TRANSIENT HEAT CONDUCTION

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Chapter 4 Transient Heat Conduction 
Chapter 4 
TRANSIENT HEAT CONDUCTION 
 
Lumped System Analysis 
 
4-1C In heat transfer analysis, some bodies are observed to behave like a "lump" whose entire body 
temperature remains essentially uniform at all times during a heat transfer process. The temperature of such 
bodies can be taken to be a function of time only. Heat transfer analysis which utilizes this idealization is 
known as the lumped system analysis. It is applicable when the Biot number (the ratio of conduction 
resistance within the body to convection resistance at the surface of the body) is less than or equal to 0.1. 
 
4-2C The lumped system analysis is more likely to be applicable for the body cooled naturally since the 
Biot number is proportional to the convection heat transfer coefficient, which is proportional to the air 
velocity. Therefore, the Biot number is more likely to be less than 0.1 for the case of natural convection. 
 
4-3C The lumped system analysis is more likely to be applicable for the body allowed to cool in the air 
since the Biot number is proportional to the convection heat transfer coefficient, which is larger in water 
than it is in air because of the larger thermal conductivity of water. Therefore, the Biot number is more 
likely to be less than 0.1 for the case of the solid cooled in the air 
 
4-4C The temperature drop of the potato during the second minute will be less than 4 ° since the 
temperature of a body approaches the temperature of the surrounding medium asymptotically, and thus it 
changes rapidly at the beginning, but slowly later on. 
C
 
4-5C The temperature rise of the potato during the second minute will be less than 5 since the 
temperature of a body approaches the temperature of the surrounding medium asymptotically, and thus it 
changes rapidly at the beginning, but slowly later on. 
°C
 
4-6C Biot number represents the ratio of conduction resistance within the body to convection resistance at 
the surface of the body. The Biot number is more likely to be larger for poorly conducting solids since such 
bodies have larger resistances against heat conduction. 
 
4-7C The heat transfer is proportional to the surface area. Two half pieces of the roast have a much larger 
surface area than the single piece and thus a higher rate of heat transfer. As a result, the two half pieces will 
cook much faster than the single large piece. 
 
4-8C The cylinder will cool faster than the sphere since heat transfer rate is proportional to the surface 
area, and the sphere has the smallest area for a given volume. 
 
4-9C The lumped system analysis is more likely to be applicable in air than in water since the convection 
heat transfer coefficient and thus the Biot number is much smaller in air. 
 
4-10C The lumped system analysis is more likely to be applicable for a golden apple than for an actual 
apple since the thermal conductivity is much larger and thus the Biot number is much smaller for gold. 
 4-1 
 
Chapter 4 Transient Heat Conduction 
 
4-11C The lumped system analysis is more likely to be applicable to slender bodies than the well-rounded 
bodies since the characteristic length (ratio of volume to surface area) and thus the Biot number is much 
smaller for slender bodies. 
 
4-12 Relations are to be obtained for the characteristic lengths of a large plane wall of thickness 2L, a very 
long cylinder of radius and a sphere of radius r ro o
Analysis Relations for the characteristic lengths of a large plane wall of thickness 2L, a very long cylinder 
of radius and a sphere of radius are ro ro
 
L V
A
LA
A
L
L V
A
r h
r h
r
L V
A
r
r
r
c wall
c cylinder
o
o
o
c sphere
o
o
o
,
,
,
/
= = =
= = =
= = =
surface
surface
surface
2
2
2 2
4 3
4 3
2
3
2
π
π
π
π
 
2ro 2ro
2L 
 
 
 
 
 
4-13 A relation for the time period for a lumped system to reach the average temperature ( ) to 
be obtained. 
/ is 
/
T Ti + ∞ 2 
Analysis The relation for time period for a lumped system to reach the average temperature 
can be determined as 
( )T Ti + ∞ 2
 
b
0.693
b
2 ln ==⎯→⎯−=−
=⎯→⎯=−
−⎯→⎯=−
−+
⎯→⎯=−
− −−
∞
∞−
∞
∞∞−
∞
∞
tbt
ee
TT
TT
e
TT
T
TT
e
TT
TtT btbt
i
ibt
i
i
bt
i
2ln
2
1
)(2
2)(
 T∞
Ti
 4-2 
 
Chapter 4 Transient Heat Conduction 
4-14 The temperature of a gas stream is to be measured by a thermocouple. The time it takes to register 99 
percent of the initial ΔT is to be determined. 
Assumptions 1 The junction is spherical in shape with a diameter of D = 0.0012 m. 2 The thermal 
properties of the junction are constant. 3 The heat transfer coefficient is constant and uniform over the 
entire surface. 4 Radiation effects are negligible. 5 The Biot number is Bi < 0.1 so that the lumped system 
analysis is applicable (this assumption will be verified). 
Properties The properties of the junction are given to be k = °35 W / m. C, , and 
. 
ρ = 8500 kg / m3
Cp = °320 J / kg. C
Analysis The characteristic length of the junction and the Biot number are 
 
L V
A
D
D
D
Bi hL
k
c
c
= = = = =
= = ° ° = <
surface
2
 m
6
 m
 W / m . C m
 W / m. C
 
π
π
3
2
6
6
0 0012 0 0002
65 0 0002
35
0 00037 01
/ . .
( )( . )
( )
. .
 
Since , the lumped system analysis is applicable. 
Then the time period for the thermocouple to read 99% of the 
initial temperature difference is determined from 
 < 0.1Bi
 
T t T
T T
b hA
C V
h
C L
T t T
T T
e e t
i
p p c
i
bt t
( )
.
)(
.
( )
. ( . )
−
− =
= = = ° ° =
−
− = ⎯ →⎯ = ⎯ →⎯ =
∞
∞
∞
∞
− −
0 01
65
8500 320
01195
0 01 0 1195
ρ ρ
 W / m . C
( kg / m J / kg. C)(0.0002 m)
 s
2
3
-1
 s-1 38.5 s
 
Gas 
h, T∞ Junction
D 
T(t) 
 4-3 
 
Chapter 4 Transient Heat Conduction 
4-15E A number of brass balls are to be quenched in a water bath at a specified rate. The temperature of 
the balls after quenching and the rate at which heat needs to be removed from the water in order to keep its 
temperature constant are to be determined. 
Assumptions 1 The balls are spherical in shape with a radius of r0 = 1 in. 2 The thermal properties of the 
balls 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 thermal conductivity, density, and specific heat of the brass balls are given to be k = 64.1 
Btu/h.ft.°F, ρ = 532 lbm/ft3, and Cp = 0.092 Btu/lbm.°F. 
Analysis (a) The characteristic length and the 
Biot number for the brass balls are 
 
1.0 01820.0
)FBtu/h.ft. 1.64(
)ft 02778.0)(F.Btu/h.ft 42(
ft 02778.0
6
ft 12/2
6
6/
2
2
3
<=°
°==
=====
k
hL
Bi
D
D
D
A
VL
c
s
c π
π
 
Brass balls, 250°F 
Water bath, 120°F 
The lumped system analysis is applicable since Bi < 0.1. Then the temperature of the balls after quenching 
becomes 
 
F 166 °=⎯→⎯=−
−⎯→⎯=−
−
==°
°===
−−
∞
∞ )(
120250
120)()(
s 00858.0 h9.30
ft) F)(0.02778Btu/lbm. 092.0)(lbm/ft (532
F.Btu/h.ft 42
s) 120)(s 00858.0(
1-1-
3
2
-1
tTetTe
TT
TtT
LC
h
VC
hA
b
bt
i
cpp
s
ρρ
 
(b) The total amount of heat transfer from a ball during a 2-minute period is 
 
Btu 97.9F)166250(F)Btu/lbm. 092.0)(lbm 29.1()]([
lbm 290.1
6
ft) 12/2()lbm/ft 532(
6
3
3
3
=°−°=−=
====
tTTmCQ
DVm
ip
ππρρ
 
Then the rate of heat transfer from the ballsto the water becomes 
 Btu/min 1196=×== )Btu 97.9(balls/min) 120(ballballQnQtotal &&
Therefore, heat must be removed from the water at a rate of 1196 Btu/min in order to keep its temperature 
constant at 120 . °F
 4-4 
 
Chapter 4 Transient Heat Conduction 
4-16E A number of aluminum balls are to be quenched in a water bath at a specified rate. The temperature 
of balls after quenching and the rate at which heat needs to be removed from the water in order to keep its 
temperature constant are to be determined. 
Assumptions 1 The balls are spherical in shape with a radius of r0 = 1 in. 2 The thermal properties of the 
balls 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 thermal conductivity, density, and specific heat of the aluminum balls are k = 137 
Btu/h.ft.°F, ρ = 168 lbm/ft3, and Cp = 0.216 Btu/lbm.°F (Table A-3E). 
Analysis (a) The characteristic length and the 
Biot number for the aluminum balls are 
 
1.000852.0
)FBtu/h.ft. 137(
)ft 02778.0)(F.Btu/h.ft 42(
ft 02778.0
6
ft 12/2
6
6/
2
2
3
<=°
°==
=====
k
hL
Bi
D
D
D
A
VL
c
c π
π
 
Aluminum balls, 250°F
Water bath, 120°F 
The lumped system analysis is applicable since Bi < 0.1. Then the temperature of the balls after quenching 
becomes 
 
F152°=⎯→⎯=−
−⎯→⎯=−
−
==°
°===
−−
∞
∞ )(
120250
120)()(
s 01157.0 h66.41
ft) F)(0.02778Btu/lbm. 216.0)(lbm/ft (168
F.Btu/h.ft 42
s) 120)(s 01157.0(
1-1-
3
2
-1
tTetTe
TT
TtT
LC
h
VC
hA
b
bt
i
cpp
s
ρρ
 
(b) The total amount of heat transfer from a ball during a 2-minute period is 
 
Btu 62.8F)152250(F)Btu/lbm. 216.0)(lbm 4072.0()]([
lbm 4072.0
6
ft) 12/2()lbm/ft 168(
6
3
3
3
=°−°=−=
====
tTTmCQ
DVm
ip
ππρρ
 
Then the rate of heat transfer from the balls to the water becomes 
 Btu/min 1034=×== )Btu 62.8(balls/min) 120(ballballQnQtotal &&
Therefore, heat must be removed from the water at a rate of 1034 Btu/min in order to keep its temperature 
constant at 120 . °F
 4-5 
 
Chapter 4 Transient Heat Conduction 
4-17 Milk in a thin-walled glass container is to be warmed up by placing it into a large pan filled with hot 
water. The warming time of the milk is to be determined. 
Assumptions 1 The glass container is cylindrical in shape with a 
radius of r0 = 3 cm. 2 The thermal properties of the milk are taken to 
be the same as those of water. 3 Thermal properties of the milk are 
constant at room temperature. 4 The heat transfer coefficient is 
constant and uniform over the entire surface. 5 The Biot number in 
this case is large (much larger than 0.1). However, the lumped system 
analysis is still applicable since the milk is stirred constantly, so that 
its temperature remains uniform at all times. 
Water
60°C 
Milk 
3°C 
Properties The thermal conductivity, density, and specific heat of the 
milk at 20°C are k = 0.607 W/m.°C, ρ = 998 kg/m3, and Cp = 4.182 
kJ/kg.°C (Table A-9). 
Analysis The characteristic length and Biot number for the glass of milk are 
1.0> 076.2
)CW/m. 607.0(
)m 0105.0)(C.W/m 120(
m 01050.0
m) 03.0(2+m) m)(0.07 03.0(2
m) 07.0(m) 03.0(
22
2
2
2
2
2
=°
°==
==+==
k
hL
Bi
rLr
Lr
A
VL
c
oo
o
s
c ππ
π
ππ
π
 
