Chapter 4 : Transient Heat Conduction - Notes, Engineering, Semester Notes | EduRev

: Chapter 4 : Transient Heat Conduction - Notes, Engineering, Semester Notes | EduRev

 Page 1


Chapter 4 Transient Heat Conduction 
 4-98 
4-118 Internal combustion engine valves are quenched in a large oil bath. The time it takes for the valve 
temperature to drop to specified temperatures and the maximum heat transfer are to be determined. 
Assumptions 1 The thermal properties of the valves are constant. 2 The heat transfer coefficient is constant 
and uniform over the entire surface. 3 Depending on the size of the oil bath, the oil bath temperature will 
increase during quenching. However, an average canstant temperature as specified in the problem will be 
used.  4 The Biot number is Bi < 0.1 so that the lumped system analysis is applicable (this assumption will 
be verified). 
Properties The thermal conductivity, density, and 
specific heat of the balls are given to be k = 48 W/m. ?C, 
? = 7840 kg/m
3
, and C p = 440 J/kg. ?C. 
Analysis (a) The characteristic length of the balls and 
the Biot number are 
1 . 0 024 . 0
) C W/m. 48 (
) m 0018 . 0 )( C . W/m 650 (
m 0018 . 0
8
m) 008 . 0 ( 8 . 1
8
8 . 1
2
) 4 / ( 8 . 1
2
2
? ?
?
?
? ?
? ? ? ? ?
k
hL
Bi
D
DL
L D
A
V
L
c
s
c
?
?
 
Therefore, we can use lumped system analysis. Then the time for a final valve temperature of 400 ?C 
becomes  
 
s 7.2 ? ? ? ? ?
?
?
? ? ? ?
?
?
?
?
?
? ? ?
? ?
?
?
t e e
T T
T t T
D C
h
V C
hA
b
bt
i
p p
s
)t s 10468 . 0 (
1 -
3
2
-1
45 800
45 400 ) (
s 10468 . 0
m) C)(0.008 J/kg. 440 )( kg/m 1.8(7840
C) . W/m 650 ( 8
8 . 1
8
? ?
 
(b) The time for a final valve temperature of 200 ?C is  
s 15.1 ? ? ? ? ?
?
?
? ? ? ?
?
?
? ?
?
?
t e e
T T
T t T
bt
i
)t s 10468 . 0 (
-1
45 800
45 200 ) (
 
(c) The time for a final valve temperature of 46 ?C is  
s 63.3 ? ? ? ? ?
?
?
? ? ? ?
?
?
? ?
?
?
t e e
T T
T t T
bt
i
)t s 10468 . 0 (
-1
45 800
45 46 ) (
 
(d) The maximum amount of heat transfer from a single valve is determined from 
 
) (per valve  = J 564 , 23 C ) 45 800 )( C J/kg. 440 )( kg 0709 . 0 ( ] [
kg 0709 . 0
4
m) 10 . 0 ( m) 008 . 0 ( 8 . 1
) kg/m 7840 (
4
8 . 1
2
3
2
kJ 23.56 ? ? ? ? ? ? ?
?
?
?
?
? ? ? ?
i f p
T T mC Q
L D
V m
 
Oil 
T ? = 45 ?C 
Engine valve 
T i = 800 ?C 
Page 2


Chapter 4 Transient Heat Conduction 
 4-98 
4-118 Internal combustion engine valves are quenched in a large oil bath. The time it takes for the valve 
temperature to drop to specified temperatures and the maximum heat transfer are to be determined. 
Assumptions 1 The thermal properties of the valves are constant. 2 The heat transfer coefficient is constant 
and uniform over the entire surface. 3 Depending on the size of the oil bath, the oil bath temperature will 
increase during quenching. However, an average canstant temperature as specified in the problem will be 
used.  4 The Biot number is Bi < 0.1 so that the lumped system analysis is applicable (this assumption will 
be verified). 
Properties The thermal conductivity, density, and 
specific heat of the balls are given to be k = 48 W/m. ?C, 
? = 7840 kg/m
3
, and C p = 440 J/kg. ?C. 
Analysis (a) The characteristic length of the balls and 
the Biot number are 
1 . 0 024 . 0
) C W/m. 48 (
) m 0018 . 0 )( C . W/m 650 (
m 0018 . 0
8
m) 008 . 0 ( 8 . 1
8
8 . 1
2
) 4 / ( 8 . 1
2
2
? ?
?
?
? ?
? ? ? ? ?
k
hL
Bi
D
DL
L D
A
V
L
c
s
c
?
?
 
Therefore, we can use lumped system analysis. Then the time for a final valve temperature of 400 ?C 
becomes  
 
s 7.2 ? ? ? ? ?
?
?
? ? ? ?
?
?
?
?
?
? ? ?
? ?
?
?
t e e
T T
T t T
D C
h
V C
hA
b
bt
i
p p
s
)t s 10468 . 0 (
1 -
3
2
-1
45 800
45 400 ) (
s 10468 . 0
m) C)(0.008 J/kg. 440 )( kg/m 1.8(7840
C) . W/m 650 ( 8
8 . 1
8
? ?
 
