What is Lumped capacity analysis and obtain the expression for temperature distribution of the
same?
All solids have a finite thermal conductivity and there will be always a temperature gradient
inside the solid whenever heat is added or removed. However for solids of large thermal conductivity
with surface areas that are large in proportion to their volume like plates and thin metallic wires, the
internal resistance {L/ kA} can be assumed to be small or negligible in comparison with the
convective resistance {1/hA} at the surface. Typical examples of this type of heat flow are:
1. Heat treatment of metals
2. Time response of thermocouples and thermometers
The process in which the internal resistance is assumed negligible in comparison with its surface
resistance is called the Newtonian heating or cooling process. The temperature, in this process is
considered to be uniform at a given time. Such an analysis is called Lumped parameter analysis
because the whole solid, whose energy at any time is a function of its temperature and total heat
capacity is treated as one lump.
Let us consider a body whose initial temperature is ti throughout and which is placed suddenly in
ambient air or any liquid at a constant temperature to as shown in figure.
Q = - VCp
dT
 hAT  Ta 
dt
dT
hA
 (T  Ta)   VCp  dt
ln T  Ta   
T – Temp pf body at any time
Ta – Ambient Temp
hA
t  C1
VCp
The boundary conditions are t=0 T = Ti
C1  ln Ti  Ta
Sub C1 in the above equation
ln T  Ta   
hA
t  ln Ti  Ta 
VCp

T  Ta
hA 
 exp 
t
Ti  Ta
 VCp 
SOLVED PROBLEMS IN UNSTEADY CONDUCTION HEAT TRANSFER
24. An aluminium cube 6 cm on a side is originally at a temperature of 500C. It is suddenly immersed in
a liquid at 10C for which h is 120 W/m2K. Estimate the time required for the cube to reach a
temperature of 250C. For aluminium  = 2700 kg/m3, C = 900 J/kg K, K = 204 W/mK.
1
Given
Thickness of cube L = 6 cm = 0.06 m
Initial temperature T0 = 500C + 273 = 773 K
Final temperature T = 10C + 273 = 283 K
Intermediate temperature T = 250C + 273 = 523 K
Heat transfer co-efficient h = 120 w/m2K
Density  = 2700 kg/m3
Specific heat C = 900 J/Kg k
Thermal conductivity K = 204 W/mK
Solution
For Cube,
Characteristic length
0.06

6
Lc 
L
6
Lc  0.01 m
We know
hLc
K
Biot number
120  0.01

204
5.88
 10 3  0.1
Bi =
Bi 
Biot number value is less than 0.1. So this is lumped heat analysis type problem
For lumped parameter system,

 hA

t 

T  T
C  V   
 e  
....(1)
T0  T
[From HMT data book Page No.48]
We know,
V
Lc 
A
Characteristics length

h

t 

T-T
C L  
(1) 
 e   c 
T0  T

120

t 

523 - 283

 e  9000.012700 
773 - 283
-120
 In (0.489) =
t
900  0.01 2700

t = 144.86 s
Time required for the cube to reach 250C is 144.86 s.
2
25. A copper plate 2 mm thick is heated up to 400C and quenched into water at 30C. Find the time
required for the plate to reach the temperature of 50C. Heat transfer co-efficient is 100 W/m2K. Density
of copper is 8800 kg/m3. Specific heat of copper = 0.36 kJ/kg K.
Plate dimensions = 30  30 cm.
Given
Thickness of plate L = 2 mm = 0.002 m
Initial temperature T0 = 400C + 273 = 673 K
Final temperature T = 30C + 273 = 303 K
Intermediate temperature T = 50C + 273 = 323 K
Heat transfer co-efficient h = 100 W/m2K
Density  = 8800 kg/m3
Specific heat C= 360 J/kg k
Plate dimensions = 30  30 cm
To find
Time required for the plate to reach 50C.
[From HMT data book Page No.2]
Solution:
Thermal conductivity of the copper K = 386 W/mK
For slab,
L
Lc 
2
Characteristic length
0.002
2
=
Lc  0.001 m
We know,
hLc
K
Biot number
100  0.001

