4-1
Solutions Manual
for
Heat and Mass Transfer: Fundamentals & Applications
5th Edition
Yunus A. Cengel & Afshin J. Ghajar
McGraw-Hill, 2015
Chapter 4
TRANSIENT HEAT CONDUCTION
(Last updated August 1, 2015)
PROPRIETARY AND CONFIDENTIAL
This Manual is the proprietary property of The McGraw-Hill Companies, Inc. (“McGraw-Hill”) 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: 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.
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. 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 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.
4-2C 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.
4-3C 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.
4-4C 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.
4-5C 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.
4-6C 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.
4-7C 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
4-8C 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.
4-9C 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.
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. 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 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.
4-11C 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.
4-12 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
4-13 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 1
e bt
e bt
2(Ti T ) 2
ln 2 0.693
bt ln 2
t
b b
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. 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
4-14 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/m2 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/m2 K
b 2.128 10 4 s -1
c pV ρc p Lc (1920 kg/m3 )( 790 J/kg K)(0.01549 m)
T (t ) T
e bt
Ti T
or
1 T (t ) T 1 5
t ln
ln 2.522 10 s 7 hours
4
b Ti T 2.128 10 s4 -1
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. © 2015 The McGraw-Hill Companies, Inc. 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
5th Edition
Yunus A. Cengel & Afshin J. Ghajar
McGraw-Hill, 2015
Chapter 4
TRANSIENT HEAT CONDUCTION
(Last updated August 1, 2015)
PROPRIETARY AND CONFIDENTIAL
This Manual is the proprietary property of The McGraw-Hill Companies, Inc. (“McGraw-Hill”) 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: 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.
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. 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 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.
4-2C 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.
4-3C 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.
4-4C 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.
4-5C 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.
4-6C 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.
4-7C 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
4-8C 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.
4-9C 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.
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. 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 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.
4-11C 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.
4-12 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
4-13 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 1
e bt
e bt
2(Ti T ) 2
ln 2 0.693
bt ln 2
t
b b
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. 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
4-14 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/m2 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/m2 K
b 2.128 10 4 s -1
c pV ρc p Lc (1920 kg/m3 )( 790 J/kg K)(0.01549 m)
T (t ) T
e bt
Ti T
or
1 T (t ) T 1 5
t ln
ln 2.522 10 s 7 hours
4
b Ti T 2.128 10 s4 -1
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. © 2015 The McGraw-Hill Companies, Inc. 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.
