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Solutions Manual 
for 
Heat and Mass Transfer: Fundamentals & Applications 
Fourth Edition 
 Yunus A. Cengel & Afshin J. Ghajar 
McGraw-Hill, 2011 
 
 
 
 
Chapter 1 
INTRODUCTION AND BASIC CONCEPTS 
 
 
 
 
 
 
 
 
PROPRIETARY AND CONFIDENTIAL 
 
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otherwise, without the prior written permission of McGraw-Hill. 
 
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Thermodynamics and Heat Transfer 
 
1-1C Thermodynamics deals with the amount of heat transfer as a system undergoes a process from one equilibrium state to 
another. Heat transfer, on the other hand, deals with the rate of heat transfer as well as the temperature distribution within the 
system at a specified time. 
 
 
1-2C The description of most scientific problems involves equations that relate the changes in some key variables to each 
other, and the smaller the increment chosen in the changing variables, the more accurate the description. In the limiting case 
of infinitesimal changes in variables, we obtain differential equations, which provide precise mathematical formulations for 
the physical principles and laws by representing the rates of changes as derivatives. 
 As we shall see in later chapters, the differential equations of fluid mechanics are known, but very difficult to solve 
except for very simple geometries. Computers are extremely helpful in this area. 
 
 
1-3C (a) The driving force for heat transfer is the temperature difference. (b) The driving force for electric current flow is the 
electric potential difference (voltage). (a) The driving force for fluid flow is the pressure difference. 
 
 
1-4C The caloric theory is based on the assumption that heat is a fluid-like substance called the "caloric" which is a massless, 
colorless, odorless substance. It was abandoned in the middle of the nineteenth century after it was shown that there is no 
such thing as the caloric. 
 
 
1-5C The rating problems deal with the determination of the heat transfer rate for an existing system at a specified 
temperature difference. The sizing problems deal with the determination of the size of a system in order to transfer heat at a 
specified rate for a specified temperature difference. 
 
 
1-6C The experimental approach (testing and taking measurements) has the advantage of dealing with the actual physical 
system, and getting a physical value within the limits of experimental error. However, this approach is expensive, time 
consuming, and often impractical. The analytical approach (analysis or calculations) has the advantage that it is fast and 
inexpensive, but the results obtained are subject to the accuracy of the assumptions and idealizations made in the analysis. 
 
 
1-7C Modeling makes it possible to predict the course of an event before it actually occurs, or to study various aspects of an 
event mathematically without actually running expensive and time-consuming experiments. When preparing a mathematical 
model, all the variables that affect the phenomena are identified, reasonable assumptions and approximations are made, and 
the interdependence of these variables are studied. The relevant physical laws and principles are invoked, and the problem is 
formulated mathematically. Finally, the problem is solved using an appropriate approach, and the results are interpreted. 
 
 
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1-8C The right choice between a crude and complex model is usually the simplest model which yields adequate results. 
 Preparing very accurate but complex models is not necessarily a better choice since such models are not much use to 
an analyst if they are very difficult and time consuming to solve. At the minimum, the model should reflect the essential 
features of the physical problem it represents. 
 
 
1-9C Warmer. Because energy is added to the room air in the form of electrical work. 
 
 
1-10C Warmer. If we take the room that contains the refrigerator as our system, we will see that electrical work is supplied 
to this room to run the refrigerator, which is eventually dissipated to the room as waste heat. 
 
 
1-11C The rate of heat transfer per unit surface area is called heat flux . It is related to the rate of heat transfer by 
. 
q&
∫= A dAqQ &&
 
 
1-12C Energy can be transferred by heat, work, and mass. An energy transfer is heat transfer when its driving force is 
temperature difference. 
 
 
1-13C Thermal energy is the sensible and latent forms of internal energy, and it is referred to as heat in daily life. 
 
 
1-14C For the constant pressure case. This is because the heat transfer to an ideal gas is mcp∆T at constant pressure and 
mcv∆T at constant volume, and cp is always greater than cv. 
 
 
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1-15 A cylindrical resistor on a circuit board dissipates 1.2 W of power. The amount of heat dissipated in 24 h, the heat flux, 
and the fraction of heat dissipated from the top and bottom surfaces are to be determined. 
Assumptions Heat is transferred uniformly from all surfaces. 
Q& 
Resistor 
1.2 W 
Analysis (a) The amount of heat this resistor dissipates during a 24-hour period is 
kJ 104= Wh28.8==∆= h) W)(242.1(tQQ & (since 1 Wh = 3600 Ws = 3.6 kJ) 
(b) The heat flux on the surface of the resistor is 
 2
22
cm 764.2513.2251.0cm) cm)(2 4.0(
4
cm) 4.0(2
4
2 =+=+=+= ππππ DLDAs 
 2 W/cm0.434=== 2cm 764.2
 W2.1
s
s A
Qq
&
& 
(c) Assuming the heat transfer coefficient to be uniform, heat transfer is proportional to the surface area. Then 
the fraction of heat dissipated from the top and bottom surfaces of the resistor becomes 
 (9.1%)or 
764.2
251.0
total
basetop
total
basetop 0.091=== −−
A
A
Q
Q
 
Discussion Heat transfer from the top and bottom surfaces is small relative to that transferred from the side surface. 
 
 
 
 
1-16E A logic chip in a computer dissipates 3 W of power. The amount heat dissipated in 8 h and the heat flux on the surface 
of the chip are to be determined. 
Assumptions Heat transfer from the surface is uniform. 
Analysis (a) The amount of heat the chip dissipates during an 8-hour period is W3=Q&Logic chip 
 kWh 0.024===∆= Wh24)h 8)( W3(tQQ &
(b) The heat flux on the surface of the chip is 
 2 W/in37.5===
2in 08.0
 W3
A
Qq
&
& 
 
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1-17 A house is heated from 10°C to 22°C by an electric heater, and some air escapes through the cracks as the heated air in 
the house expands at constant pressure. The amount of heat transfer to the air and its cost are to be determined. 
Assumptions 1 Air as an ideal gas with a constant specific heats at room temperature. 2 The volume occupied by the 
furniture and other belongings is negligible. 3 The pressure in the house remains constant at all times. 4 Heat loss from the 
house to the outdoors is negligible during heating. 5 The air leaks out at 22°C. 
Properties The specific heat of air at room temperature is cp = 1.007 kJ/kg⋅°C. 
22°C
 
10°C 
AIR 
Analysis The volume and mass of the air in the house are 
 32 m 600m) )(3m 200(ght)space)(heifloor ( ===V
 kg 9.747
K) 273.15+K)(10/kgmkPa 287.0(
)m kPa)(600 3.101(
3
3
=
⋅⋅
==
RT
Pm V 
Noting that the pressure in the house remains constant during heating, the amount of heat 
that must be transferred to the air in the house as it is heated from 10 to 22°C is 
determined to be 
 kJ 9038=°−°⋅=−= C)10C)(22kJ/kg kg)(1.007 9.747()( 12 TTmcQ p 
Noting that 1 kWh = 3600 kJ, the cost of this electrical energy at a unit cost of $0.075/kWh is 
$0.19== 5/kWh)kWh)($0.07 3600/9038(energy) ofcost used)(Unit(Energy =CostEnegy 
Therefore, it will cost the homeowner about 19 cents to raise the temperature in his house from 10 to 22°C. 
 
 
 
 
1-18 A 800 W iron is left on the ironing board with its base exposed to the air. The amount of heat the iron dissipates in 2 h, 
the heat flux on the surface of the iron base, and the cost of the electricity are to be determined. 
Assumptions Heat transfer from the surface is uniform. 
Iron 
800 W Analysis (a) The amount of heat the iron dissipates during a 2-h period is 
 kWh 1.6==∆= h) kW)(2 8.0(tQQ &
(b) The heat flux on the surface of the iron base is 
 W680= W)800)(85.0(base =Q&
 2 W/m45,300=== 2
base
base
m 015.0
 W680
A
Q
q
&
& 
(c) The cost of electricity consumed during this period is 
 $0.112=× kWh)/($0.07kWh) (1.6=yelectricit ofCost 
 
 
 
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1-19 A 15 cm × 20 cm circuit board houses 120 closely spaced 0.12 W logic chips. The amount of heat dissipated in 10 h 
and the heat flux on the surface of the circuit board are to be determined. 
Assumptions 1 Heat transfer from the back surface of the board is negligible. 2 Heat transfer from the front surface is 
uniform. 
Analysis (a) The amount of heat this circuit board dissipates during a 10-h period is 
20 cm
15 cm 
Chips, 
0.12 W Q& W4.14 W)12.0)(120( ==Q&
 kWh 0.144==∆= h) kW)(10 0144.0(tQQ &
(b) The heat flux on the surface of the circuit board is 
 2m 03.0)m 2.0)(m 15.0( ==sA
 2 W/m480===
2m 03.0
 W4.14
s
s A
Q
q
&
& 
 
 
 
 
1-20 An aluminum ball is to be heated from 80°C to 200°C. The amount of heat that needs to be transferred to the aluminum 
ball is to be determined. 
Assumptions The properties of the aluminum ball are constant. 
Properties The average density and specific heat of aluminum are given to be ρ = 2700 kg/m3 and cp = 0.90 kJ/kg⋅°C. 
Analysis The amount of energy added to the ball is simply the change in its internal energy, and is 
determined from 
Metal 
ball 
 )( 12transfer TTmcUE p −=∆=
where 
 kg 77.4m) 15.0)(kg/m 2700(
66
333 ====
πρπρ Dm V 
E Substituting, 
 kJ 515=C80)C)(200kJ/kg kg)(0.90 77.4(transfer °−°⋅=E 
Therefore, 515 kJ of energy (heat or work such as electrical energy) needs to be transferred to the aluminum ball to heat it to 
200°C. 
 