386
4
Bi = 2.59  10  0.1
Bi 
Biot number value is less than 0.1. So this is lumped heat analysis type problem.
For lumped parameter system,

 hA

t 

T  T
C  V  
 e  
T0  T
……….(1)
[From HMT data book Page No.48]
We know,
V
Characteristics length Lc = A
3


h
t 

T-T
C L  
(1) 
 e   c 
T0  T

100

t 

323 - 303

 e  3600.0018800 
673 - 303

t = 92.43 s
Time required for the plate to reach 50C is 92.43 s.
26. A 12 cm diameter long bar initially at a uniform temperature of 40 C is placed in a medium at 650C
with a convective co-efficient of 22 W/m2K. Determine the time required for the center to reach 255C.
For the material of the bar, K = 20 W/mK, Density = 580 kg/m3, specific heat = 1050 J/kg K.
Given:
Diameter of bar, D = 12 cm = 0.12 m
Radius of bar, R = 6 cm = 0.06 m
Initial temperature T0 = 40C + 273 = 313 K
Final temperature T = 650C + 273 = 923 K
Intermediate temperature T = 255C + 273 = 528 K
Heat transfer co-efficient h = 22 W/m2K
Thermal conductivity K = 20 W/mK
Density  = 580 kg/m3
Specific heat C = 1050 J/kg k
Solution
For cylinder,
Lc 
Characteristic Length
0.06
=
2
R
2
Lc  0.03 m
We know,
hLc
K
Biot number
22  0.03

20
Bi 
Bi = 0.033 < 0.1
Biot number value is less than 0.1. So this is lumped heat analysis type problem.
For lumped parameter system,

 hA

t 

T  T
C  V  
 e  
T0  T
……….(1)
[From HMT data book Page No.48]
We know,
4
V
Characteristics length Lc = A


h
t 

T-T
 C  Lc   
(1) 
e
T0  T



22t



528 - 923
 e 10500.03580 
313 - 923
t = 360.8 s
Time required for the cube to reach 255C is 360.8 s.
27. A steel ball (specific heat = 0.46 kJ/kgK. and thermal conductivity = 35 W/mK) having 5 cm
diameter and initially at a uniform temperature of 450C is suddenly placed in a control environment in
which the temperature is maintained at 100C. Calculate the time required for the balls to attained a
temperature of 150C. Take h = 10W/m2K.
Given
Specific heat C = 0.46 kJ/kg K = 460 J/kg K
Thermal conductivity K = 35 W/mK
Diameter of the sphere D = 5 cm = 0.05 m
Radius of the sphere R = 0.025 m
Initial temperature T0 = 450C + 273 = 723 K
Final temperature T = 100C + 273 = 373 K
Intermediate temperature T = 150C + 273 = 423 K
Heat transfer co-efficient h = 10 W/m2K
To find
Time required for the ball to reach 150C
[From HMT data book Page No.1]
Solution
Density of steel is 7833 kg/m3
  7833 kg / m3
For sphere,
Lc 
Characteristic Length
0.025
=
3
R
3
Lc  8.33  103 m
We know,
Bi 
hLc
K
Biot number
10  8.3  10 3

35
5
Bi = 2.38  10-3 < 0.1
Biot number value is less than 0.1. So this is lumped heat analysis type problem.
For lumped parameter system,


 hA
t 

T  T
C  V  
 e  
T0  T
……….(1)
[From HMT data book Page No.48]
We know,
V
Characteristics length Lc = A

h

t 

T-T
C L  
(1) 
 e   c 
T0  T

10

t 

423 - 373
3
 e  4608.3310 7833 
723 - 373
423 - 373
10
 In

t
723 - 373 460  8.33  10 3  7833


t = 5840.54 s
Time required for the ball to reach 150C is 5840.54 s.
28. An aluminium sphere mass 5.5 kg and initially at a temperature of 290o is suddenly immersed in a
fluid at 15C with heat transfer co-efficient 58 W/m3K. Estimate the time required to cool the aluminium
to 95C. For aluminium take  = 2700 kg/m3, C = 900 J/kg K, K = 205 W/mK.
Given
Mass, m = 5.5 kg
Initial temperature T0 = 290C + 273 = 563 K
Final temperature T = 15C + 273 = 288 K
Intermediate temperature T = 95C + 273 =368 K
Heat transfer co-efficient h = 58 W/m2K
Thermal conductivity K = 205 W/mK
Density  = 2700 kg/m3
Specific heat C = 900 J/kg K.
Solution
We know,
mass
m