For the reason explained above we can use the lumped system analysis to determine how long it will take 
for the milk to warm up to 38°C: 
min 5.8s 348 ==⎯→⎯=−
−⎯→⎯=−
−
=°
°===
−−
∞
∞ tee
TT
TtT
LC
h
VC
hA
b
tbt
i
cpp
s
)s 002738.0(
1-
3
2
-1
603
6038)(
s 002738.0
m) C)(0.0105J/kg. 4182)( kg/m(998
C.W/m 120
ρρ
 
Therefore, it will take about 6 minutes to warm the milk from 3 to 38°C. 
 4-6 
 
Chapter 4 Transient Heat Conduction 
4-18 A thin-walled glass containing milk is placed into a large pan filled with hot water to warm up the 
milk. The warming time of the milk is to be determined. 
Assumptions 1 The glass container is cylindrical in shape with a 
radius of r0 = 3 cm. 2 The thermal properties of the milk are taken to 
be the same as those of water. 3 Thermal properties of the milk are 
constant at room temperature. 4 The heat transfer coefficient is 
constant and uniform over the entire surface. 5 The Biot number in 
this case is large (much larger than 0.1). However, the lumped system 
analysis is still applicable since the milk is stirred constantly, so that 
its temperature remains uniform at all times. 
Water
60°C 
Milk 
3°C 
Properties The thermal conductivity, density, and specific heat of the 
milk at 20°C are k = 0.607 W/m.°C, ρ = 998 kg/m3, and Cp = 4.182 
kJ/kg.°C (Table A-9). 
Analysis The characteristic length and Biot number for the glass of milk are 
1.0> 15.4
)CW/m. 607.0(
)m 0105.0)(C.W/m 240(
m 01050.0
m) 03.0(2+m) m)(0.07 03.0(2
m) 07.0(m) 03.0(
22
2
2
2
2
2
=°
°==
==+==
k
hL
Bi
rLr
Lr
A
VL
c
oo
o
s
c ππ
π
ππ
π
 
For the reason explained above we can use the lumped system analysis to determine how long it will take 
for the milk to warm up to 38°C: 
 
min 2.9s 174 ==⎯→⎯=−
−⎯→⎯=−
−
=°
°===
−−
∞
∞ tee
TT
TtT
LC
h
VC
hA
b
tbt
i
cpp
s
)s 005477.0(
1-
3
2
-1
603
6038)(
s 005477.0
m) C)(0.0105J/kg. 4182)( kg/m(998
C.W/m 240
ρρ
 
Therefore, it will take about 3 minutes to warm the milk from 3 to 38°C. 
 4-7 
 
Chapter 4 Transient Heat Conduction 
4-19E A person shakes a can of drink in a iced water to cool it. The cooling time of the drink is to be 
determined. 
Assumptions 1 The can containing the drink is cylindrical in shape 
with a radius of r0 = 1.25 in. 2 The thermal properties of the milk 
are taken to be the same as those of water. 3 Thermal properties of 
the milk are constant at room temperature. 4 The heat transfer 
coefficient is constant and uniform over the entire surface. 5 The 
Biot number in this case is large (much larger than 0.1). However, 
the lumped system analysis is still applicable since the milk is 
stirred constantly, so that its temperature remains uniform at all 
times. 
Milk 
3°C 
Water
32°F 
Cola 
75°F 
Properties The density and specific heat of water at room 
temperature are ρ = 62.22 lbm/ft3, and Cp = 0.999 Btu/lbm.°F 
(Table A-9E). 
Analysis Application of lumped system analysis in this case gives 
ft 04167.0
ft) 12/25.1(2+ft) ft)(5/12 12/25.1(2
ft) 12/5(ft) 12/25.1(
22 2
2
2
2
==+== ππ
π
ππ
π
oo
o
s
c rLr
Lr
A
VL 
s 406=⎯→⎯=−
−⎯→⎯=−
−
==°
°===
−−
∞
∞ tee
TT
TtT
LC
h
VC
hA
b
tbt
i
cpp
s
)s 00322.0(
1-1-
3
2
-1
3280
3245)(
s 00322.0 h583.11
ft) F)(0.04167Btu/lbm. 999.0)(lbm/ft (62.22
F.Btu/h.ft 30
ρρ 
Therefore, it will take 7 minutes and 46 seconds to cool the canned drink to 45°F. 
 4-8 
 
Chapter 4 Transient Heat Conduction 
4-20 An iron whose base plate is made of an aluminum alloy is turned on. The time for the plate 
temperature to reach 140°C and whether it is realistic to assume the plate temperature to be uniform at all 
times are to be determined. 
Assumptions 1 85 percent of the heat generated in the resistance wires is transferred to the plate. 2 The 
thermal properties of the plate are constant. 3 The heat transfer coefficient is constant and uniform over the 
entire surface. 
Properties The density, specific heat, and thermal diffusivity of the aluminum alloy plate are given to be ρ 
= 2770 kg/m3, Cp = 875 kJ/kg.°C, and α = 7.3×10-5 m2/s. The thermal conductivityof the plate can be 
determined from α = k/(ρCp)= 177 W/m.°C (or it can be read from Table A-3). 
Analysis The mass of the iron's base plate is Air 
22°C
IRON 
1000 W 
 m V LA= = = =ρ ρ ( )( . )( . ) .2770 0 005 0 03 0 4155 kg / m m m kg3 2
Noting that only 85 percent of the heat generated is transferred to the 
plate, the rate of heat transfer to the iron's base plate is 
 & .Qin W W= × =0 85 1000 850
The temperature of the plate, and thus the rate of heat transfer from the 
plate, changes during the process. Using the average plate temperature, 
the average rate of heat loss from the plate is determined from 
 W21.2=C22
2
22140)m 03.0)(C. W/m12()( 22ave plate,loss °⎟⎠
⎞⎜⎝
⎛ −+°=−= ∞TThAQ& 
Energy balance on the plate can be expressed as 
 E E E Q t Q t E mC Tpin out plate in out plate plate − = → − = =Δ Δ Δ Δ& & Δ
Solving for Δt and substituting, 
 Δ Δt mC T
Q Q
p= −
° − °
−
plate
in out
= kg J / kg. C C
(850 21.2) J / s
= & &
( . )( )( )0 4155 875 140 22 51.8 s 
which is the time required for the plate temperature to reach 140 . To determine whether it is realistic to 
assume the plate temperature to be uniform at all times, we need to calculate the Biot number, 
°C
 
1.0<00034.0
)CW/m. 0.177(
)m 005.0)(C.W/m 12(
m 005.0
2
=°
°==
====
k
hL
Bi
L
A
LA
A
VL
c
s
c
 
It is realistic to assume uniform temperature for the plate since Bi < 0.1. 
Discussion This problem can also be solved by obtaining the differential equation from an energy balance 
on the plate for a differential time interval, and solving the differential equation. It gives 
⎟⎟⎠
⎞
⎜⎜⎝
⎛ −−+= ∞ )exp(1)( in tmC
hA
hA
Q
TtT
p
&
 
Substituting the known quantities and solving for t again gives 51.8 s. 
 4-9 
 
Chapter 4 Transient Heat Conduction 
4-21 "!PROBLEM 4-21" 
 
"GIVEN" 
E_dot=1000 "[W]" 
L=0.005 "[m]" 
A=0.03 "[m^2]" 
T_infinity=22 "[C]" 
T_i=T_infinity 
h=12 "[W/m^2-C], parameter to be varied" 
f_heat=0.85 
T_f=140 "[C], parameter to be varied" 
 
"PROPERTIES" 
rho=2770 "[kg/m^3]" 
C_p=875 "[J/kg-C]" 
alpha=7.3E-5 "[m^2/s]" 
 
"ANALYSIS" 
V=L*A 
m=rho*V 
Q_dot_in=f_heat*E_dot 
Q_dot_out=h*A*(T_ave-T_infinity) 
T_ave=1/2*(T_i+T_f) 
(Q_dot_in-Q_dot_out)*time=m*C_p*(T_f-T_i) "energy balance on the plate" 
 
h [W/m2.C] time [s] 
5 51 
7 51.22 
9 51.43 
11 51.65 
13 51.88 
15 52.1 
17 52.32 
19 52.55 
21 52.78 
23 53.01 
25 53.24 
 
Tf [C] time [s] 
30 3.428 
40 7.728 
50 12.05 
60 16.39 
70 20.74 
80 25.12 
90 29.51 
100 33.92 
110 38.35 
120 42.8 
130 47.28 
140 51.76 
150 56.27 
160 60.8 
170 65.35 
180 69.92 
190 74.51 
 4-10 
 
Chapter 4 Transient Heat Conduction 
200 79.12 
 
5 9 13 17 21 25
51
51.45
51.9
52.35
52.8
53.25
h [W/m2-C]
tim
e 
 [s
]
 
 
 
 
 
20 40 60 80 100 120 140 160 180 200
0
10
20
30
40
50
60
70
80
Tf [C]
tim
e 
 [s
]
 
 
 4-11 
 
Chapter 4 Transient Heat Conduction 
4-22 Ball bearings leaving the oven at a uniform temperature of 900°C are exposed to air for a while before 
they are dropped into the water for quenching. The time they can stand in the air before their temperature 
falls below 850°C is to be determined. 
Assumptions 1 The bearings are spherical in shape with a radius of r0 = 0.6 cm. 2 The thermal properties of 
the bearings 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 thermal conductivity, density, and specific heat of the bearings are given to be k = 15.1 
W/m.°C, ρ = 8085 kg/m3, and Cp = 0.480 kJ/kg.°F. 
Analysis The characteristic length of the steel ball bearings and Biot number are 
 
0.1< 0166.0
)CW/m. 1.15(
)m 002.0)(C.W/m 125(
m 002.0
6
m 012.0
6
6/
2
2
3
=°
°==
=====
k
hL
Bi
D
D
D
A
VL
c
s
c π
π
 Steel balls 
 900°C 
Air, 30°C Furnace 
Therefore, the lumped system analysis is applicable. 
Then the allowable time is determined to be 
 
s 3.68=⎯→⎯=−
−⎯→⎯=−
−
=°
°===
−−
∞
∞ tee
TT
TtT
LC
h
VC
hA
b
tbt
i
cpp
s
)s 0161.0(
1-
3
2
-1
30900
30850)(
s 01610.0
m) C)(0.002J/kg. 480)( kg/m8085(
C.W/m 125
ρρ
 
The result indicates that the ball bearing can stay in the air about 4 s before being dropped into the water. 
 4-12 
 