 
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1-21 An electrically heated house maintained at 22°C experiences infiltration losses at a rate of 0.7 ACH. The amount of 
energy loss from the house due to infiltration per day and its cost are to be determined. 
Assumptions 1 Air as an ideal gas with a constant specific heats at room temperature. 2 The volume occupied by the 
furniture and other belongings is negligible. 3 The house is maintained at a constant temperature and pressure at all times. 4 
The infiltrating air exfiltrates at the indoors temperature of 22°C. 
Properties The specific heat of air at room temperature is cp = 1.007 kJ/kg⋅°C. 
5°C 
0.7 ACH 
22°C 
AIR 
Analysis The volume of the air in the house is 
 32 m 450m) )(3m 150(ght)space)(heifloor ( ===V
Noting that the infiltration rate is 0.7 ACH (air changes per hour) and thus the air in 
the house is completely replaced by the outdoor air 0.7×24 = 16.8 times per day, the 
mass flow rate of air through the house due to infiltration is 
 
kg/day 8485
K) 273.15+K)(5/kgmkPa 287.0(
day)/m 045kPa)(16.8 6.89(
)ACH(
3
3
houseair
air
=
⋅⋅
×
=
×
==
o
o
o
o
RT
P
RT
P
m
VV&
&
 
 Noting that outdoor air enters at 5°C and leaves at 22°C, the energy loss of this house per day is 
 
kWh/day 40.4=kJ/day 260,145C)5C)(22kJ/kg. 007kg/day)(1. 8485(
)( outdoorsindoorsairinfilt
=°−°=
−= TTcmQ p&&
At a unit cost of $0.082/kWh, the cost of this electrical energy lost by infiltration is 
$3.31/day== 0.082/kWh)kWh/day)($ 4.40(energy) ofcost used)(Unit(Energy =CostEnegy 
 
 
 
 
1-22 The filament of a 150 W incandescent lamp is 5 cm long and has a diameter of 0.5 mm. The heat flux on the surface of 
the filament, the heat flux on the surface of the glass bulb, and the annual electricity cost of the bulb are to be determined. 
Assumptions Heat transfer from the surface of the filament and the bulb of the lamp is uniform. 
Analysis (a) The heat transfer surface area and the heat flux on the surface of the filament are 
 2cm 785.0)cm 5)(cm 05.0( === ππDLAs
 26 W/m101.91×==== 2
2
 W/cm191
cm 785.0
 W150
s
s A
Q
q
&
& 
(b) The heat flux on the surface of glass bulb is 
 222 cm 1.201cm) 8( === ππDAs
 2 W/m7500==== 2
2
 W/cm75.0
cm 1.201
 W150
s
s A
Q
q
&
& 
(c) The amount and cost of electrical energy consumed during a one-year period is 
Q&
Lamp 
150 W 
 
 
$35.04/yr=
=×=∆=
kWh)/.08kWh/yr)($0 (438=Cost Annual
kWh/yr 438h/yr) 8kW)(365 15.0(nConsumptioy Electricit tQ&
 
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1-23 Water is heated in an insulated tube by an electric resistance heater. The mass flow rate of water through the heater is to 
be determined. 
Assumptions 1 Water is an incompressible substance with a constant specific heat. 2 The kinetic and potential energy 
changes are negligible, ∆ke ≅ ∆pe ≅ 0. 3 Heat loss from the insulated tube is negligible. 
Properties The specific heat of water at room temperature is cp = 4.18 kJ/kg·°C. 
Analysis We take the tube as the system. This is a control volume since mass crosses the system boundary during the 
process. We observe that this is a steady-flow process since there is no change with time at any point and thus 
, there is only one inlet and one exit and thus 0 and 0 CVCV =∆=∆ Em mmm &&& == 21 , and the tube is insulated. The energy 
balance for this steady-flow system can be expressed in the rate form as 
)(
0)peke (since 
 0
12ine,
21ine,
energies etc. potential, 
kinetic, internal,in change of Rate
(steady) 0
system
mass and work,heat,by 
nsferenergy tranet of Rate
TTcmW
hmhmW
EEEEE
p
outinoutin
−=
≅∆≅∆=+
=→=∆=−
&&
&&&
&&
444 344 21
&
43421
&&15°C
WATER
60°C
5 kWThus, 
kg/s 0.0266=
°−°⋅
=
−
=
C)1506)(CkJ/kg 4.18(
kJ/s 5
)( 12
e,in
TTc
W
m
p
&
& 
 
 
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1-24 A resistance heater is to raise the air temperature in the room from 7 to 25°C within 15 min. The required power rating 
of the resistance heater is to be determined. 
Assumptions 1 Air is an ideal gas since it is at a high temperature and low pressure relative to its critical point values of -
141°C and 3.77 MPa. 2 The kinetic and potential energy changes are negligible, ∆ke ≅ ∆pe ≅ 0. 3 Constant specific heats at 
room temperature can be used for air. This assumption results in negligible error in heating and air-conditioning applications. 
4 Heat losses from the room are negligible. 
Properties The gas constant of air is R = 0.287 kPa⋅m3/kg⋅K (Table A-1). Also, cp = 1.007 kJ/kg·K for air at room 
temperature (Table A-15). 
Analysis We observe that the pressure in the room remains constant during this process. Therefore, some air will leak out as 
the air expands. However, we can take the air to be a closed system by considering the air in the room to have undergone a 
constant pressure expansion process. The energy balance for this steady-flow system can be expressed as 
)()( 1212,
,
energies etc. potential, 
kinetic, internal,in Change
system
mass and work,heat,by 
nsferenergy traNet 
TTmchhmHW
UWW
EEE
pine
bine
outin
−≅−=∆=
∆=−
∆=−
4342143421
 
4×5×6 m3
7°C 
AIR We
or )( 12,, TTmctW avgpine −=∆&
The mass of air is 
 
kg149.3
K)K)(280/kgmkPa(0.287
)mkPa)(120(100
m120654
3
3
1
1
3
=
⋅⋅
==
=××=
RT
P
m
V
V
 
Using cp value at room temperature, the power rating of the heater becomes 
 Wk 3.01=×−°⋅= s) 60C/(157)C)(25kJ/kg kg)(1.007 (149.3,
o&
ineW
 
 
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1-25 Liquid water is to be heated in an electric teapot. The heating time is to be determined. 
Assumptions 1 Heat loss from the teapot is negligible. 2 Constant properties can be used for both the teapot and the water. 
Properties The average specific heats are given to be 0.7 kJ/kg·K for the teapot and 4.18 kJ/kg·K for water. 
Analysis We take the teapot and the water in it as the system, which is a closed system (fixed mass). The energy balance in 
this case can be expressed as 
teapotwatersystemin
energies etc. potential, 
kinetic, internal,in Change
system
mass and work,heat,by 
nsferenergy traNet 
UUUE
EEE outin
∆+∆=∆=
∆=−
4342143421
 
Then the amount of energy needed to raise the temperature of 
water and the teapot from 15°C to 95°C is 
 
kJ 3.429
C)15C)(95kJ/kg kg)(0.7 (0.5C)15C)(95kJ/kg kg)(4.18 (1.2
)()( teapotwaterin
=
°−°⋅+°−°⋅=
∆+∆= TmcTmcE
The 1200-W electric heating unit will supply energy at a rate of 1.2 kW or 1.2 kJ per second. Therefore, the time needed for 
this heater to supply 429.3 kJ of heat is determined from 
min 6.0=====∆ s 358
kJ/s 2.1
kJ 3.429
nsferenergy tra of Rate
nsferredenergy tra Total
transfer
in
E
E
t
&
 
Discussion In reality, it will take more than 6 minutes to accomplish this heating process since some heat loss is inevitable 
during heating. Also, the specific heat units kJ/kg · °C and kJ/kg · K are equivalent, and can be interchanged. 
 
 
 
 
1-26 It is observed that the air temperature in a room heated by electric baseboard heaters remains constant even though the 
heater operates continuously when the heat losses from the room amount to 9000 kJ/h. The power rating of the heater is to be 
determined. 
 Assumptions 1 Air is an ideal gas since it is at a high temperature and low pressure relative to its critical point values of -
141°C and 3.77 MPa. 2 The kinetic and potential energy changes are negligible, ∆ke ≅ ∆pe ≅ 0. 3 The temperature of the 
room remains constant during this process. 
Analysis We take the room as the system. The energy balance in this case reduces to 
We
 AIR 
outine
outine
outin
QW
UQW
EEE
=
=∆=−
∆=−
,
,
energies etc. potential, 
kinetic, internal,in Change
system
mass and work,heat,by 
nsferenergy traNet 
0
4342143421
 since ∆U = mcv∆T = 0 for isothermal processes of ideal gases. Thus, 
 kW 2.5=⎟
⎠
⎞
⎜
⎝
⎛==
kJ/h 3600
kW 1kJ/h 0009, outine QW && 
 
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1-27 A room is heated by an electrical resistance heater placed in a short duct in the room in 15 min while the room is losing 
heat to the outside, and a 300-W fan circulates the air steadily through the heater duct. The power rating of the electric heater 
and the temperature rise of air in the duct are to be determined. 
 Assumptions 1 Air is an ideal gas since it is at a high temperature and low pressure relative to its critical point values of -
141°C and 3.77 MPa. 2 The kinetic and potential energy changes are negligible, ∆ke ≅ ∆pe ≅ 0. 3 Constant specific heats at 
room temperature can be used for air. This assumption results in negligible error in heating and air-conditioning 
applications. 3 Heat loss from the duct is negligible. 4 The house is air-tight and thus no air is leaking in or out of the room. 
Properties The gas constant of air is R = 0.287 kPa.m3/kg.K (Table A-1). Also, cp = 1.007 kJ/kg·K for air at room 
temperature (Table A-15) and cv = cp – R = 0.720 kJ/kg·K. 
Analysis (a) We first take the air in the room as the system. This is a constant volume closed system since no mass crosses 
the system boundary. The energy balance for the room can be expressed as 
)()()( 1212outinfan,ine,
outinfan,ine,
energies etc. potential, 
kinetic, internal,in Change
system
mass and work,heat,by 
nsferenergy traNet 
TTmcuumtQWW
UQWW
EEE
v
outin
−≅−=∆−+
∆=−+
∆=−
&&&
4342143421
 
5×6×8 m3
We
300 W
200 kJ/min
The total mass of air in the room is 
 
kg 284.6
)K 288)(K/kgmkPa 0.287(
)m 240)(kPa 98(
m 240m 865
3
3
1
1
33
=
⋅⋅
==
=××=
RT
P
m
V
V
 
Then the power rating of the electric heater is determined to be 
 
kW 4.93=×°−°⋅+−=
∆−−= +
s) 60C)/(181525)(CkJ/kg 0.720)(kg 284.6()kJ/s 0.3()kJ/s 200/60(
/)( 12infan,outine, tTTWQW mcv&&&
(b) The temperature rise that the air experiences each time it passes through the heater is determined by applying the energy 
balance to the duct, 
TcmhmWW
hmQhmWW
EE
p
outin
∆=∆=+
≅∆≅∆+=++
=
&&&&
&&&&&
&&
infan,ine,
2
0
out1infan,ine, 0)peke (since 
Thus, 
 C6.2°=
⋅
+
=
+
=∆
)KkJ/kg 1.007)(kg/s 50/60(
kJ/s)0.34.93(infan,ine,
pcm
WW
T
&
&&
 
 
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1-28 The resistance heating element of an electrically heated house is placed in a duct. The air is moved by a fan, and heat is 
lost through the walls of the duct. The power rating of the electric resistance heater is to be determined. 
Assumptions 1 Air is an ideal gas since it is at a high temperature and low pressure relative to its critical point values of -
141°C and 3.77 MPa. 2 The kinetic and potential energy changes are negligible, ∆ke ≅ ∆pe ≅ 0. 3 Constant specific heats at 
room temperature can be used for air. This assumption results in negligible error in heating and air-conditioning applications. 
Properties Thespecific heat of air at room temperature is cp = 1.007 kJ/kg·°C (Table A-15). 
Analysis We take the heating duct as the system. This is a control volume since mass crosses the system boundary during the 
process. We observe that this is a steady-flow process since there is no change with time at any point and thus 
. Also, there is only one inlet and one exit and thus 0 and 0 CVCV =∆=∆ Em mmm &&& == 21 . The energy balance for this 
steady-flow system can be expressed in the rate form as 
 
