CHAPTER 5
Transient Conduction:
Spatial Effects and the Role of Analytical Solutions
Prepared by: Prof. Dr. Ş. Birgül Tantekin-Ersolmaz
Prof. Dr. Hüsnü Atakül
Transient Heat Conduction: Spatial Effects
 In many transient heat transfer problems the Biot number is larger than
0.1, and lumped system can not be assumed.
 In these cases the temperature within the body changes appreciably from
point to point as well as with time.
 It is constructive to first consider the variation of temperature with time and
position in one-dimensional problems of rudimentary configurations.
a plane wall of thickness 2L  1-D
a long cylinder of radius r0
a sphere of radius r0.
Notice: All three cases possess geometric and thermal symmetry.
@ t = 0 T = Ti
@ t > 0 T = T(t,x) or T(t,r)
Heat transfer takes place between the environment and
these bodies by convection (radiation is neglected).
2
Solution to the Heat Equation for a Large Plane Wall
with Symmetrical Convection Conditions
•
•
•
•
•
•
A plane wall of thickness 2L.
Initially at a uniform temperature of Ti.
At time t=0, the wall is immersed in a fluid
at temperature T∞.
Constant heat transfer coefficient h.
Constant thermophysical properties.
No heat generation.
Temperature profiles at
different time values
•
•
The height and the width of the wall are
large relative to its thickness  1-D
approximation is valid.
There is thermal symmetry about the
midplane passing through x=0.
3
For constant properties, the differential equation representing the wall case is:
One-dimensional transient heat
conduction equation (0≤ x ≤ L)
Boundary conditions:
@ t 0
x0
@ t 0
xL
 2T 1 T

2
x
 t
T
0
x
or
T
-k
 h T  T 
x
T (0, t )
0
x
T ( L, t )
k
 hT ( L, t )  T 
x
Initial condition:
@t 0
0xL
Partial differential eqn.
T  Ti

or
T ( x,0)  Ti
Solution involves infinite series
For convenience, solution is presented in tabular graphical form. However, we
need to reduce the number of parameters to make graphical presentation
practical.
Eight independent variables: T  T  x, t , T i , T  , L, k ,  , h 
How may the functional dependence be simplified?
4
 Let’s define dimensionless quantities:
T  T
:  *
Ti  T
Dimensionless temperature
Dimensionless distance from center
:
Dimensionless time: (Fourier number) :
x
L
t
t*  2  Fo
L
x* 
Insert into the DE and BCs, then the problem becomes:
BC1: @ t*  0
x*  0
BC2: @ t*  0
x*  1
IC:
@ t*  0
 *
0
x *
 *
 Bi  *  0
x *
0  x*  1
 2 * 

2
x *
t *
*  1
Dimensionless heat transfer coefficient (Biot number)  Bi 
Before nondimensionalization 
After nondimensionalization 
x, t, L, k, , h, Ti, T
x*, t*, Bi
hL
k solid
5
 Non-dimensionalization of Heat Equation and IC/BCs:
 *  f  x * , Fo, Bi 
 Exact Solution:

   C n exp   n2 Fo  cos  n x * 
*
n 1
Cn 
4sin  n
2 n  sin  2 n 
 n tan  n  Bi
See Appendix B.3 for first four roots (eigenvalues  1 ,...,  4 ) of  n tan  n  Bi
 Approximate analytical (one-term) solution:
The exact solution involves infinite series which are difficult to deal
with. However, the terms in the series solution after the first term can
be neglected if t* (Fo) > 0.2 and the error would be only < 2 %.
 *   Cn exp  Fo cos( n x*)

n 1
2
n
The terms in the summation decline
rapidly as n and thus n increases
6
 The One-Term Approximation  Valid for Fo  0.2  :
•
Variation of midplane temperature (x*= 0) with time  Fo  :
 o* 
•
T o  T    C exp  2 Fo
 1 
1
T

T
 i 
Table 5.1  C 1 and  1 as a function of Bi
Variation of temperature with location (x*) and time  Fo :
 *   o* cos  1 x * 
•
Change in thermal energy storage with time:
E st  Q
(5.46a)
Q o   cV T i  T  
 sin  1 * 
Q  Q o 1 
o 
1