(b) The time for a final valve temperature of 200 ?C is  
s 15.1 ? ? ? ? ?
?
?
? ? ? ?
?
?
? ?
?
?
t e e
T T
T t T
bt
i
)t s 10468 . 0 (
-1
45 800
45 200 ) (
 
(c) The time for a final valve temperature of 46 ?C is  
s 63.3 ? ? ? ? ?
?
?
? ? ? ?
?
?
? ?
?
?
t e e
T T
T t T
bt
i
)t s 10468 . 0 (
-1
45 800
45 46 ) (
 
(d) The maximum amount of heat transfer from a single valve is determined from 
 
) (per valve  = J 564 , 23 C ) 45 800 )( C J/kg. 440 )( kg 0709 . 0 ( ] [
kg 0709 . 0
4
m) 10 . 0 ( m) 008 . 0 ( 8 . 1
) kg/m 7840 (
4
8 . 1
2
3
2
kJ 23.56 ? ? ? ? ? ? ?
?
?
?
?
? ? ? ?
i f p
T T mC Q
L D
V m
 
Oil 
T ? = 45 ?C 
Engine valve 
T i = 800 ?C 
Chapter 4 Transient Heat Conduction 
 4-99 
4-119 A watermelon is placed into a lake to cool it. The heat transfer coefficient at the surface of the 
watermelon and the temperature of the outer surface of the watermelon are to be determined. 
Assumptions 1 The watermelon is a homogeneous spherical object. 2 Heat conduction in the watermelon is 
one-dimensional because of symmetry about the midpoint. 3 The thermal properties of the watermelon are 
constant. 4 The heat transfer coefficient is constant and uniform over the entire surface. 5 The Fourier 
number is ? > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are 
applicable (this assumption will be verified). 
Properties The properties of the watermelon are given to be k = 0.618 W/m. ?C, ? = 0.15 ?10
-6
 m
2
/s, ? = 995 
kg/m
3
 and C p = 4.18 kJ/kg. ?C. 
Analysis The Fourier number is 
  
? ?
252 . 0
m) 10 . 0 (
s/min 60 min)  40 60 (4 /s) m 10 15 . 0 (
2
2 6
2
?
? ? ? ?
?
?
? ?
?
o
r
t
 
which is greater than 0.2. Then the one-term solution can be 
written in the form 
   
) 252 . 0 (
1 1
0
sph 0,
2
1
2
1
25 . 0
15 35
15 20
? ? ? ? ?
?
?
? ?
?
?
? ? ? ?
?
?
? ? e A e A
T T
T T
i
 
It is determined from Table 4-1 by trial and error that this equation is satisfied when Bi = 10, which  
corresponds to   ?
1 1
2 8363 19249 ? ? . .   and   A . Then the heat transfer coefficient can be determined from 
 C . W/m 61.8
2
? ?
?
? ? ? ? ? ?
) m 10 . 0 (
) 10 )( C W/m. 618 . 0 (
o
o
r
kBi
h
k
hr
Bi 
The temperature at the surface of the watermelon is 
 
C 15.5 ? ? ? ? ? ?
?
?
?
?
?
?
?
?
? ?
? ? ? ?
?
?
) , ( 0269 . 0
15 35
15 ) , (
8363 . 2
) rad 8363 . 2 sin(
) 9249 . 1 (
/
) / sin( ) , (
) , (
) 252 . 0 ( ) 8363 . 2 (
1
1
1
2 2
1
t r T
t r T
e
r r
r r
e A
T T
T t r T
t r
o
o
o o
o o
i
o
sph o
 
Water 
melon 
Ti = 35 ?C 
Lake 
15 ?C 
Page 3


Chapter 4 Transient Heat Conduction 
 4-98 
4-118 Internal combustion engine valves are quenched in a large oil bath. The time it takes for the valve 
temperature to drop to specified temperatures and the maximum heat transfer are to be determined. 
Assumptions 1 The thermal properties of the valves are constant. 2 The heat transfer coefficient is constant 
and uniform over the entire surface. 3 Depending on the size of the oil bath, the oil bath temperature will 
increase during quenching. However, an average canstant temperature as specified in the problem will be 
used.  4 The Biot number is Bi < 0.1 so that the lumped system analysis is applicable (this assumption will 
be verified). 
Properties The thermal conductivity, density, and 
specific heat of the balls are given to be k = 48 W/m. ?C, 
? = 7840 kg/m
3
, and C p = 440 J/kg. ?C. 
Analysis (a) The characteristic length of the balls and 
the Biot number are 
1 . 0 024 . 0
) C W/m. 48 (
) m 0018 . 0 )( C . W/m 650 (
m 0018 . 0
8
m) 008 . 0 ( 8 . 1
8
8 . 1
2
) 4 / ( 8 . 1
2
2
? ?
?
?
? ?
? ? ? ? ?
k
hL
Bi
D
DL
L D
A
V
L
c
s
c
?
?
 
Therefore, we can use lumped system analysis. Then the time for a final valve temperature of 400 ?C 
becomes  
 
s 7.2 ? ? ? ? ?
?
?
? ? ? ?
?
?
?
?
?
? ? ?
? ?
?
?
t e e
T T
T t T
D C
h
V C
hA
b
bt
i
p p
s
)t s 10468 . 0 (
1 -
3
2
-1
45 800
45 400 ) (
s 10468 . 0
m) C)(0.008 J/kg. 440 )( kg/m 1.8(7840
C) . W/m 650 ( 8
8 . 1
8
? ?
 
(b) The time for a final valve temperature of 200 ?C is  
s 15.1 ? ? ? ? ?
?
?
? ? ? ?
?
?
? ?
?
?
t e e
T T
T t T
bt
i
)t s 10468 . 0 (
-1
45 800
45 200 ) (
 
(c) The time for a final valve temperature of 46 ?C is  
s 63.3 ? ? ? ? ?
?
?
? ? ? ?
?
?
? ?
?
?
t e e
T T
T t T
bt
i
)t s 10468 . 0 (
-1
45 800
45 46 ) (
 
(d) The maximum amount of heat transfer from a single valve is determined from 
 
) (per valve  = J 564 , 23 C ) 45 800 )( C J/kg. 440 )( kg 0709 . 0 ( ] [
kg 0709 . 0
4
m) 10 . 0 ( m) 008 . 0 ( 8 . 1
) kg/m 7840 (
4
8 . 1
2
3
2
kJ 23.56 ? ? ? ? ? ? ?
?
?
?
?
? ? ? ?
i f p
T T mC Q
L D
V m
 