)(
0)peke (since 
 0
12infan,outine,
2out1infan,ine,
energies etc. potential, 
kinetic, internal,in change of Rate
(steady) 0
system
mass and work,heat,by 
nsferenergy tranet of Rate
TTcmWQW
hmQhmWW
EEEEE
p
outinoutin
−+−=
≅∆≅∆+=++
=→=∆=−
&&&&
&&&&&
&&
444 344 21
&
43421
&&
We
300 W
250 W 
Substituting, the power rating of the heating element is determined to be 
 
kW 2.97=
°°⋅−= C)C)(5kJ/kg 7kg/s)(1.00 (0.6+kW) 3.0()kW 25.0(ine,W&
 
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1-29 Air is moved through the resistance heaters in a 900-W hair dryer by a fan. The volume flow rate of air at the inlet and 
the velocity of the air at the exit are to be determined. 
Assumptions 1 Air is an ideal gas since it is at a high temperature and low pressure relative to its critical point values of -
141°C and 3.77 MPa. 2 The kinetic and potential energy changes are negligible, ∆ke ≅ ∆pe ≅ 0. 3 Constant specific heats at 
room temperature can be used for air. 4 The power consumed by the fan and the heat losses through the walls of the hair 
dryer are negligible. 
Properties The gas constant of air is R = 0.287 kPa.m3/kg.K (Table A-1). Also, cp = 1.007 kJ/kg·K for air at room 
temperature (Table A-15). 
Analysis (a) We take the hair dryer as the system. This is a control volume since mass crosses the system boundary during 
the process. We observe that this is a steady-flow process since there is no change with time at any point and thus 
, and there is only one inlet and one exit and thus 0 and 0 CVCV =∆=∆ Em mmm &&& == 21 . The energy balance for this 
steady-flow system can be expressed in the rate form as 
)(
0)peke (since 
 0
12ine,
2
0
1
0
infan,ine,
energies etc. potential, 
kinetic, internal,in change of Rate
(steady) 0
system
mass and work,heat,by 
nsferenergy tranet of Rate
TTcmW
hmQhmWW
EEEEE
p
out
outinoutin
−=
≅∆≅∆+=++
=→=∆=−
&&
&&&&&
&&
444 344 21
&
43421
&&
 
T2 = 50°C 
A2 = 60 cm2
P1 = 100 kPa
T1 = 25°C 
We = 900 W 
·
Thus, 
( )
kg/s 0.03575
C)2550)(CkJ/kg 1.007(
kJ/s 0.9
12
e,in
=
°−°⋅
=
−
=
TTc
W
m
p
&
&
 
Then, 
 
/sm 0.0306 3===
=
⋅⋅
==
)/kgm 0.8553)(kg/s 0.03575(
/kgm 0.8553
kPa 100
)K 298)(K/kgmkPa 0.287(
3
11
3
3
1
1
1
vV
v
m
P
RT
&&
 
(b) The exit velocity of air is determined from the conservation of mass equation, 
 
m/s 5.52=
×
==⎯→⎯=
=
⋅⋅
==
− 24
3
2
2
222
2
3
3
2
2
2
m1060
)/kgm 0.9270)(kg/s 0.03575(1
/kgm 0.9270
kPa 100
)K 323)(K/kgmkPa 0.287(
A
m
VVAm
P
RT
v
v
v
&
&
 
 
 
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preparation. If you are a student using this Manual, you are using it without permission. 
1-14
 
1-30 The ducts of an air heating system pass through an unheated area, resulting in a temperature drop of the air in the duct. 
The rate of heat loss from the air to the cold environment is to be determined. 
Assumptions 1 Air is an ideal gas since it is at a high temperature and low pressure relative to its critical point values of -
141°C and 3.77 MPa. 2 The kinetic and potential energy changes are negligible, ∆ke ≅ ∆pe ≅ 0. 3 Constant specific heats at 
room temperature can be used for air. This assumption results in negligible error in heating and air-conditioning applications. 
Properties The specific heat of air at room temperature is cp = 1.007 kJ/kg·°C (Table A-15). 
Analysis We take the heating duct as the system. This is a control volume since mass crosses the system boundary during the 
process. We observe that this is a steady-flow process since there is no change with time at any point and thus 
. Also, there is only one inlet and one exit and thus 0 and 0 CVCV =∆=∆ Em mmm &&& == 21 . The energy balance for this 
steady-flow system can be expressed in the rate form as 
 
)(
0)peke (since 
 0
21out
21
energies etc. potential, 
kinetic, internal,in change of Rate
(steady) 0
system
mass and work,heat,by 
nsferenergy tranet of Rate
TTcmQ
hmQhm
EEEEE
p
out
outinoutin
−=
≅∆≅∆+=
=→=∆=−
&&
&&&
&&
444 344 21
&
43421
&&
90 kg/min AIR 
Q
·
Substituting, 
 ( )( )( ) kJ/min272 C3CkJ/kg 1.007kg/min 90out =°°⋅=∆= TcmQ p&& 
 
 
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1-15
 
1-31E Air gains heat as it flows through the duct of an air-conditioning system. The velocity of the air at the duct inlet and 
the temperature of the air at the exit are to be determined. 
Assumptions 1 Air is an ideal gas since it is at a high temperature and low pressure relative to its critical point values of -
222°F and 548 psia. 2 The kinetic and potential energy changes are negligible, ∆ke ≅ ∆pe ≅ 0. 3 Constant specific heats at 
room temperature can be used for air. This assumption results in negligible error in heating and air-conditioning applications. 
Properties The gas constant of air is R = 0.3704 psia·ft3/lbm·R (Table A-1E). Also, cp = 0.240 Btu/lbm·R for air at room 
temperature (Table A-15E). 
Analysis We take the air-conditioning duct as the system. This is a control volume since mass crosses the system boundary 
during the process. We observe that this is a steady-flow process since there is no change with time at any point and thus 
, there is only one inlet and one exit and thus 0 and 0 CVCV =∆=∆ Em mmm &&& == 21 , and heat is lost from the system. The 
energy balance for this steady-flow system can be expressed in the rate form as 
 
)(
0)peke (since 
 0
12in
21in
energies etc. potential, 
kinetic, internal,in change of Rate
(steady) 0
system
mass and work,heat,by 
nsferenergy tranet of Rate
TTcmQ
hmhmQ
EEEEE
p
outinoutin
−=
≅∆≅∆=+
=→=∆=−
&&
&&&
&&
444 344 21
&
43421
&&
450 ft3/min AIR D = 10 in
2 Btu/s 
(a) The inlet velocity of air through the duct is determined from 
 ft/min825 
ft) 12/5(
/minft 450
2
3
2
1
1
1
1 ====
ππrA
V
VV &&
 
(b) The mass flow rate of air becomes 
 
slbm 5950lbm/min 35.7
/lbmft 12.6
minft 450
/lbmft 6.12
psia 15
)R 510)(R/lbmftpsia 0.3704(
3
3
1
1
3
3
1
1
1
/.
/
====
=
⋅⋅
==
v
V
v
&
&m
P
RT
 
Then the exit temperature of air is determined to be 
 F64.0°=
°⋅
+°=+=
)FBtu/lbm 0.240)(lbm/s 0.595(
Btu/s 2
F50in12
pcm
Q
TT
&
&
 
 
 
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1-16
 
1-32 A classroom is to be air-conditioned using window air-conditioning units. The cooling load is due to people, lights, 
and heat transfer through the walls and the windows. The number of 5-kW window air conditioning units required is to be 
determined. 
Assumptions There are no heat dissipating equipment (such as computers, TVs, or ranges) in the room. 
Analysis The total cooling load of the room is determined from 
 gainheatpeoplelightscooling QQQQ &&&& ++=
Room 
50 people 
10 bulbs 
12,000 kJ/hwhere 
Qcool
·
kW .333kJ/h 12,000
kW 5kJ/h 000,18kJ/h 36005
kW 1W 10010
gainheat
people
lights
==
==×=
=×=
Q
Q
Q
&
&
&
 
Substituting, 
kW .33933.351cooling =++=Q& 
Thus the number of air-conditioning units required is 
 units2 87.1
kW/unit 5
kW 9.33
⎯→⎯= 
 
 
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1-17
 
Heat Transfer Mechanisms 
 
1-33C Diamond is a better heat conductor. 
 
 
1-34C The thermal conductivity of a material is the rate of heat transfer through a unit thickness of the material per unit area 
and per unit temperature difference. The thermal conductivity of a material is a measure of how fast heat will be conducted in 
that material. 
 
 
1-35C The mechanisms of heat transfer are conduction, convection and radiation. Conduction is the transfer of energy from 
the more energetic particles of a substance to the adjacent less energetic ones as a result of interactions between the particles. 
Convection is the mode of energy transfer between a solid surface and the adjacent liquid or gas which is in motion, and it 
involves combined effects of conduction and fluid motion. Radiation is energy emitted by matter in the form of 
electromagnetic waves (or photons) as a result of the changes in the electronic configurations of the atoms or molecules. 
 
 
1-36C In solids, conduction is due to the combination of the vibrations of the molecules in a lattice and the energy transport 
by free electrons. In gases and liquids, it is due to the collisions of the molecules during their random motion. 
 
 
1-37C The parameters that effect the rate of heat conduction through a windowless wall are the geometry and surface area 
of wall, its thickness, the material of the wall, and the temperature difference across the wall. 
 
 
1-38C Conduction is expressed by Fourier's law of conduction as 
dx
dTkAQ −=cond& where dT/dx is the temperature gradient, 
k is the thermal conductivity, and A is the area which is normal to the direction of heat transfer. 
 Convection is expressed by Newton's law of cooling as where h is the convection heat 
transfer coefficient, A
)(conv ∞−= TThAQ ss&
s is the surface area through which convection heat transfer takes place, Ts is the surface temperature 
and T∞ is the temperature of the fluid sufficiently far from the surface. 
 Radiation is expressed by Stefan-Boltzman law as where )( 4surr
4
rad TTAQ ss −= εσ& ε is the emissivity of surface, As 
is the surface area, Ts is the surface temperature, Tsurr is the average surrounding surface temperature and 
 is the Stefan-Boltzman constant. 428 K W/m1067.5 ⋅×= −σ
 
 
1-39C Convection involves fluid motion, conduction does not. In a solid we can have only conduction. 
 
 
1-40C No. It is purely by radiation. 
 
 
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1-18
 
1-41C In forced convection the fluid is forced to move by external means such as a fan, pump, or the wind. The fluid 
motion in natural convection is due to buoyancy effects only. 
 
 
1-42C Emissivity is the ratio of the radiation emitted by a surface to the radiation emitted by a blackbody at the same 
temperature. Absorptivity is the fraction of radiation incident on a surface that is absorbed by the surface. The Kirchhoff's 
law of radiation states that the emissivity and the absorptivity of a surface are equal at the same temperature and wavelength. 
 
 
1-43C A blackbody is an idealized body which emits the maximum amount of radiation at a given temperature and which 
absorbs all the radiation incident on it. Real bodies emit and absorb less radiation than a blackbody at the same temperature. 
 
