03 kg/hr

m2 = 0.4 kg/hr


u1)/(3600)()006.0)(/(8979.0

)/)(03.0(4223 hrsmmkg

hrkg = 0.328 m/s

and

u2)/(3600)()006.0)(/(8979.0

)/)(4.0(4223 hrsmmkg

hrkg = 4.38 m/s


ReD1 = = 78.1 and

ReD2 = = 1,043.3

Thus, the flow is laminar in both tubes. For laminar fully developed flow through tubes with uniform surface temperature, the Nusselt number is given by

NuD = hD/k = 3.66 or

h = 3.66 k/D (f) where

NuD = Nusselt number k = thermal conductivity of air = 0.03261 W/m-oC

(iii) Computations. Applying (f) to the two tubes and noting that D and k are the same for both tubes

h1 = h2 = 3.66 (0.03261)(W/m-oC)/ 0.006(m) = 19.9 W/m2-oC

Substituting into equation (b)

q

q

1

2

19 9

19 91

2

2

. ( / )

. ( / )

W m C

W m C

o

o


(iv) Checking. Dimensional check: Computations showed that equations (c), (e) and (f) are dimensionally correct.

Quantitative check: The value of h is within the range given in Table 1.1 for force convection of gases.

(5) Comments. It is surprising that although the velocity in tube 2 is over 13 times greater than that in tube 1, the heat flux is the same for both tubes. This is due to the fact that for laminar fully developed flow, the Nusselt number is independent of the Reynolds number.

PROBLEM 8.24

Two concentric tubes of diameters 2.5 cm and 6.0 cm are used as a heat exchanger. Air flows

through the inner tube with a mean velocity of 2 m/s and mean temperature of 190oC. Water

flows in the annular space between the two tubes with a mean velocity of 0.5 m/s and a mean

temperature of 30oC. Determine the inside and outside heat transfer coefficients.

(1) Observations. (i) This is an internal forced convection problem. (ii) The geometry consists of two concentric tubes. (iii) Air flows in the inner tube while water flows in the annular space between the two tubes. (iv) The Reynolds number should be computed for both fluids to establish if the flow is laminar or turbulent. (v) Convection resistance depends on the heat transfer coefficient.

(2) Problem Definition. Determine the air side and water side heat transfer coefficients.

(3) Solution Plan. Compute the Reynolds numbers for air in the tube and for water in the annular space to establish if the flow is laminar or turbulent. Use analytic solutions or correlation equations to determine the heat transfer coefficients.

(4) Plan Execution.

(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3) constant properties, (4) axisymmetric flow, (5) uniform surface temperature, (6) fully developed velocity and temperature, (7) negligible wall conduction resistance, (8) negligible wall thickness and (9) smooth tubes.

(ii) Analysis. To determine hi and ho, the two Reynolds numbers are calculated first to establish if the flow is laminar or turbulent. For air flow inside the inner tube the Reynolds number is

aiaDi DuRe / (a)

where

iD = diameter of inner tube = 2.5 cm = 0.025 m

DiRe = Reynolds number for air flow through inner tube

au = mean air velocity in inner tube = 2 m/s

a = kinematic viscosity of air = 33.34 10-6 m2/s


DiRe =)s/m(1034.33

)m(025.0)s/m(226

= 1499.7

Thus, the flow is laminar in the inner tube. The Nusselt number for fully developed laminar flow in tubes depends on surface boundary condition. Assuming uniform surface temperature, the Nusselt number is given by equation (7.57)

DiNu =a

ii

k

Dh = 3.66 (b)

where

ka = thermal conductivity of air = 0.03718 W/m-oC

DiNu = average Nusselt number

The Reynolds number for water flow in the annular space between the two tubes is defined as

air

hi

ho

water

air

DiDo

PR0BLEM 8.24 (continued)

w

ewDe

DuRe (c)

where

De = equivalent diameter of the annular space, m

DeRe = Reynolds number for water flow through the annular space

wu = mean water velocity in the annular space = 0.5 m/s

w = kinematic viscosity of water = 0.8012 10-6 m2/s


P

ADe

4 (d)

where

A = flow area, m2

P = wet perimeter, m

The flow area and wet perimeter for the annular space are given by

A = 4/)( 22io DD (e)

and

)( oi DDP (f)

where

Do = diameter of outer tube = 6 cm = 0.06 m

Substituting (e) and (g) into (d)

ioe DDD (g)

Using (h), equation (d) becomes

eDRe =)s/m(108012.0

)m)(025.006.0)(s/m(5.026

= 21,842

Therefore, the flow in the annular space is turbulent. The Nusselt number for turbulent flow through channels is given by the Gnielinski equation (8.17a)

)1()8/(7.121

)1000)(8/(2/32/1

w

weD

w

eoDe

Prf

PrRef

k

DhNu 3/2)/(1 LDe (h)

Valid for valid for 0< De/L <1 developing or fully developed turbulent flow through channels

2300 < ReDe < 5 106

0.5 <Pr < 2000 properties at Tm (i)

where

f = friction factor kw = thermal conductivity of water = 0.615 W/m-oC

wPr = Prandtl number of water = 5.42

PR0BLEM 8.24 (continued)

DeNu = Nusselt number in turbulent flow

For a smooth tube f is given by

f = (0.79lnReDe 1.64)–2 (j)

For fully developed flow, set De/L = 0 in (h)

)1()8/(7.121

)1000()8/(

3/22/1 Prf

PrRef

k

hDNu eD

w

eeD (k)

where the average and local heat transfer coefficients, oh and ho are identical in the fully

developed region.

(iii) Computations. Equation (b) gives hi

hi = 3.66 (0.03718)(W/m-oC)/0.025(m) = 5.44 W/m2-oC

Equation (k) gives f

f = (0.79 ln 21,842 64.1 )-2 = 0.02557


w

eoD

k

DhNu

e)142.5(8/02557.07.121

42.5)1000842,21)(8/02557.0(3/22/1

= 144.6

and

oh = 144.6 (0.615)(W/m-oC)/(0.06-0.025)(m) = 2541 W/m2-oC

(iv) Checking. Dimensional check: Computations showed that the units of equations (a)- (c) and (k) are dimensionally correct.

Quantitative check: The values hi for air and ho for water are within the approximate range shown in Table 1.1 for forced convection.

(5) Comments. Resistance to heat transfer is inversely proportional to the heat transfer coefficient. Thus, the total resistance in this example is dominated by air side convection resistance. The contribution to the total resistance of the much larger water side heat transfer coefficient is insignificant and therefore can be neglected.

PROBLEM 8.25

A heat exchanger consists of a tube and square duct. The tube is

placed co-axially inside the duct. Hot water flows through the tube

while cold water passes through the duct. The inside and outside

diameters are 5 cm and 5.2 cm, respectively. The side of the duct is

10 cm. At a section far away from the inlet the mean hot water

temperature is 90oC and the mean cold water temperature is 30

oC.

The mean hot water velocity is 1.32 m/s and the mean cold water

velocity is 0.077 m/s. Determine the inside and outside heat transfer

coefficients.

(1) Observations. (i) This is an internal forced convection problem. (ii) The geometry consists of a tube concentrically placed inside a square duct,. (iii) Water flows in the tube and the duct. (iv) The Reynolds number should be computed for the two fluids to establish if the flow is laminar or turbulent. (v) Far away from the inlet the velocity and temperature may be assumed fully developed.

(2) Problem Definition. Determine the heat transfer coefficient for the flow inside the tube and for the flow in the duct.

(3) Solution Plan. Compute the Reynolds number for the flow in the tube and in the duct to establish if the flow is laminar or turbulent. Use analytic solutions or correlation equations to determine the heat transfer coefficients.

(4) Plan Execution.

(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3) constant properties, (4) axisymmetric flow, (5) uniform surface temperature, (6) fully developed velocity and temperature and (7) smooth channels.

(ii) Analysis. To determine ih and oh , the two Reynolds numbers are calculated first to establish

if the flow is laminar or turbulent. For water flow inside the tube the Reynolds number is

i

iiDi

DuRe (a)

where

iD = inside tube diameter = 5 cm = 0.05 m

DiRe = Reynolds number for flow in tube

iu = mean water velocity in tube = 1.32 m/s

i = kinematic viscosity of water in tube = 6103264.0 /sm2


DiRe = 5

2610022.2

)/sm(103264.0

)m(05.0)m/s(32.1

Thus the flow is turbulent in the tube. The Nusselt number for fully developed turbulent flow is given by (set D/L = 0 in eq. 8.17a)

iDoD

S

cold water

waterhot


)1()8/(7.121

)1000)(8/(2/32/1

i

iiD

i

iiD

Prf

PrRef

k

DhNu i (b)

Valid for

fully developed turbulent flow through channels

1/0 LDi

2300 < < 5 106

51052300iDRe prop

0.5 < iPr < 2000

properties at miT

where

f = friction factor

ik = conductivity of water in tube = 0.6727 CW/m o

iPr = Prandtl number of water in tube = 1.97

miT = mean temperature of water in tube = 90 Co

For a smooth tube f is given by

f = (0.79ln iDRe 1.64)–2 (c)

The Reynolds number for fluid flow in the annular duct space is defined as

o

eoDe

DuRe (d)

where

eD = equivalent diameter of the duct annular space, m

DeRe = Reynolds number for water flow through the duct

ou = mean water velocity through the duct = 0.077 m/s

o = kinematic viscosity of water in duct = 6108012.0 /sm2


P

ADe

4 (e)

where

A = flow area, m2

P = wet perimeter, m

The flow area and wet perimeter for the annular duct space are given by

A = 4/22oDS (f)

and

SDP o 4 (g)

where

S 10 cm = 0.1 m

Substituting (f) and (g) into (e)


SD

DSD

o

oe

4

4 22

(h)

Using (h)

05592.0)m)(1.0(4)m)(052.0(

)m()052.0()m()1.0(4 2222

eD m

eDRe =)/sm(108012.0

)m(05592.0)m/s(077.026

= 5,374

Therefore, the flow in the duct space is turbulent. The Nusselt number for fully developed

turbulent flow through channels is given by equation (b) with iD replaced by eD

)1()8/(7.121

)1000)(8/(2/32/1

o

oeD

o

eoDe

Prf

PrRef

k

DhNu (i)

properties at moT

where

ok = conductivity of water in duct = 0.615 CW/m o

oPr = Prandtl number of water in duct = 5.42

moT = mean temperature of water in duct = 30 Co

(iii) Computations. For the flow through the tube, equation (c) gives f

01558.064.110022.2ln79.025f


i

iiD

k

DhNu i

)197.1(8/01558.07.121

97.1)100010022.2)(8/01558.0(3/22/1

5

= 584.6

Thus

CW/m7865m)(05.0

)CW/m(6727.06.5846.584 o2

o

i

ii

D

kh

For the flow in the annular duct space equation (c) gives f

03777.064.15374ln79.02

f

Applying (b) to the annular duct space and setting ei DD

o

ooD

k

DhNu e 74.39

)142.5(8/03777.07.121

42.5)10005374)(8/03777.0(3/22/1

Thus

CW/m1.437m)(05592.0

)CW/m(615.074.3974.39 o2

o

e

oo

D

kh


(iv) Checking. Dimensional check: Computations showed that the units of equations (a), (b), and (h) are dimensionally correct.

Quantitative check: The values hi for and ho for water are within the range shown in Table 1.1 for forced convection of liquids.

(5) Comments. Since convection resistance to heat transfer is inversely proportional to the heat transfer coefficient, The total resistance in this example is dominated by the hot water convection resistance in the tube.

PROBLEM 8.26

In designing an air conditioning system for a pizza restaurant an estimate of the heat added to

the kitchen from the door of the pizza oven is needed. The rectangular door is 50 cm 120 cm

with its short side along the vertical direction. Door surface temperature is 110oC. Ambient air

and surroundings temperatures are 20oC and 24

oC, respectively. Door surface emissivity is 0.08.

Estimate the heat loss from the door.

(1) Observations. (i) Heat is lost from the door to the surroundings by free convection and radiation. (ii) To determine the rate of heat loss, the door can by modeled as a vertical plate losing heat by free convection to an ambient air and by radiation to a large surroundings. (iii) Newton’s law of cooling gives the rate of heat transfer by convection and Stefan-Boltzmann relation gives the heat loss by radiation.


(3) Solution Plan. Apply Newton's law of cooling to the door. Use correlation equations to determine the average heat transfer coefficient. Apply Stefan-Boltzmann relation, equation (1.12), to determine the heat transfer by radiation

(4) Plan Execution.

(i) Assumptions. (1) steady state, (2) two-dimensional, (3) constant properties (except in buoyancy), (4) uniform surface temperature, (5) quiescent ambient fluid, (6) door is in the closed position at all times, (7) surroundings is at uniform temperature and (8) the door is small compared to the surroundings.

(ii) Analysis. Total heat transfer q is given by

rc qqq (a)

where

cq heat transfer by convection, W

rq heat transfer by radiation, W

Application of Newton's law of cooling to the surface gives

cq = h A ( sT - T ) (b)

where



sT = surface temperature = 110oC = 283.15 K


Surface area is given by A = LW (c)

where

L = door height = 50 cm = 0.5 m

gT

W

L

sT


W = door width = 120 cm = 1.2 m

The average heat transfer coefficient h is determined from correlation equations for free convection over a vertical plate. The Rayleigh number RaL is calculated first to determine the

appropriate correlation equation for h . The Rayleigh number is defined as

RaL = PrLTTg s

2

3

(d)

where


L = door side in the direction of gravity = 50 cm = 0.5 m Pr = Prandtl number RaL = Rayleigh number




Tf=

2

TTs = 2

)C)(20110( o

= 65oC


k = 0.02887 W/m-oCPr = 0.7075

= 19.4 10-6 m2/s

For ideal gases, the coefficient of thermal expansion is given by

= 002957.015.273)C(65

1

)(

1

of KT

(1/K)

Substituting into (d) gives

RaL = 7075.0)/s(m)10(19.4

)(mC)(0.5)20)()(110/sC)9.81(m/0.002957(1

2426

33o2o

= 0.61349 109

Since RaL

< 109, the flow is laminar. Thus, the appropriate equation for the average Nusselt

number is given by (8.25b)

4/14/1

2/1 953.4884.4435.2LL Ra

PrPr

Pr

k

LhuN (e)

Valid for vertical plate constant surface temperature Ts

laminar free convection 104 < RaL < 109

0 < Pr < properties at Tf (f)

where


LuN = average Nusselt number

Radiation heat transfer is given by equation (1.12)

)( 44sursr TTAq (g)

where

surT surroundings temperature = 24 Co = 24 + 273.15 = 297.15 K

surface emissivity = 0.08

Stefan-Boltzmann constant = 428 KW/m1067.5

(iii) Computations. Substitution into (e) gives

k

LhuN L

4/19

4/1

2/1)1061349.0(

)7075.0(953.4)7075.0(884.4435.2

7075.0 = 81.07


h = 84.02k

L = 81.07 (0.02887)(W/m-oC)/0.5(m) = 4.68 W/m2-oC

Equation (b) gives

cq = 4.68 (W/m2-oC) (0.5)(m)(1.2)(m) (110 )20 (oC) = 252.7 W

Equation (g) gives

)08.0(rq )KW/m(1067.5 428 (0.5)(m)(1.2)(m) )K()15.297()K()15.383( 4444 37.4 W


q = 252.7 + 37.4 =290.1 W

(iii) Checking. Dimensional check: Computations showed that equations (b), (d), (e) and (g) are dimensionally consistent.


Validity of correlation equation (d): The conditions listed in (f) are met.

(5) Comments. (i) Equation (8.26a) applies to both laminar and turbulent flow. However, it is

less accurate than (8.25b) or (e) above. Equation (8.26a) gives h = 6.37 W/m2-oC and q = 344.2 W. This is 31% higher than that found using the laminar free convection equation (e). (ii) Although radiation accounts for only 13% of the total heat added to the room, the contribution of radiation is minimized due to the low surface emissivity of the door. (iii) Opening and closing the door results in transient effects not accounted for in the above model. In addition, when the door is open radiation from the interior of oven may be significant.

PROBLEM 8.27

To compare the rate of heat transfer by radiation with that by free convection, consider the

following test case. A vertical plate measuring 12 cm 12 cm is maintained at a uniform

surface temperature of 125oC. The ambient air and the surroundings are at 25

oC. Compare the

two modes of heat transfer for surface emissivities of 0.2 and 0.9.

(1) Observations. (i) This is a free convection and radiation problem. (ii) The geometry is a vertical plate. (iii) Surface temperature is uniform. (iv) Newton’s law of cooling gives convection heat transfer rate while Stefan-Boltzmann law gives radiation heat transfer rate. (v) The Rayleigh number should be computed to determine if the flow is laminar or turbulent. (vi) Since radiation heat transfer is considered in this problem, all temperatures should be expressed in degrees kelvin. (vii) The fluid is air.

(2) Problem Definition. Determine heat transfer rate by free convection and by radiation from a vertical plate in air.

(3) Solution Plan. Apply Newton's law of cooling to determine the rate of heat loss by convection. Apply Stefan-Boltzmann radiation law to determine the rate of heat loss by radiation. Compute the Rayleigh number and select an appropriate correlation equations to obtain the average heat transfer coefficient.

(4) Plan Execution.

(i) Assumptions. (1) Steady state, (2) two-dimensional, (3) constant properties (except in buoyancy), (4) uniform surface temperature, (5) the plate is a small surface enclosed by a much larger surface at a uniform temperature and (6) quiescent ambient.

(ii) Analysis. Application of Newton's law of cooling to the vertical plate gives

qc = h A (Ts - T ) (a) where

A = surface area of vertical side, m2

h = average heat transfer coefficient, W/m2-oC or W/m2-K

qc = convection heat transfer rate, W Ts = surface temperature = 125(oC) + 273.15 = 398.15 K

T = ambient temperature = 25(oC) + 273.13 = 298.15 K

surface area is A = L2 (b)where

L = side of square plate = 12 cm = 0.12 m

The average heat transfer coefficient h is determined from correlation equations for free convection over vertical plates. The Rayleigh number RaL is calculated first to determine the


RaL =g T T Ls

3

2 Pr (c)

where

g

T

L

sTL

qrqc

surroundings at Tsur



Pr = Prandtl number Ra

L = Rayleigh number




Tf=

T Ts

2 =

( . . )( )29815 39815

2

K = 348.15K

Appendix C gives air properties at Tf = 348.15(K) - 273.15 = 75oC

k = 0.02957 W/m-oCPr = 0.7065

= 20.41 10-6 m2/s


= 002872.015.348

1

)K(

1

fT(1/K)


RaL

= 7065.0)s/m()1041.20(

)m()12.0)(C)(25125)(s/m(81.9)C/1(002872.02426

33o2o

= 8.257 610

Since RaL

< 109, the flow is laminar. Thus the appropriate equation for the average Nusselt

number is given by equation (8.25b)

4/14/1

2/1 953.4884.4435.2LL Ra

PrPr

Pr

k

LhuN (d)



0 < Pr < properties at Tf (e)

where

NuL = average Nusselt number

Radiation heat loss qr is given by the Stefan-Boltzmann law. Assuming that the plate is a small surface which is surrounded by a much larger surface, qr is given by equation (1.15)

qr = A ( 44surs TT ) (f)

where

qr = radiation heat loss, W Tsur = surroundings temperature = 25(oC) + 273.13 = 298.15 K


= emissivity


(iii) Computations. Convection heat loss. Substitution into (d) gives

k

LhuN L

4/16

4/1

2/1)10257.8(

)7065.0(953.4)7065.0(884.4435.2

7065.0 = 27.6


h = 27.6 k

L = 27.6(0.02957)(W/m-oC) / 0.12(m) = 6.8 W/m2-oC

Substituting into (a) and using (b) gives the rate of heat loss by convection

qc = 6.8 (W/m2-oC) (0.12)(m)(0.12)(m) (125 -25)(oC) = 9.79 W

Radiation heat loss. Equation (f) is used to determine qr. For = 0.2:

qr = 0.2 5.67 10-8 (W/m2-K4) 0.12(m) 0.12(m) ( 39815 298154 4. . )(K4) = 2.81 W

For = 0.9:

qr = 0.9 5.67 10-8 (W/m2-K4) 0.12(m) 0.12(m) ( 39815 298154 4. . )(K4) = 12.66 W

(iv) Checking. Dimensional check: Computations showed that equations (a)-(d) and (f) are dimensionally consistent.


Limiting check: For Ts = T = Tsur, both convection and radiation vanish. Setting Ts = T = Tsur

in (a) and (f) gives qc = qr = 0.


(5) Comments. (i) When compared with free convection, radiation heat loss can be significant

and in general should not be neglected. (ii) The magnitude of is the same whether it is

expressed in units of degree Celsius or kelvin. The reason is that is measured in terms of degree change. One degree change on the Celsius scale is equal to one degree change on the kelvin scale. This is also true of units of heat transfer coefficient and specific heat. (iii) Because temperature in the Stefan-Boltzmann radiation law must be expressed in degrees kelvin, care should be exercised in using the correct units when carrying radiation computations.

PROBLEM 8.28

A sealed electronic package is designed to be cooled by free convection.

The package consists of components which are mounted on the inside

surfaces of two cover plates measuring 10 cm 10 cm each. Because the

plates are made of high conductivity material, surface temperature may

be assumed uniform. The maximum allowable surface temperature is

70oC. Determine the maximum power that can be dissipated in the

package without violating design constraints. Ambient air temperature is 20

oC. Neglect radiation.

(1) Observations. (i) This is a free convection problem. (ii) The power dissipated in the electronic package is transferred to the ambient fluid by free convection. (iii) As the power is increased, surface temperature increases. (iv) The maximum power dissipated corresponds to the maximum allowable surface temperature. (v) Surface temperature is related to surface heat transfer by Newton’s law of cooling. (vi) The problem can be modeled as free convection over a vertical plate. (vii) The Rayleigh number should be computed to determine if the flow is laminar or turbulent. (viii) The fluid is air.



(4) Plan Execution.



P = q = h A (Ts - T ) (a) where


h = average heat transfer coefficient, W/m2-oCP = power dissipated in package, W q = heat transfer from the surface to the ambient air, W T

s= surface temperature = 70oC



A = 2 LW (b)where

L = package height = 10 cm = 0.1 m

gT

components

air

g

T

W

L

sT


W = package width = 10 cm = 0.1 m



RaL=g T T Ls

3

2 Pr (c)

where


L = plate dimension in the direction of gravity = 10 cm = 0.1 m Pr = Prandtl number RaL = Rayleigh number




Tf=

T Ts

2 =

( )( )70 20

2

oC = 45oC

Appendix C gives air properties at this temperature

k = 0.02746 W/m-oCPr = 0.7095

= 17.44 10-6 m2/s


= 003143.015.273)C(45

1

)K(

1o

fT(1/K)


RaL = 7095.0)s/m()1044.17(

)m()1.0)(C)(2070)(s/m(81.9)C/1(003143.02426

33o2o

= 3.5962 610

Since RaL < 109, the flow is laminar. Thus, the appropriate equation for the average Nusselt number is (8.25b)

4/14/1

2/1 953.4884.4435.2LL Ra

PrPr

Pr

k

LhuN (d)




where


LNu = average Nusselt number


k

LhuN

L

4/16

4/1

2/1105962.3

)7095.0(953.4)7095.0(884.4435.2

7095.0 = 22.44


h = 22.44k

L = 22.44(0.02746)(W/m-oC)/0.1(m) = 6.16 W/m2-oC


P = q = 6.16 (W/m2-oC)2(0.1)(m)(0.1)(m) (70 - 20)(oC) = 6.16W

(iv) Checking. Dimensional check: Computations showed that equations (a)-(d) are dimensionally consistent.


Qualitative check: Increasing the allowable surface temperature Ts should increase the maximum

power P. According to equation (a), q is directly proportional to Ts. Furthermore, h increases when Ts is increased.




(5) Comments. (i) The model used to solve this problem is conservative due to both neglecting radiation and heat loss from the top and bottom surfaces. (ii) The maximum power dissipated is relatively small, indicating the limitation of free convection in air as a cooling mode for such

applications. (iii) The magnitude of is the same whether it is expressed in units of degree

Celsius or kelvin. The reason is that is measured in terms of degree change. One degree change on the Celsius scale is equal to one degree change on the kelvin scale. This is also true of units of heat transfer coefficient and specific heat.

PROBLEM 8.29

Assume that the electronic package of Problem 8.28 is to be used in an undersea application.

Determine the maximum power that can be dissipated if the ambient water temperature is 10oC.

(1) Observations. (i) This is a free convection problem. (ii) The power dissipated in the electronic package is transferred to the ambient fluid by free convection. (iii) As the power is increased, surface temperature increases. (iv) The maximum power dissipated corresponds to the maximum allowable surface temperature. (v) Surface temperature is related to surface heat transfer by Newton’s law of cooling. (vi) The problem can be modeled as free convection over a vertical plate. (vii) The Rayleigh number should be computed to determine if the flow is laminar or turbulent. (viii) The fluid is water.



(4) Plan Execution.



P = q = h A (Ts - T ) (a) where


h = average heat transfer coefficient, W/m2-oCP = power dissipated in package, W q = heat transfer from the surface to the water, W T

s= surface temperature = 70oC

T = ambient water temperature = 10oC


A = 2 LW (b)where

L = package height = 10 cm = 0.1 m W = package width = 10 cm = 0.1 m

gT

components

water

g

T

W

L

sT




RaL=g T T Ls

3

2 Pr (c)

where


Pr = Prandtl number RaL = Rayleigh number




Tf=

T Ts

2 =

2

)C)(1070( o

= 40oC


k = 0.6286 W/m-oCPr = 4.34

= 0.000389 1/K

= 0.6582 10-6 m2/s


RaL = 34.4)s/m()106582.0(

)m()1.0)(C)(1070)(s/m(81.9)C/1(000389.02426

33o2o

= 2.294 910

Since RaL > 109, the flow is turbulent. Thus, the appropriate equation for the average Nusselt number is (8.26a)

k

LhuN L

2

8/279/16

6/1

/0.4921

0.3870.825

Pr

RaL (d)

Valid for vertical plate uniform surface temperature Ts

laminar, transition, and turbulent

10 1 < RaL <1012


where




k

LhuN L

2

8/279/16

6/19

34.4/0.4921

]1040.387[2.290.825 = 191.7


h = 191.7L

k = 191.7(0.6286)(W/m-oC)/0.1(m) = 1205 W/m2-oC


P = q = 1205(W/m2-oC)2(0.1)(m)(0.1)(m) (70 - 10)(oC) = 1446 W

(iv) Checking. Dimensional check: Computations showed that equations (a), (b), (c) and (d) are dimensionally consistent.

Quantitative check: The magnitude of h is within the approximate range given in Table 1.1 for free convection of liquids.

Qualitative check: Increasing the allowable surface temperature Ts should increase the maximum

power P. According to equation (a), q is directly proportional to Ts. Furthermore, h increases when Ts is increased.




(5) Comments. (i) The model used to solve this problem is conservative due to both neglecting heat loss from the top and bottom surfaces. (ii) The maximum power dissipated is relatively

large, indicating the effectiveness of water as a free convection medium. (iii) The magnitude of

is the same whether it is expressed in units of degree Celsius or kelvin. The reason is that is measured in terms of degree change. One degree change on the Celsius scale is equal to one degree change on the kelvin scale. This is also true of units of heat transfer coefficient and specific heat.

PROBLEM 8.30

A plate 20 cm high and 25 cm wide is placed vertically in water at 29.4oC. The plate is

maintained at 70.6oC. Determine the free convection heat transfer rate from each half.

(1) Observations. (i) This is a free convection problem. (ii) The surface is maintained at uniform temperature. (iii) The heat transfer coefficient decreases with distance from the leading edge of the plate. (iv) The heat transfer rate from the lower half 1 is greater than that from the upper half 2. (v) Total heat transfer from each half can be determined using the average heat transfer coefficient. (vi) Heat transfer from the upper half is equal to the heat transfer from the entire plate minus heat transfer from the lower half.

(2) Problem Definition. Determine the heat transfer rate from each vertical rectangle.

(3) Solution Plan. Since the heat transfer coefficient changes with distance, the average heat transfer coefficient should be used in Newton's law of cooling. The Rayleigh number should be computed to determine if the flow is laminar or turbulent.

(4) Plan Execution.

(i) Assumptions. (1) steady state, (2) two-dimensional, (3) constant properties (except in buoyancy), (4) uniform surface temperature, (5) negligible radiation and (6) quiescent fluid.

(ii) Analysis. Application of Newton's law to rectangle 1

2/)(11 HwTThq s (a)

where

1h = average heat transfer coefficient for rectangle 1, CW/m o2

H = plate height = 20 cm = 0.2 m

1q = heat transfer rate from rectangle 1, W

sT = surface temperature = 70.6 Co

T = ambient temperature = 29.4oC

w = plate width = 25 cm = 0.25 m

The heat transfer rate from rectangle 2 is given by

1212 qqq (b)

where

2q = heat transfer rate from rectangle 2, W

21q = total heat transfer rate from rectangles 1 and 2, W

Application of Newton’s law of cooling to the entire plate gives 21q

HwTThq s )(2121 (c)

where

21h = average heat transfer coefficient for rectangles 1 and 2, CW/m o2

g

w

H

0

x1

2T


To determine the average heat transfer coefficient for free convection over a vertical plate, the Rayleigh number is computed to establish if the flow is laminar or turbulent. The Rayleigh number for rectangle 1 is defined as

PrHTTg

Ra s2

3

1

)2/)(( (d)

where

g = gravitational acceleration = 9.81 2m/s

Pr = Prandtl number

1Ra = Rayleigh number for rectangle 1




2/)( TTT sf (e)


502/)4.296.70(fT Co


k = 0.6405 CW/m o

Pr = 3.57 310462.0 1/K 6105537.0 /sm2


9

2426

33o23

1 10174.257.3)/sm()105537.0(

)m()2/2.0)(C)(4.296.70()m/s(81.9)K/1(10462.0Ra

Since the Rayleigh number is greater than 910 , it follows that the flow turbulent. Similarly, the

Rayleigh number for the entire plate, 21Ra is

PrHTTg

Ra s2

3

21

)( (f)


9

2426

33o23

21 103947.1757.3)/sm()105537.0(

)m()2.0)(C)(4.296.70()m/s(81.9)K/1(10462.0Ra

Thus the flow is turbulent over the entire upper half of the plate. The applicable correlation equation for the average Nusselt number for a vertical plate of height H/2 is

2

27/816/9

6/111

2/

)/492.0(1

387.0825.0

)2/(

Pr

Ra

k

HhNuH (g)


Valid for vertical plate

constant surface temperature Ts

laminar, transition, and turbulent 12

2/1 1010 HRa

0 < Pr < properties at Tf

Similarly, the average Nusselt number for the entire plate is

2

27/816/9

6/12121

)/492.0(1

387.0825.0

Pr

Ra

k

HhNu H (h)

(iii) Computations. Substituting into (g)

81.185)57.3/492.0(1

10174.2387.0825.0

)2/(2

27/816/9

6/191

2/k

HhNu H

Solving for 1h

CW/m1190m))(2/2.0(

)CW/m(6405.081.185

2/81.185 o2

o

1H

kh


2/m)(0.2)(m)(25.0)C()4.296.70)(CW/m(1190 oo21q = 1225.7 W

Similarly, (h) gives the average heat transfer coefficient for the entire plate

58.358)57.3/492.0(1

103947.17387.0825.0

2

27/816/9

6/1921

k

HhNu H

Solving for 21h

CW/m3.1148m)(2.0

)CW/m(6405.058.35858.358 o2

o

21H

kh


W5.3652.2(m)0m)(25.0C)()4.296.70)(CW/m(3.1148 oo221q

Substituting into (b) gives the heat transfer rate from rectangle 2

W8.1139)W(7.1225)W(5.23652q

(iii) Checking. Dimensional check: Computations showed that units of equations (a)-(d), (g)

and (h) are dimensionally consistent.

Limiting check: If TTs , no free convection takes place and consequently 0211 qq .


Setting TTs in (a) and (c) gives the anticipate result.

Qualitative check: The heat transfer rate from the leading rectangle 1 should be higher than that from the trailing rectangle 2. This is in agreement with the results obtained.

(5) Comments. (i) More heat is transferred from rectangle 1 than rectangle 2. This was predicted prior to solving the problem analytically. However, the difference between the two heat transfer rates is not very significant. (ii) The average heat transfer coefficients are slightly outside the range given in Table 1.1 for free convection in liquids. It should be remembered that values listed in Table 1.1 are for typical applications. Exceptions should be expected.

PROBLEM 8.31

Consider laminar free convection from a vertical plate at uniform surface temperature. Two 45

triangles are drawn on the plate as shown.

[a] Explain why free convection heat transfer from triangle 1 is greater than that from the

triangle 2. [b] Determine the ratio of the heat transfer from two triangles.

(1) Observations. (i) This is a free convection problem. (ii) The surface is maintained at uniform temperature. (iii) The heat transfer coefficient decreases with distance from the leading edge of the plate. (iv) The width of each triangle changes with distance from the leading edge.

(2) Problem Definition. Examine the variation of local heat transfer coefficient with distance and determine the heat transfer rate from each triangle.

(3) Solution Plan.

[a] Formulate an equation for )(xh for laminar free convection over a flat plate.

[b] Since the heat transfer coefficient and area change with distance, Newton's law of cooling should be applied to an infinitesimal area of each triangle. Integration over the area gives the total heat transfer rate.

(4) Plan Execution.

(i) Assumptions. (1) Laminar flow, (2) steady state, (3) two-dimensional, (4) constant properties (except in buoyancy), (5) uniform surface temperature, (6) quiescent fluid and (7) no radiation.

(ii) Analysis.

[a] The heat transfer coefficient for laminar free convection over a vertical plate is given by equation (8.25a)

4/14/1

2/1 953.4884.4435.24

3xx Ra

PrPr

Pr

k

hxNu (a)

and

Rax = Pr)(

2

3xTTg s (b)

where


h = local heat transfer coefficient, W/m2-oCk = thermal conductivity, W/m-oCPr = Prandtl number Rax = local Rayleigh number x = vertical distance from the leading edge of plate, m



Valid for

g

T1

2


vertical plate constant surface temperature Ts

laminar free convection 104 < Rax < 109

0 < Pr < properties at Tf (c)

Substituting (b) into (a) and solving for h

h(x) = C x-1/4 (d)

where the constant C is defined as

C =

4/1

22/1

2

953.4884.4435.24

3

PrPr

TTgPrk s (e)

According to (d), )(xh decreases with distance along the plate x.

It is infinite at .0x Triangle 1 has its wide base starting at

0x while triangle 2 has its apex at 0x . Since heat transfer

rate is proportional to the product of heat transfer coefficient and area, it follows that triangle 1 transfers more heat than triangle 2.