Chapter 4 Transient Heat Conduction 
4-23 A number of carbon steel balls are to be annealed by heating them first and then allowing them to 
cool slowly in ambient air at a specified rate. The time of annealing and the total rate of heat transfer from 
the balls to the ambient air are to be determined. 
Assumptions 1 The balls are spherical in shape with a radius of r0 = 4 mm. 2 The thermal properties of the 
balls 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 thermal conductivity, density, and specific heat of the balls are given to be k = 54 W/m.°C, 
ρ = 7833 kg/m3, and Cp = 0.465 kJ/kg.°C. 
Analysis The characteristic length of the balls and the Biot number are 
 
1.0 0018.0
)CW/m. 54(
)m 0013.0)(C.W/m 75(
m 0013.0
6
m 008.0
6
6/
2
2
3
<=°
°==
=====
k
hL
Bi
D
D
D
A
VL
c
s
c π
π
 Steel balls 
 900°C 
Air, 35°C Furnace 
Therefore, the lumped system analysis is applicable. 
Then the time for the annealing process is 
determined to be 
 
min 2.7s 163 ==⎯→⎯=−
−⎯→⎯=−
−
=°
°===
−−
∞
∞ tee
TT
TtT
LC
h
VC
hA
b
bt
i
cpp
s
)ts 01584.0(
1-
3
2
-1
35900
35100)(
s 01584.0
m) C)(0.0013J/kg. 465)( kg/m(7833
C.W/m 75
ρρ
 
The amount of heat transfer from a single ball is 
 
m V D
Q mC T Tp f i
= = = =
= − = ° − ° =
ρ ρ π π
3
6
7833 0 008 0 0021
0 0021 465 900 100
( ) ( . .
[ ] ( . )( )( )
 kg / m m)
6
 kg
 kg J / kg. C C 781 J = 0.781 kJ (per ball)
3
3
 
Then the total rate of heat transfer from the balls to the ambient air becomes 
 W543==×== kJ/h953,1) kJ/ball781.0( balls/h)2500(ballQnQ &&
 
 
 4-13 
 
Chapter 4 Transient Heat Conduction 
4-24 
"!PROBLEM 4-24" 
 
"GIVEN" 
D=0.008 "[m]" 
"T_i=900 [C], parameter to be varied" 
T_f=100 "[C]" 
T_infinity=35 "[C]" 
h=75 "[W/m^2-C]" 
n_dot_ball=2500 "[1/h]" 
 
"PROPERTIES" 
rho=7833 "[kg/m^3]" 
k=54 "[W/m-C]" 
C_p=465 "[J/kg-C]" 
alpha=1.474E-6 "[m^2/s]" 
 
"ANALYSIS" 
A=pi*D^2 
V=pi*D^3/6 
L_c=V/A 
Bi=(h*L_c)/k "if Bi < 0.1, the lumped sytem analysis is applicable" 
b=(h*A)/(rho*C_p*V) 
(T_f-T_infinity)/(T_i-T_infinity)=exp(-b*time) 
m=rho*V 
Q=m*C_p*(T_i-T_f) 
Q_dot=n_dot_ball*Q*Convert(J/h, W) 
 
 
 
 
Ti [C] time [s] Q [W] 
500 127.4 271.2 
550 134 305.1 
600 140 339 
650 145.5 372.9 
700 150.6 406.9 
750 155.3 440.8 
800 159.6 474.7 
850 163.7 508.6 
900 167.6 542.5 
950 171.2 576.4 
1000 174.7 610.3 
 
 
 
 
 4-14 
 
Chapter 4 Transient Heat Conduction 
500 600 700 800 900 1000
120
130
140
150
160
170
180
250
300
350
400
450
500
550
600
650
Ti [C]
tim
e 
 [s
]
Q
 [
W
]
time
heat
 
 
 4-15 
 
Chapter 4 Transient Heat Conduction 
4-25 An electronic device is on for 5 minutes, and off for several hours. The temperature of the device at 
the end of the 5-min operating period is to be determined for the cases of operation with and without a heat 
sink. 
Assumptions1 The device and the heat sink are isothermal. 2 The thermal properties of the device and of 
the sink are constant. 3 The heat transfer coefficient is constant and uniform over the entire surface. 
Properties The specific heat of the device is given to be Cp = 850 J/kg.°C. The specific heat of the 
aluminum sink is 903 J/kg.°C (Table A-19), but can be taken to be 850 J/kg.°C for simplicity in analysis. 
Analysis (a) Approximate solution 
This problem can be solved approximately by using an average temperature 
for the device when evaluating the heat loss. An energy balance on the 
device can be expressed as 
 E E E E Q t E t mC Tpin out generation device out generation device − + = ⎯ →⎯ − + =Δ Δ Δ& & Δ
Electronic 
device 
30 W 
or, )(
2generation ∞∞
∞ −=Δ⎟⎠
⎞⎜⎝
⎛ −+−Δ TTmCtTTThAtE ps& 
Substituting the given values, 
 C)25)(CJ/kg. 850)(kg 02.0()s 605(C
2
25)m 0005.0)(C. W/m12()s 605)(J/s 30( o22 °−°=×⎟⎠
⎞⎜⎝
⎛ −°−× TT 
which gives T = 527.8°C 
 If the device were attached to an aluminum heat sink, the temperature of the device would be 
C)25)(CJ/kg. 850(kg)02.020.0()s 605(C
2
25)m0085.0)(C. W/m12()s 605)(J/s 30( 22 °−°×+=×°⎟⎠
⎞⎜⎝
⎛ −°−× TT 
which gives T = 69.5°C 
Note that the temperature of the electronic device drops considerably as a result of attaching it to a heat 
sink. 
(b) Exact solution 
This problem can be solved exactly by obtaining the differential equation from an energy balance on the 
device for a differential time interval dt. We will get 
 
pp
s
mC
E
TT
mC
hA
dt
TTd generation)(
)( &=−+− ∞∞ 
It can be solved to give 
 ⎟⎟⎠
⎞
⎜⎜⎝
⎛ −−+= ∞ )exp(1)( generation tmC
hA
hA
E
TtT
p
s
s
&
 
Substituting the known quantities and solving for t gives 527.3°C for the first case and 69.4°C for the 
second case, which are practically identical to the results obtained from the approximate analysis. 
 
 4-16 
 
Chapter 4 Transient Heat Conduction 
 
Transient Heat Conduction in Large Plane Walls, Long Cylinders, and Spheres 
 
4-26C A cylinder whose diameter is small relative to its length can be treated as an infinitely long cylinder. 
When the diameter and length of the cylinder are comparable, it is not proper to treat the cylinder as being 
infinitely long. It is also not proper to use this model when finding the temperatures near the bottom or top 
surfaces of a cylinder since heat transfer at those locations can be two-dimensional. 
 
4-27C Yes. A plane wall whose one side is insulated is equivalent to a plane wall that is twice as thick and 
is exposed to convection from both sides. The midplane in the latter case will behave like an insulated 
surface because of thermal symmetry. 
 
4-28C The solution for determination of the one-dimensional transient temperature distribution involves 
many variables that make the graphical representation of the results impractical. In order to reduce the 
number of parameters, some variables are grouped into dimensionless quantities. 
 
4-29C The Fourier number is a measure of heat conducted through a body relative to the heat stored. Thus 
a large value of Fourier number indicates faster propagation of heat through body. Since Fourier number is 
proportional to time, doubling the time will also double the Fourier number. 
 
4-30C This case can be handled by setting the heat transfer coefficient h to infinity since the 
temperature of the surrounding medium in this case becomes equivalent to the surface temperature. 
∞
 
4-31C The maximum possible amount of heat transfer will occur when the temperature of the body reaches 
the temperature of the medium, and can be determined from Q mC T Tp imax ( )= −∞ . 
 
4-32C When the Biot number is less than 0.1, the temperature of the sphere will be nearly uniform at all 
times. Therefore, it is more convenient to use the lumped system analysis in this case. 
 
4-33 A student calculates the total heat transfer from a spherical copper ball. It is to be determined whether 
his/her result is reasonable. 
Assumptions The thermal properties of the copper ball are constant at room temperature. 
Properties The density and specific heat of the copper ball are ρ = 8933 kg/m3, and Cp = 0.385 kJ/kg.°C 
(Table A-3). 
Analysis The mass of the copper ball and the maximum amount of heat transfer from the copper ball are 
 
kJ 1064C)25200)(CkJ/kg. 385.0)(kg 79.15(][
kg 79.15
6
m) 15.0(
)kg/m 8933(
6
max
3
3
3
=°−°=−=
=⎥⎥⎦
⎤
⎢⎢⎣
⎡=⎟⎟⎠
⎞
⎜⎜⎝
⎛==
∞TTmCQ
DVm
ip
ππρρ
 
Copper 
ball, 200°C 
Q 
Discussion The student's result of 4520 kJ is not reasonable since it is 
greater than the maximum possible amount of heat transfer. 
 4-17
Chapter 4 Transient Heat Conduction 
4-34 An egg is dropped into boiling water. The cooking time of the egg is to be determined. √ 
Assumptions 1 The egg is spherical in shape with a radius of r0 = 2.75 cm. 2 Heat conduction in the egg is 
one-dimensional because of symmetry about the midpoint. 3 The thermal properties of the egg are 
constant. 4 The heat transfer coefficient is constant and uniform over the entire surface. 4 The Fourier 
number is τ > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are 
applicable (this assumption will be verified). 
Properties The thermal conductivity and diffusivity of the eggs are given to be k = 0.6 W/m.°C and α = 
0.14×10-6 m2/s. 
Analysis The Biot number for this process is 
 Bi hr
k
o= = ° ° =
( )( . )
( . )
.1400 0 0275
0 6
64 2 W / m . C m
 W / m. C
2
 
Water
97°C
Egg 
Ti = 8°C 
The constants λ1 and A1 corresponding to this Biot number 
are, from Table 4-1, 
 9969.1 and 0877.3 11 == Aλ 
Then the Fourier number becomes 
 2.0198.0)9969.1(
978
9770 22
1 )0877.3(
1
0
,0 ≈=⎯→⎯=−
−⎯→⎯=−
−= −−
∞
∞ τθ ττλ eeA
TT
TT
i
sph 
Therefore, the one-term approximate solution (or the transient temperature charts) is applicable. Then the 
time required for the temperature of the center of the egg to reach 70°C is determined to be 
 min 17.8==×=α
τ= − s 1068/s)m 1014.0(
m) 0275.0)(198.0(
26
22
ort 
 
 
 4-18
Chapter 4 Transient Heat Conduction 
4-35 
"!PROBLEM 4-35" 
"GIVEN" 
D=0.055 "[m]" 
T_i=8 "[C]" 
"T_o=70 [C], parameter to be varied" 
T_infinity=97 "[C]" 
h=1400 "[W/m^2-C]" 
 
"PROPERTIES" 
k=0.6 "[W/m-C]" 
alpha=0.14E-6 "[m^2/s]" 
 
"ANALYSIS" 
Bi=(h*r_o)/k 
r_o=D/2 
"From Table 4-1 corresponding to this Bi number, we read" 
lambda_1=1.9969 
A_1=3.0863 
(T_o-T_infinity)/(T_i-T_infinity)=A_1*exp(-lambda_1^2*tau) 
time=(tau*r_o^2)/alpha*Convert(s, min) 
 
To [C] time [min] 
50 39.86 
55 42.4 
60 45.26 
65 48.54 
70 52.38 
75 57 
80 62.82 
85 70.68 
90 82.85 
95 111.1 
 