Can the foregoing results be used for a plane wall that is well insulated on
one side and convectively heated or cooled on the other?
Can the foregoing results be used if an isothermal condition  T s  T i  is
instantaneously imposed on both surfaces of a plane wall or on one surface
of a wall whose other surface is well insulated?
7
8
Radial Systems (Long Cylinders and Spheres)
 Long Cylinders or Spheres Heated or Cooled by Convection:
Similar analysis could be made for cylinder and sphere:
T (r , t )  T
* 
Ti  T
r* 
r
r0
t* 
t
Bi 
r02
Cylinder:
1    *   *
 r *

r * r *  r *  t *
Sphere:
1   2  *   *
 r *

2
r * r * 
r *  t *
hr0
k
 One-Term Approximations: C 1 ,  1  Table 5.1
 *cly  C1 exp   Fo  J 0 ( 1r*)
2
1
Long Rod
 *sph
sin( 1r*)
 C1 exp   Fo 
 1r *
2
1
Sphere
Once the Bi number is known the temperature anywhere in the medium
can be determined.
9
Radial Systems (Long Cylinders and Spheres)
 Variation of midplane temperature (x*= 0) with time
 Fo  :
 *0,cly  C1 exp   12 Fo 
 *0,sph  C1 exp   12 Fo 
 Change in thermal energy storage with time:
Maximum Q is achieved in the case of t  
Qmax  VC p (Ti  T )
 Q 
J1 ( 1 )

  1  2 *0,cyl
1
 Qmax cly
 Q 
sin  1   1 cos  1

  1  3 *0, sph
3
Q

1
 max  sph
10
Bessel Functions of the First Kind
11
Graphical Representation of the One-Term Approximation
The Heisler Charts, Section 5 S.1
 The solution of the transient temperature for a large plane wall, long cylinder,
and sphere are also presented in graphical form for t > 0.2 known as the
transient temperature charts (Heisler Charts)
 There are three charts associated with each geometry:
 the temperature T0 at the center of the geometry at a given time t.
 the temperature at other locations at the same time in terms of T0.

 Use



the total amount of heat transfer up to the time t.
of these charts is limited to the conditions we have assumed:
initially the body is at uniform temperature
temperature of the environment is constant and h is constant and uniform
no heat generation
 Graphical Representations: (in Supplemental Material)
Plane Wall:
Figs. 5 S.41– 5 S.3
Long Rod:
Figs. 5 S.4 – 5 S.6
Sphere:
Figs. 5 S.7 – 5 S.9
12
Graphical Representation of the One-Term Approximation
The Heisler Charts for Plane Wall
• Midplane Temperature:
13
• Temperature Distribution:
• Change in Thermal Energy Storage:
14
Significance of Biot Number
• When 1/ Bi = k/hL = 0  Bi  and therefore h  
Surfaces of the plate are at the ambient temperature.
• When 1/Bi >> 1  k is large
Uniform temperature distribution and lumped system
analysis is valid.
15
Significance of Fourier Number
t
kL2 (1 / L) t
Fo  t*  2 
L
C p L3 / t t
q
=
qconducted
The rate at which heat is conducted
across L of a body of volume L3
The rate at which heat is stored
in a body of volume L3
What does a large Fo number
mean?
16
Example 1: Steel Pipeline
Consider a steel pipeline (AISI 1010) that is 1 m in diameter and has a wall
thickness of 40 mm. The pipe is heavily insulated on the outside, and before the
initiation of flow, the walls of the pipe are at a uniform temperature of -20C.
With the initiation of flow, hot oil at 60C is pumped through the pipe creating a
convective surface condition corresponding to h=500W/m2.K at the inner surface
of the pipe.
a) At t=8 min, what is the temperature of the exterior pipe surface covered by
insulation?
b) What is the heat flux q (W/m2) to the pipe from the oil at t=8 min?
c) How much energy per meter of pipe length has been transferred from the oil
to the pipe at t=8 min?
17
Known: Wall subjected to sudden change in convective surface condition.
Find: • Temperature of exterior pipe surface after 8 minutes.
• Heat flux to the wall at 8 minutes.
• Energy transferred to pipe per unit length after 8 minutes
Schematic:
Properties:
From the related tables:
Tavg = (Ti+ T )/2 =293 K  300 K
k=63.9 W/m.K
Cp=434 J/kg.K
=7823 kg/m3
 = 18.8x10-6 m2/s
Assumptions: • Pipe wall can be approximated as plane wall, since the
thickness is much less than the diameter.
• Outer surface of pipe is adiabatic.
• Constant properties.
18
Analysis:
First, we need to compute the Biot and Fourier numbers at t=8 minutes
with Lc = L:
hLc  500  0.04 
Bi 

 0.313  0.1
k
63.9
 t 18.8 10   8  60 
Fo 

 5.64  0.2
6
L2
 0.04  2
With Bi = 0.313, use of the lump capacitance method is inappropriate.
However, since Fo > 0.2 and transient conditions in the insulated pipe
wall of thickness L correspond to those in a plane wall of thickness 2L
experiencing the same surface condition, the desired results may be
obtained from the one-term approximation for a plane wall.
The midplane temperature can be
obtained from:
 *0, wall
T0  T

 C1 exp   12 Fo 
Ti  T
C1 and 1 are functions of Bi number  Table 5.1
C1= 1.047, 1 = 0.531
19
 0, wall  (1.047) exp (0.531) 2 (5.64)   0.214
T0  T
 0.214  T0  42.9 C
Ti  T
Heat transfer to the inner surface at x = L is by convection, and at any
time t the heat flux may be obtained from Newton’s law of cooling.