[b] Application of Newton's law to an infinitesimal area gives

dq = (Ts - T )h(x) dA (f) where


q = heat transfer rate, W T

s= surface temperature, oC


The area dA for each triangle is dA1 = b1(x) dx (g)

anddA2 = b2 (x) dx (h)

Similarity of triangles give b1(x) and b2(x)

b1(x) = B(1 - x

H) (i)

and

b2(x) = Bx

H (j)

where

B = base of triangle, m H = height of triangle, m

Substituting (i) into (g) and (j) into (h)

dA1 = B(1 - x

H)dx (k)

and

x

dx

dx

b1(x)

b2(x)

g

T

12

H

B


dA2 = Bx

H dx (m)

Applying (f) to triangle 1, substituting (d) and (k) into (f) and integrating from x = 0 to x = Hgives the total heat transfer from triangle 1

q1 = dxxHxTTBCdqH

s

H

0

4/1

01 /1 = (16/21) B C (Ts -T ) 4/3H (n)

Similarly, applying (f) to triangle 2, substituting (d) and (m) into (f) and integrating from x = 0 to x = H gives the total heat transfer, q2, from triangle 2

q2 = dxxHxTTBCdqH

s

H

0

4/1

02 / = (4/7) B C (Ts - T ) 4/3H (o)

Taking the ratio of (n) and (o) q

q

1

2

= 4

3

(iii) Checking. Dimensional check: To check the units of (d), units of C in equation (e) is determined first

C =

4/1

2422/1

os

2o

o )s/m(555.2(

)C)()(s/m(g)C/1(

Cm

W

Pr)Pr

TTPrk

2

= W/m7/4-oC

Thus units of h(x) in (d) are

h(x) = C (W/m7/4-oC) x-1/4(m)-1/4 = W/m2-oC

Examining units of q1 in (n)

q1 = B(m) C(W/m7/4-oC) (Ts -T ) (oC) 4/3H (m3/4) = W

Limiting check: If = 0 or g = 0 or Ts = T , no free convection takes place and consequently q1

= q2 = 0. Any of these limiting cases give C = 0. Thus, according to (n) and (o), q1 = q2 = 0.

(5) Comments. (i) More heat is transferred from triangle 1 than triangle 2. This was predicted prior to solving the problem analytically. (ii) The result applies to any right angle triangles and is not limited to 45o triangles. (iii) Heat transfer from a surface of fixed area depends on its orientation relative to the leading edge. (iv) This problem illustrates how integration is used to account for variations in element area and heat transfer coefficient. The same approach can be applied if surface temperature and/or ambient temperature vary over a surface area.

PROBLEM 8.32

A vertical plate measuring 21 cm 21 cm is at a uniform surface temperature of 80oC. The

ambient air temperature is 25oC. Determine the free convection heat flux at 1 cm, 10 cm and 20

cm from the lower edge.

(1) Observations. (i) This is a free convection problem over a vertical plate. (ii) The surface is maintained at uniform temperature. (iii) Local heat flux is determined by Newton’s law of cooling. (iv) Heat flux depends on the local heat transfer coefficient. (v) Free convection heat transfer coefficient for a vertical plate decreases with distance from the leading edge. Thus, the flux also decreases. (vi) The Rayleigh number should be computed to select an appropriate Nusselt number correlation equation. (vii) The fluid is air.

(2) Problem Definition. Determine the local heat transfer coefficient for free convection over a vertical plate at uniform surface temperature.

(3) Solution Plan. Apply Newton’s law of cooling, compute the Rayleigh number and select an appropriate Nusselt number correlation equation.

(4) Plan Execution.

(i) Assumptions. (1) Steady state, (2) two-dimensional, (3) constant properties (except in buoyancy), (4) uniform surface temperature, (5) quiescent fluid and (6) negligible radiation.

(ii) Analysis. Application of Newton's law gives

q = h(x) (Ts - T ) (a)

where

h(x) = local heat transfer coefficient, W/m2-oCq = local heat flux, W

Ts



The local heat transfer coefficient is determined from correlation equations for free convection over a vertical plate. The Rayleigh number Rax is calculated to select an appropriate correlation equation for h. The Rayleigh number is defined as

Rax = PrxTTg s

2

3

(b)

where


Pr = Prandtl number Rax = local Rayleigh number x = distance from leading edge, m




Tf= 2/)( TTs = 2/)C)(2580( o = 52.5oC


g

T

W

L

sT

x


k = 0.02799 W/m-oCPr = 0.709

= 18.165 10-6 m2/s


= 003071.015.273)C(5.52

1

)(

1

of KT

(1/K)

Substituting into (b) to evaluate the Rayleigh number at the trailing end x = L = 20 cm = 0.2 m

RaL = 709.0)/()10165.18(

)()2.0)()(2580)(/(81.9)/1(003071.02426

332

sm

mCsmC oo

= 2.8482 107

Since RaL

< 109, the flow is laminar over the region of interest. Thus, the appropriate equation

for the local Nusselt number xNu is given by (8.25a)

4/14/1

2/1 953.4884.4435.24

3xx Ra

PrPr

Pr

k

hxNu (c)



0 < Pr < properties at Tf (d)

(iii) Computations. To evaluate the flux at x = L = 20 cm = 0.2 m, the heat transfer coefficient at this location is computed using (c)

k

hLNuL

4/17

4/1

2/1108482.2

)709.0(953.4)709.0(884.4435.2

709.0

4

3 = 28.2

Solving the above for h(L)

h(L)= 28.2 k

L = 28.2(0.02799)(W/m-oC)/0.2(m) = 3.95 W/m2-oC


q = 3.95(W/m2-oC)(80 25 )(oC) = 217.3 W/m2

The same procedure is followed to determine the flux at x = 1 cm and x = 10 cm. Results for the three locations are tabulated.

x (cm) Rax Nux h(x)(W/m2-oC) q (W/m2)

1 0.356 104 2.98 8.34 458.7

10 0.356 107 16.78 4.7 258.3

20 2.8482 107 28.2 3.95 217.3


(iv) Checking. Dimensional check: Computations showed that equations (a), (b) and (c) are dimensionally consistent.

Quantitative check: The magnitudes of h are approximately within the range given in Table 1.1 for free convection of gases.

Limiting check: The flux should vanish for Ts = T . Setting Ts = T in (a) gives 0q .

Limitations on correlation equation (c): The conditions listed in (d) are met except at x = 1 cm where the Rayleigh number is below the lower limit. The accuracy of the computed flux at the location is in doubt.

Validity of correlation equation (c): The conditions listed in (d) are met.

(5) Comments. (i) This problem illustrates the importance of verifying the applicability of correlation equations to specific cases. It should be noted that correlation equations usually do not suddenly break down outside the limits of their applicability. Instead, their accuracy begins to deteriorate. (ii) The heat transfer literature should be consulted for applicable correlation

equations whenever the need arises. (iii) The magnitude of is the same whether it is expressed

in units of degree Celsius or kelvin. The reason is that is measured in terms of degree change. One degree change on the Celsius scale is equal to one degree change on the kelvin scale. This is also true of units of heat transfer coefficient and specific heat.

PROBLEM 8.33

200 square chips measuring 1 cm 1 cm each are mounted on both sides of

a thin vertical board measuring 10 cm 10 cm. The chips dissipate 0.035

W each. Assume uniform surface heat flux. Determine the maximum

surface temperature in air at 22oC. Neglect heat exchange by radiation.

(1) Observations. (i) This is a free convection problem over a vertical plate. (ii) The power dissipated in the chips is transferred to the air by free convection. (iii) This problem can be modeled as free convection over a vertical plate with constant surface heat flux. (iv) Surface temperature increases as the distance from the leading edge is increased. Thus, the maximum surface temperature occurs at the top end of the plate (trailing end). (v) Newton’s law of cooling relates surface temperature to heat flux and heat transfer coefficient. (vi) The fluid is air.


(3) Solution Plan. Apply the analysis of surface temperature distribution of a vertical plate with uniform surface heat flux in free convection.

(4) Plan Execution.


(ii) Analysis. Surface temperature distribution for a vertical plate with uniform surface flux is given by equation (8.29a)

T x TPr Pr

Pr g

q

kxs

s4 9 101 2 4 1 5/

/

(a)

Valid for: vertical plate uniform surface flux qs

laminar, 104 < RaL < 109

0 < Pr < bwhere




RaL = Rayleigh number at x = LTs = surface temperature, oC


= thermal diffusivity, W/m2-s



g

T


The Rayleigh number is defined as

RaL = PrLTTg s

2

3

(c)

where

L = vertical side of plate = 10 cm = 0.1 m

If all dissipated power in the chip leaves the surface, conservation of energy gives

sq = P/A (d)

where

A = chip surface area = 1 cm2 = 0.0001 m2

P = power dissipated in chip = 0.035 W

Properties are evaluate at the film temperature defined as

2

)2/( TLTT s

f (e)

where Ts(L/2) is surface temperature at the mid-point. However, since Ts(x)is unknown, an iterative procedure is required to obtain a solution. An assumed value for Ts(L/2) is used to calculate the film temperature at which properties are determined. Equation (a) is then used to calculate Ts(L/2). If the calculated value does not agree with the assumed temperature, the procedure is repeated until a satisfactory agreement is obtained.

(iii) Computations. Equation (d) gives surface flux

sq = 0.035(W)/0.0001(m2) = 350 W/m2

Assume Ts (L/2) = 58oC. Equation (e) gives

Tf = (58 + 22)(oC)/2 = 40oC


cp = 1006.8 J/kg- oCk = 0.0271 W/m-oCPr = 0.71

= 16.96 10-6 m2/s

= 1.1273 kg/m3


= 003193.015.273)C(40

1

15.273

1o

fT1/K

Thermal diffusivity is defined as

pc

ks/m10877.23

)Ckg/J(8.1006)m/kg(1273.1

)Cm/W(0271.0 26

o3

o

Substituting into (a) and letting x = L/2 = 0.1(m)/2 = 0.05 m

5/14

o

2

2o

26262/1o )m(05.0

)Cm/W(0271.0

)m/W(350

)s/m(81.9)C/1(003193.0

)s/m(1096.16)s/m(10877.23

71.0

)71.0(10)71.0(94)C(22/2LTs


Ts(L/2) = 76.3oC


Tf = 50oCcp = 1007.4 J/kg- oCk = 0.02781 W/m-oCPr = 0.709

= 25.27 10-6 m2/s

= 0.0030945 1/K

= 17.92 10-6 m2/s

= 1.0924 kg/m3

Substituting into (a) gives Ts(L/2) = 76.8oC. This is close to the assumed value of 78oC.

Surface temperature at the trailing end is now computed by evaluating (a) at x = L = 0.1 m

5/14

o

2

2o

26262/1o )m(1.0

)Cm/W(02781.0

)m/W(350

)s/m(81.9)C/1(0030945.0

)s/m(1092.17)s/m(1027.25

709.0

)709.0(10)709.0(94)C(22/2LTs

Ts(L) = 84.9oC

The condition on the Rayleigh number in equation (b) is verified next. Substituting into (c)

RaL = 709.0)s/m()1092.17(

)m()1.0)(C)(229.84)(s/m(81.9)C/1(0030945.02426

33o2o

= 4.22 106

This satisfies the condition on RaL given in equation (b).

(iv) Checking. Dimensional check: Equations (a), (c) and (d) are dimensionally consistent..


h(L/2) = ])2/(/[ TLTq ss = 350(W/m2)/(76.8 - 22)(oC) = 6.39 W/m2-oC


Validity of correlation equation (a): The conditions listed in (b) are met.

(5) Comments. (i) Surface temperature is determined without calculating the heat transfer coefficient. This is possible because equation (a) combines the correlation equation for the heat transfer coefficient and Newton's law of cooling to eliminate h and obtain an equation for surface

temperature in terms of surface heat flux. (ii) The magnitude of is the same whether it is

expressed in units of degree Celsius or kelvin. The reason is that is measured in terms of degree change. One degree change on the Celsius scale is equal to one degree change on the kelvin scale. This is also true of units of heat transfer coefficient and specific heat.

PROBLEM 8.34

An apparatus is designed to determine surface emissivity of

materials. The apparatus consists of an electrically heated

cylindrical sample (disk) of diameter D and thickness . the

disk is insulated along its heated side and rim. It is placed

horizontally with its heated surface facing down in a large

chamber whose surface is maintained at uniform temperature

.surT The sample is cooled by free convection and radiation

from its upper surface. To determine the emissivity of a

sample, measurements are made of the diameter D, electric

power input P, surface temperature sT , surroundings

temperature surT and ambient temperature .T Determine

the emissivity of a sample using the following data:

D = 12 cm, 0.5 cm, P 13.2 W, C98osT , C27o

surT , C22oT

(1) Observations. (i) Power supply to the disk is lost from the surface to the surroundings by free convection and radiation. (ii) To determine the rate of heat loss, the disk can by modeled as a horizontal plate losing heat by free convection to an ambient air and by radiation to a large surroundings. (iii) Newton’s law of cooling gives the rate of heat transfer by convection and Stefan-Boltzmann relation gives the heat loss by radiation. (iv) Free convection correlations give the heat transfer coefficient. (v) Conservation of energy at the surface gives the emissivity, if it is the only unknown.


(3) Solution Plan. Apply conservation of energy at the exposed surface. Use Newton's law of cooling Stefan-Boltzmann relation, equation. Use correlation equations to determine the average heat transfer coefficient for free convection from a horizontal surface.

(4) Plan Execution.

(i) Assumptions. (1) steady state, (2) constant properties (except in buoyancy), (3) uniform surface temperature, (4) quiescent ambient fluid, (5) the heated surface and the rim are perfectly insulated, (6) surroundings is at uniform temperature and (7) the disk is small compared to the surroundings.

(ii) Analysis. Conservation of energy at the surface gives

rc qqP (a)

where

P = electric energy power supplied to the disk = 13.2 W

cq heat transfer by convection, W

rq heat transfer by radiation, W

Application of Newton's law of cooling to the surface gives

cq = h A ( sT - T ) (b)

+ -

D

T

surT

sTg


where



sT = surface temperature = 98oC = 371.15 K

T = ambient air temperature = 22oC = 295.15 K


2

4DA (c)

where

D = disk diameter = 12 cm = 0.12 m


)( 44sursr TTAq (d)

where

surT surroundings temperature = 27 + 273.15 = 300.15 K

surface emissivity


(c) and (d) into (a)

P = h A ( sT - T ) + )( 44surs TTA (e)

Solving (e) for

)(

)()/(44

surs

s

TT

TThAP (f)

The average heat transfer coefficient h for a heated plate facing up is determined from (8.29)

4/154.0 LL Rak

LhuN for 75 10210 LRa (8.29a)

3/114.0 LL Rak

LhuN for 107 103102 LRa (8.29b)

(8.29c)

The Rayleigh number RaL is defined as

RaL = PrLTTg s

2

3

(g)

where


L = characteristic length for the horizontal disk, m

horizontal plate hot surface up or cold surface down

properties, except , at fT

at fT for liquids, T for gases


Pr = Prandtl number



The length L is defined as

4

4/2D

D

D

p

AL (i)


Tf=

2

TTs = 2

)C)(2298( o

= 60oC

at this temperature air properties are

k = 0.02852 W/m-oCPr = 0.708

= 18.9 10-6 m2/s

For ideal gases, the coefficient of thermal expansion for a horizontal surface is given by

= 0033881.015.273)C(22

1

)(

1

oKT(1/K)

Equation (i) gives L

L = 0.12 (m)/4 =0.03 m


RaL = 708.0)/s(m)10(18.9

)(mC)(0.03)22)()(98/sC)9.81(m1/0.0033881(2426

33o2o

= 1.3518 510

Thus equation (8.29a) is applicable. Radiation heat transfer is given by equation (1.12)

)( 44sursr TTAq (j)

where

surT surroundings temperature = 27 Co = 27 + 273.15 = 300.15 K


(iii) Computations. Substitution into (8.29a) gives

k

LhuN L

4/15 )103518.1(54.0 = 10.35


h = 10.35k

L = 10.35 (0.02852)(W/m-oC)/0.03(m) = 9.84 W/m2-oC

Surface area is given by (c)


)m()12.0(4

22A = 0.01131 2m

Equation (f) gives

681.0)](K(300.15)[(371.15)105.67

295.15)C)(371.159.84(W/m).01131(m[13.2(W)/04448

o22

(iii) Checking. Dimensional check: Computations showed that equations (c), (f) and (g) are dimensionally consistent.


Validity of correlation equation (8.29c): The conditions listed in (8.29c) are met.

(5) Comments. (i) Disk thickness is not needed for the determination of . (ii) It is important

to provide good insulation of the heated surface and the rim. (iii) According to equation (f), increasing the operating surface temperature minimizes the error in the measurement of the surroundings temperature.

PROBLEM 8.35

It is desired to increase the heat by free convection from a wide vertical plate without increasing

its surface temperature. Increasing the height of the plate is ruled out because of the limited

vertical space available. It is suggested that a taller plate can be accommodated in the same

vertical space by tilting it 45o. Explore this suggestion and make appropriate recommendations.

Assume laminar flow.

(1) Observations. (i) This is a free convection problem. (ii) The geometry is a flat plate. (iii) Heat transfer from two plates is to be compared. One plate is vertical and the other is inclined. Both plates fit in the same vertical space. Thus, the inclined plate is longer than the vertical plate. (iv) Both plates are maintained at uniform surface temperature. (v) Heat transfer depends on surface area and average heat transfer coefficient.

(2) Problem Definition. Determine the average heat transfer coefficient for a vertical plate and an inclined plate under free convection conditions.

(3) Solution Plan. Apply Newton’s law of cooling and use appropriate correlation equations for the average Nusselt number to determine the average heat transfer coefficient for each orientation.

(4) Plan Execution.

(i) Assumptions. (1) Steady state, (2) two-dimensional, (3) constant properties (except in buoyancy), (4) uniform surface temperature, (5) negligible radiation and (6) quiescent fluid.


q = h A (Ts - T ) = h LS(Ts - T ) (a) where


h = average heat transfer coefficient, W/m2-oCL = plate length, m q = heat transfer rate, W S = plate width, m Ts = surface temperature, oC


Applying (a) to the two plates and using the subscripts i and v to refer to the inclined and vertical

plates, respectively

qi = hi iL S (Ts - T ) (b)

qv= vh vL S (Ts - T ) (c)

Taking the ratio of (e) and (f)

vq

qi =vv Lh

Lh ii (d)

The average Nusselt number for laminar free convection flow over a vertical plate of height vL

is


4/14/1

2/1)(

953.4884.4435.2 v

vvv LL Ra

PrPr

Pr

k

LhNu (e)

where

PrLTTg

Ras

L 2

3)(v

v (f)

where


vh = local heat transfer coefficient for vertical plate, W/m2-oC


vLRa = Rayleigh number for vertical plate

vL = length of vertical plate, m





0 < Pr < properties at Tf (g)

Solving (a) and (b) for vh4/1

v

vL

gCh (h)

where C is defined as4/1

22/1

2

)953.4884.4435.2(

)(

PrPr

TTPrkC s (i)

Applying (h) to the inclined plate4/1

cos

i

iL

gCh (j)

Substitute (h) and (j) into (d)

vq

qi = 4/1

4/3

)(cosvL

Li (k)

However, the length of the inclined plate is given by

Li = cos

vL (m)

Combining (k) and (m)


vq

qi =2/1)(cos

1(n)

(iii) Computations. For a plate inclined at 45o equation (m) gives

vq

qi = 2/1o )45(cos

1 = 1.19

(iv) Checking. Dimensional check: Equation (n) is dimensionless.

Limiting check: For the special case of = 0, the heat transfer from the two plates should be

identical. Setting = 0 in (n) gives

vq

qi = 1

(5) Comments. (i) The inclined plate loses more heat than the vertical plate. The increase in heat transfer rate is due to an increase in area rather than heat transfer coefficient. In fact if the two plates have the same length, equation (k) shows that the inclined plate loses less heat than the vertical plate. (ii) According to equation (n), the greater the angle of inclination, the more heat will transfer from the plate. (iii) The above analysis assumes that conditions (g) are satisfied

by both plates. (iv) The magnitude of is the same whether it is expressed in units of degree


PROBLEM 8.36

Estimate the free convection heat transfer rate from five sides of a cubical

ceramic kiln. Surface temperature of each side is assumed uniform at

70oC and the ambient air temperature is 20

oC. Each side measures 48

cm.

(1) Observations. (i) This is a free convection problem. (ii) The kiln has four vertical sides and a horizontal top. (iii) All surfaces are at the same uniform temperature. (iv) Newton’s law of cooling gives the heat transfer rate. (v) The sides can be modeled as vertical plates and the top as a horizontal plate. (vi) The fluid is air.

(2) Problem Definition. Determine the average heat transfer coefficient for a vertical plate and for a horizontal plate.

(3) Solution Plan. Apply Newton’s law of cooling and use appropriate correlation equations for the average Nusselt number to determine the average heat transfer coefficients.

(4) Plan Execution.

(i) Assumptions. (1) Steady state, (2) constant properties (except in buoyancy), (3) uniform surface temperature, (4) quiescent fluid, (5) no heat loss from the bottom surface and (6) negligible radiation.

(ii) Analysis. Heat transfer from the five sides is given by

ts qqq 4 (a)

where

q = heat transfer rate from five surfaces, W qs = heat transfer rate from each of four sides, W qt = heat transfer rate from the top, W

Application of Newton's law of cooling gives

)( TTAhq sss (b)

and

)( TTAhq stt (c)

where

A = surface area of one side of cube, m2

sh = average heat transfer coefficient for a vertical side, W/m2-oC

th = average heat transfer coefficient for top surface, W/m2-oC



Surface area of each side is given by A = L2 (d)

where

L = length of each square side = 48 cm = 0.48 m

T

g


The average heat transfer coefficient is determined from correlation equations for free convection over a vertical plate. The Rayleigh number RaL is calculated first to select an


RaL = PrLTTg s

2

3)( (e)

where


Pr = Prandtl number RaL = Rayleigh number




Tf=

2

TTs = 2

)C)(2070( o

= 45oC


k = 0.02746 W/m-oCPr = 0.7095

= 17.44 10-6 m2/s

For ideal gases the coefficient of thermal expansion for the sides is given by

s = 003143.015.273)C(45

1

)K(

1o

fT(1/K)

Substituting into (e) gives

RaL = 7095.0)s/m()1044.17(

)m()48.0)(C)(2070)(s/m(81.9)C/1(003143.02426

33o2o

= 0.3977 109

Since RaL

< 109, the flow is laminar. Thus, the appropriate equation for the average Nusselt

number for a vertical plate is given by (8.25b)

4/14/1

2/1 953.4884.4435.2LL Ra

PrPr

Pr

k

LhuN (f)

This equation is valid for vertical plate constant surface temperature Ts


0 < Pr < properties at Tf (g)where

LuN = average Nusselt number


Correlation equations for a heated horizontal plate is given by equation (8.29)

4/154.0 LL RauN for 75 10210 LRa (8.29a)

3/114.0 LL RauN for 107 103102 LRa (8.29b)

(8.29c)

Thus the Rayleigh number for the top surface must be determined. It is based on the characteristic length L defined in equation (8.31) as

p

ALt (h)

where p is perimeter. Using (h)

12.04

)m(48.0

44

2 L

L

LLt m

For a horizontal plat in air t is evaluated at T

t = 003411.015.273)C(20

1

)(

1oKT

(1/K)

(e) gives

RaL = 6

2426

33o2o

10744.67095.0)/s(m)10(17.44

)(mC)(0.12)20)()(70/sC)9.81(m/0.003411(1

Therefore, (8.29a) is applicable.

(iii) Computations. Substituting into (f) gives sh

k

LhuN s

L

4/19

4/1

2/1103977.0

)7095.0(953.4)7095.0(884.4435.2

7095.0 = 72.77

Solving the above for sh

sh = 72.77k

L = 72.77 (0.02746)(W/m-oC)/0.48(m) = 4.16 W/m2-oC

Equations (b) and (d) give the heat transfer rate from each side

sq = 4.16 (W/m2-oC) (0.48)2(m2)(70 )20 (oC) = 47.9 W

Heat transfer coefficient for the top surface is computed from equation (8.29a)





4/16 )10744.6(54.0k

LhuN t

L = 27.52

Solving for th

th = 27.520.12(m)

C)m0.02746(W/52.27

o

cL

k = 6.3 W/m2-oC

Equations (c) and (d) give the heat transfer rate from the top surface

tq = 6.3(W/m2-oC) (0.48)2(m2)(70 )20 (oC) = 72.6 W

Equation (a) gives the total heat loss from the five surfaces

q = 4(47.9)(W) + 72.6(W) = 264.2 W

(iv) Checking. Dimensional check: Computations showed that equations (a)-(f) and (h) are dimensionally consistent.

Quantitative check: Heat transfer coefficients are within the approximate range of Table 1.1.

Validity of correlation equations (f) and (h): The conditions listed in (g) and (i) are met.

(5) Comments. (i) Modeling each side as a free standing vertical plate is an approximation if the kiln rests on a flat surface. (ii) Heat loss from the top surface is considerably more than form a side surface. (iii) Depending on surface emissivity, radiation loss may be appreciable. (iv) The

magnitude of is the same whether it is expressed in units of degree Celsius or kelvin. The

reason is that is measured in terms of degree change. One degree change on the Celsius scale is equal to one degree change on the kelvin scale. This is also true of units of heat transfer coefficient and specific heat.

POBLEM 8.37

Determine the surface temperature of a single burner electric stove when its power supply is 70

W. The diameter of the burner is 18 cm and its emissivity is 0.32. The ambient air temperature is

30 Co and the surroundings temperature is 25 Co .

(1) Observations. (i) Heat transfer from the surface is by free convection and radiation. (ii) The burner can be modeled as a horizontal disk with its heated side facing down. (iii) Newton’s law of cooling gives heat transfer by convection and Stefan-Boltzmann relations gives heat transfer by radiation. (iv) Both convection and radiation depend on surface temperature. (v) If the burner is well insulated at the bottom heated surface and its rim, then the electric power supply is equal to surface heat transfer.

(2) Problem Definition. Determine the average heat transfer coefficient for a for a horizontal plate.

(3) Solution Plan. Apply Newton’s law of cooling and use appropriate correlation equations for the average Nusselt number to determine the average heat transfer coefficients. Use Stefan-Boltzmann relation to determine heat loss by radiation.

(4) Plan Execution.

(i) Assumptions. (1) Steady state, (2) constant properties (except in buoyancy), (3) uniform surface temperature, (4) quiescent fluid, (5) no heat loss from the bottom surface and rim, (6) uniform surroundings temperature and (7) surface of burner is small compared to surroundings area.

(ii) Analysis. Based on assumption (5), conservation of energy at the surface gives

rc qqP (a)

where

P = power supply to burner = 70 W

cq convection heat transfer rate, W

rq radiation heat transfer rate, W


)( TTAhq sc (b)

where

A = surface area of one side of cube, m2





A = 2

4D (c)

where


g

D

T

sT

+ -

cqrq


The average heat transfer coefficient is determined from correlation equations for free convection over a horizontal plate given by (8.29)

4/154.0 LL RauN for 75 10210 LRa (8.29a)

3/114.0 LL RauN for 107 103102 LRa (8.29b)

(8.29c)

The Rayleigh number RaL should be calculated to select an appropriate correlation equation from the above. The Rayleigh number is defined as

RaL = PrLTTg s

2

3)( (d)

where


L = characteristic lengthPr = Prandtl number RaL = Rayleigh number



The characteristic length L is defined in equation (8.31) as

p

AL (e)

where p is perimeter. Examination of (d) shows that the Rayleigh number depends on surface

temperature .sT However, sT is unknown. This suggest that the an iterative procedure must be

used to solve the problem.


)( 44sursr TTAq (f)

where

surT surroundings temperature = 25 + 273.15 = 298.15 K

surface emissivity = 0.3


(b) and (f) into (a)

P = h A ( sT - T ) + )( 44surs TTA (g)

Equation (g) is used to determine sT using an iterative procedure. A value for sT is assumed,

properties of air are determined, (d) is used to compute the Rayleigh number and an equation is





selected for the free convection Nusselt number, and finally (g) is used to calculate the dissipated power P. If the calculated P is not equal to 70 W, the procedure is repeated until a satisfactory agreement is obtained.

(iii) Computations. Assume C190osT . Using (e)

m045.04

(m)18.0

4

4/2 D

D

DL


Tf=

2

TTs = 2

C)()30190( o

= 110oC


k = 0.03194 W/m-oCPr = 0.704

= 24.1 10-6 m2/s

For a horizontal plate in air is evaluated at T

= 0032987.015.273)C(30

1

)(

1oKT

(1/K)

(d) gives

RaL = 5

2426

33o2o

107188.5704.0)/s(m)10(24.1

)(mC)(0.045)30)()(190/sC)9.81(m1/0.0032987(

Therefore, (8.29a) is applicable.

4/15 )107188.5(54.0k

LhuN L = 14.85

Solving for h

h = 14.85k

L = 14.85(0.03194)(W/m-oC)/0.045(m) = 10.54 W/m2-oC


222 m025447.04/)m18.0(4/DA


)](K273.15)(25273.15))[(190K)(W/m100.3(5.67

C)30)()(190(mC)0.0254470.54(W/m1

444428

o2o2P

P = 59.4 W

This calculated value is not close to the given power 70P W. The procedure is repeated for an

assume value C.210osT The following result is obtained at this assume surface temperature:


C120ofT

k = 0.03261 W/m-oCPr = 0.703

= 25.19 10-6 m2/s0032987.0

5108806.5LRa

95.14k

LhuN L

h =10.83 W/m2-oC

64.49cq W

17.20rq

P = (49.64 + 20.17) = 69.8 W

This is close to the given value of P = 70 W.

(iv) Checking. Dimensional check: Computations showed that equations (d,) (f) and (g) are dimensionally consistent.

Quantitative check: Heat transfer coefficient is within the approximate range of Table 1.1.

Validity of correlation equations (8.29a): The conditions listed in (8.29c) are met.

(5) Comments. (i) Radiation loss is appreciable and thus can not be neglected. (ii) In practice some of the electric power supplied to the burner is lost through the bottom side and rim. This has the effect of lowering surface temperature. Thus the model used to solve the problem overestimates surface temperature.

PROBLEM 8.38

A test apparatus is designed to determine surface emissivity of material. Samples are machined

into disks of diameter D. A sample disc is heated electrically on one side and allowed to cool off

on the opposite side. The heated side and rim are well insulated. The disk is first placed

horizontally with its exposed surface facing up in a large chamber. At steady state the exposed

surface temperature is measured. The procedure is repeated, without changing the power

supplied to the disk, with the exposed surface facing down. Ambient air temperature in the

chamber is recorded.

[a] Show that surface emissivity is given by

)(

)()(41

42

2211

ss

ss

TT

TThTTh

where subscripts 1 and 2 refer to the exposed surface facing up and down, respectively, and

h average heat transfer coefficient,

CW/m o2

sT surface temperature,= K

T ambient temperature, K

Stefan-Boltzmann constant, 42 KW/m

[b] Calculate the emissivity for the following case:

D 12 cm, KCT os 15.5332601 , 15.5733002 CT o

s K, 15.29320 CT o K

(1) Observations. (i) Heat transfer from the surface is by free convection and radiation. (ii) The sample can be modeled as a horizontal disk with its heated side facing down or up. (iii) Newton’s law of cooling gives heat transfer by convection and Stefan-Boltzmann relation gives heat transfer by radiation. (iv) Radiation depends on surface emissivity. (v) If the disk is well insulated at the heated surface and its rim, then the electric power supply is equal to surface heat transfer. (vi) Since the electric power is the same for both orientations, it follows that surface heat transfer rate is also the same. (vii) Each orientation has its own Nusselt number correlation equation.

(2) Problem Definition. Determine surface heat transfer by convection and radiation for two horizontal orientations of a plate: heated side facing up and heated side facing down.

(3) Solution Plan. Apply Newton’s law of cooling and use appropriate correlation equations for the average Nusselt number to determine the average heat transfer coefficients. Use Stefan-Boltzmann relation to determine heat loss by radiation. Equate total surface heat transfer rate for both orientations.

(4) Plan Execution.

(i) Assumptions. (1) Steady state, (2) constant properties (except in buoyancy), (3) uniform surface temperature, (4) quiescent fluid, (5) no heat loss from the heated power supply surface and rim, (6) uniform surroundings temperature and (7) surface of disk is small compared to surroundings area.

surT

-

D

T sT

+

g

1 2

+

-

TsTD


(ii) Analysis. Based on assumption (5), conservation of energy at the surface gives

rc qqP (a)

where

P power supply to disk, W

cq convection heat transfer rate, W

rq radiation heat transfer rate, W


)( TTAhq sc (b)

where






)( 44sursr TTAq (c)

where

surT surroundings temperature

surface emissivity


(b) and (c) into (a)

P = h A ( sT - T ) + )( 44surs TTA (d)

This result is applied to orientation 1 (heated surface facing up) and orientation 2 (heated surface facing down), noting that P is the same for both

P = 1h A ( 1sT - T ) + )( 441 surs TTA (e)

P = 2h A ( 2sT - T ) + )( 442 surs TTA (f)

Equating (e) and (f) and solving for gives

)(

)()(41

42

2211

ss

ss

TT

TThTTh(g)

[b] The average heat transfer coefficient is determined from correlation equations for free convection over a horizontal plate. For a plate with the heated side facing up (orientation 1), use (8.29)

4/111 54.0 LL RauN for 7

15 10210 LRa (8.29a)

3/111 14.0 LL RauN for 10

17 103102 LRa (8.29b)

PROBLEM 8.38(continued)

(8.29c)

The Rayleigh 1LRa number is defined as

12

1

31 )(

1 PrLTTg

Ra sL (h)

where


L = characteristic length

1Pr = Prandtl number

RaL1 = Rayleigh number


1v = kinematic viscosity, m2/s

For the average heat transfer coefficient for horizontal plate with the heated side facing down (orientation 2), use (8.30)

4/122 27.0 LL RauN for 10

25 103103 LRa (8.30a)

(8.30b)

The Rayleigh number 2LRa is given by

222

32 )(

2 PrLTTg

Ra sL

v

(i)

The characteristic length L is defined as

44perimeter

areasurface 2 D

D

DL (j)

where


(iii) Computations. Consider orientation 1 first. Equation (j) gives

m035.04

)m(14.0L

Properties of air are determined at the film temperature 1fT defined as




horizontal plate hot surface down or cold surface up




2

11

TTT s

f = 2

C)()20260( o

= 140oC


1k = 0.03394 W/m-oC

1Pr = 0.702

1v = 27.44 10-6 m2/s

For a horizontal plate in air is evaluated at T

= 0034112.015.273)C(20

1

)(

1

oKT(1/K)

Equation (h) gives

RaL1 = 5

2426

33o2o

1021.3702.0)/s(m)10(27.44

)(mC)(0.035)20)()(260/sC)9.81(m1/0.0034112(

Substituting into (8.29a)

85.12)1021.3(54.0 4/15

1

11

k

LhuN L

Solving for 1h

1h = 14.85 46.12m)(035.0

C)(W/m-03394.0 o1

L

k W/m2-oC

Consider orientation 2. Properties of air are determined at the film temperature 2fT defined as

2

22

TTT s

f = 2

C)()20300( o

= 160oC


2k = 0.03525 W/m-oC

2Pr = 0.701

2v = 29.75 10-6 m2/s

Equation (I) gives

RaL2 = 5

2426

33o2o

10182.3701.0)/s(m)10(29.75

)(mC)(0.035)20)()(300/sC)9.81(m1/0.0034112(


41.6)10182.3(27.0 4/15

2

22

k

LhuN L

Solving for 2h


2h = 6.41 46.6m)(035.0

C)(W/m-03525.0 o2

L

k W/m2-oC


769.0]K)()15.273260(K)()15.273300[()K(W/m1067.5

C))(20300(C)(W/m46.6C))(20260(C)(W/m46.124444428

oo2oo2

(iv) Checking. Dimensional check: Computations showed that equations (g)-(j) are dimensionally consistent.

Quantitative check: Heat transfer coefficients are within the approximate range of Table 1.1.

Validity of correlation equations (8.29a) and (8.30a): The conditions listed in (8.29c) and (8.30b) are met.