Oil 
T ? = 45 ?C 
Engine valve 
T i = 800 ?C 
Chapter 4 Transient Heat Conduction 
 4-99 
4-119 A watermelon is placed into a lake to cool it. The heat transfer coefficient at the surface of the 
watermelon and the temperature of the outer surface of the watermelon are to be determined. 
Assumptions 1 The watermelon is a homogeneous spherical object. 2 Heat conduction in the watermelon is 
one-dimensional because of symmetry about the midpoint. 3 The thermal properties of the watermelon are 
constant. 4 The heat transfer coefficient is constant and uniform over the entire surface. 5 The Fourier 
number is ? > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are 
applicable (this assumption will be verified). 
Properties The properties of the watermelon are given to be k = 0.618 W/m. ?C, ? = 0.15 ?10
-6
 m
2
/s, ? = 995 
kg/m
3
 and C p = 4.18 kJ/kg. ?C. 
Analysis The Fourier number is 
  
? ?
252 . 0
m) 10 . 0 (
s/min 60 min)  40 60 (4 /s) m 10 15 . 0 (
2
2 6
2
?
? ? ? ?
?
?
? ?
?
o
r
t
 
which is greater than 0.2. Then the one-term solution can be 
written in the form 
   
) 252 . 0 (
1 1
0
sph 0,
2
1
2
1
25 . 0
15 35
15 20
? ? ? ? ?
?
?
? ?
?
?
? ? ? ?
?
?
? ? e A e A
T T
T T
i
 
It is determined from Table 4-1 by trial and error that this equation is satisfied when Bi = 10, which  
corresponds to   ?
1 1
2 8363 19249 ? ? . .   and   A . Then the heat transfer coefficient can be determined from 
 C . W/m 61.8
2
? ?
?
? ? ? ? ? ?
) m 10 . 0 (
) 10 )( C W/m. 618 . 0 (
o
o
r
kBi
h
k
hr
Bi 
The temperature at the surface of the watermelon is 
 
C 15.5 ? ? ? ? ? ?
?
?
?
?
?
?
?
?
? ?
? ? ? ?
?
?
) , ( 0269 . 0
15 35
15 ) , (
8363 . 2
) rad 8363 . 2 sin(
) 9249 . 1 (
/
) / sin( ) , (
) , (
) 252 . 0 ( ) 8363 . 2 (
1
1
1
2 2
1
t r T
t r T
e
r r
r r
e A
T T
T t r T
t r
o
o
o o
o o
i
o
sph o
 
Water 
melon 
Ti = 35 ?C 
Lake 
15 ?C 
Chapter 4 Transient Heat Conduction 
 4-100 
4-120 Large food slabs are cooled in a refrigeration room. Center temperatures are to be determined for 
different foods. 
Assumptions 1 Heat conduction in the slabs is one-dimensional since the slab is large relative to its 
thickness and there is thermal symmetry about the center plane. 3 The thermal properties of the slabs are 
constant. 4 The heat transfer coefficient is constant and uniform over the entire surface.  5 The Fourier 
number is ? > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are 
applicable (this assumption will be verified). 
Properties The properties of foods are given to be k = 0.233 W/m. ?C and ? = 0.11 ?10
-6
 m
2
/s for margarine, 
k = 0.082 W/m. ?C and ? = 0.10 ?10
-6
 m
2
/s for white cake, and k = 0.106 W/m. ?C and ? = 0.12 ?10
-6
 m
2
/s for 
chocolate cake. 
Analysis (a) In the case of margarine, the Biot number is  
365 . 5
) C W/m. 233 . 0 (
) m 05 . 0 )( C . W/m 25 (
2
?
?
?
? ?
k
hL
Bi 
The constants ?
1 1
 and A corresponding to this Biot number are, 
from Table 4-1,  
 2431 . 1 and      3269 . 1
1 1
? ? A ? 
The Fourier number is            2 . 0 9504 . 0
m) 05 . 0 (
s/h) 600 3 h /s)(6 m 10 11 . 0 (
2
2 6
2
? ?
? ?
? ?
?
L
t ?
? 
Therefore, the one-term approximate solution (or the transient temperature charts) is applicable. Then the 
temperature at the center of the box if the box contains margarine becomes 
 
C 7.0 ? ? ? ? ? ?
?
?
? ?
?
?
? ?
? ? ? ?
?
?
) , 0 ( 233 . 0
0 30
0 ) , 0 (
) 2431 . 1 (
) , 0 (
) , 0 (
) 9504 . 0 ( ) 3269 . 1 (
1
2 2
1
t T
t T
e e A
T T
T t T
t
i
wall
 
(b) Repeating the calculations for white cake,  
 ? ? ? ?
?
?
? ? 24 . 15
) C W/m. 082 . 0 (
) m 05 . 0 )( C . W/m 25 (
2
k
hL
Bi 2661 . 1 and      4641 . 1
1 1
? ? A ?   
 2 . 0 864 . 0
m) 05 . 0 (
s/h) 600 3 h /s)(6 m 10 10 . 0 (
2
2 6
2
? ?
? ?
? ?
?
L
t ?
? 
 
C 6.0 ? ? ? ? ? ?
?
?
? ?
?
?
? ?
? ? ? ?
?
?
) , 0 ( 199 . 0
0 30
0 ) , 0 (
) 2661 . 1 (
) , 0 (
) , 0 (
) 864 . 0 ( ) 4641 . 1 (
1
2 2
1
t T
t T
e e A
T T
T t T
t
i
wall
 
(c) Repeating the calculations for chocolate cake, 
 ? ? ? ?
?
?
? ? 79 . 11
) C W/m. 106 . 0 (
) m 05 . 0 )( C . W/m 25 (
2
k
hL
Bi 2634 . 1 and     4356 . 1
1 1
? ? A ? 
 2 . 0 0368 . 1
m) 05 . 0 (
s/h) 600 3 h /s)(6 m 10 12 . 0 (
2
2 6
2
? ?
? ?
? ?
?
L
t ?
? 
 