50 55 60 65 70 75 80 85 90 95
30
40
50
60
70
80
90
100
110
120
To [C]
tim
e 
 [m
in
]
 
 
 4-19
Chapter 4 Transient Heat Conduction 
4-36 Large brass plates are heated in an oven. The surface temperature of the plates leaving the oven is to 
be determined. 
Assumptions 1 Heat conduction in the plate is one-dimensional since the plate is large relative to its 
thickness and there is thermal symmetry about the center plane. 3 The thermal properties of the plate are 
constant. 4 The heat transfer coefficient is constant and uniform over the entire surface. 5 The Fourier 
number is τ > 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 brass at room temperature are given to be k = 110 W/m.°C, α = 33.9×10-6 m2/s 
Analysis The Biot number for this process is 
 Bi hL
k
= = ° ° =
(80 )( . )
( )
. W / m . C m
 W / m. C
2 0015
110
0 0109 
The constants λ1 and A1 corresponding to this Biot 
number are, from Table 4-1, 
 λ1 101039 10018= =. . and A 
The Fourier number is 
 τ α= = × × = >
−t
L2
6339 10 10
0 015
90 4 0 2( .
( .
. . m / s)( min 60 s / min)
 m)
2
2 
Plates 
25°C
Furnace, 
700°C 
Therefore, the one-term approximate solution (or the transient temperature charts) is applicable. Then the 
temperature at the surface of the plates becomes 
 
C 445 °=⎯→⎯=−
−
==λ=−
−=θ −τλ−
∞
∞
),(378.0
70025
700),(
378.0)1039.0cos()0018.1()/cos(
),(
),( )4.90()1039.0(11
22
1
tLTtLT
eLLeA
TT
TtxT
tL
i
wall
 
 
Discussion This problem can be solved easily using the lumped system analysis since Bi < 0.1, and thus 
the lumped system analysis is applicable. It gives 
C 448 °=°+°=−+=→=−
−
=°⋅⋅×
°⋅=====
°⋅⋅×=×
°⋅==→=
−−∞∞−
∞
∞ s) 600)(s 001644.0(
1-
36
2
36
26-
1-
C)700-(25C700)()( 
)(
s 001644.0
C)s/m W10245.3m)( 015.0(
C W/m80
)/()(
Cs/m W10245.3
/m 1033.9
C W/m110 
eeTTTtTe
TT
TtT
kL
h
LC
h
CLA
hA
VC
hAb
s
kC
C
k
bt
i
bt
i
ppp
p
p
αρρρ
αρρα
 
which is almost identical to the result obtained above. 
 
 4-20
Chapter 4 Transient Heat Conduction 
4-37 "!PROBLEM 4-37" 
 
"GIVEN" 
L=0.03/2 "[m]" 
T_i=25 "[C]" 
T_infinity=700 "[C], parameter to be varied" 
time=10 "[min], parameter to be varied" 
h=80 "[W/m^2-C]" 
 
"PROPERTIES" 
k=110 "[W/m-C]" 
alpha=33.9E-6 "[m^2/s]" 
 
"ANALYSIS" 
Bi=(h*L)/k 
"From Table 4-1, corresponding to this Bi number, we read" 
lambda_1=0.1039 
A_1=1.0018 
tau=(alpha*time*Convert(min, s))/L^2 
(T_L-T_infinity)/(T_i-T_infinity)=A_1*exp(-lambda_1^2*tau)*Cos(lambda_1*L/L) 
 
T∞ [C] TL [C] 
500 321.6 
525 337.2 
550 352.9 
575 368.5 
600 384.1 
625 399.7 
650 415.3 
675 430.9 
700 446.5 
725 462.1 
750 477.8 
775 493.4 
800 509 
825 524.6 
850 540.2 
875 555.8 
900 571.4 
 
time [min] TL [C] 
2 146.7 
4 244.8 
6 325.5 
8 391.9 
10 446.5 
12 491.5 
14 528.5 
16 558.9 
18 583.9 
20 604.5 
22 621.4 
24 635.4 
26 646.8 
28 656.2 
 4-21
Chapter 4 Transient Heat Conduction 
30 664 
 
500 550 600 650 700 750 800 850 900
300
350
400
450
500
550
600
T∞ [C]
T L
 [
C
]
 
 
 
 
0 5 10 15 20 25 30
100
200
300
400
500
600
700
time [min]
T L
 [
C
]
 
 
 
 
 4-22
Chapter 4 Transient Heat Conduction 
4-38 A long cylindrical shaft at 400°C is allowed to cool slowly. The center temperature and the heat 
transfer per unit length of the cylinder are to be determined. 
Assumptions 1 Heat conduction in the shaft is one-dimensional since it is long and it has thermal symmetry 
about the center line. 2 The thermal properties of the shaft are constant. 3 The heat transfer coefficient is 
constant and uniform over the entire surface. 4 The Fourier number is τ > 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 stainless steel 304 at room temperature are given to be k = 14.9 W/m.°C, ρ = 
7900 kg/m3, Cp = 477 J/kg.°C, α = 3.95×10-6 m2/s 
Analysis First the Biot number is calculated to be 
 705.0
)CW/m. 9.14(
)m 175.0)(C.W/m 60( 2 =°
°==
k
hr
Bi o 
Steel shaft 
Ti = 400°C 
Air 
T∞ = 150°C 
The constants λ1 and A1 corresponding to this 
Biot number are, from Table 4-1, 
 λ1 110935 11558= =. . and A 
The Fourier number is 
 1548.0
m) 175.0(
s) 60/s)(20m 1095.3(
2
26
2 =
××==
−
L
tατ 
which is very close to the value of 0.2. Therefore, the one-term approximate solution (or the transient 
temperature charts) can still be used, with the understanding that the error involved will be a little more 
than 2 percent. Then the temperature at the center of the shaft becomes 
 
C 390 °=⎯→⎯=−
−
===−
−= −−
∞
∞
0
0
)1548.0()0935.1(
1
0
,0
9605.0
150400
150
9605.0)1558.1(
22
1
T
T
eeA
TT
TT
i
cyl
τλθ
 
The maximum heat can be transferred from the cylinder per meter of its length is 
 
kJ 638,90C)150400)(CkJ/kg. 477.0)(kg 1.760(][
kg 1.760)]m 1(m) 175.0()[kg/m 7900(
max
232
=°−°=−=
=π=ρπ=ρ=
∞ ip
o
TTmCQ
LrVm
Once the constant = 0.4689 is determined from Table 4-2 corresponding to the constant J1 λ1 =1.0935, the 
actual heat transfer becomes 
 
kJ 16,015==
=⎟⎠
⎞⎜⎝
⎛
−
−−=λ
λ
⎟⎟⎠
⎞
⎜⎜⎝
⎛
−
−−=⎟⎟⎠
⎞
⎜⎜⎝
⎛
∞
∞
)kJ 638,90(177.0
177.0
0935.1
4689.0
150400
15039021
)(
21
1
110
max
Q
J
TT
TT
Q
Q
icyl
 
 
 
 4-23
Chapter 4 Transient Heat Conduction 
4-39 
"!PROBLEM 4-39" 
 
"GIVEN" 
r_o=0.35/2 "[m]" 
T_i=400 "[C]" 
T_infinity=150 "[C]" 
h=60 "[W/m^2-C]" 
"time=20 [min], parameter to be varied" 
 
"PROPERTIES" 
k=14.9 "[W/m-C]" 
rho=7900 "[kg/m^3]" 
C_p=477 "[J/kg-C]" 
alpha=3.95E-6 "[m^2/s]" 
 
"ANALYSIS" 
Bi=(h*r_o)/k 
"From Table 4-1 corresponding to this Bi number, we read" 
lambda_1=1.0935 
A_1=1.1558 
J_1=0.4709 "From Table 4-2, corresponding to lambda_1" 
tau=(alpha*time*Convert(min, s))/r_o^2 
(T_o-T_infinity)/(T_i-T_infinity)=A_1*exp(-lambda_1^2*tau) 
 
L=1 "[m], 1 m length of the cylinder is considered" 
V=pi*r_o^2*L 
m=rho*V 
Q_max=m*C_p*(T_i-T_infinity)*Convert(J, kJ) 
Q/Q_max=1-2*(T_o-T_infinity)/(T_i-T_infinity)*J_1/lambda_1 
 
 
 
 
time [min] To [C] Q [kJ] 
5 425.9 4491 
10 413.4 8386 
15 401.5 12105 
20 390.1 15656 
25 379.3 19046 
30 368.9 22283 
35 359 25374 
40 349.6 28325 
45 340.5 31142 
50 331.9 33832 
55 323.7 36401 
60 315.8 38853 
 
 
 
 4-24
Chapter 4 Transient Heat Conduction 
0 10 20 30 40 50 60
300
320
340
360
380
400
420
440
0
5000
10000
15000
20000
25000
30000
35000
40000
time [min]
T o
 [
C
]
Q
 [
kJ
]
temperature
heat
 
 
 4-25
Chapter 4 Transient Heat Conduction 
4-40E Long cylindrical steel rods are heat-treated in an oven. Their centerline temperature when they leave 
the oven is to be determined. 
Assumptions 1 Heat conduction in the rods is one-dimensional since the rods are long and they have 
thermal symmetry about the center line. 2 The thermal properties of the rod are constant. 3 The heat 
transfer coefficient is constant and uniform over the entire surface. 4 The Fourier number is τ > 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 AISI stainless steel rods are given to be k = 7.74 Btu/h.ft.°F, α = 0.135 ft2/h. 
Analysis The time the steel rods stays in the oven can be determined from 
 t = = =length
velocity
 ft
 ft / min
 min = 180 s30
10
3 
Oven, 1700°F 
 
Steel rod, 85°F 
 
 The Biot number is 
 4307.0
)FBtu/h.ft. 74.7(
)ft 12/2)(F.Btu/h.ft 20( 2 =°
°==
k
hr
Bi o 
The constants λ1 and A1 corresponding to this Biot number are, from Table 4-1, 
 λ1 10 8784 10995= =. . and A 
The Fourier number is 
 τ α= = =t
ro
2
0135
2 12
0 243( .
( /
. ft / h)(3 / 60 h)
 ft)
2
2 
Then the temperature at the center of the rods becomes 
 θ λ τ0 0 1 0 8784 0 24312 210995 0 912, ( . ) ( . )( . ) .cyl
i
T T
T T
A e e= −− = = =
∞
∞
− − 
 F228°=⎯→⎯=−
−
0
0 912.0
170085
1700
T
T
 
