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TRANSCRIPT
q
T1
T2
k
x A
MODULE I
BASICS OF HEAT TRANSFER
While teaching heat transfer, one of the first questions students commonly ask is the difference between heat and temperature. Another common question concerns the difference between the subjects of heat transfer and thermodynamics. Let me begin this chapter by trying to address these two questions.
1.1 Difference between heat and temperature In heat transfer problems, we often interchangeably use the terms heat and temperature. Actually, there is a distinct difference between the two. Temperature is a measure of the amount of energy possessed by the molecules of a substance. It manifests itself as a degree of hotness, and can be used to predict the direction of heat transfer. The usual symbol for temperature is T. The scales for measuring temperature in SI units are the Celsius and Kelvin temperature scales. Heat, on the other hand, is energy in transit. Spontaneously, heat flows from a hotter body to a colder one. The usual symbol for heat is Q. In the SI system, common units for measuring heat are the Joule and calorie.
1.2 Difference between thermodynamics and heat transfer Thermodynamics tells us:
• how much heat is transferred (δQ) • how much work is done (δW) • final state of the system
Heat transfer tells us: • how (with what modes) δQ is transferred • at what rate δQ is transferred • temperature distribution inside the body
1.3 Modes of Heat Transfer • Conduction: An energy transfer across a system boundary due to a temperature difference
by the mechanism of intermolecular interactions. Conduction needs matter and does not require any bulk motion of matter.
Thermodynamics Heat transfer complementary
moving fluid T∞
q
Ts
Ts>T∞
Conduction rate equation is described by the Fourier Law:
T kA q ∇ − = ρ
where: q = heat flow vector, (W) k = thermal conductivity, a thermodynamic property of the material.
(W/m K) A = Cross sectional area in direction of heat flow. (m 2 ) ∇T = Gradient of temperature (K/m)
= ∂T/∂x i + ∂T/∂y j + ∂T/∂z k Note: Since this is a vector equation, it is often convenient to work with one component of the vector. For example, in the x direction:
qx = k Ax dT/dx
In circular coordinates it may convenient to work in the radial direction: qr = k Ar dT/dr
• Convection: An energy transfer across a system boundary due to a temperature difference by the combined mechanisms of intermolecular interactions and bulk transport. Convection needs fluid matter.
Newton’s Law of Cooling: q = h As ∆T
where: q = heat flow from surface, a scalar, (W) h = heat transfer coefficient (which is not a thermodynamic property of
the material, but may depend on geometry of surface, flow characteristics, thermodynamic properties of the fluid, etc. (W/m 2 K)
As = Surface area from which convection is occurring. (m 2 ) ∆T = = − ∞ T T S Temperature Difference between surface and coolant. (K)
Convection
Free or natural convection (induced by buoyancy forces)
Forced convection (induced by external means)
May occur with phase change (boiling, condensation)
q”co n v. q”r a d. Ts u r
Area = A Ts
Table 1. Typical values of h (W/m 2 K)
• Radiation: Radiation heat transfer involves the transfer of heat by electromagnetic radiation that arises due to the temperature of the body. Radiation does not need matter.
Emissive power of a surface: E=σεTs
4 (W/ m 2 )
where: ε = emissivity, which is a surface property (ε = 1 is black body) σ = Steffan Boltzman constant = 5.67 x 10 8 W/m 2 K 4 . Ts = Absolute temperature of the surface (K)
The above equation is derived from Stefan Boltzman law, which describes a gross heat emission rather than heat transfer. The expression for the actual radiation heat transfer rate between surfaces having arbitrary orientations can be quite complex, and will be dealt with in Module 9. However, the rate of radiation heat exchange between a small surface and a large surrounding is given by the following expression:
q = ε∙σ∙A∙(Ts 4 – Tsur
4 )
where: ε = Surface Emissivity A= Surface Area Ts = Absolute temperature of surface. (K) Tsur = Absolute temperature of surroundings.(K)
Free convection gases: 2 25 liquid: 50 – 100
Forced convection gases: 25 250 liquid: 50 20,000
Boiling/Condensation 2500 100,000
1.4 Thermal Conductivity, k
As noted previously, thermal conductivity is a thermodynamic property of a material. From the State Postulate given in thermodynamics, it may be recalled that thermodynamic properties of pure substances are functions of two independent thermodynamic intensive properties, say temperature and pressure. Thermal conductivity of real gases is largely independent of pressure and may be considered a function of temperature alone. For solids and liquids, properties are largely independent of pressure and depend on temperature alone.
k = k (T)
Table 2 gives the values of thermal conductivity for a variety of materials.
Material Thermal Conductivity, W/m K Copper 401 Silver 429 Gold 317 Aluminum 237 Steel 60.5 Limestone 2.15 Bakelite 1.4 Water 0.613 Air 0.0263
Let us try to gain an insight into the basic concept of thermal conductivity for various materials. The fundamental concept comes from the molecular or atomic scale activities. Molecules/atoms of various materials gain energy through different mechanisms. Gases, in which molecules are free to move with a mean free path sufficiently large compared to their diameters, possess energy in the form of kinetic energy of the molecules. Energy is gained or lost through collisions/interactions of gas molecules.
Lattice vibration may be transferred between molecules as nuclei attract/repel each other.
Table 2. Thermal Conductivities of Selected Materials at Room Temperature.
Kinetic energy transfer between gas molecules.
Solids, on the other hand, have atoms/molecules which are more closely packed which cannot move as freely as in gases. Hence, they cannot effectively transfer energy through these same mechanisms. Instead, solids may exhibit energy through vibration or rotation of the nucleus. Hence, the energy transfer is typically through lattice vibrations.
Another important mechanism in which materials maintain energy is by shifting electrons into higher orbital rings. In the case of electrical conductors the electrons are weakly bonded to the molecule and can drift from one molecule to another, transporting their energy in the process. Hence, flow of electrons, which is commonly observed in metals, is an effective transport mechanism, resulting in a correlation that materials which are excellent electrical conductors are usually excellent thermal conductors.
Module 1: Worked out problems
Problem 1: A cold storage consists of a cubical chamber of dimension 2m x 2m x 2m, maintained at 10°C inside temperature. The outside wall temperature is 35°C. The top and side walls are covered by a low conducting insulation with thermal conductivity k = 0.06 W/mK. There is no heat loss from the bottom. If heat loss through the top and side walls is to be restricted to 500W, what is the minimum thickness of insulation required?
Solution:
Known: Dimensions of the cold storage, inner and outer surfaces temperatures, thermal conductivity of the insulation material.
To find: Thickness of insulation needed to maintain heat loss below 500W.
Schematic:
Assumptions: (1) perfectly insulted bottom, (2) onedimensional conduction through five walls of areas A=4m 2 , (3) steadystate conditions
Analysis: Using Fourier’s law, the heat rate is given by
total ' ' A
L T k A . q q
∆ = =
Solving for L and recognizing that A total =5*W 2
mm m L W
m C K m W L
q TW k L
108 108 . 0 500
4 * 45 * . / 06 . 0 * 5
5
2 0
2
= =
=
∆ =
insulation
Problem 2: A square silicon chip is of width W=5mm on a side and of thickness t=1mm. The chip is mounted in a substrate such that there is no heat loss from its side and back surfaces. The top surface is exposed to a coolant. The thermal conductivity of the chip is 200W/m.K. If 5W are being dissipated by the chip, what is the temperature difference between its back and front surfaces?
Known: Dimensions and thermal conductivity of a chip. Power dissipated on one surface.
Find: temperature drop across the chip
Schematic:
Assumptions: (1) steadystate conditions, (2) constant properties, (3) uniform dissipation, (4) negligible heat loss from back and sides, (5) onedimensional conduction in chip.
Analysis: All of the electrical power dissipated at the back surface of the chip is transferred by conduction through the chip. Hence, Fourier’s law,
C T m K m W
W m kW P t T
t T kA q P
0
2 2
0 . 1 ) 005 . 0 ( . / 200
5 * 001 . 0 .
= ∆
= = ∆
∆ = =
Comments: for fixed P, the temperature drop across the chip decreases with increasing k and W, as well as with decreasing t.
Problem 3: Air flows over a rectangular plate having dimensions 0.5 m x 0.25 m. The free stream temperature of the air is 300°C. At steady state, the plate temperature is 40C. If the convective heat transfer coefficient is 250 W/m 2 .K, determine the heat transfer rate from the air to one side of the plate.
Known: air flow over a plate with prescribed air and surface temperature and convection heat transfer coefficient.
Find: heat transfer rate from the air to the plate
Schematic:
Assumptions: (1) temperature is uniform over plate area, (2) heat transfer coefficient is uniform over plate area
Analysis: the heat transfer coefficient rate by convection from the airstreams to the plate can be determined from Newton’s law of cooling written in the form,
) T T ( hA A . q q s ' ' − = = ∞
where A is the area of the plate. Substituting numerical values,
W 8125 q C ) 40 300 ( m ) 50 . 0 * 25 . 0 ( * K . m / W 250 q 0 2 2
= − =
Comments: recognize that Newtown’s law of cooling implies a direction for the convection heat transfer rate. Written in the form above, the heat rate is from the air to plate.
Problem 4 : A sphere of diameter 10 mm and emissivity 0.9 is maintained at 80°C inside an oven with a wall temperature of 400°C. What is the net transfer rate from the oven walls to the object?
Known: spherical object maintained at a prescribed temperature within a oven.
Find: heat transfer rate from the oven walls to the object
Schematic:
Assumptions: (1) oven walls completely surround spherical object, (2) steadystate condition, (3) uniform temperature for areas of sphere and oven walls, (4) oven enclosure is evacuated and large compared to sphere.
Analysis: heat transfer rate will be only due to the radiation mode. The rate equation is
) T T ( A q 4s
4 sur s rad − σ ε =
Where As=πD 2 , the area of the sphere
W 04 . 3 q K ] ) 273 80 ( ) 273 400 [( K . m / W 10 * 67 . 5 * m ) 10 * 10 ( * 9 . 0 q
rad
4 4 4 2 8 2 2 3 rad
=
+ − + π = − −
Discussions:
(1) this rate equation is applicable when we are calculating the net heat exchange between a small object and larger surface that completely surrounds the smaller one.
(2) When performing radiant heat transfer calculations, it is always necessary to have temperatures in Kelvin (K) units.
Problem 5: A surface of area 0.5m 2 , emissivity 0.8 and temperature 150 0 C is placed in a large, evacuated chamber whose walls are maintained at 25 0 C. Find the rate at which radiation is emitted by the surface? What is the net rate of radiation exchange between the surface and the chamber walls?
Known: Area, emissivity and temperature of a surface placed in a large, evacuated chamber of prescribed temperature.
Find: (a) rate of surface radiation emission, (b) net rate of radiation exchange between the surface and chamber walls.
Schematic:
Assumptions: (1) area of the enclosed surface is much less than that of chamber walls.
Analysis (a) the rate at which radiation is emitted by the surface is emitted
W 726 qemit ] K ) 273 150 [( K . m / W 10 * 67 . 5 ) m 5 . 0 ( 8 . 0 qemit
Ts A A q q 4 4 2 8 2
4 . emit emit
= + =
σ ε = = −
(b) The net rate at which radiation is transferred from the surface to the chamber walls is
W 547 q ) K 298 ( ) K 423 [( K . m / W 10 * 67 . 5 ) m 5 . 0 ( 8 . 0 q
) 4 Tsurr 4 Ts ( A q 4 4 4 2 8 2
= − =
− σ ε = −
Problem 6: A solid aluminium sphere of emissivity ε, initially at a high temperature, is cooled by convection and radiation in a chamber having walls at a lower temperature. Convective cooling is achieved with a gas passing through the chamber. Write a differential equation to predict the variation of sphere temperature with time during the cooling process.
Known: Initial temperature, diameter and surface emissivity of a solid aluminium sphere placed in a chamber whose walls are maintained at lower temperature. Temperature and convection coefficient associated with gas flow over the sphere.
Find: equation which could be used to determine the aluminium temperature as a function of time during the cooling process.
Schematic:
Assumptions: (1) at any time t, the temperature T of the sphere is uniform, (2) constant properties; (3) chamber walls are large relative to sphere.
Analysis: applying an energy balance at an instant of time to a control volume about the sphere, it follows that
out .
st .
E E − =
Identifying the heat rates out of the CV due to convection and radiation, the energy balance has the form
)] T T ( ) T T ( h [ cD 6
dt dT
)] T T ( ) T T ( h [ Vc A
dt dT
) q q ( ) VcT ( dt d
4 surr
4
4 surr
4
rad conv
− εσ + − ρ
=
− εσ + − ρ
− =
+ − = ρ
∞
∞
Where A=πD 2 , V=πD 3 /6 and A/V=6/D for the sphere.
Problem 7: An electronic package dissipating 1 kW has a surface area 1m 2 . The package is mounted on a spacecraft, such that the heat generated is transferred from the exposed surface by radiation into space. The surface emissivity of the package is 1.0. Calculate the steady state temperature of the package surface for the following two conditions:
(a) the surface is not exposed to the sun (b) The surface is exposed to a solar flux of 750W/m 2 and its absorptivity to solar
radiation is 0.25?
Known: surface area of electronic package and power dissipation by the electronics. Surface emissivity and absorptivity to solar radiation. Solar flux.
Find: surface temperature without and with incident solar radiation.
Schematic:
Assumptions: steady state condition
Analysis: applying conservation of energy to a control surface about the compartment, at any instant
0 E E E g . .
out in .
= + − It follows that, with the solar input,
4 1
s
S ' '
s S s
4 s s S
' ' s S
emit ' '
s S ' '
s S
A P q A
T
0 P T A q A
0 P q A q A
εσ + α
=
= + εσ − α
= + − α
In the shade ( 0 q ' ' = )
K 364 K . m / W 10 * 67 . 5 * 1 * m 1
W 1000 T 4 1
4 2 8 2 s =
=
−
In the sun,
K 380 K . m / W 10 * 67 . 5 * 1 * m 1 W 1000 m / W 750 * m 1 * 25 . 0 T
4 1
4 2 8 2
2 2
s =
+ =
−
MODULE 2
ONE DIMENSIONAL STEADY STATE HEAT CONDUCTION
2.1 Objectives of conduction analysis: The primary objective is to determine the temperature field, T(x,y,z,t), in a body (i.e. how temperature varies with position within the body) T(x,y,z,t) depends on: - Boundary conditions - Initial condition - Material properties (k, cp, )
- Geometry of the body (shape, size)
Why we need T (x, y, z, t)? - To compute heat flux at any location (using Fourier’s eqn.) - Compute thermal stresses, expansion, deflection due to temp. Etc. - Design insulation thickness - Chip temperature calculation - Heat treatment of metals 2.2 General Conduction Equation Recognize that heat transfer involves an energy transfer across a system boundary. The analysis for such process begins from the 1st Law of Thermodynamics for a closed system:
dE
dtQ W
system
in out
The above equation essentially represents Conservation of Energy. The sign convention on work is such that negative work out is positive work in.
dE
dtQ W
system
in in
The work in term could describe an electric current flow across the system boundary and through a resistance inside the system. Alternatively it could describe a shaft turning across the system boundary and overcoming friction within the system. The net effect in either case would cause the internal energy of the system to rise. In heat transfer we generalize all such terms as “heat sources”.
dE
dtQ Q
system
in gen
The energy of the system will in general include internal energy, U, potential energy, ½ mgz, or kinetic energy, ½ mv2. In case of heat transfer problems, the latter two terms could often be neglected. In this case,
E U m u m c T T V c T Tp ref p r ef
where Tref is the reference temperature at which the energy of the system is defined as zero. When we differentiate the above expression with respect to time, the reference temperature, being constant, disappears:
c VdT
dtQ Qp
systemin gen
Consider the differential control element shown below. Heat is assumed to flow through the element in the positive directions as shown by the 6 heat vectors.
qz+z
qx
qy
qy+y
qx+x
z y
qz
x
In the equation above we substitute the 6 heat inflows/outflows using the appropriate sign:
c x y zdT
dtq q q q q q Qp
systemx x x y y y z z z g
en
Substitute for each of the conduction terms using the Fourier Law:
x
x
Tzyk
xx
Tzyk
x
Tzyk
t
Tzyxc
systemp
k x z
T
yk x z
T
y yk x z
T
yy
k x yT
zk x y
T
z zk x y
T
zz
q x y z
where q is defined as the internal heat generation per unit volume. The above equation reduces to:
c x y zdT
dt xk y z
T
xxp
system
y
k x zT
yy
z
k x yT
zz q x y z
Dividing by the volume (xyz),
c
dT
dt xk
T
x yk
T
y zk
T
zqp
system
which is the general conduction equation in three dimensions. In the case where k is independent of x, y and z then
c
k
dT
dt
T
x
T
y
T
z
q
kp
system
2
2
2
2
2
2
Define the thermodynamic property, , the thermal diffusivity:
k
cp
Then 1 2
2
2
2
2
2
dT
dt
T
x
T
y
T
z
q
ksystem
or, : 1
2
dT
dtT
q
ksystem
The vector form of this equation is quite compact and is the most general form. However, we often find it convenient to expand the spatial derivative in specific coordinate systems: Cartesian Coordinates
k
q
z
T
y
T
x
TT
a
2
2
2
2
2
21
Circular Coordinates
Z R y Θ x
k
q
z
TT
rr
Tr
rr
T
a
2
2
2
2
2
111
Spherical Coordinates
Z φ r y x
k
q
z
T
r
T
rr
Tr
rr
T
a
sin
sin
1
sin
11122
2
22
2
2
In each equation the dependent variable, T, is a function of 4 independent variables, (x,y,z,τ); (r, ,z,τ); (r,φ,θ,τ) and is a 2nd order, partial differential equation. The solution of such equations will normally require a numerical solution. For the present, we shall simply look at the simplifications that can be made to the equations to describe specific problems.
Steady State: Steady state solutions imply that the system conditions are not changing with time. Thus 0/ T .
One dimensional: If heat is flowing in only one coordinate direction, then it follows that there is no temperature gradient in the other two directions. Thus the two partials associated with these directions are equal to zero.
Two dimensional: If heat is flowing in only two coordinate directions, then it follows that there is no temperature gradient in the third direction. Thus, the partial derivative associated with this third direction is equal to zero.
No Sources: If there are no volumetric heat sources within the system then the term,
0
q . Note that the equation is 2nd order in each coordinate direction so that integration will result in 2 constants of integration. To evaluate these constants two boundary conditions will be required for each coordinate direction. 2.3 Boundary and Initial Conditions
• The objective of deriving the heat diffusion equation is to determine the temperature distribution within the conducting body.
• We have set up a differential equation, with T as the dependent variable. The solution will give us T(x,y,z). Solution depends on boundary conditions (BC) and initial conditions (IC).
• How many BC’s and IC’s ?
- Heat equation is second order in spatial coordinate. Hence, 2 BC’s needed for each coordinate.
* 1D problem: 2 BC in x-direction
* 2D problem: 2 BC in x-direction, 2 in y-direction
* 3D problem: 2 in x-dir., 2 in y-dir., and 2 in z-dir.
- Heat equation is first order in time. Hence one IC needed.
2.4 Heat Diffusion Equation for a One Dimensional System q
T1 q T2
Ax
L y x
z Consider the system shown above. The top, bottom, front and back of the cube are insulated, so that heat can be conducted through the cube only in the x direction. The internal heat generation per unit volume is q (W/m3). Consider the heat flow through a differential element of the cube.
qx qx+x
From the 1st Law we write for the element:
stgenoutin EEEE )( (2.1)
(2.2)
t
EqxAqq xxxx
)(
(2.3) x
TkAq xx
(2.4) x
x
qqq x
xxx
(2.5)
(2.6)
t
Txcq
x
Tk
x
t
TxAcqxAx
x
Tk
xA
x
TkA
x
TkA
Thermal inertia Internal heat generation
Longitudinal conduction
If k is a constant, then (2.7) t
T
t
T
k
c
k
q
x
T
1
2
2
• For T to rise, LHS must be positive (heat input is positive)
• For a fixed heat input, T rises faster for higher
• In this special case, heat flow is 1D. If sides were not insulated, heat flow could be 2D, 3D.