C 4.5 ? ? ? ? ? ?
?
?
? ?
?
?
? ?
? ? ? ?
?
?
) , 0 ( 149 . 0
0 30
0 ) , 0 (
) 2634 . 1 (
) , 0 (
) , 0 (
) 0368 . 1 ( ) 4356 . 1 (
1
2 2
1
t T
t T
e e A
T T
T t T
t
i
wall
 
Air 
T ? = 0 ?C 
Margarine,  T i = 30 ?C 
Page 4


Chapter 4 Transient Heat Conduction 
 4-98 
4-118 Internal combustion engine valves are quenched in a large oil bath. The time it takes for the valve 
temperature to drop to specified temperatures and the maximum heat transfer are to be determined. 
Assumptions 1 The thermal properties of the valves are constant. 2 The heat transfer coefficient is constant 
and uniform over the entire surface. 3 Depending on the size of the oil bath, the oil bath temperature will 
increase during quenching. However, an average canstant temperature as specified in the problem will be 
used.  4 The Biot number is Bi < 0.1 so that the lumped system analysis is applicable (this assumption will 
be verified). 
Properties The thermal conductivity, density, and 
specific heat of the balls are given to be k = 48 W/m. ?C, 
? = 7840 kg/m
3
, and C p = 440 J/kg. ?C. 
Analysis (a) The characteristic length of the balls and 
the Biot number are 
1 . 0 024 . 0
) C W/m. 48 (
) m 0018 . 0 )( C . W/m 650 (
m 0018 . 0
8
m) 008 . 0 ( 8 . 1
8
8 . 1
2
) 4 / ( 8 . 1
2
2
? ?
?
?
? ?
? ? ? ? ?
k
hL
Bi
D
DL
L D
A
V
L
c
s
c
?
?
 
Therefore, we can use lumped system analysis. Then the time for a final valve temperature of 400 ?C 
becomes  
 
s 7.2 ? ? ? ? ?
?
?
? ? ? ?
?
?
?
?
?
? ? ?
? ?
?
?
t e e
T T
T t T
D C
h
V C
hA
b
bt
i
p p
s
)t s 10468 . 0 (
1 -
3
2
-1
45 800
45 400 ) (
s 10468 . 0
m) C)(0.008 J/kg. 440 )( kg/m 1.8(7840
C) . W/m 650 ( 8
8 . 1
8
? ?
 
(b) The time for a final valve temperature of 200 ?C is  
s 15.1 ? ? ? ? ?
?
?
? ? ? ?
?
?
? ?
?
?
t e e
T T
T t T
bt
i
)t s 10468 . 0 (
-1
45 800
45 200 ) (
 
(c) The time for a final valve temperature of 46 ?C is  
s 63.3 ? ? ? ? ?
?
?
? ? ? ?
?
?
? ?
?
?
t e e
T T
T t T
bt
i
)t s 10468 . 0 (
-1
45 800
45 46 ) (
 
(d) The maximum amount of heat transfer from a single valve is determined from 
 
) (per valve  = J 564 , 23 C ) 45 800 )( C J/kg. 440 )( kg 0709 . 0 ( ] [
kg 0709 . 0
4
m) 10 . 0 ( m) 008 . 0 ( 8 . 1
) kg/m 7840 (
4
8 . 1
2
3
2
kJ 23.56 ? ? ? ? ? ? ?
?
?
?
?
? ? ? ?
i f p
T T mC Q
L D
V m
 
Oil 
T ? = 45 ?C 
Engine valve 
T i = 800 ?C 
Chapter 4 Transient Heat Conduction 
 4-99 
4-119 A watermelon is placed into a lake to cool it. The heat transfer coefficient at the surface of the 
watermelon and the temperature of the outer surface of the watermelon are to be determined. 
Assumptions 1 The watermelon is a homogeneous spherical object. 2 Heat conduction in the watermelon is 
one-dimensional because of symmetry about the midpoint. 3 The thermal properties of the watermelon are 
constant. 4 The heat transfer coefficient is constant and uniform over the entire surface. 5 The Fourier 
number is ? > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are 
applicable (this assumption will be verified). 
Properties The properties of the watermelon are given to be k = 0.618 W/m. ?C, ? = 0.15 ?10
-6
 m
2
/s, ? = 995 
kg/m
3
 and C p = 4.18 kJ/kg. ?C. 
Analysis The Fourier number is 
  
? ?
252 . 0
m) 10 . 0 (
s/min 60 min)  40 60 (4 /s) m 10 15 . 0 (
2
2 6
2
?
? ? ? ?
?
?
? ?
?
o
r
t
 
which is greater than 0.2. Then the one-term solution can be 
written in the form 
   
) 252 . 0 (
1 1
0
sph 0,
2
1
2
1
25 . 0
15 35
15 20
? ? ? ? ?
?
?
? ?
?
?
? ? ? ?
?
?
? ? e A e A
T T
T T
i
 
It is determined from Table 4-1 by trial and error that this equation is satisfied when Bi = 10, which  
corresponds to   ?
1 1
2 8363 19249 ? ? . .   and   A . Then the heat transfer coefficient can be determined from 
 C . W/m 61.8
2
? ?
?
? ? ? ? ? ?
) m 10 . 0 (
) 10 )( C W/m. 618 . 0 (
o
o
r
kBi
h
k
hr
Bi 
The temperature at the surface of the watermelon is 
 
C 15.5 ? ? ? ? ? ?
?
?
?
?
?
?
?
?
? ?
? ? ? ?
?
?
) , ( 0269 . 0
15 35
15 ) , (
8363 . 2
) rad 8363 . 2 sin(
) 9249 . 1 (
/
) / sin( ) , (
) , (
) 252 . 0 ( ) 8363 . 2 (
1
1
1
2 2
1
t r T
t r T
e
r r
r r
e A
T T
T t r T
t r
o
o
o o
o o
i
o
sph o
 