(5) Comments. (i) The determination of emissivity using this method does not require measuring the surroundings temperature. However, surroundings temperature must be uniform and the same for both orientations. (ii) Emissivity can be determined without knowing the power supply as long as it is the same for both orientations. (iii) The disk need not be perfectly insulated as long as heat loss from the insulated surfaces is the same for both orientations. (iv). Based on the calculated emissivity, heat transfer from the surface q is given by (b), (c) and (d). However, to compute heat transfer by radiation the surroundings temperature must be known.

Assume C20osurT . Thus, for orientation 1:

46.03C)(20))(260(m0.14/2)C)12.46(W/m o22o2cq W

)K]()15.27320()15.273260)[(m()2/14.0()K(W/m)1067.5(769.0 44422428rq = 49.27W

03.46q W + 49.27 W = 95.3 W

Similarly, for orientation 2

84.27cq W

46.67rq W

q 95.3 W

This confirms that the electric power supply to both positions is the same.

POBLEM 8.39

A hot water tank of diameter 65 cm and height 160 cm loses heat by free convection. Estimate

the free convection heat loss from its cylindrical and top surfaces. Assume a surface temperature

of 50oC and an ambient air temperature of 20

oC.

(1) Observations. (i) This is a free convection problem. (ii) Heat is transferred from the cylindrical surface and top surface of tank to the ambient air. (iii) Under certain conditions a vertical cylindrical surface can be modeled as a vertical plate. (iv) Newton’s law of cooling gives the heat transfer rate from tank. (v) The fluid is air.

(2) Problem Definition. Determine the average heat transfer coefficient for a vertical cylindrical surface and a heated horizontal surface facing up.

(3) Solution Plan. Apply Newton’s law of cooling and use appropriate correlation equations for the average Nusselt number to determine the average heat transfer coefficient for a vertical cylinder and a heated horizontal plate.

(4) Plan Execution.

(i) Assumptions. (1) Steady state, (2) constant properties (except in buoyancy), (3) uniform surface temperature, (4) quiescent fluid and (6) negligible radiation.

(ii) Analysis. The total heat transfer from tank is

q = qc + qt (a) where

q = total rate of heat transfer from tank, W qc = rate of heat transfer from cylindrical surface, W qt = rate of heat transfer from top surface, W

Applying Newton’s law of cooling to the two surfaces

qc = hc Ac (Ts - T ) (b)

and

qt = ht At (Ts - T ) (c)

where

Ac = area of cylindrical surface, m2

At = area of top surface, m2

hc = average heat transfer coefficient for the cylindrical surface, W/m2-oC

ht = average heat transfer coefficient for the top surface, W/m2-oC



The two areas Ac and At are

Ac = D L (d) and

At = D2/4 (e)

where

D

L

qc

qt

g

T

qc


D = tank diameter = 0.65 m L = tank height = 1.6 m

The heat transfer coefficients are determined form correlation equations for the Nusselt number. To determine if the cylindrical surface can be modeled as vertical plate the following criterion is checked

4/1)(

35

LGrL

D (f)

where the Grashof number GrL is defined as

GrL =2

3LTTg s (g)

where





Tf = 2/)( TTs = (50 + 20)(oC)/2 = 35oC


k = 0.02674 W/m-oCPr = 0.711

= 16.485 10-6 m2/s

For an ideal gas is given by

=)K(

1

fT =

15.273C35

1o

= 0.003245 1/K


GrL =2226

33o2o

)s/m()10485.16(

)m()6.1)(C(2050)s/m(81.9)C/1(003245.0 = 14.394 109

The right hand side of (f) is

(35/GrL)1/4 = (35/14.394 109)1/4 = 0.00702

The left hand side of (f) is

D/L = 0.65(m)/1.6(m) = 0.406

Equation (f) is satisfied and thus, the cylindrical surface can be treated as a vertical plate.

The Rayleigh number RaL is computed next to select an appropriate correlation equation for the average Nusselt number for the cylindrical surface

RaL = GrL Pr = 14.394 109 0.711 = 10.234 109

For this Rayleigh number the correlation equation is given by (8.26a)


2

27/816/9

6/1

)/492.0(1

387.0825.0

Pr

Ra

k

LhuN Lc

L (h)

valid for vertical plate uniform surface temperature Ts

laminar, transition, and turbulent

10 1 < RaL <1012

0 < Pr < properties at Tf (i) where


Heat loss from the top surface is modeled as a horizontal heated plate facing up. The Rayleigh umber RaD for the top is computed to select an appropriate correlation equation for the Nusselt number. The Rayleigh number is defined as

RaD = 2

3)( DTTg s Pr

RaD = 711.0)s/m()10485.16(

)m()65.0)(C(2050)s/m(81.9)C/1(003245.02226

33o2o

= 0.6862 109

Thus, the appropriate equation for this Raleigh number is given by (8.31b)

DNu =k

Dht = 0.15 (RaD)1/3 (j)

valid for96 106.1108 LRa (k)

(iii) Computations. For the cylindrical surface, equation (h) gives

2

27/816/9

6/19

711.0/492.01

)10234.10(387.0825.0

k

LhuN c

L = 254.2

Solving the above for ch

ch = 254.2 k/L = 254.2(0.02674)(W/m-oC)/1.6(m) = 4.25 W/m2-oC

Equation (i) gives the heat transfer coefficient for the top surface

DNu =k

Dht = 0.15 (0.6862 109)1/3 = 132.3

Solving for ht

ht = 132.3 k/D = 132.3 (0.02674)(W/m-oC)/0.65(m) = 5.44 W/m2-oC

The surface areas are calculated from (d) and (e)


Ac = 0.65(m) 1.6(m) = 3.267 m2

and

At = (0.65)2(m2)/4 = 0.3318 m2

Equations (b) and (c) give the rate of heat transfer from the two surfaces

qc = 4.25(W/m2-oC) 3.267(m2)(50 - 20)(oC) = 416.5 W

and

qt = 5.44 (W/m2-oC) 0.3318(m2) (50 - 20)(oC) = 54.1 W

The total heat loss from the tank is

q = 416.5 (W) + 54.1 (W) = 470.6 W

(iv) Checking. Dimensional check: Computations showed that equations (b)-(h) and (j) are dimensionally consistent.

Quantitative check. The computed values of the heat transfer coefficient are representative of values listed in Table 1.1 for free convection of gases.

Validity of correlation equations (h) and (j): The conditions lis ted in (i) and (k) are met.

(5) Comments. (i) It is important to check the Rayleigh and Grashof number s to determine the applicable correlation equations for the Nusselt number. (ii) The heat loss from the tank is

significant. To conserve energy the tank should be insulated. (iii) The magnitude of is the

same whether it is expressed in units of degree Celsius or kelvin. The reason is that is measured in terms of degree change. One degree change on the Celsius scale is equal to one degree change on the kelvin scale. This is also true of units of heat transfer coefficient and specific heat.

PROBLEM 8.40

Hot gases from a furnace are discharged through a round horizontal duct 30 cm in diameter.

The average surface temperature of a 3 m duct section is 180oC. Estimate the free convection

heat loss from the duct to air at 25oC.

(1) Observations. (i) This is a free convection problem. (ii) The geometry is a horizontal round duct. (iii) Heat is transferred from duct surface to the ambient air. (iv) According to Newton’s law of cooling, the rate of heat transfer depends on the heat transfer coefficient, surface area and surface and ambient temperatures.

(2) Problem Definition. Determine the free convection heat transfer coefficient for a horizontal round duct (cylinder).

(3) Solution Plan. Apply Newton’s law of cooling. Use Nusselt number correlation equations to determine the average heat transfer coefficient for a horizontal cylinder in free convection.

(4) Plan Execution.

(i) Assumptions. (1) Steady state, (2) two-dimensional (neglect end effects), (3) constant properties (except in buoyancy), (4) uniform surface temperature, (5) quiescent fluid and (6) negligible radiation.

(ii) Analysis. Applying Newton’s law of cooling to the duct

q = h A (Ts - T ) = h DL(Ts - T ) (a) where


D = duct diameter = 3 cm = 0.3 m

h = average heat transfer coefficient,W/m2-oC

L = duct length = 3 m q = heat transfer rate, W Ts = surface temperature = 180oC


The heat transfer coefficient is obtained from correlation equations for the Nusselt number for a horizontal cylinder in free convection given by equation (8.35a)

2

27/816/9

6/1

/559.01

)(387.060.0

Pr

Ra

k

DhNu D

D (b)

valid for: horizontal cylinder uniform surface temperature or flux

10 5 < RaD < 1012

properties at Tf (c)where



Pr = Prandtl number RaD = Rayleigh number


D

L

sT

T

g

T


RaD =2

3)( DTTg s Pr (d)

where





= K15.273

1

fT (e)

Properties are determined at the film temperature Tf, defined as

Tf = (Ts + T )/2 (f)

(iii) Computations. The film temperature is calculated using (f)

Tf = (180 +25)(oC)/2 = 102.5oC


k = 0.03144W/m-oCPr = 0.704

= 1/(102.5 + 273.15)(K) = 0.0026621(1/K)

= 23.29 10-6 m2/s


RaD = 2226

33o2o

)s/m()1029.23(

)m()3.0)(C)(25180)(s/m(81.9)C/1(0026621.0 0.704 = 1.418 108

Equation (b) is applicable for this Rayleigh number. Substituting into (b)

2

27/816/9

6/18

D

)704.0/559.0(1

)10418.1(387.060.0

k

DhuN =66.86

Solving for h

h = 66..86 k/D = 66.86(0.03144)(W/m-oC)/0.3(m) = 7.01 W/m2-oC

Equation (a) gives

q = 7.01(W/m2-oC) 0.3(m)2.5(m) ( 25180 )(oC) = 2560 W

(iv) Checking. Dimensionless check: Computations showed that equations (a), (b), (d) and (e) are dimensionally consistent.

Quantitative check: The value of the heat transfer coefficient is within the approximate range given in Table 1.1 for free convection of gases.

Validity of correlation equation (b): The conditions listed in (c) are met.

(5) Comments. Since surface temperature is relatively high, radiation heat loss may be significant.

PROBLEM 8.41

A 6 m long horizontal steam pipe has a surface temperature of 120oC. The diameter of the pipe

is 8 cm. It is estimated that if the pipe is covered with a 2.5 cm thick insulation material its

surface temperature will drop to 40oC. Determine the free convection heat loss from the pipe

with and without insulation. The ambient air temperature is 20oC.

(1) Observations. (i) This is a free convection problem. (ii) The geometry is a horizontal pipe. (iii) Heat is transferred from pipe surface to the ambient air. (iv) Adding insulation material reduces heat loss from pipe. (v) According to Newton’s law of cooling, the rate of heat transfer depends on the heat transfer coefficient, surface area and surface and ambient temperatures. (vi) Heat transfer coefficient and surface area change when insulation is added. (vii) The fluid is air.

(2) Problem Definition. Determine the free convection heat transfer coefficient for a horizontal pipe (cylinder) with and without insulation.

(3) Solution Plan. Apply Newton’s law of cooling. Use correlation equations for the Nusselt number to determine the average heat transfer coefficient for a horizontal cylinder in free convection.

(4) Plan Execution.


(ii) Analysis. Applying Newton’s law of cooling to the bare and insulated pipe

qo = oh Ao (Tso - T ) = oh DoL(Tso - T ) (a)

and

qi = ih Ai(Tsi - ) = ih DiL(Tsi - T ) (b)

where the subscripts i and o refer to insulated and bare pipes, respectively, and


Di = insulation diameter = 8 cm + 2(2.5)(cm) = 13 cm = 0.13 m Do = pipe diameter = 8 cm = 0.08 m

ih = average heat transfer coefficient for insulated pipe, W/m2-oC

oh = average heat transfer coefficient for bare pipe, W/m2-oC

L = pipe length = 6 m Tsi = surface temperature of insulated pipe = 40oCTso = surface temperature of bare pipe = 120oC


The heat transfer coefficient is obtained from correlation equations for the Nusselt number for a horizontal cylinder in free convection given by equation (8.35a)

2

27/816/9

6/1

/559.01

387.060.0

Pr

Ra

k

DhuN D

D (c)

D

L

sT

T

g

PROBLEM 8.41 (continued)Valid for

horizontal cylinder uniform surface temperature or flux

10 5 < RaD < 1012


D = diameter, m k = thermal conductivity, W/m-oC




RaD =2

3DTTg s Pr (e)

whereg = gravitational acceleration = 9.81 m/s2




= )(K)15.273(

1

fT (f)

Properties are determined at the film temperature Tf

Tf = (Ts + T )/2 (g)

(iii) Computations. Consider the bare pipe first. Film temperature is given by (g)

Tf = (120 +20)(oC)/2 = 70oC

Properties at this temperature are

k = 0.02922 W/m-oCPr = 0.707

= 1/(70 + 273.15) K = 0.002914 1/K

= 19.9 10-6 m2/s


oDRa =2226

33o2o

)s/m()109.19(

)m()08.0)(C)(20120)(s/m(81.9)C/1(002914.0 0.707 = 0.261 107

Equation (c) is applicable for this Rayleigh number. Substituting into (c)

2

27/816/9

6/17

)707.0/559.0(1

)10261.0(387.060.0

k

DhuN oo

oD = 19.08

Solving for oh


oh = 19.08k/Do = 19.08(0.02922)(W/m-oC)/0.08(m) = 6.97 W/m2-oC

Equation (a) gives

qo = 6.97(W/m2-oC) 0.08(m)6(m) (120-20)(oC) = 1051.1 W

Consider next the insulated pipe. Film temperature is calculated using (g)

Tf = (40 +20)(oC)/2 = 30oC


k = 0.02638 W/m-oCPr = 0.712

= 1/(30 + 273.15) K = 0.003299 1/K (ideal gas)

= 16.01 10-6 m2/s


iDRa =2226

33ooo

)s/m()1001.16(

)m(13.0)C(2040)C/1(81.9)C/1(003299.0 0.712 = 0.395 107

Equation (c) is applicable for this Rayleigh number. Substituting into (c)

2

27/816/9

6/17

)712.0/559.0(1

)10395.0(387.060.0

k

DhuN ii

Di= 21.53

Solving for ih

ih = 21.53k/Di = 21.53(0.02638)(W/m-oC)/0.13(m) = 4.37 W/m2-oC

Equation (b) gives

qi = 4.37(W/m2-oC) 0.13(m)6(m) (40-20)(oC) = 214.2 W

(iv) Checking. Dimensionless check: Computations showed that equations (a)-(c) and (e) are dimensionally consistent.

Quantitative check: The value of the heat transfer coefficient is within the approximate range given in Table 1.1 for free convection of gases.

Validity of correlation equation (c): The conditions listed in (d) are met.

(5) Comments. (i) Heat loss from the pipe is significantly reduced when insulation is added. The reduction is due to a decrease in surface temperature and heat transfer coefficient. (ii) The



PROBLEM 8.42

An electric wire dissipates 0.6 W/m while suspended horizontally in air at 20oC. Determine its

surface temperature if the diameter is 0.1 mm. Neglect radiation.

(1) Observations. (i) This is a free convection problem. (ii) The geometry is a horizontal wire (cylinder). (iii) Under steady state conditions the power dissipated in the wire is transferred to the surrounding air. (iv) According to Newton’s law of cooling, surface temperature is determined by the heat transfer rate, heat transfer coefficient, surface area and ambient temperature. (v) The fluid is air.

(2) Problem Definition. Determine the free convection heat transfer coefficient for a horizontal wire.


(4) Plan Execution.

(i) Assumptions. (1) Steady state, (2) two-dimensional (neglect end effects), (3) constant properties (except in buoyancy), (4) uniform surface flux and temperature, (5) no axial conduction, (6) quiescent fluid and (7) negligible radiation.

(ii) Analysis. Applying Newton’s law of cooling to the wire

q = h A (Ts - T ) = h DL(Ts - T )Solving the above for Ts

Ts = T + hD

Lq / (a)

where


D = wire diameter = 0.1 mm = 0.0001 m

h = average heat transfer coefficient, W/m2-oCL = wire length, m q/L = power dissipated per unit length = 0.6 W/m Ts = surface temperature, oC


The heat transfer coefficient is obtained from correlation equations for the Nusselt number for a horizontal cylinder in free convection, given by equation (8.35a)

2

27/816/9

6/1

)/559.0(1

387.060.0

Pr

Ra

k

DhuN D

D (b)

Valid for horizontal cylinder uniform surface temperature or flux

10 5 < RaD < 1012



Lq/ sT Tg





RaD =g T T Ds

3

2 Pr (d)

where





Tf = (Ts + T )/2 (e)

(iii) Computations. Examining (a), (b) and (c) shows that h depends on Ts through the Rayleigh number. Thus, equation (a) must be solved for Ts by trial and error. A value for Ts is assumed, properties are determined based on the assumed temperature, equation (d) is used to

calculate RaD, (b) is used to determine h and (a) is used to calculated Ts. If the calculated temperature is not close to the assumed value, the procedure is repeated until a satisfactory agreement is obtained.

Assume Ts = 200oC. Using (e)

Tf = (200 + 20)(oC)/2 = 110oC


k = 0.03194 W/m-oCPr = 0.704

= 1/(Tf + 273.15) K = 1/(110 + 273.15) K = 0.00261 (1/K) (ideal gas)

= 24.1 10-6 (m2/s)


RaD = 2226

33o2o

)s/m()101.24(

)m(0001.0)C(20200)s/m(81.9)C/1(00261.0 0.704 = 5.5863 10-3

Equation (b) can be used for this Rayleigh number. Substituting into (b)

2

27/816/9

6/1

)704.0/559.0(1

0055863.0387.060.0

k

DhuN D = 0.5406

Solving for h

h = 0.5406 k/D = 0.5406(0.03194)(W/m-oC)/0.0001(m) = 172.7 W/m2-oC

Equation (a) gives

Ts = 20(oC) +Cm/W7.172m0001.0

m/W6.0o2

= 31.1oC


This calculated value of Ts is significantly different from the assumed value. Repeating the

procedure with a new assumed value of Ts = 30oC gives h = 127.84 W/m2-oC and a calculated Ts

= 34.9oC. Since this is close to the assumed value, it follows that the estimated surface temperature is 34.9oC.

(iv) Checking. Dimensionless check: Computations showed that equations (a), (b) and (d) are dimensionally consistent.

Limiting check: If no energy is dissipated in the wire, its temperature should be the same as the

ambient temperature. Setting q = 0 in (a) gives Ts = T .

Quantitative check: The value of the heat transfer coefficient departs significantly from typical free convection values listed in Table 1.1. A review of the computations revealed no errors in the analysis or calculations. The departure from the range of h in Table 1.1 is due to the unusual nature of this application in which the diameter is very small.


(5) Comments. (i) Solution by trial and error was necessary because properties depend on the unknown surface temperature and equations (a), (b) and (d) can not be solved explicitly for Ts.(ii) Table 1.1 should be used as a guide only, keeping in mind that exceptions to values listed should be expected. (iii) Taking radiation into consideration has the effect of reducing surface

temperature. (iv) The magnitude of is the same whether it is expressed in units of degree


PROBLEM 8.43

The diameter of a 120 cm long horizontal section of a neon sign is 1.5 cm. Estimate the surface

temperature in air at 25oC if 12 watts are dissipated in the section. Neglect radiation heat loss.

(1) Observations. (i) This is a free convection problem. (ii) The geometry is a horizontal tube. (iii) Under steady state conditions the power dissipated in the neon tube is transferred to the surrounding air. (iv) According to Newton’s law of cooling, surface temperature is determined by the heat transfer rate, heat transfer coefficient, surface area and ambient temperature. (v) The fluid is air.

(2) Problem Definition. Determine the free convection heat transfer coefficient for a horizontal tube.


(4) Plan Execution.

(i) Assumptions. (1) Steady state, (2) two-dimensional (neglect end effects), (3) constant properties (except in buoyancy), (4) uniform surface temperature, (5) no axial conduction, (6) quiescent fluid and (7) negligible radiation.

(ii) Analysis. Applying Newton’s law of cooling to the wire

q = h A (Ts - T ) = h DL(Ts - T )Solving the above for Ts

Ts = T + hDL

q (a)

where


D = tube diameter = 1.5 cm = 0.015 m

h = average heat transfer coefficient, W/m2-oCL = wire length = 120 cm = 1.2 m q = power dissipated in wire per unit length = heat transfer rate from wire =12 W Ts = surface temperature, oC



2

27/816/9

6/1

)/559.0(1

387.060.0

Pr

Ra

k

DhuN D

D (b)

valid for horizontal cylinder uniform surface temperature or flux

10 5 < RaD < 1012







RaD =g T T Ds

3

2 Pr (d)

where





Tf = (Ts + T )/2 (e)

(iii) Computations. Examining (a), (b) and (c) shows that h depends on Ts through the Rayleigh number. Thus, equation (a) must be solved for T- by trial and error. A value for Ts is assumed, properties are determined based on the assumed temperature, equation (d) is used to

calculate RaD, (b) is used to determine h and (a) is used to calculated Ts. If the calculated temperature is not close to the assumed value, the procedure is repeated until a satisfactory agreement is obtained.

Assume Ts = 75oC. Using (e)

Tf = (25 + 75)(oC)/2 = 50oC

Air properties at this temperature are given in Appendix C

k = 0.02781 W/m-oCPr = 0.709

= 1/(50oC + 273.15) = 0.0030945 1/K (ideal gas)

= 17.92 10-6 m2/s


RaD =0 0030945 1 9 81 75 25 0 015

17 92 100 709

2 3 3

6 2 4 2

. ( / ) . ( / )( )( )( . ) ( )

( . ) ( / ).

o oC m s C m

m s= 1.131 104

and

Nuh D

kD

2

27/816/9

6/14

709.0/559.01

)10131.1(387.060.0 = 4.503


h = )m(015.0

)CW/m(02781.0503.4

o

= 8.35 W/m2-oC

Substituting into (a) gives the surface temperature


Ts = 25oC + )Cm/W(35.8)m(2.1)m(015.0

)W(12o2

= 50.4oC

This is not close to the assumed value of 75oC. The procedure is repeated for Ts = 50oC and 55oC. The tabulated results below shows that Ts 53.4oC.

Assumed Ts (oC) h (W/m2-oC) Calculated Ts (

oC)

75 8.348 50.4

50 7.16 54.6

55 7.46 53.4


Limiting check: If no power is dissipated (neon sign is off), surface temperature should be the

same as the ambient temperature. Setting q = 0 in (a) gives Ts = T .

Quantitative check: The value of the heat transfer coefficient is within the range of values listed in Table 1.1 for free convection of gases.


(5) Comments. (i) Solution by trial and error was necessary because properties depend on the unknown surface temperature and equations (a), (b) and (d) can not be solved explicitly for Ts.(ii) Taking radiation into consideration has the effect of reducing surface temperature. (iii) The



PROBLEM 8.44

An air conditioning duct passes horizontally a distance of 2.5 m through the attic of a house. The

diameter is 30 cm and the average surface temperature is 10oC. The average ambient air

temperature in the attic during the summer is 42oC. Duct surface emissivity is 0.1. Estimate the

rate of heat transfer to the cold air in the duct.

(1) Observations. (i) This is a free convection problem. (ii) The geometry is a round horizontal round duct. (iii) Heat is transferred from the ambient air to the duct. (iv) According to Newton’s law of cooling, the rate of heat transfer to the surface depends on the heat transfer coefficient, surface area and surface and ambient temperatures. (v) The fluid is air.

(2) Problem Definition. Determine the average free convection heat transfer coefficient for a horizontal round duct.


(4) Plan Execution.


(ii) Analysis. Applying Newton’s law of cooling to the round duct

q = h A(T - Ts) = h DL(T - Ts) (a) where


D = duct diameter = 0.3 m


L = duct length = 2.5 m q = rate of heat transfer, oCTs = surface temperature = 10oC



2

27/816/9

6/1

/559.01

387.060.0

Pr

Ra

k

DhuN D

D (b)

valid for: horizontal cylinder uniform surface temperature or flux

10 5 < RaD < 1012




Pr = Prandtl number

D

L

sTTgq


RaD = Rayleigh number


RaD =2

3)( DTTg s Pr (d)

where





Tf = (Ts + T )/2 (e)

For an ideal gas, is given by

= K))(15.273(

1

fT (f)

(iii) Computations. The film temperature is calculated first using (e)

Tf = (10 +42)(oC)/2 = 26oC


k = 0.02608W/m-oCPr = 0.7124

= 1/(26 + 273.15) K = 0.003343 (1/K) = 0.003343(1/oC)

= 15.64 10-6 (m2/s)


RaD = 2226

33o2o

)s/m()1064.15(

)m(3.0)C(1042)s/m(81.9)C/1(003343.0 0.7124 = 8.2522 107

Equation (b) is applicable for this Rayleigh number. Substituting into (b)

2

27/816/9

6/17

7124.0/559.01

102522.8387.060.0

k

DhuN D = 53.37

Solving for h

h = 53.37 k/D = 53.37(0.02608)(W/m-oC)/0.3(m) = 4.64 W/m2-oC

Equation (a) gives

q = 4.64 (W/m2-oC) 0.3(m) 2.5(m) (42 - 10)(oC) = 349.8 W


Limiting check: No heat will transfer to the duct if the ambient temperature is the same as

surface temperature. Setting T = Ts in (a) gives q = 0.


Quantitative check: The value of the heat transfer coefficient within the approximate range given in Table 1.1 for free convection of gases.


(5) Comments. (i) Heat added to the cold air in the duct is significant. At $0.15/Kw-Hr it represents a cost of $38/month. (ii) In practice, air conditioning and heating duct passing through

unoccupied areas are insulated. (iii) The magnitude of is the same whether it is expressed in

units of degree Celsius or kelvin. The reason is that is measured in terms of degree change. One degree change on the Celsius scale is equal to one degree change on the kelvin scale. This is also true of units of heat transfer coefficient and specific heat.

PROBLEM 8.45

Estimate the surface temperature of a light bulb if its capacity is 150 W and the ambient air is at

23oC. Model the bulb as a sphere of diameter 9 cm. Neglect radiation.

(1) Observations. (i) This is a free convection and radiation problem. (ii) The geometry is a sphere. (iii) Under steady state conditions the power dissipated in the bulb is transferred to the surroundings by free convection and radiation and through the base by conduction. (iv) According to Newton’s law of cooling and Stefan-Boltzmann radiation law, heat loss from the surface depends on surface temperature. (v) The ambient fluid is air.

(2) Problem Definition. Determine the free convection heat transfer coefficient and radiation heat loss for the sphere.

(3) Solution Plan. Apply Newton’s law of cooling and Stefan-Boltzmann law. Use Nusselt number correlation equations to determine the average heat transfer coefficient for a sphere in free convection.

(4) Plan Execution.

(i) Assumptions. (1) Steady state, (2) constant properties (except in buoyancy), (3) uniform surface temperature, (4) no conduction through the bulb base, (5) all bulb power is transmitted to the surroundings by surface convection and radiation, (6) filament does not radiate to the surroundings directly. It radiates to the bulb surface, (7) bulb is a small surface surrounded by a much larger surface, (8) bulb is spherical, (9) surroundings is at the ambient air temperature and (10) quiescent fluid.

(ii) Analysis. Application of conservation of energy to the bulb gives

P = qc + qr (a) where

P = bulb power capacity = 150 W qc = heat transfer by convection, Wqr = heat transfer by radiation, W

Applying Newton’s law of cooling to the sphere

qc = h A (Ts - T ) = h D2 )( TTs (b)

Application of Stefan-Boltzmann law

)( 442sursr TTDq (c)

where


D = bulb diameter = 9 cm = 0.09 m

h = average heat transfer coefficient, W/m2-oCP = bulb power = 150 W qc = convection heat transfer rate, W qr = radiation heat transfer rate, W Ts = surface temperature, K Tsur = surroundings temperature = 23oC + 273.15 = 296.15 K

T = ambient air temperature = 23oC + 273.15 = 296.15 K = emissivity of glass = 0.94


sTTg

rqcq

D


Substituting (b) and (c) into (a)

P = h D2 )( TTs + )( 442

surs TTD (d)

The heat transfer coefficient is obtained from correlation equations for the Nusselt number for a sphere in free convection, given by equation (8.36a)

9/416/9

4/1

469.01

)(589.02

Pr

Ra

k

DhuN D

D (e)

Valid for: sphere uniform surface temperature or flux RaD < 1011

Pr > 0.7 properties at Tf (f)

where





RaD =g T T Ds

3

2 Pr (g)

where



= kinematic viscosity of air, m2/s


Tf = (Ts + T )/2 (h)

Equation (d) can not be solved explicitly for Ts because air properties depend Ts, the heat transfer

coefficient depends on Ts through the Rayleigh number, and the non-linear 4sT term in radiation.

Thus, equation (d) must be solved for Ts by trial and error. A value for Ts is assumed, properties are determined based on the assumed temperature, equation (g) is used to calculate RaD, (e) is

used to determine h and (d) is used to calculate P. If the calculated power is not close to the given value of 150 W, the procedure is repeated until a satisfactory agreement is obtained.

(iii) Computations.

Assume Ts = 177oC + 273.15 = 450.15 K. Using (h)

Tf = (296.15 + 450.15)(K)/2 = 373.15 K = 100oC


k = 0.03127 W/m-oCPr = 0.704

= 1/(100oC + 273.15) = 0.00268 1/K (ideal gas)


= 23.02 10-6 m2/s

Substituting into (g) and (e)

RaD = 704.0)s/m()1002.23(

)m()09.0)(K)(15.29615.450)(s/m(81.9)K/1(00268.02426

332

= 3.921 106

and

Nuh D

kD 9/416/9

4/16

704.0/469.01

)10921.3(589.02 = 22.21


h = )m(09.0

)Cm/W(03127.021.22

o

= 7.72 W/m2-oC


P = 7.72(W/m2-oC) (0.09)2(m2)(177 )23 (oC) +

0.94(5.67 10-8)(W/m2-K4) (0.09)2(m2)[(450.15)4 4)15.296( ](K4)

P = 30.25 (W) + 45.26 (W) = 75.51 W

This is not close to the given power of 150 W. The procedure is repeated for other assumed values of Ts and the results are tabulated below. At Ts = 535.15 K the calculated power is 151.46 W. This is close enough to the given power.

Assume Ts

KTf

KRaD h

W/m2-oC

qc

Wqr

Wqc +qr

W

450.15 373.13 3.921 106 7.72 30.25 45.26 75.51

530.15 413.15 3.776 106 8.3 49.42 96.7 146.12

535.15 415.65 3.757 106 8.33 50.66 100.8 151.46

(iv) Checking. Dimensionless check: Computations showed that equations (b)-(e) and (g) are dimensionally consistent.

Limiting check: If no power is dissipated (bulb is off), surface temperature should be the same as ambient temperature. Examination of (d) shows that the sum of the two terms will vanish only if

each term vanishes. Setting each term equal to zero gives Ts = Tsur = T .

Quantitative check: The value of the heat transfer coefficient is within the range of values listed in Table 1.1 for free convection of gases.

Validity of correlation equation (e): The conditions listed in (f) are met.

(5) Comments. (i) Solution by trial and error was necessary because equation (d) can not be solved explicitly for Ts. (ii) Radiation accounts for 66% of the total heat loss from the bulb. (iii) The estimated surface temperature of 535 K or 262oC appears to be high. Direct radiation between the filament and surroundings acts to lower surface temperature.

PROBLEM 8.46

A sphere of radius 2.0 cm is suspended in a very large water bath at

25oC. The sphere is heated internally using an electric coil. Determine

the rate of electric power that must be supplied to the sphere so that its

average surface temperature is 85oC. Neglect radiation.

(1) Observations. (I) At steady state, power supply to the sphere must be equal to the heat loss from the surface. (ii) Heat loss from the surface is by free convection. (iii) The surface is maintained at uniform temperature.

(2) Problem Definition. Determine the free convection heat transfer rate from the surface of a sphere.

(3) Solution Plan. Apply Newton’s law of cooling to the sphere. Use an appropriate correlation equation to determine the heat transfer coefficient.

(4) Plan Execution.

(i) Assumptions. (1) Steady state, (2) constant properties (except in buoyancy), (3) uniform surface temperature, (4) negligible radiation and (5) quiescent fluid.

(ii) Analysis. Conservation of energy for the sphere

qP (a)

where

P = power supply to the sphere, W

q = surface heat transfer rate from the sphere, W

Application of Newton's law of cooling

)( TTAhq s (b)

where

A = surface area of sphere



T = ambient fluid temperature = 25 Co

Surface area of a sphere is 2DA (c)

where

D = Diameter of sphere = 4 cm = 0.04 m

Substituting (c) into (b) and (a)

)( TTDhP s (d)

The average heat transfer coefficient is determined from correlation equations. For free convection over a sphere the appropriate equation is given in equation (8.36)

water


9/416/9

4/1

)/469.0(1

589.02

Pr

Ra

k

DhNu D

D (e)

valid for uniform surface temperature

1110DRe

7.0Pr

properties at fT


PrDTTg

Ra sD 2

3)( (f)

where

g = gravitational acceleration = 9.81 2m/s


Pr = Prandtl number

DRa = Rayleigh number




2/)( TTT sf (g)

(iii) Computations. Substituting into (g)

552/)2585(fT Co


k = 0.6458 CW/m o

Pr = 3.27 310496.0 1/K 6105116.0 /sm2


8

2426

33o23

1 1033436.227.3)/sm()105116.0(

)m()04.0)(C)(2585()m/s(81.9)K/1(10496.0Ra


02.66)27.3/469.0(1

1033436.2589.02

9/416/9

4/18

k

DhNuD

Solving for h

CW/m9.1065m)(04.0

)CW/m(6458.002.6602.66 o2

o

D

kh



)C(2585)(m()04.0)(CW/m()9.1065( o22o2P = 321.5 W

(iii) Checking. Dimensional check: Computations showed that units of equations (d)-(f) are dimensionally consistent.

Limiting check: If TTs , no free convection takes place and consequently 0P . Setting

TTs in (d) gives the anticipate result.

(5) Comments. The average heat transfer coefficient is slightly outside the range given in Table 1.1 for free convection in liquids. It should be remembered that values listed in Table 1.1 are for typical applications. Exceptions should be expected.

PROBLEM 8.47

A fish tank at a zoo is designed to maintain water

temperature at 4 Co . Fish are viewed from outdoors

through a glass window L = 1.8 m high and w = 3 m wide.

The average ambient temperature during summer months

is 26 Co . To reduce water cooling load it is proposed to

create an air enclosure over the entire window using a

pexiglass plate. Estimate the reduction in the rate of heat

transfer to the water if the air gap thickness is 6 cm.

Neglect radiation. Assume that the cold side of the

enclosure is at the same temperature as the water and the

warm side is at ambient temperature.

(1) Observations. (i) Heat is transferred from the ambient air to the water in the fish tank. (ii) Adding an air enclosure reduces the rate of heat transfer. (iii) To estimate the reduction in cooling load, heat transfer from the ambient air to the water with and without the enclosure must be determined. (iv) Neglecting the thermal resistance of glass, the resistance to heat transfer form the air to the water is primarily due to the air side free convection heat transfer coefficient. (v) Installing an air cavity introduces an added thermal resistance. (vi) The problem can be modeled as a vertical plate and as a vertical rectangular enclosure. (vii) The outside surface temperature of the enclosure is unknown. (viii) Newton’s law of cooling gives the heat transfer rate. (ix) The Rayleigh number should be determined for both vertical plate and rectangular enclosure so that appropriate correlation equations for the Nusselt number are selected. However, since the outside surface temperature of the enclosure is unknown, the Rayleigh number can not be determined. The problem must be solved using an iterative procedure.