2.5 One Dimensional Steady State Heat Conduction
The plane wall:
The differential equation governing heat diffusion is: 0
dx
dTk
dx
d
With constant k, the above equation may be integrated twice to obtain the general solution:
21)( CxCxT
where C1 and C2 are constants of integration. To obtain the constants of integration, we apply the boundary conditions at x = 0 and x = L, in which case
1,)0( sTT and 2,)( sTLT
Once the constants of integration are substituted into the general equation, the temperature distribution is obtained:
1,1,2, )()( sss TL
xTTxT
The heat flow rate across the wall is given by:
kAL
TTTT
L
kA
dx
dTkAq
ssssx
/
2,1,2,1,
Thermal resistance (electrical analogy):
Physical systems are said to be analogous if that obey the same mathematical equation. The above relations can be put into the form of Ohm’s law:
V=IRelec
Using this terminology it is common to speak of a thermal resistance:
thermqRT
A thermal resistance may also be associated with heat transfer by convection at a surface. From Newton’s law of cooling,
)( TThAq s
the thermal resistance for convection is then
hAq
TTR s
convt
1,
Applying thermal resistance concept to the plane wall, the equivalent thermal circuit for the plane wall with convection boundary conditions is shown in the figure below
The heat transfer rate may be determined from separate consideration of each element in the network. Since qx is constant throughout the network, it follows that
Ah
TT
kAL
TT
Ah
TTq ssss
x2
2,2,2,1,
1
1,1,
/1//1
In terms of the overall temperature difference , and the total thermal resistance Rtot,
the heat transfer rate may also be expressed as 2,1, TT
tot
x R
TTq 2,1,
Since the resistance are in series, it follows that
AhkA
L
AhRR ttot
21
11
Composite walls:
Thermal Resistances in Series: Consider three blocks, A, B and C, as shown. They are insulated on top, bottom, front and back. Since the energy will flow first through block A and then through blocks B and C, we say that these blocks are thermally in a series arrangement.
The steady state heat flow rate through the walls is given by:
TUA
Ahk
L
k
L
k
L
Ah
TT
R
TTq
C
C
B
B
A
Atx
21
2,1,2,1,
11
where AR
Utot
1 is the overall heat transfer coefficient. In the above case, U is expressed as
21
111
hk
L
k
L
k
L
h
U
C
C
B
B
A
A
Series-parallel arrangement:
The following assumptions are made with regard to the above thermal resistance model:
1) Face between B and C is insulated.
2) Uniform temperature at any face normal to X.
1-D radial conduction through a cylinder:
One frequently encountered problem is that of heat flow through the walls of a pipe or through the insulation placed around a pipe. Consider the cylinder shown. The pipe is either insulated on the ends or is of sufficient length, L, that heat losses through the ends is negligible. Assume no heat sources within the wall of the tube. If T1>T2, heat will flow outward, radially, from the inside radius, R1, to the outside radius, R2. The process will be described by the Fourier Law.
T2
T1
R1
R2
L
The differential equation governing heat diffusion is: 01
dr
dTr
dr
d
r
With constant k, the solution is
The heat flow rate across the wall is given by:
kAL
TTTT
L
kA
dx
dTkAq
ssssx
/
2,1,2,1,
Hence, the thermal resistance in this case can be expressed as: kLrr
2
ln2
1
Composite cylindrical walls:
Critical Insulation Thickness :
hLrkLR ir
r
tot )2(
1
2
)ln(
0
0
Insulation thickness : ro-ri
Objective : decrease q , increase totR
Vary ro ; as ro increases, first term increases, second term decreases.
This is a maximum – minimum problem. The point of extrema can be found by setting
00
dr
dRtot
02
1
2
12
0
ohLrLkr
or,
or, h
kr 0
In order to determine if it is a maxima or a minima, we make the second derivative zero:
at h
kr 00
2
2
o
tot
dr
Rd
h
krooo
tot
hLrLkrdr
Rd
0
222
2 1
2
1
0
2 3
2
Lk
h
Minimum q at ro =(k/h) = rcr (critical radius)
1-D radial conduction in a sphere:
k
rrR
rr
TTk
dr
dTkAq
TTTrT
dr
dTkr
dr
d
r
condt
ssr
rr
rrsss
4
/1/1
/1/1
4
)(
01
21,
21
2,1,
/1
/12,1,1,
22
21
1
2.6 Summary of Electrical Analogy
System Current Resistance Potential Difference
Electrical I R V Cartesian Conduction q
kA
L T
Cylindrical Conduction q
kL
rr
2
ln1
2
T
Conduction through sphere q
k
rr
4
/1/1 21 T
Convection q
1
h As T
2.7 One-Dimensional Steady State Conduction with Internal Heat Generation
Applications: current carrying conductor, chemically reacting systems, nuclear reactors.
Energy generated per unit volume is given by V
Eq
Plane wall with heat source: Assumptions: 1D, steady state, constant k, uniform q
dx
dTkq
TT
L
xTT
L
x
k
LqT
CC
CxCxk
qT
TTLx
TTLxk
q
dx
Td
x
ssss
s
s
:fluxHeat
221
2:solutionFinal
andfindtoconditionsboundaryUse2
:Solution
,
,:cond.Boundary
0
1,2,1,2,2
22
21
212
2,
1,
2
2
Note: From the above expressions, it may be observed that the solution for temperature is no longer linear. As an exercise, show that the expression for heat flux is no longer independent of x. Hence thermal resistance concept is not correct to use when there is internal heat generation.
Cylinder with heat source: Assumptions: 1D, steady state, constant k, uniform q
Start with 1D heat equation in cylindrical co-ordinates
s
s
Tr
rr
k
qr
dr
dTr
TTrr
k
q
20
22
0
0
14
)
0,0
,:
0
TS
dr
dTr
dr
d
r
(:olution
cond.Boundary
1
Exercise: Ts may not be known. Instead, T∞ and h may be specified. Eliminate Ts, using T∞ and h.
MODULE 2: Worked-out Problems Problem 1: The steady-state temperature distribution in a one–dimensional slab of thermal conductivity 50W/m.K and thickness 50 mm is found to be T= a+bx2, where a=2000C, b=-20000C/ m2, T is in degrees Celsius and x in meters. (a) What is the heat generation rate in the slab? (b) Determine the heat fluxes at the two wall faces. From the given temperature distribution and the heat fluxes obtained, can you comment on the heat generation rate? Known: Temperature distribution in a one dimensional wall with prescribed thickness and thermal conductivity. Find: (a) the heat generation rate, q in the wall, (b) heat fluxes at the wall faces and relation to q.
Schematic:
Assumptions: (1) steady-state conditions, (2) one –dimensional heat flow, (3) constant properties. Analysis: (a) the appropriate form of heat equation for steady state, one dimensional condition with constant properties is
3520.
2
.
.
m/W100.2K.m/W50)m/CC2000(2q
bk2]bx2[dx
dk
)bxa(dx
d
dx
dkq
dx
dT
dx
dKq
(b) The heat fluxes at the wall faces can be evaluated from Fourier’s law,
x
''x dx
dTk)x(q
Using the temperature distribution T(x) to evaluate the gradient, find
dx
dk)x(q ''
x [a+bx2]= -2kbx.
The flux at the face, is then x=0
2x
''
20''x
x''
m/W000,10)L(q
m050.0)m/C2000(K.m/W502kbL2)l(q,LatX
0)0(q
Comments: from an overall energy balance on the wall, it follows that
0EEE g
.
out
.
in
.
0Lq)L(q)0(q.
x''''
x
352"
x"x
.
m/W100.2m050.0
0m/w00,10
L
)0(q)L(qq
Problem 2: Consider a solar pond having three distinct layers of water-salt solution. The top and bottom layers are well mixed with salt. These layers are subjected to natural convention, but the middle layer is stationary. With this arrangement, the top and bottom surfaces of the middle layer is maintained at uniform temperature T1 and T2, where T1>T2. Solar radiation is absorbed in the middle layer in the form q=Ae-mx , resulting in the following temperature distribution in the central layer
CBxeka
AxT mx
2)(
In the above equation, k is the thermal conductivity, and the constants A (W/m3), a (1/m), B (K/m) and C(K) are also known. Obtain expressions for the interfacial heat flux from the bottom layer to the middle layer, and from the middle layer to the top layer. Are the conditions are steady or transient? Next, obtain an expression for the rate at which thermal energy is generated in the entire middle layer, per unit surface area.
Known: Temperature distribution and distribution of heat generation in central layer of a solar pond.
Find: (a) heat fluxes at lower and upper surfaces of the central layer, (b) whether conditions are steady or transient (c) rate of thermal energy generation for the entire central layer.
Schematic:
Assumptions: (1) central layer is stagnant, (2) one-dimensional conduction, (3)constant properties. Analysis (1) the desired fluxes correspond to conduction fluxes in the central layer at the lower and upper surfaces. A general form for the conduction flux is
Bkm
AkqqBe
km
Akqq
Hence
Bekm
Akq
xconduml
Lxcondl
mxcond
")0(
""0(
"
"
(b) Conditions are steady if T/t=0. Applying the heat equation,
t
T1
k
q
t
T.
2
2
t
ek
Ae
k
A mxmx
1
Hence conditions are steady since
t
T
=0 (for all 0<=x<=L)
For the central layer, the energy generation is
)1()1(0
.
00
".
mLmLLmxg
LLg
ea
Ae
a
Ae
a
AE
mxdxeAqdxE
Alternatively, from an overall energy balance,
g
."1
"2 Eqq =0 = (-q”cond(x=0))-(q”cond(x=L))
"
g
.
E = "1q - "
2q
)1(.
mLmLg e
m
ABe
km
AKB
km
AkE
Comments: Conduction is the negative x-direction, necessitating use of minus signs in the above energy balance.
Problem 3: Consider 1D heat transfer across a slab with thermal conductivity k and thickness L. The steady state temperature is of the form T=Ax3+Bx2+Cx+D. Find expressions for the heat generation rate per unit volume in the slab and heat fluxes at the two wall faces (i.e. x=0, L). Known: steady-state temperature distribution in one-dimensional wall of thermal conductivity, T(x)=Ax3+Bx2+CX+d. Find: expressions for the heat generation rate in the wall and the heat fluxes at the two wall faces(x=0, L). Assumptions: (1) steady state conditions, (2) one-dimensional heat flow, (3) homogeneous medium. Analysis: the appropriate form of the heat diffusion equation for these conditions is
0k
q
dx
Td.
2
2
Or 2
2.
dx
Tdkq
Hence, the generation rate is
]B2Ax6[kq
]0CBx2Ax3[dx
dk
dx
dT
dx
dq
.
2.
which is linear with the coordinate x. The heat fluxes at the wall faces can be evaluated from Fourier’s law,
]CBx2Ax3[kdx
dTkq 2"
x
Using the expression for the temperature gradient derived above. Hence, the heat fluxes are:
Surface x=0; (0)=-kC "xq
Surface x=L; "xq (L) = -K [3AL2+2BL+C]
COMMENTS: (1) from an over all energy balance on the wall, find
BkL2AkL3E
0E]CBL2AL3)[K()kC()L(q)0(q
0EEE
2''
g
.
g
.2
x"
x"
g
.
out
.
in
.
From integration of the volumetric heat rate, we can also find
BkL2AkL3E
]Bx2Ax3[kdx]B2Ax6[kdx)x(qE
2g
''.
L
0
L0
2.
L0
''
g
.
Problem 4: Consider a one dimensional system of mass M with constant properties and no internal heat generation as shown in the figure below. The system is initially at a uniform temperature Ti. The electrical heater is suddenly switched ON, resulting in a uniform heat flux q”o at the surface x=0 . The boundaries at x=L and elsewhere are perfectly insulated.
(a) Set up the differential equation along with the boundary and initial conditions for the
temperature T(x,t). (b) Sketch the temperature variation with x for the initial condition (t<=0) and for several
times after the heater is switched ON. Comment whether a steady-state temperature distribution will ever be reached.
(c) For any given time, sketch the heat flux variation with x. Choose the following planes: x=0, x=L/2, and x=L.
(d) After time te, the heater power is switched off. Assuming no heat loss, derive an expression determine Tf , the final uniform temperature, as a function of the relevant parameters.
Known: one dimensional system, initially at a uniform temperature Ti, is suddenly exposed to a uniform heat flux at one boundary while the other boundary is insulated. Find: (a) proper form of heat diffusion equation; identify boundary and initial conditions, (b) sketch temperature distributions for following conditions: initial condition (t<=0), several times after heater is energized ;will a steady-state condition be reached?, (c) sketch heat flux for x=0, L/2, L as a function of time, (d) expression for uniform temperature, Tf, reached after heater has been switched off the following an elapsed time , te, with the heater on.]
Schematic: Assumptions: (1) one dimensional conduction, (2) no internal heat generation, (3) constant properties. Analysis: (a) the appropriate form of the heat equation follows. Also the appropriate boundary and initial conditions are:
t
T1
x
T2
2
Initial condition: T(x, 0) =Ti uniform temperature
Boundary conditions: x=0 0t)x/Tkq 0"o
x=L 0)x/T L Insulated
(b) The temperature distributions are as follows:
No steady-state condition will be reached since is constant. in
..
outin
.
EandEE
(c) The heat flux as a function of time for positions x=0, L/2 and L appears as:
( d) If the heater is energized until t=to and then switched off, the system will eventually reach a uniform temperature , Tf. Perform an energy balance on the system, for an interval of time t=te,
.
stin
.
EE es"os
"0
t0inin tAqdtAqQE e )TT(McE ifst
It follows that = OR es"o tAq )TT(Mc if if TT
Mc
tAq es"o
Problem 5:
A 1–m-long metal plate with thermal conductivity k=50W/m.K is insulated on its sides. The top surface is maintained at 1000C while the bottom surface is convectively cooled by a fluid at 200C. Under steady state conditions and with no volumetric heat generation, the temperature at the midpoint of the plate is measured to be 850C. Calculate the value of the convection heat transfer coefficient at the bottom surface. Known: length, surface thermal conditions, and thermal conductivity of a Plate. Plate midpoint temperature. Find: surface convection coefficient
Schematic: Assumptions: (1) one-dimensional, steady conduction with no generation, (2) Constant properties
Analysis: for prescribed conditions, is constant. Hence,
2/L
TTq 21"
cond = 2
0
m/W1500k.m/W50/m5.0
C15
K.m/W30h
m/W1500W/K.m)h/102.0(
C30
)h/1()k/L(
TT"q
2
22
01
Comments: The contributions of conduction and convection to the thermal resistance are
W/K.m033.0
h
1R
W/K.m02.0K
LR
2cond,t
"
2cond,t
"
Problem 6:
The wall of a building is a multi-layered composite consisting of brick (100-mm layer), a 100-mm layer of glass fiber(paper faced. 28kg/m2), a 10-mm layer of gypsum plaster (vermiculite), and a 6-mm layer of pine panel. If hinside is 10W/m2.K and houtside is 70W/m2.K, calculate the total thermal resistance and the overall coefficient for heat transfer. Known: Material thickness in a composite wall consisting of brick, glass fiber, and vermiculite and pine panel. Inner and outer convection coefficients. Find: Total thermal resistance and overall heat transfer coefficient.
Schematic:
Kb Kgl
Brick GypsumglassPine panel,Kp
100mm 10mm 6mm
hi=10W/m2.Kh0=70W/m2.K
0h
1
b
b
K
L
gl
lg
k
L
gy
gy
k
Lp
p
K
L
ih
1
Assumptions: (1) one dimensional conduction, (2) constant properties, (3) negligible contact resistance.
Properties: T= 300K: Brick, kb=1.3 W/m.K: Glass fiber (28kg/m3), kg1= 0.038W/m.K: gypsum, kgy=0.17W/m.K: pine panel, kp=0.12W/m.K. Analysis: considering a unit surface Area, the total thermal resistance
W/K..m93.2R
W/K.m)1.00500.00588.06316.20769.00143.0(R
W
K.m
10
1
12.0
006.0
17.0
01.0
038.0
1.0
3.1
1.0
70
1R
h
1
K
L
k
L
k
L
K
L
h
1R
2"tot
2"tot
2"tot
ip
p
gy
gy
1g
1g
B
B
0
"tot
The overall heat transfer coefficient is
.K.m/W341.0U
)W/K.m93.2(R
1
AR
1U
2
12
tot"
tot
Comments: an anticipated, the dominant contribution to the total resistance is made by the insulation.
Problem 7:
The wall of an oven is a composite of the following layers. Layers A has a thermal conductivity kA=20W/m.K, and layer C has a thermal conductivity kC=50W/m.K. The corresponding thicknesses are LA=0.30m and LC=0.15m, respectively. Layer B is sandwiched between layers A and C, is of known thickness, LB=0.15m, but unknown thermal conductivity kB. Under steady-state operating conditions, the outer surface temperature is measured to be Ts,0=200C. Measurements also tell us that the inner surface temperature Ts,i is 6000C and the oven air temperature is T =8000C. The inside convection coefficient h is known to be 25W/m2.K. Find the value of kB.
Known: Thickness of three material which form a composite wall and thermal conductivities of two of the materials. Inner and outer surface temperatures of the composites; also, temperature and convection coefficient associated with adjoining gas.
Find: value of unknown thermal conductivity, kB.
Schematic:
Ts,0=200C
LA=0.3m LB=LC=0.15m kA=20W/m.K kC=50W/m.K
KC
T=8000C h= 25W/m2.K
KB KA Ts,i=6000C
LA LB LC Assumptions: (1) steady state conditions, (2) one-dimensional conduction, (3) constant properties, (4) negligible contact resistance, (5) negligible radiation effects.
Analysis: Referring to the thermal circuit, the heat flux may be expressed as
2
B
B
0
C
C
B
B
A
A
0,si,s"
m/WK/15.0018.0
580
K.m/W50
m15.0
K
m15.0
018.0
m3.0C)20600(
K
L
K
L
K
L
TTq
The heat flux can be obtained from
2''
02i,s
"
m/W5000q
C)600800(K.m/W25)TT(hq
Substituting for heat flux,
.K.m/W53.1K
098.0018.05000
580018.0
q
580
K
15.0
B
"B
Comments: In an over, radiation effects are likely to have a significant influence on the net heat flux at the inner surface of the oven.
Problem 8:
A steam pipe of 0.12 m outside diameter is insulated with a 20-mm-thick layer of calcium silicate. If the inner and outer surfaces of the insulation are at temperatures of Ts,1=800 K and Ts,2=490 K, respectively, what is the heat loss per unit length of the pipe?
Known: Thickness and surface temperature of calcium silicate insulation on a steam pipe. Find: heat loss per unit pipe length.
Schematic:
Ts,1=800K
D2=0.16m
Steam
Ts,2=490K D1=0.12m
Calcium silicate insulation
Assumptions: (steady state conditions, (2) one-dimensional conduction, (3) constant properties.
Properties: calcium silicate (T=645K): k=0.089W/m.K
Analysis: The heat per unit length is
m/W603q
)m12.0/m16.0ln(
K)490800)(K.m/W089.0(2q
)D/Dln(
)TT(K2
q
'r
'r
12
2,s1,s
L
r'r
Comments: heat transferred to the outer surface is dissipated to the surroundings by convection and radiation.
Problem 9:
A cylindrical nuclear fuel rod of 0.1m dia has a thermal conductivity k=0.0W/m.K and generates uniformly 24,000W/m3. This rod is encapsulated within another cylinder having an outer radius of 0.2m and a thermal conductivity of 4W/m.K. The outer surface is cooled by a coolant fluid at 1000C, and the convection coefficient between the outer surface and the coolant is 20W/m2.K. Find the temperatures at the interface between the two cylinders and at the outer surface.
Known: A cylindrical rod with heat generation is cladded with another cylinder whose outer surface is subjected to a convection process. Find: the temperature at the inner surfaces, T1, and at the outer surface, Tc.
Schematic:
Assumptions: (1) steady-state conditions, (2) one-dimensional radial conduction, (3), negligible contact resistance between the cylinders. Analysis: The thermal circuit for the outer cylinder subjected to the convection process is
o
'2
2
1o'1
r2h
1R
k2
r/rlnR
Using the energy conservation requirement, on the inner cylinder,
g
.
out
.
EE Find that
21
.
