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PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
Solutions Manual
for
Heat and Mass Transfer: Fundamentals & Applications
Fourth Edition
Yunus A. Cengel & Afshin J. Ghajar
McGraw-Hill, 2011
Chapter 4
TRANSIENT HEAT CONDUCTION
PROPRIETARY AND CONFIDENTIAL
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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 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.
4-3C 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.
4-4C 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.
4-5C 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
4-6C The temperature drop of the potato during the second minute will be less than 4°C 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.
4-7C The temperature rise of the potato during the second minute will be less than 5°C 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.
4-8C 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.
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4-13 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
r 0 = 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.
Milk 3 ° C
Water 70 ° C
Properties The thermal conductivity, density, and specific heat of the
milk at 20°C are k = 0.598 W/m.°C, ρ = 998 kg/m 3 , and c p = 4.
kJ/kg.°C (Table A-9).
Analysis The characteristic length and Biot number for the glass of milk are
( 0. 598 W/m.C)
( 240 W/m.C)( 0. 0105 m)
- 01050 m 2 ( 0. 03 m)(0.07m)+ 2 ( 0. 03 m)
( 0. 03 m) ( 0. 07 m)
2 2
2
2
2
2
2
k
hL Bi
rL r
r L
A
L
c
o o
o
s
c
V 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:
= ⎯⎯→ = 135 s = 2.25 min −
− −
∞
∞ e e t T T
Tt T
c L
h
c
hA b
bt t
i
p p c
s
( 0. 005477 s )
2
3 70
- 005477 s (998kg/m )( 4182 J/kg.C)(0.0105m)
240 W/m .C
ρ V ρ
Therefore, it will take 135 s to warm the milk from 3 to 38°C.
4-14 A relation for the time period for a lumped system to reach the average temperature ( T (^) i + T ∞)/ 2 is to be obtained.
Analysis The relation for time period for a lumped system to reach the average temperature can be determined
as
( T (^) i + T ∞)/ 2
b
b
ln 2 − =− ⎯⎯→ = =
− −
∞
∞
−
∞
∞
∞ −
∞
∞
bt t
e e T T
T T
e T T
T
T T
e T T
Tt T
bt bt
i
i
bt
i
i bt
i
ln 2
T ∞
Ti
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4-15 The temperature of a gas stream is to be measured by a thermocouple. The time it takes to register 99 percent of the
initial ∆ T 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. 3 ρ = 8500 kg/m cp = 320 J/kg.°C
Analysis The characteristic length of the junction and the Biot number are
Junction D T ( t )
35 W/m.C
( 110 W/m.C)( 0. 0002 m)
0002 m 6
0012 m
6
2
2
3
surface
k
hL Bi
D
D
D
A
L
c
c π
V π
Gas
Since , the lumped system analysis is applicable. Then the time period^ h, T ∞
for the thermocouple to read 99% of the initial temperature difference is
determined from
Bi < 0.
= ⎯⎯→ = ⎯⎯→ = 22.8 s −
− −
∞
∞
∞
∞
e e t T T
Tt T
c L
h
c
hA b
T T
Tt T
bt t
i
p p c
i
( 0. 2022 s )
2
- 01
- 2022 s ( 8500 kg/m )( 320 J/kg.C)(0.0002m)
110 W/m .C
ρ V ρ
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4-17E 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 ro = 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/ft
3 , and c p = 0.216 Btu/lbm.°F (Table A-3E).
Analysis ( a ) The characteristic length and the Biot number for
the aluminum balls are
Water bath, 120°F
Aluminum balls, 250 °F
( 137 Btu/h.ft.F)
( 42 Btu/h.ft .F)( 0. 02778 ft)
- 02778 ft 6
2 / 12 ft
6
2
2
3
k
hL Bi
D
D
D
A
L
c
c
V π
The lumped system analysis is applicable since Bi < 0.1. Then the temperature of the balls after quenching becomes
= ⎯⎯→ = 152 ° F
− −
∞
∞ () 250 120
- 66 h 0. 01157 s (168lbm/ft )( 0. 216 Btu/lbm.F)(0.02778ft)
42 Btu/h.ft .F
( 0. 01157 s)( 120 s)
2
Tt e T T
Tt T
c L
h
c
hA b
bt
i
p p c
s
ρ V ρ
( b ) The total amount of heat transfer from a ball during a 2-minute period is
[ ()] ( 0. 4072 lbm)( 0. 216 Btu/lbm.F)( 250 152 )F 8. 62 Btu
- 4072 lbm 6
( 2 / 12 ft) ( 168 lbm/ft ) 6
3 3
3
Q mc T T t
D
m
p i
ρV ρ
Then the rate of heat transfer from the balls to the water becomes
Q & (^) total = n &ball Q ball=( 120 balls/min)×( 8. 62 Btu)= 1034 Btu/min
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.