 
1-44C No. Such a definition will imply that doubling the thickness will double the heat transfer rate. The equivalent but 
“more correct” unit of thermal conductivity is W⋅m/m2⋅°C that indicates product of heat transfer rate and thickness per unit 
surface area per unit temperature difference. 
 
 
1-45C In a typical house, heat loss through the wall with glass window will be larger since the glass is much thinner than a 
wall, and its thermal conductivity is higher than the average conductivity of a wall. 
 
 
1-46C The house with the lower rate of heat transfer through the walls will be more energy efficient. Heat conduction is 
proportional to thermal conductivity (which is 0.72 W/m.°C for brick and 0.17 W/m.°C for wood, Table 1-1) and inversely 
proportional to thickness. The wood house is more energy efficient since the wood wall is twice as thick but it has about one-
fourth the conductivity of brick wall. 
 
 
1-47C The rate of heat transfer through both walls can be expressed as 
 
)(88.2
m 25.0
)C W/m72.0(
)(6.1
m 1.0
)C W/m16.0(
21
21
brick
21
brickbrick
21
21
wood
21
woodwood
TTA
TT
A
L
TT
AkQ
TTA
TT
A
L
TT
AkQ
−=
−
°⋅=
−
=
−=
−
°⋅=
−
=
&
&
 
where thermal conductivities are obtained from Table A-5. Therefore, heat transfer through the brick wall will be larger 
despite its higher thickness. 
 
 
1-48C The thermal conductivity of gases is proportional to the square root of absolute temperature. The thermal conductivity 
of most liquids, however, decreases with increasing temperature, with water being a notable exception. 
 
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1-19
 
1-49C Superinsulations are obtained by using layers of highly reflective sheets separated by glass fibers in an evacuated 
space. Radiation heat transfer between two surfaces is inversely proportional to the number of sheets used and thus heat loss 
by radiation will be very low by using this highly reflective sheets. At the same time, evacuating the space between the layers 
forms a vacuum under 0.000001 atm pressure which minimize conduction or convection through the air space between the 
layers. 
 
 
1-50C Most ordinary insulations are obtained by mixing fibers, powders, or flakes of insulating materials with air. Heat 
transfer through such insulations is by conduction through the solid material, and conduction or convection through the air 
space as well as radiation. Such systems are characterized by apparent thermal conductivity instead of the ordinary thermal 
conductivity in order to incorporate these convection and radiation effects. 
 
 
1-51C The thermal conductivity of an alloy of two metals will most likely be less than the thermal conductivities of both 
metals. 
 
 
 
1-52 Two surfaces of a flat plate are maintained at specified temperatures, and the rate of heat transfer through the plate is 
measured. The thermal conductivity of the plate material is to be determined. 
PlateAssumptions 1 Steady operating conditions exist since the surface temperatures of the plate 
remain constant at the specified values. 2 Heat transfer through the plate is one-
dimensional. 3 Thermal properties of the plate are constant. 
Q 
Analysis The thermal conductivity is determined directly from the steady one-dimensional 
heat conduction relation to be 
C W/m0.125 °⋅=
°−
=
−
=→
−
=
C0)(80
m) (0.02) W/m500(
)(
)/( 
2
21
21
TT
LAQk
L
TT
kAQ
&
& 0°C80°C 
 
 
 
1-53 Four power transistors are mounted on a thin vertical aluminum plate that is cooled by a fan. The temperature of the 
aluminum plate is to be determined. 
Assumptions 1 Steady operating conditions exist. 2 The entire plate is nearly isothermal. 3 Thermal properties of the wall are 
constant. 4 The exposed surfacearea of the transistor can be taken to be equal to its base area. 5 Heat transfer by radiation is 
disregarded. 6 The convection heat transfer coefficient is constant and uniform over the surface. 
Analysis The total rate of heat dissipation from the aluminum plate and the total heat transfer area are 
 
2m 0484.0m) m)(0.22 22.0(
 W48 W124
==
=×=
sA
Q&
12 W
Ts
Disregarding any radiation effects, the temperature of the 
aluminum plate is determined to be 
 C64.7°=
°⋅
+°=+=⎯→⎯−= ∞∞ )m 0484.0)(C W/m25(
 W48C25)( 22
s
sss hA
QTTTThAQ
&
& 
 
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1-20
 
1-54 The convection heat transfer coefficient heat transfer between the surface of a pipe carrying superheated vapor and the 
surrounding is to be determined. 
Assumptions 1 Steady operating conditions exist. 2 Heat transfer by radiation is not considered. 3 Rate of heat loss from the 
vapor in the pipe is equal to the heat transfer rate by convection between pipe surface and the surrounding. 
Properties The specific heat of vapor is given to be 2190 J/kg · °C. 
Analysis The surface area of the pipe is 
2m 571.1)m 10)(m 05.0( === ππDLAs 
The rate of heat loss from the vapor in the pipe can be 
determined from 
 W19710
J/s 19710C )30(C)J/kg 2190)(kg/s 3.0(
)( outinloss
=
=°°⋅=
−= TTcmQ p&&
 
With the rate of heat loss from the vapor in the pipe assumed equal to the heat transfer rate by convection, the heat transfer 
coefficient can be determined using the Newton’s law of cooling: 
)(convloss ∞−== TThAQQ ss&& 
Rearranging, the heat transfer coefficient is determined to be 
C W/m157 2 °⋅=
°−
=
−
=
∞ C )20100)(m 571.1(
 W19710
)( 2
loss
TTA
Q
h
ss
&
 
Discussion By insulating the pipe surface, heat loss from the vapor in the pipe can be reduced. 
 
 
 
 
1-55 An electrical resistor with a uniform temperature of 90 °C is in a room at 20 °C. The heat transfer coefficient by 
convection is to be determined. 
Assumptions 1 Steady operating conditions exist. 2 Radiation heat transfer is negligible. 3 No hot spot exists on the resistor. 
Analysis The total heat transfer area of the resistor is 
222 m 01276.0)m 15.0)(m 025.0(4/)m 025.0(2)4/(2 =+=+= ππππ DLDAs 
The electrical energy converted to thermal energy is transferred by convection: 
 W30)V 6)(A 5(conv === IVQ& 
From Newton’s law of cooling, the heat transfer by convection is given as 
)(conv ∞−= TThAQ ss& 
Rearranging, the heat transfer coefficient is determined to be 
C W/m33.6 2 °⋅=
°−
=
−
=
∞ C )2090)(m 01276.0(
 W30
)( 2
conv
TTA
Q
h
ss
&
 
Discussion By comparing the magnitude of the heat transfer coefficient determined here with the values presented in Table 
1-5, one can conclude that it is likely that forced convection is taking place rather than free convection. 
 
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1-21
 
1-56 The inner and outer surfaces of a brick wall are maintained at specified temperatures. 
The rate of heat transfer through the wall is to be determined. 
Assumptions 1 Steady operating conditions exist since the surface temperatures of the wall 
remain constant at the specified values. 2 Thermal properties of the wall are constant. 
0.3 m 
Brick wall 
Properties The thermal conductivity of the wall is given to be k = 0.69 W/m⋅°C. 
Analysis Under steady conditions, the rate of heat transfer through the wall is 
8°C 26°C 
 W1159=°−×°⋅=∆=
m 0.3
C)8(26)m 7C)(4W/m (0.69 2cond L
TkAQ& 
 
 
 
 
 
1-57 The inner and outer surfaces of a window glass are maintained at specified temperatures. The amount of heat transfer 
through the glass in 5 h is to be determined. 
Assumptions 1 Steady operating conditions exist since the surface temperatures of the glass remain constant at the specified 
values. 2 Thermal properties of the glass are constant. 
Glass Properties The thermal conductivity of the glass is given to be k = 0.78 W/m⋅°C. 
Analysis Under steady conditions, the rate of heat transfer through the glass by 
conduction is 
 W 4368
m0.005
C3)(10
)m 2C)(2W/m (0.78 2cond =
°−
×°⋅=
∆
=
L
TkAQ& 
3°C 10°C Then the amount of heat transfer over a period of 5 h becomes 
0.5 cm kJ 78,620=×=∆= s) 3600kJ/s)(5 (4.368cond tQQ &
If the thickness of the glass doubled to 1 cm, then the amount of heat transfer will go down by half to 39,310 kJ. 
 
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1-22
 
1-58 Prob. 1-57 is reconsidered. The amount of heat loss through the glass as a function of the window glass 
thickness is to be plotted. 
Analysis The problem is solved using EES, and the solution is given below. 
 
"GIVEN" 
L=0.005 [m] 
A=2*2 [m^2] 
T_1=10 [C] 
T_2=3 [C] 
k=0.78 [W/m-C] 
time=5*3600 [s] 
 
"ANALYSIS" 
Q_dot_cond=k*A*(T_1-T_2)/L 
Q_cond=Q_dot_cond*time*Convert(J, kJ) 
 
 
 
0.002 0.004 0.006 0.008 0.01
0
50000
100000
150000
200000
250000
300000
350000
400000
L [m]
Q
co
nd
 [
kJ
]
 
L [m] Qcond [kJ] 
0.001 393120 
0.002 196560 
0.003 131040 
0.004 98280 
0.005 78624 
0.006 65520 
0.007 56160 
0.008 49140 
0.009 43680 
0.01 39312 
 
 
 
 
 
 
 
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1-23
 
1-59 Heat is transferred steadily to boiling water in the pan through its bottom. The inner surface temperature of the bottom 
of the pan is given. The temperature of the outer surface is to be determined. 
Assumptions 1 Steady operating conditions exist since the surface temperatures of the pan remain constant at the specified 
values. 2 Thermal properties of the aluminum pan are constant. 
Properties The thermal conductivity of the aluminum is given to be k = 237 W/m⋅°C. 
Analysis The heat transfer area is 
 A = π r2 = π (0.075 m)2 = 0.0177 m2
105°C
1400 W 
Under steady conditions, the rate of heat transfer through the bottom 
of the pan by conduction is 
 
L
TT
kA
L
TkAQ 12
−
=
∆
=& 
0.4 cm
 Substituting, 
 
m 0.004
C105
)m C)(0.0177W/m (237W 1400 22
°−
°⋅=
T
 
which gives 
 T2 = 106.33°C 
 
 
 
 
1-60E The inner and outer surface temperatures of the wall of an electrically heated home during a winter night are 
measured. The rate of heat loss through the wall that night and its cost are to be determined. 
Assumptions 1 Steady operating conditions exist since the surface temperatures of the wall remain constant at the specified 
values during the entire night. 2 Thermal properties of the wall are constant. 
Properties The thermal conductivity of the brick wall is given to be k = 0.42 Btu/h⋅ft⋅°F. 
Analysis (a) Noting that the heat transfer through the wall is by conduction and the surface area of the wall is 
, the steady rate of heat transfer through the wall can be determined from 2ft 200=ft 10ft 20 ×=A
 Btu/h 3108=°−°=
−
=
ft 1
F)2562()ft F)(200Btu/h.ft. 42.0( 221
L
TT
kAQ& 
1 ft 
Q 
Brick
or 0.911 kW since 1 kW = 3412 Btu/h. 
(b) The amount of heat lost during an 8 hour period and its cost are 
 kWh 7.288h) kW)(8 911.0( ==∆= tQQ &
25°F 62°F 
 
$0.51=
/kWh)kWh)($0.07 (7.288=
 energy) ofcost it energy)(Un of(Amount =Cost 
Therefore, the cost of the heat loss through the wall to the home owner that night is $0.51. 
 