Density  = volume V
m
 V=

=
5.5
2700
V  2.037  10 3 m3
6
We know,
V
4
 R3
3
Volume of sphere
3V 3  2.03  103
R3 

4
4
R  0.0786 m

For sphere,
R
Lc 
3
Characteristic Length
0.0786
=
3
Lc  0.0262 m
We know,
hLc
K
Biot number
58  0.0262

205
Bi 
Bi = 7.41  10-3 < 0.1
Biot number value is less than 0.1. So this is lumped heat analysis type problem.
For lumped parameter system,

 hA

t 

T  T
C  V  
 e  
T0  T
……….(1)
[From HMT data book Page No.48]
We know,
V
Characteristics length Lc = A


h
t 

T-T
C L  
(1) 
 e   c 
T0  T

58

t 

368 - 288

 e  9000.02622700 
563 - 288
58
 368 - 288 
 In 

t

 563 - 288  900  0.0262  2700

t = 1355.36 s
Time required to cool the aluminium to 95C is 1355.6 s.
7
29. Alloy steel ball of 2 mm diameter heated to 800C is quenched in a bath at 100C. The material
properties of the ball are K = 205 kJ/m hr K,  = 7860 kg/m3, C = 0.45 kJ/kg K, h = 150 KJ/ hr m2 K.
Determine (i) Temperature of ball after 10 second and (ii) Time for ball to cool to 400C.
Given
Diameter of the ball D = 12 mm = 0.012 m
Radius of the ball R = 0.006m
Initial temperature T0 = 800C + 273 = 1073 K
Final temperature T = 100C + 273 = 373 K
Thermal conductivity K = 205 kJ/m hr K
205  1000J

3600 s mK
 56.94 W / mK
[ J/s = W]
Density  = 7860 kg/m3
Specific heat C = 0.45 kJ/kg K
= 450 J/kg K
Heat transfer co-efficient h = 150 kJ/hr m2 K
150  1000J

3600 s m2K
 41.66 W / m2K
Solution
Case (i) Temperature of ball after 10 sec.
For sphere,
Characteristic Length
R
3
0.006
=
3
Lc 
Lc  0.002 m
We know,
hLc
K
Biot number
41.667  0.002

56.94
Bi 
Bi = 1.46  10-3 < 0.1
Biot number value is less than 0.1. So this is lumped heat analysis type problem.
For lumped parameter system,

 hA

t 

T  T
C  V  
 e  
T0  T
……….(1)
[From HMT data book Page No.48]
We know,
8
V
Characteristics length Lc = A

h

t 

T-T
C L  
(1) 
 e   c 
T0  T

..........(2)
41.667

10 

T - 373

 e  4500.0027860 
1073 - 373

T = 1032.95 K
Case (ii) Time for ball to cool to 400C
T = 400C + 273 = 673 K

h

t 

T-T
C L  
(2) 
 e   c 
T0  T

41.667
.......(2)

t 

673 - 373
 e  4500.0027860 
1073 - 373
41.667
 673 - 373 
 In 

t

1073 - 373  450  0.002  7860


t = 143.849 s
30. A large wall 2 cm thick has uniform temperature 30C initially and the wall temperature is suddenly
raised and maintained at 400C. Find
The temperature at a depth of 0.8 cm from the surface of the wall after 10 s.
Instantaneous heat flow rate through that surface per m2 per hour.
Take  = 0.008 m2/hr, K = 6 W/mC.
Given
Thickness L = 2 cm = 0.02 m
Initial temperature Ti = 30C + 273 = 303 K
Surface temperature T0 = 400C + 273 = 673 K
Thermal diffusivity  = 0.008 m2/h
= 2.22  10-6 m2/s
Thermal conductivity K = 6 W/mC.
Case (i)
Depth  0.8 cm = 0.8  10-2 m
= 0.008 m
Time t = 10 s
Case (ii)
Time t = 1 h = 3600 s
Solution
9
In this problem heat transfer co-efficient h is not given. So take it as . i.e. h  .
We know that,
hLc
Biot number Bi = K
h=
Bi  

Bi value is . So this is semi infinite solid type problem.
Case (i)
For semi infinite solid.
Tx  T0
 x 
 erf 

Ti  T0
 2 at 
[From HMT data book Page No. 50]
Tx  T0

 erf (X) .......(1)
Ti  T0
Where,
x
X
2 at
Put x = 0.008 m, t = 10 s,  = 2.22  10-6 m2/s.