Water 
melon 
Ti = 35 ?C 
Lake 
15 ?C 
Chapter 4 Transient Heat Conduction 
 4-100 
4-120 Large food slabs are cooled in a refrigeration room. Center temperatures are to be determined for 
different foods. 
Assumptions 1 Heat conduction in the slabs is one-dimensional since the slab is large relative to its 
thickness and there is thermal symmetry about the center plane. 3 The thermal properties of the slabs are 
constant. 4 The heat transfer coefficient is constant and uniform over the entire surface.  5 The Fourier 
number is ? > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are 
applicable (this assumption will be verified). 
Properties The properties of foods are given to be k = 0.233 W/m. ?C and ? = 0.11 ?10
-6
 m
2
/s for margarine, 
k = 0.082 W/m. ?C and ? = 0.10 ?10
-6
 m
2
/s for white cake, and k = 0.106 W/m. ?C and ? = 0.12 ?10
-6
 m
2
/s for 
chocolate cake. 
Analysis (a) In the case of margarine, the Biot number is  
365 . 5
) C W/m. 233 . 0 (
) m 05 . 0 )( C . W/m 25 (
2
?
?
?
? ?
k
hL
Bi 
The constants ?
1 1
 and A corresponding to this Biot number are, 
from Table 4-1,  
 2431 . 1 and      3269 . 1
1 1
? ? A ? 
The Fourier number is            2 . 0 9504 . 0
m) 05 . 0 (
s/h) 600 3 h /s)(6 m 10 11 . 0 (
2
2 6
2
? ?
? ?
? ?
?
L
t ?
? 
Therefore, the one-term approximate solution (or the transient temperature charts) is applicable. Then the 
temperature at the center of the box if the box contains margarine becomes 
 
C 7.0 ? ? ? ? ? ?
?
?
? ?
?
?
? ?
? ? ? ?
?
?
) , 0 ( 233 . 0
0 30
0 ) , 0 (
) 2431 . 1 (
) , 0 (
) , 0 (
) 9504 . 0 ( ) 3269 . 1 (
1
2 2
1
t T
t T
e e A
T T
T t T
t
i
wall
 
(b) Repeating the calculations for white cake,  
 ? ? ? ?
?
?
? ? 24 . 15
) C W/m. 082 . 0 (
) m 05 . 0 )( C . W/m 25 (
2
k
hL
Bi 2661 . 1 and      4641 . 1
1 1
? ? A ?   
 2 . 0 864 . 0
m) 05 . 0 (
s/h) 600 3 h /s)(6 m 10 10 . 0 (
2
2 6
2
? ?
? ?
? ?
?
L
t ?
? 
 
C 6.0 ? ? ? ? ? ?
?
?
? ?
?
?
? ?
? ? ? ?
?
?
) , 0 ( 199 . 0
0 30
0 ) , 0 (
) 2661 . 1 (
) , 0 (
) , 0 (
) 864 . 0 ( ) 4641 . 1 (
1
2 2
1
t T
t T
e e A
T T
T t T
t
i
wall
 
(c) Repeating the calculations for chocolate cake, 
 ? ? ? ?
?
?
? ? 79 . 11
) C W/m. 106 . 0 (
) m 05 . 0 )( C . W/m 25 (
2
k
hL
Bi 2634 . 1 and     4356 . 1
1 1
? ? A ? 
 2 . 0 0368 . 1
m) 05 . 0 (
s/h) 600 3 h /s)(6 m 10 12 . 0 (
2
2 6
2
? ?
? ?
? ?
?
L
t ?
? 
 
C 4.5 ? ? ? ? ? ?
?
?
? ?
?
?
? ?
? ? ? ?
?
?
) , 0 ( 149 . 0
0 30
0 ) , 0 (
) 2634 . 1 (
) , 0 (
) , 0 (
) 0368 . 1 ( ) 4356 . 1 (
1
2 2
1
t T
t T
e e A
T T
T t T
t
i
wall
 
Air 
T ? = 0 ?C 
Margarine,  T i = 30 ?C 
Chapter 4 Transient Heat Conduction 
 4-101 
4-121 A cold cylindrical concrete column is exposed to warm ambient air during the day. The time it will 
take for the surface temperature to rise to a specified value, the amounts of heat transfer for specified values 
of center and surface temperatures are to be determined.  
Assumptions 1 Heat conduction in the column is one-dimensional since it is long and it has thermal 
symmetry about the center line. 2 The thermal properties of the column are constant. 3 The heat transfer 
coefficient is constant and uniform over the entire surface. 4 The Fourier number is ? > 0.2 so that the one-
term approximate solutions (or the transient temperature charts) are applicable (this assumption will be 
verified). 
Properties The properties of concrete are given to be k = 0.79 W/m. ?C, ? = 5.94 ?10
-7
 m
2
/s, ? = 1600 kg/m
3
 
and C p = 0.84 kJ/kg. ?C 
Analysis  (a) The Biot number is  
 658 . 2
) C W/m. 79 . 0 (
) m 15 . 0 )( C . W/m 14 (
2
?
?
?
? ?
k
hr
Bi
o
 
The constants ?
1 1
 and A corresponding to this 
Biot number are, from Table 4-1, 
  3915 . 1   and   7240 . 1
1 1
? ? ? A 
Once the constant J
0
=0.3841 is determined from Table 4-2 
corresponding to the constant ?
1
, the Fourier number is 
determined to be  
   6253 . 0 ) 3841 . 0 ( ) 3915 . 1 (
28 16
28 27
) / (
) , ( 2 2
1
) 7240 . 1 (
1 0 1
? ? ? ? ? ?
?
?
? ? ? ? ?
?
?
? ? ? ? ?
?
?
e r r J e A
T T
T t r T
o o
i
o
 
which is above the value of 0.2. Therefore, the one-term approximate solution (or the transient temperature 
charts) can be used. Then the time it will take for the column surface temperature to rise to 27 ?C becomes 
 hours 6.6 ? ?
?
?
?
?
?
?
s 685 , 23
/s) m 10 94 . 5 (
m) 15 . 0 )( 6253 . 0 (
2 7
2 2
o
r
t 
(b) The heat transfer to the column will stop when the center temperature of column reaches to the ambient 
temperature, which is 28 ?C. That is, we are asked to determine the maximum heat transfer between the 
ambient air and the column. 
 