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1-24
 
1-61 The thermal conductivity of a material is to be determined by ensuring one-dimensional heat conduction, and by 
measuring temperatures when steady operating conditions are reached. 
Assumptions 1 Steady operating conditions exist since the temperature readings do not change with time. 2 Heat losses 
through the lateral surfaces of the apparatus are negligible since those surfaces are well-insulated, and thus the entire heat 
generated by the heater is conducted through the samples. 3 The apparatus possesses thermal symmetry. 
Analysis The electrical power consumed by the heater and converted to heat is 
Q 
 W66)A 6.0)(V 110( === IWe V&
The rate of heat flow through each sample is 
 W33
2
 W66
2
=== e
W
Q
&
& 3 
Then the thermal conductivity of the sample becomes 
3 
 
C W/m.98.5 °=
°
=
∆
=⎯→⎯
∆
===
)C8)(m 001257.0(
m) W)(0.0333(=
m 001257.0
4
)m 04.0(
4
2
2
22
TA
LQk
L
TkAQ
DA
&
&
ππ
 
 
 
 
 
1-62 The thermal conductivity of a material is to be determined by ensuring one-dimensional heat conduction, and by 
measuring temperatures when steady operating conditions are reached. 
Assumptions 1 Steady operating conditions exist since the temperature readings do not change with time. 2 Heat losses 
through the lateral surfaces of the apparatus are negligible since those surfaces are well-insulated, and thus the entire heat 
generated by the heater is conducted through the samples. 3 The apparatus possesses thermal symmetry. 
Analysis For each sample we have 
Q& Q&
A
L L 
 
 
C87482
m 01.0m) 1.0m)( 1.0(
 W5.122/25
2
°=−=∆
==
==
T
A
Q&
Then the thermal conductivity of the material becomes 
 C W/m.0.781 °=
°
=
∆
=⎯→⎯
∆
=
)C8)(m 01.0(
m) W)(0.0055.12(
2TA
LQ
k
L
TkAQ
&
& 
 
 
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1-25
 
1-63 The thermal conductivity of a material is to be determined by ensuring one-dimensional heat conduction, and by 
measuring temperatures when steady operating conditions are reached. 
Assumptions 1 Steady operating conditions exist since the temperature readings do not change with time. 2 Heat losses 
through the lateral surfaces of the apparatus are negligible since those surfaces are well-insulated, and thus the entire heat 
generated by the heater is conducted through the samples. 3 The apparatus possesses thermal symmetry. 
Analysis For each sample we have 
Q&Q&
A
L L 
 
 
C87482
m 01.0m) 1.0m)( 1.0(
 W102/20
2
°=−=∆
==
==
T
A
Q&
Then the thermal conductivity of the material becomes 
 C W/m0.625 °⋅=
°
=
∆
=⎯→⎯
∆
=
)C8)(m 01.0(
m) W)(0.00510(
2TA
LQ
k
L
TkAQ
&
& 
 
 
 
 
1-64 The thermal conductivity of a refrigerator door is to be determined by measuring the 
surface temperatures and heat flux when steady operating conditions are reached. 
q& 
L = 3 cm 
7°C 
Assumptions 1 Steady operating conditions exist when measurements 
are taken. 2 Heat transfer through the door is one dimensional since 
the thickness of the door is small relative to other dimensions. 
Analysis The thermal conductivity of the door material is determined 
directly from Fourier’s relation to be 
15°C
 C W/m0.120 °⋅=
°−
=
∆
=⎯→⎯
∆
=
C)715(
m) )(0.03 W/m32( 2
T
Lqk
L
Tkq
&
& 
 
 
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1-26
 
1-65 The rate of radiation heat transfer between a person and the surrounding surfaces at specified temperatures is to be 
determined in summer and in winter. 
Assumptions 1 Steady operating conditions exist. 2 Heat transfer by convection is not considered. 3 The person is 
completely surrounded by the interior surfaces of the room. 4 The surrounding surfaces are at a uniform temperature. 
Properties The emissivity of a person is given to be ε = 0.95 
Analysis Noting that the person is completely enclosed by the surrounding surfaces, the net rates of radiation heat transfer 
from the body to the surrounding walls, ceiling, and the floor in both cases are: 
(a) Summer: Tsurr = 23+273=296 
Qrad
Tsurr
 W84.2=
]KK) (296273)+)[(32m )(1.6.K W/m1067.5)(95.0(
)(
4442428
4
surr
4
rad
−×=
−=
−
TTAQ ssεσ&
 
(b) Winter: Tsurr = 12+273= 285 K 
 W177.2=
]KK) (285273)+)[(32m )(1.6.K W/m1067.5)(95.0(
)(
4442428
4
surr
4
rad
−×=
−=
−
TTAQ ssεσ&
 
Discussion Note that the radiation heat transfer from the person more than doubles in winter. 
 
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1-27
 
1-66 Prob. 1-65 is reconsidered. The rate of radiation heat transfer in winter as a function of the temperature of the 
inner surface of the room is to be plotted. 
Analysis The problem is solved using EES, and the solution is given below. 
 
"GIVEN" 
T_infinity=(20+273) [K] 
T_surr_winter=(12+273) [K] 
T_surr_summer=(23+273) [K] 
A=1.6 [m^2] 
epsilon=0.95 
T_s=(32+273) [K] 
 
"ANALYSIS" 
sigma=5.67E-8 [W/m^2-K^4] "Stefan-Boltzman constant" 
Q_dot_rad_summer=epsilon*sigma*A*(T_s^4-T_surr_summer^4) 
Q_dot_rad_winter=epsilon*sigma*A*(T_s^4-T_surr_winter^4) 
 
 
Tsurr, winter 
[K] 
Qrad, winter 
[W] 
281 208.5 
282 200.8 
283 193 
284 185.1 
285 177.2 
286 169.2 
287 161.1 
288 152.9 
289 144.6 
290 136.2 
291 127.8 
281 283 285 287 289 291
120
130
140
150
160
170
180
190
200
210
Tsurr,winter [K]
Q
ra
d,
w
in
te
r 
[W
]
 
 
 
 
 
 
 
 
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1-28
 
1-67 A person is standing in a room at a specified temperature. The rate of heat transfer between a person and the 
surrounding air by convection is to be determined. 
Qconv
Room air
TairAssumptions 1 Steady operating conditions exist. 2 Heat transfer by radiation is 
not considered. 3 The environment is at a uniform temperature. 
Analysis The heat transfer surface area of the person is 
 As = πDL= π(0.3 m)(1.70 m) = 1.602 m2
Under steady conditions, the rate of heat transfer by convection is 
 W 205=°−°⋅=∆= C)18)(34m C)(1.602W/m (8 22conv ThAQ s&
 
 
 
 
 
1-68 Hot air is blown over a flat surface at a specified temperature. The rate of heat transfer from the air to the plate is to be 
determined. 
Assumptions 1 Steady operating conditions exist. 2 Heat transfer by radiation is 
not considered. 3 The convection heat transfer coefficient is constant and uniform 
over the surface. 
30°C
80°C 
Air 
Analysis Under steady conditions, the rate of heat transfer by convection is 
 W22,000 C30))(80m 4C)(2W/m (55 22conv =°−×°⋅=∆= ThAQ s&
 
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1-29
 
1-69 Prob. 1-68 is reconsidered. The rate of heat transfer as a function of the heat transfer coefficient is to be plotted. 
Analysis The problem is solved using EES, and the solution is given below. 
 
"GIVEN" 
T_infinity=80 [C] 
A=2*4 [m^2] 
T_s=30 [C] 
h=55 [W/m^2-C] 
 
"ANALYSIS" 
Q_dot_conv=h*A*(T_infinity-T_s) 
 
 
h 
[W/m2.C] 
Qconv 
[W] 
20 8000 
30 12000 
40 16000 
50 20000 
60 24000 
70 28000 
80 32000 
90 36000 
100 40000 
20 30 40 50 6070 80 90 100
5000
10000
15000
20000
25000
30000
35000
40000
h [W/m2-C]
Q
co
nv
 [
W
]
 
 
 
 
 
 
 
 
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preparation. If you are a student using this Manual, you are using it without permission. 
1-30
 
1-70 A spacecraft in space absorbs solar radiation while losing heat to deep space by thermal radiation. The surface 
temperature of the spacecraft is to be determined when steady conditions are reached. 
Assumptions 1 Steady operating conditions exist since the surface temperatures of the wall remain constant at the specified 
values. 2 Thermal properties of the wall are constant. 
Properties The outer surface of a spacecraft has an emissivity of 0.8 and an absorptivity of 0.3. 
α = 0.3 
ε = 0.8 
 . 
Qrad
 950 W/m2
Analysis When the heat loss from the outer surface of the spacecraft by radiation equals 
the solar radiation absorbed, the surface temperature can be determined from 
]K) (0)[KW/m 10(5.670.8) W/m950(3.0
)(
444282
4
space
4
solar
radabsorbedsolar
−⋅×××=××
−=
=
−
sss
ss
TAA
TTAQ
QQ
εσα &
&&
 
Canceling the surface area A and solving for Ts gives 
 Ts = 281.5 K
 
 
 
 
1-71 The heat generated in the circuitry on the surface of a 5-W silicon chip is conducted to the ceramic substrate. The 
temperature difference across the chip in steady operation is to be determined. 
Assumptions 1 Steady operating conditions exist. 2 Thermal properties of the chip are constant. 
Properties The thermal conductivity of the silicon chip is given to 
be k = 130 W/m⋅°C. 
Analysis The temperature difference between the front and back 
surfaces of the chip is 
2m 000036.0m) m)(0.006 006.0( ==A 
C0.53°=
°⋅
==∆
∆
=
)m 6C)(0.00003 W/m130(
m) 0005.0 W)(5(
2kA
LQT
L
TkAQ
&
&
 
5 W 
Chip 
6 × 6 × 0.5 mm
Ceramic 
substrate 
Q& 
 
 
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1-72 An electric resistance heating element is immersed in water initially at 20°C. The time it will take for this heater to 
raise the water temperature to 80°C as well as the convection heat transfer coefficients at the beginning and at the end of the 
heating process are to be determined. 
Assumptions 1 Steady operating conditions exist and thus the rate of heat loss from the wire equals the rate of heat 
generation in the wire as a result of resistance heating. 2 Thermal properties of water are constant. 3 Heat losses from the 
water in the tank are negligible. 
Properties The specific heat of water at room temperature is c = 4.18 kJ/kg⋅°C (Table A-9). 
Analysis When steady operating conditions are reached, we have . This is also equal to the rate of 
heat gain by water. Noting that this is the only mechanism of energy transfer, the time it takes to raise the water temperature 
from 20°C to 80°C is determined to be 
 W800generated == EQ &&
h 6.53==°−°⋅=
−
=∆
−=∆
−=
s 510,23
J/s 800
C20)C)(80J/kg kg)(4180 (75)(
)(
)(
in
12
12in
12in
Q
TTmc
t
TTmctQ
TTmcQ
&
& water 
800 W 120°C
The surface area of the wire is 
2m 0.00628 = m) m)(0.4 005.0(ππ == DLAs 
 The Newton's law of cooling for convection heat transfer is expressed as . Disregarding any heat transfer 
by radiation and thus assuming all the heat loss from the wire to occur by convection, the convection heat transfer 
coefficients at the beginning and at the end of the process are determined to be 
)( ∞−= TThAQ ss&
 
C W/m3185
C W/m1274
2
2
°⋅=
°−
=
−
=
°⋅=
°−
=
−
=
∞
∞
C)80120)(m (0.00628
 W800
)(
C)20120)(m (0.00628
 W800
)(
2
2
2
2
1
1
TTA
Q
h
TTA
Q
h
ss
ss
&
&
 
Discussion Note that a larger heat transfer coefficient is needed to dissipate heat through a smaller temperature difference for 
a specified heat transfer rate. 
 
