me2135 12/13 tutorial solutions 1-10

7/28/2019 ME2135 12/13 Tutorial Solutions 1-10

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1

ME2135 FLUID MECHANICS II – 2012/13

Part 2: Viscous Flow

Tutorial Solutions

1. (a) The flow is unsteadybecause time t appears in the velocity components.The flow is 3-D because all three velocity components are non-zero.

ka jaia

kdt

dw j

dt

dvi

dt

du

dt

Vda

zyx

where:

xtx

xzytttxx

z

uw

y

uv

x

uu

t

u

dt

du

Dt

Duax

2

2

164

0402444

ytty

xztyttxty

z

vw

y

vv

x

vu

t

v

dt

dv

Dt

Dvay

4

22

44

0422044

zxzxt

xxzytztx

z

ww

y

wv

x

wu

t

w

dt

dw

Dt

Dwaz

2

2

1616

4402440

kzxtxz jyttyixtxdt

Vda 242 161644164

At 0,1,1,, zyx :

k jttita 014414 32 Answer



2

(b) Recalling that V = u i + v j + wk

where: zcybxau 111

zcybxav 222

zcybxaw 333

For incompressible flow:

0 V – continuity

or 0

z

w

y

v

x

u

Substituting u, vand w into the continuity will yield:

0321 cba Answer

(c) Again applying the continuity condition:

0 zw

yv

xu – for incompressible flow in Cartesian coordinates

Hence: 022

z

b

y

vayax

x

that is:

axy

v

y

vax 202

Integrating partially with respect to y gives:

tzxf axytzyxv ,,2,,, Answer



3

(d) A 2-D velocity field:

jyxixyxV 222

xyxu 22and yxv 2

jdt

dvi

dt

du

dt

Vda

yx aa

yyxxxyx

y

uv

x

uu

t

u

Dt

Du

dt

duax

2212

22

yxxyx

y

vv

x

vu

t

v

Dt

Dv

dt

dvay

2222

At 2,1, yx : iax 10 and jay 8 Answer

For steady, 2-D flow, the rate of change of pressure is:

yyxxxyx

y

pv

x

pu

t

p

Dt

Dp

dt

dp

829 222

At 2,1, yx : units46Dt

Dp Answer



4

sing

dzds

s

V

streamline

2. Bernoulli’s Equation is for ideal fluids, that is, viscosity ν = 0

For ν = 0, the Navier-Stokes Equations become:

1

1 1

1

Bx

B By

Bz

Du Pf

Dt x

DV Dv Pf P f Dt Dt y

Du Pf

Dt z

Apply Euler’s Equation along a streamline in a steady, incompressible plane

flow sVV , where s= unit vector.

sindsdz

ss

VV

t

V

s

VV

t

V

Dt

VD

0 (for steady flow)

in s-direction: s

P

gs

V

V

1

sin

s

P

s

zg

1

that is: 01

s

zg

s

P

s

VV

or: 02

2

gz

PV

s

or: gzPV

2

2

constant (Bernoulli’s Equation)

(Euler’s Equation)



5

3.

Let the flow be parallel to the x-axis.

For a steady, 2-D, parallel flow (in Cartesian coordinate system):

2

2

y

u

dx

dP

Since 0xu (from Continuity equation, that is, u is independent of x):

2

2

2

2

dy

ud

y

u

dx

dP

dy

ud

2

2

with the following boundary conditions:

0u at by (no-slip)

0dy

duat 0y (symmetry of flow)

Integrating with respect to ygives:

Aydx

dP

dy

du

Applying the symmetry of flowboundary condition:

ydx

dP

dy

duA 0

Integrating again with respect to ygives:

By

dx

dPu

2

2

y

x

L1 2

u

U 2b=h



6

Applying the no-slipboundary condition:

dx

dPbyu

b

dx

dPB 22

2

2

1

2

Noting that at 0

y , Uu

would yield: dx

dPbU

2

2

Comparing the above with the previous result, the following is obtained:

2

1

b

y

U

u Proven

The volume flow rate per unit depth of channel:

dx

dPbyb

y

dx

dPdyby

dx

dPdyuq

bbb

b

3

0

23

0

22

3

2

3

1

2

22

Since h = 2bor hb21 , then:

dx

dPhq

12

3

that is, a discharge in the positive x-direction requires a negative pressure gradient

or a pressure field that is decreasing in the flow direction.

Spatial average bulk velocity V :

dxdPh

hqV

12

2

2

1

22

112

dPV

hdx

PL

PPV

hxx

12

2

1212

NOTE: pressure drop 21 PPP 2

12

h

LVP

OR: h

L

hVhV

VL

V

P

24

12

22

212

21

But

hVRe , hence: h

L

V

P

Re

24

221

Proven



7

NOTE: The use of the no-slip boundary condition ( 0u at by ) is sufficient to

obtain the velocity profile u.