Hence at t = 480 s: qx ( L, 480s)  h T ( L, 480s)  T

Using the one-term approximation for the surface temperature:
 *( x, t ) wall
 * ( x, t ) wall
T ( x, t )  T

 C1 exp   12 Fo  cos( 1 x / L)
Ti  T


T ( L,480s )  60

 (1.047) exp  (0.531) 2 (5.64) cos(0.531)
 20i  60
T ( L,480 s)  45.2 C

qx  L, 480 s    500   45.2  60  7400 W/m 2
20
Energy transfer to the pipe wall over the 8 minute interval may be obtained
from:
 Q

 Qmax

sin  1
  1   *0, wall
1
 wall
sin  0.531
 Q 
 0.80

  1   0.214 
0.531
 Qmax  wall
Q  0.80  C pV Ti  T 
 (0.80)(7823)(434) (1)(0.04)( 20  60)
Q  2.73 107 J/m
Comments:
1. The minus sign associated with q and Q simply implies that the
direction of heat transfer is from the oil to the pipe.
2. The foregoing results could also be obtained by applying the charts.
21
The Semi-Infinite Solid
Semi-infinite medium 
an idealized body that has a single plane
surface and extends to infinity in all
directions
In such a body the temp. change in the
region close to the surface is due to the
thermal conditions on a single surface
Example:
the earth, a thick wall
Consider a semi-infinite solid that is at a uniform temperature :
@t 0
T  Ti
At t > 0 there occurs a change in one of the surfaces. There are three possibilities:
1) Surface is subject to a constant temperature: @ t  0
2) Constant heat flux imposed on the surface: @ t  0
x0
x0
T  Ts
qx  q0
3) Surface is subject to convection with the environment:
@t  0
x0
qx  h T  T 
22
The differential equation representing this semi-infinite medium is :
T
 2T
 2
t
x
@t>0
x   T = Ti
Case 1: Nondimensionalize temperature with
T ( x, t )  Ts
 ( x, t ) 
Ti  Ts


 2
 2
t
x
IC
:
T ( x,0) T i
IC
:
 ( x,0)  1
BC1
:
T (0,t )  Ts
BC1
:
 (0,t )  0
BC2
:
T (,t )  Ti
BC2
:
 (,t )  1
Solution  Combination of variables (similarity solution)
Assuming T = T() (to be verified) and using the chain rule, all derivatives in
the heat conduction equation can be transformed into the new variable.
Similarity variable: η 
x
2 αt
 PDE becomes an ODE
The PDE can be converted into an ODE by combining the two independent
variables x and t into a single variable , called the similarity variable.
23
ODE
 T 1 T

2
x
 t
2
ODE
d 2T
dT
 2
0
:
2
d
d
η
Solution of ODE:
T  C1  e
η2
d 2T
T
 2
2
d
d
BC1
:
@η  0
T  Ts
BC2
:
@η  
T  Ti
(also at t = 0)
dη  C2
0
2
erf
(
η
)

Error function 
π

η
0
e
u2
du
u is a dummy
integration variable
(This integral cannot be performed analytically)
24
2
erf
(

)

Error function 
π

η
0
e
u2
du
Complementary error function 
erfc( )  1  erf ( )
erf (0) = 0
erfc (0) = 1
erf () = 1
erfc () = 0
Case 1 Solution:
T ( x, t )  Ts
 x 
 ( x, t ) 
 erf 

Ti  Ts
 2 t 
25
2q0   t 
T ( x, t )  Ti 
 
k  
1/ 2
Case 2:
  x 2  q0x
 x 
exp 
erfc 


4

t
k
2

t




 x
 hx h 2 t  
T ( x, t )  Ti
h  t 
 x 
Case 3:
 erfc 
 exp   2  erfc 

 