 
 
 
1-73 A hot water pipe at 80°C is losing heat to the surrounding air at 5°C by natural convection with a heat transfer 
coefficient of 25 W/m2⋅°C. The rate of heat loss from the pipe by convection is to be determined. 
Assumptions 1 Steady operating conditions exist. 2 Heat transfer by radiation is 
not considered. 3 The convection heat transfer coefficient is constant and 
uniform over the surface. 
L = 10 m
D =5 cm 
80°C
Analysis The heat transfer surface area is 
 As = πDL = π (0.05 m)(10 m) = 1.571 m2
QUnder steady conditions, the rate of heat transfer by convection is 
Air, 5°C 
 W2945 C5))(80m C)(1.571W/m(25 22conv =°−°⋅=∆= ThAQ s&
 
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1-32
 
1-74 A hollow spherical iron container is filled with iced water at 0°C. The rate of heat loss from the sphere and the rate at 
which ice melts in the container are to be determined. 
Assumptions 1 Steady operating conditions exist since the surface temperatures of the wall remain constant at the specified 
values. 2 Heat transfer through the shell is one-dimensional. 3 Thermal properties of the iron shell are constant. 4 The inner 
surface of the shell is at the same temperature as the iced water, 0°C. 
Properties The thermal conductivity of iron is k = 80.2 W/m⋅°C (Table A-3). The heat of fusion of water is given to be 333.7 
kJ/kg. 
Analysis This spherical shell can be approximated as a plate of thickness 0.4 cm and area 
Iced 
water 
0°C 
5°C 
0.2 cm
 A = πD2 = π (0.2 m)2 = 0.126 m2
Then the rate of heat transfer through the shell by conduction is 
 kW 25.3==°−°⋅=∆= W 263,25
m 0.002
C0)(5
)m C)(0.126W/m (80.2 2cond L
TkAQ& 
Considering that it takes 333.7 kJ of energy to melt 1 kg of ice at 0°C, the rate at which 
ice melts in the container can be determined from 
 kg/s 0.0757===
kJ/kg 333.7
kJ/s 25.263
ice
ifh
Q
m
&
& 
Discussion We should point out that this result is slightly in error for approximating a curved wall as a plain wall. The error 
in this case is very small because of the large diameter to thickness ratio. For better accuracy, we could use the inner surface 
area (D = 19.6 cm) or the mean surface area (D = 19.8 cm) in the calculations. 
 
 
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1-33
 
1-75 Prob. 1-74 is reconsidered. The rate at which ice melts as a function of the container thickness is to be plotted. 
Analysis The problem is solved using EES, and the solution is given below. 
 
"GIVEN" 
D=0.2 [m] 
L=0.2 [cm] 
T_1=0 [C] 
T_2=5 [C] 
 
"PROPERTIES" 
h_if=333.7 [kJ/kg] 
k=k_(Iron, 25) 
 
"ANALYSIS" 
A=pi*D^2 
Q_dot_cond=k*A*(T_2-T_1)/(L*Convert(cm, m)) 
m_dot_ice=(Q_dot_cond*Convert(W, kW))/h_if 
 
 
L 
[cm] 
mice 
[kg/s] 
0.1 
0.2 
0.3 
0.4 
0.5 
0.6 
0.7 
0.8 
0.9 
1 
0.1515 
0.07574 
0.0505 
0.03787 
0.0303 
0.02525 
0.02164 
0.01894 
0.01683 
0.01515 
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
0
0.02
0.04
0.06
0.08
0.1
0.12
0.14
0.16
L [cm]
m
ic
e 
 [k
g/
s]
 
 
 
 
 
 
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1-34
 
1-76E The inner and outer glasses of a double pane window with a 0.5-in air space are at specified temperatures. The rate of 
heat transfer through the window is to be determined 
Air
Glass
Q&
Assumptions 1 Steady operating conditions exist since the surface temperatures 
of the glass remain constant at the specified values. 2 Heat transferthrough the 
window is one-dimensional. 3 Thermal properties of the air are constant. 
Properties The thermal conductivity of air at the average temperature of 
(60+48)/2 = 54°F is k = 0.01419 Btu/h⋅ft⋅°F (Table A-15E). 
Analysis The area of the window and the rate of heat loss through it are 
2m 16ft) 4(ft) 4( =×=A 
60°F
Btu/h 131=
°−
°=
−
=
ft 12/25.0
F)4860(
)ft F)(16Btu/h.ft. 01419.0( 221
L
TT
kAQ& 48°F 
 
 
 
 
 
1-77 A transistor mounted on a circuit board is cooled by air flowing over it. The transistor case temperature is not to exceed 
70°C when the air temperature is 55°C. The amount of power this transistor can dissipate safely is to be determined. 
Assumptions 1 Steady operating conditions exist. 2 Heat transfer by radiation is 
disregarded. 3 The convection heat transfer coefficient is constant and uniform 
over the surface. 4 Heat transfer from the base of the transistor is negligible. 
Power 
transistor 
Air, 
55°C 
Analysis Disregarding the base area, the total heat transfer area of the transistor 
is 
 
24
22
2
m 10037.1
cm 037.14/)cm 6.0(cm) cm)(0.4 6.0(
4/
−×=
=+=
+=
ππ
ππ DDLAs
Then the rate of heat transfer from the power transistor at specified 
conditions is 
 W0.047=°−×°⋅=−= ∞ C)5570)(m 10C)(1.037 W/m30()(
2-42TThAQ ss&
Therefore, the amount of power this transistor can dissipate safely is 0.047 W. 
 
 
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1-35
 
1-78 Prob. 1-77 is reconsidered. The amount of power the transistor can dissipate safely as a function of the 
maximum case temperature is to be plotted. 
Analysis The problem is solved using EES, and the solution is given below. 
 
"GIVEN" 
L=0.004 [m] 
D=0.006 [m] 
h=30 [W/m^2-C] 
T_infinity=55 [C] 
T_case_max=70 [C] 
 
"ANALYSIS" 
A=pi*D*L+pi*D^2/4 
Q_dot=h*A*(T_case_max-T_infinity) 
 
 
Tcase, max 
[C] 
Q 
[W] 
60 0.01555 
62.5 0.02333 
65 0.0311 
67.5 0.03888 
70 0.04665 
72.5 0.05443 
75 0.0622 
77.5 0.06998 
80 0.07775 
82.5 0.08553 
85 0.09331 
87.5 0.1011 
90 0.1089 
60 65 70 75 80 85 90
0
0.02
0.04
0.06
0.08
0.1
0.12
Tcase,max [C]
Q
 [
W
]
 
 
 
 
 
 
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1-79E A 300-ft long section of a steam pipe passes through an open space at a specified temperature. The rate of heat loss 
from the steam pipe and the annual cost of this energy lost are to be determined. 
Assumptions 1 Steady operating conditions exist. 2 Heat transfer by radiation 
is disregarded. 3 The convection heat transfer coefficient is constant and 
uniform over the surface. 
Analysis (a) The rate of heat loss from the steam pipe is 
D =4 in
280°F 
2ft 2.314ft) 300(ft) 12/4( === ππDLAs 
Btu/h 433,500≅
°−°⋅⋅=−=
Btu/h 433,540=
F)50280)(ft 2.314(F)ftBtu/h 6()( 22airpipe TThAQ ss& L=300 ft Q
Air,50°F
(b) The amount of heat loss per year is 
 Btu/yr 10798.3h/yr) 24Btu/h)(365 540,433( 9×=×=∆= tQQ &
The amount of gas consumption per year in the furnace that has an efficiency of 86% is 
 therms/yr161,44
Btu 100,000
 therm1
86.0
Btu/yr 10798.3LossEnergy Annual
9
=⎟⎟
⎠
⎞
⎜⎜
⎝
⎛×
= 
Then the annual cost of the energy lost becomes 
 
$48,576/yr=
=
therm)/10.1($) therms/yr(44,161=
energy) ofcost loss)(Unitenergy Annual(costEnergy 
 
 
 
 
1-80 A 4-m diameter spherical tank filled with liquid nitrogen at 1 atm and -196°C is exposed to convection with ambient 
air. The rate of evaporation of liquid nitrogen in the tank as a result of the heat transfer from the ambient air is to be 
determined. 
Assumptions 1 Steady operating conditions exist. 2 Heat transfer by radiation is disregarded. 3 The convection heat transfer 
coefficient is constant and uniform over the surface. 4 The temperature of the thin-shelled spherical tank is nearly equal to 
the temperature of the nitrogen inside. 
Properties The heat of vaporization and density of liquid nitrogen at 1 atm are given to be 198 kJ/kg and 810 kg/m3, 
respectively. 
Analysis The rate of heat transfer to the nitrogen tank is 
Vapor 
 222 m 27.50m) 4( === ππDAs Air 
20°C 
Q&
1 atm 
Liquid 
N2
 
 W430,271
C)]196(20)[m 27.50(C) W/m25()( 22air
=
°−−°⋅=−= TThAQ ss&
Then the rate of evaporation of liquid nitrogen in the tank is determined to be 
 kg/s 1.37===⎯→⎯=
kJ/kg 198
kJ/s 430.271
fg
fg h
Q
mhmQ
&
&&& 
 
 
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1-81 A 4-m diameter spherical tank filled with liquid oxygen at 1 atm and -183°C is exposed to convection with ambient air. 
The rate of evaporation of liquid oxygen in the tank as a result of the heat transfer from the ambient air is to be determined. 
Assumptions 1 Steady operating conditions exist. 2 Heat transfer by radiation is disregarded. 3 The convection heat transfer 
coefficient is constant and uniform over the surface. 4 The temperature of the thin-shelled spherical tank is nearly equal to 
the temperature of the oxygen inside. 
Properties The heat of vaporization and density of liquid oxygen at 1 atm are given to be 213 kJ/kg and 1140 kg/m3, 
respectively. 
Vapor Analysis The rate of heat transfer to the oxygen tank is 
Air 
20°C 
Q& 
1 atm 
Liquid O2
-183°C 
 222 m 27.50m) 4( === ππDAs
 
 W120,255
C)]183(20)[m 27.50(C). W/m25()( 22air
=
°−−°=−= TThAQ ss&
Then the rate of evaporation of liquid oxygen in the tank is determined to be 
 kg/s 1.20===⎯→⎯=
kJ/kg 213
kJ/s 120.255
fg
fg h
Q
mhmQ
&
&&& 
 
 
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1-82 Prob. 1-80 is reconsidered. The rate of evaporation of liquid nitrogen as a function of the ambient air 
temperature is to be plotted. 
Analysis The problem is solved using EES, and the solution is given below. 
 