4-18 Relations are to be obtained for the characteristic lengths of a large plane wall of thickness 2 L , a very long cylinder of
radius ro and a sphere of radius ro.
Analysis Relations for the characteristic lengths of a large plane
wall of thickness 2 L , a very long cylinder of radius ro and a
sphere of radius ro are (^) 2 r^2 ro o
2
3
surface
,
2
surface
,
surface
,
o
o
o csphere
o
o
o ccylinder
cwall
r
r
r
A
L
r
rh
r h
A
L
L
A
LA
A
L
V
V
V
2 L
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4-19 A long copper rod is cooled to a specified temperature. The cooling time is to be determined.
Assumptions 1 The thermal properties of the geometry are constant. 2 The heat transfer coefficient is constant and uniform
over the entire surface.
Properties The properties of copper are k = 401 W/m⋅ºC, ρ = 8933 kg/m
3 , and cp = 0.385 kJ/kg⋅ºC (Table A-3).
Analysis For cylinder, the characteristic length and the Biot number are
( 401 W/m.C)
( 200 W/m.C)( 0. 005 m)
005 m 4
02 m
4
2
2
surface
k
hL Bi
D
DL
D L
A
L
c
c
V π
D = 2 cm
Ti = 100 ºC
Since Bi <0.1, the lumped system analysis is applicable. Then the cooling time is determined from
= ⎯⎯→ = = 4.0 min −
− −
∞
∞ 238 s 100 20
- 01163 s (8933kg/m )( 385 J/kg.C)(0.005m)
200 W/m.C
( 0. 01163 s)
2
Tt T
c L
h
c
hA b
bt t
i
ρ p V ρ p c
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4-21 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/m 3 , c p
= 875 kJ/kg.°C, and α = 7.3× 10
2 /s. The thermal conductivity of the plate can be determined from k = αρ cp = 177 W/m.°C
(or it can be read from Table A-3).
Analysis The mass of the iron's base plate is Air
m = ρV= ρ LA =( 2770 kg/m^3 )( 0. 005 m)( 0. 03 m^2 )= 0. 4155 kg^22 °C
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 IRON
800 W Q &in = 0. 85 × 800 W= 680 W
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
22 C=21.2 W
( ) ( 12 W/m .C)( 0. 03 m ) 2 2 loss plate,ave ⎟° ⎠
Q & = hAT − T ∞ = °
Energy balance on the plate can be expressed as
E (^) in − E out=∆ E plate → Q in∆ t − Q out∆ t =∆ E plate= mcp ∆ T plate
Solving for ∆ t and substituting,
(680 21.2)J/s
( 0. 4155 kg)( 875 J/kg.C)( 140 22 )C
in out
plate 65.1s −
Q Q
mc T t
p & &
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
( 177. 0 W/m.C)
( 12 W/m.C)( 0. 005 m)
- 005 m
2
°
k
hL Bi
L
A
LA
A
L
c
s
c
V
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
( )= (^) ∞ + 1 −exp(− ) in t mc
hA
hA
Q
Tt T p
Substituting the known quantities and solving for t again gives 65.1 s.
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4-22 Prob. 4-21 is reconsidered. The effects of the heat transfer coefficient and the final plate temperature on the time
it will take for the plate to reach this temperature are to be investigated.
Analysis The problem is solved using EES, and the solution is given below.
"GIVEN"
E_dot=800 [W]
L=0.005 [m]
A=0.03 [m^2]
T_infinity=22 [C]
T_i=T_infinity
h=12 [W/m^2-C]
f_heat=0.
T_f=140 [C]
"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=hA(T_ave-T_infinity)
T_ave=1/2*(T_i+T_f)
(Q_dot_in-Q_dot_out)time=mc_p*(T_f-T_i) "energy balance on the plate"
5 9 13 17 21 25
64
65
66
67
h [W/m
2 -C]
time [s]
h [W/m
2 .C]
time [s] 5 7 9 11 13 15 17 19 21 23 25
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4-23 A body is found while still warm. The time of death is to be estimated.
