BLOCK 2
Indira Gandhi
National Open University
School of Engineering & Technology
MRW-002 HEAT TRANSFER
Conduction
Indira Gandhi
National Open University
School of Engineering & Technology
MRW – 002 Heat Transfer
Block
2
CONDUCTION UNIT 4
Governing Equations of Heat Conduction 5 UNIT 5
Numerical Methods to Solve Heat Conduction Problems 51 UNIT 6
Heat Transfer from Extended Surface 99
CONDUCTION
Present Block is devoted to conduction heat transfer. The entire block consists of three units.
Unit 4, introduces the governing equations of conduction heat transfer.
Application of Fourier’s law to steady and unsteady state conduction has been analysed. The governing equation of heat conduction is presented in rectangular, cylindrical as well as spherical coordinates. Heat transfer and determination of temperature distribution for plane wall, cylinder and sphere under steady state condition are discussed. Transit heat transfer concept is introduced with heat generation number, Biot number and Fourier number. A detail analysis of transient heat transfer by lumped capacitance method is described.
Unit 5, describes various methods for solving heat conduction equations.
Analytical and graphical methods are described. Limitation of these methods are highlighted and it is emphasized that numerical methods is used in complicated geometries, boundary conditions and variable thermal properties. Finite difference method is described both for the steady and transient conduction problems.
Unit 6, describes application of extended surfaces for heat transfer devices.
Different types of fins and their applications are cited in this unit. Formulation and solution are done for extended surfaces. Performance parameters such as fin efficiency and effectiveness are discussed in detail.
The units are supported with solved problems. All the three units include some unsolved problems and SAQ’s which will help you understanding the heat conduction mechanism.
5
Governing Equations of Heat Conduction
UNIT 4 GOVERNING EQUATIONS OF HEAT CONDUCTION
Structure
4.1 Introduction
Objectives
4.2 General Equation of Heat Conduction
4.2.1 Rectangular Coordinate System 4.2.2 Cylindrical Coordinates 4.2.3 Spherical Coordinates
4.3 Steady State Heat Conduction in Simple Geometrical Systems
4.3.1 Plane Wall 4.3.2 Cylindrical Wall 4.3.3 Spherical Shell
4.4 Transient Conduction
4.4.1 Lumped Capacitance Method 4.4.2 General Lumped Capacitance Analysis
4.5 The Semi-infinite Solid 4.6 Multi-dimensional Effects 4.7 Summary
4.8 Key Words 4.9 Answers to SAQs
1.1 INTRODUCTION
In conduction mode heat is transferred through a complex sub-microscopic mechanism that involves flow of free electrons and lattice vibration. Conduction is predominant in case of solid or liquid metals. In case of liquids and gases, once heat begins to flow, even if no external force is applied, density gradients are set up and convective currents are set in motion. Heat is then transported on a macroscopic scale as well as on a
microscopic scale with convection currents generally being the more effective. In the following sub-units conduction heat transfer has been discussed in details for steady as well as transient states.
Objectives
After studying this unit, you should be able to
formulate heat transfer by conduction in different geometries under steady state condition,
evaluate temperature at different locations in rectangular, cylindrical and spherical coordinate system,
differentiate the steady state and transient nature of heat conduction,
evaluate heat transfer by lumped capacitance method and its limitations, and
solve some problems on steady and transient heat transfer.
4.2 GENERAL EQUATION OF HEAT CONDUCTION
General equation of heat conduction is explained in three coordinate systems.
6
Conduction 4.2.1 Rectangular Coordinate System
Differential Form
Let us consider an infinitesimal volume element of sides x, y and z as shown in Figure 4.1. The considerations here will include the non-steady condition of temperature variation with time t.
Figure 4.1 : Volume Element for Determining Heat Conduction Equation
According to the Fourier heat conduction law, the heat flowing into the left most face of the element in the x-direction
x
dQ k y z dT
x
. . . (4.1)
The value of the heat flow out of the right face of the element can be obtained by expanding dQx in a Taylor series and retaining only the first two terms as an approximation :
( ) . . .
x dx x x
dQ dQ dQ x
x
. . . (4.2)
The net heat flow by conduction in the x-direction is therefore
( )
x x dx x
dQ dQ dQ x k y z T x
x x x
2 2
k T x y z x
. . . (4.3)
Similarly, the net heat flows in the y-and z-direction are
2
y y dy T2
dQ dQ k x y z
y
. . . (4.4)
2
z z dz T2
dQ dQ k x y z
z
. . . (4.5)
Here, the solid has been assumed to be isotropic and homogeneous with properties uniform in all directions. Let us consider that there is some heat source within the solid, and heat is produced internally as a result of the flow of electrical current or nuclear or chemical reactions. Let qG is the rate at which heat is generated
Z
z
dQy
y
x
dQz
x dQx + dx
y dQx + dy
dQy + dz
dQx
7
Governing Equations of Heat Conduction
internally per unit volume W3 m
. Then the total rate of heat generation in the elemental volume is qG x y z. The net heat flow owing to conduction and the heat generated within the element together will increase the internal energy of the volume element. The rate of accumulation of internal energy IE within the control volume is
. .
c T
IE x y z
t
. . . (4.6)
where c is the specific heat and is the density of the solid.