From:dx

dP

dy

ud

2

2

Aydx

dP

dy

du

Hence: BAyydxdP

dydu 2

2

For 0u at by :

220

22 b

dx

dPAbBBAb

b

dx

dP

For 0u at by :

BAbb

dx

dP

20

2

Inserting the previous result:

220

22 b

dx

dPAbAb

b

dx

dP

2and020

2b

dx

dPBAAb

2222

2

1

22by

dx

dPb

dx

dPy

dx

dPu

The average velocity in the gap V is formulated as:

h

bdx

dP

h

yby

dx

dP

h

dyu

area

rateflowvolumeV

b

b

b

b

323

3

4

2

1

32

1

dx

dPh

dx

dP

h

bV

123

2 23

Integrating between x= 1 and x= 2 :

2

1

2

1

2

12dp

V

hdx

PL

PPV

hxx

12

2

1212

L

PhV

12

2



8

4. (a) Plane Couette Flow: lower plate is stationary, upper plate moves to the right at velocity U

From: dx

dP

y

u

2

2

Aydx

dP

y

u

BAyy

dx

dPu

2

2

Boundary conditions:

0u at 0y : 0B

Uu at hy : Ahh

dx

dPU 2

2

2

12h

dx

dPU

hA

22

22 h

dx

dPU

h

yy

dx

dPu

h

y

h

yh

dx

dPy

h

U

yhydxdPy

hU

h

dx

dPU

h

yy

dx

dPu

2

22

2

22

2

1

2

1

2

1

2

1

h

y

h

y

dx

dPhy

h

Uu 1

2

2

Answer

Volume flow rate across the plates per unit z-width:

h

hh

x

hyy

dx

dPy

h

U

dyyhydxdPy

hUdyuq

0

232

0

2

0

232

1

2

2

1

dx

dPhUhqx

122

3

Answer

y

x

h

U



9

4. (b) Plane Couette Flow: lower plate moves to the right at velocity U, upper plate is stationary

From: BAyy

dx

dPu

2

2

Boundary conditions:

Uu at 0y : BU

0u at hy : UAh

h

dx

dP

20

2

dx

dPhUhdyuq

hyy

dx

dP

h

yUu

h

x

122

21

3

0

Answer

y

x

h

U



10

5.

2 212 2

2 212 2

0

u u u P u uu v f xt x y x x y

v v v P v vu v f yt x y y x y

u v

x y

// flow in the x-direction:2

20 0, 0

u uv

x x

For steady flow: ' 0st

Since 0 (given), the equation becomes:P

x

2

2

2

2

0 sin

sin

ugy

ug

y

With the boundary conditions: 1. at 0, 0y u

2. at , 0 0u u

y hy y

Integrating gives: sinu g y Ay

Applying the boundary condition 2: 0 sing y A

sinA gh

sin sinu

gy ghy

y

x

h

( )u y

g

sing

cosg



11

Integrating again gives:

2

sin sin2

yu g ghy B

Applying the boundary condition 1: 0 B

Hence:

2

sin2

g yu hy

Answer

Flow rate per unit width Q is given by:

0

2

0

2 3

0

sin2

sin2 6

h

h

h

Q udy

g yhy dy

g y yh

3

sin .3

g hQ

Answer



12

6. a) The assumptions made are:

(i) is constant

(ii) is constant

(iii) Two-dimensional (2-D) flow, since the equations of motion and

continuity do not include the azimuthal (angular) term

(iv) Body forces negligible

(v) Flow is steady no ' termst

For fully developed flow, the conditions are:

0 no flow in the r-direction

(parallel flow)

v

2

2also 0, 0

v v

x x

2

2and 0, 0

i.e. is independent of ,

hence:

u u

x x

u x

u du

r dr

Substituting the conditions to the equations of motions:

1 1

10, i.e.: is independent of

( ) only and

P d dur

x r dr dr

PP r

r

P dPP P x x dx

,r v

,x u



13

b)

Fully developed velocity profile in the annulus of concentric cylinders:

where R 1 = radius of cylindrical cooling channel, and

R 2 = radius of fuel element (solid and co-axial with the cooling channel)

From the simplified equation:

2

2

1

1So that

2

1and

2

1ln

4

d du dPr r

dr dr dx

du dP rr A

dr dx

du dP r A

dr dx r

dP ru A r B

dx

with the boundary conditions:

2

1 1 1

2

2 2 2

1(1) 0 at 0 ln

4

1(2) 0 at 0 ln

4

dPu r R R A R B

dx

dPu r R R A R B

dx

2 2 11 2

2

1ln

4

dP RR R A

dx R

2 2

1 2

1

2

1

4

ln

dPR R

dxA

R

R

fuel

elementx

r

u(r)

