Solutions Manual for Heat and Mass Transfer: Fundamentals & Applications 5th Edition Yunus A. Cengel & Afshin J. Ghajar McGraw-Hill, 2015(Chapter 3 STEADY HEAT CONDUCTION)

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Solutions Manual
for
Heat and Mass Transfer: Fundamentals & Applications
5th Edition
Yunus A. Cengel & Afshin J. Ghajar
McGraw-Hill, 2015




Chapter 3
STEADY HEAT CONDUCTION




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Steady Heat Conduction in Plane Walls


3-1C The temperature distribution in a plane wall will be a straight line during steady and one dimensional heat transfer with
constant wall thermal conductivity.




3-2C In steady heat conduction, the rate of heat transfer into the wall is equal to the rate of heat transfer out of it. Also, the
temperature at any point in the wall remains constant. Therefore, the energy content of the wall does not change during steady
heat conduction. However, the temperature along the wall and thus the energy content of the wall will change during transient
conduction.




3-3C The thermal resistance of a medium represents the resistance of that medium against heat transfer.




3-4C Yes. The convection resistance can be defined as the inverse of the convection heat transfer coefficient per unit surface
area since it is defined as Rconv  1 /(hA) .




3-5C Convection heat transfer through the wall is expressed as Q  hAs (Ts  T ) . In steady heat transfer, heat transfer rate to
the wall and from the wall are equal. Therefore at the outer surface which has convection heat transfer coefficient three times
that of the inner surface will experience three times smaller temperature drop compared to the inner surface. Therefore, at the
outer surface, the temperature will be closer to the surrounding air temperature.




3-6C The combined heat transfer coefficient represents the combined effects of radiation and convection heat transfers on a
surface, and is defined as hcombined = hconvection + hradiation. It offers the convenience of incorporating the effects of radiation in
the convection heat transfer coefficient, and to ignore radiation in heat transfer calculations.




3-7C The convection and the radiation resistances at a surface are parallel since both the convection and radiation heat
transfers occur simultaneously.




3-8C The temperature of each surface in this case can be determined from

Q  (T1  Ts1 ) / R1 s1   Ts1  T1  (Q R1 s1 )
Q  (Ts 2  T 2 ) / R s 2 2  Ts 2  T 2  (Q R s 2 2 )

where R i is the thermal resistance between the environment  and surface i.




3-9C Yes, it is.


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3-10C The blanket will introduce additional resistance to heat transfer and slow down the heat gain of the drink wrapped in a
blanket. Therefore, the drink left on a table will warm up faster.




3-11C The new design introduces the thermal resistance of the copper layer in addition to the thermal resistance of the
aluminum which has the same value for both designs. Therefore, the new design will be a poorer conductor of heat.




3-12C For a surface of A at which the convection and radiation heat transfer coefficients are hconv and hrad , the single
equivalent heat transfer coefficient is heqv  hconv  hrad when the medium and the surrounding surfaces are at the same
temperature. Then the equivalent thermal resistance will be Reqv  1 /(heqv A) .




3-13C The thermal resistance network associated with a five-layer composite wall involves five single-layer resistances
connected in series.




3-14C Once the rate of heat transfer Q is known, the temperature drop across any layer can be determined by multiplying
heat transfer rate by the thermal resistance across that layer, T  Q R
layer layer




3-15C The window glass which consists of two 4 mm thick glass sheets pressed tightly against each other will probably have
thermal contact resistance which serves as an additional thermal resistance to heat transfer through window, and thus the heat
transfer rate will be smaller relative to the one which consists of a single 8 mm thick glass sheet.




3-16 The two surfaces of a wall are maintained at specified temperatures. The rate of heat loss through the wall is to be
determined.
Assumptions 1 Heat transfer through the wall is steady since the surface temperatures
remain constant at the specified values. 2 Heat transfer is one-dimensional since any Wall
significant temperature gradients will exist in the direction from the indoors to the
outdoors. 3 Thermal conductivity is constant. L= 0.3 m
Properties The thermal conductivity is given to be k = 0.8 W/m°C.
Q
Analysis The surface area of the wall and the rate of heat loss through the wall
are 14C 2C
A  (3 m)  (6 m)  18 m 2


T  T2 (14  2)C
Q  kA 1  (0.8 W/m C)(18 m 2 )  576 W
L 0.3 m




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3-17 A person is dissipating heat at a rate of 150 W by natural convection and radiation to the surrounding air and surfaces.
For a given deep body temperature, the outer skin temperature is to be determined.
Assumptions 1 Steady operating conditions exist. 2 The heat transfer coefficient is
constant and uniform over the entire exposed surface of the person. 3 The
surrounding surfaces are at the same temperature as the indoor air temperature. 4
Heat generation within the 0.5-cm thick outer layer of the tissue is negligible.
Properties The thermal conductivity of the tissue near the skin is given to
be k = 0.3 W/m°C. Qrad
Analysis The skin temperature can be determined directly from Tskin
T  Tskin
Q  kA 1
L Qconv

QL (150 W)(0.005 m)
Tskin  T1   37C   35.5C
kA (0.3 W/m C)(1.7 m 2 )




3-18E The inner and outer surfaces of the walls of an electrically heated house remain at specified temperatures during a
winter day. The amount of heat lost from the house that day and its cost are to be determined.
Assumptions 1 Heat transfer through the walls is steady since the surface temperatures of the walls remain constant at the
specified values during the time period considered. 2 Heat transfer is one-dimensional since any significant temperature
gradients will exist in the direction from the indoors to the outdoors. 3 Thermal conductivity of the walls is constant.
Properties The thermal conductivity of the brick wall is given to be k = 0.40 Btu/hft°F.
Analysis We consider heat loss through the walls only. The total heat transfer area is

A  2(50  9  35  9)  1530 ft 2 Wall
The rate of heat loss during the daytime is
T T L
(55  45)F
Q day  kA 1 2  (0.40 Btu/h  ft  F)(1530 ft 2 )  6120 Btu/h
L 1 ft
Q
The rate of heat loss during nighttime is
T1 T2
T  T2
Q night  kA 1
L
(55  35)C
 (0.40 Btu/h  ft  F)(1530 ft 2 )  12,240 Btu/h
1 ft
The amount of heat loss from the house that night will be
Q
Q   Q  Q t  10Q day  14Q night  (10 h)(6120 Btu/h)  (14 h)(12,240 Btu/h)

t
 232,560Btu
Then the cost of this heat loss for that day becomes
Cost  (232, kWh)($0.09 / kWh)  $6.13




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