(2) Problem Definition. Determine the average free convection heat transfer coefficient for (1) vertical plate at uniform surface temperature, and (2) for a rectangular cavity at uniform surface temperatures.

(3) Solution Plan. (i) Apply Newton's law of cooling to the window with and without the enclosure. (ii) Assume an outside enclosure surface temperature and compute the Rayleigh number. (iii) Select appropriate Nusselt number correlations equations for the two cases. (iv) Check the assumed temperature using conservation of energy from the tank to the ambient air.

(4) Plan Execution.

(i) Assumptions. (1) Outside surface temperature of the window is the same as water temperature, (2) insulated top and bottom surfaces of the enclosure, (3) negligible radiation, (4) uniform surface temperatures and (5) quiescent ambient air.

(ii) Analysis and Computations. Consider first the window without the added cavity. Newton's law of cooling gives

)(11 TTAhq w (a) where

A = surface area of window and enclosure = 1.8(m) 3.0(m) = 5.4 m2

1h = average heat transfer coefficient, W/m2-oC

1q heat transfer rate to water, W


wT = cold surface temperature = outside surface temperature of window = 4oC

g

T

L


The problem reduces to determining the heat transfer coefficients 1h for the

window without the enclosure. The window is modeled as a vertical plate at uniform temperature. The Rayleigh number is calculated to determine the

appropriate correlation equation for the average heat transfer coefficient .1h


PrLTTg

Ra wL 2

3

(b)

where


L = window height = 1.8 m

Pr = Prandtl number



Air properties are evaluated at the film temperature fT defined as

C152

)C)(264(

2

ooTT

T wf


k = thermal conductivity = 0.02526 W/m-oCPr = 0.7145

= 14.64 10 6 m2/s


fT

1 (c)

where Tf in this equation is in degrees kelvin. Thus

= 1/(15 + 273.15)K = 0.0034704 1/K


10

2426

332

104562.17145.0)/sm()1064.14(

)m(8.1C))(426()m/s(81.91/K0034704.0LRa

Thus the flow is turbulent and the appropriate correlation equation is (8.22a)

2

27/816/9

6/1

/492.01

387.0825.0

Pr

Ra

k

LhuN L

L (d)

Valid for: vertical plate uniform surface temperature

laminar, transition, and turbulent 110 < RaL < 1210

0 < Pr <

properties at fT (e)

g

L

wT

T


Applying (d) to the window gives 1h

4.284

7145.0/492.01

104562.1387.0825.0

2

27/816/9

6/1101

k

LhuN L

C-W/m99.3.8(m)1

)C-W/m(02526.04.284 21h


1q = 3.99(W/m2-oC) 1.8(m)(3)(m) (26–4)(oC) = 474.1 W

Consider next the window with the added enclosure. To determine the appropriate correlation equation for a vertical rectangular cavity the aspect ratio and Rayleigh number are calculated. The aspect ration is defined as

aspect ratio = L

(f)

where

L length of rectangle = 1.8 m

width of rectangle = 0.06 m

Equation (b) gives

30(m)06.0

.8(m)1L

The Rayleigh number for the enclosure is defined as

PrTTg

Ra ws

2

3)( (g)

where sT is outside surface temperature of enclosure. This temperature is needed to determine

both the enclosure heat transfer coefficient h and the outside surface coefficient .2h An iterative

procedure is required to determine sT , h and .2h A value for sT is assumed and h and 2h are

determined from applicable correlation equations. To check the assumed sT , conservation of

energy is applied to heat transfer from the water to the air. This gives

)()(2 wss TThTTh (h)

Assume C14osT . First determine the outside heat transfer coefficient 2h using (b) and (d). Air

properties are evaluated at the film temperature at T which is the average temperature of the two vertical surfaces of the enclosure given by

C202

)C)(2614(

2

ooTT

T s


g

LT

wT sT


k = thermal conductivity = 0.02564 W/m-oCPr = 0.713

= 15.09 10 6 m2/s

= 1/(20 + 273.15)K = 0.003411 1/K


9

2426

332

103326.7713.0)/sm()1009.15(

)m(8.1C))(1426()m/s(81.91/K003411.0LRa

Thus the flow is turbulent and the appropriate correlation equation is (d)

89.228713.0/492.01

103326.7387.0825.0

2

27/816/9

6/192

k

LhuN L

C-W/m26.3.8(m)1

)C-W/m(02564.089.228 22h

To determine cavity heat transfer coefficient 2h Rayleigh number Ra is computed. Air

properties are determined at T , given by

C92

)C)(414(

2

oo

ws TTT


k = 0.02479 W/m-oCPr = 0.716

= 14.102 10 6 m2/s

= 1/(9 + 273.15)K = 0.0035442 1/K

5

2426

332

107039.2716.0)/sm()10102.14(

)m(06.0C))(414()m/s(81.91/K0035442.0Ra

Thus correlation equation (8.39a) is applicable

3.025.0012.0

42.0L

RaPrk

hNu (8.39a)

Valid for

(8.39b)


vertical rectangular enclosure

4010L

41021 Pr74 1010 Ra

properties at 2/)( hc TTT


438.330107039.20.71642.03.05012.0 25.0

k

hNu

C-W/m42.1.06(m)0

)C-W/m(02479.0438.3 2h

Use (h) to check the assume temperature

)C)(414)(CW/m(42.1?)1426)(CW/m(26.3 oo2o2

)W/m(2.14)W/m(1.39 22

Since (h) is not satisfied the procedure is repeated until a satisfactory agreement is obtained.

Assume C18osT . At this temperature the following result is obtained:

9107391.4LRa

9953.1LNu

86.22h )CW/m( o2

5107173.3Ra

723.3Nu

548.1h


)C)(418)(CW/m(548.1?)C)(1826)(CW/m(86.2 oo2oo2

)W/m(67.21)W/m(88.22 22

The heat transfer rate is given by applying Newton’s law of cooling between the ambient air and the outside surface of the enclosure

2q = 6.123)C)(1826(m)3(m)(8.1)CW/m(86.2 oo2 W

Thus the reduction in the cooling load due to the addition of the cavity is

Load reduction = W5.350W123.6-W1.474

(iv) Checking. Dimensional check: Computations showed that equations (a), (b), (d) and (8.39a) are dimensionally consistent.

Quantitative check: The magnitude of ,h 1h and 2h are in line with typical free convection

values for air given in Table 1.1.

Validity of correlation equations (d) and (8.39a): Conditions listed in equations (e) (8.39b) are satisfied.

(5) Comments. The addition of a rectangular cavity reduces the heat transfer to the water by 74%. This is a significant saving in energy.

PROBLEM 8.48

It is proposed to replace a single pane observation window with double pane. On a typical

winter day the inside and outside air temperatures are C20oiT and C10o

oT . The inside

and outside heat transfer coefficients are ih CW/m4.9 o2 and CW/m37 o2oh . The height

of the window is 28.0L m and its width is w = 3 m. The thickness of glass is t = 0.3 cm and its

conductivity is .CW/m7.0 ogk Estimate the savings in energy if the single pane window is

replaced. Note that for the single pane window there are three resistances in series and the heat

transfer rate 1q is given by

ogi

oi

hk

t

h

TTAq

11

)(1

For the double pane window, two additional resistances are added. The width of the air space in

the double pane is 3 cm. In determining the heat transfer coefficient in the cavity, assume

that enclosure surface temperatures are the same as the inside and outside air temperatures.

(1) Observations. (i) Heat is transferred from the inside to the outside. (ii) Adding an air enclosure reduces the rate of heat transfer. (iii) To estimate the savings in energy, heat transfer through the single and double pane windows must be determined. (iv) The double pane window introduces an added glass conduction resistance and a cavity convection resistance. (v) the problem can be modeled as a vertical rectangular enclosure. (vi) Newton’s law of cooling gives the heat transfer rate. (vii) The aspect ratio and Rayleigh number should be determined for the rectangular enclosure so that an appropriate correlation equation for the Nusselt number can be selected.

(2) Problem Definition. Determine the average free convection heat transfer coefficient for a rectangular cavity.

(3) Solution Plan. (i) Apply Newton's law of cooling to the single pane and double pane windows taking into consideration the multiple resistances in series. (ii) Select an appropriate Nusselt number correlation equation for the rectangular cavity.

(4) Plan Execution.

(i) Assumptions. (1) Insulated top and bottom surfaces of the enclosure, (2) negligible radiation and (3) uniform surface temperatures.

(ii) Analysis. Consider first the single pane window

ogi

oi

hk

t

h

TTAq

11

)(1 (a)

where

A = surface area of window and enclosure = 0.28(m) 3.0(m) = 0.84 m2

ih = inside heat transfer coefficient = 9.4 W/m2-oC

oh = outside heat transfer coefficient = 37 W/m2-oC

gk glass conductivity = 0.7 W/m-oC

1q heat transfer rate form single pane window, W


t glass thickness = 0.3 cm = 0.003 m

iT inside air temperature = 20oC

oT outside air temperature = -10oC

For the double pane window, two additional resistances are added in series: a glass conduction resistance and a rectangular cavity convection resistance. Equation (a) is modified to

hhk

t

h

TTAq

ogi

oi

1121

)(2 (b)

where

h rectangular cavity convection heat transfer coefficient, W/m2-oC

2q heat transfer rate from the double pane window, W

To determine cavity heat transfer coefficient h the aspect ratio and Rayleigh number Ra are

computed. The aspect ratio is given by

Aspect ratio = 33.9)m(03.0

)m(28.0L


PrTTg

Ra ch

2

3)( (c)

where


Pr = Prandtl number

cT enclosure cold surface temperature, Co

hT enclosure hot surface temperature, Co


enclosure air thickness = 3 cm = 0.03 m


Air properties are determined at T which is the average temperature of the two vertical surfaces

of the enclosure, cT and .hT Both temperatures are unknown . They can be determined by an

iterative procedure. A simpler approach is to assume that the two surfaces are at the inside and

outside temperatures. That is, ih TT and oc TT . This assumption is conservative in that it

will overestimate the heat loss from the double pane window. Using this approximation gives

C52

)C)(1020(

2

oo

oi TTT


k = 0.02448 W/m-oCPr = 0.717

= 13.75 10 6 m2/s


= 1/(5 + 273.15)K = 0.003595 1/K

5

2426

332

100834.1717.0)/sm()1075.13(

)m(03.0C))(1020()m/s(81.91/K003595.0Ra


25.028.0

2.022.0

LRa

Pr

Pr

k

hNu (8.38a)

Valid for

(8.38b)

(iii) Computations. Equation (a) gives

W02703.000429.010638.0

2.25

C)W/m(37

1

C)W/m(7.0

)m(003.0

C)W/m(4.9

1

)C)(1020)(m(84.0

o2o2o2

o

1q

1831q W


018.3333.90834.1717.2.0

22.025.0

28.0

0

0.717

k

hNu

CW/m463.2)m(03.0

)CW/m(02448.0018.3 o2

o

h

Equation (b) gives

W46.3W406.002703.000429.010638.0

2.25

C)W/m(463.2

1

C)W/m(37

1

C)W/m(7.0

)m(003.02

C)W/m(4.9

1

)C)(1020)(m(84.0

o2o2o2o2

o

2q

Thus the percent savings in energy are

%Savings = %75W186

W)3.46183(


102L

510Pr103 1010 Ra



(iv) Checking. Dimensional check: Computations showed that equations (a), (b), (c) and (8.38a) are dimensionally consistent.

Quantitative check: The magnitude of h is in line with typical free convection values for air

given in Table 1.1.

Validity of correlation equations (8.38a): Conditions listed in equation (8.38b) are satisfied.

(5) Comments. (i) Using double pane window reduces the heat loss by 75%. This is a

significant saving in energy. (ii) The assumption that ih TT and oc TT overestimates h . Thus

in fact heat loss form the double pane window is less than 46.3 W and the savings are more than 75%..

PROBLEM 8.49

To reduce heat loss form an oven, a glass door with a rectangular air

cavity is used. The cavity has a baffle at its center. The height of the

door is 65L cm and its width cm70w . The air space thickness is

cm5.1 . Estimate the heat transfer rate through the door if the

inside and outside surface temperatures of the cavity are C198o and

C42o .

(1) Observations. (i) Heat is transferred through the door from the inside to the outside. (ii) Newton’s law of cooling gives the heat transfer rate. (iii) The aspect ratio and Rayleigh number should be determined for the rectangular enclosure so that an appropriate correlation equation for the Nusselt number can be selected. (iv) The baffle divides the vertical cavity into two equal parts. This has the effect of decreasing the aspect ratio by a factor of two.


(3) Solution Plan. (i) Apply Newton's law of cooling (ii) Select an appropriate Nusselt number correlation equation for the rectangular cavity.

(4) Plan Execution.

(i) Assumptions. (1) Insulated top and bottom surfaces of the enclosure, (2) negligible radiation and (3) uniform surface temperatures.


)( ch TTAhq (a)

where

A = surface area of door = wL 0.065(m) 0.7(m) = 0.455 m2

h = enclosure heat transfer coefficient, W/m2-oC


cT cavity cold side surface temperature = 42oC

hT cavity hot side temperature = 198oC




)m)(2/65.0(L


PrTTg

Ra ch

2

3)( (b)

where


Pr = Prandtl number


enclosure air thickness = 1.5 cm = 0.015 m


L

L/2


Air properties are determined at T

C1202

)C)(19842(

2

oo

hc TTT


k = 0.03261 W/m-oCPr = 0.703

= 25.19 10 6 m2/s

= 1/(120 + 273.15)K = 0.002544 1/K

4

2426

332

104557.1703.0)/sm()1019.25(

)m(015.0C))(42198()m/s(81.91/K002544.0Ra


3.025.0012.0

42.0L

RaPrk

hNu (8.39a)

Valid for

(8.39b)

(iii) Computations. Equation (8.39a)) gives

826.1667.21104557.10.70342.03.04012.0 25.0

k

hNu

CW/m969.3)m(015.0

)CW/m(03261.0826.1 o2

o

h


W7.281)C)(42198)(m(455.0)CW/m(969.3 o2o2q

(iv) Checking. Dimensional check: Computations showed that equations (a), (b) and (8.39a) are dimensionally consistent.


given in Table 1.1.


(5) Comments. (i) Even with a double pane door the heat loss is significant. This added energy to the surroundings raises the temperature in the cooking area. (ii) Without the baffle the aspect


4010L

41021 Pr74 1010 Ra



ratio is 43.334. Although this is slightly over the limit of (8.39a), this equation can still be used without introducing significant error. With no baffle (8.39a)

CW/m224.3 o2h

8.228q W

Therefore the cavity acts to increase the heat loss and thus should be eliminated.

.

PROBLEM 8.50

The ceiling of an exhibit room is designed to provide

natural light by using an array of horizontal skylights.

Each unit is rectangular with an air gap cm5.6 thick.

The length and width of each unit are cm54L and

cm120w . On a typical day the inside and outside glass

surface temperatures are C15o and C15o . Estimate the rate of heat loss from each unit.

(1) Observations. (i) Heat is transferred through the skylight from the inside to the outside. (ii) Newton’s law of cooling gives the heat transfer rate. (iii) The aspect ratio and Rayleigh number should be determined for the rectangular enclosure so that an appropriate correlation equation for the Nusselt number can be selected.


(3) Solution Plan. (i) Apply Newton's law of cooling (ii) Select an appropriate Nusselt number correlation equation for the rectangular cavity.

(4) Plan Execution.

(i) Assumptions. (1) Insulated end surfaces of the enclosure, (2) negligible radiation, (3) uniform surface temperatures and (4) negligible glass resistance no temperature drop across the glass).


)( ch TTAhq (a)

where

A = surface area of skylight = wL 0.54(m) 1.2(m) = 0.648 m2

h = enclosure heat transfer coefficient, W/m2-oC


cT cavity cold side surface temperature = -15oC

hT cavity hot side temperature = 15oC




)m)(54.0(L


PrTTg

Ra ch

2

3)( (b)

where


Pr = Prandtl number


enclosure air thickness = 6.5 cm = 0.065 m


L

w g



C02

)C)(1515(

2

oo

hc TTT


k = 0.02408 W/m-oCPr = 0.718

= 25.19 10 6 m2/s

= 1/(0 + 273.15)K = 0.003661 1/K

6

2426

332

101922.1718.0)/sm()1031.13(

)m(065.0C))(1515()m/s(81.91/K003661.0Ra


074.03/1069.0 PrRa

k

hNu (8.41a)

Valid for

(8.41b)

(iii) Computations. Equation (8.39a)) gives

1389.7718.0101922.1069.0074.03/16

k

hNu

CW/m645.2)m(065.0

)CW/m(02408.01389.7 o2

o

h


W4.51)C)(1515)(m(648.0.0)CW/m(645.2 o2o2q

(iv) Checking. Dimensional check: Computations showed that equations (a), (b) and (8.41a) are dimensionally consistent.


given in Table 1.1.


(5) Comments. Using double pane skylight reduces heat loss to the surroundings.

horizontal rectangular enclosure heated from below

95 107103 Ra


PROBLEM 8.51

Repeat Example 8.4 using inclination angles of ,0o ,60o ,90o ,120o o150 and .175o Plot heat

transfer rate q vs. inclination angle .

(1) Observations. (i) Power requirement is equal to the heat transfer rate through the enclosure. (ii) The problem can be modeled as a rectangular cavity at specified hot and cold surface

temperatures. (iii) The inclination angle varies from o0 to .175o (iv)

Newton’s law of cooling gives the heat transfer rate. (v) The aspect ratio and critical inclination angle should be computed to determine the applicable correlation equation for the Nusselt number.

(2) Problem Definition. Determine the average free convection

heat transfer coefficient h for a rectangular enclosure at various inclination angles.

(3) Solution Plan. (i) Apply Newton's law of cooling. (ii) Compute the aspect ratio and critical inclination angle. (iii) Select an appropriate Nusselt number correlation equation for convection for a horizontal, vertical and inclined rectangular cavity.

(4) Plan Execution.

(i) Assumptions. (1) Uniform hot and cold surface temperatures and (2) insulated end surfaces, (3) negligible radiation.


)( ch TTAhqP (a)

where

A = surface area of rectangle = 0.7(m) 0.7(m) = 0.49 m2


P power requirement, W q heat transfer rate through cavity, W

hT = hot surface temperature = 27oC

cT = cold surface temperature = 23oC

The aspect ratio is defined as

aspect ratio = L

(b)

where


width of rectangle = 0.05 m

Equation (b) gives

14(m)05.0

(m)7.0L

According to Table 8.1, the critical angle is o70c . For 12/L and o700 , the

applicable correlation equation for the Nusselt number is (8.42a)

g

hT

cT

abso

rber

plat

e

L


*3/16.1*

118

)cos(

cos

)sin8.1(17081

cos

1708144.11

Ra

RaRak

hNu (8.42a)

Valid for

(8.42b)


PrTTg

Ra ch

2

3

(c)

where


Pr = Prandtl number



Water properties are evaluated at the film temperatureT defined as

2

)( ch TTT (d)

To determine the applicable correlation equations for the horizontal, vertical and inclination

angles 90175 , the Rayleigh number is computed. Equation (d) gives

C252

)C)(2327( oo

T


k = thermal conductivity = 0.6076 CW/m o

Pr = 6.13 310259.0 1/K

6108933.0 m2/s


6

2426

3323

1075898.913.6)/sm()108933.0(

)m(05.0C))(2327()m/s(81.91/K10259.0Ra

Thus the applicable correlation equation for the horizontal position is (8.41a)

074.03/1069.0 PrRa

k

hNu (8.41a)

Valid for

inclined rectangular enclosure

12/L

c0

set 0* when negative



(8.41b)

For the vertical orientation the applicable correlation equation is (8.39a)

3/1046.0 Ra

k

hNu (8.39a)

Valid for

(8.39b)

For 90175 the applicable correlation equation is

sin1)90(1 oNuk

hNu (8.45a)

Valid for

(8.45b)

(iii) Computations.

(1) Horizontal position: .0 Substituting into (8.41a)

074.03/16o

o 6.131075898.9069.0)0(

)0(k

hNu = 16.862

CW/m9.204m)(05.0

)CW/m(6076.0862.16862.16)0( o2

oo k

h

The corresponding power is given by (a)

C))(2327)(C)(0.49)(mW/m(9.204)0( o2o2oP 401.6 W

horizontal rectangular enclosure heated from below

95 107103 Ra



401L

201 Pr96 1010 Ra



all /Loo 18090



(2) Inclination angle .60 Use (8.42a)

86.10118

)60cos1075898.9(

60cos1075898.9

)60sin8.1(17081

60cos1075898.9

1708144.11

)60()60(

*3/16

o6

6.1o*

o6

oo

k

hNu

CW/m132m)(05.0

)CW/m(6076.086.1086.10)60( o2

oo k

h

Equation (a) gives the required power

C))(2327)(C)(0.49)(mW/m(132)60( o2o2oP 258.7 W

(3) Vertical orientation, .90 Use (8.39a)

825.91075898.9046.0)90(

)90(3/16

oo

k

hNu

CW/m4.119m)(05.0

)CW/m(6076.0825.9825.9)90( o2

oo k

h

C))(2327)(C)(0.49)(mW/m(4.119)90( o2o2oP 234 W


oo

o 120sin1825.91)120(

)120(k

hNu 8.643

CW/m03.105m)(05.0

)CW/m(6076.0643.8643.8)120( o2

oo k

h

C))(2327)(C)(0.49)(mW/m(03.105)120( o2o2oP 206 W


oo

150sin1825.91)150(

)150(k

hNu 5.413

CW/m8.65m)(05.0

)CW/m(6076.0413.5413.5)150( o2

oo k

h

C))(2327)(C)(0.49)(mW/m(8.65)150( o2o2oP 129 W


oo

o 175sin1825.91)175(

)175(k

hNu 1.77

CW/m5.21m)(05.0

)CW/m(6076.077.1647.8)175( o2

oo k

h


C))(2327)(C)(0.49)(mW/m(5.21)175( o2o2oP 42.1 W

Using the result of Example 8.4 and the above data, the required power at various angles is tabulated and plotted.

(iv) Checking. Dimensionalcheck: Computations showed that equations (a), (c), (8.39a), (8.41a), (8.42a) and (8.45a) are dimensionally correct.

Quantitative check: The magnitude of h is in line with typical free convection values for liquids given in Table 1.1.

Validity of correlation equations (8.39a), (8.41a), (8.42a) and (8.45a): Conditions listed in equations (8.39b), (8.41b), (8.42b) and (8.45b)are satisfied.

(5) Comments. (i) If the device is to be used continuously, the estimate power requirement is relatively high. Decreasing the temperature difference between the hot and cold surfaces will reduce the power requirement. (ii) The ambient temperature plays a role in the operation of the proposed device. The design must take into consideration changing ambient temperature. (iii) Specification of the driving motor should be based on highest power which corresponds to the horizontal orientation.

)(o P(W)

0 401.6

30 303.8

60 258.7

90 234

120 206

150 129

180 42.1

o

P(W)

030 60 90 120 150 1800

100

200

300

400

PROBLEM 8.52

A rectangular solar collector has an absorber plate of length

m5.2L and width m.0.4w A protection cover is used to

form a rectangular air enclosure of thickness cm4 to

provide insulation. Estimate the heat loss by convection from

the plate when the enclosure inclination angle is o45 and its

surfaces are at C28o and C.72o

(1) Observations. (i) The absorber plate is at a higher temperature than the ambient air. Thus heat is lost through the rectangular cavity to the atmosphere. (ii) The problem can be modeled as an inclined rectangular cavity at specified hot and cold surface temperatures. (iii) Newton’s law of cooling gives the heat transfer rate. (iv) The aspect ratio and critical inclination angle should be computed to determine the applicable correlation equation for the Nusselt number.

(2) Problem Definition. Determine the average free convection heat transfer coefficient h for an inclined rectangular enclosure.

(3) Solution Plan. (i) Apply Newton's law of cooling. (ii) Compute the aspect ratio and critical inclination angle. Select an appropriate Nusselt number correlation equation for convection in an inclined rectangular cavity.

(4) Plan Execution.

(i) Assumptions. (1) Uniform hot and cold surface temperatures, (2) insulated end surfaces and (3) negligible radiation.


)( ch TTAhq (a)

where

A = surface area of rectangle = 2.5(m) 4(m) = 10 m2


q heat transfer rate through cavity, W

hT = hot surface temperature = 72oC

cT = cold surface temperature = 28oC

The aspect ratio is defined as

aspect ratio = L

(b)

where


width of rectangle = 4 cm = 0.04 m

Equation (b) gives

5.62(m)04.0

.5(m)2L

g

hT

cT

abso

rber

plat

e

L


According to Table 8.1, the critical angle is o70c . Since 12/L and c0 , it

follows that the applicable correlation equation for the Nusselt number is

*3/16.1*

118

)cos(

cos

)sin8.1(17081

cos

1708144.11

Ra

RaRak

hNu (8.42a)

Valid for

(8.42b)


PrTTg

Ra ch

2

3

(c)

where


Pr = Prandtl number



Water properties are evaluated at the film temperatureT defined as

2

)( ch TTT (d)

(iii) Computations. Equation (d) gives

C502

)C)(2872( oo

T


k = thermal conductivity = 0.0.02781 CW/m o

Pr = 0709 0.0030946(K))15.27350/(1 1/K

61092.17 m2/s


5

2426

332

1088745.1709.0)/sm()1092.17(

)m(04.0C))(2872()m/s(81.91/K0030946.0Ra



12/L

c0

set 0* when negative



2338.4118

)45cos1088742.1(

45cos1088742.1

)45sin8.1(17081

45cos1088742.1

1708144.11

*3/15

o5

6.1o*

o5k

hNu

CW/m9435.2m)(04.0

)CW/m(02781.02338.42338.4 o2

okh

Equation (a) gives the heat transfer rate

C))(2872)(C)(10)(mW/m(9435.2 o2o2q 1295 W

(iv) Checking. Dimensional check: Computations showed that equations (a), (8.42a) and (c) are dimensionally consistent.

Quantitative check: The magnitude of h is in line with typical free convection values for air given in Table 1.1.

Validity of correlation equation (8.42a): Conditions listed in equation (8.42b) are satisfied.

(5) Comments. The rate of energy loss from the collector is significant. Increasing the thickness of the air gap will reduce the heat lost to the atmosphere.

PROBLEM 8.53

A liquid-vapor mixture at C20oiT flows inside a tube of diameter

cm4iD and length m.3L The tube is placed concentrically

inside another tube of diameter cm.6oD Surface temperature of the

outer tube is at C.10ooT Air fills the annular space. Determine the

heat transfer rate from the mixture.

(1) Observations. (i) Heat is transferred through the annular space from the outer cylinder to the inner. (ii) Newton’s law of cooling gives the heat transfer rate. (iii) The Rayleigh number should be determined for the enclosure formed by the concentric cylinders so that an appropriate correlation equation can be selected. (iv) The cylinders are horizontally oriented.

(2) Problem Definition. Determine the average free convection heat transfer coefficient annular cavity between two concentric horizontal cylinders.

(3) Solution Plan. (i) Use equation (8.46) describing heat transfer between two concentric cylinders. (ii) Select an appropriate correlation equation for this geometry.

(4) Plan Execution.

(i) Assumptions. (1) negligible radiation and (2) uniform surface temperatures.

(ii) Analysis. Heat transfer between two concentric cylinders is given by equation (8.46)

)()/ln(

2iTT

DD

kq o

io

eff (8.46)

where

iD 4 cm = 0.04 m

oD 6 cm =0.06 m

effk effective conductivity, CW/m o

q heat transfer rate per unit length of tube, W/m

iT = inner surface temperature (cold) = -20oC

oT = outer surface temperature (hot) = 10oC

Using (8.46), the total heat transfer rate from a tube of length L is

LTTDD

kq io

io

eff)(

)/ln(

2 (a)

where

L tube length = 3 m q total heat transfer rate, W

Correlation equation for the effective conductivity effk is

4/1*

861.0386.0 Ra

rP

rP

k

keff (8.47a)

where

iD

oD

iT

oT


Ra

DD

DDRa

oi

io5

5/35/33

4*

)()(

)/ln( (8.47b)

2

io DD (8.47c)

Valid for

(8.47d)

The Rayleigh number Ra in (8.47b) is defined as

PrTTg

Ra ch

2

3)( (b)

where


Pr = Prandtl number



(iii) Computations. Equation (8.47) gives

01.02

)(m)04.006.0(m


C52

)C)(1020(

2

oo

oi TTT


k = 0.023698 W/m-oCPr = 0.7195

= 12.885 10 6 m2/s

= 1/(-5 + 273.15)K = 0.0037293 1/K


4.47567195.0)/sm()10885.12(

)m(01.0C))(1515()m/s(81.91/K0037293.02426

332

Ra

Use (8.47b)

2.4554.4756

)m60.0()m04.0()01.0(

)04.0/06.0ln(5

5/35/33

4*Ra

concentric cylinders 7*2 1010 Ra

properties at 2/)( oi TTT


Thus condition (8.47d) is satisfied. Substituting into (8.47a)

465.12.4557195.0861.0

7195.0386.0

4/1

k

keff

C)(W/m)02369.0(465.1465.1 okk ffe

Equation (a) gives q

m4.48)m(3C))(2010()m04.0/m06.0ln(

C)W/m)(0347.0(2 oo

q

(iv) Checking. Dimensional check: Computations showed that equations (a), (b), (8.47a) and (8.47b) are dimensionally consistent.

Quantitative check: The magnitude of h is in line with typical free convection values for air given in Table 1.1.

Validity of correlation equations (8.47a): Conditions listed in equation (8.47d) are satisfied.

(5) Comments. The concentric annular space provides good insulation. This is indicated by the

low value of kkeff / = 1.465. A ratio of unity corresponds to pure conduction with no fluid

circulation.

PROBLEM 9.1

The speed of sound, c, in an ideal gas is given by

TRc

where is the specific heat ratio R is gas constant and T is temperature. Show that

Re

MKn

2

where M is Mach number defined as

c

VM

(1) Observations. (i) Definitions of Knudsen number, Reynolds number, and Mach number are needed. (ii) Fluid velocity appears in the definition of Reynolds number and Mach number.

(2) Problem Definition. Show that the Knudsen number can be expressed in terms of Reynolds and Mach numbers.

(3) Solution Plan. Star with the definition of Knudsen number and multiply and divide by variables to form the Reynolds and Mach numbers

(4) Plan Execution.

(i) Analysis. The Knudsen number is defined as

eDKn (1.2)

The mean free path for an ideal gas is given by

RTp 2

(9.2)

Since the Reynolds number is expressed in terms of density , use the ideal gas law to

eliminate p in (9. 2) RTp (9.31)

Substitute (9.31) into (9.2)

22 RTRT

RT(a)

(a) into (1.2)

2RTDKn

e

(b)

Multiply and divide (b) by V


RT

V

VDKn

e2 (c)

Introduce the definition of the Reynolds and Mach number

eDVRe (d)

RT

V

c

VM (e)

Substitute (d) and (e) into (c)

Re

MKn

2 (f)

(ii) Checking: Dimensional check: Both sides of (f) are dimensionless.

(5) Comments. In determining the number of governing parameters in flow through microchannels, it should be noted that the three parameters, Kn, Re and M are not independent.

PROBLEM 9.2

Reported discrepancies in experimental data on the fiction factor f are partially attributed to

errors in measurements. One of the key quantities needed to calculate f is channel diameter D.

Show that5Df

(1) Observations. (i) The definition of friction factor shows that it depends on pressure drop, diameter, length and mean velocity. (ii) Mean velocity is determined from flow rate measurements and channel flow area.

(2) Problem Definition. Determine the dependency of friction factor on diameter.

(3) Solution Plan. Starting with the definition of friction factor f, express it in terms of diameter.

(4) Plan Execution.

(i) Assumptions. Continuum.

(ii) Analysis. Friction factor f is defined as

22

1

mu

p

L

Df (9.6b)

Pressure drop is determined by measuring the pressure at the inlet and outlet chambers. If pressure drop at the inlet and outlet can be neglected, then p is independent of diameter. Mean

velocity is determine from flow rate measurements:

4

2DuAum mm (a)

where

A = flow area m = mass flow rate

Solve (a) for mu

2

14

D

mum (b)

Substitute (b) into (9.6b)

5

2

24

2

2

32162

1D

mL

pD

mp

L

Df (f)

This result shows that f is proportional to the fifth power of diameter.

(iii) Checking.

Dimensional check: (f) should be dimensionless:


unityDmL

pf )m(

)kg/s(m)(

)mkg/s()kg/m(

32

55

22

232

(5) Comments. Accurate measurements of diameter or channel spacing in microchannels is critical in obtaining accurate data on friction factor.

PROBLEM 9.3

Consider shear driven Couette flow between parallel plates separated by a distance H. The

lower plate is stationary while the upper plate moves with a velocity .su Assume that no heat is

conducted through the lower plate and that the upper plate is maintained at uniform temperature

.sT Taking into consideration dissipation, velocity slip and temperature jump, determine the

Nusselt number. Assume steady state ideal gas flow.

(1) Observations. (i) The determination of the Nusselt number requires the determination of the temperature distribution. (ii) Temperature field depends on the velocity field. (iii) The velocity field for Couette flow with a moving upper plate is give in Section 9.6.2. (iv) The solution to the energy equation gives the temperature distribution.

(2) Problem Definition. Determine the temperature distribution for Couette flow with insulated stationary plate and uniform temperature moving plate.

(3) Solution Plan. Start with the definition of the Nusselt number, use the velocity solution for Couette flow of Section 9.6.2, formulate the energy equation and boundary conditions, and solve for the temperature distribution.

(4) Plan Execution.

(i) Assumptions. (1) Constant viscosity, conductivity and specific heat, (2) infinite plates, (3) uniform boundary conditions, (4) uniform spacing between plates, (5) no variation of density along y, (6) no gravity, (7)

,0.1Tu (8) velocity field is

independent of temperature, (9) ideal gas, and (10) continuum, slip flow regime conditions apply.