R 1R 2

x

r

R 1

R 2

–



14

2 2

1 2 12 2

1 1 1

1

2

1ln

1 14and ln

4 4ln

dPR R R

dP dPdxB A R R R

dx dxR

R

which gives:

2 2 2 2

1 2 1 2 12 2

1

1 1

2 2

ln ln1

4ln ln

R R r R R RdPu r R

dx R R

R R

Answer

is maximum when 0du

udr

2 2

1 2

1

2

1 12 0

4ln

R Rdu dPr

dr dx rR

R

2 2 2 2

1 2 1 22 2

1 1

2 2

2

ln 2ln

R R R Rr r

R R

R R

2 2

1 22 2

1 2 12 2

1 21 2 2

max 1

1 1

2 2

2 2

1 2

1

2

ln2ln

1

42ln ln

occurs at

2ln

R RR R RR RR RdP

u Rdx R R

R R

R Rr

R

R

Answer



15

7 (a)

i)

Cαx bα

qμ6

αx bα

Uμ6x p

2

1

x

1

Eqn. (2.24) on Lecture Notes p. 23

with b.c.' s : 1) at x = 0, p = 0 (gauge) and 2) at x = L, p = 0 (gauge).

Substituting the b.c. 1) into Eqn. (2.24): Cα b

qμ6 bα

Uμ6 02

1

x

1

and the b.c. 2) into Eqn. (2.24): C

αL bα

qμ6

αL bα

Uμ6 0

2

1

x

1

From these two equations:αL) b(

U

b

U

αL)(b

q

b

q

11

2

1

x

2

1

x

whereL

b b α 21 Hence:

21

21x

21

2

2

x

2

1

x

b b

U b bq

b

U

b

U

b

q

b

q

ii) From Eqns. (2.25) and (2.22):21

21

21

21x

b b

b b2 b

2

bU

b b

U b bq

iii) b(x) = b1 – αx → xα b b whereL

b b α 21

21

1211

21

21

b b

L bx

L

x) b- b( b

b b

b b2

iv) The pressure distribution is given by Eqn. (2.26):

21

2

2

b bx b

x bx bUμ6x p

21

2

2mm

b b b

x b bUμ6 p b b(x)andxxat p p(x)

Substituting the expressions for b and x into the previous expression of pm ,

we will get:

2121

21m

b b

L

b b

) b b(Uμ

2

3 p



16

7 (b) Given: 500 kN mF

w

Determine: i) of the lubricant

ii) Power/width

iii) Max. pressure pm

iv) Centre of pressure xcp

Solution:

i) From:

2

1 1 2

2

2 1 21 2

6ln 2

F UL b b b

w b b bb b

Equation (2.27)

Express in terms of the others and after substituting the corresponding

numerical values will give 20.0626 Ns/m 0.0626 Pa.s (or 0.626 cP)

ii)

1

2

ln2

F wD U b

w b

Equation (2.28)

where 1 2 0.075 0.025 0.050 0.001

150 150 3

b b

L

x1.5 m/sU

150 mmL

“toe” “heel”

b1

b2

b1 = 0.075 mm

b2 = 0.025 mm

centi Poise

(c.g.s. unit)



17

3 0.0626 1.5500x10 0.001ln3

2 3 0.001 3

0.0626 4.55001.0986

6 0.001

83.333 309.476 N/m

392.81 N/m

D

w

Power consumed per width = x 589.213 W/mD w U

iii) From:

2

2

1 2

6 ( )( )

( )

U b x b xp x

b x b b

Equation (2.26)

maxp p occurs at x x and ( )b x b

where:

1 2

1 2

2(can you derive this?)

2 0.075 0.025= mm 0.0375 mm

0.075 0.025

bbb

b b

1 21

1 2

1

1 21 21

1 2

( ) where

2

b bb x b x

L

b b xb b

L

b b xbbb

b b L

Hence:

1

1 2

0.075 150mm 112.5 mm

0.075 0.025

bLx

b b



18

Therefore,

2

max 2

1 2

2 3

3

6

6 0.0626 1.5 0.0125 112.5

0.0375 0.1x10

5634x10 Pa = 5.634 MPa

U b b xp

b b b

Be careful of the units !!

iv) Using:

1 1 2 1 1 2

2

2 1 21 2

1 2 1

1 2 2

2 5ln

2

ln 2

cp

b b b b L b bL

b b bb bx

b b b

b b b

Equation (2.33)

and substituting the numerical values:

1

2

0.075 mm (inlet gap)

0.025 mm (outlet gap)

150 mm (length of bearing)

b

b

L

will give xcp = 91.088 mm (that is, centre of pressure cp is 91.1 mm from the

“toe” or the inlet)