An energy balance can be achieved on the volume element as : Rate of energy storage within the solid
= Rate of heat influx – Rate of heat outflux + Rate of heat generation
or c T ( x y z) ( x dx y dy z dz)
x y z dQ dQ dQ dQ dQ dQ
t
qG x y z
2 2 2
2 2 2 G
T T T
k x y z q x y z
x y z
. . . (4.7)
or
2 2 2
2 2 2
c T T T T G
k q
t x y z
. . . (4.8)
2 2 2
2 2
G 1 q
T T T T
z k t
x y
. . . (4.9)
where is the thermal diffusivity of the solid given by
c
k
.
If the temperature of a material is not a function of time, the system is in the steady state and does not store any energy. The steady state form of a three- dimensional conduction equation in rectangular coordinates is
2 2 2
2 2 2 qG 0
T T T
x y z k
. . . (4.10)
If the stem is in steady state and no heat is generated internally, the conduction equation simplifies to
2 2 2
2 2 2 0
T T T
x y z
. . . (4.11)
Eq. (4.11) is known as the Laplace equation. It occurs in a number of areas in addition to heat transfer, for example, in diffusion of mass or in electromagnetic fields. The operation of taking the second derivatives of the potential in a field has therefore been given a short symbol, 2, called the Laplacian operator. For the rectangular coordinate system Eq. (4.11) becomes
2 2 2
2
2 2 2 0
T T T
x y z T
. . . (4.12)
Since the operator 2 is independent of coordinate system, the above form will be useful when we study conduction in cylindrical and spherical coordinates. The heat conduction can thus be written as
2 qG 1 T
T k t
. . . (4.13)
8
Conduction Vector Method
Equation can also be derived vectorially. Let us consider a control surface S enclosing a volume V as shown in Figure 4.2.
Figure 4.2 : Heat Conduction through a Volume Element
The net rate of heat outflow across the surface S is given by .
s
q nds
where n is the normal direction. Converting the surface integral to volume integral. div
s v
q n ds q dv
. . . (4.14)Where q is the heat flux per unit area. The net rate of heat inflow to the control volume (CV) is div
v
q dv.If there is a volumetric heat source inside the CV the rate of heat generation is
G v
q dv . . . (4.15)The rate of energy accumulation within the CV is
v v
e dv c T dV
t t
. . . (4.16)e being the specific energy (J/kg). By energy balance,
G div
v v v
c T dv q dv q dV
t
. . . (4.17)where q is the heat flux per unit area. Writing the energy equation for the elemental volume dV within the CV
G div
c T dV q dV q dV t
. . . (4.18)
Since dV is now independent, it can be removed from the above equation.
Therefore,
G div
c T q q
t
. . . (4.19)
Now, divq .q . . . (4.20)
and q k T . . . (4.21)
divq ( k T) k 2T . . . (4.22) for constant k.
q ds
d
qG
q Control surface S
9
Governing Equations of Heat Conduction
Substituting in Eq. (4.19),
G 2
c T q k T
t
. . . (4.23)
Therefore, 2 qG 1 T
T k t
. . . (4.24)
Which is the same as Eq. (4.13).
4.2.2 Cylindrical Coordinates
For a general transient three-dimensional heat conduction problem in the cylindrical coordinates with T = T (r, q, z, t), let us consider an elementary volume as shown in Figure 4.3.