kJ 3990 ? ? ? ? ? ? ?
? ? ? ?? ? ? ?
?
C ) 16 28 )( C kJ/kg. 84 . 0 )( kg 8 . 395 ( ] [
kg 8 . 395 )] m 5 . 3 ( m) 15 . 0 ( )[ kg/m 1600 (
max
2 3 2
i p
o
T T mC Q
L r V m
 
(c) To determine the amount of heat transfer until the surface temperature reaches to 27 ?C, we first 
determine  
2169 . 0 ) 3915 . 1 (
) , 0 (
) 6253 . 0 ( ) 7240 . 1 (
1
2 2
1
? ? ?
?
?
? ? ? ?
?
?
e e A
T T
T t T
i
 
Once the constant J 1 = 0.5787 is determined from Table 4-2 corresponding to the constant ?
1
, the amount 
of heat transfer becomes 
 
kJ 3409 ? ?
?
? ? ? ? ?
?
?
?
?
?
?
?
?
?
?
?
?
? ?
?
?
?
?
?
?
?
?
?
?
) kJ 3990 ( 854 . 0
0854
854 . 0
7240 . 1
5787 . 0
2169 . 0 2 1
) (
2 1
max
1
1 1 0
cy l
max
Q
Q Q
J
T T
T T
Q
Q
i
 
30 cm 
Column 
16 ?C 
Air  
28 ?C 
Page 5


Chapter 4 Transient Heat Conduction 
 4-98 
4-118 Internal combustion engine valves are quenched in a large oil bath. The time it takes for the valve 
temperature to drop to specified temperatures and the maximum heat transfer are to be determined. 
Assumptions 1 The thermal properties of the valves are constant. 2 The heat transfer coefficient is constant 
and uniform over the entire surface. 3 Depending on the size of the oil bath, the oil bath temperature will 
increase during quenching. However, an average canstant temperature as specified in the problem will be 
used.  4 The Biot number is Bi < 0.1 so that the lumped system analysis is applicable (this assumption will 
be verified). 
Properties The thermal conductivity, density, and 
specific heat of the balls are given to be k = 48 W/m. ?C, 
? = 7840 kg/m
3
, and C p = 440 J/kg. ?C. 
Analysis (a) The characteristic length of the balls and 
the Biot number are 
1 . 0 024 . 0
) C W/m. 48 (
) m 0018 . 0 )( C . W/m 650 (
m 0018 . 0
8
m) 008 . 0 ( 8 . 1
8
8 . 1
2
) 4 / ( 8 . 1
2
2
? ?
?
?
? ?
? ? ? ? ?
k
hL
Bi
D
DL
L D
A
V
L
c
s
c
?
?
 
Therefore, we can use lumped system analysis. Then the time for a final valve temperature of 400 ?C 
becomes  
 
s 7.2 ? ? ? ? ?
?
?
? ? ? ?
?
?
?
?
?
? ? ?
? ?
?
?
t e e
T T
T t T
D C
h
V C
hA
b
bt
i
p p
s
)t s 10468 . 0 (
1 -
3
2
-1
45 800
45 400 ) (
s 10468 . 0
m) C)(0.008 J/kg. 440 )( kg/m 1.8(7840
C) . W/m 650 ( 8
8 . 1
8
? ?
 
(b) The time for a final valve temperature of 200 ?C is  
s 15.1 ? ? ? ? ?
?
?
? ? ? ?
?
?
? ?
?
?
t e e
T T
T t T
bt
i
)t s 10468 . 0 (
-1
45 800
45 200 ) (
 
(c) The time for a final valve temperature of 46 ?C is  
s 63.3 ? ? ? ? ?
?
?
? ? ? ?
?
?
? ?
?
?
t e e
T T
T t T
bt
i
)t s 10468 . 0 (
-1
45 800
45 46 ) (
 
(d) The maximum amount of heat transfer from a single valve is determined from 
 
) (per valve  = J 564 , 23 C ) 45 800 )( C J/kg. 440 )( kg 0709 . 0 ( ] [
kg 0709 . 0
4
m) 10 . 0 ( m) 008 . 0 ( 8 . 1
) kg/m 7840 (
4
8 . 1
2
3
2
kJ 23.56 ? ? ? ? ? ? ?
?
?
?
?
? ? ? ?
i f p
T T mC Q
L D
V m
 
Oil 
T ? = 45 ?C 
Engine valve 
T i = 800 ?C 
Chapter 4 Transient Heat Conduction 
 4-99 
4-119 A watermelon is placed into a lake to cool it. The heat transfer coefficient at the surface of the 
watermelon and the temperature of the outer surface of the watermelon are to be determined. 
Assumptions 1 The watermelon is a homogeneous spherical object. 2 Heat conduction in the watermelon is 
one-dimensional because of symmetry about the midpoint. 3 The thermal properties of the watermelon are 
constant. 4 The heat transfer coefficient is constant and uniform over the entire surface. 5 The Fourier 
number is ? > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are 
applicable (this assumption will be verified). 
Properties The properties of the watermelon are given to be k = 0.618 W/m. ?C, ? = 0.15 ?10
-6
 m
2
/s, ? = 995 
kg/m
3
 and C p = 4.18 kJ/kg. ?C. 
Analysis The Fourier number is 
  