 
 4-26
Chapter 4 Transient Heat Conduction 
4-41 Steaks are cooled by passing them through a refrigeration room. The time of cooling is to be 
determined. 
Assumptions 1 Heat conduction in the steaks is one-dimensional since the steaks arelarge relative to their 
thickness and there is thermal symmetry about the center plane. 3 The thermal properties of the steaks are 
constant. 4 The heat transfer coefficient is constant and uniform over the entire surface. 5 The Fourier 
number is τ > 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 steaks are given to be k = 0.45 W/m.°C and α = 0.91×10-7 m2/s 
Analysis The Biot number is 
 200.0
)C W/m.45.0(
)m 01.0)(C. W/m9( 2 =°
°==
k
hLBi Steaks 
25°C 
Refrigerated air 
-11°C 
The constants λ1 and A1 corresponding to this 
Biot number are, from Table 4-1, 
 λ1 1 0 4328 10311= =. . and A
The Fourier number is 
2.0601.5)4328.0cos()0311.1(
)11(25
)11(2
)/cos(
),(
2
2
1
)4328.0(
11
>=⎯→⎯=−−
−−
=−
−
−
−
∞
∞
τ
λ
τ
τλ
e
LLeA
TT
TtLT
i 
Therefore, the one-term approximate solution (or the transient temperature charts) is applicable. Then the 
length of time for the steaks to be kept in the refrigerator is determined to be 
 min 102.6==×== − s 6155/s)m 1091.0(
m) 01.0)(601.5(
27
22
α
τLt 
 4-27
Chapter 4 Transient Heat Conduction 
4-42 A long cylindrical wood log is exposed to hot gases in a fireplace. The time for the ignition of the 
wood is to be determined. 
Assumptions 1 Heat conduction in the wood is one-dimensional since it is long and it has thermal 
symmetry about the center line. 2 The thermal properties of the wood are constant. 3 The heat transfer 
coefficient is constant and uniform over the entire surface. 4 The Fourier number is τ > 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 wood are given to be k = 0.17 W/m.°C, α = 1.28×10-7 m2/s 
Analysis The Biot number is 
 00.4
)C W/m.17.0(
)m 05.0)(C. W/m6.13( 2 =°
°==
k
hr
Bi o 
The constants λ1 and A1 corresponding to this 
Biot number are, from Table 4-1, 
 4698.1 and 9081.1 11 == Aλ 
Once the constant is determined from Table 4-2 
corresponding to the constant 
J0
λ1 =1.9081, the Fourier number is 
determined to be 
10 cm Wood log, 10°C 
Hot gases 
700°C 
251.0)2771.0()4698.1(
50010
500420
)/(
),(
2
2
1
)9081.1(
101
=τ⎯→⎯=−
−
λ=−
−
τ−
τλ−
∞
∞
e
rrJeA
TT
TtrT
oo
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 length of time before the log ignites is 
 min 81.7==×=α
τ= − s 4904/s)m 1028.1(
m) 05.0)(251.0(
27
22
ort 
 4-28
Chapter 4 Transient Heat Conduction 
4-43 A rib is roasted in an oven. The heat transfer coefficient at the surface of the rib, the temperature of 
the outer surface of the rib and the amount of heat transfer when it is rare done are to be determined. The 
time it will take to roast this rib to medium level is also to be determined. 
Assumptions 1 The rib is a homogeneous spherical object. 2 Heat conduction in the rib is one-dimensional 
because of symmetry about the midpoint. 3 The thermal properties of the rib are constant. 4 The heat 
transfer coefficient is constant and uniform over the entire surface. 5 The Fourier number is τ > 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 rib are given to be k = 0.45 W/m.°C, ρ = 1200 kg/m3, Cp = 4.1 kJ/kg.°C, 
and α = 0.91×10-7 m2/s. 
Analysis (a) The radius of the roast is determined to be 
Oven 
163°C
Rib, 
4.5°C 
 
m 08603.0
4
)m 002667.0(3
4
3
3
4
m 002667.0
kg/m 1200
kg 2.3
3
3
33
3
3
===⎯→⎯=
===⎯→⎯=
πππ
ρρ
VrrV
mVVm
oo
 
The Fourier number is 
 1217.0
m) 08603.0(
60)s45+3600/s)(2m 1091.0(
2
27
2
=×××==
−
or
tατ 
which is somewhat below the value of 0.2. Therefore, the one-term approximate solution (or the transient 
temperature charts) can still be used, with the understanding that the error involved will be a little more 
than 2 percent. Then the one-term solution can be written in the form 
 )1217.0(11
0
,0
2
1
2
1 65.0
1635.4
16360 λτλθ −−
∞
∞ ==−
−⎯→⎯=−
−= eAeA
TT
TT
i
sph 
It is determined from Table 4-1 by trial and error that this equation is satisfied when Bi = 30, which 
corresponds to λ1 13 0372 19898= =. and A . . Then the heat transfer coefficient can be determined from 
 C. W/m156.9 2 °=°==⎯→⎯=
)m 08603.0(
)30)(C W/m.45.0(
o
o
r
kBih
k
hr
Bi 
This value seems to be larger than expected for problems of this kind. This is probably due to the Fourier 
number being less than 0.2. 
(b) The temperature at the surface of the rib is 
 
C 159.5 °=⎯→⎯=−
−
==−
−= −−
∞
∞
),(0222.0
1635.4
163),(
0372.3
)rad 0372.3sin()9898.1(
/
)/sin(),(
),( )1217.0()0372.3(
1
1
1
22
1
trT
trT
e
rr
rr
eA
TT
TtrT
tr
o
o
oo
oo
i
o
spho λ
λθ τλ
 
 4-29
Chapter 4 Transient Heat Conduction 
(c) The maximum possible heat transfer is 
 kJ2080C)5.4163)(C kJ/kg.1.4)( kg2.3()(max =°−°=−= ∞ ip TTmCQ 
Then the actual amount of heat transfer becomes 
 
Q
Q
Q Q
o sph
max
,
max
sin( ) cos( )
( . ) sin( . ) ( . ) cos( . )
( . )
.
. ( . )(
= − − = − − =
= = =
1 3 1 3 0 65 3 0372 3 0372 3 0372
3 0372
0 783
0 783 0 783 2080
1 1 1
1
3 3θ
λ λ λ
λ
 kJ) 1629 kJ
 
(d) The cooking time for medium-done rib is determined to be 
 
hr 3≅==×==
=⎯→⎯=−
−⎯→⎯=−
−=
−
−−
∞
∞
min 181s 866,10
/s)m 1091.0(
m) 08603.0)(1336.0(
1336.0)9898.1(
1635.4
16371
27
22
)0372.3(
1
0
,0
22
1
α
τ
τθ ττλ
o
i
sph
r
t
eeA
TT
TT
 
This result is close to the listed value of 3 hours and 20 minutes. The difference between the two results is 
due to the Fourier number being less than 0.2 and thus the error in the one-term approximation. 
 
Discussion The temperature of the outer parts of the rib is greater than that of the inner parts of the rib 
after it is taken out of the oven. Therefore, there will be a heat transfer from outer parts of the rib to the 
inner parts as a result of this temperature difference. The recommendation is logical. 
 
 
 
4-44 A rib is roasted in an oven. The heat transfer coefficient at the surface of the rib, the temperature of 
the outer surface of the rib and the amount of heat transfer when it is well-done are to be determined. The 
time it will take to roast this rib to medium level is also to be determined. 
Assumptions 1 The rib is a homogeneous spherical object. 2 Heat conduction in the rib is one-dimensional 
because of symmetry about the midpoint. 3 The thermal properties of the rib are constant. 4 The heat 
transfer coefficient is constant and uniform over the entire surface. 5 The Fourier number is τ > 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 rib are given to be k = 0.45 W/m.°C, ρ = 1200 kg/m3, Cp = 4.1 kJ/kg.°C, 
and α = 0.91×10-7 m2/s 
Analysis (a) The radius of the rib is determined to be 
Oven 
163°C
Rib, 
4.5°C 
 
m 08603.0
4
)m 00267.0(3
4
3
3
4
m 00267.0
kg/m 1200
kg 2.3
3
3
33
3
3
===⎯→⎯=
===⎯→⎯=
πππ
ρρ
VrrV
mVVm
oo
 
The Fourier number is 
 1881.0
m) 08603.0(
60)s15+3600/s)(4m 1091.0(
2
27
2
=×××==
−
or
tατ 
which is somewhat below the value of 0.2. Therefore, the one-term approximatesolution (or the transient 
temperature charts) can still be used, with the understanding that the error involved will be a little more 
than 2 percent. Then the one-term solution formulation can be written in the form 
 4-30
Chapter 4 Transient Heat Conduction 
 )1881.0(11
0
,0
2
1
2
1 543.0
1635.4
16377 λτλθ −−
∞
∞ ==−
−⎯→⎯=−
−= eAeA
TT
TT
i
sph 
It is determined from Table 4-1 by trial and error that this equation is satisfied when Bi = 4.3, which 
corresponds to 7402.1and 4900.2 11 == Aλ . Then the heat transfer coefficient can be determined from. 
 C. W/m22.5 2 °=°==⎯→⎯=
)m 08603.0(
)3.4)(C W/m.45.0(
o
o
r
kBih
k
hr
Bi 
(b) The temperature at the surface of the rib is 
 
C 142.1 °=⎯→⎯=−
−
==−
−= −−
∞
∞
),(132.0
1635.4
163),(
49.2
)49.2sin()7402.1(
/
)/sin(),(
),( )1881.0()49.2(
1
1
1
22
1
trT
trT
e
rr
rr
eA
TT
TtrT
tr
o
o
oo
oo
i
o
spho λ
λθ τλ
 
(c) The maximum possible heat transfer is 
 kJ2080C)5.4163)(C kJ/kg.1.4)( kg2.3()(max =°−°=−= ∞ ip TTmCQ 
Then the actual amount of heat transfer becomes 
 
kJ 1512===
=−−=−−=
 kJ)2080)(727.0(727.0
727.0
)49.2(
)49.2cos()49.2()49.2sin(
)543.0(31
)cos()sin(
31
max
33
1
111
,
max
QQ
Q
Q
spho λ
λλλθ
 
(d) The cooking time for medium-done rib is determined to be 
 
hr 4===×=α
τ=
=τ⎯→⎯=−
−⎯→⎯=−
−=θ
−
τ−τλ−
∞
∞
min 0.240s 403,14
/s)m 1091.0(
m) 08603.0)(177.0(
177.0)7402.1(
1635.4
16371
27
22
)49.2(
1
0
,0
22
1
o
i
sph
r
t
eeA
TT
TT
 
This result is close to the listed value of 4 hours and 15 minutes. The difference between the two results is 
probably due to the Fourier number being less than 0.2 and thus the error in the one-term approximation. 
 