dV d r r d dz . . . (4.25)
Figure 4.3 : Heat Conduction in a Cylindrical Volume Element (r, , z) By Fourier’s law,
( )
r T
dQ k r d z r
. . . (4.26)
( )
r dr r r
dQ dQ dQ dr
r
. . . (4.27)
r r dr T
dQ dQ k r d dz dr
r r
2 2
T T
kr d dz dr k d dz dr r r
. . . (4.28)
Similarly, T
dQ k dz dr
r
. . . (4.29)
d
dQ d Q k dz dr T d
r . . . (4.30)
dQθ +dθ
dQr +dr
dQz
dQr dθ
dθ θ
dQθ
z dQz +dz
r dr ds
n q
r
r
z
θ θ
10
Conduction
( )
dQ d d Q d Q r d
r
. . . (4.31)
2 2
T T T
dr r d dzc kr d dr k d dz dr
t r r
2 2
2 2
T T
k dz dr d k dr rd dz z
1 2T2 k dz dr d
r
. . . (4.32)
z
dQ k dr rd T z
. . . (4.33)
( )
z dz z z
dQ dQ dQ dz
z
. . . (4.34)
z z dz T
dQ dQ k dr r d dz
z z
2 2
k dr r d dz T z
. . . (4.35)
Rate of heat generation from an internal heat source q dr rd dzG
. . . (4.36)
Rate of energy accumulation within the CV
( ) T
dr r d c t
. . . (4.37)
By energy balance, from Eqs. (4.28)-(4.37),
2 2
2 2
T T T T
dr rd d z c k r d dr k d dz dr k dz dr d
t r r
2
2 G
k dr r d dz T q dr r d dz z
. . . (4.38)
2 2 2
2 2 2 2
1 1
T T T T T
c k k k k
t r r r r z
. . . (4.39)
or
2 2 2
2 2 2 2
1 1 G 1
G
q
T T T T T
q r r r r z k t
. . . (4.40)
or
2 2
2 2 2
1 T 1 T T qG 1 T
r r r r r z k t
. . . (4.41)
This is the general heat conduction equation in cylindrical co-ordinates. If we compare this equation with Eq. (4.13), the Laplacian is
2 2
2
2 2 2
1 T 1 T T
T r
r r r r z
. . . (4.42)
If heat flows only in radial direction, T = T (r, t), Eq. (4.41) reduces to
1 T qG 1 T
r r r r k t
. . . (4.43)
11
Governing Equations of Heat Conduction
If the temperature distribution does not vary with time, then at steady state,
1 T qG 0
r r r r k
. . . (4.44)
In this case the equation for the temperature contains only a single variable r and is therefore an ordinary differential equation.
When there is no volumetric energy generation and temperature is a function of the radius only, the steady-state conduction for cylindrical coordinates is
d dT 0 dr r dr
. . . (4.45)
4.2.3 Spherical Coordinates
For spherical coordinates, as shown in Figure 4.4 the temperature is a function of the three space coordinates r, and and time t, i.e. T = T (r, , , t). The general form of the conduction equation in spherical coordinates can be found as
Figure 4.4 : Spherical Coordinate System for the General Conduction Equation
2 2
2 2 2 2 2
1 1 1 1
sin sin sin
qG
T T T T
r r r k t
r r r
. . . (4.46)
Where the Lapacian includes the first three terms of the above equation in the spherical coordinates.
4.3 STEADY STATE HEAT CONDUCTION IN SIMPLE GEOMETRICAL SYSTEMS
We will now derive solutions to the conduction equation as obtained in the previous section for simple geometrical systems with and without heat generation.
4.3.1 Plane Wall
In Unit 2 we saw that the temperature distribution for one-dimensional steady conduction through a wall is linear. We will verify the result by simplify the more general equation (Eq. 4.13)
2 qG 1 T
T k t
. . . (4.47)
dr
y
d
x
θ z
dθ
12
Conduction
For steady state, T 0 t
. Since T is only a function of x, T 0 y
and T 0 z
. There is no internal heat generation, qG = 0. Therefore, the above equation reduces
2
2 0
d T
dx . . . (4.48)
Integrating the ordinary differential equation twice yields the linear temperature distribution
1 2
( )
T x C xC . . . (4.49)
For a wall as shown in Figure 4.5, at x = 0, T = T1 and at x = b, T = T2
Figure 4.5 : Heat Conduction Through a Plane Wall
1 2
2
T T
T x T
b
. . . (4.50)
Which agrees with the linear temperature distribution deduced by increasing Fourier’s law
k
Q kAdT
dx . . . (4.51)
Let us now consider a heat source generating throughout the system. If the thermal conductivity is constant and the heat generation is uniform, Eq. (4.13) reduces to
2
2 qG 0
d T
dx k . . . (4.52)
On integration, dT qG 1
dx k C . . . (4.53)
A second integration gives
2 1 2
( ) 2
qG
T x x C x C
k . . . (4.54)
where C1 and C2 are constants.
At x = 0, T = T1 and at x = b, T = T2 substituting in Eq. (4.52),
1 2
T C . . . (4.55)
2 2 2 1 1
qG
T b C b T
k . . . (4.56)
T1
T2
b
13
Governing Equations of Heat Conduction
2 1
1 2
qG
T T
C b
b k
. . . (4.57)
Therefore, the temperature distribution is
2 2 1
( ) 1
2 2
G G
q T T q
T x x x x T
k b k
. . . (4.58)
It may be seen that Eq. (4.49) is now modified by two terms containing the heat generation and that the temperature distribution is no longer linear. If the two surface temperatures are equal (T1 = T2), then the temperature distribution becomes
2
2 1
( ) 2
qG x x
T x b T
k b b
. . . (4.59) Which is a parabolic and symmetric about the central plane with a minimum Tmax
at 2
x b.