19

8. Using volume flow rate per unit z-width in the x-direction:

h2

U

dx

dP

12μ

hq

3

x

(a) In zone 1 of the bearing: h = h1 andlmP

x p

Hence:2

h

P

12μ

hq 1m1

1

3

xoV

l

→ for zone 1

(b) In zone 2 of the bearing: h = h2 andlmP

x

p

Hence: 2

h

P

12μ

hq 2m2

2

3

x oV

l

→ for zone 2

(c) To satisfy continuity of flow :21

xx qq

Hence: 21m1m2 hh

2

P

12μ

hP

12μ

h

33

oV

ll

and therefore:

3321

21

hh

hhμ6

m

P

oVl

(d) Load per unit width F/w = Area of the pressure distribution

= ½ x 2 l x Pm = Pm l

(e) If h2 = ½ h1 Pm = 8 μ l Vo / 3 h12

and therefore: 21

2

22

2

h3

μ8

or h3

μ2

F/woo VlVl

If h2 = h1 F/w = 0 since there is no step, and hence Pm = 0.



20

9. Velocity profile:n

1

δ

y

U

u

where n > 0

Boundary layer displacement thickness

δ

0

dyU

u1δ*

1n

δ

n

1)(n

δδ

1)n

1(

δ

δ

1δ

1)n

1(

y

δ

1ydy

δ

y1

1n

1

n

1

δ

0

1n

1

n

1

δ

0

n

1

Boundary layer momentum thickness

δ

0dyU

u1U

uθ

2)(n1)(n

nδ

2n

nδ

1n

nδ

1)

n

2(

δ

δ

1

1)

n

1(

δ

δ

1

1)

n

2(

y

δ

1

1)

n

1(

y

δ

1

dyδ

y

δ

ydy

δ

y1

δ

y

1n

2

n

2

1n

1

n

1

δ

0

1n

2

n

2

1n

1

n

1

δ

0

n

2

n

1δ

0

n

1

n

1

n

2n

θ

δH parameter shapeHence,

1Hn1δ

yyallfor Uu:layer boundary No

1.2857H;0972.09x8

7

δ

θ;

8

1

δ

δ7nFor (ii)

3H;6

1

3x2

1

δ

θ;

2

1

δ

δ1nFor (i)

n

1

NOTE: For a laminar boundary layer over a flat plate: H = 2.59



21

10. For the velocity distribution:

yba

U

u

(a) To find a and b, use the boundary conditions:

0u at 0y (at the wall) 0a

Uu at y (at the edge of boundary layer) 1b

Hence, the velocity profile is

y

U

u or

yUu .

Boundary layer displacement thickness

0

1* dyU

u

221*

0

2

0

yydyy

5.0* Answer

From Blasius solution:

344.0*907.2

72.1

5

*

Boundary layer momentum thickness

0

1 dyUu

Uu

6321

0

2

32

0

2

2

0

yydy

yydy

yy

167.0 Answer

From Blasius solution:

133.0530.7

664.0

5

(b) From the definition of local skin friction:

U

y

u

y

0

0



22

From the von Karman integral equation:

dx

dU

dx

dU

6

22

0

U

dx

dU

6

2

xU

dxU

dx

6

2

6 2

00

Uxx

U

x

x

x

U

x

12

1212

xxx Re

464.3

Re

12

Answer where:

Uxx Re

Drag per unit width: D/w

x

x

xxxx

xU

Ux

xUUxUx

U

U

dxx

U

Udx

U

x

Udx

UdxwD

Re

1

12

2

12

2

12

22

12

1212/

22

0

0000

0

2

1

2

1

xx

DxU

wDC

Re

1

464.3

4

Re

1

12

4/2

21

x

DCRe

155.1 Answer

Linear Approximation Blasius Solution 5.0* 344.0*

167.0 133.0

xx Re

464.3

xx Re

5

x

DCRe

155.1

x

DCRe

328.1



For the velocity distribution:

2

yc

yba

U

u

2

2

cybU

dy

du

(a) First, find a, band cusing the boundary conditions: 0u at 0y (no slip condition) 0a

Uu at y (at the edge of boundary layer) cb1

0dy

duat y (at the edge of boundary layer) cb 20

Hence 2b and 1c and therefore the velocity distribution (profile) is given by:

2

2

22

yy

U

u

where:

ddyy

For 00 y

1 y

Displacement thickness:

1

0

32

1

0

2

0 3211*

ddyU

u

3

1* Answer

Momentum thickness:

5

11

3

51

5

1

3

5452

22422121

1

0

44321

0

432

1

0

433221

0

22

0

d

dddyU

u

U

u

15

2 Answer Hence, shape factor 5.2

*

H

Parabolic Approx. Blasius Solution

333.0* 344.0*

133.0 133.0

xx Re

47.5

xx Re

5

x

DCRe

46.1

x

DCRe

328.1

me2135 12/13 tutorial solutions 1-10

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