Discussion The temperature of the outer parts of the rib is greater than that of the inner parts of the rib after 
it is taken out of the oven. Therefore, there will be a heat transfer from outer parts of the rib to the inner 
parts as a result of this temperature difference. The recommendation is logical. 
 4-31
Chapter 4 Transient Heat Conduction 
4-45 An egg is dropped into boiling water. The cooking time of the egg is to be determined. 
Assumptions 1 The egg is spherical in shape with a radius of r0 = 2.75 cm. 2 Heat conduction in the egg is 
one-dimensional because of symmetry about the midpoint. 3 The thermal properties of the egg are 
constant. 4 The heat transfer coefficient is constant and uniform over the entire surface. 5 The Fourier 
number is τ > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are 
applicable (this assumption will be verified). 
Properties The thermal conductivity and diffusivity of the eggs can be approximated by those of water at 
room temperature to be k = 0.607 W/m.°C, α = k Cp/ ρ = 0.146×10-6 m2/s (Table A-9). 
Analysis The Biot number is 
Water
100°C
Egg 
Ti = 8°C 
 2.36
)CW/m. 607.0(
)m 0275.0)(C.W/m 800( 2 =°
°==
k
hr
Bi o 
The constants λ1 and A1 corresponding to this 
Biot number are, from Table 4-1, 
 9925.1and 0533.3 11 == Aλ 
Then the Fourier number and the time period become 
 1633.0)9925.1(
1008
10060 22
1 )0533.3(
1
0
,0 =⎯→⎯=−
−⎯→⎯=−
−= −−
∞
∞ τθ ττλ eeA
TT
TT
i
sph 
which is somewhat below the value of 0.2. Therefore, the one-term approximate solution (or the transient 
temperature charts) can still be used, with the understanding that the error involved will be a little more 
than 2 percent. Then the length of time for the egg to be kept in boiling water is determined to be 
 min 14.1==×== − s 846/s)m 10146.0(
m) 0275.0)(1633.0(
26
22
α
τ ort 
 4-32
Chapter 4 Transient Heat Conduction 
4-46 An egg is cooked in boiling water. The cooking time of the egg is to be determined for a location at 
1610-m elevation. 
Assumptions 1 The egg is spherical in shape with a radius of r0 = 2.75 cm. 2 Heat conduction in the egg is 
one-dimensional because of symmetry about the midpoint. 3 The thermal properties of the egg and heat 
transfer coefficient are constant. 4 The heat transfer coefficient is constant and uniform over the entire 
surface. 5 The Fourier number is τ > 0.2 so that the one-term approximate solutions (or the transient 
temperature charts) are applicable (this assumption will be verified). 
Properties The thermal conductivity and diffusivity of the eggs can be approximated by those of water at 
room temperature to be k = 0.607 W/m.°C, α = k Cp/ ρ = 0.146×10-6 m2/s (Table A-9). 
Analysis The Biot number is 
Water
94.4°C
Egg 
Ti = 8°C 
 2.36
)CW/m. 607.0(
)m 0275.0)(C.W/m 800( 2 =°
°==
k
hr
Bi o 
The constants λ1 and A1 corresponding to this 
Biot number are, from Table 4-1, 
 9925.1and 0533.3 11 == Aλ 
Then the Fourier number and the time period become 
 1727.0)9925.1(
4.948
4.9460 22
1 )0533.3(
1
0
,0 =⎯→⎯=−
−⎯→⎯=−
−= −−
∞
∞ τθ ττλ eeA
TT
TT
i
sph 
which is somewhat below the value of 0.2. Therefore, the one-term approximate solution (or the transient 
temperature charts) can still be used, with the understanding that the error involved will be a little more 
than 2 percent. Then the length of time for the egg to be kept in boiling water is determined to be 
 min 14.9==×== − s 895/s)m 10146.0(
m) 0275.0)(1727.0(
26
22
α
τ ort 
 4-33
Chapter 4 Transient Heat Conduction 
4-47 A hot dog is dropped into boiling water, and temperature measurements are taken at certain time 
intervals. The thermal diffusivity and thermal conductivity of the hot dog and the convection heat transfer 
coefficient are to be determined. 
Assumptions 1 Heat conduction in the hot dog is one-dimensional since it is long and it has thermal 
symmetry about the center line. 2 The thermal properties of the hot dog are constant. 3 The heat transfer 
coefficient is constant and uniform over the entire surface. 4 The Fourier number is τ > 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 hot dog available are given to be ρ = 980 kg/m3 and Cp = 3900 J/kg.°C. 
Analysis (a) From Fig. 4-14b we have 
 15.01
1
17.0
9459
9488
==
⎪⎪⎭
⎪⎪⎬
⎫
==
=−
−=−
−
∞
∞
o
o
o
o
o
hr
k
Bi
r
r
r
r
TT
TT
 Water
94°C
Hot dog The Fourier number is determined from Fig. 4-14a to be 
 20.0
47.0
9420
9459
15.01
2
==
⎪⎪⎭
⎪⎪⎬
⎫
=−
−=−
−
==
∞
∞ o
i
o
o
r
t
TT
TT
hr
k
Bi ατ 
The thermal diffusivity of the hot dog is determined to be 
 /sm 102.017 27−×===α⎯→⎯=α
s 120
m) 011.0)(2.0(2.020.0
22
2 t
r
r
t o
o
 
(b) The thermal conductivity of the hot dog is determined from 
 CW/m. 0.771 °=°×=αρ= − C)J/kg. )(3900kg/m /s)(980m 10017.2( 327pCk
(c) From part (a) we have 15.01 ==
ohr
k
Bi
. Then, 
 m 0.00165m) 011.0)(15.0(15.0 0 === rh
k 
Therefore, the heat transfer coefficient is 
 C.W/m 467 2 °=°=⎯→⎯=
m 0.00165
C W/m.771.000165.0 h
h
k 
 4-34
Chapter 4 Transient Heat Conduction 
4-48 Using the data and the answers given in Prob. 4-43, the center and the surface temperatures of the hot 
dog 4 min after the start of the cooking and the amount of heat transferred to the hot dog are to be 
determined. 
Assumptions 1 Heat conduction in the hot dog is one-dimensional since it is long and it has thermal 
symmetry about the center line. 2 The thermal properties of the hot dog are constant. 3 The heat transfer 
coefficient is constant and uniform over the entire surface. 4 The Fourier number is τ > 0.2 so that the one-
term approximatesolutions (or the transient temperature charts) are applicable (this assumption will be 
verified). 
Properties The properties of hot dog and the convection heat transfer coefficient are given or obtained in 
P4-47 to be k = 0.771 W/m.°C, ρ = 980 kg/m3, Cp = 3900 J/kg.°C, α = 2.017×10-7 m2/s, and h = 467 
W/m2.°C. 
Analysis The Biot number is 
Water
94°C
Hot dog 
 66.6
)C W/m.771.0(
)m 011.0)(C. W/m467( 2 =°
°==
k
hr
Bi o 
The constants λ1 and A1 corresponding to this 
Biot number are, from Table 4-1, 
 5357.1and 0785.2 11 == Aλ 
The Fourier number is 
 2.04001.0
m) 011.0(
s/min) 60min /s)(4m 10017.2(
2
27
2 >=
××==
−
L
tατ 
Then the temperature at the center of the hot dog is determined to be 
 
C 73.8 °=⎯→⎯=−
−
===−
−= −−
∞
∞
0
0
)4001.0()0785.2(
1
0
,
2727.0
9420
94
2727.0)5357.1(
22
1
T
T
eeA
TT
TT
i
cylo
τλθ
 
From Table 4-2 we read =0.2194 corresponding to the constant J0 λ1 =2.0785. Then the temperature at the 
surface of the hot dog becomes 
 
C 89.6 °=⎯→⎯=−
−
===−
− −−
∞
∞
),(05982.0
9420
94),(
05982.0)2194.0()5357.1()/(
),( )4001.0()0785.2(
101
22
1
trT
trT
errJeA
TT
TtrT
o
o
oo
i
o λτλ
 
The maximum possible amount of heat transfer is 
 
[ ]
J 13,440C)2094)(CJ/kg. 3900)(kg 04657.0()(
kg 04657.0m) 125.0(m) 011.0()kg/m 980(
max
2.32
=°−°=−=
====
∞TTmCQ
LrVm
ip
o πρπρ 
From Table 4-2 we read = 0.5760 corresponding to the constant J1 λ1 =2.0785. Then the actual heat 
transfer becomes 
 kJ 11,409==⎯→⎯=−=λ
λθ−=⎟⎟⎠
⎞
⎜⎜⎝
⎛
)kJ 440,13(8489.08489.0
0785.2
5760.0)2727.0(21
)(
21
1
11
,
max
Q
J
Q
Q
cylo
cyl
 
 4-35
Chapter 4 Transient Heat Conduction 
4-49E Whole chickens are to be cooled in the racks of a large refrigerator. Heat transfer coefficient that 
will enable to meet temperature constraints of the chickens while keeping the refrigeration time to a 
minimum is to be determined. 
Assumptions 1 The chicken is a homogeneous spherical object. 2 Heat conduction in the chicken is one-
dimensional because of symmetry about the midpoint. 3 The thermal properties of the chicken are constant. 
4 The heat transfer coefficient is constant and uniform over the entire surface. 5 The Fourier number is τ > 
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 chicken are given to be k = 0.26 Btu/h.ft.°F, ρ = 74.9 lbm/ft3, Cp = 0.98 
Btu/lbm.°F, and α = 0.0035 ft2/h. 
Analysis The radius of the chicken is determined to be 
Refrigerator
T∞ = 5°F 
Chicken 
Ti = 72°F 
ft 2517.0
4
)ft 06676.0(3
4
3
3
4
ft 06676.0
lbm/ft 9.74
lbm 5
3
3
33
3
3
===⎯→⎯=
===⎯→⎯=
πππ
ρρ
VrrV
mVVm
oo
 
From Fig. 4-15b we have 
 75.11
1
75.0
545
535
==
⎪⎪⎭
⎪⎪⎬
⎫
==
=−
−=−
−
∞
∞
o
o
o
o
o
hr
k
Bi
r
r
r
x
TT
TT
 
Then the heat transfer coefficients becomes 
 F.Btu/h.ft 0.590 2 °=°==
ft) 2517.0(75.1
)FBtu/h.ft. 26.0(
75.1 or
kh 
 4-36
Chapter 4 Transient Heat Conduction 
4-50 A person puts apples into the freezer to cool them quickly. The center and surface temperatures of the 
apples, and the amount of heat transfer from each apple in 1 h are to be determined. 
Assumptions 1 The apples are spherical in shape with a diameter of 9 cm. 2 Heat conduction in the apples 
is one-dimensional because of symmetry about the midpoint. 3 The thermal properties of the apples are 
constant. 4 The heat transfer coefficient is constant and uniform over the entire surface. 5 The Fourier 
number is τ > 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 apples are given to be k = 0.418 W/m.°C, ρ = 840 kg/m3, Cp = 3.81 
kJ/kg.°C, and α = 1.3×10-7 m2/s. 
Analysis The Biot number is 
Apple 
Ti = 20°C 
Air 
T∞ = -15°C 861.0
)C W/m.418.0(
)m 045.0)(C. W/m8( 2 =°
°==
k
hr
Bi o 
The constants λ1 and A1 corresponding to this 
Biot number are, from Table 4-1, 
 2390.1 and 476.1 11 == Aλ 
The Fourier number is 
 τ α= = × × = >
−t
r0
2
713 10 3600 s
0 045
0 231 0 2( .
( .
. . m / s)(1 h / h)
 m)
2
2 
Then the temperature at the center of the apples becomes 
 C11.2°=⎯→⎯==−−
−−⎯→⎯=−
−= −−
∞
∞
0
)231.0()476.1(0
1
0
, 749.0)239.1()15(20
)15( 22
1 Te
T
eA
TT
TT
i
spho
τλθ 
The temperature at the surface of the apples is 
 
C2.7°=⎯→⎯=−−
−−
===−
−= −−
∞
∞
),(505.0
)15(20
)15(),(
505.0
476.1
)rad 476.1sin(
)239.1(
/
)/sin(),(
),( )231.0()476.1(
1
1
1
22
1
trT
trT
e
rr
rr
eA
TT
TtrT
tr
o
o
oo
oo
i
o
spho λ
λθ τλ
 