2
2
1 2 0
2 qG
dT b x
dx k b b
. . . (4.60)
or 22x 1
b b . . . (4.61)
or 2
x b . . . (4.62)
and
2
MAX 1
8 q bG
T T
k . . . (4.63)
In dimensionless form, on dividing Eq. (4.59) by Eq. (4.63) 70 x
b . . . (4.64)
1 2
max 1
( ) 4 ( )
T x T
T T
. . . (4.65)
Let us consider the case where heat is transferred from the two sides of the wall to the surrounding fluid at T (Figures 4.6-4.7). For simplicity, let us assume that both the wall surfaces are at Tw (Figure 4.7). At steady state and for one-dimensional heat flow,
Figure 4.6 : Heat Flow through a Wall with Heat Generation
∞
qG = heat generated per unit volume b/2
T1
T2
∞
x
b
14
Conduction 2
2 qG 0
dT
dx k . . . (4.66)
Let the excess temperature be T T
. . . (4.68)
So that
2 2
2 2
d d T
dx dx
. . . (4.69)
Figure 4.7 : Heat Transfer from Two Sides of a Wall having a Heat Source
Therefore,
2 2
qG
d dx k
. . . (4.70)
G 1
q
d x C
dx k
. . . (4.71)
and 2 1 2
2 qG
x C x C
k . . . (4.72)
Which is parabolic. The central plane is the plane of symmetry where the solid temperature is the maximum and d 0
dx
and is taken as the reference plane with x = 0.
At x = 0, d 0
dx
, C1 = 0 . . . (4.73)
At 2
x b,
2 2
G b
q
d b
dx k
Again, at wall
2
( )
2 2
G G
k b w w
q b q b
q k k h T T h
x k
. . . (4.74)
2
w G
q b
h . . . (4.75)
From Eq. (4.72)
2 2
2 qG
x C
k . . . (4.76)
When ,
2 w
xb
∞
∞
k T∞
T T
θo
o
b/2 b/2
T∞
h
θ
∞
T
k h
T
15
Governing Equations of Heat Conduction 2
2 4 2
w q bG
q C
k . . . (4.77)
2 2
2 2 4 2 8
G G G
w
q b q b q
C b
k h k
. . . (4.78)
Therefore, the final temperature distribution is
2
2 2 8
G G G
q q b q
x T
k h k
. . . (4.79)
2 2
( 4 )
8 2
G G
q q b
T b x T
k h
. . . (4.80)
This is the temperature distribution. At the mid-plane, x = 0 and T = Tmax.
2
8 2
G G
q q b
T b T
k h
. . . (4.81)
SAQ 1
(a) Show that the temperature profile for heat conduction through a plane wall of constant thermal conductivity is a straight line.
(b) Show that the temperature profile for heat conduction through a plane wall with a heat source and constant thermal conductivity is parabolic.
4.3.2 Cylindrical Wall
Cylinder without Heat Generation
As shown in the Figure 4.8, heat is assumed to flow only radially in a hollow pipe consisting fluid inside the heat is transferred through a hollow pipe. We want to determine the temperature distribution and the heat transfer rate in a long hollow cylinder of length L if the inside and outside surface temperatures are Ti and To
respectively and there is no internal heat generation.
(a) (b)
Figure 4.8 : Radial Heat Conduction through a Hollow Cylinder
Since the temperatures at two surfaces are constant, the temperature distribution in wall is not a function of time, and the conduction equation is given by Eq. (4.44)
d r dT 0 dr dr
. . . (4.82)
On integration
1
r dT C
dr . . . (4.83)
To
qk
L
T1
ro
r1
T1
T To
r ro
r1
dr T = T(r)
K = uniform qG = 0
16
Conduction
or dT C1
dr r . . . (4.84)
A second integration yields
1ln 2
T C rC . . . (4.85)
At rr Ti, Ti
1ln 2
i i
T C r C . . . (4.86)
At r r T0, T0
0 1ln 0 2 1ln 0 i i ln i
T C r C C r T C r . . . (4.87)
1 0
ln 0 i
i
T T
C r
r
. . . (4.88)
and 2 0
0
ln ln
i i i
i
T T
C T r
r r
. . . (4.89)
Substituting C1 and C2 in Eq. (4.87),
0 0
0 0
ln ln
ln ln
i i
i i
i i
T T T T
T r T r
r r
r r
. . . (4.90)
0 0
( ) ln
ln
i i i
i
r T r T r
T T r
r
. . . (4.91)
The rate of heat transfer by conduction
0 1
0
2 2
ln
k i
i
T T
C
Q kAdT k r L k L
dr r r
r
. . . (4.92)
0 0
2 ( )
ln
k i
i
k L T T
Q r
r
. . . (4.93)
Then the thermal resistance offered by the wall is
0 0
ln 2
i i
th k
r T T r
R Q k L
. . . (4.94)
As shown in Figure 4.9, the rate of heat conduction through a composite cylindrical wall with convection at the inside and outside surfaces is given by
2 3
1 2
1 1 2 0 3
ln ln
1 1
2 2 2 2
h c
th
i
T T
Q T
R r r
r r
h r L k L k L h k L
. . . (4.95)
17
Governing Equations of Heat Conduction
where Th and Tc are the hot and cold fluid temperatures, hi and ho are the inside and outside heat transfer coefficients and k1 and k2 are the thermal conductivities of the two walls in series.