1' rqq
The heat rate equation has the form hence,R/Tq '.
'''2
'1
'i R/Tandq)RR(qTT
Numerical values:
m/W0.754m)1.0(m/W000,24q
W/m.K0398.0m20.02K.m/W20/1R
W/m.K0276.0K.m/W42/1.0/2.0lnR
223'
2'2
'1
Hence
C13030100W/m.K0398.0m/W0.754C100T
C8.1508.50100W/m.K)cccc0276.0(m/W0.754C100T00
C
00i
Comments: knowledge of inner cylinder thermal conductivity is not needed.
Problem 10:
A steel cable having a diameter of 0.005m and an electrical resistance of 6*10-4 /m carries an electrical current of 700 A. The surrounding temperature of the cable is 300°C, and the effective coefficient associated with heat loss by convection and radiation between the cable and the environment is approximately 25W/m2.K.
(a) If the cable is uncoated, what is its surface temperature? (b) If a very thin coating of electrical insulation is applied to the cable, with a contact resistance of 0.02m2K/W, what are the insulation and cable surface temperatures? (c) What thickness of this insulation (k=0.5W/m.K) will yields the lowest value of the maximum insulation temperature? What is the value of the maximum temperature when the thickness is used? Known: electric current flow, resistance, diameter and environmental conditions associated with a cable. Find: (a) surface temperature of bare cable, (b) cable surface and insulation temperatures for a thin coating of insulation, (c) insulation thickness which provides the lowest value of the maximum insulation temperature. Corresponding value of this temperature.
Schematic:
T∞
Ts
Eg
q
Assumptions: (1) steady-state conditions, (2) one-dimensional conduction in r, (3) constant properties.
Analysis: (a) the rate at which heat is transferred to the surroundings is fixed by the rate of heat generation in the cable. Performing an energy balance for a
control surface about the cable, it follows that or, for the bare cable,
qEg
.
/10 4 .m/W294)m6()A700(RIwithq).TT)(LD(hLRI 2'e
2'si
'e
2
It follows that
C7.778T
)m005.0()K.m/W25(
m/W294C30
Dh
qTT
0s
20
i
'
s
(b) With thin coating of insulation, there exists contact and convection resistances to heat transfer from the cable. The heat transfer rate is determined by heating within the cable, however, and therefore remains the same,
h
1R
)TT(Dq
LDh
1
LD
RTT
LDh
1R
TTq
c,t
si'
ii
c,t
s
ic,t
s
And solving for the surface temperature, find
C1153T
C30W
K.m04.0
W
K.m02.0
)m005.0(
m/W294T
h
1R
D
qT
0s
022
c,ti
'
s
The insulation temperature is then obtained from
e,t
s
R
TTq
Or
C7.778T
)m005.0(W
K.m02.0
m
W294
C1153LD
RqC1153qRTT
0i
2
0
i
c,t"
0c,tsi
(c) The maximum insulation temperature could be reduced by reducing the resistance to heat transfer from the outer surface of the insulation. Such a reduction is possible Di<Dcr.
m02.0K.m/W25
K.m/W5.0
h
kr
2cr
Hence, Dcr =0.04m> Di =0.005m. To minimize the maximum temperature, which exists at the inner surface of the insulation, add insulation in the amount.
mt
mDDDDt icri
0175.02
)005.004.0(
220
The cable surface temperature may then be obtained from
C318.2T
π(0.005m)W
.Km0.02
m
W294
C692.5LπD
RqTqRTT
/RTTq , that grecognizin
C692.5T
2.25m.K/W
C30T
0.32)m.K/W0.66(1.27
C30T
m
W294
hence,
π(0.04m).Km
W25
1
)2ππ(0.5W/.
005)ln(0.04/0.
π(0.005m)
.K/W0.02m
C30T
hππ
1
2ππ
)/Dln(D
πD
R
TTq
0i
2
0
i
"ct,
sc,t,si
c,t,is
0s
0s
0s
2
2
0s
rc,
irc,
i
"ct,
s'`
Comments: use of the critical insulation in lieu of a thin coating has the effect of reducing the maximum insulation temperature from 778.70C to 318.20C. Use of the critical insulation thickness also reduces the cable surface temperatures to 692.50C from 778.70C with no insulation or fro 11530C with a thin coating.
MODULE 6
CONVECTION 6.1 Objectives of convection analysis: Main purpose of convective heat transfer analysis is to determine:
- flow field - temperature field in fluid
- heat transfer coefficient, h
How do we determine h ? Consider the process of convective cooling, as we pass a cool fluid past a heated wall. This process is described by Newton’s law of Cooling:
q=h·A·(TS-T)
q” y y
U T T(y)
Ts
u(y) U
Near any wall a fluid is subject to the no slip condition; that is, there is a stagnant sub layer. Since there is no fluid motion in this layer, heat transfer is by conduction in this region. Above the sub layer is a region where viscous forces retard fluid motion; in this region some convection may occur, but conduction may well predominate. A careful analysis of this region allows us to use our conductive analysis in analyzing heat transfer. This is the basis of our convective theory. At the wall, the convective heat transfer rate can be expressed as the heat flux.
TThy
Tkq s
y
fconv
0
y
T(y)
T U
Ts
Hence,
TT
y
Tk
hs
y
f
0
But 0
yy
T depends on the whole fluid motion, and both fluid flow and heat transfer
equations are needed The expression shows that in order to determine h, we must first determine the temperature distribution in the thin fluid layer that coats the wall. 2.2 Classes of Convective Flows
• extremely diverse • several parameters involved (fluid properties, geometry, nature of flow, phases etc) • systematic approach required • classify flows into certain types, based on certain parameters • identify parameters governing the flow, and group them into meaningful non-
dimensional numbers • need to understand the physics behind each phenomenon
Common classifications: A. Based on geometry: External flow / Internal flow B. Based on driving mechanism Natural convection / forced convection / mixed convection C. Based on number of phases Single phase / multiple phase D. Based on nature of flow Laminar / turbulent
Forced convection (induced by external means)
Convection
Free or natural convection (induced by buoyancy forces) May occur
with phase change (boiling, condensation)
Table 6.1. Typical values of h (W/m2K)
Free convection gases: 2 - 25 liquid: 50 – 100 Forced convection gases: 25 - 250 liquid: 50 - 20,000 Boiling/Condensation 2500 -100,000
2.3 How to solve a convection problem ?
• Solve governing equations along with boundary conditions • Governing equations include
1. conservation of mass 2. conservation of momentum 3. conservation of energy
• In Conduction problems, only (3) is needed to be solved. Hence, only few parameters are involved
• In Convection, all the governing equations need to be solved. large number of parameters can be involved 2.4 FORCED CONVECTION: external flow (over flat plate) An internal flow is surrounded by solid boundaries that can restrict the development of its boundary layer, for example, a pipe flow. An external flow, on the other hand, are flows over bodies immersed in an unbounded fluid so that the flow boundary layer can grow freely in one direction. Examples include the flows over airfoils, ship hulls, turbine blades, etc
U
U < U
U
• Fluid particle adjacent to the solid surface is at rest • These particles act to retard the motion of adjoining layers • boundary layer effect
Inside the boundary layer, we can apply the following conservation principles: Momentum balance: inertia forces, pressure gradient, viscous forces, body forces Energy balance: convective flux, diffusive flux, heat generation, energy storage
2.5 Forced Convection Correlations Since the heat transfer coefficient is a direct function of the temperature gradient next to the wall, the physical variables on which it depends can be expressed as follows: h=f(fluid properties, velocity field ,geometry,temperature etc.) As the function is dependent on several parameters, the heat transfer coefficient is usually expressed in terms of correlations involving pertinent non-dimensional numbers.
Forced convection: Non-dimensional groupings
• Nusselt No. Nu = hx / k = (convection heat transfer strength)/ (conduction heat transfer strength)
• Prandtl No. Pr = / = (momentum diffusivity)/ (thermal diffusivity) • Reynolds No. Re = U x / = (inertia force)/(viscous force)
Viscous force provides the dampening effect for disturbances in the fluid. If dampening is strong enough laminar flow Otherwise, instability turbulent flow critical Reynolds number For forced convection, the heat transfer correlation can be expressed as
Nu=f (Re, Pr)
The convective correlation for laminar flow across a flat plate heated to a constant wall temperature is:
x
U
Nux = 0.323·Rex½ · Pr1/3
where
Nux hx/k
Rex (Ux)/
Pr cP/k
Physical Interpretation of Convective Correlation
The Reynolds number is a familiar term to all of us, but we may benefit by considering what the ratio tells us. Recall that the thickness of the dynamic boundary layer, , is proportional to the distance along the plate, x.
Rex (Ux)/ (U)/ = (U2)/( U/)
The numerator is a mass flow per unit area times a velocity; i.e. a momentum flow per unit area. The denominator is a viscous stress, i.e. a viscous force per unit area. The ratio represents the ratio of momentum to viscous forces. If viscous forces dominate, the flow will be laminar; if momentum dominates, the flow will be turbulent.
Physical Meaning of Prandtl Number The Prandtl number was introduced earlier.
If we multiply and divide the equation by the fluid density, , we obtain:
Pr (/)/(k/cP) = /
The Prandtl number may be seen to be a ratio reflecting the ratio of the rate that viscous forces penetrate the material to the rate that thermal energy penetrates the material. As a consequence the Prandtl number is proportional to the rate of growth of the two boundary layers:
/t = Pr1/3
Physical Meaning of Nusselt Number The Nusselt number may be physically described as well.
Nux hx/k
If we recall that the thickness of the boundary layer at any point along the surface, , is also a function of x then
Nux h/k (/kA)/(1/hA)
We see that the Nusselt may be viewed as the ratio of the conduction resistance of a material to the convection resistance of the same material.
Students, recalling the Biot number, may wish to compare the two so that they may distinguish the two.
Nux hx/kfluid Bix hx/ksolid
The denominator of the Nusselt number involves the thermal conductivity of the fluid at the solid-fluid convective interface; The denominator of the Biot number involves the thermal conductivity of the solid at the solid-fluid convective interface.
Local Nature of Convective Correlation Consider again the correlation that we have developed for laminar flow over a flat plate at constant wall temperature
Nux = 0.323·Rex½ · Pr1/3
To put this back into dimensional form, we replace the Nusselt number by its equivalent, hx/k and take the x/k to the other side:
h = 0.323·(k/x)Rex½ · Pr1/3
Now expand the Reynolds number
h = 0.323·(k/x)[(Ux)/]½ · Pr1/3
We proceed to combine the x terms:
h = 0.323·k[(U)/( x)]½ · Pr1/3
And see that the convective coefficient decreases with x½.
We see that as the boundary layer thickens, the convection coefficient decreases. Some designers will introduce a series of “trip wires”, i.e. devices to disrupt the boundary layer, so that the buildup of the insulating layer must begin anew. This will result in regular “thinning” of the boundary layer so that the convection coefficient will remain high.
Averaged Correlations If one were interested in the total heat loss from a surface, rather than the temperature at a point, then they may well want to know something about average convective coefficients.
U
x Hydrodynamic Boundary Layer,
Convection Coefficient, h.
Thermal Boundary Layer, t
x
Average Convection Coefficient, hL
U
Local Convection Coefficient, hx.
The desire is to find a correlation that provides an overall heat transfer rate:
Q = hLA[Twall-T] = h T T dA h T T dxx wall x
L
wall [ ] [ ]0
where hx and hL, refer to local and average convective coefficients, respectively. Compare the second and fourth equations where the area is assumed to be equal to A = (1L):
hLL[Twall-T] = h T T dxx
L
wall0 [ ]
Since the temperature difference is constant, it may be taken outside of the integral and cancelled:
hLL= h dxx
L
0 This is a general definition of an integrated average. Proceed to substitute the correlation for the local coefficient.
hLL= 0 3230
0
1/3. P
.5
k
x
U xdx
L
r
Take the constant terms from outside the integral, and divide both sides by k.
hLL/k = 0 3231
0
1/3
0
0. Pr
.5 .5
U
xdx
L
Integrate the right side.
hLL/k = 0 32305
0
1/3
0
0
. Pr.
.5 .5
U x
L
The left side is defined as the average Nusselt number, NuL. Algebraically rearrange the right side.
NuL = 3
15.0
5.03
15.0
Pr646.0Pr5.0
323.0
LU
LU
The term in the brackets may be recognized as the Reynolds number, evaluated at the end of the convective section. Finally,
315.0 PrRe646.0 L NuL =
This is our average correlation for laminar flow over a flat plate with constant wall temperature.
Reynolds Analogy In the development of the boundary layer theory, one may notice the strong relationship between the dynamic boundary layer and the thermal boundary layer. Reynold’s noted the strong correlation and found that fluid friction and convection coefficient could be related. This is known as the Reynolds Analogy.
Conclusion from Reynold’s analogy: Knowing the frictional drag, we know the Nusselt Number. If the drag coefficient is increased, say through increased wall roughness, then the convective coefficient will also increase. Turbulent Flow We could develop a turbulent heat transfer correlation in a manner similar to the von Karman analysis. It is probably easier, having developed the Reynolds analogy, to follow that course. The local fluid friction factor, Cf, associated with turbulent flow over a flat plate is given as:
Cf = 0.0592/Rex0.2
Substitute into the Reynolds analogy:
(0.0592/Rex0.2)/2 = Nux/RexPr1/3
Rearrange to find
Nux = 0.0296Rex0.8Pr1/3
Local Correlation Turbulent Flow Flat Plate.
In order to develop an average correlation, one would evaluate an integral along the plate similar to that used in a laminar flow:
Laminar Region Turbulent region
hLL = h dx h dx h dxx
L
x la ar
L
x turbulentL
Lcrit
crit0 0 , min ,
Note: The critical Reynolds number for flow over a flat plate is 5105; the critical Reynolds number for flow through a round tube is 2000.
The result of the above integration is:
Nux = 0.037(Rex
0.8 – 871)Pr1/3
Note: Fluid properties should be evaluated at the average temperature in the boundary layer, i.e. at an average between the wall and free stream temperature.
Tprop = 0.5(Twall+ T)
2.6 Free convection Free convection is sometimes defined as a convective process in which fluid motion is caused by buoyancy effects.
Heated boundary layer
T < Tboundry. layer < Tw < boundry. layer
Tw Velocity Profiles
Compare the velocity profiles for forced and natural convection shown below:
Forced Convection Free Convection
U = 0 U > 0
Coefficient of Volumetric Expansion The thermodynamic property which describes the change in density leading to buoyancy in the Coefficient of Volumetric Expansion, .
1
TP Const .
Evaluation of
Liquids and Solids: is a thermodynamic property and should be found from
Property Tables. Values of are found for a number of engineering fluids in Tables given in Handbooks and Text Books.
Ideal Gases: We may develop a general expression for for an ideal gas from the ideal gas law:
P = RT
Then, = P/RT
Differentiating while holding P constant:
d
dT
P
R T
R T
R T TP Const
.
2 2
Substitute into the definition of
1
Tabs
Grashof Number Because U is always zero, the Reynolds number, [UD]/, is also zero and is no longer suitable to describe the flow in the system. Instead, we introduce a new parameter for natural convection, the Grashof Number. Here we will be most concerned with flow across a vertical surface, so that we use the vertical distance, z or L, as the characteristic length.
Grg T L
3
2
Just as we have looked at the Reynolds number for a physical meaning, we may consider the Grashof number:
Ideal Gas
Grg T L
g T LL
U
UL
Buoyant ForceArea
MomentumArea
ViscousForceArea
2 3
2
3
22
2
2
2
2
( ) ( )max
max
Free Convection Heat Transfer Correlations The standard form for free, or natural, convection correlations will appear much like those for forced convection except that (1) the Reynolds number is replaced with a Grashof number and (2) the exponent on Prandtl number is not generally 1/3 (The von Karman boundary layer analysis from which we developed the 1/3 exponent was for forced convection flows):
Nux = CGrxmPrn Local Correlation
NuL = CGrL
mPrn Average Correlation
Quite often experimentalists find that the exponent on the Grashof and Prandtl numbers are equal so that the general correlations may be written in the form:
Nux = C[GrxPr]m Local Correlation
NuL = C[GrLPr]m Average Correlation
This leads to the introduction of the new, dimensionless parameter, the Rayleigh number, Ra: Rax = GrxPr
RaL = GrLPr
So that the general correlation for free convection becomes:
Nux = CRaxm Local Correlation
NuL = CRaL
m Average Correlation
Laminar to Turbulent Transition
Just as for forced convection, a boundary layer will form for free convection. The boundary layer, which acts as a thermal resistance, will be relatively thin toward the leading edge of the surface resulting in a relatively high convection coefficient. At a Rayleigh number of about 109 the flow over a flat plate will become transitional and finally become turbulent. The increased turbulence inside the boundary layer will enhance heat transfer leading to relative high convection coefficients because of better mixing.
Tu
rbu
len
t F
low
L
amin
ar F
low
Ra < 109 Laminar flow. [Vertical Flat Plate] Ra > 109 Turbulent flow. [Vertical Flat Plate]
Generally the characteristic length used in the correlation relates to the distance over which the boundary layer is allowed to grow. In the case of a vertical flat plate this will be x or L, in the case of a vertical cylinder this will also be x or L; in the case of a horizontal cylinder, the length will be d.
Critical Rayleigh Number
Consider the flow between two surfaces, each at different temperatures. Under developed flow conditions, the interstitial fluid will reach a temperature between the temperatures of the two surfaces and will develop free convection flow patterns. The fluid will be heated by one surface, resulting in an upward buoyant flow, and will be cooled by the other, resulting in a downward flow.
If the surfaces are placed closer together, the flow patterns will begin to interfere:
T1
Q
T2
Q
L
T1
Q
T2
Q
L
T1
Q
T2
Q
L
Free Convection Inside an Enclosure With Complete Flow
Interference (Channel flow limit)
Free Convection Inside an Enclosure (boundary layer limit)
Note that for enclosures it is customary to develop correlations which describe the overall (both heated and cooled surfaces) within a single correlation.
Free Convection Inside an Enclosure With Partial Flow
Interference
In the case of complete flow interference, the upward and downward forces will cancel, canceling circulation forces. This case would be treated as a pure convection problem since no bulk transport occurs. The transition in enclosures from convection heat transfer to conduction heat transfer occurs at what is termed the “Critical Rayleigh Number”. Note that this terminology is in clear contrast to forced convection where the critical Reynolds number refers to the transition from laminar to turbulent flow.
Racrit = 1000 (Enclosures With Horizontal Heat Flow)
Racrit = 1728 (Enclosures With Vertical Heat Flow)
The existence of a Critical Rayleigh number suggests that there are now three flow regimes: (1) No flow, (2) Laminar Flow and (3) Turbulent Flow. In all enclosure problems the Rayleigh number will be calculated to determine the proper flow regime before a correlation is chosen.
MODULE 6: Worked-out Problems Problem 1:
For laminar free convection from a heated vertical surface, the local convection coefficient may be expressed as hx=Cx -1/4, where hx is the coefficient at a distance x from the leading edge of the surface and the quantity C, which depends on the fluid properties, is independent of x.
Obtain an expression for the ratio , where is the average coefficient between the leading edge (x=0) and the x location. Sketch the
variation of hx and with x.
xx hh /
xh
xh
Schematic:
Boundary layer,hx=C x
-1/4 w here C is a constant
Ts
x
Analysis: It follows that average coefficient from 0 to x is given by
x1/4
x
x
0
1/4x
0
xx
h3
4Cx
3
4x3/4
x
C
3
4h
dxxx
Cdxh
x
1h
Hence 3
4
h
h
x
x
The variation with distance of the local and average convection coefficient is shown in the sketch.
xx hh
xx hh3
4
4/1Cxhx
Comments: note that x =4/3, independent of x. hence the average
coefficients for an entire plate of length L is =4/3L, where hL is the local coefficient at x=L. note also that the average exceeds the local. Why?
hhx /
Lh
Problem 2:
Experiments to determine the local convection heat transfer coefficient for uniform flow normal to heated circular disk have yielded a radial Nusselt number distribution of the form
n
oD r
ra1
k
h(r)DNu
Where n and a are positive. The Nusselt number at the stagnation point is correlated in terms of the Reynolds number (ReD=VD/) and Prandtl
0.361/2Do Pr0.814Re
k
0)Dh(rNu
Obtain an expression for the average Nusselt number, , corresponding to heat transfer from an isothermal disk. Typically boundary layer development from a stagnation point yields a decaying convection coefficient with increasing distance from the stagnation point. Provide a plausible for why the opposite trend is observed for the disk.
kDhNuD /
Known: Radial distribution of local convection coefficient for flow normal to a circular disk.