(ii) Analysis. The Nusselt number for flow between parallel plates is defined as

k

HhNu

2 (a)

The heat transfer coefficient h for channel flow is defined as

sm TT

y

HTk

h

)(

(b)

(b) into (a)

sm TT

y

HT

HNu

)(

2 (9.19)

where

k thermal conductivity of fluid

suy

xHu

sT


T fluid temperature function (variable)

mT fluid mean temperature

sT plate temperature

The mean temperature mT , as defined in Section 6.6.2, is

H

pmp dyTucWTmc

0

(9.21)

where

pc = specific heat

m = mass flow rate

T = temperature distribution u = velocity distribution

W = plate width

= density

The velocity distribution is given in Section 9

)(21

1Kn

H

y

Knu

u

s

(9.14)

2

suWHm (9.16)

(9.16) into (9.21) H

sm dyTu

HuT

0

2 (9.22)

Temperature distribution is governed by the energy equation. Based pm the above assumptions, energy equation (2.15) simplifies to

02

2

y

Tk (9.23)

where2

y

u (9.24)

(9.24) into (9.23)2

2

2

dy

du

kdy

Td(9.25)

Note that T is independent of x. Substitute (9.14) into (9.25)

2

2

2

)21( KnH

u

kdy

Td s (c)

Defining the constant as


2

)21( KnH

u

k

s (d)

Substituting (d) into (c)

2

2

dy

Td (e)

This energy equation requires two boundary conditions. They are:

0)0(

dy

dT (f)

The second boundary condition is at y = H. Plate temperature is specified at this boundary. However, the boundary condition must be associated with the fluid at y = H and not the plate. Knowing plate temperature, temperature jump condition (9.11) gives fluid temperature T(H). For

yHn and 1T , (9.11) gives

dy

HdT

PrHTTs

)(

1

2)( (9.20)

Solve for )(HT

dy

HdT

PrTHT s

)(

1

2)( (g)

Integration of (e) gives

DCyyT 2

2 (h)

where C and D are constants of integration. Application of boundary conditions (f) and (g) gives the two constants:

0C (i)

and

22

1

2

2H

Pr

KnHTD s (j)


sTHPr

KnHyT 2

22

1

2

22(k)

To determine the Nusselt number using (9.19), equation (k) is used to formulate dy

HdT )( and

.mT Differentiating (k)

Hdy

HdT )( (l)

mT is determined by substituting (9.14) and (k) into (9.22)


dyDyKnH

y

KnHT

H

m )2

)(()21(

2 2

0

(m)

where D is defined in (j). Evaluating the integral, gives

DKnHHKn

Tm22

6

1

8

1

21

2

Substituting (j) into the above

sm THPr

KnHKnHH

KnT 2

222

1

2

26

1

8

1

21

2

or

sm THPr

KnKnHH

KnT 222

1

2

3

2

4

1

21

1 (n)

Using (l) and (n) into (9.19) gives the Nusselt number

222

2

1

2

3

2

4

1

21

1

2

HPr

KnKnHH

Kn

HNu

This simplifies to

Pr

KnKnKn

KnNu

)21(

1

8

3

81

)21(8 (o)

(iii) Checking.

Dimensional check: (i) Noting that units of are 2o C/m , each term in (n) has units of

temperature. (ii) The Nusselt number in (o) is dimensionless.

Limiting check: Setting Kn = 0 in (o) gives

8oNu (p)

This is the correct value of Nusselt number for macrochannel flow

(5) Comments.

(i) The Nusselt number is independent of the Reynolds number. This is also the case with macrochannel flows.

(ii) Unlike macrochannels, the Nusselt number depends on the fluid.

(iii) The Knudsen number in (o) represents the effect of rarefaction while the third term in the denominator represents the effect of temperature jump. Both act to reduce the Nusselt number.


(iv) If dissipation is neglected ( )0 , equation (k) gives the corresponding temperature solution

as

sTT

Thus, the temperature is uniform and no heat transfer takes place.

PROBLEM 9.4

A large plate moves with constant velocity su

parallel to a stationary plate separated by a

distance H. An ideal gas fills the channel

formed by the plates. The stationary plate is

at temperature oT and the moving plate is at

temperature sT . Assume laminar flow and

take into consideration dissipation and

velocity slip and temperature jump:

(a) Show that temperature distribution is given by

H

y

Pr

Kn

Pr

Kn

TT

H

y

H

y

Pr

Kn

Knk

uTT o

oss

1

2

1

221

1

2

)21(2 2

2

2

2

(b) Determine the heat flux at the plates.

(1) Observations. (i) Temperature distribution depends on the velocity field. (ii) The velocity field for Couette flow with a moving upper plate is give in Section 9.6.2. (iii) The solution to the energy equation gives the temperature distribution. (iv) Two temperature boundary conditions must be specified. (v) Temperature distribution and Fourier’s law give surface heat flux.

(2) Problem Definition. Determine the temperature distribution for Couette flow with specified surface temperature on both plates.

(3) Solution Plan. Use the velocity solution for Couette flow of Section 9.6.2, formulate the energy equation and boundary conditions, and solve for the temperature distribution.

(4) Plan Execution.

(i) Assumptions. (1) Constant viscosity, conductivity and specific heat, (2) infinite plates, (3) uniform boundary conditions, (4) uniform spacing between plates, (5) no variation of density

along y, (6) no gravity, (7) ,0.1Tu (8) the velocity field is independent of temperature,

and (9), ideal gas, and (10) continuum, slip flow regime conditions apply.

(ii) Analysis.

(a) Temperature distribution is governed by the energy equation. Based pm the above assumptions, energy equation (2.15) simplifies to

02

2

y

Tk (9.23)

2

y

u (9.24)

where

k = thermal conductivity of fluid

u = axial velocity

= fluid viscosity

suy

xHu

sT

oT


(9.24) into (9.23)2

2

2

dy

du

kdy

Td(9.25)

The velocity distribution is given in Section 9

)(21

1Kn

H

y

Knu

u

s

(9.14)

2

2

2

)21( KnH

u

kdy

Td s (a)


2

)21( KnH

u

k

s (b)

Substituting (b) into (a)

2

2

dy

Td (c)

This energy equation requires two boundary conditions. The temperature of each plate is specified. However, the boundary conditions must be associated with the fluid at y = H and not the plates. Knowing plate temperature, temperature jump condition (9.11) gives fluid temperature. At ,0y fluid temperature T(0) is obtained from (9.11). At n = y = 0 and for

1T , equation (9.11) gives

dy

dT

PrTT o

)0(

1

2)0( (d)

Similarly, at n = H – y and 1T , (9.11) gives

dy

HdT

PrTHT s

)(

1

2)( (e)

Integration of (c) gives

DCyyT 2

2 (h)

where C and D are constants of integration. Application of boundary conditions (d) and (e) gives the two constants:

Pr

KnHH

TTHPr

KnH

Cos

1

22

1

2

2

22

(i)

and


HCPr

KnTD o

1

2(j)

Substitute (i) and (j) into (h)

oos T

Pr

Kn

H

y

Pr

Kn

TT

Pr

Kn

H

y

H

yHT

1

2

1

41

1

2

2 2

22

(k)

Using the definition of in (b) into (k) and rearranging, gives

H

y

Pr

Kn

Pr

Kn

TT

H

y

H

y

Pr

Kn

Knk

uTT o

oss

1

2

1

41

1

2

)21(2 2

2

2

2

(l)

This result can be written in dimensionless form as

H

y

Pr

Kn

Pr

KnH

y

H

y

Pr

Kn

TTKnk

u

TT

TT

oo

o

s

s

s 1

2

1

41

1

1

2

)()21(2 2

2

2

2

(m)

(b) Heat flux .q Application of Fourier’s law at y = 0

dy

dTkq

)0()0( (n)

(l) into (m)

HPr

KnH

TTk

KnH

uq oss

1

4

)(

)21(2)0(

2

2

(o)

(o) is written in dimensionless form as

Pr

KnTTKn

u

H

TTk

q

oo s

s

s

1

41

1

)()21(2)(

)0(2

2

(p)

Similarly, at y = H

dy

HdTkHq

)()( (q)

(l) into (q)

HPr

KnH

TTk

KnH

uHq oss

1

4

)(

)21(2)(

2

2

(r)

Written in dimensionless form, (r) becomes


Pr

KnTTKn

u

H

TTk

Hq

oo s

s

s

1

41

1

)()21(2)(

)(2

2

(s)

(iii) Checking.

Dimensional check: (i) Noting that units of are 2o C/m , each term in (k) has units of

temperature. (ii) Each term in (o) and (r) has units of heat flux. (iii) each term in (m), (p) and (s) is dimensionless.

Limiting check: (i) Since the velocity profile is linear, dissipation is uniform. Thus, if oTTs ,

heat flux at each plate should be equal in magnitude and opposite in direction. Setting oTTs in

(o) and (r) gives

2)21(2)0(

2

KnH

uq s (t)

2)21(2)(

2

KnH

uHq s (u)

(ii) If dissipation is neglected, temperature distribution should be linear. Setting 0 in (l) gives

H

y

Pr

Kn

Pr

Kn

TTTT o

os

1

2

1

41

(s)

(iii) If dissipation and rarefaction are neglected, temperature distribution should be the same as one dimensional conduction. Setting Kn = 0 in (l) gives

H

yTTTT oo s )( (v)

(5) Comments.

(i) The Knudsen number in (l), (o) and (r) represents the effect of rarefaction while the Prandtl number terms represents the effect of temperature jump.

(ii) The solution is governed by two parameters:

Dissipation parameter = )()21( 2

2

oTTKnk

u

s

s (w)

Temperature jump parameter =Pr

Kn

1 (x)

PROBLEM 9.5

Consider Couette flow between two parallel plates separated by a distance H. The lower

plate moves with velocity 1su and the upper plate moves in the opposite direction with

velocity .2su The channel is filled with ideal gas. Assume velocity slip conditions,

determine the mass flow rate. Under what condition will the net flow rate be zero?

(1) Observations. (i) To determine mass flow rate it is necessary to determine the velocity distribution. (ii) Velocity slip takes place at both boundaries of the flow channel. (iii) Because plates move in opposite directions, the fluid moves in both directions. This makes it possible for the net flow rate to be zero.

(2) Problem Definition. Determine the velocity distribution in the channel.

(3) Solution Plan. Apply the Navier-Stokes equations and formulate the velocity slip boundary conditions. Follow the analysis of Section 9.6.2 and Example 9.1.

(4) Plan Execution.

(i) Assumptions. (1) Steady state, (2) laminar flow, (3) one-dimensional (no variation with axial distance x and normal distance z), (4) slip flow regime (0.001 < Kn < 0.1), (5) ideal gas, (6) constant viscosity (7) negligible lateral variation of density, (8) the velocity

accommodation coefficient is equal to unity, ,0.1u (9) continuum, slip flow regime

conditions apply, (10) ideal gas, and (11) no gravity.

(ii) Analysis. Mass flow rate is given The flow rate, m , for a channel of width W is

given by H

dyuWm

0

(9.15)

whereu axial velocity

density

To determine u we follow the analysis of Section 9.6.2. The axial component of the Navier-Stokes equations for Couette flow between parallel plates is given by (9.12)

02

2

dy

ud(9.12)

Boundary conditions for (9.12) are given by (9.10)

n

xuuxu

u

us

)0,(2)0,( (9.10)

Applying (9.10) to the lower surface, 0yn , and setting 1u

dy

duuu s

)0()0( 1 (a)

y

x

H

1su

2su

u


For the upper surface, n = H – y, (9.10) gives

dy

HduuHu s

)()( 2 (b)

The solution to (9.12) is ByAu (c)

Boundary conditions (a) and (b) give the two constants of integration A and B

)21(

21

KnH

uuA ss ,

)21(

)( 211

Kn

KnuuuB ss

s (d)

where Kn is the Knudsen number, defined as

HKn (9.13)


)()21(

)( 211

H

yKn

Kn

uuuu ss

s (e)

Substituting (e) into (9.15) and noting that is assumed constant along y, gives

H

dyKnH

y

Kn

uuuWm ss

s

0

211 )(

21 (f)


212

ss uuWH

m (g)

Examination of this result shows that the net mass flow rate is zero when the two velocities are the same. That is m = 0 for

21 ss uu (h)

(iii) Checking.

Dimensional check: Equation (g) has the correct units for mass flow rate.

Limiting Check: For the special case of stationary lower plate, ,02su equation (g)

reduces to

12

suWH

m (i)

This agrees with (9.17) of Section 9.6.2.

Boundary conditions check: Solution (d) satisfies boundary conditions (a) and (b).

(5) Comments. (i) The effect of slip is to decrease fluid velocity at the upper and lower surfaces. (ii) Because the velocity distribution is linear, slip velocity is the same for both plates. (iii) The mass flow rate is independent of Knudsen number.

PROBLEM 9.6

Determine the frictional heat generated by the fluid in Example 9.1.

(1) Observations. (i) In Example 9.1, Couette flow between parallel plates is used to model the flow in the channel between the shaft and housing. (ii) At steady state, the heat generated due to friction (dissipation) is equal to the net heat conducted from the channel. (iii) Since no heat is transferred to the shaft, the net heat leaving the channel is equal to the heat conducted through the housing surface. (iv) Velocity slip takes place at both boundaries of the flow channel. (v) To determine heat transfer rate it is necessary to determine fluid temperature distribution. This requires the determination of the velocity field.

(2) Problem Definition. Determine the velocity and temperature distribution in the channel.

(3) Solution Plan. (i) Model channel flow as Couette flow between parallel plates. (ii) Apply Fourier’s law at the housing surface to determine heat leaving the channel. (iii) Apply the Navier-Stokes equations and formulate the velocity slip boundary conditions. Follow the analysis of Section 9.6.2 and Example 9.1. (iv) Use the energy equation to determine the temperature distribution

(4) Plan Execution.

(i) Assumptions. (1) Steady state, (2) laminar flow, (3) one-dimensional (no variation with axial distance x and normal distance z), (4) slip flow regime (0.001 < Kn < 0.1), (5) ideal gas, (6) constant viscosity (7) negligible lateral variation of density, (8) the velocity accommodation

coefficients are equal to unity, ,0.1Tu (9) continuum, slip flow regime conditions apply,

(10) ideal gas, and (11) no gravity.

(ii) Analysis. Apply Fourier’s law at the housing surface (upper plate)

dy

HdTkAHq

)()( (a)

where

A surface area k fluid thermal conductivity

q heat transfer rate

T fluid temperature distribution

Surface area A for a shaft of length L is

LHRA )(2 (b)

(b) into (a)

dy

HdTLHRkHq

)()(2)( (c)

y

x

H

1su

2su

u

2T


Flow Field. To determine u we follow the analysis of Section 9.6.2. The axial component of the Navier-Stokes equations for Couette flow between parallel plates is given by (9.12)

02

2

dy

ud(9.12)

Boundary conditions for (9.12) are given by (9.10)

n

xuuxu

u

us

)0,(2)0,( (9.10)

Applying (9.10) to the lower surface, 0yn , and setting 1u

dy

duuu s

)0()0( 1 (d)

For the upper surface, n = H – y, (9.10) gives

dy

HduuHu s

)()( 2 (e)

The solution to (9.12) is ByAu (f)

Boundary conditions (a) and (b) give the two constants of integration A and B

)21(

21

KnH

uuA ss ,

)21(

)( 211

Kn

KnuuuB ss

s (g)

where Kn is the Knudsen number, defined as

HKn (9.13)

Substituting (g) into (f)

)()21(

)( 211

H

yKn

Kn

uuuu ss

s (h)

Temperature Field. Temperature distribution is governed by the energy equation. Based pm the above assumptions, energy equation (2.15) simplifies to

02

2

y

Tk (9.23)

where2

y

u (9.24)

(9.24) into (9.23)2

2

2

dy

du

kdy

Td(9.25)


Note that T is independent of x. Substitute (h) into (9.25)

2

21

2

2

)21( KnH

uu

kdy

Td ss (i)


2

2

)21( KnH

uu

k

ss (j)


2

2

dy

Td (k)

This energy equation requires two boundary conditions. They are:

0)0(

dy

dT (l)

The second boundary condition is at y = H. Plate temperature is specified as 2T at this boundary.

However, the boundary condition must be associated with the fluid at y = H and not the plate. Knowing plate temperature, temperature jump condition (9.11) gives fluid temperature T(H). For

yHn and 1T , (9.11) gives

dy

HdT

PrHTT

)(

1

2)(2 (9.20)

Solve for )(HT

dy

HdT

PrTHT

)(

1

2)( 2 (m)

Integration of (k) gives

DCyyT 2

2 (n)

where C and D are constants of integration. Application of boundary conditions (l) and (m) gives the two constants:

0C (o)

and

22

21

2

2H

Pr

KnHTD (p)

Substituting into (n)

22

22

1

2

22TH

Pr

KnHyT (q)

Rewriting (q) in dimensionless form

Pr

Kn

H

y

H

TT

1

21

2

12

2

2

2


Using the definition of in the above

Pr

Kn

H

y

Kn

uu

k

TT

ss1

21

2

1

)21(

2

2

2

2

2 (r)

Frictional Heat. Using (r) to form the temperature gradient at Hy and substituting into (c)

gives the heat generated by fluid friction

2

2

)21()(2)(

Kn

uu

H

LHRHq ss (s)

(iii) Checking.

Dimensional check: Each term in (h) has units of velocity. Each term in (q) has units of temperature. Each term in (r) is dimensionless. q(H) in (s) is expressed in watts.

Governing equations check: Velocity solution (h) satisfies (9.12). Temperature solution (r) satisfies (9.25).

Boundary conditions check: Velocity solution (h) satisfies boundary conditions (d) and (e). Temperature solution (r) satisfies boundary conditions (l) and (m).

Limiting checks:

(i) If the two plates are stationary )0( 21 ss uu , there is no fluid motion, gas temperature is

uniform, and there is no frictional energy. Setting 021 ss uu in (h) gives u = 0. Setting 0

in (q) gives .2TT Setting 021 ss uu in (s) gives q(H) = 0.

(5) Comments. (i) The velocity field is governed by a single parameter, Kn. The temperature

field is governed by the parameter .1

2

Pr

Kn

(ii) Equation (s) shows that the narrower the gap H between the rotor and housing, the greater is the frictional energy. In addition, frictional energy decreases as the Knudsen number is increased.

PROBLEM 9.7

Consider shear driven Couette flow between

parallel plates. The upper plate moves with velocity

su and is maintained at uniform temperature .sT

The lower plate is heated with uniform flux .oq The

fluid between the two plates is an ideal gas. Taking

into consideration velocity slip, temperature jump,

and dissipation, determine the temperature of the

lower plate.

(1) Observations. (i) To determine the temperature of the lower plate, fluid temperature distribution must be known. (ii) Temperature distribution depends on the velocity field. (iii) The velocity field for Couette flow with a moving upper plate is given in Section 9.6.2. (iv) The solution to the energy equation gives the temperature distribution. (v) Two temperature boundary conditions must be specified.

(2) Problem Definition. Determine the temperature distribution for Couette flow with uniform flux at the lower plate and specified temperature at the upper plate.

(3) Solution Plan. Use the velocity solution for Couette flow of Section 9.6.2, formulate the energy equation and boundary conditions, and solve for the temperature distribution.

(4) Plan Execution.

(i) Assumptions. (1) Steady state, (2) laminar flow, (3) Constant viscosity, conductivity and specific heat, (4) infinite plates, (5) uniform boundary conditions, (6) uniform spacing between

plates, (7) no variation of density and pressure along y, (8) no gravity, (7) ,0.1Tu (9) the

velocity field is independent of temperature, (10), ideal gas, and (11) continuum, slip flow regime conditions apply.

(ii) Analysis. Surface temperature is related to fluid temperature through temperature jump condition (9.11)

n

xT

PrTxT

T

Ts

)0,(

1

22)0,( (9.11)

Solving the above for plate temperature at y = 0, oT , and assuming 1T , gives

n

xT

PrxTTo

)0,(

1

2)0,( (a)

where )0,(xT fluid temperature at the lower plate n = y . Thus fluid temperature distribution is

needed to determine oT . Temperature distribution is governed by the energy equation. Based on

the above assumptions, energy equation (2.15) simplifies to (see Section 9.6.2)

02

2

y

Tk (9.23)

where

suy

xHu

sT

oq


2

y

u (9.24)

where

k = thermal conductivity of fluid

u = axial velocity

= fluid viscosity

Noting that velocity and temperature are independent of axial distance, (9.24) into (9.23)2

2

2

dy

du

kdy

Td(9.25)

The velocity distribution is given in Section 9.6.2

)(21

1Kn

H

y

Knu

u

s

(9.14)

2

2

2

)21( KnH

u

kdy

Td s (b)


2

)21( KnH

u

k

s (c)

Substituting (c) into (b)

2

2

dy

Td (d)

This energy equation requires two boundary conditions. Heat flux is specified at the lower plate. Fourier’s law gives

oqdy

dTk

)0( (e)

Surface temperature is specified at the upper plate. However, this boundary condition must be associated with the fluid at y = H and not the plate. Knowing plate temperature, temperature

jump condition (9.11) gives fluid temperature. Thus at n = H – y and 1T , (9.11) gives

dy

HdT

PrTHT s

)(

1

2)( (f)

Integration of (d) gives

DCyyT 2

2 (g)

where C and D are constants of integration. Application of boundary conditions (e) and (f) gives the two constants:


k

qC o , s

o TPr

KnH

Pr

Kn

k

HqD

1

41

21

21

2

(h)

Substitute (h) into (g)

22

21

41

21

21 yy

k

q

Pr

KnH

Pr

Kn

k

HqTT oo

s (i)

Using the definition of in (c), the above becomes

2

222

)21(2

1

1

41

)21(21

21

H

y

Kn

u

ky

k

q

Pr

Kn

Kn

u

kPr

Kn

k

HqTT soso

s (j)

Surface temperature of the lower plate is determined by substituting (j) into (a) and setting y = 0

Pr

Kn

Kn

u

kk

HqTT s

so

o1

41

)21(2 2

2

(k)

(iii) Checking.

Dimensional check: (i) Noting that units of are 2o C/m , each term in (j) has units of

temperature.

Limiting check: (i) If dissipation is neglected, temperature distribution should be linear. Setting 0 in (j) gives

yk

q

Pr

Kn

k

HqTT oo

s1

21 (l)

(ii) If dissipation and rarefaction are neglected, the process reduces to pure conduction. Setting 0Kn in (j) gives

H

y

k

qTT o

s 1 (m)

This is the one dimensional conduction solution to the problem.

(5) Comments. (i) The Knudsen number in (j) and (l) represents the effect of rarefaction while the Prandtl number term represents the effect of temperature jump.

(ii) Fluid temperature adjacent to the lower surface is obtained by setting y = 0 in (j)

Pr

Kn

Kn

u

kPr

Kn

k

HqTT so

s1

41

)21(21

21)0(

2

(n)


To examine the difference between plate and fluid temperature at y = 0, (n) is subtracted from (k)

k

Hq

Pr

KnTT o

o1

2)0( (o)

This result shows that departure of plate temperature from fluid temperature at y = 0 increases with increasing heat flux and rarefaction.

(iii) To identify the governing parameters, solution (j) is expressed in dimensionless form

2

222

)21(21

41

)21(21

21

H

y

Kn

u

Hqh

y

Pr

Kn

Kn

u

HqPr

Kn

k

Hq

TT s

o

s

oo

s (p)

This result shows that temperature distribution is governed by two parameters:

Pr

Kn

1

2 and

2

)21(2 Kn

u

Hq

s

o

(q)

PROBLEM 9.8

Pressure distribution in Poiseuille flow between parallel plates is given by

L

x

p

pKn

p

p

p

pKnKn

p

xp

o

io

o

i

o

ioo

o

)1(12)1(66)(

2

22

(9.35)

This equation was derived in Section 9.6.3 using the continuity equation to determine the y

velocity component v. An alternate approach to derive (9.35) is based on the condition that for

steady state the flow rate is invariant with axial distance x. That is

022/

0

H

udyWdx

d

dx

dm

where W is channel width. Derive (9.35) using this approach.

(1) Observations. (i) To use the proposed approach, the solution to the axial velocity distribution must be know. (ii) The velocity distribution for Poiseuille flow between parallel plates is given by equation (9.30) of Section 9.6.3.

(2) Problem Definition. Determine the mass flow rate.

(3) Solution Plan. Use the solution to the axial velocity for Poiseuille flow, equation (9.30), to determine the mass flow rate.

(4) Plan Execution.

(i) Assumptions. (1) Steady state, (2) laminar flow, (3) constant viscosity (4) infinite plates, (5) uniform spacing between plates, (6) no variation of density and pressure along y, (7) no

gravity, (8) ,0.1u (9) the velocity field is independent of temperature, (10) ideal gas, and

(11) continuum, slip flow regime conditions apply.

(ii) Analysis. To derive pressure solution (9.35), it is proposed to use the following conservation of mass equation

02

2/

0

H

udyWdx

d

dx

dm(a)

where W = plate width. The solution to the axial velocity u is given by (9.30)

2

22

4)(418 H

ypKn

dx

dpHu (9.30)

Substituting (9.30) into (a) and noting that varies along x and is assumed constant along y, we

obtain

04)(418

2

2/

02

22H

dyH

ypKn

dx

dpHW

dx

d

dx

dm

This result simplifies to


04)(41

2/

02

2H

dyH

ypKn

dx

dp

dx

d

dx

dm (b)

Evaluating the integral in (b)

0)(23

pKnHH

dx

dp

dx

d(c)

To proceed, the density and Knudsen number in (c) must be expressed in terms of pressure. Ideal gas law (9.31) gives

RT

p (d)

The Knudsen number is expressed in terms of pressure in (9.33)

pRT

HHKn

1

2 (9.33)

(d) and (9.33) into (c)

01

22

3 pRT

H

dx

dp

RT

p

dx

d

Assuming isothermal flow, the above simplifies to

01

2

2

3

1

pRT

Hdx

dpp

dx

d(e)

Integrating (e) once

Cp

RTHdx

dpp )

1

2

2

3

1

Rewriting the above

Cdx

dpRT

Hdx

pdp

2

2

3

1(f)

Integrating again

DCxpRTH

p 26

1 2 (g)

The boundary conditions on p are

ipp )0( , opLp )( (h)

Here L is channel length. Equation (g) and (h) are identical to (o) and (q) of Section 9.6.3. Thus the solution to p is the same for both, given in (9.35)

L

x

p

pKn

p

p

p

pKnKn

p

xp

o

io

o

i

o

ioo

o

)1(12)1(66)(

2

22

(9.35)

(4) Comments. This approach for determining p(x) is simpler than that used in Section 9.6.3 where it was necessary to first determine the normal velocity component v.

PROBLEM 9.9

One of the factors affecting mass flow rate through a microchannels is channel height H. To

examine this effect, consider air flow through two microchannels. Both channels have the same

length, inlet pressure and temperature and outlet pressure. The height of one channel is double that of the other. Compute the mass flow ratio for the following case:

m,51H m,102H kPa,420ip kPa,105op C30 oiT

(1) Observations. (1) This is a pressure driven microchannel Poiseuille flow between parallel plates. (ii) The solution to mass flow rate through microchannels is given in Section 9.6.3. (iii) Channel height affects the Knudsen number.

(2) Problem Definition. Determine the mass flow rate for microchannel Poiseuille flow between parallel plates.

(3) Solution Plan. Apply the mass flow solution, equation (9.39), to two channels having different heights and take their ratio.

(4) Plan Execution.

(i) Assumptions. (1) Steady state, (2) laminar flow, (3) constant viscosity (4) infinite plates, (5) uniform spacing between plates, (6) no variation of density and pressure along y, (7) no

gravity, (8) ,0.1u (9) the velocity field is independent of temperature, (10) ideal gas, (11)

continuum, slip flow regime conditions apply, and (12) fully developed flow.

(ii) Analysis. The mass flow rate through Poiseuille flow microchannels is given by equation (9.39):

)( 112124

12

223

o

io

o

io

p

pKn

p

p

LRT

pHWm (9.39)

Apply (9.39) to two channels having heights 1H and 2H , and outlets Knudsen numbers 1oKn

and 2oKn

)( 112124

112

2231

1

o

io

o

io

p

pKn

p

p

LRT

pHWm (a)

)( 112124

122

2232

2

o

io

o

io

p

pKn

p

p

LRT

pHWm (b)

Take the ratio of (b) and (a)

)(

)(

1121

1121

1

2

2

2

3

1

2

1

2

o

io

o

i

o

io

o

i

p

pKn

p

p

p

pKn

p

p

H

H

m

m (c)

Equation (9.34) gives the outlet Knudsen number


RTpH

Kno

o2

(9.34)

Apply (9.34) to the two channels

RTpH

Kno

o21

1 (d)

RTpH

Kno

o22

2 (e)

(iii) Computations. The following date is given

m105m5 -61H

m1001m10 -62H

m-s

kg000,105kPa105

2op

m-s

kg000,420kPa420

2ip

C30 ooi TT


712.0Pr

K-s

m287

K-kg

J287

2

2

R

m-s

kg1065.18

6

Substituting into (d) and (e)

01313.0)K)(303)(K-s

m)(287(

2)

m-s

kg)(000,105((m)105

)m-s

kg(1065.18

2

2

2

6-

6

1oKn

006566.0)K)(303)(K-s

m)(287(

2)

m-s

kg)(000,105((m)1001

)m-s

kg(1065.18

2

2

2

6-

6

2oKn



878.7

1105

420)01313.0(121

105

420

1105

420)006566.0(121105

420

5

10

)(

)(

2

2

3

1

2

m

m

(4) Checking.

Dimensional check: Computations showed that units of Kn in (d) and (e) are dimensionless.

Limiting check: If 21 HH , the mass ratio should be unity. Setting 21 HH in (d) and (e) gives

21 oo KnKn

Setting 21 HH and 21 oo KnKn in (c) gives

21 mm

(5) Comments. (i) The effect of channel size on mass flow rate is significant. (ii) Setting

021 oo KnKn in (c) gives the mass ratio for macrochannels

81

2

m

m

This indicates that the effect of rarefaction on the mass ratio is 1.5%.

PROBLEMT 9.10

A micro heat exchanger consists of rectangular channels of height m,25H width

m,600W and length mm.10L Air enters the channels at temperature C20 oiT and

pressure kPa.420ip The outlet pressure is kPa.105op The air is heated with uniform

surface heat flux .W/m1100 2sq Taking into

consideration velocity slip and temperature jump,

assume fully developed conditions, compute the

following:

(a) Mass flow rate, m.

(b) Mean outlet temperature, .moT

(c) Heat transfer coefficient at the outlet, ).(Lh

(d) Surface temperature at the outlet, ).(LTs

(1) Observations. (i) This is a pressure driven microchannel Poiseuille flow. (ii) Since channel height is much smaller than channel width, the rectangular channel can be modeled as Poiseuille flow between parallel plates. (iii) Channel surface is heated with uniform flux. (iv) The solution to mass flow rate, temperature distribution, and Nusselt number for fully developed Poiseuille channel flow with uniform surface flux is presented in Section 9.6.3.

(2) Problem Definition. Determine flow and heat transfer characteristics of fully developed Poiseuille channel flow with uniform surface heat flux.

(3) Solution Plan. Apply the analysis and results of Section 9.6.3.

(4) Plan Execution.

(i) Assumptions. (1) Steady state, (2) laminar flow, (3) constant viscosity (4) infinite plates, (5) uniform spacing between plates, (6) no variation of density and pressure along y, (7)

,0.1Tu (8) no gravity, (9) the velocity field is independent of temperature, (10) ideal

gas, (11) continuum, slip flow regime conditions apply, and (12) fully developed flow.

(ii) Analysis.

(a) Mass flow rate m. Equation (9.39) gives the mass flow rate through the channel:

)( 112124

12

223

o

io

o

io

p

pKn

p

p

LRT

pHWm (9.39)

where

m-61052m25heightchannelH

oKn outlet Knudsen number = )( opKn

L channel length = 0.01 m

q

q

H

W

L

PROBLEMT 9.10 (continued)

ip inlet pressure m-kg/s000,420kPa420 2

m-kg/s000,105kPa105pressureoutlet 2op

2W/m1100sq

m10600 widthchannel 6W

K-s/m287K-J/kg287constantgas 22R

m-kg/s1017.18 6


oooo

pRT

HHpKnKn

1

2)( (9.34)

where moo TT is the mean outlet temperature.

(b) Mean outlet temperature, .moT The local mean temperature )(xTm is given by equation

(9.60):

mimp

sm Tx

Huc

qxT

2)( (9.60)

where

CJ/kg3.998heatspecific opc

2W/m1100fluxheatsurfacesq

C20eratureinlet tempmean oimi TT

s

mity,mean velocmu

3kg/mdensity,

The mean outlet temperature is obtained by setting x = L in (9.60)

mimp

smmo TL

Huc

qLTT

2)( (a)

The product mu is determined from the mass flow rate using the continuity equation:

HW

mum (b)

(b) into (a)

mip

smmo TLW

mc

qLTT

2)( (c)

CW/m02564.0 ok

C20 ooi TT


(c) Heat transfer coefficient at the outlet, ).(Lh The Nusselt number is used to determine the

heat transfer coefficient. Nusselt number for channel flow is defined as

k

hHNu

2 (d)

where

Cm

W02564.0tyconcuctivithermal

ok

Applying (d) at the outlet, x = L and solving for h(L)

)(2

)( LNuH

kLh (e)

The Nusselt number is given by (9.64)

KnPr

1KnKn

KnKn

Kn

Nu

1

2

560

13

40

13)(

)61(

1

48

5

2

1

)61(

3

2

2

(9.64)

where

Kn local Knudsen number 0.713numberPrandtlPr

= specific heat ratio = 1.4

Evaluation (9.64) at x = L where oKnKn

zzzKnPr

1KnKn

KnKn

Kn

Nu

oooo

oo 1

2

560

13

40

13)(

)61(

1

48

5

2

1

)61(

3

2

2

(f)

(d) Surface temperature at the outlet, ).(LTs Surface temperature distribution is given by

(9.63):

)(1

2

48

5

2

1

)61(

3)( xgKn

Prk

HqKn

Knk

HqxT ss

s (9.63)

where g(x) is given by (9.62):

560

13

40

13)(

)61(

32)( 2

2KnKn

Knk

Hqx

Huc

qTxg s

mp

smi (9.62)

Substituting (b) into (9.62), setting x = L and oKnKn , gives

560

13

40

13)(

)61(

32)( 2

2 oos

p

smi KnKn

Knk

HqL

mc

WqTLg (g)


To determine surface temperature at the outlet, ),(LTs the Knudsen number in (9.62) and (9.63) is

evaluated at outlet pressure and g(x) is evaluated at x = L.

(iii) Computations.

(a) Mass flow rate m. Equation (9.39) for m is based on the assumption that the flow is

isothermal. Since the outlet temperature oT is not yet determined, as a first approximation we

assume K293io TT in (9.34)

002516.0K))(293)(K-s/m)(287(2)m-kg/s)(000,105(m)(1052

)m-kg/s(1017.18 22

26-

6

oKn


)( 1105000

420000002516.0(121

105000

420000

K))(293)(K-s/m(287m))(01.0()m-kg/s(1017.18

m-kg/s)000,105((m))1025((m)10600

24

12

226

222366 3

m

kg/s1025126.4 6m

(b) Mean outlet temperature, .moT Equation (c) gives

mip

smmo TLW

mc

qLTT

2)( (d)

)C(20m))(10m)(600)(01.0()kg/s)(1025126.4)(CJ/kg(3.998

)W/m()1100(2)( o6-

6-o

2

LTT mmo

C11.23 omoT

(c) Heat transfer coefficient at the outlet, ).(Lh Setting oKnKn , and substituting into (9.64),

gives the Nusselt number at the outlet

(0.713)

)002516.0(

14.1

)4.1(2

560

13)002516.0(

40

13)002516.0(

)002516.0(61

1

48

5)002516.0(

2

1

)002516.0(61

3

2)(

2

LNu

14.8)(LNu

Equation (e) gives h(L)

CW/m417414.8)(m)6-1052(2

)CW/m(02564.0)( o2

o

Lh

(d) Surface temperature at the outlet, ).(LTs Use (f) to compute g(L)


560

13)002516.0(

40

13)002516.0(

)002516.061)(CW/m(02564.0

)(m)-61052)(W/m()1100(3

(m)01.0)(kg/s)-61025126.4)(CJ/kg(3.998

)(m)10600)(W/m()1100(2C)(20)(

2

2o

2

o

62oLg

04.23)(Lg Co


)002516.0()713.0)(CW/m(02564.0

)(m)-61052)(W/m()1100(

14.1

)4.1(2

48

5)002516.0(

2

1

)002516.061)(CW/m(02564.0

)(m)-61052)(W/m()1100(3)(

o

2

o

2

LTs

C373.23)( oLTs

(iv) Checking.

Dimensional check: computations showed that equations (9.34), (9.39), (9.60), (9.62) and (9.63) are dimensionally correct.