Figure 4.9 : Temperature Distribution in a Composite Cylinder with Convection at the Interior and Exterior Surfaces
Now Q U A T 0 0 ( h Tc) . . . (4.96)
where Uo is the overall heat transfer coefficient based on outside area
o 2 o
A r L given by
0 0 2 3
1 2
1 2 0 0
1 1
ln ln
1 1
2 2
th
i i
U A R r r
r r
h A k L k L h A
. . . (4.97)
where ri = r1 and ro = r3. To find the wall temperature T1, T2 and T3, we can use
1 1
h
i i
T T Q R Q
h A . . . (4.98)
1 2 1
2 1
2 ln Q k L T T
r r
. . . (4.99)
2 3
o o
T T Q
h A . . . (4.100)
Cylinder with Heat Generation
Let us consider the Figure 4.10, a long solid cylinder of radius R with internal heat generation, such as an electric coil in which heat is generated as a result of electric current in the wire or a cylindrical nuclear fuel element in which heat is generated by nuclear fission.
The one-dimensional heat conduction equation in cylindrical coordinates is
1 d dT qG 0
r dr r dr k
. . . (4.101)
q rG
d r dT
dr dr k
. . . (4.102)
Tc, ∞
T2
T1
r1 = n r2
r3 = r0
hc,o
Th, ∞
T1 T2
T3
Th ∞ T1 T2 T3
Tc, ∞
hc,1
1
hc1 2r1L 1
hc,02r0L ln(r2/r1)
2kAL ln(r3/r2)
2kBL q B
L T3
Th, ∞ AL
18
Conduction On integration,
2
2 1
q rG
r dT C
dr k . . . (4.103)
1
2 q rG C dT
dr k r . . . (4.104)
Figure 4.10 : Temperature Distribution in a Cylindrical Rod with Heat Generation
By second integration,
2
1ln 2
4 q rG
T C r C
k . . . (4.105)
At 0, dT 0
r dr (with origin at the centre line of the cylinder). But from Eq. (4.104), dT
dr , which is impossible.
, 1
2
G r R
q R C r R dT
dr k R
. . . (4.106)
Heat generated in the cylindrical rod qG R L2 .
This heat is conducted to the surface and then converted away.
2 2
G
r R
q R L k L R dT
dr
. . . (4.107)
2
G
r R
q R dT
dr k
. . . (4.108)
From Eqs. (4.104) and (4.108), C1 = 0
2 2
4 qG
T r C
k . . . (4.109)
At rR T, Tw
2 4 2
w G
C T q R
k . . . (4.110)
Substituting in Eq. (4.105)
qG
hc
L
T∞
q hc
T
T∞
q
O
R r
T
19
Governing Equations of Heat Conduction
2 2
4 4
G G
w
q q
T r T R
k k
. . . (4.111)
or ( ) ( 2 2) 2 1 22
4 4
G G
w W
q q R r
T r R r T T
k k R
. . . (4.112)
This is the maximum temperature variation along the wall radius. The maximum temperature occurs at r = 0
2
max 4
G w
T q R T
k . . . (4.113)
In dimensionless form Eq. (4.112) becomes
2
max
( ) w 1
w
T r T r
T T R
. . . (4.114)
For a hollow cylinder with uniformly distributed heat source and specified surface temperature T = Ti at r = ri and T = T0 at r = r0, Eq. (4.105) gives
2 1ln 2
4
i G i i
T q r C r C
k . . . (4.115)
2 1ln 2
4
o G o o
T q r C r C
k . . . (4.116)
Evaluating C1 and C2 from the above two equations and substituting in Eq. (4.105).