T  Ti
k  
k  
 2 t 
 k
 2 t
26
Figure 5.9 Variation of temperature with position and time in a semiinfinite solid initially at Ti subjected to convection BC.
As h   
h t

k
When is this realized?
 the surface temperature Ts becomes equal to
the fluid temperature T  the problem
reduces to Case 1
27
Summary of Semi-Infinite Solid
 A solid that is initially of uniform temperature Ti and is assumed to
extend to infinity from a surface at which thermal conditions are
altered.
 Special Cases:
Case 1: Change in Surface Temperature (Ts)
T  0, t   T s  T  x,0   T i
T  x, t   T s
 x 
 erf 

Ti Ts
 2 t 
qs 
k T s  T i 
 t
28
Case 2:
Constant Heat Flux
2 qo  t /  
T  x, t   T i 
k
1
 qs  qo 
 x2 
exp  

 4 t 
q x
 x 
 o erfc 

k
 2 t 
2
Case 3: Surface Convection  h, T  
T
k
x
x 0
 h T   T  0, t 
T  x, t   T i
 x 
 erfc 

T Ti
 2 t 
 x
h  t 

 hx h 2 t  
 exp  

erfc 


2
k
k
k
2

t


 
 

29
Example 2: Susceptibility to fire of a thick oak wall
We wish to determine the susceptibility to fire of a thick oak wall. The
wall is initially at 25°C and suddenly exposed to combustion products at
800°C for which the heat transfer coefficient is 20 W/m2.K.
Determine the time of exposure required for the surface to reach 400°C,
the ignition temperature of the wood.
Known: Thick oak wall, initially at uniform temperature of 25C, is suddenly
exposed to combustion products at 800C with a convection
coefficient of 20 W/m2K.
Find:
Schematic:
a) Time of exposure
required for the surface
to reach an ignition
temperature of 400C.
b) Temperature
distributionat time
t = 325 s.
30
Assumptions:
•
•
•
•
Oak wall can be treated as semi-infinite solid
1-D conduction
Constant properties
Negligible radiation
Properties: From Table A-3, Oak, cross grain (300 K):  = 545 kg/m3,
Cp = 2385 J/kgK, k = 0.17 W/mK,
  k  C p  1.31107 m 2 /s
Analysis: This situation corresponds to case 3.
 x
 hx h 2 t  
T ( x, t )  Ti
h  t 
 x 
 erfc 

 
  exp  k  k 2  erfc 
T  Ti
k
2

t
2

t



 

 
a) Time for the surface to reach 400C:
T (0, t )  Ti 400  25

 0.48
T  Ti
800  25
h 2 t
 0.75
2
k
 0.75k 
 t   1/ 2 
 h 
x
2 t
0
2
2


0.75

0.17
W/m

K
  310 s
t 
1/ 2

7
2
 20 W/m  K  1.3110 m /s  


31
b) Temperature distribution at time t = 325 s.
h  t 
k
20 W/m  K 1.31107 m 2 /s 
1/ 2
1/ 2

0.17 W/m  K
 325 s 
1/ 2
 0.768
32
Objects with Constant Surface Temperatures or
Surface Heat Fluxes
 Transient response of a variety of objects to a step change in surface
temperature or heat flux can be unified by defining the dimensionless
conduction heat rate:
Semi-infinite solid
qs L c
q* 
k T s  T i 
where Lc is a characteristic length that depends on the geometry of
the object.
 Consider the variation of q* with time, or Fo, for
– Interior heat transfer: Heat transfer inside objects such as plane
walls, cylinders, or spheres,
– Exterior heat transfer: Heat transfer in an infinite medium
surrounding an embedded object.
33
 When q* is plotted versus Fo, we see that:
– All objects behave the same as a semi-infinite solid for short
times.
– q* approaches a steady state for exterior objects.
– q* does not reach a steady state for interior objects, but
decreases continually with time (Fo).
Constant Ts
Constant qs
Why do all objects behave the same as a semi-infinite solid for short times?
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Approximate Solutions for Objects with
Constant Ts or qs
 Easy-to-use approximate solutions for q*(Fo) are presented in Table 5.2 for
all the cases presented.
 As an example of the use of Table 5.2, consider:
– Infinite cylinder initially at Ti has constant heat flux imposed at its surface.
– Find its surface temperature as a function of time.
 Look in Table 5.2b for constant surface heat flux, Interior Cases, Infinite
cylinder.
– Length scale is Lc = ro, the cylinder radius.
– Exact solution for q*(Fo) is a complicated infinite series.
– Approximate solution is given by:
1  
q* 
 for Fo  0.2
2 Fo 8
or
1

q*   2 Fo  
4

1
for Fo  0.2
 It is then a simple matter to find Ts from the definition, q* 
qs L c
k T s  T i 
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