Surface temperature check: Application of Newton’s law at the outlet gives

moss TLTLhq )()(

Solving for )(LTs

mos

s TLh

qLT

)()(

Using this equation to compute )(LTs , gives

C23.374C11.23)CW/m(4174

)W/m()1100()( oo

o2

2

LTs

This is close to the value determined above.

(5) Comments. (i) The flow field solution (velocity, mass flow rate and pressure) is based on the assumption that the flow is isothermal. Computation showed that the outlet temperature is

C11.23 omoT . Since the outlet is C11.3 o above the inlet temperature, it follows that the

assumption of isothermal flow is reasonable. (ii) The heat transfer coefficient at the outlet is high compared to values for air encountered in typical macrochannels applications. (iii) The Nusselt

number for slip theory for fully developed macrochannel flow is obtained by setting 0oKn in

(f). This gives

235.8oNu

Thus macrochannel theory overestimates the Nusselt number if applied to microchannels.

PROBLEM 9.11

Rectangular microchannels are used to remove heat

from a device at uniform surface heat flux. The height,

width, and length of each channel are m,29.6H

m,90W and mm,10L respectively. Using air

at C20oiT as the coolant fluid, determine the mass

flow rate and the variation of Nusselt number along

the channel. Inlet and outlet pressure are 410ip

kPa, 105op kPa. Assume steady state fully

developed slip flow and temperature jump conditions.

(1) Observations. (i) The problem can be modeled as pressure driven Poiseuille flow between two parallel plates with uniform surface flux. (ii) Assuming fully developed velocity and temperature, the analysis of Section 9.6.3 gives the mass flow rate and Nusselt number. (iii) The Nusselt number depends on the Knudsen number, Kn. Since Kn varies along the channel due to pressure variation, it follows that pressure distribution along the channel must be determined.

(2) Problem Definition. Determine the flow and temperature fields for fully developed Poiseuille flow with uniform surface flux.

(3) Solution Plan. Apply the results of Section 9.6.3 for the mass flow rate, pressure distribution, and Nusselt number.

(4) Plan Execution.

(i) Assumptions. (1) Steady state, (2) laminar flow, (3) two-dimensional (no variation along the width W), (4) slip flow regime (0.001 < Kn < 0.1), (5) ideal gas, (6) constant viscosity, conductivity and specific heats, (7) negligible lateral variation of density and pressure, (8) the

accommodation coefficients are equal to unity, ,0.1Tu (9) negligible dissipation, (10)

uniform surface flux, (11) negligible axial conduction, and (12) no gravity.

(ii) Analysis. Assuming isothermal flow, the results of Section 9.6.3 give the mass flow rate as

)( 112124

12

223

o

io

o

i

o

o

p

pKn

p

p

LRT

pHWm (9.39)

The Knudsen number at the exit, oKn is

oo

oo RT

pHH

pKn

2

)( (9.34)

where the temperature oT at the outlet is assumed to be the same as inlet temperature and the

viscosity is based on inlet temperature.

The Nusselt number, ,Nu is given by

x

yHL

Wsq

sqm


KnPr

1KnKn

KnKn

Kn

Nu

1

2

560

13

40

13)(

)61(

1

48

5

2

1

)61(

3

2

2

(9.64)

The local Knudsen number, ,Kn depends on the local pressure p(x) according to

RTpHH

Kn2

(9.33)

Equation (9.35) gives )(xp

L

x

p

pKn

p

p

p

pKnKn

p

xp

o

io

o

i

o

ioo

o

)1(12)1(66)(

2

22

(9.35)

Thus, (9.35) is used to determine p(x), (9.33) to determine ),(xKn and (9.64) to determine the

variation of the Nusselt number along the channel.

(iii) Computations. Air properties are determined at .C20o To compute p(x), ),(xKn

and ,Nu the following data is used

m29.6H

ip m/skg10420 23

op m/skg10105 23

713.0Pr

Ks/m287Kkg/J287 22R

C20 ooi TTT

m90W

4.1

m/skg1017.18 6


(K)K)(293.15)-s/m(2872m)/skg(10105m)(1029.6

m)/skg(1017.18 22

236

6

oKn

01.0oKn

Using (9.39) and noting that 4/ oi pp

)12(01.0121)4(93.15(K)2K)s/m)287(m(01.0m)/skg(1017.18

)m/skg()10105)(m()1090(m)(1090

24

1 2

226

242233366

m

kg/s1003776.1 12m


Axial pressure variation is obtain from (9.35)

L

x

p

xp

o

)41(01.012)4(1)401.06(01.06)( 22

L

x

p

xp

o

36.154836.1606.0)(

(a)

Equation (a) is used to tabulate pressure variation with x/L. Equations (9.33) and (9.64) are used to compute the corresponding Knudsen and Nusselt numbers.

(iii) Checking. Dimensional check: Units for equations (9.33), (9.35), (9.39) and (9.64) are consistent.

Limiting check: No-slip macrochannel Nusselt number is obtained by setting 0Kn in (9.64). This gives Nu =

8.235. This agrees with the value given in Table 6.2.

(5) Comments. (i) To examine the effect of rarefaction and compressibility on the mass flow

rate, equation (9.41) is used to calculate omm / :

56.2)01.01214(2

1121

2

1o

o

i

o

Knp

p

m

m

This shows that incompressible no-slip theory will significantly underestimate the mass flow

rate. If rarefaction is neglected )0( oKn , the above gives

5.2)14(2

11

2

1

o

i

o p

p

m

m

Thus, compressibility plays a dominant role in the mass flow rate.

(ii) No-slip Nusselt number for fully developed Poiseuille flow between parallel plates with uniform surface heat flux is Nu = 8.235. Thus, no-slip theory overestimates the Nusselt number if applied to microchannels.

(iii) It should be noted that the equations used to compute ,m ),(xp and Nu are based on the

assumptions of isothermal conditions in the determination of the flow field. This is a reasonable approximation for typical applications.

Knopp/ Nux/L

1.0

0.20.40.60.8

0

1.000

0.02504.03.6023.1552.6361.988

0.002780.003170.003790.005290.0100

8.1418.1308.1158.0928.0367.862

PROBLEM 9.12

A micro heat exchanger consists of rectangular

channels of height m,7.6H width m,400W

and length mm.8L Air enters the channels at

temperature C30 oiT and pressure kPa.510ip

The outlet pressure is kPa.102op Channel surface

is at uniform temperature C.50osT Assume fully

developed flow and temperature, compute:

(a) Mass flow rate, m.(b) Heat transfer coefficient at the inlet, ),0(h and outlet, ).(Lh

(c) Mean outlet temperature, .moT

(d) Surface heat flux at the outlet, ).(Lqs

(1) Observations. (i) This is a pressure driven microchannel Poiseuille flow. (ii) Since channel height is much smaller than channel width, the rectangular channel can be modeled as Poiseuille flow between parallel plates. (iii) Channel surface is maintained at uniform temperature. (iv) The solution to velocity, pressure, and mass flow rate is presented in Section 9.63. (v) The solution to the temperature distribution and Nusselt number for fully developed Poiseuille channel flow with uniform surface temperature is presented in Section 9.6.4. (vi) Surface heat flux is determined using Newton’s law.

(2) Problem Definition. Determine flow and heat transfer characteristics of fully developed Poiseuille channel flow with uniform surface temperature.

(3) Solution Plan. To determine velocity, mass flow rate and pressure, apply the analysis and results of Section 9.6.3. To determine temperature and Nusselt number, apply the analysis and results of Section 9.6.4.

(4) Plan Execution.

(i) Assumptions. (1) Steady state, (2) laminar flow, (3) constant viscosity (4) infinite plates, (5) uniform spacing between plates, (6) no variation of density and pressure along y, (7)

,0.1Tu (8) no gravity, (9) the velocity field is independent of temperature, (10) ideal

gas, (11) continuum, slip flow regime conditions apply, and (12) fully developed flow.

(ii) Analysis.

(a) Mass flow rate m. Equation (9.39) gives the mass flow rate through the channel:

)( 112124

12

223

o

io

o

io

p

pKn

p

p

LRT

pHWm (9.39)

where

m

H

W

L

sT

sT


m-6107.6m7.6heightchannelH





m10400 widthchannel 6W


m-s

kg1065.18

6


oooo

pRT

HHpKnKn

1

2)( (9.34)

where moo TT is the mean outlet temperature.

(b) Heat transfer coefficient at inlet, )0(h and outlet, ).(Lh The Nusselt number is used to

determine the heat transfer coefficient. Nusselt number for channel flow is defined as

k

hHNu

2 (a)

where

Cm

W02638.0tyconductivithermal

ok

Applying (a) at the inlet, x = 0 and solving for h(0)

)0(2

)0( NuH

kh (b)

Similarly, at the outlet, (a) gives

)(2

)( LNuH

kLh (c)

The Nusselt number for air at various Peclet numbers and Knudsen numbers is given in Fig. 9.11. The Peclet number is defined as

RePrPe (d)


HuRe m 2

(e)

Here is density and mu is the mean velocity obtained from continuity


HW

mum (f)

(f) into (e)

W

mRe

2 (g)

The Knudsen number at the inlet is given by

iiii

pRT

HHpKnKn

1

2)( (h)

(c) Mean outlet temperature, .moT The local mean temperature )(xTm for channel flow at

uniform surface temperature is given by equation (6.13):

][exp)()( xcm

hPTTTxT

psmism (6.13)

where


P = channel perimeter = m10134.810)4007.6(2)(2 66WH


C50 osT

m/sity,mean velocmu

3kg/mdensity,

h is the average heat transfer along the channel between inlet and section x, defined in (6.12)

x

dxxhx

h

0

)(1

(6.12)

The mean outlet temperature is obtained by setting x = L in (6.13)

][exp)()( Lcm

hPTTTLTT

psmismmo (i)

(d) Surface heat flux at the outlet, ).(Lqs Application of Newton’s law at the outlet gives

surface heat flux ))(( moss TTLhq (j)

(iii) Computations.

(a) Mass flow rate m. Equation (9.39) for m is based on the assumption that the flow is


assume K303io TT in (9.34)


01009.0K))(303)(K-s/m)(287(2)m-kg/s)(000,102(m)(107.6

)m-kg/s(1065.18 22

26-

6

oKn


)( 1102000

510000)01009.0(121

102000

510000

K))(303)(K-s/m(287m))(008.0()m-kg/s(10657.18

m-kg/s)000,102((m))107.6((m)10400

24

12

226

222366 3

m

kg/s10098368.0 6m

(b) Heat transfer coefficient at the outlet, )0(h and ).(Lh To use Fig. 9.11 for the determination

of the Nusselt number, the Prandtl and Reynolds number, Peclet number, and Knudsen number are needed. Air properties give

712.0Pr

(g) gives the Reynolds number

372.26(m)10m)400-(kg/s1065.18

)(kg/s10098368.0)2(

66

6

Re

Thus the Peclet number is

777.18712.0372.26Pe

At this Peclet number the curve corresponding to Pe gives the approximate Nusselt number

for this case. The Knudsen number at the inlet is computed using (h)

00218.0K))(303)(K-s/m)(287(2)m-kg/s)(000,105(m)(107.6

)m-kg/s(1065.18 22

26-

6

iKn

At this value of Knudsen number Fig. 9.11 gives

5.7)0(Nu

Substitute into (b)

Cm

W765,14)5.7(

7.6(2

)CW/m(02638.0)0(

o2

o

)(m)-610h

At the outlet where 01009.0oKn , Fig. 9.11 gives

25.7)(LNu

(b) gives

Cm

W273,14)25.7(

7.6(2

)CW/m(02638.0)(

o2

o

)(m)-610Lh


(c) Mean outlet temperature, .moT The average heat transfer coefficient, h , is needed to

determine .moT Using (6.12) to determine h requires the numerical integration of the local heat

transfer coefficient. However, since the change in h between inlet and outlet is very small, the

arithmetical mean can be used to approximate h . Thus

Cm

W519,14

2

765,14273,14

2

)()0(

o2

Lhhh

][ m)(008.0)kg/s)(10.098368)0()CJ/kg(006.41

)CW/m(519,14)m(10134.8)C)(5030()C(50

6-o

o26oo expmoT

C00215.50 omoT

(d) Surface heat flux at the outlet, ).(Lqs Use (j) to compute ).(Lqs

2

o2

m

W7.305000215.50)CW/m(273,14)(Lqs

(iv) Checking.

Dimensional check: computations showed that equations (9.34), (9.39), (6.13), and (j) are dimensionally correct.


C50omoT . To improve the solution, an iterative procedure can be followed by repeating the

computation assuming an arithmetical average of mean temperature in the channel equal to

C.40C)]/250(C)(30[ ooo (ii) The heat transfer coefficient at the outlet is high compared to

values for air encountered in typical macrochannels applications. (iii) The Nusselt number for no-slip theory and negligible axial conduction is obtained from Fig. 9.11 at Kn = 0 and Pe .This gives

5407.7oNu

This is close to the inlet Nusselt number when rarefaction is included, indicating a small rarefaction effect. (iv) Unlike the Nusselt number for fully developed flow in macrochannels, the Nusselt number is not constant along microchannels.

PROBLEM 9.13

Consider isothermal Poiseuille flow of gas in a microtube of radius .or Taking into

consideration velocity slip, show that the axial velocity is given by

2

22

414

o

oz

r

rKn

dz

dprv

(1) Observations. (i) Cylindrical coordinates should be used to solve this problem. (ii) The axial component of the Navier-Stokes equations must be solved to determine the

axial velocity .zv (iii) The procedure

and simplifying assumptions used in the solution of the corresponding Couette flow between parallel plates, detailed in Section 9.6.2, can be applied to this case.

(2) Problem Definition. Solve the axial component of the Navier-Stokes equations of motion.

(3) Solution Plan. Start axial component of the governing Navier-Stokes equations of motion, introduce simplifying assumptions, write down the slip velocity boundary conditions and solve the governing equation.

(4) Plan Execution.

(i) Assumptions. (1) Steady state, (2) laminar flow, (3) two-dimensional (axial and radial), (4) slip flow regime (0.001 < Kn < 0.1), (5) ideal gas, (6) constant viscosity (7) negligible radial variation of density and pressure, (8) negligible gravity, (9) the velocity

accommodation coefficients is equal to unity, ,0.1u (10) isothermal flow, (11) the

dominant viscous force is r

vr

rr

z1, and (12) negligible inertia forces;

z

vv

v

r

v

r

vv zzz

zr

(ii) Analysis. Following the analysis of Section 9.6.2, we begin with the axial component of the Navier-Stokes equations

2

2

2

2

2

11

z

vv

rr

vr

rrz

pg

t

v

z

vv

v

r

v

r

vv

zzzz

zzzzzr

(2.11z)

Introducing the above assumptions, this equation simplifies to

z

p

r

vr

rr

z 11 (a)

z

orr r


The boundary conditions are

0),0(

r

zvz (b)

r

vzru z

o ),( (c)

Since pressure is assumed independent of r, this equation can be integrated directly to

give the axial velocity zv . Thus

21

2

ln4

1CrC

r

z

pvz

(d)

Application of boundary conditions (b) and (c) give

01C ,o

o

rdz

dprC

24

12

2 (e)

Susbtituting (e) into (d)

2

2

0

2

241

4o

oz

r

r

rz

prv (f)

Introducing the definition of Knudsen number for tube flow

orKn

2 (g)

Substituting (g) into (f)

2

22

414

o

oz

r

rKn

z

prv (h)

(iii) Checking. Dimensional check: (h) is dimensionally correct.

Governing equation check: (h) satisfies (a).

Boundary conditions check: (h) satisfies conditions (b) and (c).

(5) Comments. It is important to note the assumptions leading to solution (h).

PROBLEM 9.14

Consider fully developed isothermal Poiseuille flow through a microtube. Follow the analysis of Section 9.6.3 and use the continuity equation in cylindrical coordinates to derive the following:

(a) The radial velocity component rv

)(24

1

2

11

4 3

33

pKnr

r

r

r

r

r

dz

dpp

zp

rv

ooo

or

where Kn(p) is the local Knudsen number.

(b) The local pressure p(z)

L

z

p

pKn

p

p

p

pKnKn

p

zp

o

io

o

i

o

ioo

o

)1(16)1(88)(

2

22

(9.78)

where ip is inlet pressure, op outlet pressure, and oKn is the outlet Knudsen number.

(1) Observations. (i) This a pressure driven Poiseuille flow through a microtube. (ii) The procedure for determining the radial velocity component and axial pressure distribution is identical to that for slip Poiseuille flow between parallel plates. (iii) The solution to the axial velocity is given by equation (9.74). (iv) Continuity equation gives the radial velocity component. (v) Axial pressure is determined by setting the radial velocity component equal to zero at the surface. (vi) Cylindrical coordinates should be used to solve this problem.

(2) Problem Definition. Determine the radial velocity component of slip flow through a tube.

(3) Solution Plant. Use the continuity equation in radial coordinates and the solution to the axial velocity component to determine the radial component. Set the radial component equal to zero at the surface to determine the axial pressure distribution.

(i) Assumptions. (1) Steady state, (2) laminar flow, (3) two-dimensional, (4) slip flow regime (0.001 < Kn < 0.1), (5) ideal gas, (6) constant viscosity, conductivity, and specific heats, (7) negligible radial variation of density and pressure, (8) negligible dissipation (9) negligible gravity, and (10) The velocity accommodation coefficient is equal to unity, .0.1u

(ii) Analysis.

(a) To determine the radial velocity component, we follow the derivation of Section 9.6.3 for the

analogous problem of Poiseuille flow between parallel plates. The axial velocity zv for tube flow

is derived in Section 9.6.5

2

22

414

o

oz

r

rKn

dz

dprv (9.74)

The Knudsen number, Kn, for the flow through tubes is defined as

RTprr

Knoo 2

1

22(a)

The radial component is determined using the continuity equation for compressible flow in cylindrical coordinates


011

zrzr

rrrt

vvv (2.4)

This simplifies to

01

zr vz

vrrr

(b)

Using the ideal gas law, (9.31), to express in terms of pressure and rearranging, the above is

written as

01

zr vpz

vrprr

Substituting (9.74) into the above

)(2

22

414

1

o

or

r

rKn

dz

dpp

z

rvrp

rr (c)

Integration of (c) gives the radial velocity component .rv Boundary conditions on rv are

0),0( zvr (d)

0),( zrv or (e)

Multiplying (c) by ,rdr integrating with respect to r and using boundary condition (d)

r

rdrr

rKn

dz

dpp

z

rv

vrpr

o

or

r

00

)(2

22

414

Evaluating the integrals and noting that the integrand on the right hand side is a function of zonly, yields

)(2

2

42

24

42

2

o

or

r

rrKn

r

dz

dpp

zd

drvrp

Solving for rv

)(2

2

42

2

1

4

42

2

o

or

r

rrKn

r

dz

dpp

zd

d

pr

rv (f)

(b) To determine axial pressure distribution, (f) is applied to boundary condition (e)

)( 24

110 Kn

dz

dpp

zd

d

p

Using (a) to eliminate Kn

024

1)( RT

prdz

dpp

dz

d

o

(g)

Integration with respect to z twice gives

DCzpRTr

po 28

1 2


where C and D are constants of integration. Noting that pressure is positive, the solution to this quadratic equation is

)(82

162

42

2

DCzRTr

RTr

poo

(h)

The boundary conditions on pressure are

ipp )0( and opLp )( (i)

Using (i),the constants C and D are determined

)(2

)(8

1 22io

oio ppRT

Lrpp

LC (j)

io

i pRTr

pD2

)8

1 2 (k)

Substituting (j) and (k) into (h)

o

i

ooo

i

o

i

ooo

i

oo

ooo

p

pRT

prp

p

L

z

p

pRT

prp

p

p

RT

r

RTprp

zp

22

8)1(

2

81

2

16

24

)(

2

2

2

2

22

2 (l)

Using (a), this result is expressed in terms of the Knudsen number at the outlet

L

z

p

pKn

p

p

p

pKnKn

p

zp

o

io

o

i

o

ioo

o

)1(16)1(88)(

2

22

(m)

(iii) Checking. Dimensional check: Each term in (h) has units of velocity. Each term in (l) is dimensionless.

(5) Comments. Unlike fully developed Poiseuille flow in macrochannels, the radial velocity component does not vanish (streamlines are not parallel) and axial pressure distribution is not linear.

PROBLEM 9.15

Taking into consideration velocity slip, show that the mass flow rate for laminar, fully developed

isothermal Poiseuille flow in a microtube is give by

)( 116116 2

224

o

io

o

ioo

p

pKn

p

p

LRT

prm (9.798a)

(1) Observations. (i) Cylindrical coordinates should be used to solve this problem. (i) Axial velocity component is needed to determine mass flow rate. (iii) Equation (9.74) gives the axial velocity for this case. (iv) Since axial velocity vary with radial distance, mass flow rate requires integration of the axial velocity over the flow cross section area. (v) The procedure and simplifying assumptions used in the solution of the corresponding Couette flow between parallel plates, detailed in Section 9.6.2, can be applied to this problem.

(2) Problem Definition. Integration of the axial velocity over flow cross section.

(3) Solution Plan. Formulate the flow rate integral, use the axial velocity for the Poiseuille flow through tubes and carry out the integration of over tube radius.

(4) Plan Execution.

(i) Assumptions. (1) Steady state, (2) laminar flow, (3) two-dimensional (axial and radial), (4) slip flow regime (0.001 < Kn < 0.1), (5) ideal gas, (6) constant viscosity, (7) negligible radial variation of density, (8) negligible gravity, (9) the velocity accommodation coefficients is

equal to unity, ,0.1u (9) isothermal flow, (10) the dominant viscous force is r

vr

rr

z1,

and (11) negligible inertia forces; z

vv

v

r

v

r

vv zzz

zr

(ii) Analysis. The mass flow rate is given by o

z

r

rdrvm

0

2 (a)

Based on the above assumptions, the axial velocity component is given in Section 9.6.2 as

2

22

414

o

oz

r

rKn

dz

dprv (9.74)

Substituting (9.74) into (a) and recalling that density is assumed constant along the radial

distance r, gives o

o

o

r

rdrr

rKn

dz

dprm

0

2

22

412

(b)

Evaluating the integral in (b)

Kndz

dprm o 81

8

4

(c)


The density of an ideal gas is given by (9.31)

RT

p (9.31)

The Knudsen number for tube flow is

RTpr

Kno 2

1

2 (d)

Substituting (9.31) and (d) into (c)

RTr

pdz

dp

RT

rm

o

o

2

4

8

4

(e)

This result gives the mass flow rate in terms of pressure. The solution to the pressure distribution )(zp is

L

z

p

pKn

p

p

p

pKnKn

p

zp

o

io

o

i

o

ioo

o

)1(16)1(88)(

2

22

(9.78)

where oKn is the Knudsen number at the discharge. Evaluating (d) at the discharges where

opp

RTpr

Knoo

o2

1

2 (f)

Rewriting (9.78) as

CzBAzp )( (g)

where

oo pKnA 8 (h)

28 ioo ppKnB (i)

)1(1612

22

o

io

o

io

p

pKn

p

p

L

pC (j)

Differentiating (g)

2/1)(2

CzBC

dz

dp (k)

Substituting (g) and (k) into (e)

RTr

CzBACzBC

RT

rm

o

o

2

4)(

28

2/14

(l)

This simplifies to

CRT

rm o

4

16 (m)

Substituting (j) into (m)


)1(16116 2

224

o

io

o

ioo

p

pKn

p

p

LRT

prm (9.79a)

(iii) Checking. Dimensional check: Equation (9.79a) has the correct mass flow units of kg/s.

Limiting check: If ,oi pp no flow takes place and thus the mass flow rate should be zero.

Setting oi pp in (9.78a) gives m = 0.

(5) Comments. The Knudsen number in (9.79a) represents the effect of rarefaction. Neglecting

rarefaction ( )0oKn , (9.79a) reduces to

116 2

224

o

ioo

p

p

LRT

prm (n)

If both rarefaction and compressibility are neglected, the flow rate is given by (9.79b)

18

24

o

ioo

p

p

LRT

prm (9.70b)

PROBLEM 9.16

Pressure distribution for fully developed Poiseuille flow through tubes is given by

L

z

p

pKn

p

p

p

pKnKn

p

zp

o

io

o

i

o

ioo

o

)1(16)1(88)(

2

22

(9.78)

Derive this equation using the condition that, for steady state, the mass flow rate is invariant

with axial distance z. That is

02

0

o

z

r

drrvdz

d

dz

dm

(1) Observations. (i) To use the proposed approach, the solution to the axial velocity distribution must be known. (ii) The velocity distribution for Poiseuille flow through tubes is given by equation (9.74) of Section 9.6.5. (iii) Cylindrical coordinates should be used to solve this problem.

(2) Problem Definition. Determine the mass flow rate.

(3) Solution Plan. Use the solution to the axial velocity for Poiseuille flow, equation (9.74), to determine the mass flow rate.

(4) Plan Execution.

(i) Assumptions. (1) Steady state, (2) laminar flow, (3) constant viscosity (4) no radial

variation of density and pressure, (5) no gravity, (6) ,0.1u (7) the velocity field is

independent of temperature, (8) ideal gas, and (9) continuum, slip flow regime conditions apply.

(ii) Analysis. To derive pressure solution (9.78), it is proposed to use the following conservation of mass equation

02

0

o

z

r

drrvdz

d

dz

dm (a)

The solution to the axial velocity zv is given by (9.743)

2

22

414

o

oz

r

rKn

dz

dprv (9.74)


0)(414

02

22o

o

or

rrdr

rpKn

zd

dpr

zd

d

zd

dm(b)

Noting that density and pressure p vary along z and are assumed constant along r, and that the

viscosity is constant, (b) is written as


0)(410

2

2o

o

r

rrdr

rpKn

zd

dp

zd

d

zd

dm(c)

Evaluating the integral in (c)

0)(242

222

ooo rpKn

rr

zd

dp

zd

d

zd

dm

This simplifies to

0)(24

1pKn

zd

dp

zd

d(d)

To proceed, the density and Knudsen number in (c) must be expressed in terms of pressure. Ideal gas law (9.31) gives

RT

p (9.31)

The Knudsen number for tube flow is

RTpr

Kno 2

1

2 (e)

(9.31) and (e) into (d)

01

24

1

pRT

rdz

dp

RT

p

zd

d

o

Since the flow is assumed isothermal, the above simplifies to

01

24

1

pRT

rdz

dpp

zd

d

o

(f)

Integrating (f) once

1

1

24

1C

pRT

rdz

dpp

o

Rewriting the above as

124

1C

zd

dpRT

rzd

pdp

o

Integrating again

212

28

1CzCpRT

rp

o

(g)

The boundary conditions on p are

ipp )0( , opLp )( (h)

Here L is tube length. Boundary conditions (h) give 1C and 2C

)(2

)(8

1 221 io

oio ppRT

Lrpp

LC (i)


io

i pRTr

pC28

1 22 (j)

The solution to quadratic equation (g) is

)(8162

4212

2

CzCRTr

RTr

p

oo

(k)

Substituting (i) and (j) into (k), normalizing the pressure by ,op and introducing the Knudsen

number (e), we obtain

L

z

p

pKn

p

p

p

pKnKn

p

zp

o

io

o

i

o

ioo

o

)1(16)1(88)(

2

22

(9.78)

(iii) Checking. Dimensional check: Each term in (k) has units of pressure.

Limiting check: If oi pp , axial velocity will vanish and pressure should be uniform throughout

the tube. Setting oi pp in (9.78) gives

1188)(

ooo

KnKnp

zp

(4) Comments. This approach for determining p(z) is simpler than that proposed in Section 9.6.4 where it is necessary to first determine the radial velocity component .rv

PROBLEMT 9.17

Air is heated in a microtube of radius m5or and length mm.2L Inlet temperature and

pressure are C20 oiT and Outlet pressure is kPa.100op Uniform surface flux,

,W/m1500 2sq is used to heat the air. Taking into consideration velocity slip and

temperature jump and assuming fully developed flow and temperature, compute:

(a) Mass flow rate, m.

(b) Mean outlet temperature, .moT

(c) Heat transfer coefficient at outlet, ).(Lh

(d) Surface temperature at the outlet, ).(LTs

(1) Observations. (i) This is a pressure driven Poiseuille flow through a microtube. (ii) Tube surface is heated with uniform flux. (iii) The solution to mass flow rate, temperature distribution and Nusselt number for fully developed Poiseuille flow through a tube with uniform surface flux is presented in Section 9.6.5.

(2) Problem Definition. Determine flow and heat transfer characteristics of fully developed Poiseuille flow through a tube with uniform surface heat flux.

(3) Solution Plan. Apply the analysis and results of Section 9.6.5.

(4) Plan Execution.

(i) Assumptions. (1) Steady state, (2) laminar flow, (3) constant conductivity, specific heat

and viscosity, (4) no radial variation of density and pressure, (5) ,0.1Tu (6) no gravity,

(7) the velocity field is independent of temperature, (8) ideal gas, (9) continuum, slip flow regime conditions apply, and (10) fully developed flow.

(ii) Analysis.

(a) Mass flow rate m. Equation (9.79a) gives the mass flow rate through the tube:

)( 116116 2

224

o

io

o

i

o

oo

p

pKn

p

p

LRT

prm (9.79a)

where





or tube radius = m-6105m5


sq

sq

z

orr r


C20oio TT

m-kg/s1017.18 6

The outlet Knudsen number is given by

o

o

oo

oop

RTrr

pKnKn1

222)( (a)

(b) Mean outlet temperature, .moT Conservation of energy between inlet and outlet, gives

mi

p

soom T

mc

qLrT

2 (b)

where


2W/m1500fluxheatsurfacesq


(c) Heat transfer coefficient at the outlet, ).(Lh The Nusselt number is used to determine the

heat transfer coefficient. Nusselt number for channel flow is defined as

k

hrNu o2

(c)

where

Cm

W02564.0tyconcuctivithermal

ok

Applying (c) at the outlet, z = L, and solving for h(L)

)(2

)( LNur

kLh

o

(d)

The Nusselt number is given by (9.98)

KnPr

KnKnKn

KnKn

Nu1

1

4

24

7

3

1416

)81(

1

16

3

)81(

4

2

2

2)(

(9.98)

where

Kn local Knudsen number 0.713numberPrandtlPr

= specific heat ratio = 1.4

Evaluation (9.98) at z = L where oKnKn

ooo

o

o KnPr

KnKnKn

KnKn

Nu1

1

4

24

7

3

1416

)81(

1

16

3

)81(

4

2

2

2)(

(e)


(d) Surface temperature at the outlet, ).(LTs Surface temperature distribution is given by

(9.97):

)(1

4

16

3

)81(

4)( zgKn

Prk

rqKn

Knk

rqzT oo ss

s (9.97)

where g(z) is given by (9.96):

24

7

3

1416

)81(

2)( 2

2KnKn

Knk

rqz

vrc

qTzg os

mzop

smi (9.96)

where zmv is the mean velocity. Continuity equation gives zmv in terms of mass flow rate m

2o

zmr

mv (f)

To determine surface temperature at the outlet, ),(LTs the Knudsen number in (9.96) and (9.97) is

evaluated at outlet pressure and g(z) is evaluated at z = L. Equation (9.97) becomes

)(1

4

16

3

)81(

4)( LgKn

Prk

rqKn

Knk

rqLT o

oo

o

o sss (g)

Using (f), and setting z = L and oKnKn in (9.96) gives g(L)

24

7

3

1416

)81(

2)( 2

2 oo

o

os

p

somi KnKn

Knk

rqL

mc

qrTLg (h)

(iii) Computations.

(a) Mass flow rate m. Equation (9.79a) for m is based on the assumption that the flow is


assume K293io TT in (9.79a). The outlet Knudsen number is computed using (a)

0066054.0K))(293)(K-s/m)(287(2)m-kg/s)(000,100(m)(1010

)m-kg/s(1017.18 22

26-

6

oKn


)( 1100000

6000000066054.0(161

100000

600000

K))(293)(K-s/m(287m))(002.0()m-kg/s(1017.18

m-kg/s)000,100((m))105((m)10600

16

2

226

2223366

m

kg/s10426.1 8m

(b) Mean outlet temperature, .moT Equation (b) gives

)C(20)kg/s)(10426.1)(CJ/kg(3.998

m))(002.0(m)(105)W/m()1500(2 o

-8o

62

moT

C6204.26 omoT


(c) Heat transfer coefficient at the outlet, ).(Lh Substituting into (e), gives the Nusselt number

at the outlet

(0.713)

)0044054.0(

14.1

)4.1(2

24

7)0066054.0(

3

14)0066054.0(16

)0066054.0(81

1

16

30066054.0

)0066054.0(81

4

2)(

2

LNu

278.4)(LNu

Equation (d) gives h(L)

CW/m969,10278.4)(m)-6105(2

)CW/m(02564.0)( o2

o

Lh

(d) Surface temperature at the outlet, ).(LTs Use (h) to compute g(L)

24

7)0066054.0(

3

14)0066054.0(16

)0066054.081)(CW/m(02564.0

)(m)6-105)(W/m(1500

)(kg/s)8-10426.1)(CJ/kg(3.998

)(m)002.0(m))(105()W/m()1500(2C)(20)(

2

2o

2

o

62oLg

53507.26)(Lg Co

Substitute into (g)

C)(04.23)0066054.0()713.0)(CW/m(02564.0

)(W/m()1500(

14.1

)4.1(4

16

3)0066054.0(

)0066054.081)(CW/m(02564.0

)(W/m()1500(4)( o

o

2

o

2 )(m)-6105)(m)-6105LTs

C7571.26)( oLTs

(iv) Checking.

Dimensional check: computations showed that equations (9.79a), (d), (g), and (h) are dimensionally correct.

Surface temperature check: Application of Newton’s law at the outlet gives

moss TLTLhq )()(

Solving for )(LTs

mos

s TLh

qLT

)()(

Using this equation to compute )(LTs , we obtain


C26.7572C6204.26)CW/m(10969

)W/m()1500()( oo

o2

2

LTs

This is close to the value determined above.


C62.26 omoT . Since the outlet is C62.6 o above the inlet temperature, it follows that the

assumption of isothermal flow is reasonable. (ii) The heat transfer coefficient at the outlet is very high compared to values for air encountered in typical macrochannels applications. (iii) In computing moT and sT , results are presented showing four decimal points. This is done to avoid

errors where temperature differences are small. (iv) The Nusselt number for slip theory for fully

developed macrotube flow is obtained by setting 0oKn in (e). This gives

364.4oNu

Thus macrochannel theory overestimates the Nusselt number if applied to microchannels.

PROBLEM 9.18

Determine the axial variation of the Nusselt number and heat transfer coefficient of the microtube in Problem 9.17.

(1) Observations. (i) The problem is a pressure driven Poiseuille flow through microtube with uniform surface heat flux. (ii) The Nusselt number depends on the Knudsen number, Kn. Since Kn

varies along the tube due to pressure variation, it follows that pressure distribution along the tube must be determined. (iii) Assuming fully developed velocity and temperature, the analysis of Section 9.6.5 gives axial pressure and Nusselt number variation along tube. (iv) The definition of Nusselt number gives the heat transfer coefficient.

(2) Problem Definition. Determine the flow and temperature fields for fully developed Poiseuille tube flow with uniform surface flux.

(3) Solution Plan. Apply the results of Section 9.6.5 for pressure and Nusselt number distribution.

(4) Plan Execution.