Find: Expression for average Nusselt number. Schematic:
Assumptions: Constant properties.
Analysis: The average convection coefficient is
o
o
s
r
0
no
2n2
3o
no0
r
02o
A
ss
2)r(n
ar
2
r
r
kNuoh
]2π2πr)a(r/r[1NuD
k
πr
1h
hdAA
1h
Where Nuo is the Nusselt number at the stagnation point (r=0).hence,
0.361/2D
o
r
o
2n2
o
Pr2)]0.841Re2a/(n[1NuD
2)]2a/(n[1NuNuD
ro
r
2)(n
a
2
r/ro2Nu
k
DhNuD
o
Comments: The increase in h(r) with r may be explained in terms of the sharp turn, which the boundary layer flow must take around the edge of the disk. The boundary layer accelerates and its thickness decreases as it makes the turn, causing the local convection coefficient to increase.
Problem 3: In a flow over a surface, velocity and temperature profiles are of the forms
u(y)=Ay+By2-Cy3 and T(y)=D+Ey+Fy2-Gy3
Where the coefficients A through G are constants. Obtain expressions for friction coefficients Cf and the convection coefficient h in terms of u∞, T∞ and appropriate profile coefficients and fluid properties. Known: form of the velocity and temperature profiles for flow over a surface. Find: expressions for the friction and convection coefficients. Schematic:
u∞,T∞
T(y)=D+Ey+Fy2-Gy3
Ts=T(0)=D
U(y)=Ay-By2-Cy
3
y y
Analysis: The shear stress at the wall is
.]32[ 02
0
ACyByAy
uy
y
s
Hence, the friction coefficient has the form,
2f
22s
f
u
2AνC
ρu
2Aμ
/2ρu
τC
The convection coefficient is
TD
Ekh
TD
]3Gy2Fy[Ek
TT
y)T/(kh
f
0y2
f
s
0yf
Comments: It is a simple matter to obtain the important surface parameters from knowledge of the corresponding boundary layers profile. However is rarely simple matter to determine the form of the profile.
Problem 4: In a particular application involving airflow over a heated surface, the boundary layer temperature distribution may be approximated as
v
yuPrexp1
TT
TT
s
s
Where y is the distance normal to the surface and the Prandtl number, Pr=cp/k=0.7, is a dimensionless fluid property. If T∞ =400K, Ts=300K, and u∞/v=5000m-1, what is the surface heat flux? Known: Boundary layer temperature distribution Find: Surface heat flux. Schematic:
)Prexp(1)(
yu
TT
TyT
s
s
Properties: Air ( ): k = 0.0263 W/m.k kTS 300
Analysis: Applying the Fourier’s law at y=0, the heat flux is
0y
s
0y
"s ν
yuPrexp
ν
uPr)Tk(T
y
Tkq
ν
u)PrTk(Tq s
"s
1/m0.7x5000.K(100K)0.02063w/mq"s
Comments: (1) Negligible flux implies convection heat transited surface (2) Note use of k at to evaluate from Fourier’s law. sT "
sq
Problem 5:
Consider a lightly loaded journal bearing using oil having the constant properties =10-2 kg/s-m and k=0.15W/m. K. if the journal and the bearing are each mentioned at a temperature of 400C, what is the maximum temperature in the oil when the journal is rotating at 10m/s?
Known: Oil properties, journal and bearing temperature, and journal speed for lightly loaded journal bearing. Find: Maximum oil temperature. Schematic:
Assumptions: (1) steady-state conditions, (2) Incompressible fluid with constant properties, (3) Clearances is much less than journal radius and flow is Couette. Analysis: The temperature distribution corresponds to the result obtained in the text example on Couette flow.
22
o L
y
L
yU
2k
μTT(y)
The position of maximum temperature is obtained from
22
L
2y
L
1U
2k
μ0
dy
dT
y=L/2. Or, The temperature is a maximum at this point since hence 0.T/dyd 22
22
2 U8k
μTo
4
1
2
1U
2k
μToT(l/2)axTm,
C40.83T
0.15W/m.K8
/s)kg/s.m(10m10C40T
max
22
max
Comments: Note that Tmax increases with increasing µ and U, decreases with increasing k, and is independent of L.
Problem 6: Consider two large (infinite) parallel plates, 5mm apart. One plate is stationary, while the other plate is moving at a speed of 200m/s. both plates are maintained at 27C. Consider two cases, one for which the plates are separated by water and the other for which the plates are separated by air.
For each of the two fluids, which is the force per unit surface area required to maintain the above condition? What is the corresponding requirement? What is the viscous dissipation associated with each of the two fluids? What is the maximum temperature in each of the two fluids?
Known: conditions associated with the Couette flow of air or water. Find: (a) Force and power requirements per unit surface area, (2) viscous dissipation,(3) maximum fluid temperature. Schematic:
Assumptions: (1) Fully developed Couette flow, (2) Incompressible fluid with constant properties. Properties: Air (300K); µ=184.6*10-7 N.s/m2, k=26.3*10-3W/m.K; water (300K): µ=855*10—6N.s/m2,k=0.613W/m.K Analysis: (a) the force per unit area is associated with the shear stress. Hence, with the linear velocity profile for Couette flow τ=µ (du/dy) = µ (U/L).
34.2N/m0.005m
200m/sN.s/m10855 τ : Water
0.738N/m0.005m
200m/sN.s/m10184.6 τ : Air
226-water
227-air
With the required power given by P/A= .U
6840W/m200m/s)(34.2N/m(P/A) :Water
147.6W/m200m/s)(0.738N/m(P/A) :Air22
water
22air
(b) The viscous dissipation is .henceμ(U/L)μ(du/dy)μ 22
W/m101.370.005m
200m/sN.s/m10855 (μμφ : Water
W/m102.950.005m
200m/sN.s/m10184.6 (μμφ : Air
362
26-water
342
27-air
The location of the maximum temperature corresponds to ymax=L/2. Hence Tmax=To+µU2/8k and
C34.0K80.613W/m.
(200m/s)2N.s/m10855C27(T : Water
C30.5K0.0263W/m.8
(200m/s)2_N.s/m10184.6C27)(T : Air
26-
watermax)
2-7
airmax
Comments: (1) the viscous dissipation associated with the entire fluid layer, ),(LA must equal the power, P.
(2) Although aterw )( >> air )( , kwater>>kair. Hence, Tmax,water Tmax,air .
Problem 7: A flat plate that is 0.2m by 0.2 m on a side is orientated parallel to an atmospheric air stream having a velocity of 40m/s. the air is at a temperature of T=20C, while the plate is maintained at Ts=120C. The sir flows over the top and bottom surfaces of the plate, and measurement of the drag force reveals a value of 0.075N. What is the rate of heat transfer from both sides of the plate to the air?
Known: Variation of hx with x for flow over a flat plate. Find: Ratio of average Nusselt number for the entire plate to the local Nusselt number at x=L. Schematic:
Analysis: The expressions for the local and average Nusselt number are
2.NuL
NuL
andk
2CL
k
(L)2CLNuL
Hence
2CLLL
2Cdxx
L
Cdxh
L
1h
wherek
LhNu
k
CL
k
)L(CL
k
LhNu
-
1/21/2-
1/21/2L
0
1/2L
0
xL
LL
1/21/2L
L
Comments: note the manner in which is defined in terms of . Also note that
-NuL Lh
L
0
xL dxNuL
1Nu
Problem 8:
For flow over a flat plate of length L, the local heat transfer coefficient hx is known to vary as x-1/2, where x is the distance from the leading edge of the plate. What is the ratio of the average Nusslet number for the entire plate to the local Nusslet number at x=L (NuL)?
Known: Drag force and air flow conditions associated with a flat plate. Find: Rate of heat transfer from the plate. Schematic:
Assumptions: (1) Chilton-Colburn analogy is applicable. Properties: Air(70C,1atm): =1.018kg/m3, cp=1009J/kg.K, pr=0.70, =20.22*10-6m2/s. Analysis: the rate of heat transfer from the plate is
)T(T(L)h2q s2
Where may be obtained from the Chilton-Colburn analogy, h
240Wq
C20)(120.K)(0.3m)2(30W/mq
is rateheat The
.K30W/mh
2/3.70)9J/kg.K)(0)40m/s(100(1.018kg/m105.76h
Prcρu2
Ch
hence,
105.76/2(40m/s)1.018kg/m
(0.2m)(0.075N/2)
2
1
/2ρu
τ
2
1
2
C
Prcρu
htPrS
2
Cj
22
2-
34-
2/3p
f-
4
23
2
2
sf
2/3
p
2/3fH
Comments: Although the flow is laminar over the entire surface ( , the pressure gradient is zero and the Chilton-Colburn analogy is applicable to average, as well as local, surface conditions. Note that the only contribution to the drag force is made by the surface shear stress.
)100.4/1022.20/2.0/40/Re 526 smmsmLuL
Problem 9:
Consider atmospheric air at 25C in parallel flow at 5m/s over both surfaces of 1-m-long flat plate maintained at 75C. Determine the boundary layer thickness, the surface shear stress, and the heat flux at the trailing edge. Determine the drag force on the plate and the total heat transfer from the plate, each per unit width of the plate.
Known: Temperature, pressure, and velocity of atmospheric air in parallel flow over a Plate of prescribed length and temperature. Find: (a) Boundary layer thickness, surface shear stress and heat flux at trailing edges, (b) drag force and total heat transfer flux per unit width of plate.
Schematic:
Fluid
∞=5m/sT∞=25C
P∞=1 atm L=1m
x)
Ts=75°C
Assumptions: (1) Critical Reynolds number is 5*105, (2) flow over top and bottom surfaces Properties: (Tf=323K, 1atm) Air: =1.085kg/m3,=18.2*10-6m2/s,k=0.028W/m.K,pr=0.707 Analysis: (a) calculate the Reynolds number to know the nature of flow
526L 102.75/sm1018.2
m1m/s5
ν
LuRe
Hence the flow is laminar, and at x=L
22Ls,
1/2523
1/2L
2Ls,
1/251/2L
0.0172N/m.s0.0172kg/mτ
)1050.664/(2.7(5m/s)m
kg
2
1.085/2)0.664Re(ρρτ
9.5mm)101m/(2.7555LReδ
Using the correct correlation,
22s
2L
1/31/21/31/2L
LL
C)217W/m25C.K(754.34W/m)ThL(Ts(L)q"
.K4.34W/m8W/m.K)/1m155.1(0.02h
hence,
155.1(0.707)105)0.332(2.75Pr0.332Rek
hLNu
(b) The drag force per unit area plate width is where the factor of two is
included to account for both sides of the plate. Hence with LsLD
2'
868W/mC25).K(75/m2(1m)8.68W)T(T2Lhq
.K8.68W/m2hLh with also
0.0686N/m3N/m2(1m)0.034D
0.0343N/mτ
is drag The
)101.328(2.75/2)(5m/s)(1.085kg/m/2)1.328Re(ρρτ
2s
-
L"
2-
L
2'
2Ls
1/25231/2L
2Ls
Problem 10: Engine oil at 100C and a velocity of 0.1m/s flows over both surfaces of a 1-m-long flat plate maintained at 20C. Determine
a. The velocity and thermal boundary thickness at the trailing edge. b. The local heat flux and surface shear stress at the trailing edge. c. The total drag force and heat transfer per unit area width of the plate.
Known: Temperature and velocity of engine oil Temperature and length of flat plate. Find: (a) velocity and thermal boundary thickness at the trailing edge, (b) Heat flux and surface shear stress at the trailing edge, (c) total drag force and heat transfer per unit plate width.
Schematic:
Assumptions: engine oil (Tf=33K): =864kg/m3,=86.1*106m2/s, k=0.140W/m /K, Pr=1081. Analysis: (a) calculate the Reynolds number to know the nature of flow
1161/sm10*86.1
1m0.1m/s
ν
LuReL
26-
Hence the flow is laminar at x=L, and
0.0143m)0.147(1081δPrδ
0.147m)5(1m)(11615LReδ1/31/3
t
1/21/2L
(b) The local convection coefficient and heat flux at x=L are
22s
21/31/31/2LL
1300W/mC100).K(2016.25W/m)ThL(Tq"
.K16.25W/m)(1081)0.332(11611m
0.140W/m.KPr0.3325LRe
L
kh
Also the local shear stress is
22Ls
1/223
1/2L
2Ls
0.0842N/m.s0.0842kg/mτ
)0.664(1161(0.1m/s)2
864kg/m/2)0.664Re(ρρτ
(c) With the drag force per unit width given by where the factor of 2 is
included to account for both sides of the plate, is follows that LsLD
2'
5200W/mC100).K(20/m2(1m)32.5W)T(Th2Lq
that follows alsoit .K,32.5W/m2hh with
0.673N/m)1.328(1161(0.1m/s)/m2(1m)864kg/2)1.328Re(ρρτ
2sL
_
2LL
_
1/2231/2L
2Ls
Comments: Note effect of Pr on (/t).
Problem 11:
Consider water at 27C in parallel flow over an isothermal, 1-m-long, flat plate with a velocity of 2m/s. Plot the variation of the local heat transfer coefficient with distance along the plate. What is the value of the average coefficient?
Known: velocity and temperature of air in parallel flow over a flat plate of prescribed length.
Find: (a) variation of local convection coefficient with distance along the plate, (b) Average convection coefficient.
Schematic:
Assumptions: (1) Critical Reynolds number is 5*105. Properties: Water (300K): =997kg/m3,=855*10-6N.s/m2, =/=0.858*10-6m2/s, k=0.613W/m. K, Pr=5.83 Analysis: (a) With
526L 102.33/sm100.858
1m2m/s
ν
LuRe
Boundary layer conditions are mixed and
4055 4240 4491 4871 5514 .K)(W/mh
1.0 0.8 0.6 0.4 0.0215 x(m)
xPrν
u0.0296kPr0.0296Re
x
khx 0.215m,xfor
1206 1768 hx(W/m2.K)
0.215 0.1 x(m)
xPrν
u0.332kPr0.332Re
x
khx 0.215m,x for
0.215m)10/2.33101m(5)/ReL(Rex
x
0.21/3
4/5
1/34/5x
1/21/3
1/2
1/31/2x
65Lcx,c
2
The Spatial variation of the local convection coefficient is shown above (b) The average coefficient is
.K4106W/mh
871(5.83))1033[0.0379(2.1m
0.613W/m.K871)Pr(0.037Re
L
kh
2L
_
1/34/561/34/5LL
_
Problem 12: A circular cylinder of 25-mm diameter is initially at 150C and is quenched by immersion in a 80C oil bath, which moves at a velocity of 2m.s in cross flow over the cylinder. What is the initial rate of heat loss unit length of the cylinder?
Known: Diameter and initial temperature of a circular cylinder submerged in an oil bath f prescribed temperature and velocity. Find: initial rate of heat loss unit per length.
Schematic:
Assumptions: (1) Steady-state conditions, (2) uniform surface temperature. Properties: Engine oil (T=353K): =38.1*10-6m2/s, k=0.138W/m. K, Pr=501; (Ts=423K): Prs=98. Analysis: The initial heat loss per unit length is
8.8W/mC80)0.025m(150π.K1600W/mq'
.K1600W.mh
98
501(501)0.26(1312)
0.025m
0.138W/m.K)1/4/Pr(PrPrCRe
D
kh
hence . 7.4 table from 0.6m and 0.26C Find
1312/sm1038.1
m)2m/s(0.025
ν
VDRe
when relation. Zhukauskas the from computed be may where
)TπD(Thq
2
2_
1/40.370.6
snm
D
_
26D
s
_
_
h
Comments: Evaluating properties at the film temperature, Tf=388K(=14.0*10-6m2/s, k=0.135W/m. K, Pr=196), find ReD=3517.
Problem 13: An uninsulated steam pipe is used to transport high-temperature steam from one building to one another. The pipe is 0.5-m diameter, has a surface temperature of 150C, and is exposed to ambient air at -10C.the air moves in cross flow over the pipe with a velocity of 5m.s.What is the heat loss per unit length of pipe?
Known: Diameter and surface temperature of uninsulated steam pipe. Velocity and temperature of air in cross flow. Find: Heat loss per unit length.
Schematic:
V=5m/sT∞=263K
Air D=0.5mm
Ts=150 C
Assumptions: (1) steady-state conditions, (2) uniform surface temperature Properties: Air (T=263K, 1atm):=12.6*10-6m2/s, k=0.0233W/m. K, Pr=0.72;(Ts=423K, 1atm); Prs=0.649. Analysis: the heat loss per unit length is
)TπD(Thq s
_
.K1600W.mh
98
501(501)0.26(1312)
0.025m
0.138W/m.K)1/4/Pr(PrPrCRe
D
kh
hence . 7.4 table from 0.6m and 0.26C Find
1312/sm1038.1
m)2m/s(0.025
ν
VDRe
when relation. Zhukauskas the from computed be may where
2_
1/40.370.6
snm
D
_
26D
_
h
Hence the heat rate is
4100W/mC10)(-0.5m(150π.K16.3W/mq' 2 Comments: Note that q’Dm, in which case the heat loss increases significantly with increasing D.
Problem 14:
Atmospheric air at 25C and velocity of 0.5m/s flows over a 50-W incandescent bulb whose surface temperature is at 140C. The bulb may be approximated as a sphere of 50-mm diameter. What is the rate of heat loss by convection to the air?
Known: Conditions associated with airflow over a spherical light bulb of prescribed diameter and surface temperature.
Find: Heat loss by convection.
Schematic:
Assumptions: (1) steady-state conditions, (2) uniform surface temperature.
roperties: Air (Tf=25C, 1atm): =15.71*10 m /s, k=0.0261W/m. /K.Pr=.
Analysis:
-6 2P
0.71,=183.6*10-7N.s/m2; Air (Ts=140C, 1atm): =235.5*10-7N.s/m2
)TπD(Thq s _
10.3WC25)(140π(0.05m).Km
W11.4q
is rateheat the and
.K11.4W/mh
235.5
183.6](0.71)0.06(1591)[0.4(1591)2
0.05m
K0.0261W/m.h
hence
1591/sm1015.71
0.05m0.5m/s
ν
VDRe
Where
])(μμ/)Pr0.06Re(0.4Re[2D
kh
when relation. Whitaker the from computed be may where
22
2_
1/40.42/31/2
_
26D
1/4s
0.42/3D
1/2D
_
_
h
Comments: (1) The low value of suggests that heat transfer by free convection may be significant and hence that the total loss by convection exceeds 10.3W
_
h
(2) The surface of the bulb also dissipates heat to the surroundings by radiation. Further, in an actual light bulb, there is also heat loss by conduction through the socket. (3) The Correlation has been used its range of application (/s )<1.
Problem 15:
Water at 27 C flows with a mean velocity of 1m/s through a 1-km-long cast iron pipe at 0.25 m inside diameter.
(a) Determine the pressure drop over the pipe length and the
corresponding pump power requirement, if the pipe surface is clean.
(b) If the pipe surface roughness is increased by 25% because of contamination, what is the new pressure drop and pump power requirement.
Known: Temperature and velocity of water in a cast iron pipe of prescribed dimensions. Find: pressure drops and power requirement for (a) a clean surface and (b) a surface with a 25% larger roughness.
Schematic:
Assumptions: (1) Steady, fully developed flow. Properties: Water (300K):=1000 kg/m3, =855*10-6 N.s/m2. Analysis: (a) from eq.8.22, the pressure drop is
L2D
ρufΔp
2m
e=2.6*10-4 m for clean cast iron; hence e/D=1.04*10-3. With
5326
mD 102.92
/1000kg/mN.s/m10855
0.25m1m/s
ν
DuRe
0.42barN/m104.2P
.mkg/s104.21000m2(0.25m)
(1m/s)(1000kg/m0.021P
, hence 0.021.fthat 8.3 fig from find
24
2423
)
The pump power requirement is
2.06kW1m/s/4)m0.25(πN/m104.2P
/4)uΔp(πDVΔp.P
2224
m2
.