X=
0.008
2 2.22  10-6  10
X = 0.848
X = 0.848, corresponding erf (X) is 0.7706

erf (X) = 0.7706
[Refer HMT data book Page No.52]
(1) 
Tx -T0
 0.7706
Ti  T0
Tx - 673
 0.7706
303 - 673
T - 673
 x
 0.7706
- 370


Tx = 387.85 K
Case (ii)
10
Instantaneous heat flow
qx 
K  T0  Ti 
a t
e
  x2 


 4 t 
[From HMT data book Page No.50]
t = 3600 s (Given)
 qx 
6 (673  303)
  2.22  10-6  3600
e


 (0.008)2


6
 42.2210 3600 
qx  13982.37 W / m2
Intermediate temperature Tx = 387.85 K
Heat flux qx = 13982.37 W/m2.
31. A large cast iron at 750C is taken out from a furnace and its one of its surface is suddenly lowered
and maintained at 45C. Calculate the following:
The time required to reach the temperature 350C at a depth of 45 mm from the surface.
Instantaneous heat flow rate at a depth of 45 mm and on surface after 30 minutes.
Total heat energy after 2 hr for ingot,
Take  = 0.06 m2/hr, K = 48.5 W/mK.
Given
Initial temperature Ti = 750C + 273 = 1023 K
Surface temperature T0 = 45C + 273 = 318 K
Intermediate temperature Tx = 350C + 273 = 623 K
Depth x = 45 mm = 0.045 m
Thermal diffusivity  = 0.06 m2/hr = 1.66  10-5 m2/s
Thermal conductivity K = 48.5 W/mK.
Solution
In this problem heat transfer co-efficient h is not given. So take it as , i.e. h  .
We know that,

hLc
Biot number Bi = K
h=
Bi  
Bi value is . So this is semi infinite solid type problem.
1. For semi infinite solid.
11
Tx  T0
 x 
 erf 

Ti  T0
 2 at 
[From HMT data book Page No. 50]

Tx  T0
 erf (X) where,
Ti  T0
x
X
2 at

623  318
 erf (X)
1023  318

0.432 = erf (X)

erf (X) = 0.432
erf (X) = 0.432, corresponding X is 0.41
X  0.41

We know
X


x
2 at
0.045
0.41 =
2 1.66  10-5  t
(0.045)2
2
(0.41) 
(2)2  1.66  105  t

t = 181.42 s
Time required to reach 350C is 181.42 s.
2. Instantaneous heat flow
qx 
K  T0  Ti 
a t
e
  x2 


 4 t 
[From HMT data book Page No.50]
t = 30 minutes (Given)
t = 1800 s
 qx 
48.5 (318  1023)
  1.66  10  1800
-5
e


 (0.045)2


5
 41.6610 1800 
qx  109725.4 W / m2
[Negative sign shows that heat lost from the ingot].
12
3. Total heat energy
q  2K[T0  Ti ]
t

7200
  1.66  10 5
[Time is given, 2 hr = 7200 s]
 2  48.5(318  1023) 
q  803.5  106 J/ m2
[Negative sign shows that heat lost from the ingot]
32. A large steel plate 5 cm thick is initially at a uniform temperature of 400C. It is suddenly exposed on
both sides to a surrounding at 60C with convective heat transfer co-efficient of 285 W/m2K. Calculate
the centre line temperature and the temperature inside the plate 1.25 cm from themed plane after 3
minutes.
Take K for steel = 42.5 W/mK,  for steel = 0.043 m2/hr.
Given
Thickness L = 5 cm = 0.05 m
Initial temperature Ti = 400C + 273 = 673 K
Final temperature T = 60C + 273 = 333 K
Distance x = 1.25 mm = 0.0125 m
Time t = 3 minutes = 180 s
Heat transfer co-efficient h = 285 W/m2K
Thermal diffusivity  = 0.043 m2/hr
= 1.19  10-5 m2/s.
Thermal conductivity K = 42.5 W/mK.
Solution
For Plate :
Characteristic Length
Lc 
=
L
2
0.05
2
Lc  0.025 m
We know,
hLc
K
Biot number
285  0.025