(i) Assumptions. (1) Steady state, (2) laminar flow, (3) two-dimensional (no angular variation), (4) slip flow regime (0.001 < Kn < 0.1), (5) ideal gas, (6) constant viscosity and conductivity, (7) negligible radial variation of density and pressure, (8) the accommodation

coefficients are assumed equal to unity, ,0.1Tu (9) negligible dissipation, (10) uniform

surface flux, (11) negligible axial conduction, and (12) no gravity.

(ii) Analysis. The Nusselt number for tube flow heat is defined as

k

hrNu o2

Solving the above for the heat transfer coefficient h

or

kNuh

2 (a)

The Nusselt number, ,Nu is given by (9.99)

KnPr

KnKnKn

KnKn

Nu1

1

4

24

7

3

1416

)81(

1

16

3

)81(

4

2

2

2)(

(9.98)

The local Knudsen number, ,Kn depends on the local pressure p(x) according to

pRT

rrpKnKn

oo

1

222)( (b)

sq

sq

z

orr r


Evaluating (b) at the outlet

ooooo

pRT

rrpKnKn

1

222)( (c)

Equation (9.78) gives )(zp

L

z

p

pKn

p

p

p

pKnKn

p

zp

o

io

o

i

o

ioo

o

)1(16)1(88)(

2

22

(9.78)

Thus, (9.78) is used to determine p(z), (b) to determine ),( pKn (c) gives oKn , (9.98) to determine

the variation of the Nusselt number along the tube, and (a) the heat transfer coefficient.

(iii) Computations. Air properties are determined at .C20o To compute p(x), ),(xKn







C20oio TTT

m-kg/s1017.18 6

713.0Pr

Ks/m287Kkg/J287 22R

C20 ooi TTT

4.1

m/skg1017.18 6

Substituting into(c)

0066054.0K))(293)(K-s/m)(287(2)m-kg/s)(000,100(m)(1010

)m-kg/s(1017.18 22

26-

6

oKn

Using (9.78) and noting that 6/ oi pp gives axial pressure variation

L

z

p

xp

o

)61(0066054.016)6(1)60066054.08(0066054.08)( 22

L

z

p

xp

o

528432.3563691.360528432.0)(

(d)


The local Knudsen number Kn(p) is computed by taking the ratio of (b) and (c)

oo

o

pppp

KnpKn

/

0066054.0

/)( (e)

Using (d), (e) and Nusselt number, and (9.98), results are tabulated below.

(iii) Checking. Dimensional check: Computations showed that units of equations (a), (b), and (9.98) are consistent.

Limiting check: No-slip macrochannel Nusselt number is obtained by setting 0Kn in (9.98).

This gives Nu = 4.364. This agrees with the value given in Table 6.2.

(5) Comments. (i) No-slip Nusselt number for fully developed Poiseuille flow between parallel plates with uniform surface heat flux is Nu = 4.364 (Table 6.2). Thus, no-slip theory overestimates the Nusselt number if applied to microtubes.

(ii) The Nusselt number for fully developed flow is constant along channels. This example shows that in microtubes the Nusselt number vary slightly along the tube.

(iii) It should be noted that the equations used to compute ),(xp and Nu are based on the

assumptions of isothermal conditions in the determination of the flow field. Determining the outlet temperature will give an indication of the validity of this assumption.

6.0

5.3814

z/Lopp/ Kn Nu

1.0

0.2

0.40.6

0.8

0

4.68273.8612

2.8132

1.0

0.001109

0.0012275

0.0014106

0.0017107

0.002348

0.0066054

4.3486

4.3462

4.3429

4.3344

4.2780

)CW/m( o2h

11,150

11,144

11,134

11,113

10,969

4.3501 11,153

PROBLEM 9.19

A micro heat exchanger uses micro tubes of

radius m3or and length mm.6L Inlet air

temperature and pressure are C20 oiT and

pressure kPa.600ip Outlet pressure is

kPa.100op Each tube is maintained at

uniform surface temperature C.60 osT Taking

into consideration velocity slip and temperature

jump and assuming fully developed flow and

temperature, determine the following:

(a) Heat transfer coefficient at the inlet, ),0(h and outlet, ).(Lh

(b) Mean outlet temperature .moT

(1) Observations. (i) This is a pressure driven Poiseuille flow through a tube at uniform surface temperature. (ii) Since the flow field is assumed independent of temperature, it follows that the velocity, mass flow rate and pressure distribution for tubes at uniform surface flux, presented in Section 9.6.6, are applicable to tubes at uniform surface temperature.. (iii) The heat transfer coefficient can be determined if the Nusselt number is known. (iv) The variation of the Nusselt number with Knudsen number for air is shown in Fig. 9.16. (v) The determination of Knudsen number at the inlet and outlet and Fig. 9.16 establish the Nusselt number at these locations. (vi) The use of Fig. 9.16 requires the determination of the Peclet number. (vii) Mean temperature variation along the tube is given by equation (6.13). Application of this equation requires the determination of the average heat transfer coefficient.

(2) Problem Definition. Determine the Nusselt number at the inlet and outlet and the average hat transfer coefficient.

(3) Solution Plan. Compute the Knudsen number at the inlet and outlet, compute the Peclet number, and use Fig. 9.16 to determine the Nusselt number. Use (6.13) to compute the outlet temperature.

(4) Plan Execution.

(i) Assumptions. (1) Steady state, (2) laminar flow, (3) two-dimensional (no angular variation, (4) slip flow regime (0.001 < Kn < 0.1), (5) ideal gas, (6) constant viscosity, conductivity and specific heats, (7) negligible radial variation of density and pressure, (8) the

accommodation coefficients are assumed to be equal to unity, ,0.1Tu (9) negligible

dissipation, (10) uniform surface temperature, and (11) negligible gravity.

(ii) Analysis.

(a) Heat transfer coefficient at inlet, )0(h and outlet, ).(Lh The Nusselt number is used to

determine the heat transfer coefficient. Nusselt number for tube flow is defined as

k

hrNu o2

(a)

where

z

orr r

sTsT


h heat transfer coefficient, CW/m o2

CW/m02564.0tyconductivithermal ok

Nu Nusselt number

or = tube radius = m103 6

Applying (a) at the inlet, z = 0 and solving for h(0)

)0(2

)0( Nur

kh

o

(b)

Similarly, at the outlet, (a) gives

)(2

)( LNur

kLh

o

(c)

L tube length = 0.006 m

The Nusselt number for air at various Peclet numbers and Knudsen numbers is given in Fig. 9.16. The Peclet number is defined as

RePrPe (d)

where

Pr = 0.713


omz rvRe

2 (e)

where

zmv mean velocity, m/s

viscosity = 18.17 CW/m o

density, 3kg/m

Continuity gives

2o

zmr

mu (f)

(f) into (e)

or

mRe

2 (g)

The mass flow rate m is given by equation (9.79a)

)( 116116 2

224

o

io

o

ioo

p

pKn

p

p

LRT

prm (9.798a)

where




K-s/m287K-J/kg287 22R

C20ooi TTT

The Knudsen number at the inlet and outlet is given by

ii

ooii

pRT

rrpKnKn

1

222)( (h)

oo

ooo

pRT

rrpKnKn

1

222)( 0 (i)

(b) Mean outlet temperature, .moT The local mean temperature )(xTm for channel flow at

uniform surface temperature is given by equation (6.13):

][exp)()( zpcm

hPTTTzT smism (6.13)

where


P = channel perimeter = or2 , m


C60 osT

h is the average heat transfer along the tube between inlet and outlet, defined in (6.12)

L

zdzhL

h

0

)(1

(6.12)

An accurate method for computing h requires the determination of the variation of local heat

transfer coefficient, h(z), and evaluating the integral in (6.12) numerically. An approximate approach is to use the arithmetical average. This is justified if h(z) is linear or its change is small.

(iii Computations.

The inlet and outlet Knudson numbers are computed first

001835.0K))(293)(K-s/m)(287(2)m-kg/s)(000,600)(m)(103(2

)m-kg/s(1017.18 22

26-

6

iKn

01101.0K))(293)(K-s/m)(287(2)m-kg/s)(000,100)(m)(103(2

)m-kg/s(1017.18 22

26-

6

oKn


Substitute into (9.79a)

)( 1100000

600000)01101.0(121

100000

600000

K))(293)(K-s/m(287m))(006.0()m-kg/s(1017.18

m-kg/s)000,10((m))103((m)

16

2

226

22236 3

m

910622154.0m kg/s

(g) gives the Reynolds number

7266.7(m)10m)6-(kg/s1017.18

)(kg/s10622154.0)4(66

6

Re


181.5713.07266.7Pe

At the above values of iKn and Pe, Fig. 9.16 gives

75.3)0(Nu

Substitute into (b)

Cm

W910,15)75.3(

)(m)6-103(2

)CW/m(02546.0)0(

o2

o

h

At the outlet Fig. 9.16 gives

68.3)(LNu

(b) gives

Cm

W620,15)68.3(

)(m)6-103(2

)CW/m(02546.0)(

o2

o

Lh

Since the variation of h between inlet and outlet is small, using the arithmetical average heat transfer coefficient in (6.13) is justified

Cm

W765,15)620,15910,15(

2

1)()0(

2

1o2

Lhhh

substituting into (6.13) and setting z = L

][ m)(006.0)kg/s(10622154.0)CJ/kg(998.3

)CW/m(765,15)m)(103(2)C)(6020()C(60

9-o

o26oo expTmo

C60omoT

(iv) Checking.


Dimensional check: computations showed that equations (b), (c), (9.79a), and (6.13) are dimensionally correct.


C60omoT . To improve the solution, an iterative procedure can be followed by repeating the

computation assuming an arithmetical average of mean temperature in the channel equal to

C.40C)]/260(C)(20[ ooo (ii) The heat transfer coefficients are very high compared to values

for air encountered in typical macrotubes applications.

PROBLEM 9.20

Air enters a microtube at temperature C20 oiT and pressure kPa.600ip Outlet pressure is

kPa.100op Tube radius is m1or and its length is mm.2.1L The surface is maintained

at uniform temperature C.40 osT Taking into

consideration velocity slip and temperature jump

and assuming fully developed conditions,

determine the variation along the tube of the following:

(a) Nusselt number, ).(zNu

(b) Heat transfer coefficient, ).(zh

(c) Mean temperature, ).(zTm

(1) Observations. (i) The problem is a pressure driven Poiseuille flow through microtube at uniform surface temperature. (ii) The Nusselt number depends on the Knudsen number, Kn.Since Kn varies along the tube due to pressure variation, it follows that pressure distribution must be determined. (iii) Assuming fully developed velocity and temperature, the analysis of Section 9.6.6 gives axial pressure and Nusselt number variation along the tube. (iv) The definition of Nusselt number gives the heat transfer coefficient. (v) The variation of the Nusselt number with Knudsen number and Peclet number for air is shown in Fig. 9.16. (vii) Mean temperature variation along the tube is given by equation (6.13). Application of this equation requires the determination of the average heat transfer coefficient.

(2) Problem Definition. Determine the flow and temperature fields for fully developed Poiseuille tube flow with uniform surface temperature.

(3) Solution Plan. Apply the results of Section 9.6.6 for pressure and Nusselt number distribution. Use Fig. 9.16 to determine Nusselt number variation along the tube.

(4) Plan Execution.

(i) Assumptions. (1) Steady state, (2) laminar flow, (3) two-dimensional (no angular variation), (4) slip flow regime (0.001 < Kn < 0.1), (5) ideal gas, (6) constant viscosity, conductivity, and specific heat, (7) negligible radial variation of density and pressure, (8) the

accommodation coefficients are assumed equal to unity, ,0.1Tu (9) negligible

dissipation, (10) uniform surface temperature, and (11) no gravity.

(ii) Analysis.

(a) Nusselt number ),(zNu and (b) Heat transfer coefficient, ).(zh The Nusselt number for

tube flow is defined as

k

hrNu o2

Solving the above for the heat transfer coefficient h

or

kNuh

2 (a)

z

orr r

sTsT


The Nusselt number for air at various Peclet numbers and Knudsen numbers is given in Fig. 9.16. The local Knudsen number, ,Kn depends on the local pressure p(z) according to

pRT

rrpKnKn

oo

1

222)( (b)

Equation (9.78) gives )(zp

L

z

p

pKn

p

p

p

pKnKn

p

zp

o

io

o

i

o

ioo

o

)1(16)1(88)(

2

22

(9.78)

The Peclet number is defined as RePrPe (c)

where

Pr = 0.713


omz rvRe

2 (d)

where

zmv mean velocity, m/s

viscosity = skg/m1018.17 -6

density, 3kg/m

Continuity gives

2o

zmr

mu (e)

(e) into (d)

or

mRe

2 (f)

The mass flow rate m is given by equation (9.79a)

)( 116116 2

224

o

io

o

ioo

p

pKn

p

p

LRT

prm (9.79a)

where



K-s/m287K-J/kg287 22R

C20ooi TTT


(b) Mean outlet temperature, .mT The local mean temperature )(zTm for channel flow at

uniform surface temperature is given by equation (6.13)

][exp)()( Lpcm

hPTTTzT smism (6.13)

where

pc specific heat = CJ/kg98.39 o

P = channel perimeter = or2 , m


C40 osT

h is the average heat transfer along the tube between inlet and location z, defined in (6.12)

z

zdzhL

h

0

)(1

(6.12)

An accurate method for computing h requires the determination of the variation of local heat

transfer coefficient, h(z), and evaluating the integral in (6.12) numerically. An approximate approach is to use the arithmetical average. This is justified if h(z) is linear or its change is small.

Thus, (9.78) is used to determine p(z), (b) to determine ),( pKn (c) gives the Peclet number, Fig.

9.16 gives the variation of the Nusselt number along the tube, (a) the heat transfer coefficient, and (6.13) the mean temperature.

(iii) Computations. Air properties are determined at .C20o To compute p(z), ),(zKn




op outlet pressure m-kg/s000,100kPa100 2



C20oio TTT

m-kg/s1017.18 6

713.0Pr

Ks/m287Kkg/J287 22R

C20 ooi TTT

4.1

m/skg1017.18 6

Evaluating (b) at the outlet


033027.0K))(293)(K-s/m)(287(2)m-kg/s)(000,100(m)(102

)m-kg/s(1017.18 22

26-

6

oKn

Noting that 6/ oi pp , (9.79a) gives m

)16)(033027.0(161)6(K))(293)(K-s/m(287m))(0012.0()m-kg/s(1017.18

m-kg/s)000,100((m))101(

16

2

226

22246

m

kg/s1010403.0m

Substituting into (f) gives the Reynolds number

412.1(m)10m)2-(kg/s1017.18

)(kg/s10403.0)4(66

10

Re


007.1713.0412.1Pe

Using (9.78) and noting that 6/ oi pp gives axial pressure variation

L

z

p

xp

o

)61(033027.016)6(1)6033027.08(033027.08)( 22

L

z

p

xp

o

64216.37240402.39264216.0)(

(g)

The local Knudsen number Kn(p) is given by

oo

o

pppp

KnpKn

/

033027.0

/)( (h)

With 1Pe , equations (g), (h), Fig. 9.16, and (a) are used to compute and tabulate pressure,

Knudsen number, Nusselt number, and heat transfer coefficient, respectively, as functions of z/L.

6.0

5.3814

z/L opp/ Kn Nu

1.0

0.2

0.40.6

0.8

0

4.68273.8612

2.8132

1.0

)CW/m( o2h

0.0055505

0.0061536

0.0070972

0.0086528

0.0119812

0.033027

4.01

4.0

3.99

3.97

3.95

3.75

51,410

51,280

51,150

51,900

50,640

48,080

C)( ozT

40

40

40

40

40

40


Since the variation of h between inlet and outlet is small, using the arithmetical average heat transfer coefficient in (6.13) is justified. For determining the outlet temperature, we set

Cm

W745,49)080,48410,51(

2

1)()0(

2

1o2

Lhhh

To determine the highest mean temperature, (6.13) is applied at z = L

][ m)(0012.0)kg/s(10403.0)CJ/kg(998.3

)CW/m(745,49)m)(101(2)C)(4020()C(40

10-o

o26oo expmoT

C40omoT

Application (6.13) at other values of z gives C40)( ozTm . At z = 0, (6.13) gives C20omoT .

(iii) Checking. Dimensional check: Computations showed that units of equations (a), (f), (g), (h), and (9.79a) are dimensionally consistent.

(5) Comments. (i) No-slip Nusselt number for fully developed Poiseuille flow through tubes at uniform surface temperature is Nu = 3.656 (Table 6.2). Thus, no-slip theory overestimates the Nusselt number if applied to microtubes.

(ii) The Nusselt number for fully developed flow in macrotubes is constant. This example shows that for microtubes the Nusselt number varies along the tube.

(iii) It should be noted that the equations used to compute ),(xp and Nu are based on the

assumptions of isothermal conditions in the determination of the flow field. Determining the outlet temperature will give an indication of the validity of this assumption.

(v) The fluid equilibrates with surface temperature very close to the inlet and remains essentially at constant temperature, equal to throughout the tube. At a distance mz 1 , fluid temperature

increases to 39.98 C.o

Problem 1.1

Heat is removed from the surface by convection. Therefore, Newton's law of cooling is applicable.

Ambient temperature and heat transfer coefficient are uniform

Surface temperature varies along the rectangle.

Problem 1.2

Heat is removed from the surface by convection. Therefore, Newton's law of cooling may be helpful.

Ambient temperature and surface temperature are uniform.

Surface area and heat transfer coefficient vary along the triangle.

Problem 1.3

Heat flux leaving the surface is specified (fixed).

Heat loss from the surface is by convection and radiation.

Convection is described by Newton's law of cooling.

Changing the heat transfer coefficient affects temperature distribution.

Surface temperature decreases as the heat transfer coefficient is increased.

Surface temperature gradient is described by Fourier’s law

Ambient temperature is constant.

Problem 1.3

Metabolic heat leaves body at the skin by convection and radiation.


Fanning increases the heat transfer coefficient and affects temperature distribution, including surface temperature.


Surface temperature is described by Newton’s law of cooling.


Problem 1.4

Metabolic heat leaves body at the skin by convection and radiation.


Fanning increases the heat transfer coefficient and affects temperature distribution, including surface temperature.


Surface temperature is described by Newton’s law of cooling.

) Ambient temperature is constant.

Problem 1.5

Melting rate of ice depends on the rate of heat added at the surface.

Heat is added to the ice from the water by convection.

Newton's law of cooling is applicable.

Stirring increases surface temperature gradient and the heat transfer coefficient. An increase in gradient or h increases the rate of heat transfer.

Surface temperature remains constant equal to the melting temperature of ice.

Water temperature is constant.

Problem 1.6

This problem is described by cylindrical coordinates.

For parallel streamlines 0vvr .

Axial velocity is independent of axial and angular distance.

Problem 1.7


Streamlines are concentric circles. Thus the velocity component in the radial direction

vanishes ( ).0rv

For one-dimensional flow there is no motion in the z-direction ( 0zv ).

The -velocity component, , depends on distance r and time t.v

Problem 1.8

This problem is described by Cartesian coordinates.

For parallel streamlines the y-velocity component 0v .

For one-dimensional flow there is no motion in the z-direction (w = 0).

The x-velocity component depends on distance y and time t.

Problem 1.9

This problem is described by Cartesian coordinates.

For parallel streamlines the y-velocity component 0v .

For one-dimensional flow there is no motion in the z-direction (w = 0).

The x-velocity component depends on distance y only.

Problem 1.10

Heat flux leaving the surface is specified (fixed).

Heat loss from the surface is by convection and radiation


Changing the heat transfer coefficient affects temperature distribution.


Surface temperature gradient is described by Fourier’s law


Problem 1.11

Heat is removed from the surface by convection. Therefore, Newton's law of cooling is applicable.

Ambient temperature and heat transfer coefficient are uniform.

Surface temperature varies along the area.

The area varies with distance x.

Problem 2.2

The fluid is incompressible.

Radial and tangential velocity components are zero.

Streamlines are parallel.

Cylindrical geometry.

Problem 2.3


axial velocity is invariant with axial distance.

Plates are parallel.

Cartesian geometry.

Problem 2.4


Radial and tangential velocity components are zero.

Streamlines are parallel.

Cylindrical geometry.

Problem 2.5

Shearing stresses are tangential surface forces.

xy and yx are shearing stresses in a Cartesian coordinate system.

Tangential forces on an element result in angular rotation of the element.

If the net external torque on an element is zero its angular acceleration will vanish.

Problem 2.6

Properties are constant.

Cartesian coordinates.

Parallel streamlines: no velocity component in the y-direction.

Axial flow: no velocity component in the z-direction.

The Navier-Stokes equations give the three momentum equations.

Problem 2.7


Cylindrical coordinates.

Parallel streamlines: no velocity component in the r-direction.

Axial flow: no velocity component in the -direction.

No variation in the -direction. The Navier-Stokes equations give the three momentum

equations.

Problem 2.8



Two dimensional flow (no velocity component in the z-direction

The Navier-Stokes equations give two momentum equations.

Problem 2.9


Cylindrical coordinates.

Two dimensional flow (no velocity component in the -direction.

The Navier-Stokes equations give two momentum equations.

Problem 2.10

Motion in energy consideration is represented by velocity components.

Fluid nature is represented by fluid properties.

Problem 2.11





Problem 2.12





The fluid is an ideal gas.

Problem 2.13

This is a two-dimensional free convection problem.

The flow is due to gravity.

The flow is governed by the momentum and energy equations. Thus the governing equations are the Navier-Stokes equations of motion and the energy equation.

The geometry is Cartesian.

Problem 2.15

The flow is due to gravity.

For parallel streamlines the velocity component v = 0 in the y-direction.

Pressure at the free surface is uniform (atmospheric).



Problem 2.16

This is a forced convection problem.

Flow properties (density and viscosity) are constant.

Upstream conditions are uniform (symmetrical)

The velocity vanishes at both wedge surfaces (symmetrical).

Surface temperature is asymmetric.

Flow field for constant property fluids is governed by the Navier-Stokes and continuity equations.

If the governing equations are independent of temperature, the velocity distribution over the wedge should be symmetrical with respect to x.


Problem 2.18



Axial flow (no motion in the z-direction).

Parallel streamlines means that the normal velocity component is zero.

Specified flux at the lower plate and specified temperature at the upper plate.

Problem 2.19

The geometry is cylindrical.

No variation in the axial and angular directions.


Problem 2.20

The geometry is cylindrical.

No variation in the angular direction.


Parallel streamlines means that the radial velocity component is zero.

Problem 2.21

The geometry is cylindrical. (ii)

No variation in the axial and angular directions.


Problem 2.22


The same fluid flows over both spheres.

Sphere diameter and free stream velocity affect the Reynolds number which in turn affect the heat transfer coefficient.

Problem 2.23

This is a free convection problem.

The average heat transfer coefficient h depends on the vertical length L of the plate.

L appears in the Nusselt number as well as the Grashof number.

Problem 2.24


The same fluid flows over both spheres.

Sphere diameter and free stream velocity affect the Reynolds number which in turn affect the heat transfer coefficient. (iv) Newton’s law of cooling gives the heat transfer

Problem 2.25

Dissipation is important when the Eckert number is high compared to unity.

If the ratio of dissipation to conduction is small compared to unity, it can be neglected.

Problem 2.26

The plate is infinite.

No changes take place in the axial direction (infinite plate).

This is a transient problem.

Constant properties.


Problem 2.27

The plate is infinite.

No changes take place in the axial direction (infinite plate).

This is a transient problem.

Constant properties.


Gravity is neglected. Thus there is no free convection.

The fluid is stationary.

Problem 3.1

Moving plate sets fluid in motion in the x-direction.

Since plates are infinite the flow field does not vary in the axial direction x.

The effect of pressure gradient is negligible.

The fluid is incompressible (constant density).

Use Cartesian coordinates.

Problem 3.2



The effect of pressure gradient must be included.


Using Fourier’s law, Temperature distribution gives surface heat flux of the moving plate.


Problem 3.3





Problem 3.4

Moving plates set fluid in motion in the positive and negative x-direction.



The fluid is stationary at the center plane y = 0.

Symmetry dictates that no heat is conducted through the center plane.


Problem 3.5

Fluid motion is driven by axial pressure drop.

For a very long tube the flow field does not vary in the axial direction z.


Heat is generated due to viscous dissipation. It is removed from the fluid by convection at the surface.

The Nusselt number is a dimensionless heat transfer coefficient.

To determine surface heat flux and heat transfer coefficient requires the determination of temperature distribution.

Temperature distribution depends on the velocity distribution.

Use cylindrical coordinates.

Problem 3.6

Fluid motion is driven by axial pressure drop.




Problem 3.7

Fluid motion is driven by axial motion of the rod. Thus motion is not due to pressure gradient.



Heat is generated due to viscous dissipation. It is removed from the fluid by conduction at the surface.

The Nusselt number is a dimensionless heat transfer coefficient.

To determine the heat transfer coefficient require the determination of temperature distribution.



Problem 3.8

Fluid motion is driven by gravity.

No velocity and temperature variation in the axial direction.


Heat is generated due to viscous dissipation.



Problem 3.9

Fluid motion is driven by gravity.

No velocity and temperature variation in the axial direction.


Heat is generated due to viscous dissipation.


the inclined surface is at specified temperature and the free surface exchanges heat by convection with the ambient.


Problem 3.11

Fluid motion is driven by shaft rotation

The housing is stationary.

Axial variation in velocity and temperature are negligible for a very long shaft.

Velocity and temperature do not vary with angular position.


Heat generated by viscous dissipation is removed from the oil at the housing.

No heat is conducted through the shaft.

The maximum temperature occurs at the shaft.

Heat flux at the housing is determined from temperature distribution and Fourier’s law of conduction.


Problem 3.12

Fluid motion is driven by sleeve rotation

The shaft is stationary.


Velocity and temperature do not vary with angular position.


Heat generated by viscous dissipation is removed from the oil at the housing.

No heat is conducted through the shaft.

The maximum temperature occurs at the shaft. (ix) Use cylindrical coordinates.

Problem 3.13

Fluid motion is driven by shaft rotation


Velocity, pressure and temperature do not vary with angular position.


Heat generated by viscous dissipation is conducted radially.

The determination of surface temperature and heat flux requires the determination of temperature distribution in the rotating fluid.


Problem 3.14

Axial pressure gradient sets fluid in motion.


The flow field is determined by solving the continuity and Navier-Stokes equations.

Energy equation gives the temperature distribution.

Fourier’s law and temperature distribution give surface heat flux.

Axial variation of temperature is neglected.

Problem 4.3

This is forced convection flow over a streamlined body.

Viscous (velocity) boundary layer approximations can be made if the Reynolds number Rex > 100.

Thermal (temperature) boundary layer approximations can be made if the Peclet number Pex = Rex Pr > 100.

The Reynolds number decreases as the distance along the plate is decreased.

Problem 4.4

The surface is streamlined.

The fluid is water.

Inertia and viscous effects can be estimated using scaling.

If a viscous term is small compared to inertia, it can be neglected.

Properties should be evaluated at the film temperature .2/)( TTT sf

Problem 4.5


The fluid is water.

Convection and conduction effects can be estimated using scaling.

If a conduction term is small compared to convection, it can be neglected.

The scale for Lt / depends on whether t or .t

Properties should be evaluated at the film temperature .2/)( TTT sf

Problem 4.6

The fluid is air.

Dissipation and conduction can be estimated using scaling.

Dissipation is negligible if the Eckert number is small compared to unity.

Problem 4.7


The fluid is air.

Problem 4.9

This is a forced convection problem over a flat plate.

At the edge of the thermal boundary layer, the axial velocity is Vu .

Blasius solution gives the distribution of the velocity components u(x,y) and v(x,y).

Scaling gives an estimate of v(x,y).

Problem 4.11

This is a laminar boundary layer flow problem.

Blasius solution gives the velocity distribution for the flow over a semi-infinite flat plate. (iii) A solution for the boundary layer thickness depends on how the thickness is defined.

Problem 4.12

Since the flow within the boundary layer is two-dimensional the vertical velocity component does not vanish. Thus stream lines are not parallel.

Blasius solution is valid for laminar boundary layer flow over a semi-infinite plate.

The transition Reynolds number from laminar to turbulent flow is .5105

Boundary layer approximations are valid if the Reynolds number is greater than 100.

Problem 4.13

This is an external flow problem over a flat plate.

Blasius’s solution for the velocity distribution and wall shearing stress is assumed to be applicable.

Of interest is the value of the local stress at the leading edge of the plate.

Problem 4.14

This is an external flow problem over a flat plate.

The force needed to hold the plate in place is equal to the total shearing force by the fluid on the plate.

Integration of wall shear over the surface gives the total shearing force.

Blasius’s solution for the velocity distribution and wall shearing stress is assumed to be applicable.

Problem 4.16

This is an external forced convection problem for flow over a flat plate.

Of interest is the region where the upstream fluid reaches the leading edge of the plate.

The fluid is heated by the plate.

Heat from the plate is conducted through the fluid in all directions.

Pohlhausen’s solution assumes that heat is not conducted upstream from the plate and therefore fluid temperature at the leading edge is the same as upstream temperature.

Problem 4.18

This is a forced convection problem over a flat plate.

At the edge of the thermal boundary layer, fluid temperature is TT .

Pohlhausen’s solution gives the temperature distribution in the boundary layer.

The thermal boundary layer thickness t increases with distance from the leading edge.

t depends on the Prandtl number.

Problem 4.19


Pohlhausen’s solution for the temperature distribution and heat transfer coefficient is assumed to be applicable.

Of interest is the value of the local heat flux at the leading edge of the plate.

Knowing the local transfer coefficient and using Newton’s law, gives the heat flux

Problem 4.20


Pohlhausen’s solution for the temperature distribution is assumed to be applicable.

Of interest is the value of the normal temperature gradient at the surface.

Problem 4.22`

This is an external forced convection problem over two flat plates.

Both plates have the same surface area.

For flow over a flat plate, the heat transfer coefficient h decreases with distance from the leading edge.

Since the length in the flow direction is not the same for the two plates, the average heat transfer coefficient is not the same. It follows that the total heat transfer rate is not the same.

The flow over a flat plate is laminar if the Reynolds number is less than 5 105.

Problem 4.23



Of interest is the value of the heat transfer rate from a section of the plate at a specified location and of a given width.

Newton’s law of cooling gives the heat transfer rate.

Problem 4.24


Of interest is the variation of the local heat transfer coefficient with free stream velocity and distance from the leading edge.

Pohlhausen's solution applies to this problem.

Problem 4.25

This is an external flow problem.

At the edge of the thermal boundary layer, ty , fluid temperature approaches free

stream temperature. That is, TT and .1)/()(*ss TTTTT

According to Pohlhausen's solution, Fig. 4.6, the thermal boundary layer thickness

depends on the Prandtl number, free stream velocity V , kinematic viscosity and location x.

Problem 4.26


The Reynolds number and Peclet number should be checked to determine if the flow is laminar and if boundary layer approximations are valid.

Pohlhausen's solution is applicable if 100 < Rex < 100 105 and Pex = Rex Pr > 100.

Thermal boundary layer thickness and heat transfer coefficient vary along the plate.

Newton’s law of cooling gives local heat flux. (vi) The fluid is water.

Problem 4.27

This is an external forced convection problem over a flat plate.

Increasing the free stream velocity, increases the average heat transfer coefficient. This in turn causes surface temperature to drop.

Based on this observation, it is possible that the proposed plan will meet design specification.

Since the Reynolds number at the downstream end of the package is less than 500,000, it follows that the flow is laminar throughout.

Increasing the free stream velocity by a factor of 3, increases the Reynolds number by a factor of 3 to 330,000. At this Reynolds number the flow is still laminar.

The power supplied to the package is dissipated into heat and transferred to the surroundings from the surface.

Pohlhausen's solution can be applied to this problem.

The ambient fluid is unknown.

Problem 4.28

This is an external forced convection problem of flow over a flat plate.

Convection heat transfer from a surface can be determined using Newton’s law of cooling.

The local heat transfer coefficient changes along the plate. The total heat transfer rate can be determined using the average heat transfer coefficient.

For laminar flow, Pohlhausen's solution gives the heat transfer coefficient.

For two in-line fins heat transfer from the down stream fin is influenced by the upstream fin. The further the two fins are apart the less the interference will be.

Problem 4.29



Knowing the heat transfer coefficient, the local Nusselt number can be determined.

the Newton’s law of cooling gives the heat transfer rate.

Pohlhausen’s solution gives the thermal boundary layer thickness.

Problem 4.30



The local heat transfer coefficient changes along the plate.

For each triangle the area changes with distance along the plate.

The total heat transfer rate can be determined by integration along the length of each triangle.

Pohlhausen's solution may be applicable to this problem.

Problem 4.31


Heat transfer rate can be determined using Newton’s law of cooling.


The area changes with distance along the plate.

The total heat transfer rate can be determined by integration along the length of the triangle.


Problem 4.32





The total heat transfer rate can be determined by integration along over the area of the semi-circle.

Pohlhausen's solution gives the heat transfer coefficient.

Problem 4.33







Problem 4.34


This problem involves determining the heat transfer rate from a circle tangent to the leading edge of a plate






Problem 4.36

The flow field for this boundary layer problem is simplified by assuming that the axial velocity is uniform throughout the thermal boundary layer.

Since velocity distribution affects temperature distribution, the solution for the local Nusselt number can be expected to differ from Pohlhausen’s solution.

The Nusselt number depends on the temperature gradient at the surface.

Problem 4.37

The flow field for this boundary layer problem is simplified by assuming that the axial velocity varies linearly in the y-direction.

Since velocity distribution affects temperature distribution, the solution for the local Nusselt number can be expected to differ from Pohlhausen’s solution.

The Nusselt number depends on the temperature gradient at the surface.

Problem 4.38

The flow and temperature fields for this boundary layer problem are simplified by assuming that the axial velocity and temperature do not vary in the x-direction.

The heat transfer coefficient depends on the temperature gradient at the surface.

Temperature distribution depends on the flow field.

The effect of wall suction must be taken into consideration.

Problem 4.39

This is a forced convection flow over a plate with variable surface temperature.

The local heat flux is determined by Newton’s law of cooling.

The local heat transfer coefficient and surface temperature vary with distance along the plate. The variation of surface temperature and heat transfer coefficient must be such that Newton’s law gives uniform heat flux.

The local heat transfer coefficient is obtained from the local Nusselt number.

Problem 4.40


The Reynolds number should be computed to determine if the flow is laminar or turbulent.

The local heat transfer coefficient and surface temperature vary with distance along the plate.

The local heat transfer coefficient is obtained from the solution to the local Nusselt number.

The determination of the Nusselt number requires determining the temperature gradient at the surface.

Problem 4.41


The Reynolds number should be computed to determine if the flow is laminar or turbulent.

Newton’s law of cooling gives the heat transfer rate from the plate.

The local heat transfer coefficient and surface temperature vary with distance along the plate. Thus determining the total heat transfer rate requires integration of Newton’s law along the plate.

The local heat transfer coefficient is obtained from the local Nusselt number.

Problem 4.42

This is an external forced convection problem of flow over a flat plate


The local heat transfer coefficient and surface temperature vary along the plate.

For each triangle the area varies with distance along the plate.

The total heat transfer rate can be determined by integration along the length of each triangle.

Problem 4.43

This is a forced convection boundary layer flow over a wedge.

Wedge surface is maintained at uniform temperature.

The flow is laminar.

The fluid is air.

Similarity solution for the local Nusselt number is presented in Section 4.4.3.

The Nusselt number depends on the Reynolds number and the dimensionless temperature gradient at the surface ./)0( dd (vii) Surface temperature gradient depends on wedge

angle.

Problem 4.44

This is a forced convection boundary layer flow over a wedge.