Assumptions 1 The body can be modeled as a 30-cm-diameter, 1.70-m-long cylinder. 2 The thermal properties of the body
and the heat transfer coefficient are constant. 3 The radiation effects are negligible. 4 The person was healthy(!) when he or
she died with a body temperature of 37°C.
Properties The average human body is 72 percent water by mass, and thus we can assume the body to have the properties of
water at the average temperature of (37 + 25)/2 = 31°C; k = 0.617 W/m · °C, ρ = 996 kg/m
3 , and cp = 4178 J/kg·°C (Table
A–9).
Analysis The characteristic length and the Biot number are
- 617 W/m.C
( 8 W/m .C)( 0. 0689 m)
- 0689 m 2 ( 0. 15 m)( 1. 7 m) 2 ( 0. 15 m)( 0. 15 m)
( 0. 15 m) ( 1. 7 m)
2 2
2
2
2
2
2
surface
k
hL Bi
rL r
r L
A
L
c
o o
o c
V π
Therefore, lumped system analysis is not applicable. However, we can still use it to get a “rough” estimate of the time of
death. Then,
= − × ⎯⎯→ = = 12.2 h −
= ×
− −
∞
∞
−
exp[( 2. 79 10 s )] 43 , 860 s 37 20
- 79 10 s (996kg/m)( 4178 J/kg.C)(0.0689m)
8 W/m.C
5 - 1
5 - 1 3
2
e t t T T
Tt T
c L
h
c
hA b
bt
i
ρ p V ρ p c
Therefore, as a rough estimate, the person died about 12 h before the body was found, and thus the time of death is 5 AM.
Discussion This example demonstrates how to obtain “ball park” values using a simple analysis. A similar analysis is used in
practice by incorporating constants to account for deviation from lumped system analysis.
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4-24 The ambient temperature in the oven necessary to heat the steel rods from 20°C to 450°C within 10 minutes is to be
determined.
Assumptions 1 Thermal properties are constant. 2 Convection heat transfer coefficient is uniform. 3 Heat transfer by
radiation is negligible.
Properties The properties of the steel rods are given as ρ = 7832 kg/m
3 , cp = 434 J/kg · K, and k = 63.9 W/m · K.
Analysis For a cylindrical rod, the characteristic length and the Biot number are
00625 m 4
025 m
4
2 = = = = =
D
DL
D L
A
L
s
c
V π
- 9 W/m K
( 20 W/m K)( 0. 00625 m) 2 = < ⋅
k
hL Bi
c
Since Bi < 0.1, the lumped system analysis is applicable. Then the ambient temperature in the oven is
4 1 3
2
- 414 10 s ( 7832 kg/m)( 434 J/kgK)( 0. 00625 m)
20 W/m K -
p p c
s ρc L
h
c
hA b
− = × ⋅
ρ V
bt
i
e T T
T t T −
∞
∞
−
or
= 1016 ° C
−
− ×
− ×
−
−
∞ 1
( 20 C) 450 C
( 9. 41410 )( 600 )
( 9. 41410 )( 600 )
4
4
e
e
e
Te Tt T bt
bt i
Discussion By increasing the ambient temperature in the oven, the time required to heat the steel rods to the desired
temperature would be reduced.
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4-26 The time required to cool a brick from 1100°C to a temperature difference of 5°C from the ambient air temperature is to
be determined.
Assumptions 1 Thermal properties are constant. 2 Convection heat transfer coefficient is uniform. 3 Heat transfer by
radiation is negligible.
Properties The properties of the brick are given as ρ = 1920 kg/m
3 , cp = 790 J/kg · K, and k = 0.90 W/m · K.
Analysis For a brick, the characteristic length and the Biot number are
[ ]
- 01549 m 2 ( 0. 203 0. 102 ) 2 ( 0. 102 0. 057 ) 2 ( 0. 203 0. 057 )m
( 0. 203 0. 102 0. 057 )m 2
3
× + × + ×
× ×
s
c A
L
V
- 90 W/m K
( 5 W/m K)( 0. 01549 m)
2 = < ⋅
k
hL Bi c
Since Bi < 0.1, the lumped system analysis is applicable. Then the time required to cool a brick from 1100°C to a
temperature difference of 5°C from the ambient air temperature is
4 1 3
2
- 128 10 s ( 1920 kg/m)( 790 J/kgK)( 0. 01549 m)
5 W/m K -
p p c
s
ρc L
h
c
hA b
− = × ⋅
ρ V
bt
i
e T T
T t T −
∞
∞
−
or
⎥= × =^7 hours ⎦
× −
− ∞
∞
- 522 10 s 1100 30
ln
- 128 10 s
ln
4 - 1 Ti T
Tt T
b
t
Discussion In practice, it takes days to cool bricks coming out of kilns, since they are being burned and cooled in bulk.