42.5
 Bi  0.1675
Bi 
0.1 < Bi < 100, So this is infinite solid type problem.
Infinite Solids
13
Case (i)
[To calculate centre line temperature (or) Mid plane temperature for infinite plate, refer HMT data
book Page No.59 Heisler chart].
t
X axis  Fourier number = 2
Lc
=
1.19  10-5  180
(0.025)2
X axis  Fourier number = 3.42
Curve 

hLc
K
285  0.025
 0.167
42.5
Curve 
hLc
 0.167
K
X axis value is 3.42, curve value is 0.167, corresponding Y axis value is 0.64
Y axis =
T0  T
 0.64
Ti  T
T0  T
 0.64
Ti  T
T0  T
 0.64
T

T
i



T0  333
 0.64
673  333
 T0  550.6 K
Center line temperature T0  550.6 K
Case (ii)
Temperature (Tx) at a distance of 0.0125 m from mid plane
[Refer HMT data book Page No.60, Heisler chart]
hL
X axis  Biot number Bi  c  0.167
K
x 0.0125
Curve 

 0.5
Lc
0.025
X axis value is 0.167, curve value is 0.5, corresponding Y axis value is 0.97.
14
Tx  T
 0.97
T0  T
Y axis =
Tx  T
 0.97
T0  T

Tx  T
 0.97
T0  T

Tx  333
 0.97
550.6  333

Tx  544 K
Temperature inside the plate 1.25 cm from the mid plane is 544 K.
33. A 10 cm diameter apple approximately spherical in shape is taken from a 20C environment and
placed in a refrigerator where temperature is 5C and average heat transfer coefficient is 6 W/m2K.
Calculate the temperature at the centre of the apple after a period of 1 hour. The physical properties of
apple are density = 998 kg/m3. Specific heat = 4180 J/kg K, Thermal conductivity = 0.6 W/mK.
Given:
Diameter of sphere D = 10 cm = 0.10 m
Radius of sphere R = 5 cm = 0.05 m
Initial temperature Ti = 20C + 273 = 293 K
Final temperature T = 5C + 273 = 278 K
Time t = 1 hour = 3600 s
Density  = 998 kg/m3
Heat transfer co-efficient h = 6 W/m2K
Specific heat C = 4180 J/kg K
Thermal conductivity K = 0.6 W/mK
K
0.6

C 998  4180
Thermal diffusivity  =
  1.43  107 m2 / s.
Solution
For Sphere,
Lc 
Characteristic Length
0.05
=
3
R
3
Lc  0.016 m
We know,
Biot number
Bi 
hLc
K
15

6  0.016
0.6
 Bi  0.16
0.1 < Bi < 100, So this is infinite solid type problem.
Infinite Solids
[To calculate centre line temperature for sphere, refer HMT data book Page No.63].
X axis =
t
R2
1.43  10-7  3600
=
(0.05)2
X axis = 0.20
Curve 

hR
K
6  0.05
 0.5
0.6
Curve  0.5
X axis value is 0.20, curve value is 0.5, corresponding Y axis value is 0.86.
T0  T
 0.86
Ti  T

Y axis =

T0  T
 0.86
Ti  T

T0  278
 0.86
293  278

T0  290.9 K
Center line temperature T0 = 290.9 K.
34. A long steel cylinder 12 cm diameter and initially at 20C is placed into furnace at 820C with h =
140 W/m2K. Calculate the time required for the axis temperature to reach 800C. Also calculate the
corresponding temperature at a radius of 5.4 cm at that time. Physical properties of steel are K = 21
W/mK,  = 6.11  10-6 m2/s.
Given:
Diameter of cylinder D = 12 cm = 0.12 m
Radius of sphere R = 6 cm = 0.06 m
Initial temperature Ti = 20C + 273 = 293 K
Final temperature T = 820C + 273 = 1093 K
Heat transfer co-efficient h = 140 W/m2K
16


(or)
 T0  800C  273  1073 K
Centre line temperature 

Intermediate radius r = 5.4 cm = 0.054 m
Thermal diffusivity  = 6.11  10-6 m2/s.
Thermal conductivity K = 21W/mK
Axis temperature
To find
Time (t) required for the axis temperature to reach 800C.
Corresponding temperature (Tt) at a radius of 5.4 cm.
Solution
For Cylinder,
Characteristic Length
Lc  0.03 m
Lc 
R 0.06