"GIVEN" 
D=4 [m] 
T_s=-196 [C] 
T_air=20 [C] 
h=25 [W/m^2-C] 
 
"PROPERTIES" 
h_fg=198 [kJ/kg] 
 
"ANALYSIS" 
A=pi*D^2 
Q_dot=h*A*(T_air-T_s) 
m_dot_evap=(Q_dot*Convert(J/s, kJ/s))/h_fg 
 
 
Tair 
[C] 
mevap 
[kg/s] 
0 
2.5 
5 
7.5 
10 
12.5 
15 
17.5 
20 
22.5 
25 
27.5 
30 
32.5 
35 
37.5 
40 
1.244 
1.26 
1.276 
1.292 
1.307 
1.323 
1.339 
1.355 
1.371 
1.387 
1.403 
1.418 
1.434 
1.45 
1.466 
1.482 
1.498 
0 5 10 15 20 25 30 35 40
1.2
1.25
1.3
1.35
1.4
1.45
1.5
Tair [C]
m
ev
ap
 [
kg
/s
]
 
 
 
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1-83 A person with a specified surface temperature is subjected to radiation heat transfer in a room at specified wall 
temperatures. The rate of radiation heat loss from the person is to be determined. 
Assumptions 1 Steady operating conditions exist. 2 Heat transfer by convection is disregarded. 3 The emissivity of the 
person is constant and uniform over the exposed surface. 
Properties The average emissivity of the person is given to be 0.5. 
Analysis Noting that the person is completely enclosed by the surrounding surfaces, the net rates of radiation heat transfer 
from the body to the surrounding walls, ceiling, and the floor in both cases are 
(a) Tsurr = 300 K 
32°C
Qrad
Tsurr
 
 W26.7=
]KK) (300273)+)[(32m )(1.7.K W/m1067.5)(5.0(
)(
44424284
surr
4
rad
−×=
−=
−
TTAQ ssεσ&
(b) Tsurr = 280 K 
 
 W121=
]KK) (280273)+)[(32m )(1.7.K W/m1067.5)(5.0(
)(
4442428
4
surr
4
rad
−×=
−=
−
TTAQ ssεσ&
Discussion Note that the radiation heat transfer goes up by more than 4 times as the temperature of the surrounding surfaces 
drops from 300 K to 280 K. 
 
 
 
 
1-84 A circuit board houses 80 closely spaced logic chips on one side, each dissipating 0.06 W. All the heat generated in the 
chips is conducted across the circuit board. The temperature difference between the two sides of the circuit board is to be 
determined. 
Assumptions 1 Steady operating conditions exist. 2 Thermal properties of the board are constant. 3 All the heat generated in 
the chips is conducted across the circuit board. 
Properties The effective thermal conductivity of the board is given to 
be k = 16 W/m⋅°C. 
ChipsAnalysis The total rate of heat dissipated by the chips is Q& 
 W8.4 W)06.0(80 =×=Q&
Then the temperature difference between the front and back surfaces of the board is 
 2m 0216.0m) m)(0.18 12.0( ==A
 C0.042°=
°⋅
==∆⎯→⎯
∆
=
)m C)(0.0216 W/m16(
m) 003.0 W)(8.4(
2kA
LQ
T
L
TkAQ
&
& 
Discussion Note that the circuit board is nearly isothermal. 
 
 
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1-85 A sealed electronic box dissipating a total of 120 W of power is placed in a vacuum chamber. If this box is to be cooled 
by radiation alone and the outer surface temperature of the box is not to exceed 55°C, the temperature the surrounding 
surfaces must be kept is to be determined. 
Assumptions 1 Steady operating conditions exist. 2 Heat transfer by convection is disregarded. 3 The emissivity of the box is 
constant and uniform over the exposed surface. 4 Heat transfer from the bottom surface of the box to the stand is negligible. 
Properties The emissivity of the outer surface of the box is given to be 0.95. 
Analysis Disregarding the base area, the total heat transfer area of the electronic box is 
 2m 65.0)m 5.0)(m 2.0(4m) m)(0.5 5.0( =×+=sA
The radiation heat transfer from the box can be expressed as 
 [ ]4surr42428
4
surr
4
rad
)K 27355()m 65.0)(K W/m1067.5)(95.0( W120
)(
T
TTAQ ss
−+⋅×=
−=
−
εσ&
120 W 
ε = 0.95 
Ts =55°C 
which gives Tsurr = 300.4 K = 27.4°C. Therefore, the temperature of the surrounding surfaces must be less than 27.4°C. 
 
 
 
 
1-86E Using the conversion factors between W and Btu/h, m and ft, and K and R, the Stefan-Boltzmann constant 
 is to be expressed in the English unit, . 428 K W/m1067.5 ⋅×= −σ 42 RftBtu/h ⋅⋅
Analysis The conversion factors for W, m, and K are given in conversion tables to be 
 
R 1.8 =K 1
ft 3.2808 = m 1
Btu/h 3.41214 = W 1
 Substituting gives the Stefan-Boltzmann constant in the desired units, 
 42 RftBtu/h 0.171 ⋅⋅=×⋅=
42
42
R) (1.8ft) 2808.3(
Btu/h 3.412145.67=K W/m67.5σ 
 
 
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1-87E Using the conversion factors between W and Btu/h, m and ft, and °C and °F, the convection coefficient in SI units is 
to be expressed in Btu/h⋅ft2⋅°F. 
Analysis The conversion factors for W and m are straightforward, and are given in conversion tables to be 
 
ft 3.2808 = m 1
Btu/h 3.41214 = W 1
The proper conversion factor between °C into °F in this case is 
 F1.8=C1 °°
since the °C in the unit W/m2⋅°C represents per °C change in temperature, and 1°C change in temperature corresponds to a 
change of 1.8°F. Substituting, we get 
 FftBtu/h 1761.0
F) (1.8ft) 2808.3(
Btu/h 3.41214=C W/m1 2
2
2 °⋅⋅=
°
°⋅ 
which is the desired conversion factor. Therefore, the given convection heat transfer coefficient in English units is 
 FftBtu/h 3.87 2 °⋅⋅=°⋅⋅×°⋅= FftBtu/h 0.176122=C W/m22 22h
 
 
 
 
1-88 An aircraft flying under icing conditions is considered. The temperature of the wings to prevent ice from forming on 
them is to be determined. 
Assumptions 1 Steady operating conditions exist. 2 Heat transfer coefficient is constant. 
Properties The heat of fusion and the density of ice are given to be 333.7 kJ/kg and 920 kg/m3, respectively. 
Analysis The temperature of the wings to prevent ice from forming on them is determined to be 
C34.1°=
°⋅
+°=+=
C W/m150
J/kg) 00m/s)(333,7 )(0.001/60kg/m 920(
C0
2
3
icewing h
Vh
TT if
ρ
 
 
 
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1-89 The convection heat transfer coefficient for heat transfer from an electrically heated wire to air is to be determined by 
measuring temperatures when steady operating conditions are reached and the electric power consumed. 
Assumptions 1 Steady operating conditions exist since the temperature readings do not change with time. 2 Radiation heat 
transfer is negligible. 
Analysis In steady operation, the rate of heat loss from the wire equals the rate of heat generation in the wire as a result of 
resistance heating. That is, 
 W330= A) V)(3 110(generated === IEQ V&&
D =0.2 cm
 Q 
180°C 
The surface area of the wire is 
 2m 0.01319 = m) m)(2.1 002.0(ππ == DLAs
L = 2.1 m 
The Newton's law of cooling for convection heat transfer is expressed as Air, 20°C 
 )( ∞−= TThAQ ss&
Disregarding any heat transfer by radiation, the convection heat transfer coefficient is determined to be 
 C W/m156 2 °⋅=
°−
=
−
=
∞ C)20180)(m (0.01319
 W330
)( 21 TTA
Qh
s
&
 
Discussion If the temperature of the surrounding surfaces is equal to the air temperature in the room, the value obtained 
above actually represents the combined convection and radiation heat transfer coefficient. 
 
 
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1-43
 
1-90 Prob. 1-89 is reconsidered. The convection heat transfer coefficient as a function of the wire surface temperature 
is to be plotted. 
Analysis The problem is solved using EES, and the solution is given below. 
 
"GIVEN" 
L=2.1 [m] 
D=0.002 [m] 
T_infinity=20 [C] 
T_s=180 [C] 
V=110 [Volt] 
I=3 [Ampere] 
 
"ANALYSIS" 
Q_dot=V*I 
A=pi*D*L 
Q_dot=h*A*(T_s-T_infinity) 
 
 
Ts 
[C] 
h 
[W/m2.C] 
100 
120 
140 
160 
180 
200 
220 
240 
260 
280 
300 
312.6 
250.1 
208.4 
178.6 
156.3 
138.9 
125.1 
113.7 
104.2 
96.19 
89.32 
100 140 180 220 260 300
50
100
150
200
250
300
350
Ts [C]
h 
[W
/m
2 -
C]
 
 
 
 
 
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Simultaneous Heat Transfer Mechanisms 
 
1-91C All three modes of heat transfer can not occur simultaneously in a medium. A medium may involve two of them 
simultaneously. 
 
 
1-92C (a) Conduction and convection: No. (b) Conduction and radiation: Yes. Example: A hot surface on the ceiling. (c) 
Convection and radiation: Yes. Example: Heat transfer from the human body. 
 
 
1-93C The human body loses heat by convection, radiation, and evaporation in both summer and winter. In summer, we can 
keep cool by dressing lightly, staying in cooler environments, turning a fan on, avoiding humid places and direct exposure to 
the sun. In winter, we can keep warm by dressing heavily, staying in a warmer environment, and avoiding drafts.1-94C The fan increases the air motion around the body and thus the convection heat transfer coefficient, which increases the 
rate of heat transfer from the body by convection and evaporation. In rooms with high ceilings, ceiling fans are used in 
winter to force the warm air at the top downward to increase the air temperature at the body level. This is usually done by 
forcing the air up which hits the ceiling and moves downward in a gently manner to avoid drafts. 
 