We can obtain the temperature distribution as
2 2 2 2
ln
( ) ( ) ( )
4 4
ln
G o G
o o i o i o
o i
r
q r q
T r r r r r T T T
k r k
r
. . . (4.117)
If a solid cylinder is immersed in a fluid at T and the convection heat transfer coefficient is hc and T = Tw at r = R, the heat conduction from the cylindrical is equal to the rate of convection at the surface, or
( )
c w
r R
k dT h T T
dr
. . . (4.118)
From Eq. (4.108)
( )
2
c w G
r R
h T T q R
dT
dr k k
. . . (4.119)
2
w G
c
T T q R
h
. . . (4.120)
From Eq. (4.112)
2 2
( ) 1 2
4 2
G G
c
q R r q R
T r T
k R h
. . . (4.121)
In dimensionless form, ( ) 2
2 1
4
G c
c
q R h R
T r T r
T h T k R
. . . (4.122)
20
Conduction and maximum temperature is
max 1 2
4
G c
c
T q R h R
T h T k
. . . (4.123)
There are two dimensionless parameters in the above equation which are important in conduction, viz. G
c
q R
h T the heat generation number, and h Rc k the Biot number, which appears in problems with simultaneous conduction and convection.
The Biot number is the ratio of conduction resistance to convection resistance or
1
o
k c o
i c
c
r
R k h r
B R k
h
. . . (4.124)
The limits of Biot numbers are
i 0
B when k ro 0
R k or, 1
c c
R h
Bi when 1
c 0
c
R h or, k ro
R k
The Biot number approaches zero when the conductivity of solids very large (k ) or the convection coefficient of heat transfer is very low (hc 0), i.e.
when the solid is practically isothermal and the temperature change is mostly caused in the fluid by convection at the interface. On the contrary, the Biot number approaches infinity when the thermal resistance predominates (k 0) or the convection resistance is very low (hc 0).
SAQ 2
(a) Show that the maximum temperature in a cylindrical rod with heat generation kW3
G m q
is given by max 1 2
4
G c
c
T q R h R
T h T k
. (b) What are the limiting values of Biot number?
(c) What is heat generation number?
4.3.3 Spherical Wall
As shown in Figure 4.11 a hollow sphere with uniform temperature at the inner and outer surfaces, the temperature distribution without heat generation in the steady state can be obtained by simplifying Eq. (4.46). Under these conditions the temperature is only a function of the radius r, and the conduction equation in the spherical coordinates is
2 2
1 d dT 0
dr r dr r
. . . (4.125)
On integration, and putting at r = ri, T = Ti and r = ro, T = To
( ) i ( o i) o 1 i
o i
r r
T r T T T
r r r
. . . (4.126)
21
Governing Equations of Heat Conduction
Figure 4.11 : Heat Conduction in a Hollow Sphere without Heat Generation
The rate of heat transfer through the spherical wall is 4 2
( )
4
i o
k o i
o i
T T Q k r dT
r r dr
k r r
. . . (4.127)
The thermal resistance for spherical wall is then 4
o i
th
o i
r r
R k r r
. . . (4.128)
4.4 TRANSIENT CONDUCTION
Many heat transfer problems are time dependent. Such unsteady, or transient, problems typically arise when the boundary conditions of a system are changed. For example, if the surface temperature of a system is altered, the temperature at each point in the system will also begin to change. The changes will continue to occur until a steady-state
temperature distribution is reached. Consider a hot metal billet that is removed from a furnace and exposed to a cool air stream. Energy is transferred by convection and radiation from its surface to the surroundings. Energy transfer by conduction also occurs from the interior of the metal to the surface, and the temperature at each point in the billet decreases until a steady-state condition is reached. Such time dependent effects occur in many industrial heating and cooling processes. In the following sub-units, transient heat transfer analysis has been discussed.
4.4.1 Lumped Capacitance Method
A simple, yet common, transient conduction problem is one for which a solid
experiences a sudden change in its thermal environment. Consider a hot metal forging that is initially at a uniform temperature Ti and is quenched by immersing it in a liquid of lower temperature T < Ti as shown in Figure 4.12. If the quenching is said to begin at time t = 0, the temperature of the solid will decrease for time t > 0, until it eventually reaches T. This reduction is due to convection heat transfer at the solid-liquid interface.
The essence of the lumped capacitance method is the assumption that the temperature of the solid is spatially uniform at any instant during the transient process. This assumption implies that temperature gradients within the solid are negligible.
From Fouriers law, heat conduction in the absence of a temperature gradient implies the existence of infinite thermal conductivity. Such a condition is clearly impossible.
However, although the condition is never satisfied exactly, it is closely approximated if the resistance to conduction within the solid is small compared with the resistance to heat transfer between the solid and its surroundings. For now we assume that this is, in fact, the case.