The maximum possible heat transfer is 
 
[ ] kJ 76.42C)15(20)CkJ/kg. 81.3)(kg 3206.0()(
kg 3206.0m) 045.0(
3
4)kg/m 840(
3
4
max
3.33
=°−−°=−=
=⎥⎦
⎤⎢⎣
⎡ π=πρ=ρ=
∞TTmCQ
rVm
ip
o 
Then the actual amount of heat transfer becomes 
kJ 17.2===
=−−=λ
λλ−λθ−=
kJ) 76.42)(402.0(402.0
402.0
)476.1(
)rad 476.1cos()476.1()rad 476.1sin()749.0(31
)cos()sin(
31
max
33
1
111
,
max
QQ
Q
Q
spho 
 
 
 4-37
Chapter 4 Transient Heat Conduction 
4-51 
"!PROBLEM 4-51" 
 
"GIVEN" 
T_infinity=-15 "[C]" 
"T_i=20 [C], parameter to be varied" 
h=8 "[W/m^2-C]" 
r_o=0.09/2 "[m]" 
time=1*3600 "[s]" 
 
"PROPERTIES" 
k=0.513 "[W/m-C]" 
rho=840 "[kg/m^3]" 
C_p=3.6 "[kJ/kg-C]" 
alpha=1.3E-7 "[m^2/s]" 
 
"ANALYSIS" 
Bi=(h*r_o)/k 
"From Table 4-1 corresponding to this Bi number, we read" 
lambda_1=1.3525 
A_1=1.1978 
tau=(alpha*time)/r_o^2 
(T_o-T_infinity)/(T_i-T_infinity)=A_1*exp(-lambda_1^2*tau) 
 
(T_r-T_infinity)/(T_i-T_infinity)=A_1*exp(-
lambda_1^2*tau)*Sin(lambda_1*r_o/r_o)/(lambda_1*r_o/r_o) 
 
V=4/3*pi*r_o^3 
m=rho*V 
Q_max=m*C_p*(T_i-T_infinity) 
Q/Q_max=1-3*(T_o-T_infinity)/(T_i-T_infinity)*(Sin(lambda_1)-
lambda_1*Cos(lambda_1))/lambda_1^3 
 
 
 
 
Ti [C] To [C] Tr [C] Q [kJ] 
2 -1.658 -5.369 6.861 
4 -0.08803 -4.236 7.668 
6 1.482 -3.103 8.476 
8 3.051 -1.97 9.283 
10 4.621 -0.8371 10.09 
12 6.191 0.296 10.9 
14 7.76 1.429 11.7 
16 9.33 2.562 12.51 
18 10.9 3.695 13.32 
20 12.47 4.828 14.13 
22 14.04 5.961 14.93 
24 15.61 7.094 15.74 
26 17.18 8.227 16.55 
28 18.75 9.36 17.35 
30 20.32 10.49 18.16 
 
 
 
 4-38
Chapter 4 Transient Heat Conduction 
0 5 10 15 20 25 30
-10
-5
0
5
10
15
20
25
Ti [C]
T o
 [
C
]
T0
Tr
 
 
 
 
 
0 5 10 15 20 25 30
6
8
10
12
14
16
18
20
Ti [C]
Q
 [
kJ
]
 
 
 4-39
Chapter 4 Transient Heat Conduction 
4-52 An orange is exposed to very cold ambient air. It is to be determined whether the orange will freeze in 
4 h in subfreezing temperatures. 
Assumptions 1 The orange is spherical in shape with a diameter of 8 cm. 2 Heat conduction in the orange 
is one-dimensional because of symmetry about the midpoint. 3 The thermal properties of the orange are 
constant, and are those of water. 4 The heat transfer coefficient is constant and uniform over the entire 
surface. 5 The Fourier number is τ > 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 orange are approximated by those of water at the average temperature of 
about 5°C, k = 0.571 W/m.°C and (Table A-9). α ρ= = × = × −k Cp/ . / ( ) .0 571 1000 4205 0136 10 6 m / s2
Analysis The Biot number is 
Orange 
Ti= 15°C 
Air 
T∞ = -15°C Bi hr
k
o= = ° ° = ≈
( )( . )
( . )
. .15 0 04
0 571
1051 10 W / m . C m
 W / m. C
2
 
The constants λ1 and A1 corresponding to this 
Biot number are, from Table 4-1, 
 λ1 1 15708 12732= =. . and A
The Fourier number is 
 2.0224.1
m) 04.0(
s/h) 6003 h/s)(4m 10136.0(
2
26
2 >=
××==
−
L
tατ 
Therefore, the one-term approximate solution (or the transient temperature charts) is applicable. Then the 
temperature at the surface of the oranges becomes 
 
C 5.2 - °=⎯→⎯=−−
−−
===−
−= −−
∞
∞
),(0396.0
)6(15
)6(),(
0396.0
5708.1
) rad5708.1sin()2732.1(
/
)/sin(),(
),( )224.1()5708.1(
1
1
1
22
1
trT
trT
e
rr
rr
eA
TT
TtrT
tr
o
o
oo
oo
i
o
spho λ
λθ τλ
 
which is less than 0°C. Therefore, the oranges will freeze. 
 4-40
Chapter 4 Transient Heat Conduction 
4-53 A hot baked potato is taken out of the oven and wrapped so that no heat is lost from it. The time the 
potato is baked in the oven and the final equilibrium temperature of the potato after it is wrapped are to be 
determined. 
Assumptions 1 The potato is spherical in shape with a diameter of 8 cm. 2 Heat conduction in the potato is 
one-dimensional because of symmetry about the midpoint. 3 The thermal properties of the potato are 
constant. 4 The heat transfer coefficient is constant and uniform over the entire surface. 5 The Fourier 
number is τ > 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 potato are given to be k = 0.6 W/m.°C, ρ = 1100 kg/m3, Cp = 3.9 kJ/kg.°C, 
and α = 1.4×10-7 m2/s. 
Oven 
T∞ = 170°C
Potato 
T0 = 70°C 
Analysis (a) The Biot number is 
 Bi hr
k
o= = ° ° =
( )( . )
( . )
.25 0 04
0 6
167 W / m . C m
 W / m. C
2
 
The constants λ1 and A1 corresponding to this 
Biot number are, from Table 4-1, 
 λ1 118777 14113= =. . and A 
Then the Fourier number and the time period become 
 2.0203.0)4113.1(69.0
17025
17070 22
1 )8777.1(
1
0
,0 >=⎯→⎯==−
−⎯→⎯=−
−= −−
∞
∞ τθ ττλ eeA
TT
TT
i
sph 
The baking time of the potatoes is determined to be 
 t
ro= = × = =−
τ
α
2
7
0 203 0 04
14 10
2320( . )( .
( .
 m)
 m / s)
 s
2
2 38.7 min 
(b) The maximum amount of heat transfer is 
 
kJ 166.76C)25170)(CkJ/kg. 900.3)(kg 295.0()(
kg 295.0m) 04.0(
3
4)kg/m 1100(
3
4
max
3.33
=°−°=−=
=⎥⎦
⎤⎢⎣
⎡ π=πρ=ρ=
∞ ip
o
TTmCQ
rVm
 
Then the actual amount of heat transfer becomes 
 
kJ 87.5===
=−−=−−=
 kJ)76.166)(525.0(525.0
525.0
)8777.1(
)8777.1cos()8777.1()8777.1sin(
)69.0(31
)cos()sin(
31
max
33
1
111
,
max
QQ
Q
Q
spho λ
λλλθ
 
The final equilibrium temperature of the potato after it is wrapped is 
 C101°=°+°=+=⎯→⎯−= )CkJ/kg. 9.3)(kg 295.0(
kJ 87.5C25)(
p
ieqvieqvp mC
QTTTTmCQ 
 4-41
Chapter 4 Transient Heat Conduction 
4-54 The center temperature of potatoes is to be lowered to 6°C during cooling. The cooling time and if 
any part of the potatoes will suffer chilling injury during this cooling process are to be determined. 
Assumptions 1 The potatoes are spherical in shape with a radius of r0 = 3 cm. 2 Heat conduction in the 
potato is one-dimensional in the radial direction because of the symmetry about the midpoint. 3 The 
thermal properties of the potato are constant. 4 The heat transfer coefficient is constant and uniform over 
the entire surface. 5 The Fourier number is τ > 0.2 so that the one-term approximate solutions (or the 
transient temperature charts) are applicable (this assumption will be verified). 
Properties The thermal conductivity and thermal diffusivity of potatoes are given to be k = 0.50 W/m⋅°C 
and α = 0.13×10-6 m2/s. 
Analysis First we find the Biot number: 
 Bi W / m . C) m
0.5 W / m C
2= = ° ° =
hr
k
0 19 0 03 114( ( . )
.
. 
From Table 4-1 we read, for a sphere, λ1 = 1.635 
and A1 = 1.302. Substituting these values into the 
one-term solution gives 
 θ τλ τ τ0 1 63512 26 225 2 1302 0 753=
−
− = →
−
− = →
∞
∞
− −T T
T T
A e eo
i
 =. .(1. ) 
Potato 
Ti = 25°C 
Air 
2°C 
4 m/s 
which is greater than 0.2 and thus the one-term solution is applicable. Then the cooling time becomes 
h 1.45==×==⎯→⎯= s 5213s/m 100.13
m) 03.0)(753.0( 26-
22
0
2
0 α
τατ rt
r
t 
The lowest temperature during cooling will occur on the surface (r/r0 = 1), and is determined to be 
 
001
0010
001
001
0
0
01
01
1 /
)/sin(
=
/
)/sin()(
 
/
)/sin()( 2
1
rr
rr
TT
TT
rr
rr
TT
TrT
rr
rr
eA
TT
TrT
iii λ
λ
λ
λθλ
λτλ
∞
∞
∞
∞−
∞
∞
−
−=−
−→=−
−
 
Substituting, C4.44=)( 
1.635
 rad)635.1sin(
225
26
225
2)(
0
0 °⎯→⎯⎟⎠
⎞⎜⎝
⎛
−
−=−
−
rT
rT
 
which is above the temperature range of 3 to 4 °C for chilling injury for potatoes. Therefore, no part of 
the potatoes will experience chilling injury during this cooling process. 
Alternative solution We could also solve this problem using transient temperature charts as follows: 
 15a)-4 (Fig. 75.0
t
=
174.0
225
26
877.0
m)C)(0.03.W/m(19
CW/m.50.0
Bi
1
2
o2
o
=
⎪⎪⎭
⎪⎪⎬
⎫
=−
−=−
−
===
∞
∞ o
i
o
o
r
TT
TT
hr
k
ατ 
Therefore, t
r
s
s= = × = ≅−
τ
α
0
2 2
6
0 75 0 03
013 10
5192
( . )( . )
. /m 2
1.44 h 
The surface temperature is determined from 
 