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4-27 The satellite shell temperature after 5 minutes of reentry is to be determined
Assumptions 1 Thermal properties are constant. 2 Convection heat transfer coefficient is uniform. 3 Heat transfer is uniform
over the outer surface of the shell. 4 Heat transfer is limited to the shell only. 5 Heat transfer by radiation is negligible.
Properties The properties of stainless steel are given as ρ = 8238 kg/m 3 , cp = 468 J/kg · K, and k = 13.4 W/m · K.
Analysis For a spherical shell, the characteristic length and the Biot number are
[ ] m 0. 00995 m 6 ( 4 )
/ 6 ( 2 ) 4 [ 4 2 ( 0. 01 )]
2
3 3
2
3 3
D
D D L
A
L
s
c π
V π
- 4 W/m K
( 130 W/m K)( 0. 00995 m) 2 = < ⋅
k
hL Bi c
Since Bi < 0.1, the lumped system analysis is applicable. Then the shell temperature after 5 minutes of reentry is
1 3
2
- 003389 s ( 8283 kg/m)( 468 J/kg K)( 0. 00995 m)
130 W/m K -
p p c
s ρc L
h
c
hA b = ⋅
ρ V
bt
i
e T T
T t T −
∞
∞
−
or
∞
− T t = T − T ∞ e + T bt () ( i )
= ° − ° + ° = 801 ° C
− ( 5 min) ( 10 C 1250 C) 1250 C
( 0. 003389 )( 300 ) T e
Discussion The analysis to this problem has been simplified by assuming the shell temperature to be uniform during the
reentry.
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4-29 Prob. 4-28 is reconsidered. The effect of the initial temperature of the balls on the annealing time and the total
rate of heat transfer is to be investigated.
Analysis The problem is solved using EES, and the solution is given below.
"GIVEN"
D=0.008 [m]
T_i=900 [C]
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^
V=pi*D^3/
L_c=V/A
Bi=(h*L_c)/k "if Bi < 0.1, the lumped sytem analysis is applicable"
b=(hA)/(rhoc_p*V)
(T_f-T_infinity)/(T_i-T_infinity)=exp(-b*time)
m=rho*V
Q=mc_p(T_i-T_f)
Q_dot=n_dot_ballQConvert(J/h, W)
Ti [C]
time [s]
Q [W] 500 127.4 271. 550 134 305. 600 140 339 650 145.5 372. 700 150.6 406. 750 155.3 440. 800 159.6 474. 850 163.7 508. 900 167.6 542. 950 171.2 576. 1000 174.7 610.
500 600 700 800 900 1000
120
130
140
150
160
170
180
250
300
350
400
450
500
550
600
650
T
i
[C]
tim e
[s]
Q
[
W
]
tim e
heat
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
4-30 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.
Assumptions 1 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 c p = 850 J/kg.°C. The specific heat of the aluminum sink is 903
J/kg.°C (Table A-3), 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
Electronic device, 18 W
E (^) in − E out+ E generation=∆ E device⎯⎯→− Q out∆ t + E generation∆ t = mcp ∆ T device
or, ( ) 2
generation ∞ ∞
∞ ∆ = − ⎟
∆ − T t mc T T
T T
E^ & t hAs p
Substituting the given values,
C( 5 60 s) ( 0. 02 kg)( 850 J/kg.C)( 25 ) C 2
( 18 J/s)( 5 60 s) ( 12 W/m.C)( 0. 0004 m ) 2 2 o ⎟ × = ° − ° ⎠
× − ° T
T
which gives T = 329.7 ° C
If the device were attached to an aluminum heat sink, the temperature of the device would be
( 0. 20 0. 02 )kg ( 850 J/kg.C)( 25 ) C
C( 5 60 s) 2
( 18 J/s)( 5 60 s) ( 12 W/m.C)( 0. 0084 m ) 2 2
= + × ° − °
⎟° ×
× − °
T
T
which gives T = 51.7 ° 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
p p
s mc
E
T T
mc
hA
dt
d T T generation ( )
∞
∞
It can be solved to give
( )= (^) ∞ + 1 −exp(− )
generation t mc
hA
hA
E
Tt T p
s
s
Substituting the known quantities and solving for t gives 329.6 ° C for the first case and 51.7 ° C for the second case, which are
practically identical to the results obtained from the approximate analysis.