Wedge surface is maintained at uniform temperature.

The flow is laminar.

The average Nusselt number depends on the average heat transfer coefficient..

Similarity solution for the local heat transfer coefficient is presented in Section 4.4.3.

Problem 4.45

Newton’s law of cooling gives the heat transfer rate from a surface.

Total heat transfer from a surface depends on the average heat transfer coefficient h .

Both flat plate and wedge are maintained at uniform surface temperature.

Pohlhausen’s solution gives h for a flat plate.

Similarity solution for the local heat transfer coefficient for a wedge is presented in Section 4.4.3.

Problem 4.46

The flow field for this boundary layer problem is simplified by assuming that axial velocity within the thermal boundary layer is the same as that of the external flow.

Since velocity distribution affects temperature distribution, the solution for the local Nusselt number differs from the exact case of Section 4.4.3.

The local Nusselt number depends the local heat transfer coefficient which depends on the temperature gradient at the surface.

Problem 4.47

The flow field for this boundary layer problem is simplified by assuming that axial velocity within the thermal boundary layer varies linearly with the normal distance.

Since velocity distribution affects temperature distribution, the solution for the local Nusselt number differs from the exact case of Section 4.4.3.

The local Nusselt number depends on the local heat transfer coefficient which depends on the temperature gradient at the surface.

Problem 5.1

In general, to determine the Nusselt number it is necessary to determine the velocity and temperature distribution

Fluid velocity for 1Pr is assumed to be uniform, Vu . This represents a

significant simplification.

Problem 5.2

In general, to determine the Nusselt number it is necessary to determine the velocity and temperature distribution.

Fluid velocity for 1Pr is assumed to be linear, )/(yVu .

Problem 5.3

The velocity is assumed to be uniform, ,Vu throughout the thermal boundary layer.

A leading section of length is unheated.ox

at , surface heat flux is uniform. oxx

The determination of the Nusselt number requires the determination of the temperature distribution.

Surface temperature is unknown.

The maximum surface temperature for a uniformly heated plate occurs at the trailing end.

Problem 5.4

The velocity distribution is known.

Surface temperature is uniform.


Newton’s law of cooling gives the heat transfer rate. This requires knowing the local heat transfer coefficient.

Problem 5.5

The velocity distribution is known

Total heat transfer is equal to heat flux times surface area.

Heat flux is given. However, the distance x = L at which 2/Ht is unknown.

Problem 5.6

The determination of the Nusselt number requires the determination of the velocity and temperature distributions.

Velocity is assumed uniform.

Surface temperature is variable.

Newton’s law of cooling gives surface heat flux. This requires knowing the local heat transfer coefficient.

Problem 5.7


Velocity is assumed linear.



Problem 5.8


Velocity is assumed uniform.



Problem 5.9


Surface heat flux is variable. It decreases with distance x.


Newton’s law of cooling gives surface temperature. This requires knowing the local heat

transfer coefficient. (v) .1/t

Problem 5.10




Surface heat flux is variable. It increases with distance x.

Surface temperature is unknown. Since flux increases with x and heat transfer coefficient decreases with x, surface temperature is expected to increase with x. Thus maximum surface temperature is at the trailing end x = L.

Newton’s law of cooling gives surface temperature. This requires knowing the local heat transfer coefficient.

Problem 5.11




Surface heat flux is variable. It increases with distance x.

Surface temperature is unknown. Since flux increases with x and heat transfer coefficient decreases with x, surface temperature is expected to increase with x. Thus maximum surface temperature is at the trailing end x = L.

Newton’s law of cooling gives surface temperature. This requires knowing the local heat transfer coefficient.

Problem 5.12


Velocity variation with y is negligible.

Conservation of mass requires that radial velocity decrease with radial distance r.


Problem 5.13


Velocity variation with y is negligible.

Conservation of mass requires that radial velocity decrease with radial distance r.

Surface heat flux is uniform


Problem 5.14




The plate is porous.

Fluid is injected through the plate with uniform velocity.

The plate is maintained at uniform surface temperature.

A leading section of the plate is insulated.

Problem 5.15




The plate is porous.

Fluid is injected through the plate with uniform velocity.

The plate is heated with uniform surface flux

Surface temperature is unknown, (vii) A leading section of the plate is insulated.

Problem 5.16

There are two thermal boundary layers in this problem.

The upper and lower plates have different boundary conditions. Thus, temperature distribution is not symmetrical.

The lower plate is at uniform temperature while heat is removed at uniform flux along the upper plate.

Fluid velocity is assumed uniform throughout the channel.

Problem 6.1

This is an internal forced convection problem.

Scaling gives estimates of and hL .tL

Exact solutions for and are available for laminar flow through channels. hL tL

Exact solutions for depend on channel geometry and surface boundary conditions. tL

Problem 6.2


Scaling gives estimates of andhL

Exact solutions for and are available for laminar flow through channels. hL tL

Exact solutions for depend on channel geometry and surface boundary conditions. tL

Problem 6.3

This is an internal force convection problem.

The channel is a long tube.

The surface is maintained at a uniform temperature.

Since the tube section is far away from the entrance, the velocity and temperature can be assumed fully developed.

Tube diameter, mean velocity and inlet, outlet and surface temperatures are known. The length is unknown.

The fluid is air.

Problem 6.4

This is an internal force convection in a tube.

The surface is heated at uniform flux.

Surface temperature increases along the tube and is unknown.

The flow is assumed laminar and fully developed.

The heat transfer coefficient for fully developed flow through channels is constant.

According to Newton’s law of cooling, surface temperature is related to mean fluid temperature, surface heat flux and heat transfer coefficient.

Problem 6.5

This is an internal force convection in a tube.


Surface temperature increases along the tube and is unknown.

The flow is assumed laminar and fully developed.

The heat transfer coefficient for fully developed flow through channels is constant.

According to Newton’s law of cooling, surface temperature is related to mean fluid temperature, surface heat flux and heat transfer coefficient.

Problem 6.6

This is an internal forced convection problem in a tube.


Surface temperature changes along the tube and is unknown.

The Reynolds number should be checked to determine if the flow is laminar or turbulent.

If hydrodynamic and thermal entrance lengths are small compared to tube length, the flow can be assumed fully developed throughout.

For fully developed flow, the heat transfer coefficient is uniform.

The length of the tube is unknown.

The fluid is water.

Problem 6.7


The channel is a tube.


Entrance effect is important in this problem.

The average Nusselt number for a tube of length L depends on the average heat transfer coefficient over the length.

Problem 6.8


The fluid is heated at uniform wall flux.

Surface temperature changes with distance along the channel. It reaches a maximum value at the outlet.

The Reynolds and Peclet numbers should be checked to establish if the flow is laminar or turbulent and if this is an entrance or fully developed problem.

The channel has a square cross-section.

Application of Newton’s law of cooling at the outlet relates outlet temperature to surface temperature, surface flux and heat transfer coefficient.

Application of conservation of energy gives a relationship between heat added, inlet temperature, outlet temperature, specific heat and mass flow rate.

Problem 6.9

This is an internal forced convection problem in tubes.

The flow is laminar and fully developed.

The surface is maintained at uniform temperature.

All conditions are identical for two experiments except the flow rate through one is half that of the other.

The total heat transfer rate depends on the outlet temperature.

Problem 6.10


The channel has a rectangular cross section.


The Reynolds and Peclet numbers should be checked to establish if the flow is laminar or turbulent and if entrance effects can be neglected.

Channel length is unknown.

The fluid is air.

Problem 6.11


The channel is a rectangular duct.


The velocity and temperature are fully developed.


Duct size, mean velocity and inlet, outlet and surface temperatures are known. The

length is unknown. (vii) Duct length depends on the heat transfer coefficient.

The fluid is water.

Problem 6.12

This is an internal forced convection problem in a channel.


Surface temperature changes along the channel. It reaches a maximum value at the outlet.


Velocity and temperature profiles become fully developed far away from the inlet.

The heat transfer coefficient is uniform for fully developed flow.

The channel has a square cross section.

tube length is unknown. (ix) The fluid is air.

Problem 6.13

This is an internal forced convection problem in tubes.

The flow is laminar and fully developed.


All conditions are identical for two tubes except the diameter of one is twice that of the other.

The total heat transfer in each tube depends on the outlet temperature.

Problem 6.14


Equation (6.3) gives scaling estimate of the thermal entrance length.

Equation (6.20b) gives scaling estimate of the local Nusselt number.

The Graetz problem deals with laminar flow in the entrance of a tube at uniform surface temperature.

Graetz solutions gives the thermal entrance length (distance to reach fully developed temperature) and local Nusselt number.

Problem 6.15


Equation (6.20b) gives scaling estimate of the local Nusselt number.

The Graetz problem deals with laminar flow in the entrance of a tube at uniform surface temperature.

Problem 6.16


The velocity is fully developed.

The temperature is developing.

Surface is maintained at uniform temperature.

The Reynolds number should be computed to establish if flow is laminar or turbulent.

Tube length is unknown.

The determination of tube length requires determining the heat transfer coefficient.

Problem 6.17




Surface is maintained at uniform temperature.


Outlet mean temperature is unknown.

The determination of outlet temperature requires determining the heat transfer coefficient.

Since outlet temperature is unknown, air properties can not be determined. Thus a trial and error procedure is needed to solve the problem.

Problem 6.18


The velocity is fully developed and the temperature is developing.

The surface is heated with uniform flux.

The Reynolds number should be computed to establish if the flow is laminar or turbulent.

Compute thermal entrance length to determine if it can be neglected.

Surface temperature varies with distance from entrance. It is maximum at the outlet. Thus surface temperature at the outlet is known.

Analysis of uniformly heated channels gives a relationship between local surface temperature, heat flux and heat transfer coefficient.

The local heat transfer coefficient varies with distance form the inlet.

Knowing surface heat flux, the required power can be determined.

Newton’s law of cooling applied at the outlet gives outlet temperature.

Problem 6.19

This is an internal forced convection problem in a rectangular channel.

The velocity is fully developed and the temperature is developing.



Compute entrance lengths to determine if they can be neglected

Surface flux varies with distance from entrance. It is minimum at outlet.

Newton’s law gives surface flux in terms of the local heat transfer coefficient and

the local mean temperature

)(xh

)(xTm .

The local and average heat transfer coefficient decrease with distance form the inlet.

The local mean temperature depends on the local average heat transfer coefficient

).(xh (x) Surface temperature is unknown.

Problem 7.2

This is an external free convection problem over a vertical plate.

The Rayleigh number should be computed to determine if the flow is laminar or turbulent.

The solution for laminar flow is given in Section 7.4

For laminar flow, Fig.7.2 gives the viscous boundary layer thickness and Fig. 7.3 gives

the thermal boundary layer thickness t .


Equation (7.23) gives the average heat transfer coefficient h . (vii) The fluid is water.

Problem 7.3

This is an external free convection problem for flow over a vertical plate.

Laminar flow solution for temperature distribution for a plate at uniform surface temperature is given in Fig. 7.3 .

The dimensionless temperature gradient at the surface is given in Table 7.1.

The solution depends on the Prandtl number.

Problem 7.4


Heat is lost from the door to the surroundings by free convection and radiation.

To determine the rate of heat loss, the door can by modeled as a vertical plate losing heat by free convection to an ambient air.

As a first approximation, radiation can be neglected.

Newton’s law of cooling gives the rate of heat transfer.


For laminar flow the solution of Section 7.4 is applicable.

Problem 7.5

This is a free convection and radiation problem.

The geometry is a vertical plate.


Newton’s law of cooling gives convection heat transfer rate while Stefan-Boltzmann law gives radiation heat transfer rate.



Since radiation heat transfer is considered in this problem, all temperatures should be expressed

Problem 7.6


The power dissipated in the electronic package is transferred to the ambient fluid by free convection.

As the power is increased, surface temperature increases.

The maximum power dissipated corresponds to the maximum allowable surface temperature.

Surface temperature is related to surface heat transfer by Newton’s law of cooling.

The problem can be modeled as free convection over a vertical plate.



The fluid is air.

Problem 7.7









The fluid is water.

Problem 7.8



Newton’s law of cooling determines the heat transfer rate.

Heat transfer rate depends on the heat transfer coefficient.

The heat transfer coefficient decreases with distance from the leading edge of the plate.

The width of each triangle changes with distance from the leading edge.


Problem 7.9

This is a free convection problem over a vertical plate.


Local heat flux is determined by Newton’s law of cooling.

Heat flux depends on the local heat transfer coefficient

Free convection heat transfer coefficient for a vertical plate decreases with distance from the leading edge. Thus, the flux also decreases.

The Rayleigh number should be computed to determine if the flow is laminar or turbulent. For

Laminar flow the solution of Section 7.4 is applicable.

The fluid is air.

Problem 7.10


The power dissipated in the chips is transferred to the air by free convection.

This problem can be modeled as free convection over a vertical plate with constant surface heat flux.

Surface temperature increases as the distance from the leading edge is increased. Thus, the maximum surface temperature occurs at the top end of the plate (trailing end).


For laminar flow the analysis of Section 7.5 gives surface temperature distribution.

The fluid is air.

Properties depend on the average surface temperature . Since is unknown, the sT sT

problem must be solved by trail and error.

Problem 7.11


The power dissipated in the chips is transferred to the air by free convection




For laminar flow the analysis of Section 7.5 gives surface temperature distribution.

The fluid is air.

Properties depend on the average surface temperature . Since is unknown, the

problem must be solved by trail and error.

sT sT

Problem 7.12

This is a free convection problem over a vertical plate at uniform surface temperature.


The integral method can be used to determine the velocity and temperature distribution.

Application of the integral method reduces to determining the velocity and temperature boundary layer thickness.

Problem 7.13

This is a free convection problem over a vertical plate at uniform surface heat flux.


The integral method can be used to determine the velocity and temperature distribution.

Application of the integral method reduces to determining the velocity and temperature boundary layer thickness.

Problem 8.1

This is an external forced convection problem.

The geometry can be modeled as a flat plate.


Newton’s law of cooling gives heat transfer rate from the surface to the air.

The average heat transfer coefficient must be determined.

The Reynolds number should be evaluated to establish if the flow is laminar, turbulent or mixed.

Analytic or correlation equations give the heat transfer coefficient.

Problem 8.2


The geometry can be modeled as a flat plate.


To determine the heat flux at a given location, the local heat transfer coefficient must be determined.

The average heat transfer coefficient is needed to determine the total heat transfer rate.

Newton’s law of cooling gives surface flux and total heat transfer rate.

The Reynolds number should be checked to establish if the flow is laminar, turbulent or mixed.


Problem 8.3


Surface temperature is assumed uniform.

The heat transfer coefficient in turbulent flow is greater than that in laminar flow. Thus higher heat transfer rates can be sustained in turbulent flow than laminar flow.


Heat loss from the surface is approximately equal to the power dissipated in the package.

Newton’s law of cooling gives a relationship between heat transfer rate, surface area, heat transfer coefficient, surface temperature and ambient temperature.

The fluid is air.

Problem 8.4


The geometry is a flat plate.





If the flow is laminar throughout, heat transfer from the first half should be greater than that from the second half.

Second half heat transfer can be obtained by subtracting first half heat rate from the heat transfer from the entire plate.

The fluid is water.

Problem 8.5

The chip is cooled by forced convection.

This problem can be modeled as a flat plate with an unheated leading section.

Newton's law of cooling can be applied to determine the rate of heat transfer between the chip and the air.

Check the Reynolds number to establish if the flow is laminar or turbulent.

Problem 8.6

Heat transfer from the collector to the air is by forced convection.

This problem can be modeled as a flat plate with an unheated leading section.

Newton's law of cooling can be applied to determine the rate of heat transfer between the collector and air.

The heat transfer coefficient varies along the collector.


Problem 8.7


The flow is over a flat plate.


Plate orientation is important.

Variation of the heat transfer coefficient along the plate affects the total heat transfer.

The heat transfer coefficient for laminar flow decreases as the distance from the leading edge is increased. However, at the transition point it increases and then decreases again.

Higher rate of heat transfer may be obtained if the wide side of a plate faces the flow. On the other hand, higher rate may be obtained if the long side of the plate is in line with the flow direction when transition takes place

The fluid is water.

Problem 8.8


The flow is over a flat plate.

The problem can be modeled as flow over a flat plate with uniform surface heat flux.

Surface temperature varies with distance along plate. The highest surface temperature is at the trailing end.

Tripping the boundary layer at the leading edge changes the flow from laminar to turbulent. This increases the heat transfer coefficient and lowers surface temperature.

Newton’s law of cooling gives surface temperature.

Problem 8.9


The flow is normal to a tube.


Tube length is unknown.

Newton’s law of cooling can be used to determine surface area. Tube length is related to surface area.

The fluid is water.

Problem 8.10

Heat is removed by the water from the steam causing it to condense.

The rate at which steam condenses inside the tube depends on the rate at which heat is removed from the outside surface.

Heat is removed from the outside surface by forced convection.

This is an external forced convection problem of flow normal to a tube. (v) Newton’s law of cooling gives the rate of heat loss from the surface.

Problem 8.11

Electric power is dissipated into heat and is removed by the water.

This velocity measuring concept is based on the fact that forced convection heat transfer is affected by fluid velocity.

Velocity affects the heat transfer coefficient which in term affects surface temperature.

Newton’s law of cooling relates surface heat loss to the heat transfer coefficient, surface area and surface temperature.

This problem can be modeled as external flow normal to a cylinder.

The fluid is water.

Problem 8.12


The flow is normal to a rod.

Surface heat transfer rate per unit length is known. However, surface temperature is unknown.

In general, surface temperature varies along the circumference. However, the rod can be assumed to have a uniform surface temperature.

This problem can be modeled as forced convection normal to a rod with uniform surface flux or temperature.

Newton’s law of cooling gives surface temperature.

The fluid is air.

Problem 8.13

Electric power is dissipated into heat and is removed by the fluid.

This velocity measuring instrument is based on the fact that forced convection heat transfer is affected by fluid velocity.

velocity affects the heat transfer coefficient which in term affects surface temperature and heat flux.

Newton’s law of cooling relates surface heat loss to the heat transfer coefficient, surface area and surface temperature.

This problem can be modeled as external flow normal to a cylinder.

The fluid is air.

Problem 8.14

The sphere cools off as it drops. Heat loss from the sphere is by forced convection.

The height of the building can be determined if the time it takes the sphere to land is known.

Time to land is the same as cooling time.

Transient conduction determines cooling time.

If the Biot number is less than 0.1, lumped capacity method can be used to determine transient temperature.

Cooling rate depends on the heat transfer coefficient.

Problem 8.15

The electric energy dissipated inside the sphere is removed from the surface as heat by forced convection.

This problem can be modeled as external flow over a sphere.

Newton’s law of cooling relates heat loss from the surface to heat transfer coefficient, surface area and surface temperature. (iv) The fluid is air.

Problem 8.16

The sphere cools off as it drops. Heat loss from the sphere is by forced convection.

This is an external flow problem with a free stream velocity that changes with time.

This is a transient conduction problem. The cooling time is equal to the time it takes the sphere to drop to street level.

If the Biot number is less than 0.1, lumped capacity method can be used to determine transient temperature.

Cooling rate depends on the heat transfer coefficient.

Problem 8.17



The outside surface is maintained at a uniform temperature.

Neglecting tube thickness resistance means that the inside and outside surface temperatures are identical.

Fluid temperature is developing.

Inlet and outlet temperatures are known.

The Reynolds number should be determined to establish if the flow is laminar or turbulent.

The required tube length depends on the heat transfer coefficient.

The fluid is water.

Problem 8.18






The outlet temperature is unknown.

The Reynolds number should be checked to establish if the flow is laminar or turbulent.

The fluid is air.

Problem 8.19

This is an internal forced convection problem

Tube surface is maintained at uniform temperature.


The length of tube is unknown.

The temperature is developing. However, depending on tube length relative to the thermal entrance length, temperature may be considered fully developed throughout.


The fluid is water.

Problem 8.20


Tube surface is maintained at a uniform temperature.

The velocity and temperature are developing. Thus, entrance effects may be important.

The outlet temperature is unknown.

The fluid is air.

Problem 8.21



The section of interest is far away from the inlet. This means that flow and temperature

can be assumed fully developed and the heat transfer coefficient uniform.

It is desired to determine the surface flux at this section. Newton’s law of cooling gives a relationship between local flux, surface temperature and heat transfer coefficient.


The fluid is water.

Problem 8.22


Both velocity and temperature are fully developed.



Mean velocity, mean inlet and outlet temperatures and tube diameter are known.

The fluid is air.

Problem 8.23


The surface of each tube is maintained at uniform temperature which is the same for both.

The velocity and temperature are fully developed. Thus, the heat transfer coefficient is uniform.

Air flows through each tube at different rates.


Surface heat flux depends on the heat transfer coefficient.

Problem 8.24


The geometry consists of two concentric tubes.

Air flows in the inner tube while water flows in the annular space between the two tubes.

The Reynolds number should be computed for both fluids to establish if the flow is laminar or turbulent.

Convection resistance depends on the heat transfer coefficient.

Problem 8.25


The geometry consists of a tube concentrically placed inside a square duct,.

Water flows in the tube and the duct.

The Reynolds number should be computed for the two fluids to establish if the flow is laminar or turbulent.

Far away from the inlet the velocity and temperature may be assumed fully developed.

Problem 8.26

Heat is lost from the door to the surroundings by free convection and radiation.

To determine the rate of heat loss, the door can by modeled as a vertical plate losing heat by free convection to an ambient air and by radiation to a large surroundings.

Newton’s law of cooling gives the rate of heat transfer by convection and Stefan-Boltzmann relation gives the heat loss by radiation.

Problem 8.27

This is a free convection and radiation problem.

The geometry is a vertical plate.


Newton’s law of cooling gives convection heat transfer rate while Stefan-Boltzmann law gives radiation heat transfer rate.


Since radiation heat transfer is considered in this problem, all temperatures should be expressed in degrees kelvin.

The fluid is air.

Problem 8.28








The fluid is air.

Problem 8.29








The fluid is water.

Problem 8.30




The heat transfer rate from the lower half 1 is greater than that from the upper half 2.

Total heat transfer from each half can be determined using the average heat transfer coefficient.

Heat transfer from the upper half is equal to the heat transfer from the entire plate minus heat transfer from the lower half.

Problem 8.31




The width of each triangle changes with distance from the leading edge.

Problem 8.32



Local heat flux is determined by Newton’s law of cooling.

Heat flux depends on the local heat transfer coefficient.

Free convection heat transfer coefficient for a vertical plate decreases with distance from the leading edge. Thus, the flux also decreases.

The Rayleigh number should be computed to select an appropriate Nusselt number correlation equation.

The fluid is air.

Problem 8.33


The power dissipated in the chips is transferred to the air by free convection.



Newton’s law of cooling relates surface temperature to heat flux and heat transfer coefficient.

The fluid is air.

Problem 8.34

Power supply to the disk is lost from the surface to the surroundings by free convection and radiation.

To determine the rate of heat loss, the disk can by modeled as a horizontal plate losing heat by free convection to an ambient air and by radiation to a large surroundings.

Newton’s law of cooling gives the rate of heat transfer by convection and Stefan-Boltzmann relation gives the heat loss by radiation.

Free convection correlations give the heat transfer coefficient.

Conservation of energy at the surface gives the emissivity, if it is the only unknown.

Problem 8.35


The geometry is a flat plate.

Heat transfer from two plates is to be compared. One plate is vertical and the other is inclined. Both plates fit in the same vertical space. Thus, the inclined plate is longer than the vertical plate.

Both plates are maintained at uniform surface temperature.

Heat transfer depends on surface area and average heat transfer coefficient.

Problem 8.36


The kiln has four vertical sides and a horizontal top.

All surfaces are at the same uniform temperature.


The sides can be modeled as vertical plates and the top as a horizontal plate.

The fluid is air.

Problem 8.37

Heat transfer from the surface is by free convection and radiation.

The burner can be modeled as a horizontal disk with its heated side facing down.

Newton’s law of cooling gives heat transfer by convection and Stefan-Boltzmann relations gives heat transfer by radiation.

Both convection and radiation depend on surface temperature.

If the burner is well insulated at the bottom heated surface and its rim, then the electric power supply is equal to surface heat transfer.

Problem 8.38

Heat transfer from the surface is by free convection and radiation.

The sample can be modeled as a horizontal disk with its heated side facing down or up.

Newton’s law of cooling gives heat transfer by convection and Stefan-Boltzmann relation gives heat transfer by radiation.

Radiation depends on surface emissivity.

If the disk is well insulated at the heated surface and its rim, then the electric power supply is equal to surface heat transfer.

Since the electric power is the same for both orientations, it follows that surface heat transfer rate is also the same.

Each orientation has its own Nusselt number correlation equation.

Problem 8.39


Heat is transferred from the cylindrical surface and top surface of tank to the ambient air.

Under certain conditions a vertical cylindrical surface can be modeled as a vertical plate.

Newton’s law of cooling gives the heat transfer rate from tank.

The fluid is air.

Problem 8.40


The geometry is a horizontal round duct.

Heat is transferred from duct surface to the ambient air.

According to Newton’s law of cooling, the rate of heat transfer depends on the heat transfer coefficient, surface area and surface and ambient temperatures.

Problem 8.41


The geometry is a horizontal pipe.

Heat is transferred from pipe surface to the ambient air.

Adding insulation material reduces heat loss from pipe.

According to Newton’s law of cooling, the rate of heat transfer depends on the heat transfer coefficient, surface area and surface and ambient temperatures

Heat transfer coefficient and surface area change when insulation is added.

The fluid is air.

Problem 8.42


The geometry is a horizontal wire (cylinder).

Under steady state conditions the power dissipated in the wire is transferred to the surrounding air.

According to Newton’s law of cooling, surface temperature is determined by the heat transfer rate, heat transfer coefficient, surface area and ambient temperature.

The fluid is air.

Problem 8.43


The geometry is a horizontal tube.

Under steady state conditions the power dissipated in the neon tube is transferred to the surrounding air.

According to Newton’s law of cooling, surface temperature is determined by the heat transfer rate, heat transfer coefficient, surface area and ambient temperature.

The fluid is air.

Problem 8.44


The geometry is a round horizontal round duct.

Heat is transferred from the ambient air to the duct.

According to Newton’s law of cooling, the rate of heat transfer to the surface depends on the heat transfer coefficient, surface area and surface and ambient temperatures.

The fluid is air.

Problem 8.45

This is a free convection and radiation problem

The geometry is a sphere.

Under steady state conditions the power dissipated in the bulb is transferred to the surroundings by free convection and radiation and through the base by conduction.

According to Newton’s law of cooling and Stefan-Boltzmann radiation law, heat loss from the surface depends on surface temperature.

The ambient fluid is air.

Problem 8.46

At steady state, power supply to the sphere must be equal to the heat loss from the surface

Heat loss from the surface is by free convection.


Problem 8.47

Heat is transferred from the ambient air to the water in the fish tank.

Adding an air enclosure reduces the rate of heat transfer.

To estimate the reduction in cooling load, heat transfer from the ambient air to the water with and without the enclosure must be determined.

) Neglecting the thermal resistance of glass, the resistance to heat transfer form the air to the water is primarily due to the air side free convection heat transfer coefficient.

Installing an air cavity introduces an added thermal resistance.

The problem can be modeled as a vertical plate and as a vertical rectangular enclosure.

The outside surface temperature of the enclosure is unknown.


The Rayleigh number should be determined for both vertical plate and rectangular enclosure so that appropriate correlation equations for the Nusselt number are selected. However, since the outside surface temperature of the enclosure is unknown, the Rayleigh number can not be determined. The problem must be solved using an iterative procedure.

Problem 8.48

Heat is transferred from the inside to the outside.

Adding i an air enclosure reduces the rate of heat transfer.

To estimate the savings in energy, heat transfer through the single and double pane windows must be determined.

The double pane window introduces an added glass conduction resistance and a cavity convection resistance.

the problem can be modeled as a vertical rectangular enclosure.

Newton’s law of cooling gives the heat transfer rate

The aspect ratio and Rayleigh number should be determined for the rectangular enclosure so that an appropriate correlation equation for the Nusselt number can be selected.

Problem 8.49

Heat is transferred through the door from the inside to the outside.



The baffle divides the vertical cavity

Problem 8.50

Heat is transferred through the skylight from the inside to the outside.



Problem 8.51

Power requirement is equal to the heat transfer rate through the enclosure.

The problem can be modeled as a rectangular cavity at specified hot and cold surface temperatures.

The inclination angle varies from too0 .175o


The aspect ratio and critical inclination angle should be computed to determine the applicable correlation equation for the Nusselt number.

Problem 8.52

The absorber plate is at a higher temperature than the ambient air. Thus heat is lost through the rectangular cavity to the atmosphere

The problem can be modeled as an inclined rectangular cavity at specified hot and cold surface temperatures.


The aspect ratio and critical inclination angle should be computed to determine the applicable correlation equation for the Nusselt number.

Problem 8.53

Heat is transferred through the annular space from the outer cylinder to the inner.


The Rayleigh number should be determined for the enclosure formed by the concentric cylinders so that an appropriate correlation equation can be selected.

The cylinders are horizontally oriented.

Problem 9.1

Definitions of Knudsen number, Reynolds number, and Mach number are needed.

Fluid velocity appears in the definition of Reynolds number and Mach number.

Problem 9.2

The definition of friction factor shows that it depends on pressure drop, diameter, length and mean velocity.

Mean velocity is determined from flow rate measurements and channel flow area.

Problem 9.3


Temperature field depends on the velocity field.

The velocity field for Couette flow with a moving upper plate is give in Section 9.6.2.

The solution to the energy equation gives the temperature distribution.

Problem 9.4

Temperature distribution depends on the velocity field.

The velocity field for Couette flow with a moving upper plate is give in Section 9.6.2.


Two temperature boundary conditions must be specified.

Temperature distribution and Fourier’s law give surface heat flux.

Problem 9.5

To determine mass flow rate it is necessary to determine the velocity distribution.

Velocity slip takes place at both boundaries of the flow channel.

Because plates move in opposite directions, the fluid moves in both directions. This makes it possible for the net flow rate to be zero.

Problem 9.6

Model channel flow as Couette flow between parallel plates.

Apply Fourier’s law at the housing surface to determine heat leaving the channel.

Apply the Navier-Stokes equations and formulate the velocity slip boundary conditions. Follow the analysis of Section 9.6.2 and Example 9.1.

Use the energy equation to determine the temperature distribution

Problem 9.7

To determine the temperature of the lower plate, fluid temperature distribution must be known.

Temperature distribution depends on the velocity field.

The velocity field for Couette flow with a moving upper plate is given in Section 9.6.2.


Two temperature boundary conditions must be specified.

Problem 9.8

To use the proposed approach, the solution to the axial velocity distribution must be know.

The velocity distribution for Poiseuille flow between parallel plates is given by equation (9.30) of Section 9.6.3.

Problem 9.9

This is a pressure driven microchannel Poiseuille flow between parallel plates.

The solution to mass flow rate through microchannels is given in Section 9.6.3.

Channel height affects the Knudsen number.

Problem 9.10

This is a pressure driven microchannel Poiseuille flow.

Since channel height is much smaller than channel width, the rectangular channel can be modeled as Poiseuille flow between parallel plates.

Channel surface is heated with uniform flux.

The solution to mass flow rate, temperature distribution, and Nusselt number for fully developed Poiseuille channel flow with uniform surface flux is presented in Section 9.6.3.

Problem 9.11

The problem can be modeled as pressure driven Poiseuille flow between two parallel plates with uniform surface flux.

Assuming fully developed velocity and temperature, the analysis of Section 9.6.3 gives the mass flow rate and Nusselt number.

The Nusselt number depends on the Knudsen number, Kn. Since Kn varies along the channel due to pressure variation, it follows that pressure distribution along the channel must be determined.

Problem 9.12

This is a pressure driven microchannel Poiseuille flow.

Since channel height is much smaller than channel width, the rectangular channel can be modeled as Poiseuille flow between parallel plates.

Channel surface is maintained at uniform temperature.

The solution to velocity, pressure, and mass flow rate is presented in Section 9.63.

The solution to the temperature distribution and Nusselt number for fully developed Poiseuille channel flow with uniform surface temperature is presented in Section 9.6.4.

Surface heat flux is determined using Newton’s law.

Problem 9.13

Cylindrical coordinates should be used to solve this problem.

The axial component of the Navier-Stokes equations must be solved to determine the

axial velocity .zv

The procedure and simplifying assumptions used in the solution of the corresponding Couette flow between parallel plates, detailed in Section 9.6.2, can be applied to this case.

Problem 9.14

This a pressure driven Poiseuille flow through a microtube.

The procedure for determining the radial velocity component and axial pressure distribution is identical to that for slip Poiseuille flow between parallel plates.

The solution to the axial velocity is given by equation (9.74).

Continuity equation gives the radial velocity component.

Axial pressure is determined by setting the radial velocity component equal to zero at the surface.


Problem 9.15


Axial velocity component is needed to determine mass flow rate.

Equation (9.74) gives the axial velocity for this case.

Since axial velocity vary with radial distance, mass flow rate requires integration of the axial velocity over the flow cross section area.

The procedure and simplifying assumptions used in the solution of the corresponding Couette flow between parallel plates, detailed in Section 9.6.2, can be applied to this problem.

Problem 9.16

To use the proposed approach, the solution to the axial velocity distribution must be known.

The velocity distribution for Poiseuille flow through tubes is given by equation (9.74) of Section 9.6.5.


Problem 9.17

This is a pressure driven Poiseuille flow through a microtube.

Tube surface is heated with uniform flux.

The solution to mass flow rate, temperature distribution and Nusselt number for fully developed Poiseuille flow through a tube with uniform surface flux is presented in Section 9.6.5.

Problem 9.18

The problem is a pressure driven Poiseuille flow through microtube with uniform surface heat flux.

The Nusselt number depends on the Knudsen number, Kn. Since Kn varies along the tube due to pressure variation, it follows that pressure distribution along the tube must be determined.

Assuming fully developed velocity and temperature, the analysis of Section 9.6.5 gives axial pressure and Nusselt number variation along tube

The definition of Nusselt number gives the heat transfer coefficient.

Problem 9.19

This is a pressure driven Poiseuille flow through a tube at uniform surface temperature.

Since the flow field is assumed independent of temperature, it follows that the velocity, mass flow rate and pressure distribution for tubes at uniform surface flux, presented in Section 9.6.6, are applicable to tubes at uniform surface temperature..

The heat transfer coefficient can be determined if the Nusselt number is known.

The variation of the Nusselt number with Knudsen number for air is shown in Fig. 9.16.

The determination of Knudsen number at the inlet and outlet and Fig. 9.16 establish the Nusselt number at these locations.

The use of Fig. 9.16 requires the determination of the Peclet number.

Mean temperature variation along the tube is given by equation (6.13). Application of this equation requires the determination of the average heat transfer coefficient.

Problem 9.20

The problem is a pressure driven Poiseuille flow through microtube at uniform surface temperature.

The Nusselt number depends on the Knudsen number, Kn. Since Kn varies along the tube due to pressure variation, it follows that pressure distribution must be determined.

Assuming fully developed velocity and temperature, the analysis of Section 9.6.6 gives axial pressure and Nusselt number variation along the tube.

The definition of Nusselt number gives the heat transfer coefficient.

The variation of the Nusselt number with Knudsen number and Peclet number for air is shown in Fig. 9.16.

Mean temperature variation along the tube is given by equation (6.13). Application of this equation requires the determination of the average heat transfer coefficient.

heat convection by latif m. jiji - solutions

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