(b) Increasing by 25% it follows that =3.25*10-4m and e/D=0.0013. With ReD unchanged, from fig 8.3, it follows that f 0.0225. Hence
2.21kW)P/f(fP bar 0.45)Δ/f(fΔP 1122122 1p Comments: (1) Note that L/D=4000>>(xfd,h/D) 10 for turbulent flow and the assumption of fully developed conditions is justified. (2) Surface fouling results in increased surface and increases operating costs through increasing pump power requirements.
Problem 16:
Consider flow in a circular tube. Within the test section length (between 1 and 2) a constant heat flux q”s is maintained.
(a) For the two cases identified, sketch, qualitatively, the surface
temperature Ts(x) and the fluid mean temperature Tm(x) as a function of distance along the test section x. in case A flow is hydro dynamically and thermally fully developed. In case B flow is not developed.
(b) Assuming that the surface flux q”s and the inlet mean temperature Tm,1 are identical for both cases, will the exit mean temperature Tm,2 for case A be greater than, equal to , or less than Tm,2 for case B? Briefly explain why?
Known: internal flow with constant surface heat flux, . "
sq
Find: (a) Qualitative temperature distributions (x), under developing and fully developed flow, (b) exit mean temperature for both situations. Schematic:
Flow
1 2
q”s=constant
Assumptions: (a) Steady-state conditions, (b) constant properties, (c) incompressible flow. Analysis: Based upon the analysis, the constant surface heat flux conditions,
constantdx
dTm
Hence, regardless of whether the hydrodynamic or thermal boundary layer is fully developed, is follows that Tm(x) is linear T m,2 will be the same for all flow conditions. The surface heat flux can be written as
(x)]Th[Tq ms"s
Under fully developed flow and thermal conditions, h=hfd is a constant. When flow is developing h> hfd . Hence, the temperature distributions appear as below.
Problem 17:
A thick-walled, stainless steel (AISI 316) pipe of inside and outside diameter Di=20mm and Do=40m is heated electrically to provide a
uniform heat generation rate of . The outer surface of the pipe is insulated while water flows through the pipe at a rate of
36.
/10 mWq
skgm /1.0.
(a) If the water inlet temperature is Tm,1=20C and the desired outlet temperature is Tm,o=40C, what is the required pipe length
(b) What are the location and value of the maximum pipe temperature?
Known: Inner and outer diameter of a steel pipe insulated on the outside and experiencing uniform heat generation. Flow rate and inlet temperature of water flowing through the pipe. Find: (a) pipe length required to achieve desired outlet temperature, (b) location and value of maximum pipe temperature. Schematic:
Assumptions: (1) steady-state conditions, (2) constant properties, (3) negligible kinetic energy, potential energy and flow work changes, (4) one-dimensional radial conduction in pipe wall, (5) outer surface is adiabatic.
Properties: Stainless steel 316 (T 400K): k=15W/m.k; water ( ); cp=4178J/kg. K, k=0.617W/m. K,=803*10-6N.s/m2, Pr=5.45
303KT.
m
Analysis: (a) performing an energy balance for a control volume about the inner tube, it follows that
8.87mL
](0.02)4m(ππ/4)[(0.W/m10
C)20178(J/kg.K(0.1kg/s)4
)LD(ππ/4)(q
)T(TmcL
)LD(ππ/4)(qq)T(Tcm
22362i
20
.
.
im,om,p
2i
20
.
im,om,p
.
(b) The maximum wall temperature exists at the pipe exit (x=L) and the insulated surface (r=ro). The radial temperature distribution in the wall is of the form
Tnr 2k
qr -r
4k
qC Cnr
2k
qrr
4k
q-T)T(r :rr
2k
qrC
r
Cr
2k
q0
dr
dT ;rr
; conditions boundary the gconsiderin
CnrCr4k
qT(r)
si
.2o2
i
.
22i
.2o2
i
.
si1
.2o
1o
1o
.
rro
212
.
o
The temperature distribution and the maximum wall temperature (r=ro) are
)Th(T4D
)D(Dq
LπD
)LDq(π(π/4)q
from followsit exit,at wall the of temperture surafaceinner the Ts, where
Tr
rn
2k
qr )r-(r
4k
q-)T(rT
Tr
rn
2k
qr )r-(r
4k
qT(r)
om,si
2i
2o
.
i
.2i
2o"
s
si
o
.2o2
i2o
.
omaxw,
si
.2o2
i2
.
Where h is the local convection coefficient at the exit. With
7928N.s/m103π(0.02m)80
0.1kg/s4
μπD
m4Re
26i
.
D
The flow is turbulent and, with (L/Di)=(8.87m/0.02m)=444>>(xfd/D)10,it is also fully developed. Hence, from the Dittus-Boelter correlation,
.K1840W/m5.45)0.023(79280.02m
0.617W/m.K)pr(0.023Re
D
kh 20.44/50.44/5
Di
Hence the inner surface temperature of the wall at the exit is
C52.4C48.20.01
0.02n
215W/m.K
(0.02)W/m10](0.01)[(0.02)
415W/m.K
W/m10T
and
C48.2C40.K(0.02m)180W/m4
](0.02m)[(0.04m)W/m10T
4D
)D(Dq
23622
36
maxw,
2
2236
om,i
2i
2o
.
s
T
Comments: The physical situation corresponds to a uniform surface heat flux, and Tm increases linearly with x. in the fully developed region, Ts also increases linearly with x.
T
xfd
Ts(x)
Tm(x)
x
Problem 18: The surface of a 50-mm diameter, thin walled tube is maintained thin walled tube is maintained at 100C. In one case air is cross flow very the tube with a temperature of 25C and a velocity of 30m/s. In another case air is in fully developed flow through the tube with a temperature of 25C and a mean velocity of 30m/s. compare the heat flux from the tube to the air for the two cases.
Known: surface temperature and diameter of a tube. Velocity and temperature of air in cross flow. Velocity and temperature of air in fully developed internal flow. Find: convection heat flux associated with the external and internal flows. Schematic:
Air
Um=30m/sTm=25°C
V=30m/sT =25 C
Ts=100 C
D=0.05m
Air
Assumptions: (1) steady-state conditions, (2) uniform cylinder surface temperature, (3) fully developed internal flow Properties: Air (298K): =15.71*10-6m2/s, k=0.0261W/m.K, Pr=0.71 Analysis: for the external and internal flow
426
1055.9/1071.15
05./30Re
sm
msmDuVD mD
From the Zhukauskas relation for the external flow, with C=0.26 and m=0.6
223)1()71.0()1055.9(26.0)PrPr(Pr/Re 4/137.06.044/1_
smDD CuN
Hence, the convection coefficient and heat rate are
232"
2__
/1073.8)25100(./4.116)(
./4.11622305.0
./0261.0
mWCkmWTThq
KmWm
KmWNuD
D
kh
s
Using the Dittus-Boelter correlation, for the internal flow, which is Turbulent,
232
__
4.05/4244.05/4_
/1058.7)25100(./101)(
.2/10119305.0
./0261.0
193)71.0()1055.9(023.0PrRe023.0
mWCKmWTTh
KmWKmW
NuD
kh
NU
ms
D
DD
"q
isflux heat the and
Comments: Convection effects associated with the two flow conditions are comparable.
Problem 19: Cooling water flows through he 25.4 mm diameter thin walled tubes of a stream condenser at 1m/s, and a surface temperature of 350K is maintained by the condensing steam. If the water inlet temperature is 290 K and the tubes are 5 m long, what is the water outlet temperature? Water properties may be evaluated at an assumed average temperature of 300K Known: Diameter, length and surface temperature of condenser tubes. Water velocity and inlet temperature. Find: Water outlet temperature. Schematic:
L=5m
D=0.0254m
Tm,i=290KUm=1m/s
Ts=350K
Assumptions: (1) Negligible tube wall conduction resistance, (2) Negligible kinetic energy, potential energy and flow work changes. Properties: Water (300K):=997kg/m3, cp=4179J/kg.K, =855*10-6kg/s.m,k=0.613W/m.K, Pr=5.83 Analysis:
29,618kg/s.m10855
54m(1m/s)0.02997kg/m
μ
DρuRe
]h)cm(ππDL)exp[T(TTT
6
3m
D
_
p
.
im,ssom,
The flow is turbulent. Since L/D=197, it is reasonable to assume fully developed flow throughout the tube. Hence
323KK)(4179J/kg.(0.505kg/s
.K)5m(4248W/mπ(0.0254m)(60K)exp350KT
0.505kg/s4m)m/s)(0.025(1/m(ππ/4)997k/4)(ππρum
With
.K4248W/m4m)/m.K/0.025176(0.613WNuD(k/D)h
2
om,
232m
.
2_
Comments: The accuracy of the calculations may be improved slightly by
reevaluating properties at . 306.5KT_
m
Problem 20: The air passage for cooling a gas turbine vane can be approximated as a tube of 3-mm diameter and 75-mm length. If the operating temperature of the vane is 650C, calculate the outlet temperature of the air if it enters the tube at 427C and 0.18kg/h. Known: gas turbine vane approximation as a tube of prescribed diameter and length maintained at an known surface temperature. Air inlet temperature and flow rate. Find: outlet temperature of the air coolant. Schematic:
hkgm /18.0.
Assumptions: (1) Steady-state conditions, (2) negligible Kinetic and potential energy changes.
Properties: Air (assume : cp=1094J/kg. K, k=0.0563 W/m.
K, =363.7*10-7 N.s/m2, Pr=0.706; Pr=0.706;(Ts=650C=923K, 1atm): =404.2*10-7N.s/m2.
)1,780_
atmKTm
Analysis: For constant wall temperatures heating,
.
_
,
, exp
pims
oms
mc
hPL
TT
TT
Where P=D. for flow in circular passage,
N.s/m103.7π(0.03m)36
/3600s/h)0.18kg/h(14
πDμ
m4Re
7
.
D
The flow is laminar, and since L/D=75mm/3mm=25, the Sieder-Tate correlation including combined entry length fields.
.K87.5W/m10404.2
10363.7
25
0.7065841.86
0.003m
K0.0563W/m.h
μ
μ
L/D
PrRe1.86
k
DhNu
2
0.14
7
71/3_
0.14
s
1/3
D
__
D
Hence, the air outlet temperature is
1094J/kg.K)kg/s(0.18/3600
.K87.5W/m0.075mπ(0.003m)exp
C427)(650
T650 2om,
Tm,o=578C
Comments: (1) based upon the calculations for Tm,o=578C, =775K which is in good agreement with our assumption to evaluate the thermo physical properties.
_
mT
Problem 21 A household oven door of 0.5-m height and 0.7-m width reaches an average surface temperature of 32C during operation. Estimate the heat loss to the room with ambient air at 22C. If the door has an emissivity of 1.0 and the surroundings are also at 22C, comment on the heat loss by free convection relative to that by radiation.
Known: Oven door with average surface temperature of 32C in a room with ambient temperature at 22C. Find: Heat loss to the room. Also, find effect on heat loss if emissivity of door is unity and the surroundings are at 22C.
Schematic:
0.7m
L=
0.5m
Assumptions: (1) Ambient air in quiescent, (2) surface radiation effects are negligible. Properties: Air (Tf=300K, 1atm): =15.89*10-6 m2/s, k=0.0263W/m. K, =22.5*10-6 m2/s, Pr=0.707, =1/Tf=3.33*10-3K-1
Analysis: the heat rate from the oven door surface by convection to the ambient air is
)T(TAhq ss
_
Where can be estimated from the free convection correlation for a vertical plate,
_
h
2
8/279/16
1/6L
__`
(0.492/Pr)1
0.387Ra0.825
k
LhLNu
The Rayleigh number,
4
2626
322s
L 101.142/sm1022.5/sm1015.89
m322)K0.52(1/300k))39.8m/s
να
)LT(TgRa
Substituting numerical values into equation, find
.K3.34W/m63.5
0.5m
K0.0263W/m.LNu
Lh
63.5(0.492/Pr)1
)1020.387(1.140.825
k
LhLNu
2_`_
2
8/279/16
1/88_
_`
kL
The heat rate using equation is
11.7W22)K(320.7)m.k(0.53.34W/mq 22
Heat loss by radiation, assuming =1 is
21.4W]k22)(27332)[(273.KW/m5.67100.7)m1(0.5q
)Tσ(TεAq4244282
rad
4sur
4ssrad
Note that heat loss by radiation is nearly double that by free convection. Comments: (1) Note the characteristics length in the Rayleigh number is the height of the vertical plate (door).
Problem 22
An Aluminum alloy (2024) plate, heated to a uniform temperature of 227C, is allowed to cool while vertically suspended in a room where the ambient air and surroundings are at 27C. The late is 0.3 m square with a thickness of 15 mm and an emissivity of 0.25.
a. Develop an expression for the time rate of change of the plate temperature assuming the temperature to be uniform at any time.
b. Determine the initial rate of cooling of the plate temperature is 227C.
c. Justify the uniform plate temperature assumption. Known: Aluminum plate alloy (2024) at uniform temperature of 227C suspended in a room where the ambient air and the surroundings are at 27C
Find: (1) expression for the time rate of change of the plate, (2) Initial
rate of cooling (K/s) when the plate temperature is 227C. (3) justify the
uniform plate temperature assumption.
Schematic:
Properties: Aluminium alloy 2024 (T=500K):=2270 kg/m3,
k=186W/m.K, c=983 J/kg.K; Air (Tf=400K, 1atm):=26.41*10-
6m2/s,k=0.0338 W/m.K,=38.3*10-6 m2/s, Pr=0.690.
Analysis :( a) from an energy balance on the plate considering free
convection and radiation exchange . ..
stout EE
)Tεσ(T)T(Th[c
2
dt
dTor
dt
dTρA)Tσ(Tε2A)T(T2Ah 4
sur4ssL
.
cs4sur
4ssss
_
L
Where Ts is the plate temperature assumed to be uniform at any time.
(b) To evaluate (dt/dx), estimate . Find first the Rayleigh number _
Lh
82626
322s
L 101.308/sm10.33/sm1026.41
0.3m27)K27(1/400k)(29.8m/s
να
)LT(TgRa
8
Substituting numerical values, find
5.55)
4/99/16
82
4/99/16
1/4L
_`
90)(0.492/0.61
1080.670(1.300.68
(0.492/Pr)1
0.670Ra0.68LNu
0.099K/s
)K300.K)(500W/m100.25(5.6727)K.K(2276.25W/m983J/kg.K0.015m2770Kg/m
2
dt
dT
.K6.25W/mK/0.3m0.0338W/m.55.5k/LNuh
44282
3
2_`
L
_
(c) The uniform temperature assumption is justified if the Biot number
criterion is satisfied. With )/A.(A)(V/AL sssc
_
1.0/ k
and . Using the linearized radiation coefficient
relation find,
___
, hBihhh totradconvtot
.K3.86W/m3002)K300)(5002)(500.K8W/m100.25(5.67)T)(TTεσ(Th 23422
sur2
ssursrad
_
Hence Bi=(6.25+3.86) W/m2.k (0.015m)/186W/m. K=8.15*10-4. Since
Bi<<0.1, the assumption is appropriate.
Problem 23 The ABC Evening News Report in news segment on hypothermia research studies at the University of Minnesota claimed that heat loss from the body is 30 times faster in 1 0C water than in air at the same temperature. Is that a realistic statement?
Known: Person, approximated as a cylinder, experiencing heat loss in water or air at 10C.
Find: Whether heat loss from body in water is 30 times that in air. Assumptions: (1) Person can be approximated as a vertical cylinder of diameter D=0.3 m and length L=1.8m, at 25C, (2) Loss is only from the lateral surface.
Properties: Air (_
T =(25+10) C/2=290K, 1atm); K=0.0293 W/m. K, ==19.91*10-6m2/s, =28.4*10-6 m2/s; Water (290K); k=0.598 W/m.K; =vf=1.081*10-6m2/s, =k vf /cp =1.431*10-7m2/s, f=174*10-6K-1.
Analysis: in both water (wa) an air(a), the heat loss from the lateral surface of the cylinder approximating the body is
)TDL(Thq s
_
Where Ts and T are the same for both situations. Hence,
_
a
wa
_
a
wa
h
h
q
q
Vertical cylinder in air:
9
2626
323
L 105.228/sm28.410/sm10*19.91
10)k(1.8m)5(1/290K)(29.8m/s
να
gββΔTRa
.K2.82W/mh 173.4)100.1(5.228CRak
LhNu
1/3,n and 0.1C with
2L
_1/39n
L
__
LL
Vertical cylinder in water.
112726
3162
L 109.643/sm101.431/sm101.081
10)K(1.8m)(25K101749.8m/sRa
.K328W/mh 978.9)100.1(9.643CRak
hLNu
1/3,n and 0.1C with
2L
_1/311n
L
__
L
Hence, from this analysis we find
30 of claim the with poorly compares which 117.K2.8W/m
.K328W/m
q
q2
2
a
wa
Problem 24 In a study of heat losses from buildings, free convection heat transfer from room air at 305 K to the inner surface of a 2.5-m-high wall at 295 K is simulated by performing laboratory experiments using water in a smaller test cell. In the experiments the water and the inner surface of the test cell are maintained at 300 and 290K, respectively. To achieve similarity between conditions in the room and the test cell, what is the required test cell height? If the average Nusselt number for the wall may be correlated exclusively in terms of the Rayleigh number, what is the ratio f the average convection coefficient for the room wall to the average coefficient for the test cell wall? Known: Air temperature and wall temperature and height for a room. Water temperature and wall temperature for a simulation experiment. Find: Required test cell height for similarity. Ratio and height convection coefficient for the two cases. Schematic:
L a=
2.5m L
w
Assumptions: (1) Air and water are quiescent; (2) Flow conditions correspond to free convection boundary layer development on an isothermal vertical plate, (3) constant properties. Properties: Air (Tf =300K, 1 atm); K=0.0293 W/m.K, ==15.9*10-6m2/s, =22.5*10-6 m2/s; =1/Tf=3.33*10-3K-1, k=0.0263W/m.K; water (Tf=295K):ρ=998kg/m3, =959*10-6N.s/m2, cp=4181 J/kg.K, =227.5*10-
6K-1, k=0.606 W/m.K; Hence =/ ρ =9.61/s, =k / ρ cp =1.45*10-7m2/s.
Analysis: Similarity requires that RaL,a=RaL,w where
mmL
oHL
L
hence
w
a
airw
a
w
45.0)179.0(5.2
10*228.0
10*33.3
10*5.22*9.15
10*45.1*61.9
)(
)(
,
)L-T(TgRa
3
3
12
143/1
2
3s
L
3
w
a
a
w
w
_
_
a
_
wL,
_
aL,wL,aL,
10*7.810.606
0.0263
2.5
0.45
k
k
L
L
h
h
.henceNuNu that followsit ,RaRa if
Comments: Similitude allows us to obtain valuable information for one system by performing experiments for a smaller system and a different fluid.
