4-1
Solutions Manual for
Heat and Mass Transfer: Fundamentals & Applications
6th Edition in SI Units
Yunus A. Çengel, Afshin J. Ghajar
McGraw-Hill, 2020
Chapter 4
TRANSIENT HEAT CONDUCTION
PROPRIETARY AND CONFIDENTIAL
This Manual is the proprietary property of McGraw-Hill Education
and protected by copyright and other state and federal laws. By
opening and using this Manual the user agrees to the following
restrictions, and if the recipient does not agree to these
restrictions, the Manual should be promptly returned unopened
to McGraw-Hill Education: This Manual is being provided only to
authorized professors and instructors for use in preparing for the
classes using the affiliated textbook. No other use or distribution
of this Manual is permitted. This Manual may not be sold and
may not be distributed to or used by any student or other third
party. No part of this Manual may be reproduced, displayed or
distributed in any form or by any means, electronic or otherwise,
without the prior written permission of McGraw-Hill Education.
PROPRIETARY MATERIAL. © 2020 McGraw-Hill Education. Limited distribution permitted only to teachers and educators for course preparation. If
you are a student using this Manual, you are using it without permission.
, 4-2
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 in air than in water since the convection heat transfer
coefficient and thus the Biot number is much smaller in air.
4-3C 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.
4-4C 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.
4-5C 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.
4-6C 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
4-7C The temperature drop of the potato during the second minute will be less than 4C 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 The temperature rise of the potato during the second minute will be less than 5C 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.
4-9C 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.
PROPRIETARY MATERIAL. © 2020 McGraw-Hill Education. Limited distribution permitted only to teachers and educators for course preparation. If
you are a student using this Manual, you are using it without permission.
, 4-3
4-10C 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.
4-11 Relations are to be obtained for the characteristic lengths of a large plane wall of thickness 2L, 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 2L, a very long cylinder of radius ro and a
sphere of radius ro are 2ro
2ro
V 2 LA
Lc , wall L
Asurface 2A
V ro2 h ro
Lc ,cylinder
Asurface 2ro h 2
V 4ro ro 2L
Lc , sphere
Asurface 4ro 2 3
4-12 A relation for the time period for a lumped system to reach the average temperature (Ti T ) / 2 is to be obtained.
Analysis The relation for time period for a lumped system to reach the average temperature (Ti T ) / 2 can be determined
as
Ti T
T T
T (t ) T
e bt
2 e bt
Ti T Ti T
Ti
Ti T bt 1 bt
e e
2(Ti T ) 2
ln 2 0.693
bt ln 2
t
b b
PROPRIETARY MATERIAL. © 2020 McGraw-Hill Education. Limited distribution permitted only to teachers and educators for course preparation. If
you are a student using this Manual, you are using it without permission.
, 4-4
4-13 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 = 1920 kg/m3, cp = 790 J/kg ∙ K, and k = 0.90 W/m ∙ K.
Analysis For a brick, the characteristic length and the Biot number are
V (0.203 0.102 0.057) m 3
Lc 0.01549 m
As 2(0.203 0.102) 2(0.102 0.057) 2(0.203 0.057) m 2
hLc (5 W/m 2 K )(0.01549 m)
Bi 0.0861 0.1
k 0.90 W/m K
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
hAs h 5 W/m 2 K
b 2.128 10 4 s -1
c pV ρc p Lc (1920 kg/m )(790 J/kg K )(0.01549 m)
3
T (t ) T
e bt
Ti T
or
1 T (t ) T 1 5
t ln 4 -1
ln 2.522 10 s 7 hours
4
b Ti T 2.128 10 s 1100 30
Discussion In practice, it takes days to cool bricks coming out of kilns, since they are being burned and cooled in bulk.
PROPRIETARY MATERIAL. © 2020 McGraw-Hill Education. Limited distribution permitted only to teachers and educators for course preparation. If
you are a student using this Manual, you are using it without permission.