T0
r0
T1
T = T(r) K = uniform qG = 0 r1
qk
22
Conduction In neglecting temperature gradients within the solid, we can no longer consider the problem from within the framework of the heat equation. Instead, the transient temperature response is determined by formulating an overall energy balance on the solid. This balance must relate the rate of heat loss at the surface to the rate of change of the internal energy. Thus, with respect to the Figure 4.12,
out st
E E
. . . (4.134)
or s ( ) dT
h A T T Vc
dt
. . . (4.135)
Figure 4.12 : Cooling of a Hot Metal Forging
Introducing the temperature difference T T
. . . (4.136)
And recognizing that d dT
dt dt
, it follows that
s
Vc d h A dt
. . . (4.137)
Separating variables and integrating from the initial condition, for which t = 0 and T (0) = Ti, we then obtain
i 0
t s
Vc d
h A dt
. . . (4.138)where i Ti T . . . (4.139)
Evaluating the integrals it follows that ln i
s
Vc t
h A
. . . (4.140)
or exp s
i i c
h A T T
T T V t
. . . (4.141)
Eq. (4.140) may be used to determine the time required for the solid to reach some temperature T, or, conversely, Eq. (4.141) may be used to compute the temperature reached by the solid at same time t.
The foregoing results indicate that the difference between the solid and fluid
temperatures must decay exponentially to zero as t approaches infinity. This behaviour is shown in Figure 4.13. From Eq. (4.141) it is also evident that the quantity
s
Vc h A
may be interpreted as a thermal time constant s. This time constant may be expressed as
t < 0 T = Ti
Liquid
t 0 T = T(t)
Eout = qconvenction
Esi
T∞ Y1
Ti
T(t)
23
Governing Equations of Heat Conduction
1 ( )
t t t
s
Vc R C h A
. . . (4.142)
Figure 4.13 : Transient Temperature Response of Lumped Capacitance Solids for Different Thermal Time Constantss
where Rt is the resistance to convection heat transfer and Ct is the lumped thermal capacitance of the solid. Any increase in Rt or Ct will cause a solid to respond more slowly to changes in its thermal environment and will increase the time required to reach thermal equilibrium ( = 0). This behavior is analogous to the voltage decay that occurs when a capacitor is discharged through a resistor in an electrical RC circuit.
To determine the total energy transfer Q occurring up to some time t, we simply write
0 0
t t
Q
q dt h As
dt . . . (4.143)Substituting for from Eq. (4.140) and integrating, we obtain
( ) i 1 exp
t
Q Vc t
. . . (4.144) The quantity Q is, of course, related to the change in the internal energy of the solid, and
from Eq. (4.134)
Q Est
. . . (4.145)
For quenching Q is positive and the solid experiences a decrease in energy. Eqs. (4.140), (4.141) and (4.144) also apply to situations where the solid is heated ( < 0), in which case Q is negative and the internal energy of the solid increases.
Validity of the Lumped Capacitance Method
From the foregoing results it is easy to see why there is a strong preference for using the lumped capacitance method. It is certainly the simplest and most convenient method that can be used to solve transient conduction problems.
Hence, it is important to determine under what conditions it may be used with reasonable accuracy.
As shown in Figure 4.14, to develop a suitable criterion consider steady-state conduction through the plane wall of area A.
Although we are assuming steady-state condition, this criterion is readily extended to transient processes. One surface is maintained at a temperature Ts, 1 and the other surface is exposed to a fluid of temperature T < Ts, 1. The temperature of
t = Vc
hAs = Rt Ct
0.368
0 t. 1 t. 2 t. 3 t. 4
T T
T T
1 1
1
t
24
Conduction this surface will be some intermediate value, Ts, 2, for which T < Ts, 2 < Ts, 1. Hence, under steady-state conditions the surface energy balance is
,1 ,2 ,2
( s s ) ( s )
kA T T hA T T
L . . . (4.146)
where k is the thermal conductivity of the solid. Rearranging, we then obtain
,1 , 2 cond
, 2 1 conv
s s
s
L
T T kA R hL Bi
T T R k
hA
. . . (4.147)
Figure 4.14 : Effect of Biot Number on Steady-State Temperature Distribution in a Plane Wall with Surface Convection
The quantity hL k
is a dimensionless parameter appearing in Eq. (4.147). It is termed the Biot number, and it plays a fundamental role in conduction problems that involve surface convection effects. According to Eq. (4.147) and as illustrated in Figure 4.14, the Biot number provides a measure of the temperature drop in the solid relative to the temperature difference between the surface and the fluid. Note especially the conditions corresponding to Bi < < 1. The results suggest that, for these conditions, it is reasonable to assume a uniform temperature distribution across a solid at any time during a transient process. This result may also be associated with interpretation of the Biot number as a ratio of thermal resistances, Eq. (4.147). If Bi < < 1, the resistance to conduction within the solid is much less than the resistance to convection across the fluid boundary layer. Hence, the assumption of a uniform temperature distribution is reasonable.
We have introduced the Biot number because of its significance to transient conduction problems. Consider the plane wall of Figure 4.15, which is initially at a uniform temperature Ti and experiences convection cooling when it is immersed in a fluid of T < Ti. The problem may be treated as one-dimensional in x, and we are interested in the temperature variation with position and time, T (x, t). This variation is a strong function of the Biot number, and three conditions are shown in Figure 4.15.