Problem 25
square plate of pure aluminum, 0.5 m on a side and 16 mm hick, is
nown: Initial temperature and dimensions of an aluminum plate.
ind: (a) initial cooling rate, (2) validity of assuming negligible temperature
chematic:
Ainitially at 300C and is suspended in a large chamber. The walls of the chamber are maintained at 27C, as is the enclosed air. If the surface emissivity of the plate temperature during the cooling process? Is it reasonable to assume a uniform plate temperature during the cooling process? KConditions of the plate surroundings. Fgradients in the plate during the cooling process. S
L
ssumptions: (1) plate temperature is uniform; (2) chamber air is quiescent,
roperties: Aluminum (573k); k=232W/m.k, cp=1022J/kg.K, ρ=2702 kg/m3:
nalysis: (a) performing an energy balance on the plate,
A(3) Plate surface is diffuse-gray, (4) Chamber surface is much larger than that of plate, (5) Negligible heat transfer from edges. PAir (Tf=436K, 1 atm):υ=30.72*10-6m2/s,α=44.7*10-6m2/s,k=0.0363W/m.K, Pr=0.687, β=0.00229K-1. A
psurss
pstsurss
wcTTTThAdtdT
dtdTVcETTTThAq
/)]4()([2/
]/[)]4()([2
4_
.4
_
8
2626
323s
L 105.58/sm017.44/sm10*30.72
)K(0.5m)721(300-0.00229K9.8m/s)L-T(TgRa
KmWh
Ra
L
kh L
./8.5
])687.0/492.0(1[
)1058.5(670.068.0
5.0
0363.0
]Pr)/492.0(1[
670.068.0
2_
9/416/9
4/18
9/416/9
4/1_
ence the initial cooling rate is H
sKdt
dT
kkgJmmkg
KKKmWCKmW
dt
dT
/136.0
./1022)016.0(/2702
])300()573[(./1067.525.0)27300(./8.523
444282
) To check the Validity of neglecting temperature gradients across the
nd the assumption is excellent.
omments: (1) Longitudinal (x) temperature gradients are likely to me more
(bplate thickness, calculate
22" ./0.11273//)14131583()/( /)2/( kmWKmWTTqhwherekwhBi itoteffeff
A
4-103.832W/m.K(0.008m)/2(11W/m2.K)Bi hence
Csevere than those associated with the plate thickness due to the variation of h with x. (2) Initially q”conv q”rad.
Problem 26
eer in cans 150 mm long and 60 mm in diameter is initially at 27C and is
nown: Dimensions and temperature of beer can in refrigerator
ind: orientation which maximum cooling rate.
Schematic:
Bto be cooled by placement in a refrigerator compartment at 4C. In the interest of maximizing the cooling rate, should the cans be laid horizontally or vertically in the compartment? As a first approximation, neglect heat transfer from the ends. Kcompartment. F
Air,T∞=4°CP=1 atmD=0.06m
Horizontal, (h) Vertical, (V)Ts=27 C
L=150m
m
Assumptions: (1) End effects are negligible, (2) Compartment air is
roperties: Air (Tf=288.5K, 1 atm): υ=14.87*10-6 m2/s, k=0.0254W/m.K,
nalysis: The ratio of cooling rates may be expressed as
quiescent, (3) constant properties. Pα=21.0*10-6m2/s, Pr=0.71, β=1/Tf=3.47*10-3K-1. A
h
v
s
s
h
v
qh
v
h
h
TT
TT
DL
DL
h
hq_
_
_
_
)(
)(
For the vertical surface, find
639L
39326-26-
-1-323s
L
1044.8)15.0(105.2Ra
L105.2L/s)m10/s)(21m10(14.87
C)(23K103.479.8m/s)L-T(TgRa
Using the correct correlation
KmWm
KmW
L
kNuhh
NuL
LvL ./03.515.0
./0254.07.29 hence,
7.29])71.0/492.0(1[
)1044.8(387.0825.0
2___
2
27/816/9
6/16
Using the correct correlation,
97.018.5
03.5,
./18.506.0
./0254.024.12
24.12])71.0/559.0(1[
)104.5(387.060.0
2___
2
27/861/9
6/15_
h
v
DhD
D
q
qhence
KmWm
KmW
D
kNuhh
Nu
Comments: in view of the uncertainties associated with equations and the neglect of end effects, the above result is inconclusive. The cooling rates are are approximately the same.
MODULE I
RADIATION HEAT TRANSFER Radiation Definition
Radiation, energy transfer across a system boundary due to a T, by the mechanism of photon emission or electromagnetic wave emission.
Because the mechanism of transmission is photon emission, unlike conduction and convection, there need be no intermediate matter to enable transmission. The significance of this is that radiation will be the only mechanism for heat transfer whenever a vacuum is present. Electromagnetic Phenomena. We are well acquainted with a wide range of electromagnetic phenomena in modern life. These phenomena are sometimes thought of as wave phenomena and are, consequently, often described in terms of electromagnetic wave length, . Examples are given in terms of the wave distribution shown below:
10-5 10-4 10-3 10-2 10-1 10-0 101 102 103 104 105
Wavelength, , m
Thermal Radiation
UV X Rays
Visi
ble
Ligh
t, 0.
4-0.
7m
Microwave radiation
One aspect of electromagnetic radiation is that the related topics are more closely associated with optics and electronics than with those normally found in mechanical engineering courses. Nevertheless, these are widely encountered topics and the student is familiar with them through every day life experiences. From a viewpoint of previously studied topics students, particularly those with a background in mechanical or chemical engineering, will find the subject of Radiation Heat Transfer a little unusual. The physics background differs fundamentally from that found in the areas of continuum mechanics. Much of the related material is found in courses more closely identified with quantum physics or electrical engineering, i.e. Fields and Waves. At this point, it is important for us to recognize that since the subject arises from a different area of physics, it will be important that we study these concepts with extra care.
Stefan-Boltzman Law Both Stefan and Boltzman were physicists; any student taking a course in quantum physics will become well acquainted with Boltzman’s work as he made a number of important contributions to the field. Both were contemporaries of Einstein so we see that the subject is of fairly recent vintage. (Recall that the basic equation for convection heat transfer is attributed to Newton.)
Eb = Tabs4
where: Eb = Emissive Power, the gross energy emitted from an
ideal surface per unit area, time.
= Stefan Boltzman constant, 5.6710-8 W/m2K4
Tabs = Absolute temperature of the emitting surface, K.
Take particular note of the fact that absolute temperatures are used in Radiation. It is suggested, as a matter of good practice, to convert all temperatures to the absolute scale as an initial step in all radiation problems.
You will notice that the equation does not include any heat flux term, q”. Instead we have a term the emissive power. The relationship between these terms is as follows. Consider two infinite plane surfaces, both facing one another. Both surfaces are ideal surfaces. One surface is found to be at temperature, T1, the other at temperature, T2. Since both temperatures are at temperatures above absolute zero, both will radiate energy as described by the Stefan-Boltzman law. The heat flux will be the net radiant flow as given by:
q" = Eb1 - Eb2 = T14 - T2
4
Plank’s Law While the Stefan-Boltzman law is useful for studying overall energy emissions, it does not allow us to treat those interactions, which deal specifically with wavelength, . This problem was overcome by another of the modern physicists, Max Plank, who developed a relationship for wave-based emissions.
Eb = ()
We plot a suitable functional relationship below:
Area = Eb
Wavelength, , m
Eb
, W/m
2 m
We haven’t yet defined the Monochromatic Emissive Power, Eb. An implicit definition is provided by the following equation:
E Eb b
0 d We may view this equation graphically as follows:
Area of the curve = Eb
Wavelength, , m
Eb
, W/m
2 m
A definition of monochromatic Emissive Power would be obtained by differentiating the integral equation:
EdE
db
b
The actual form of Plank’s law is:
EC
eb
TC
1
52
1
where: C1 = 2hco
2 = 3.742108 Wm4/m2 C2 = hco/k = 1.439104 mK Where: h, co, k are all parameters from quantum physics. We need
not worry about their precise definition here. This equation may be solved at any T, to give the value of the monochromatic emissivity at that condition. Alternatively, the function may be substituted into the integral to find the Emissive power for any temperature. While performing this integral by hand is difficult, students may readily evaluate the integral through one of several computer programs, i.e. MathCad, Maple, Mathmatica, etc.
E E db b
0
E E db b
04T
d
Emission Over Specific Wave Length Bands Consider the problem of designing a tanning machine. As a part of the machine, we will need to design a very powerful incandescent light source. We may wish to know how much energy is being emitted over the ultraviolet band (10-4 to 0.4 m), known to be particularly dangerous.
E Eb bm
m0 0001 0 4 0 001
0 4. . .
.
With a computer available, evaluation of this integral is rather trivial. Alternatively, the text books provide a table of integrals. The format used is as follows:
E
E
E d
E d
E d
E d
E d
E dF Fb
b
bm
m
b
b
m
b
b
m
b
0 001 0 40 0 4 0 0 0001
0 001
0 4
0
0
0 4
0
0
0 0001
0
. .( . ) ( .
.
. .
.
.
)
Referring to such tables, we see the last two functions listed in the second column. In the first column is a parameter, T. This is found by taking the product of the absolute temperature of the emitting surface, T, and the upper limit wave length, . In our example, suppose that the incandescent bulb is designed to operate at a temperature of 2000K. Reading from the table:
., m T, K T, mK F(0) 0.0001 2000 0.2 0
0.4 2000 600 0.000014 F(0.40.0001m) = F(00.4m)- F(00.0001m) 0.000014 This is the fraction of the total energy emitted which falls within the IR band. To find the absolute energy emitted multiply this value times the total energy emitted:
EbIR = F(0.40.0001m)T4 = 0.0000145.6710-820004 = 12.7 W/m2 Solar Radiation The magnitude of the energy leaving the Sun varies with time and is closely associated with such factors as solar flares and sunspots. Nevertheless, we often choose to work with an average value. The energy leaving the sun is emitted outward in all directions so that at any particular distance from the Sun we may imagine the energy being dispersed over an imaginary spherical area. Because this area increases with the distance squared, the solar flux also decreases with the distance squared. At the average distance between Earth and Sun this heat flux is 1353 W/m2, so that the average heat flux on any object in Earth orbit is found as:
Gs,o = Sc·f·cos θ
Where Sc = Solar Constant, 1353 W/m2 f = correction factor for eccentricity in Earth Orbit,
(0.97<f<1.03) θ = Angle of surface from normal to Sun. Because of reflection and absorption in the Earth’s atmosphere, this number is significantly reduced at ground level. Nevertheless, this value gives us some opportunity to estimate the potential for using solar energy, such as in photovoltaic cells.
Some Definitions In the previous section we introduced the Stefan-Boltzman Equation to describe radiation from an ideal surface.
Eb = σ·Tabs4
This equation provides a method of determining the total energy leaving a surface, but gives no indication of the direction in which it travels. As we continue our study, we will want to be able to calculate how heat is distributed among various objects. For this purpose, we will introduce the radiation intensity, I, defined as the energy emitted per unit area, per unit time, per unit solid angle. Before writing an equation for this new property, we will need to define some of the terms we will be using. Angles and Arc Length We are well accustomed to thinking of an angle as a two dimensional object. It may be used to find an arc length:
α L = r·α
L
Solid Angle We generalize the idea of an angle and an arc length to three dimensions and define a solid angle, Ω, which like the standard angle has no dimensions. The solid angle, when multiplied by the radius squared will have dimensions of length squared, or area, and will have the magnitude of the encompassed area.
A = r2·dΩ
r
Projected Area The area, dA1, as seen from the prospective of a viewer, situated at an angle θ from the normal to the surface, will appear somewhat smaller, as cos θ·dA1. This smaller area is termed the projected area.
Aprojected = cos θ·Anormal Intensity The ideal intensity, Ib, may now be defined as the energy emitted from an ideal body, per unit projected area, per unit time, per unit solid angle.
ddA
dqI
1cos
dA1·cos θ
θ
dA1
Spherical Geometry Since any surface will emit radiation outward in all directions above the surface, the spherical coordinate system provides a convenient tool for analysis. The three basic coordinates shown are R, φ, and θ, representing the radial, azimuthal and zenith directions. In general dA1 will correspond to the emitting surface or the source. The surface dA2 will correspond to the receiving surface or the target. Note that the area proscribed on the hemisphere, dA2, may be written as: φ
R·sin θ
R θ
dA1
dA2
Δφ
][])sin[(2 dRdRdA
or, more simply as:
]sin22 ddRdA
Recalling the definition of the solid angle,
dA = R2·dΩ we find that:
dΩ = R2·sin θ·dθ·dφ
Real Surfaces Thus far we have spoken of ideal surfaces, i.e. those that emit energy according to the Stefan-Boltzman law:
Eb = σ·Tabs4
Real surfaces have emissive powers, E, which are somewhat less than that obtained theoretically by Boltzman. To account for this reduction, we introduce the emissivity, ε.
bE
E
so that the emissive power from any real surface is given by:
E = ε·σ·Tabs4
Receiving Properties Targets receive radiation in one of three ways; they absorption, reflection or transmission. To account for these characteristics, we introduce three additional properties:
Absorptivity, α, the fraction of incident radiation absorbed.
Transmitted Radiation
Absorbed Radiation
Incident Radiation, G
Reflected Radiation
Reflectivity, ρ, the fraction of incident radiation reflected.
Transmissivity, τ, the fraction of incident radiation transmitted.
We see, from Conservation of Energy, that:
α + ρ + τ = 1
In this course, we will deal with only opaque surfaces, τ = 0, so that:
Opaque Surfaces α + ρ = 1
Relationship Between Absorptivity,α, and Emissivity,ε
Consider two flat, infinite planes, surface A and surface B, both emitting radiation toward one another. Surface B is assumed to be an ideal emitter, i.e. εB = 1.0. Surface A will emit radiation according to the Stefan-Boltzman law as: EA = εA·σ·TA
4
and will receive radiation as:
GA = αA·σ·TB4
The net heat flow from surface A will be:
q΄΄ = εA·σ·TA4 - αA·σ·TB
4
Now suppose that the two surfaces are at exactly the same temperature. The heat flow must be zero according to the 2nd law. If follows then that:
αA = εA
Sur
face
A, T
A
Sur
face
B, T
B
Because of this close relation between emissivity, ε, and absorptivity, α, only one property is normally measured and this value may be used alternatively for either property. Let’s not lose sight of the fact that, as thermodynamic properties of the material, α and ε may depend on temperature. In general, this will be the case as radiative properties will depend on wavelength, λ. The wave length of radiation will, in turn, depend on the temperature of the source of radiation. The emissivity, ε, of surface A will depend on the material of which surface A is composed, i.e. aluminum, brass, steel, etc. and on the temperature of surface A.
The absorptivity, α, of surface A will depend on the material of which surface A is composed, i.e. aluminum, brass, steel, etc. and on the temperature of surface B. In the design of solar collectors, engineers have long sought a material which would absorb all solar radiation, (α = 1, Tsun ~ 5600K) but would not re-radiate energy as it came to temperature (ε << 1, Tcollector ~ 400K). NASA developed an anodized chrome, commonly called “black chrome” as a result of this research. Black Surfaces Within the visual band of radiation, any material, which absorbs all visible light, appears as black. Extending this concept to the much broader thermal band, we speak of surfaces with α = 1 as also being “black” or “thermally black”. It follows that for such a surface, ε = 1 and the surface will behave as an ideal emitter. The terms ideal surface and black surface are used interchangeably. Lambert’s Cosine Law: A surface is said to obey Lambert’s cosine law if the intensity, I, is uniform in all directions. This is an idealization of real surfaces as seen by the emissivity at different zenith angles:
0 .2 .4 .6 .8 1.0
Dependence of Emissivity on Zenith Angle, Typical Metal.
0 .2 .4 .6 .8 1.0
Dependence of Emissivity on Zenith Angle, Typical Non-Metal.
The sketches shown are intended to show is that metals typically have a very low emissivity, ε, which also remain nearly constant, expect at very high zenith angles, θ. Conversely, non-metals will have a relatively high emissivity, ε, except at very high zenith angles. Treating the emissivity as a constant over all angles is generally a good approximation and greatly simplifies engineering calculations. Relationship Between Emissive Power and Intensity By definition of the two terms, emissive power for an ideal surface, Eb, and intensity for an ideal surface, Ib.
hemisphere
bb dIE cos
Replacing the solid angle by its equivalent in spherical angles:
2
0
2
0sincos ddIE bb
Integrate once, holding Ib constant:
2
0sincos2
dIE bb
Integrate a second time. (Note that the derivative of sin θ is cos θ·dθ.)
bbb IIE
2
0
2
2
sin2
Eb = π·Ib
Radiation Exchange During the previous lecture we introduced the intensity, I, to describe radiation within a particular solid angle.
ddA
dqI
1cos
This will now be used to determine the fraction of radiation leaving a given surface and striking a second surface. Rearranging the above equation to express the heat radiated:
ddAIdq 1cos
Next we will project the receiving surface onto the hemisphere surrounding the source. First find the projected area of surface dA2, dA2·cos θ2. (θ2 is the angle between the normal to surface 2 and the position vector, R.) Then find the solid angle, Ω, which encompasses this area. dA2
Substituting into the heat flow equation above:
22211 coscos
R
dAdAIdq
To obtain the entire heat transferred from a finite area, dA1, to a finite area, dA2, we integrate over both surfaces:
dA2·cos θ2
R
2 1
22211
21
coscosA A R
dAdAIq
To express the total energy emitted from surface 1, we recall the relation between emissive power, E, and intensity, I.
qemitted = E1·A1 = π·I1·A1
View Factors-Integral Method Define the view factor, F1-2, as the fraction of energy emitted from surface 1, which directly strikes surface 2.
1
22211
2121
2 1
coscos
AIR
dAdAI
q
qF
A A
emitted
after algebraic simplification this becomes:
2 1
22121
121
coscos1A A R
dAdA
AF
Example Consider a diffuse circular disk of diameter D and area Aj and a plane diffuse surface of area Ai << Aj. The surfaces are parallel, and Ai is located at a distance L from the center of Aj. Obtain an expression for the view factor Fij.
dr dAj
dAi
D
dAi
RL
R θi
θj
Aj
The view factor may be obtained from:
2 1
22121
121
coscos1A A R
dAdA
AF
Since dAi is a differential area
1
2121
21
coscosA R
dAF
Substituting for the cosines and the differential area:
12
2
21
2
A R
drrRL
F
After simplifying:
1
4
2
21
2A R
drrLF
Let ρ2 L2 + r2 = R2. Then 2·ρ·dρ = 2·r·dr.
1
4
2
21
2A
dLF
After integrating,
2
022
22
221
1
22
2
D
AL
LLF
Substituting the upper & lower limits
22
22
0222
221 4
1
4
4
DL
D
LDLLF
D
This is but one example of how the view factor may be evaluated using the integral method. The approach used here is conceptually quite straight forward; evaluating the integrals and algebraically simplifying the resulting equations can be quite lengthy. Enclosures In order that we might apply conservation of energy to the radiation process, we must account for all energy leaving a surface. We imagine that the surrounding surfaces act as an enclosure about the heat source which receive all emitted energy. Should there be an opening in this enclosure through which energy might be lost, we place an imaginary surface across this opening to intercept this portion of the emitted energy. For an N surfaced enclosure, we can then see that:
11
,
N
jjiF
This relationship is known as the
“Conservation Rule”.
Example: Consider the previous problem of a small disk radiating to a larger disk placed directly above at a distance L. From our conservation rule we have:
The view factor was shown to be given by the relationship: 2
22
2
21 4 DL
DF
Here, in order to provide an enclosure, we will define an imaginary surface 3, a truncated cone intersecting circles 1 and 2.
1
3
3,12,11,11
, FFFFN
jji
Since surface 1 is not convex F1,1 = 0. Then:
22
2
31 41
DL
DF
Reciprocity We may write the view factor from surface i to surface j as:
j iA A
jijijii R
dAdAFA
2
coscos
Similarly, between surfaces j and i:
j iA A
ijijijj R
dAdAFA
2
coscos
Comparing the integrals we see that they are identical so that:
ijjjii FAFA
This relationship is known as
“Reciprocity”.
Example: Consider two concentric spheres shown to the right. All radiation leaving the outside of surface 1 will strike surface 2. Part of the radiant energy leaving the inside surface of object 2 will strike surface 1, part will return to surface 2. To find the fraction of energy leaving surface 2 which strikes surface 1, we apply reciprocity:
2
1
2
12,1
2
11,22,111,22 D
D
A
AF
A
AFFAFA
Associative Rule Consider the set of surfaces shown to the right: Clearly, from conservation of energy, the fraction of energy leaving surface i and striking the combined surface j+k will equal the fraction of energy emitted from i and striking j plus the fraction leaving surface i and striking k.
kijikji FFF )( Radiosity We have developed the concept of intensity, I, which let to the concept of the view factor. We have discussed various methods of finding view factors. There remains one additional concept to introduce before we can consider the solution of radiation problems. Radiosity, J, is defined as the total energy leaving a surface per unit area and per unit time. This may initially sound much like the definition of emissive power, but the sketch below will help to clarify the concept.
1 2
i
j k
This relationship is known as the
“Associative Rule”.