? ?
252 . 0
m) 10 . 0 (
s/min 60 min)  40 60 (4 /s) m 10 15 . 0 (
2
2 6
2
?
? ? ? ?
?
?
? ?
?
o
r
t
 
which is greater than 0.2. Then the one-term solution can be 
written in the form 
   
) 252 . 0 (
1 1
0
sph 0,
2
1
2
1
25 . 0
15 35
15 20
? ? ? ? ?
?
?
? ?
?
?
? ? ? ?
?
?
? ? e A e A
T T
T T
i
 
It is determined from Table 4-1 by trial and error that this equation is satisfied when Bi = 10, which  
corresponds to   ?
1 1
2 8363 19249 ? ? . .   and   A . Then the heat transfer coefficient can be determined from 
 C . W/m 61.8
2
? ?
?
? ? ? ? ? ?
) m 10 . 0 (
) 10 )( C W/m. 618 . 0 (
o
o
r
kBi
h
k
hr
Bi 
The temperature at the surface of the watermelon is 
 
C 15.5 ? ? ? ? ? ?
?
?
?
?
?
?
?
?
? ?
? ? ? ?
?
?
) , ( 0269 . 0
15 35
15 ) , (
8363 . 2
) rad 8363 . 2 sin(
) 9249 . 1 (
/
) / sin( ) , (
) , (
) 252 . 0 ( ) 8363 . 2 (
1
1
1
2 2
1
t r T
t r T
e
r r
r r
e A
T T
T t r T
t r
o
o
o o
o o
i
o
sph o
 
Water 
melon 
Ti = 35 ?C 
Lake 
15 ?C 
Chapter 4 Transient Heat Conduction 
 4-100 
4-120 Large food slabs are cooled in a refrigeration room. Center temperatures are to be determined for 
different foods. 
Assumptions 1 Heat conduction in the slabs is one-dimensional since the slab is large relative to its 
thickness and there is thermal symmetry about the center plane. 3 The thermal properties of the slabs are 
constant. 4 The heat transfer coefficient is constant and uniform over the entire surface.  5 The Fourier 
number is ? > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are 
applicable (this assumption will be verified). 
Properties The properties of foods are given to be k = 0.233 W/m. ?C and ? = 0.11 ?10
-6
 m
2
/s for margarine, 
k = 0.082 W/m. ?C and ? = 0.10 ?10
-6
 m
2
/s for white cake, and k = 0.106 W/m. ?C and ? = 0.12 ?10
-6
 m
2
/s for 
chocolate cake. 
Analysis (a) In the case of margarine, the Biot number is  
365 . 5
) C W/m. 233 . 0 (
) m 05 . 0 )( C . W/m 25 (
2
?
?
?
? ?
k
hL
Bi 
The constants ?
1 1
 and A corresponding to this Biot number are, 
from Table 4-1,  
 2431 . 1 and      3269 . 1
1 1
? ? A ? 
The Fourier number is            2 . 0 9504 . 0
m) 05 . 0 (
s/h) 600 3 h /s)(6 m 10 11 . 0 (
2
2 6
2
? ?
? ?
? ?
?
L
t ?
? 
Therefore, the one-term approximate solution (or the transient temperature charts) is applicable. Then the 
temperature at the center of the box if the box contains margarine becomes 
 
C 7.0 ? ? ? ? ? ?
?
?
? ?
?
?
? ?
? ? ? ?
?
?
) , 0 ( 233 . 0
0 30
0 ) , 0 (
) 2431 . 1 (
) , 0 (
) , 0 (
) 9504 . 0 ( ) 3269 . 1 (
1
2 2
1
t T
t T
e e A
T T
T t T
t
i
wall
 
(b) Repeating the calculations for white cake,  
 ? ? ? ?
?
?
? ? 24 . 15
) C W/m. 082 . 0 (
) m 05 . 0 )( C . W/m 25 (
2
k
hL
Bi 2661 . 1 and      4641 . 1
1 1
? ? A ?   
 2 . 0 864 . 0
m) 05 . 0 (
s/h) 600 3 h /s)(6 m 10 10 . 0 (
2
2 6
2
? ?
? ?
? ?
?
L
t ?
? 
 
C 6.0 ? ? ? ? ? ?
?
?
? ?
?
?
? ?
? ? ? ?
?
?
) , 0 ( 199 . 0
0 30
0 ) , 0 (
) 2661 . 1 (
) , 0 (
) , 0 (
) 864 . 0 ( ) 4641 . 1 (
1
2 2
1
t T
t T
e e A
T T
T t T
t
i
wall
 
(c) Repeating the calculations for chocolate cake, 
 ? ? ? ?
?
?
? ? 79 . 11
) C W/m. 106 . 0 (
) m 05 . 0 )( C . W/m 25 (
2
k
hL
Bi 2634 . 1 and     4356 . 1
1 1
? ? A ? 
 2 . 0 0368 . 1
m) 05 . 0 (
s/h) 600 3 h /s)(6 m 10 12 . 0 (
2
2 6
2
? ?
? ?
? ?
?
L
t ?
? 
 
C 4.5 ? ? ? ? ? ?
?
?
? ?
?
?
? ?
? ? ? ?
?
?
) , 0 ( 149 . 0
0 30
0 ) , 0 (
) 2634 . 1 (
) , 0 (
) , 0 (
) 0368 . 1 ( ) 4356 . 1 (
1
2 2
1
t T
t T
e e A
T T
T t T
t
i
wall
 