1
0 877
1
0 60
0
0
Bi
Fig. 4 -15b)
= =
=
⎫
⎬
⎪⎪
⎭
⎪⎪
−
− =
∞
∞
k
hr
r
r
T r T
T T
.
( )
. ( 
which gives T T T Tsurface o= + − = + − =∞ ∞0 6 2 0 6 6 2 4 4. ( ) . ( ) . º C 
The slight difference between the two results is due to the reading error of the charts. 
 4-42
Chapter 4 Transient Heat Conduction 
4-55E The center temperature of oranges is to be lowered to 40°F during cooling. The cooling time and if 
any part of the oranges will freeze during this cooling process are to be determined. 
Assumptions 1 The oranges are spherical in shape with a radius of r0 =1.25 in = 0.1042 ft. 2 Heat 
conduction in the orange is one-dimensional in the radial direction because of the symmetry about the 
midpoint. 3 The thermal properties of the orange are constant. 4 The heat transfer coefficient is constant 
and uniform over the entire surface. 5 The Fourier number is τ > 0.2 so that the one-term approximate 
solutions (or the transient temperature charts) are applicable (this assumption will be verified). 
Properties The thermal conductivity and thermal diffusivity of oranges are given to be k = 0.26 Btu/h⋅ft⋅°F 
and α = 1.4×10-6 ft2/s. 
Analysis First we find the Biot number: 
 Bi Btu / h.ft . F) ft
0.26 Btu / h.ft. C
2= = ° ° =
hr
k
0 4 6 125 12 1843( . ( . / ) . 
From Table 4-1 we read, for a sphere, λ1 = 1.9569 and A1 = 
1.447. Substituting these values into the one-term solution gives 
 θ τλ τ τ0 1 956912 240 2578 25 1447 0 426=
−
− = →
−
− = →
∞
∞
− −T T
T T
A e eo
i
 =. .(1. ) 
Orange 
D = 2.5 in 
85% water 
Ti = 78°F 
Air 
25°F 
1 ft/s 
which is greater than 0.2 and thus the one-term solution is applicable. 
Then the cooling time becomes 
 τ α τα= → = = × = =
t
r
t r
0
2
0
2 0 426 125 12 3302 ft)
1.4 10 ft s
 s
2
-6 2
( . )( . /
/
55.0 min 
The lowest temperature during cooling will occur on the surface (r/r0 = 1), and is determined to be 
 
T r T
T T
A e r r
r r
T r T
T T
r r
r r
T T
T T
r r
r ri i i( ) sin( / )
/
( ) sin( / )
/
sin( / )
/
−
− = →
−
− =
−
−
∞
∞
− ∞
∞
∞
∞
1
1 0
0
0
0
1 0 0
0 0
0 1 0
0 0
1
2λ τ 0λ
λ θ
λ
λ
λ
λ1 1 = 1 
Substituting, F32.1=)( 
1.9569
 rad)9569.1sin(
2578
2540
2578
25)(
0
0 °⎯→⎯⎟⎠
⎞⎜⎝
⎛
−
−=−
−
rT
rT
 
which is above the freezing temperature of 31 °C for oranges . Therefore, no part of the oranges will 
freeze during this cooling process. 
Alternative solution We could also solve this problem using transient temperature charts as follows: 
 15a)-4(Fig. 43.0
283.0
2578
2540
543.0
ft) F)(1.25/12.ºBtu/h.ft (4.6
FBtu/h.ft.º0.261
2
2
=α=τ
⎪⎪⎭
⎪⎪⎬
⎫
=−
−=−
−
===
∞
∞ o
i
o
o
r
t
TT
TT
rh
k
Bi
 
Therefore, min 55.5s 3333
/sft101.4
5/12ft)(0.43)(1.2
26
22
==×== −α
τ ort 
The lowest temperature during cooling will occur on the surface (r/r0 =1) of the oranges is determined to be 
 15b)4(Fig. 0.45
)(
1
543.01
−=−
−
⎪⎪⎭
⎪⎪⎬
⎫
=
==
∞
∞
TT
TrT
r
r
rh
k
Bi
o
o
o 
which gives T T T Tsurface o= + − = + − =∞ ∞0 45 25 0 45 40 25. ( ) . ( ) 31.8º F 
The slight difference between the two results is due to the reading error of the charts. 
 4-43
Chapter 4 Transient Heat Conduction 
4-56 The center temperature of a beef carcass is to be lowered to 4°C during cooling. The cooling time 
and if any part of the carcass will suffer freezing injury during this cooling process are to be determined. 
Assumptions 1 The beef carcass can be approximated as a cylinder with insulated top and base surfaces 
having a radius of r0 = 12 cm and a height of H = 1.4 m. 2 Heat conduction in the carcass is one-
dimensional in the radial direction because of the symmetry about the centerline. 3 The thermal properties 
of the carcass are constant. 4 The heat transfer coefficient is constant and uniform over the entire surface. 
5 The Fourier number is τ > 0.2 so that the one-term approximate solutions (or the transient temperature 
charts) are applicable (this assumption will be verified). 
Properties The thermal conductivity and thermal diffusivity of carcass are given to be k = 0.47 W/m⋅°C 
and α = 0.13×10-6 m2/s. 
Analysis First we find the Biot number: 
 Bi W / m . C) m
0.47 W / m C
2= = ° ° =
hr
k
0 22 012 562( ( . )
.
. 
From Table 4-1 we read, for a cylinder, λ1 = 2.027 and A1 = 1.517. 
Substituting these values into the one-term solution gives 
 θ τλ τ τ0 1 2 02712 24 637 6 1517 0 456=
−
− = →
− −
− − = →
∞
∞
− −T T
T T
A e eo
i
 =( )
( )
. .( . ) 
Beef 
37°C 
Air 
-6°C 
1.8 m/s 
which is greater than 0.2 and thus the one-term solution is applicable. 
Then the cooling time becomes 
 h 14.0==×==→= s 558,50s/m 100.13
m) 12.0)(456.0( 26-
22
0
2
0 α
τατ rt
r
t 
The lowest temperature during cooling will occur on the surface (r/r0 = 1), and is determined to be 
 
T r T
T T
A e J r r T r T
T T
J r r T T
T T
J r r
i i i
( ) ( / ) ( ) ( / ) ( /−− = → )
−
− =
−
−
∞
∞
− ∞
∞
∞
∞
1 0 1 0
0
0 0 1 0
0
0 1 0 01
2λ τ λ θ λ = λ 
Substituting, C-3.9=)( 0485.02084.02326.0)(
)6(37
)6(4
)6(37
)6()(
010
0 °⎯→⎯=×=⎟⎟⎠
⎞
⎜⎜⎝
⎛
−−
−−=−−
−−
rTJ
rT λ 
which is below the freezing temperature of -1.7 °C. Therefore, the outer part of the beef carcass will freeze 
during this cooling process. 
Alternative solution We could also solve this problem using transient temperature charts as follows: 
 )14a4(Fig. 0.4
23.0
)6(37
)6(4
178.0
m) C)(0.12W/m².º (22
CW/m.º0.471
2
−=α=τ
⎪⎪⎭
⎪⎪⎬
⎫
=−−
−−=−
−
===
∞
∞ o
i
o
o
r
t
TT
TT
rh
k
Bi
 
Therefore, h12.3s44,308
/sm100.13
m) (0.4)(0.12
26
22
≅=×== −α
τ ort 
The surface temperature is determined from 
 14b)4(Fig. 17.0
)(
1
178.01
−=−
−
⎪⎪⎭
⎪⎪⎬
⎫
=
==
∞
∞
TT
TrT
r
r
rh
k
Bi
o
o
o 
which gives T T T Tsurface o= + − C= − + − − = −∞ ∞017 6 017 4 6 4 3. ( ) . [ ( )] . º 
The difference between the two results is due to the reading error of the charts. 
 4-44
Chapter 4 Transient Heat Conduction 
4-57 The center temperature of meat slabs is to be lowered to -18°C during cooling. The cooling time and 
the surface temperature of the slabs at the end of the cooling process are to be determined. 
Assumptions 1 The meat slabs can be approximated as very large plane walls of half-thickness L = 11.5 
cm. 2 Heat conduction in the meat slabs is one-dimensional because of the symmetry about the centerplane. 
3 The thermal properties of the meat slabs are constant. 4 The heat transfer coefficient is constant and 
uniform over the entire surface. 5 The Fourier number is τ > 0.2 so that the one-term approximate 
solutions (or the transient temperature charts) are applicable (this assumption will be verified). 6 The phase 
change effects are not considered, and thus the actual cooling time will be much longer than the value 
determined. 
Properties The thermal conductivity and thermal diffusivity of meat slabs are given to be k = 0.47 W/m⋅°C 
and α = 0.13×10-6 m2/s. These properties will be used for both fresh and frozen meat. 
Analysis First we find the Biot number: Air 
-30°C 
1.4 m/s 
Meat 
7°C 
 Bi W / m . C) m
0.47 W / m C
2
= = ° ° =
hr
k
0 20 0115 4 89( ( . )
.
. 
From Table 4-1 we read, for a plane wall, λ1 = 1.308 and 
A1=1.239. Substituting these values into the one-term solution 
gives 
 θ τλ τ τ0 1 30812 218 307 30 1239 0 783=
−
− = →
− − −
− − = →
∞
∞
− −T T
T T
A e eo
i
 =( )
( )
. .(1. ) 
which is greater than 0.2 and thus the one-term solution is applicable. 
Then the cooling time becomes 
 τ α τα= → = = × = =
t
L
t L2
2 0 783 0115 79 650 m)
0.13 10 m s
 s
2
-6 2
( . )( .
/
, 22.1 h 
The lowest temperature during cooling will occur on the surface (x/L = 1), and is determined to be 
 
T x T
T T
A e x L T L T
T T
L L T T
T Ti i
( )
cos( / )
( )
cos( / ) cos( )
−
− = →
−
− =
−
−
∞
∞
− ∞
∞
∞
∞
1 1 0 1
0
1
1
2λ τ λ θ λ =
i
λ 
Substituting, 
 C26.9°−=⎯→⎯=×=⎟⎟⎠
⎞
⎜⎜⎝
⎛
−−
−−−=−−
−−
)( 08425.02598.03243.0)cos(
)30(7
)30(18
)30(7
)30()(
1 LT
LT λ 
which is close the temperature of the refrigerated air. 
Alternative solution We could also solve this problem using transient temperature charts as follows: 
 13a)4(Fig. 75.0
324.0
)30(7
)30(18
0.204
m) C)(0.115W/m².º (20
CW/m.º0.471
2
−=α=τ
⎪⎪⎭
⎪⎪⎬
⎫
=−−
−−−=−
−
===
∞
∞ L
t
TT
TT
hL
k
Bi
i
o
 
Therefore, t ro= = × = ≅−
τ
α
2
76 300(0.75)(0.115 m)
0.13 10 m / s
s 21.2 h
2
6 2 , 
The surface temperature is determined from 
 13b)4(Fig. 22.0
)(
1
204.01
−=−
−
⎪⎪⎭
⎪⎪⎬
⎫
=
==
∞
∞
TT
TxT
L
x
hL
k
Bi
o
 
which gives T T T Tsurface o= + − = − + − − − = −∞ ∞0 22 30 0 22 18 30. ( ) . [ ( )] 27.4º C 
The slight difference between the two results is due to the reading error of the charts. 
 4-45
Chapter 4 Transient Heat Conduction 
4-58E The center temperature of meat slabs is to be lowered to 36°F during 12-h of cooling. The average 
heat transfer coefficient during this cooling process is to be determined. 
Assumptions 1 The meat slabs can be approximated as very large plane walls of half-thickness L = 3-in. 2 
Heat conduction in the meat slabs is one-dimensional because of symmetry about the centerplane. 3 The 
thermal properties of the meat slabs are constant.