ε·Eb ρ·G G J ≡ ε·Eb + ρ·G
Net Exchange Between Surfaces Consider the two surfaces shown. Radiation will travel from surface i to surface j and will also travel from j to i.
qi→j = Ji·Ai· Fi→j Jj
likewise, qj→i = Jj·Aj· Fj→j Ji
The net heat transfer is then:
qj→i (net) = Ji·Ai· Fi→j - Jj·Aj· Fj→j
From reciprocity we note that F1→2·A1 = F2→1·A2 so that
qj→i (net) = Ji·Ai· Fi→j - Jj· Ai· Fi→j = Ai· Fi→j·(Ji – Jj) Net Energy Leaving a Surface The net energy leaving a surface will be the difference between the energy leaving a surface and the energy received by a surface:
ε·Eb ρ·G G
q1→ = [ε·Eb – α·G]·A1
Combine this relationship with the definition of Radiosity to eliminate G.
J ≡ ε·Eb + ρ·G G = [J - ε·Eb]/ρ
q1→ = ε·Eb – α·[J - ε·Eb]/ρ·A1
Assume opaque surfaces so that α + ρ = 1 ρ = 1 – α, and substitute for ρ.
q1→ = ε·Eb – α·[J - ε·Eb]/(1 – α)·A1
Put the equation over a common denominator:
111 11
1A
JEA
EJEq bbb
If we assume that α = ε then the equation reduces to:
JEA
AJE
q bb
111
11
Electrical Analogy for Radiation We may develop an electrical analogy for radiation, similar to that produced for conduction. The two analogies should not be mixed: they have different dimensions on the potential differences, resistance and current flows. Equivalent
Current Equivalent Resistance
Potential Difference
Ohms Law I R ΔV Net Energy
Leaving Surface q1→
A1
Eb - J
Net Exchange Between Surfaces
qi→j 211
1
FA J1 – J2
Count the number of surfaces. (A surface must be at a “uniform”
factor resistances, 1/Ai·Fi→j.
Alternate Procedure for Developing Networks
temperature and have uniform properties, i.e. ε, α, ρ.) Draw a radiosity node for each surface. Connect the Radiosity nodes using view Connect each Radiosity node to a grounded battery, through a surface
resistance, A
1 .
This procedure should lead to exactly the same circuit as we obtain
implifications to the Electrical Network
Insulated surfaces. In steady state heat transfer, a surface cannot
lectrically cannot flow ot grounded.
urface 3 is not grounded so that the battery and surface wing.
lack surfaces: A black, or ideal surface, will have no surface
previously. S
receive net energy if it is insulated. Because the energy cannot be stored by a surface in steady state, all energy must be re-radiated back into the enclosure. Insulated surfaces are often termed as re-radiating surfaces. Ethrough a battery if it is n
Sresistance serve no purpose and are removed from the dra
Bresistance:
R3
3
31
A
01
111
AA
In this case the nodal Radiosity and emissive power will be equal. This result gives some insight into the physical meaning of a black surface. Ideal surfaces radiate at the maximum possible level. Non-black surfaces will have a reduced potential, somewhat like a battery with a corroded terminal. They therefore have a reduced potential to cause heat/current flow.
Large surfaces: Surfaces having a large surface area will behave as black surfaces, irrespective of the actual surface properties:
011
A
Physically, this corresponds to the characteristic of large surfaces that as they reflect energy, there is very little chance that energy will strike the smaller surfaces; most of the energy is reflected back to another part of the same large surface. After several partial absorptions most of the energy received is absorbed.
Solution of Analogous Electrical Circuits.
Large Enclosures
Consider the case of an object, 1, placed inside a large enclosure, 2. The system will consist of two objects, so we proceed to construct a circuit with two radiosity nodes.
1/(A1F1→2) J1 J2
Now we ground both Radiosity nodes through a surface resistance.
1/(A1F1→2)
J J1 2
(1-1)/(1A1) (1-
Since A2 is large, R2 = 0. The view factor, F1→2 = 1
Sum the series resistances:
RSeries = (1-1)/(1A1) + 1/A1 = 1/(1A1)
Ohm’s law:
i = V/R or by analogy:
q = Eb/RSeries = 1A1(T14 – T2
4)
You may recall this result from Thermo I, where it was introduced to solve this type of radiation problem.
Networks with Multiple Potentials
Eb1 T14
Eb2 T24
2)/( A2) 2
R1 R12 R2
(1-1)/(1A1) 1/(A1F1→2) J1 J2
Eb1 T14 R1 R12 Eb2 T2
4
Systems with 3 or more grounded potentials will require a slightly different solution, but one which students have previously encountered in the Circuits course.
J2 J3
211
1
FA
R12 R13
The procedure will be to apply Kirchoff’s law to each of the Radiosity junctions.
03
1
i
iq
In this example there are three junctions, so we will obtain three equations. This will allow us to solve for three unknowns.
Radiation problems will generally be presented on one of two ways:
o The surface net heat flow is given and the surface temperature is to be found.
o The surface temperature is given and the net heat flow is to be found.
Returning for a moment to the coal grate furnace, let us assume that we know (a) the total heat being produced by the coal bed, (b) the temperatures of the water walls and (c) the temperature of the super heater sections.
Apply Kirchoff’s law about node 1, for the coal bed:
013
13
12
12113121
R
JJ
R
JJqqqq
Similarly, for node 2:
Eb1
J1
Eb3
023
23
12
21
2
2223212
R
JJ
R
JJ
R
JEqqq b
(Note how node 1, with a specified heat input, is handled differently than node 2, with a specified temperature.
And for node 3:
023
32
13
31
3
3332313
R
JJ
R
JJ
R
JEqqq b
The three equations must be solved simultaneously. Since they are each linear in J, matrix methods may be used:
3
3
2
2
1
3
2
1
231332313
231312212
13121312
11111
11111
1111
R
ER
Eq
J
J
J
RRRRR
RRRRR
RRRR
b
b
The matrix may be solved for the individual Radiosity. Once these are known, we return to the electrical analogy to find the temperature of surface 1, and the heat flows to surfaces 2 and 3.
Surface 1: Find the coal bed temperature, given the heat flow:
25.0
1111
1
14
1
1
111
JRqT
R
JT
R
JEq b
Surface 2: Find the water wall heat input, given the water wall temperature:
2
24
2
2
222 R
JT
R
JEq b
Surface 3: (Similar to surface 2) Find the water wall heat input, given the water wall temperature:
3
34
3
3
333 R
JT
R
JEq b
Module 9: Worked out problems 1. A spherical aluminum shell of inside diameter D=2m is evacuated
and is used as a radiation test chamber. If the inner surface is coated with carbon black and maintained at 600K, what is the irradiation on a small test surface placed in the chamber? If the inner surface were not coated and maintained at 600K, what would the irradiation test?
Known: Evacuated, aluminum shell of inside diameter D=2m, serving as a radiation test chamber. Find: Irradiation on a small test object when the inner surface is lined with carbon black and maintained at 600K.what effect will surface coating have? Schematic:
Assumptions: (1) Sphere walls are isothermal, (2) Test surface area is small compared to the enclosure surface. Analysis: It follows from the discussion that this isothermal sphere is an enclosure behaving as a black body. For such a condition, the irradiation on a small surface within the enclosure is equal to the black body emissive power at the temperature of the enclosure. That is
24281
41
/7348)600(./1067.5
)(
mWKKmWG
TTEG ssb
The irradiation is independent of the nature of the enclosure surface coating properties. Comments: (1) The irradiation depends only upon the enclosure surface temperature and is independent of the enclosure surface properties. (2) Note that the test surface area must be small compared to the enclosure surface area. This allows for inter-reflections to occur such that the radiation field, within the enclosure will be uniform (diffuse) or isotropic. (3) The irradiation level would be the same if the enclosure were not evacuated since; in general, air would be a non-participating medium.
2 Assuming the earth’s surface is black, estimate its temperature if the sun has an equivalently blackbody temperature of 5800K.The diameters of the sun and earth are 1.39*109 and 1.29*107m, respetively, and the distance between the sun and earth is 1.5*1011m.
Known: sun has an equivalently blackbody temperature of 5800K. Diameters of the sun and earth as well as separation distances are prescribed. Find: Temperature of the earth assuming the earth is black. Schematic:
Assumptions: (1) Sun and earth emit black bodies, (2) No attenuation of solar irradiation enroute to earth, and (3) Earth atmosphere has no effect on earth energy balance.
Analysis: performing an energy balance on the earth
4/1
422
,,
..
4/
)4/(
)(..
0
Se
eeSe
ebseSpe
outin
GT
TDGD
TEAGA
EE
Where As,p and Ae,s are the projected area and total surface
area of the earth, respectively. To determine the irradiation GS at the earth’s surface, perform an energy bounded by the spherical surface shown in sketch
2
22711
42829
2,
42
..
/5.1377
]2/1029.1105.1[4
)5800(./1067.5)1039.1(
]2/[4.
0
mWG
Gm
KKmWm
GDRTD
EE
S
S
SeeSSS
outin
Substituting numerical values, find
KKmWmWTe 279)./1067.54//5.1377( 4/14282
Comments:
(1) The average earth’s temperature is greater than 279 K since the effect of the atmosphere is to reduce the heat loss by radiation.
(2) Note carefully the different areas used in the earth energy
balance. Emission occurs from the total spherical area, while solar irradiation is absorbed by the projected spherical area.
3 The spectral, directional emissivity of a diffuse material at
2000K has the following distribution.
Determine the total, hemispherical emissivity at 2000K.Determine the emissive power over the spherical range 0.8 t0 2.5 m and for the directions 030. Known: Spectral, directional emissivity of a diffuse material at 2000K. Find: (1) The total, hemispherical emissivity, (b) emissive power over the spherical range 0.8 t0 2.5 m and for the directions 030. Schematic:
Assumptions: (1) Surface is diffuse emitter.
Analysis: (a) Since the surface is diffuse,, is independent of direction; from Eq. , =
)(/),()()( ,
0
TEdTET bb
bbbb EdEEdET /)2000,()/)2000,())( ,
5.1
0
2,
5.1
0
1
Written now in terms of F (0), with F (01.5) =0.2732 at T=1.5*2000=3000m.K, find
636.0]2732.01[8.02732.02.01)2000( )5.10(2)5.10(1 FFK
(b) For the prescribed spectral and geometric limits,
write can we,E to equal and direction of
inde is nothing and, diffuse, is surface the Since ).,(),,( where
sincos),(
b/,
,,,,,
,
5.2
8.0
2
0
6/
0
,
bbe
b
ITII
dddTIE
)(
),(
)(
),()(
sincos
5.2
5.1
,2
5.1
8.0
,12
0
6/
0 TE
dTE
TE
dTETE
ddEb
b
b
b
b
Or in terms F (0) values,
][][2
sin)5.10()5.20(2)8.00()5.10(1
46/
0
22
0
FFFFT
E
From table KmT
KmT
.500020005.2
.160020008.0
6337
0197
.0
.0
)5.20(
)8.00(
F
F
2248
2
482
/89.76339.0/20001067.5(25.0
]2732.06337.80.0[]0197.02732.0[2.020001067.5
2
6/sin2
mWmWE
m
WE
4. A diffusely emitting surface is exposed to a radiant source causing the irradiation on the surface to be 1000W/m2.The intensity for emission is 143W/m2.sr and the reflectivity of the surface is 0.8.Determine the emissive power ,E(W/m2),and radiosity ,J(W/m2),for the surface. What is the net heat flux to the surface by the radiation mode?
Known: A diffusely emitting surface with an intensity due to emission of Is=143W/m2.sr and a reflectance =0.8 is subjected to irradiation=1000W/m2. Find: (a) emissive power of the surface, E (W/m2), (b) radiosity, J (W/m2), for the surface, (c) net heat flux to the surface. Schematic:
Assumptions: (1) surface emits in a diffuse manner. Analysis: (a) For a diffusely emitting surface, Is() =Ie is a constant independent of direction. The emissive power is
22 /449./143 mWsrmWsrIE e Note that has units of steradians (sr).
(b) The radiosity is defined as the radiant flux leaving the surface by emission and reflection,
222 /1249/10008.0/449 mWmWmWGEJ (c) The net radiative heat flux to the surface is determined from a radiation balance on the surface.
222"
",
",
"
/249/1249/1000 mWmWmWJGq
qqq
net
outradinradnet
Comments: No matter how the surface is irradiated, the intensity of the reflected flux will be independent of direction, if the surface reflects diffusely.
5. Radiation leaves the furnace of inside surface temperature 1500K through an aperture 20mm in diameter. A portion of the radiation is intercepted by a detector that is 1m from the aperture, as a surface area of 10-5m2, and is oriented as shown. If the aperture is open, what is the rate at which radiation leaving the furnace is intercepted by the detector? If the aperture is covered with a diffuse, semitransparent material of spectral transmissivity =0.8 for 2m and =0 for >2m, what is the rate at which radiation leaving the furnace is intercepted by the detector?
Known: Furnace wall temperature and aperture diameter. Distance of detector from aperture and orientation of detector relative to aperture. Find: Rate at which radiation leaving the furnace is intercepted by the detector, (b) effect of aperture window of prescribed spectral transmissivity on the radiation interception rate. Schematic:
Assumptions: (1) Radiation emerging from aperture has characteristics of emission from a black body, (2) Cover material is diffuse, (3) Aperture and detector surface may be approximated as infinitesimally small.
Analysis: (a) the heat rate leaving the furnace aperture and intercepted by the detector is
aase wAIq cos
hence
srm
m
r
A
r
Aw
srmWTTE
I
sn
as
ffbe
.70710.0)1(
45cos10cos.
./1014.9)1500(1067.5)(
52
25
2
2
2
24484
WsrmsrmWq 45224 1076.110707.030cos]4/)02.0([./1014.9
(b) With the window, the heat rate is
wall. furnace the by emitted radiation to window the of vitytransmissi the is where
)cos( 1
aaae wAIq
W100.384W101.760.218q
find0.218,0.80.273 with hence
0.273.m)2F(0 table from m.K,30001500Km2T with
8.0)/(8.0
)(
4-4-
)20(
2
0
,
0
,
0
.
0
0
mbb
b
fb
FdEE
dE
dTE
dG
dG
6.A horizontal semitransparent plate is uniformly irradiated from above and below, while air at T=300K flows over the top and bottom surfaces. providing a uniform convection heat transfer coefficient of h=40W/m2.K.the total, hemispherical absorptivity of the plate to the irradiation is 0.40.Under steady-state conditions measurements made with radiation detector above the top surface indicate a radiosity(which includes transmission, as well as reflection and emission) of J=5000W/m2,while the plate is at uniform temperature of T=350K
Determine the irradiation G and the total hemispherical emissivity of the plate. Is the plate gray for the prescribed conditions? Known: Temperature, absorptivity, transmissivity, radiosity and convection conditions for a semi-transparent plate. Find: Plate irradiation and total hemispherical emissivity. Schematic:
Plate,T=350K,
h
Th= W/m2.K
q”conv
q”conv
G
GJ=5000W/m2
J
Assumptions: From an energy balance on the plate
JqG
EE
conv
outin
222 "
.
Solving for the irradiation and substituting numerical values,
G=40W/m2.K (350-300) K+5000W/m2=7000W/m2 From the definition of J
GEGGEJ )1( Solving for the emissivity and substituting numerical values,
94.0)350(./1067.5
)/7000(6.0)/5000()1(
4428
22
4
KKmW
mWmW
T
GJ
Hence
And the surface is not gray for the prescribed conditions. Comments: The emissivity may also be determined by expressing the plate energy balance as
94.0)350(4./1067.5
)50(./40)/7000(4.0
)(
222
428
22
4
"
KKmW
KKmWmW
TThGT
hence
EqG conv
7 An opaque, gray surface at 27C is exposed to irradiation of
1000W/m2, and 800W/m2 is reflected. Air at 17C flows over the surface and the heat transfer convection coefficient is 15W/m2.K.Determine the net heat flux from the surface.
Known: Opaque, gray surface at 27C with prescribed irradiation, reflected flux and convection process. Find: Net heat flux from the surface. Schematic:
Assumptions: (1) Surface is opaque and gray, (2) Surface is diffuse, (3) Effects of surroundings are included in specified irradiation. Analysis: From an energy balance on the surface, the net heat flux from the surface is
GGTTThGGEqq
EEq
refssrefconvnet
inoutnet
4""
".""
)(
22"
2444282"net
ref
/42/)10008009.91150(
10/800)27327(./1067.52.0)1727(./15q
surface the fromflux heat net the /G.G where
2.08.01)1000/800(1)/(11
mWmWq
mWKKmWKKmW
GG
net
ref
Comments: (1) For this situation, the radiosity is
22 /892/)9.91800( mWmWEGJ ref
The energy balance can be written involving the radiosity (radiation leaving the surface) and the irradiation (radiation to the surface).
22""
, /42/)1501000892( mWmWqGJq convoutnet
Note the need to assume the surface is diffuse, gray and opaque in order that Eq (2) is applicable.
8. A small disk 5 mm in diameter is positioned at the center of an isothermal, hemispherical enclosure. The disk is diffuse and gray with an emissivity of 0.7 and is maintained at 900 K. The hemispherical enclosure, maintained at 300 K, has a radius of 100 mm and an emissivity of 0.85.
Calculate the radiant power leaving an aperture of diameter 2 mm located on the enclosure as shown. Known: Small disk positioned at center of an isothermal, hemispherical enclosure with a small aperture. Find: radiant power [W] leaving the aperture. Schematic:
Disk(A1),T1=900K,Di=5mm,
Hemispherical enclosure (As)
Ts=300K,
perture( z D mmR=100mm
qap
Assumptions: (1)Disk is diffuse-gray,(2) Enclosure is isothermal and has area much larger than disk,(3) Aperture area is very small compared to enclosure area, (4) Areas of disk and aperture are small compared to radius squared of the enclosure.
Analysis: the radiant power leaving the aperture is due to radiation leaving the disk and to irradiation on the aperture from the enclosure. That is
2221 .AGqqap
The radiation leaving the disk can be written in terms of the radiosity of the disk. For the diffuse disk
431
4111111
1211121
)1()(J
is radiosity thebehavior, gray thefor with and
.cos.1
TTGTE
wAJq
b
follows angle solid ).The(TEGG
);(TEenclosure, black the ofpower emissive the is G irradiatin the Where
-123b21
3b1
2212 / RA
Combining equations. (2), (3) and (4) into eq.(1) with G2=T4
3, the radiant power is
WWq
KKmWmm
m
mKKKmWq
TAR
AATTq
ap
ap
ap
1479)144319.02.36(
)300(./1067.25)002.0(4)100.0(
)002.0(4/
45cos)005.0(4
)300)(7.01()900(7.0[./1067.51
.cos])1([1
44282
2
244428
4322
211
431
411
Comments: Note the relative magnitudes of the three radiation components. Also, recognize that the emissivity of the enclosure’s 3 doesn’t enter into the analysis. Why?
MODULE 7
HEAT EXCHANGERS 7.1 What are heat exchangers? Heat exchangers are devices used to transfer heat energy from one fluid to another. Typical heat exchangers experienced by us in our daily lives include condensers and evaporators used in air conditioning units and refrigerators. Boilers and condensers in thermal power plants are examples of large industrial heat exchangers. There are heat exchangers in our automobiles in the form of radiators and oil coolers. Heat exchangers are also abundant in chemical and process industries. There is a wide variety of heat exchangers for diverse kinds of uses, hence the construction also would differ widely. However, in spite of the variety, most heat exchangers can be classified into some common types based on some fundamental design concepts. We will consider only the more common types here for discussing some analysis and design methodologies. 7.2 Heat Transfer Considerations The energy flow between hot and cold streams, with hot stream in the bigger diameter tube, is as shown in Figure 7.1. Heat transfer mode is by convection on the inside as well as outside of the inner tube and by conduction across the tube. Since the heat transfer occurs across the smaller tube, it is this internal surface which controls the heat transfer process. By convention, it is the outer surface, termed Ao, of this central tube which is referred to in describing heat exchanger area. Applying the principles of thermal resistance,
t T
di
do
Figure 7.1: End view of a tubular heat exchanger
Rh A
rr
kl h Ao o
o
i
i i
1
2
1ln
If we define overall the heat transfer coefficient, Uc, as:
URAc
o
1
Substituting the value of the thermal resistance R yields:
1 1
U h
r rr
k
A
h Ac o
oo
i o
i i
ln
Standard convective correlations are available in text books and handbooks for the convective coefficients, ho and hi. The thermal conductivity, k, corresponds to that for the material of the internal tube. To evaluate the thermal resistances, geometrical quantities (areas and radii) are determined from the internal tube dimensions available. 7.3 Fouling Material deposits on the surfaces of the heat exchanger tubes may add more thermal resistances to heat transfer. Such deposits, which are detrimental to the heat exchange process, are known as fouling. Fouling can be caused by a variety of reasons and may significantly affect heat exchanger performance. With the addition of fouling resistance, the overall heat transfer coefficient, Uc, may be modified as:
1 1
U UR
d c
"
where R” is the fouling resistance. Fouling can be caused by the following sources:
1) Scaling is the most common form of fouling and is associated with inverse solubility salts. Examples of such salts are CaCO3, CaSO4, Ca3(PO4)2, CaSiO3, Ca(OH)2, Mg(OH)2, MgSiO3, Na2SO4, LiSO4, and Li2CO3.