Solutions Manual for
Heat and Mass Transfer: Fundamentals & Applications
6th Edition in SI Units
Yunus A. Çengel, Afshin J. Ghajar
McGraw-Hill, 2020
Chapter 4
TRANSIENT HEAT CONDUCTION
PROPRIETARY AND CONFIDENTIAL
This Manual is the proprietary property of McGraw-Hill Education
and protected by copyright and other state and federal laws. By
opening and using this Manual the user agrees to the following
restrictions, and if the recipient does not agree to these
restrictions, the Manual should be promptly returned unopened
to McGraw-Hill Education: This Manual is being provided only to
authorized professors and instructors for use in preparing for the
classes using the affiliated textbook. No other use or distribution
of this Manual is permitted. This Manual may not be sold and
may not be distributed to or used by any student or other third
party. No part of this Manual may be reproduced, displayed or
distributed in any form or by any means, electronic or otherwise,
without the prior written permission of McGraw-Hill Education.
PROPRIETARY MATERIAL. © 2020 McGraw-Hill Education. Limited distribution permitted only to teachers and educators for course preparation. If
you are a student using this Manual, you are using it without permission.
, 4-2
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 in air than in water since the convection heat transfer
coefficient and thus the Biot number is much smaller in air.
4-3C 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.
4-4C 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.
4-5C 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.
4-6C 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
4-7C The temperature drop of the potato during the second minute will be less than 4C 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 The temperature rise of the potato during the second minute will be less than 5C 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.
4-9C 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.
PROPRIETARY MATERIAL. © 2020 McGraw-Hill Education. Limited distribution permitted only to teachers and educators for course preparation. If
you are a student using this Manual, you are using it without permission.
, 4-3
4-10C 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.
4-11 Relations are to be obtained for the characteristic lengths of a large plane wall of thickness 2L, 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 2L, a very long cylinder of radius ro and a
sphere of radius ro are 2ro
2ro
V 2 LA
Lc , wall L
Asurface 2A
V ro2 h ro
Lc ,cylinder
Asurface 2ro h 2
V 4ro ro 2L
Lc , sphere
Asurface 4ro 2 3
4-12 A relation for the time period for a lumped system to reach the average temperature (Ti T ) / 2 is to be obtained.
Analysis The relation for time period for a lumped system to reach the average temperature (Ti T ) / 2 can be determined
as
Ti T
T T
T (t ) T
e bt
2 e bt
Ti T Ti T
Ti
Ti T bt 1 bt
e e
2(Ti T ) 2
ln 2 0.693
bt ln 2
t
b b
PROPRIETARY MATERIAL. © 2020 McGraw-Hill Education. Limited distribution permitted only to teachers and educators for course preparation. If
you are a student using this Manual, you are using it without permission.
, 4-4
4-13 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 = 1920 kg/m3, cp = 790 J/kg ∙ K, and k = 0.90 W/m ∙ K.
Analysis For a brick, the characteristic length and the Biot number are
V (0.203 0.102 0.057) m 3
Lc 0.01549 m
As 2(0.203 0.102) 2(0.102 0.057) 2(0.203 0.057) m 2
hLc (5 W/m 2 K )(0.01549 m)
Bi 0.0861 0.1
k 0.90 W/m K
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
hAs h 5 W/m 2 K
b 2.128 10 4 s -1
c pV ρc p Lc (1920 kg/m )(790 J/kg K )(0.01549 m)
3
T (t ) T
e bt
Ti T
or
1 T (t ) T 1 5
t ln 4 -1
ln 2.522 10 s 7 hours
4
b Ti T 2.128 10 s 1100 30
Discussion In practice, it takes days to cool bricks coming out of kilns, since they are being burned and cooled in bulk.
PROPRIETARY MATERIAL. © 2020 McGraw-Hill Education. Limited distribution permitted only to teachers and educators for course preparation. If
you are a student using this Manual, you are using it without permission.