Air 
T ? = 0 ?C 
Margarine,  T i = 30 ?C 
Chapter 4 Transient Heat Conduction 
 4-101 
4-121 A cold cylindrical concrete column is exposed to warm ambient air during the day. The time it will 
take for the surface temperature to rise to a specified value, the amounts of heat transfer for specified values 
of center and surface temperatures are to be determined.  
Assumptions 1 Heat conduction in the column is one-dimensional since it is long and it has thermal 
symmetry about the center line. 2 The thermal properties of the column are constant. 3 The heat transfer 
coefficient is constant and uniform over the entire surface. 4 The Fourier number is ? > 0.2 so that the one-
term approximate solutions (or the transient temperature charts) are applicable (this assumption will be 
verified). 
Properties The properties of concrete are given to be k = 0.79 W/m. ?C, ? = 5.94 ?10
-7
 m
2
/s, ? = 1600 kg/m
3
 
and C p = 0.84 kJ/kg. ?C 
Analysis  (a) The Biot number is  
 658 . 2
) C W/m. 79 . 0 (
) m 15 . 0 )( C . W/m 14 (
2
?
?
?
? ?
k
hr
Bi
o
 
The constants ?
1 1
 and A corresponding to this 
Biot number are, from Table 4-1, 
  3915 . 1   and   7240 . 1
1 1
? ? ? A 
Once the constant J
0
=0.3841 is determined from Table 4-2 
corresponding to the constant ?
1
, the Fourier number is 
determined to be  
   6253 . 0 ) 3841 . 0 ( ) 3915 . 1 (
28 16
28 27
) / (
) , ( 2 2
1
) 7240 . 1 (
1 0 1
? ? ? ? ? ?
?
?
? ? ? ? ?
?
?
? ? ? ? ?
?
?
e r r J e A
T T
T t r T
o o
i
o
 
which is above the value of 0.2. Therefore, the one-term approximate solution (or the transient temperature 
charts) can be used. Then the time it will take for the column surface temperature to rise to 27 ?C becomes 
 hours 6.6 ? ?
?
?
?
?
?
?
s 685 , 23
/s) m 10 94 . 5 (
m) 15 . 0 )( 6253 . 0 (
2 7
2 2
o
r
t 
(b) The heat transfer to the column will stop when the center temperature of column reaches to the ambient 
temperature, which is 28 ?C. That is, we are asked to determine the maximum heat transfer between the 
ambient air and the column. 
 
kJ 3990 ? ? ? ? ? ? ?
? ? ? ?? ? ? ?
?
C ) 16 28 )( C kJ/kg. 84 . 0 )( kg 8 . 395 ( ] [
kg 8 . 395 )] m 5 . 3 ( m) 15 . 0 ( )[ kg/m 1600 (
max
2 3 2
i p
o
T T mC Q
L r V m
 
(c) To determine the amount of heat transfer until the surface temperature reaches to 27 ?C, we first 
determine  
2169 . 0 ) 3915 . 1 (
) , 0 (
) 6253 . 0 ( ) 7240 . 1 (
1
2 2
1
? ? ?
?
?
? ? ? ?
?
?
e e A
T T
T t T
i
 
Once the constant J 1 = 0.5787 is determined from Table 4-2 corresponding to the constant ?
1
, the amount 
of heat transfer becomes 
 
kJ 3409 ? ?
?
? ? ? ? ?
?
?
?
?
?
?
?
?
?
?
?
?
? ?
?
?
?
?
?
?
?
?
?
?
) kJ 3990 ( 854 . 0
0854
854 . 0
7240 . 1
5787 . 0
2169 . 0 2 1
) (
2 1
max
1
1 1 0
cy l
max
Q
Q Q
J
T T
T T
Q
Q
i
 
30 cm 
Column 
16 ?C 
Air  
28 ?C 
Chapter 4 Transient Heat Conduction 
 4-102 
4-122 Long aluminum wires are extruded and exposed to atmospheric air. The time it will take for the wire 
to cool, the distance the wire travels, and the rate of heat transfer from the wire are to be determined. 
Assumptions 1 Heat conduction in the wires is one-dimensional in the radial direction. 2 The thermal 
properties of the aluminum are constant. 3 The heat transfer coefficient is constant and uniform over the 
entire surface. 4 The Biot number is Bi < 0.1 so that the lumped system analysis is applicable (this 
assumption will be verified). 
Properties The properties of aluminum are given to be k = 236 W/m. ?C, ? = 2702 kg/m
3
, C p = 0.896 
kJ/kg. ?C, and ? = 9.75 ?10
-5
 m
2
/s. 
Analysis (a) The characteristic length of the wire and 
the Biot number are 
   
1 . 0 00011 . 0
C W/m. 236
) m 00075 . 0 )( C . W/m 35 (
m 00075 . 0
2
m 0015 . 0
2 2
2
2
? ?
?
?
? ?
? ? ? ? ?
k
hL
Bi
r
L r
L r
A
V
L
c
o
o
o
s
c
?
?
 
Since Bi < 0.1, the lumped system analysis is applicable. Then, 
 
s 144 ? ? ? ? ?
?
?
? ? ? ?
?
?
?
?
?
? ? ?
? ?
?
?
t e e
T T
T t T
L C
h
V C
hA
b
t bt
i
c p p
s
) s 0193 . 0 (
1 -
3
2
-1
30 350
30 50 ) (
s 0193 . 0
m) C)(0.00075 J/kg. 896 )( kg/m (2702
C . W/m 35
? ?
 
(b) The wire travels a distance of 
 m 24 ? ? ? ? s) m/s)(144 60 / 10 ( length
time
length
velocity 
This distance can be reduced by cooling the wire in a water or oil bath. 
(c) The mass flow rate of the extruded wire through the air is 
 kg/min 191 . 0 m/min) 10 ( m) 0015 . 0 ( ) kg/m 2702 ( ) 4 / (
2 3 2
0
? ? ? ? ? ? ? ? V r V m
?
? 
Then the rate of heat transfer from the wire to the air becomes 
 W 856 = kJ/min 51.3 = C ) 50 350 )( C kJ/kg. 896 . 0 )( kg/min 191 . 0 ( ] ) ( [ ? ? ? ? ? ?
?
T t T C m Q
p
?
?
 
350 ?C 
10 m/min 
Air 
30 ?C 
Aluminum wire 
 
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