2) Corrosion fouling is caused by chemical reaction of some fluid constituents with the heat exchanger tube material.
3) Chemical reaction fouling involves chemical reactions in the process stream which results in deposition of material on the heat exchanger tubes. This commonly occurs in food processing industries.
4) Freezing fouling is occurs when a portion of the hot stream is cooled to near the freezing point for one of its components. This commonly occurs in refineries where paraffin frequently solidifies from petroleum products at various stages in the refining process. , obstructing both flow and heat transfer.
5) Biological fouling is common where untreated water from natural resources such as rivers and lakes is used as a coolant. Biological micro-organisms such as algae or other microbes can grow inside the heat exchanger and hinder heat transfer.
6) Particulate fouling results from the presence of microscale sized particles in solution. When such particles accumulate on a heat exchanger surface they sometimes fuse and harden. Like scale these deposits are difficult to remove.
With fouling, the expression for overall heat transfer coefficient becomes:
1 1 1
U hr
r
rr
k hR
di
i
o
o
i
o
ln( )"
7.4 Basic Heat Exchanger Flow Arrangements Two basic flow arrangements are as shown in Figure 7.2. Parallel and counter flow provide alternative arrangements for certain specialized applications. In parallel flow both the hot and cold streams enter the heat exchanger at the same end and travel to the opposite end in parallel streams. Energy is transferred along the length from the hot to the cold fluid so the outlet temperatures asymptotically approach each other. In a counter flow arrangement, the two streams enter at opposite ends of the heat exchanger and flow in parallel but opposite directions. Temperatures within the two streams tend to approach one another in a nearly linearly fashion resulting in a much more uniform heating pattern. Shown below the heat exchangers are representations of the axial temperature profiles for each. Parallel flow results in rapid initial rates of heat exchange near the entrance, but heat transfer rates rapidly decrease as the temperatures of the two streams approach one another. This leads to higher exergy loss during heat exchange. Counter flow provides for relatively uniform temperature differences and, consequently, lead toward relatively uniform heat rates throughout the length of the unit.
Fig. 7.2 Basic Flow Arrangements for Tubular Heat Exchangers. 7.5 Log Mean Temperature Differences Heat flows between the hot and cold streams due to the temperature difference across the tube acting as a driving force. As seen in the Figure 7.3, the temperature difference will vary along the length of the HX, and this must be taken into account in the analysis.
Fig. 7.3 Temperature Differences Between Hot and Cold Process Streams
From the heat exchanger equations shown earlier, it can be shown that the integrated average temperature difference for either parallel or counter flow may be written as:
2
1
21
ln LMTD
The effective temperature difference calculated from this equation is known as the log mean temperature difference, frequently abbreviated as LMTD, based on the type of mathematical average that it describes. While the equation applies
T 2
t2
T1
t1
Counter Flow
t 1
T 1
T1
T 2
t2 t 1
Temperature
Position
T2
t2
Parallel FlowT 1
T2
t 1 t2
Position
Temperature
Counter FlowT 1
T2
t 2 t 1
Position
θ1
θ2
Parallel FlowT1
T2
t1 t 2
Position
θ1 θ2
to either parallel or counter flow, it can be shown that eff will always be greater in the counter flow arrangement. Another interesting observation from the above Figure is that counter flow is more appropriate for maximum energy recovery. In a number of industrial applications there will be considerable energy available within a hot waste stream which may be recovered before the stream is discharged. This is done by recovering energy into a fresh cold stream. Note in the Figures shown above that the hot stream may be cooled to t1 for counter flow, but may only be cooled to t2 for parallel flow. Counter flow allows for a greater degree of energy recovery. Similar arguments may be made to show the advantage of counter flow for energy recovery from refrigerated cold streams. 7.6 Applications for Counter and Parallel Flows We have seen two advantages for counter flow, (a) larger effective LMTD and (b) greater potential energy recovery. The advantage of the larger LMTD, as seen from the heat exchanger equation, is that a larger LMTD permits a smaller heat exchanger area, Ao, for a given heat transfer, Q. This would normally be expected to result in smaller, less expensive equipment for a given application. Sometimes, however, parallel flows are desirable (a) where the high initial heating rate may be used to advantage and (b) where it is required the temperatures developed at the tube walls are moderate. In heating very viscous fluids, parallel flow provides for rapid initial heating and consequent decrease in fluid viscosity and reduction in pumping requirement. In applications where moderation of tube wall temperatures is required, parallel flow results in cooler walls. This is especially beneficial in cases where the tubes are sensitive to fouling effects which are aggravated by high temperature. 7.7 Multipass Flow Arrangements In order to increase the surface area for convection relative to the fluid volume, it is common to design for multiple tubes within a single heat exchanger. With multiple tubes it is possible to arrange to flow so that one region will be in parallel and another portion in counter flow. An arrangement where the tube side fluid passes through once in parallel and once in counter flow is shown in the Figure 7.4. Normal terminology would refer to this arrangement as a 1-2 pass heat exchanger, indicating that the shell side fluid passes through the unit once, the tube side twice. By convention the number of shell side passes is always listed first.
Fig. 7.4 Multipass flow arrangement The primary reason for using multipass designs is to increase the average tube side fluid velocity in a given arrangement. In a two pass arrangement the fluid flows through only half the tubes and any one point, so that the Reynold’s number is effectively doubled. Increasing the Reynolds’s number results in increased turbulence, increased Nusselt numbers and, finally, in increased convection coefficients. Even though the parallel portion of the flow results in a lower effective T, the increase in overall heat transfer coefficient will frequently compensate so that the overall heat exchanger size will be smaller for a specific service. The improvement achievable with multipass heat exchangers is substantialy large. Accordingly, it is a more accepted practice in modern industries compared to conventional true parallel or counter flow designs. The LMTD formulas developed earlier are no longer adequate for multipass heat exchangers. Normal practice is to calculate the LMTD for counter flow, LMTDcf, and to apply a correction factor, FT, such that
eff T CFF LMTD
The correction factors, FT, can be found theoretically and presented in analytical form. The equation given below has been shown to be accurate for any arrangement having 2, 4, 6, .....,2n tube passes per shell pass to within 2%.
FR
P
R P
RP R R
P R R
T
2
2
2
11
1
12 1
2 1
ln
ln1
1
where the capacity ratio, R, is defined as:
RT T
t t
1 2
2 1
The effectiveness may be given by the equation:
PX
R X
N
N
shell
shell
1 1
1
/
/
provided that R1. In the case that R=1, the effectiveness is given by:
P
P
N P No
shell o shell
1
where
Pt t
T to
2 1
1 1
and
XP R
Po
o
1
1
7.8 Effectiveness-NTU Method: Quite often, heat exchanger analysts are faced with the situation that only the inlet temperatures are known and the heat transfer characteristics (UA value) are known, but the outlet temperatures have to be calculated. Clearly, LMTH method will not be applicable here. In this regard, an alternative method known as the ε-NTU method is used.
Before we introduce this method, let us ask ourselves following question: ? conditionsinlet given for perform ExchangeHeat existing willHow
:esseffectiven Define The effectiveness, ε, is the ratio of the energy recovered in a HX to that recoverable in an ideal HX.
havecan ratecapacity heat C ,C oflesser with fluid only thethen
since and
fluid One
H.Ex. long infinitelyan for is where,
maxBA
A
,,max
maxmax
T
TCTCTcmTcmQ
TTTT
Q
BBABBAA
incinh
actual
max
min
minmax
min
max
min
min
..min
..minmaxminmax
1UA-
exp1
1UA-
exp-1
......... )( intoback Substitute
s.T'outlet contain not does which for expression want We
or,
and i.e.
C
C
CC
C
C
C
C
LMTDUAQ
TTCQ
TTC
QTCQ
incinh
incinh
Fig. 7.5 Calculation of effectiveness-NTU
min
max
min
HEx.) of (size units transfer of No. and
,
C
UANTU
C
CNTU
Charts for each Configuration Procedure: Determine Cmax, Cmin/Cmax Get UA/Cmin, from chart
incinh TTCQ ..min
U
CNTUA
C
UANTU minmax
minmax
• NTUmax can be obtained from figures in textbooks/handbooks First, however, we must determine which fluid has Cmin.
Module 7: Solved Problems
1. Deionized water flows through the inner tube of 30-mm diameter in a thin-walled concentric tube heat exchanger of 0.19-m length. Hot process water at 95C flows in the annulus formed with the outer tube of 60-mm diameter. The deionized water is to be heated from 40 to 60C at a flow rate of 5 kg/s. The thermo physical properties of the fluids are:
DEIONIZED
WATERPROCESSWATER
kg/m3)982.3
4181
0.643
548
3.56
967.1
4197
0.673
324
2.02
cp(J/kg.K
k(W/m.K
N.s/m2
pr
(a) Considering a parallel-flow configuration of the heat exchanger,
determine the minimum flow rate required for the hot process water.
(b) Determine the overall heat transfer coefficient required for the conditions of part a.
(c) Considering a counter flow configuration, determine the minimum flow rate required for the hot process water. What is the effectiveness of the exchanger for this situation?
Schematic:
Process water,h
Deionized water,c
Tc,i=40 C
Th,i=95 C
ΔT1T2mc=5kg/s
Tc,0i=60 C
Th,o
T
x
Concentric tube
PF,L=0.19mD=30mm
0
Assumptions: (1) Negligible heat loss to surroundings, (2) Negligible kinetic and potential energy changes. Analysis: (a) from overall energy balances,
)()()()( ,,
.
,,
.
icochohihh TTcmTTcmq
For a fixed term , )h will be a minimum when Th,o is a minimum.
With the parallel flow configuration, this requires that Th,o=Tc,o=60C. Hence,
ihT ,
.
(m
skgCKkgJ
CKkgJskg
TTc
TTmchm
ohihh
icocc /85.2)6095(./4197
)4060(./4181/5
)(
)()(min,
,,
.
,,.
(b)From the rate equation and the log mean temperature relation,
TTln
T-T
2
1
21,,
PFlmPFlm TTUAq
And since ΔT2=0, ΔTlm=0 so that UA=. Since A=DL is finite, U must be extremely large. Hence, the heating cannot be accomplished with this arrangement.
(c) With the CF arrangements will be a minimum when Tho is a minimum. This requires that Th,o is a minimum. This requires that Th,o
is a minimum. This requires that Th,o=Tc,i=40C. Hence, from the overall energy balance,
hm.
skgKKkgJ
KKkgJskgm /81.1
)4095(./4197
)4060(./4181/5.
For this condition, Cmin=Ch which is cooled from Th,i to Tc,i, hence =1 Comments: For the counter flow arrangement, the heat exchanger must be infinitely long.
2. Water with a flow rate of 0.05kg/s enters an automobile radiator at 400K and leaves at 330 K. The water is cooled by air in cross flow which enters at 0.75kg/s and leaves at 300K. If the overall heat transfer coefficient is 200W/m2.K, what is the required heat transfer surface area? Schematic:
Water Th,i=400Kmh=0.05kg/s
Tc,i=300Kmc=0.75kg/s
Air
Th,o=330K
Tc,o
Assumptions: (1) Negligible heat loss to surroundings and kinetic and potential energy changes, (2) Constant properties. Analysis: The required heat transfer rate is
WKKkgJskgTTcmq ohihh 732,1470)./4209(/05.0)()( ,,
.
Using the -NTU method,
700.0045,21/732,14/
045,21)100(/45.210)(/,
,/25.755
/45.210
max
,,maxmin
max
min
WWqq
and
WKKWTTCChence
KWCC
KWCC
icih
c
h
From figure, NTU1.5, hence
22min 58.1)./200(/45.2105.1)/( mKmWKWUCNTUA
Comments: (1) the air outlet temperature is
KKWWKCqTT cicoc 5.319)/25.755/732,14(300/,,
(2) Using the LMTD approach, ΔTlm=51.2 K, R=0.279 and P=0.7. Hence from fig F0.95 and
.51.1]2.51)./200(95.0/[)732,14(/ 22 mKKmWWTFUqA lm
3. Saturated steam leaves a steam turbine at a flow rate of 1.5kg/s and a pressure of 0.51 bars. The vapor is to be completely condensed to saturated liquid in a shell-and –tube heat exchanger which uses water as the coolant. The water enters the thin-walled tubes at 17C and leaves at 57C. If the overall heat transfer coefficient of 200W/m2.K, determine the required heat exchanger surface area and the water flow rate. After extended operation, fouling causes the overall heat transfer coefficient to decrease to 100W/m2.K. For the same water inlet temperature and flow rate, what is the new vapor flow rate required for complete condensation? Schematic:
Assumptions: (1) Negligible heat loss to surroundings, (2) Negligible wall conduction resistance. Properties: Table for sat.Water:
2304kJ/kg.h355K,T:bars) 0.51(p ;./4178c :)310( tsacp,
fgc KkgJKT
Analysis: (a) The required heat transfer rate is
WkgJskghmq fgh66
.
1046.3)/10304.2(/5.1
And the corresponding heat capacity rate for the water is
97.0)62.01ln()--ln(1NTU
0,/CC since
62.0)65(/400,86/1046.3])[/(,
/400,8640/1048.3)/(
maxmin
6,,min
6,,min
KKWWTTCqhence
KWKWTTqCC
icih
icocc
And
skgKkgJKWcCm
mKmWKWUCNTUA
cpcc /7.20./4178//400,86/
29.41).2/2000//400,86(97.0)/(
,
.
min
(b) using the final overall heat transfer coefficient, find Since 0,/CC maxmin
skgkgJWhqm
WKKWTT
NTU
fgh
icih
/936.0/10304.2/1016.2/
16106.265)/400,886(384.0)(Cq hence,
384.0)485.0exp(1)exp(1
66.
,,min
Comments: The significant reduction (38%) in represents a significant loss in turbine power. Periodic cleaning of condenser surfaces should be employed to minimize the adverse effects of fouling.
hm.
4. Water at 225 kg/h is to be heated from 35 to 95C by means of a concentric tube heat exchanger. Oil at 225kg/h and 210C, with a specific heat of 2095 J/kg.K, is to be used as the hot fluid. If the overall heat transfer coefficient based on the outer diameter of the inner tube if 550W/m2.K, determine the length of the exchanger if the outer diameter is 100mm. Schematic:
Assumptions: (1) Negligible heat loss to surroundings, (2) Negligible kinetic and potential energy changes, (3) Constant properties. Properties: Table for Water:
KkgJcKCT cpc ./4188:)3382/)9535(( ,
_
Analysis: From rate equation with Ao=DoL, L=q/UoDoΔTm The heat rate, q, can be evaluated from an energy balance on the cold fluid,
WKKkgJhs
hkgTTcmq icccc 705,15)3595(./4188
/3600
/225)( ,0,
.
In order to evaluate ΔTm, we need to know whether the exchanger is operating in CF or PF. From an energy balance on the hot fluid, find
CKkg
J
hs
hkgWCcmqTT hhihoh 1.90
.2095
/3600
/225/705,15210/
.
,,
Since Th,o<Tc,o it follows that HXer operation must be CF. From eq. for log mean temperature difference,
CCnTTn
TTT CFm
5.81)1.55/115(
)351.90()95210(
)/( 21
21,
Substituting numerical values, the HXer length is
mKmKmWWL 12.14.81)10.0(./550/705,15 2 Comments: The –NTU method could also be used. It would be necessary to perform the hot fluid energy balance to determining CF operation existed. The capacity rate is Cmin/Cmax=0.50. From eq. for effectiveness, and from with q evaluated from an energy balance on the hot fluid,
69.035210
1.90210
,,
,,
icih
ohih
TT
TT
Th,i=210 C
Th,o=90.1 C
Tc,o=95 CTc,i=35 C
ΔT1
ΔT2
1 2
T
x
From fig, find NTU1.5 giving
mmKm
W
K
WDUCNTUL oo 14.1)10.0(.
.55094.1305.1/.
2min
Note the good agreement by both methods.
5. Consider a very long, concentric tube heat exchanger having hot and cold water inlet temperatures of 85 and 15C. The flow rate of the hot water is twice that of the cold water. Assuming equivalent hot and cold water specifies heats; determine the hot water outlet temperature for the following modes of operation (a) Counter flow, (b) Parallel flow. Schematic:
Th,i=85 C
Tc,i=15 C
Cc
Ch=2Cc
Assumptions: (1) equivalent hot and cold water specific heats, (2) Negligible Kinetic and potential energy changes, (3) No eat loss to surroundings.
Analysis: the heat rate for a concentric tube Heat exchanger with very large surface area Operating in the counter flow mode is
)( ,,minmax icih TTCqq
Combining the above relation and rearranging, find
ihicihh
cihicih
hoh TTT
C
CTTT
C
CT ,,,,,,
min, )()(
Substituting numerical values
CCCT oh 5085)1585(2
1,
For parallel flow operation, the hot and cold outlet temperatures will be equal; that is Tc,o=Th,o. Hence
)()( ,,,, ohihhicocc TTCTTC
Setting Tc,o=Th,o and rearranging
CCT
C
CT
C
CTT
oh
h
cic
h
cihoh
7.612
11/15
2
185
1/
,
,,,
Comments: Note that while =1 for CF operation, for PF operation find = q/qmax=0.67.
6. A concentric tube heat exchanger uses water, which is available at 15°C, to cool ethylene glycol from 100 to 60°C. The water and glycol flow rates are each 0.5 kg/s. Determine the maximum possible heat transfer rate and effectiveness of the exchanger. Determine which is preferred, a parallel –flow or counter flow mode of operation? Known: Inlet temperatures and flow rate for a concentric tube heat exchanger. Find: (a) Maximum possible heat transfer rate and effectiveness, (b) Proffered mode of operation. Schematic:
Ethylene glycol
Th,i=100 C
Tho=60 C
mhi=0.5kg/s Water Tc,i=15 C mc=0.5kg/s
U,A
Assumptions: (1) Steady-state operation, (2) Negligible KE and PE changes, (3) Negligible heat loss to surroundings, (4) Fixed overall heat transfer and coefficient.
Properties: Table: Ethylene glycol ( cp=2650J/kg.K; );80_
CT in
KkgJcCTWater pm ./4178:)30(_
Analysis: (a) Using the -NTU method, find
KWKkgJskgcmCChph h /1325)./2650)(/5.0(
,
.
min
WCKWTTCxq icihma
5,.min 1013.1)15100)(/1325()(
WCKkgJskgTTcmq icihhph5
,.
.
, 1053.0)60100)(./2650(/5.0)(
47.01013.1/1053.0/ 55
max qq (b)
CKkgJskg
Ccm
qTT
cpc
icoc
4.40
./4178/5.0
1053.015
5
.
,
,,
Since Tc,o<Th,o, a parallel flow mode of operation is possible.
However, with (Cmin/Cmax) = =0.63, ),/(.
,
.
cpchp cmcm h
From fig (NTU)PF0.95, (NTU)CF0.75 Hence (ACF/APF)= (NTU) CF/ (NTU) PF (0.75/0.95)=0.79 Because of the reduced size requirement, hence capital investment, the counter flow mode of operation is preferred.