2
2
We know,

hLc
Biot number Bi = K
140  0.03

21
Bi  0.2
0.1 < Bi <100, So this is infinite solid type problem.
Infinite Solids
Case (i)
Axis temperature


(or)
 T0  800C
Centre line temperature 
To = 800C + 273 = 1073 K
Time (t) ?
[Refer HMT data book Page No.61. Heisler chart]
hR
K
140  0.06
=
 0.4
21
Curve 
Y axis =
=
T0  T
Ti  T
1073 - 1093
293 - 1093
Y axis = 0.025
17
Curve value is 0.4, Y axis 0.025, corresponding X axis value is 5.
T0  T
 0.025
Ti  T


t
5
R2
5  (0.06)2
t=
(6.11 10 -6 )
X axis =
t  2945.9 s
Case (ii)
Intermediate radius r – 5.4 cm = 0.054 m
[Refer HMT data book Page No.62]
r 0.054

 0.9
R 0.06
hR
X axis =
K
140  0.06
=
 0.4
21
Curve 
Curve value is 0.9, X axis value is 0.4, corresponding Y axis value is 0.84.
Tr  T
 0.84
T0  T

Y axis =

Tr  T
 0.84
T0  T

Tr  1093
 0.84
1073  1093

Tr  1076.2 K
Time required for the axis temperature to reach 800C is 2945.9 s.
Temperature (Tr) at a radius of 5.4 cm is 1076.2 K
2 Marks Question and Answer
24. What is meant by steady state heat conduction?
If the temperature of a body does not vary with time, it is said to be in a steady state and that type
of conduction is known as steady state heat conduction.
25. What is meant by Transient heat conduction or unsteady state conduction?
18
If the temperature of a body varies with time, it is said to be in a transient state and that type of
conduction is known as transient heat conduction or unsteady state conduction.
26. What is Periodic heat flow?
In periodic heat flow, the temperature varies on a regular basis.
Example:
1. Cylinder of an IC engine.
2. Surface of earth during a period of 24 hours.
27. What is non periodic heat flow?
In non periodic heat flow, the temperature at any point within the system varies non linearly with
time.
Examples :
1. Heating of an ingot in a furnace.
2. Cooling of bars.
28. What is meant by Newtonian heating or cooling process?
The process in which the internal resistance is assumed as negligible in comparison with its
surface resistance is known as Newtonian heating or cooling process.
29. What is meant by Lumped heat analysis?
In a Newtonian heating or cooling process the temperature throughout the solid is considered to
be uniform at a given time. Such an analysis is called Lumped heat capacity analysis.
30. What is meant by Semi-infinite solids?
In a semi infinite solid, at any instant of time, there is always a point where the effect of heating
or cooling at one of its boundaries is not felt at all. At this point the temperature remains unchanged. In
semi infinite solids, the biot number value is .
31. What is meant by infinite solid?
A solid which extends itself infinitely in all directions of space is known as infinite solid.
In semi infinite solids, the biot number value is in between 0.1 and 100.
0.1 < Bi < 100.
32. Define Biot number.
It is defined as the ratio of internal conductive resistance to the surface convective resistance.
Bi =
Bi =
Internal conductive resistance
Surface convective resistance
hLL
.
K
19
33. What is the significance of Biot number?
Biot number is used to find Lumped heat analysis, semi infinite solids and infinite solids
If Bi < 0.1 L  Lumped heat analysis
Bi =   Semi infinite solids
0.1 < Bi < 100  Infinite solids.
34. Explain the significance of Fourier number.
It is defined as the ratio of characteristic body dimension to temperature wave penetration depth in
time.
Fourier Number =
Characteristic body dimension
Temperature wave penetration
depth in time
It signifies the degree of penetration of heating or cooling effect of a solid.
35. What are the factors affecting the thermal conductivity?
1.
2.
3.
4.
5.
Moisture
Density of material
Pressure
Temperature
Structure of material
36. Explain the significance of thermal diffusivity.
The physical significance of thermal diffusivity is that it tells us how fast heat is propagated or it
diffuses through a material during changes of temperature with time.
37. What are Heisler charts?
In Heisler chart, the solutions for temperature distributions and heat flows in plane walls, long
cylinders and spheres with finite internal and surface resistance are presented. Heisler charts are nothing
but a analytical solutions in the form of graphs.
20