For Bi < < 1 the temperature gradient in the solid is small and T (x, t) T (t).
Virtually all the temperature difference is between the solid and the fluid, and the solid temperature remains nearly uniform as it decreases to T. For moderate to large values of the Biot number, however, the temperature gradients within the solid are significant. Hence, T = T (x, t). Note that for Bi > > 1, the temperature
X L
Ts. 1 T
Bi = 1
Bi >> 1 Bi << 1
qcond qconv
Ts. 2
Ts. 2
Ts. 2
T∞,h
25
Governing Equations of Heat Conduction
difference across the solid is much larger than that between the surface and the fluid.
Lumped capacitance method is a preferred method for solving transient
conduction problems. Hence, when confronted with such a problem, the very first thing that one should do is calculate the Biot number. If the following condition is satisfied
h Lc 0.1
Bi k . . . (4.148)
Figure 4.15 : Transient Temperature Distributions for different Biot Numbers in a Plane Wall Symmetrically Cooled by Convection
The error associated with using the lumped capacitance method is small. For convenience, it is customary to define the characteristic length of Eq. (4.148) as the ratio of the solids volume to surface area, c
s
L V
A . Such a definition facilitates calculation of Lc for solids of complicated shape and reduces to the half-thickness L for a plane wall of thickness 2 L as shown in Figure 4.15, to 0
2 r
for a long cylinder, and to 0 3
r for a sphere. However, if one wishes to implement the criterion in a conservative fashion, Lc should be associated with the length scale corresponding to the maximum spatial temperature difference. Accordingly, for a symmetrically heated (or cooled) plane wall of thickness 2 L, Lc would remain equal to the half-thickness L. However, for a long cylinder or sphere, Lc
would equal the actual radius r0, rather than 0 2 r or 0
3 r .
Finally, we note that, with c
s
L V
A , the exponent of Eq. (4.141) may be expressed as
2 2
s c c
c c c
h A t ht h L k t h L t
Vc c L k c L k L
. . . (4.149)
or h A ts . Bi Fo Vc
. . . (4.150)
where 2
c
Fo t L
. . . (4.151)
T∞, h
-L L
x
T∞, h
T(x, O) = T1 T(x, O) = T1
-L L -L L -L L
T∞ T∞ T∞
Bi << 1
T = T(t) Bi = 1
T = T(x, t) Bi >> 1 T = T(x, t)
T∞
26
Conduction is termed the Fourier number. It is a dimensionless time, which, with the Biot number, characterizes transient conduction problems. Substituting Eq. (4.150) into Eq. (4.141), we obtain
exp ( . )
i i
T T
Bi Fo T T
. . . (4.152)
SAQ 3
(a) What do you mean by Biot number and Fourier number?
(b) What are the conditions for validity of lumped capacitance method?
4.4.2 General Lumped Capacitance Analysis
Although transient conduction in a solid is commonly initiated by convection heat transfer to or from an adjoining fluid, other processes may induce transient thermal conditions within the solid. For example, a solid may be separated from large surroundings by a gas or vacuum. If the temperatures of the solid and surroundings differ, radiation exchange could cause the internal thermal energy, and hence the temperature, of the solid to change. Temperature changes could also be induced by applying a heat flux at a portion, or all, of the surface and/or by initiating thermal energy generation within the solid. Surface heating could, for example, be applied by attaching a film or sheet electrical heater to the surface, while thermal energy could be generated by passing an electrical current through the solid.
Figure 4.16 depicts a situation for which thermal conditions within a solid may be influenced simultaneously by convection, radiation, an applied surface heat flux, and internal energy generation.
Figure 4.16 : Control Surface for Generalized Lumped Capacitance Analysis
It is presumed that, initially (t = 0), the temperature of the solid (Ti) differs from that of the fluid, T, and the surroundings, Tsur, and that both surface and volumetric heating (qs and q) are initiated. The imposed heat flux qs and the convection-radiation heat transfer occur at mutually exclusive portions of the surface, As (h) and As(c, r), respective and convection-radiation transfer is presumed to be from the surface. Moreover, although convection and radiation have been prescribed for the same surface, the
surfaces may, in fact, differ (As, c As, r). Applying conservation of energy at any instant t
s s h, g ( conv rad) s c r( , ) dT
q A E q q A Vc
dt . . . (4.153)
or, s s h, g ( ) ( 4 sur4 ) s c r( , ) dT
q A E h T T T T A Vc
dt
. . . (4.154)
P, c, V, T(0) = Ti
q”s
As, h As(c,r)
q”conv
q”rad
Eg, Est
Surroundings T sur
T, h