 
 
 
1-95 The total rate of heat transfer from a person by both convection and radiation to the surrounding air and surfaces at 
specified temperatures is to be determined. 
Assumptions 1 Steady operating conditions exist. 2 The person is 
completely surrounded by the interior surfaces of the room. 3 The 
surrounding surfaces are at the same temperature as the air in the room. 4 
Heat conduction to the floor through the feet is negligible. 5 The convection 
coefficient is constant and uniform over the entire surface of the person. 
Qconv
32°C 
ε=0.
18°C
Qrad
Tsurr
Properties The emissivity of a person is given to be ε = 0.9. 
Analysis The person is completely enclosed by the surrounding surfaces, and 
he or she will lose heat to the surrounding air by convection and to the 
surrounding surfaces by radiation. The total rate of heat loss from the person 
is determined from 
 W128.6=]K273)+(18273)+)[(32m )(1.7.K W/m1067.5)(90.0()( 44424284surr
4
rad −×=−=
−TTAQ ssεσ&
 W119C)1832()m K)(1.7W/m (5 22conv =°−⋅=∆= ThAQ s& 
 and 
W247.6 1196.128radconvtotal =+=+= QQQ &&& 
Discussion Note that heat transfer from the person by evaporation, which is of comparable magnitude, is not considered in 
this problem. 
 
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1-96 Two large plates at specified temperatures are held parallel to each other. The rate of heat transfer between the plates is 
to be determined for the cases of still air, evacuation, regular insulation, and super insulation between the plates. 
Assumptions 1 Steady operating conditions exist since the plate temperatures remain constant. 2 Heat transfer is one-
dimensional since the plates are large. 3 The surfaces are black and thus ε = 1. 4 There are no convection currents in the air 
space between the plates. 
Properties The thermal conductivities are k = 0.00015 W/m⋅°C for super insulation, k = 0.01979 W/m⋅°C at -50°C (Table A-
15) for air, and k = 0.036 W/m⋅°C for fiberglass insulation (Table A-6). 
Analysis (a) Disregarding any natural convection currents, the rates of conduction and 
radiation heat transfer 
T2T1
[ ]
 W511=+=+=
=−⋅×=
−=
=
−
°⋅=
−
=
−
372139
 W372)K 150()K 290()m1)(K W/m1067.5(1
)(
 W139
 m 0.02
K )150290(
)m C)(1 W/m01979.0(
radcondtotal
442428
4
2
4
1rad
2221
cond
QQQ
TTAQ
L
TT
kAQ
s
&&&
&
&
εσ 
2 cm 
Q ·
(b) When the air space between the plates is evacuated, there will be radiation heat 
transfer only. Therefore, 
 W372== radtotal QQ &&
(c) In this case there will be conduction heat transfer through the fiberglass insulation 
only, 
 W252=−⋅=
−
==
 m 0.02
K )150290()m C)(1 W/m036.0( 2o21condtotal L
TT
kAQQ && 
(d) In the case of superinsulation, the rate of heat transfer will be 
 W1.05=−°⋅=
−
==
 m 0.02
K )150290(
)m C)(1 W/m00015.0( 221condtotal L
TT
kAQQ && 
Discussion Note that superinsulators are very effective in reducing heat transfer between to surfaces. 
 
 
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1-97 The outer surface of a wall is exposed to solar radiation. The effective thermal conductivity of the wall is to be 
determined. 
Assumptions 1 Steady operating conditions exist. 2 The heat transfer 
coefficient is constant and uniform over the surface. 
Properties Both the solar absorptivity and emissivity of the wall 
surface are given to be 0.8. 
Analysis The heat transfer through the wall by conduction is equal to 
net heat transfer to the outer wall surface: 
[ ]
) W/m150)(8.0(
)K 27344()K 27340()K W/m10(0.8)(5.67C)44C)(40W/m (8
m 0.25
C27)-(44
)()(
2
44428-2
solar
4
2
4
surr2
12
solarradconvcond
+
+−+⋅×+°−°⋅=
°
+−+−=
−
++=
k
qTTTTh
L
TT
k
qqqq
so αεσ
&&&&
 
44ºC 
αs = ε = 0.8 
air, 40°C 
h . 
Qrad
150 W/m2
27ºC 
Solving for k gives 
 C W/m0.961 °⋅=k
 
 
 
 
1-98E A spherical ball whose surface is maintained at a temperature of 170°F is suspended in the middle of a room at 70°F. 
The total rate of heat transfer from the ball is to be determined. 
Assumptions 1 Steady operating conditions exist since the ball surface and 
the surrounding air and surfaces remain at constant temperatures. 2 The 
thermal properties of the ball and the convection heat transfer coefficient are 
constant and uniform. 
Properties The emissivity of the ball surface is given to be ε = 0.8. 
Analysis The heat transfer surface area is 
 As = πD2 = π(2/12 ft) 2 = 0.08727 ft2
Under steady conditions, the rates of convection and radiation heat 
transfer are 
Btu/h 9.4
]R) 460+(70R) 460+)[(170RftBtu/h 10)(0.1714ft 70.8(0.0872
)(
Btu/h 9.130F70))(170ft F)(0.08727ftBtu/h (15
444282
44
rad
22
conv
=
−⋅⋅×=
−=
=°−°⋅⋅=∆=
−
oss
s
TTAQ
ThAQ
εσ&
&
 
Air 
70°F 
170°F 
D = 2 in Q 
Therefore, 
Btu/h 140.3=+=+= 4.99.130radconvtotal QQQ &&& 
Discussion Note that heat loss by convection is several times that of heat loss by radiation. The radiation heat loss can 
further be reduced by coating the ball with a low-emissivity material. 
 
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1-99 An 800-W iron is left on the iron board with its base exposed to the air at 20°C. The temperature of the base of 
the iron is to be determined in steady operation. 
Assumptions 1 Steady operating conditions exist. 2 The thermal properties of the 
iron base and the convection heat transfer coefficient are constant and uniform. 
3 The temperature of the surrounding surfaces is the same as the temperature of 
the surrounding air. 
Iron 
800 W 
Properties The emissivity of the base surface is given to be ε = 0.6. 
Analysis At steady conditions, the 800 W energy supplied to the iron will be 
dissipated to the surroundings by convection and radiation heat transfer. 
Therefore, 
 W 800radconvtotal =+= QQQ &&&
where 
K) 293(0.7K) 293()m K)(0.02 W/m(35 22conv −=−⋅=∆= sss TTThAQ& 
and 
]K) (293[100.06804
]K) (293)[KW/m 10)(5.67m 0.6(0.02)(
448
44428244
rad
−×=
−⋅×=−=
−
−
s
soss
T
TTTAQ εσ&
 
Substituting, 
]K) (293[1006804.0)K 293(7.0 W800 448 −×+−= − ss TT 
Solving by trial and error gives 
 C601°== K 874sT
Discussion We note that the iron will dissipate all the energy it receives by convection and radiation when its surface 
temperature reaches 874 K. 
 
 
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1-100 A spherical tank located outdoors is used to store iced water at 0°C. The rate of heat transfer to the iced water in the 
tank and the amount of ice at 0 that melts during a 24-h period are to be determined. °C
Assumptions 1 Steady operating conditions exist since the surface temperatures of the wall remain constant at the specified 
values. 2 Thermal properties of the tank and the convection heat transfer coefficient is constant and uniform. 3 Theaverage 
surrounding surface temperature for radiation exchange is 15°C. 4 The thermal resistance of the tank is negligible, and the 
entire steel tank is at 0°C. 
Properties The heat of fusion of water at atmospheric pressure is 
. The emissivity of the outer surface of the tank is 0.75. kJ/kg 7.333=ifh
Q& 
Iced 
water 
0°C 
0°C 
1 cm 
Air 
25°C
Analysis (a) The outer surface area of the spherical tank is 
 222 m 65.28m) 02.3( === ππDAs
Then the rates of heat transfer to the tank by convection and radiation become 
 
kW 23.1==+=+=
=−⋅×=−=
=°−°⋅=−= ∞
 W102,231614488,21
 W1614])K 273(K) 288)[(K W/m10)(5.67m 65.28)(75.0()(
 W488,21C)025)(m C)(28.65 W/m30()(
radconvtotal
44428-244
surrrad
22
conv
QQQ
TTAQ
TThAQ
ss
ss
&&&
&
&
σε
(b) The amount of heat transfer during a 24-hour period is 
 kJ 000,996,1s) 3600kJ/s)(24 102.23( =×=∆= tQQ &
Then the amount of ice that melts during this period becomes 
 kg 5980===⎯→⎯=
kJ/kg 7.333
kJ 000,996,1
if
if h
QmmhQ 
Discussion The amount of ice that melts can be reduced to a small fraction by insulating the tank. 
 
 
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1-101 The roof of a house with a gas furnace consists of a 22-cm thick concrete that is losing heat to the outdoors by 
radiation and convection. The rate of heat transfer through the roof and the money lost through the roof that night during a 
14 hour period are to be determined. 
Assumptions 1 Steady operating conditions exist. 2 The emissivity and thermal conductivity of the roof are constant. 
Properties The thermal conductivity of the concrete is given to be k = 2 W/m⋅°C. The emissivity of the outer surface of the 
roof is given to be 0.9. 
Analysis In steady operation, heat transfer from the outer surface of the roof to the surroundings by convection and radiation 
must be equal to the heat transfer through the roof by conduction. That is, 
 rad+conv gs,surroundin toroofcond roof, QQQ &&& ==
The inner surface temperature of the roof is given to be Ts,in = 15°C. Letting Ts,out denote the outer surface temperatures of 
the roof, the energy balance above can be expressed as 
)()( 4surr
4
outs,surrouts,
outs,ins, TTATTAh
L
TT
kAQ o −+−=
−
= σε& 
[ ]44outs,4282
outs,
22
outs,2
K) 255(K) 273()K W/m1067.5)(m 300)(9.0( 
C)10)(m C)(300. W/m15(
m 22.0
C15
)m 300)(C W/m2(
−+⋅×+
°−°=
−°
°⋅=
− T
T
T
Q&
 
Tsky = 255 K Q& 
Solving the equations above using an equation solver (or by trial and error) gives 
 C7.8 W19,830 °== out s, and TQ&
Then the amount of natural gas consumption during a 16-hour period is 
 therms15.11
kJ 105,500
 therm1
85.0
)s 360014)(kJ/s 83.19(
85.085.0
total
gas =⎟⎟
⎠
⎞
⎜⎜
⎝
⎛×
=
∆
==
tQQE
&
 
Finally, the money lost through the roof during that period is 
 $13.4== )therm/20.1$ therms)(15.11(lostMoney 
 
 
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1-102E A flat plate solar collector is placed horizontally on the roof of a house. The rate of heat loss from the collector by 
convection and radiation during a calm day are to be determined. 
Assumptions 1 Steady operating conditions exist. 2 The emissivity and convection heat transfer coefficient are constant and 
uniform. 3 The exposed surface, ambient, and sky temperatures remain constant. 
Properties The emissivity of the outer surface of the collector is given to be 0.9. 
Air, 70°F 
Tsky = 50°F 
Solar 
collector 
Q&Analysis The exposed surface area of the collector is 
 2ft 75ft) ft)(15 5( ==sA
Noting that the exposed surface temperature of the collector is 100°F, 
the total rate of heat loss from the collector to the environment by 
convection and radiation becomes 
 
Btu/h 3551
])R 46050(R) 460100)[(RftBtu/h 10)(0.1714ft 75)(9.0()(
Btu/h 5625F)70100)(ft F)(75Btu/h.ft 5.2()(
44428-244
surrrad
22
conv
=
+−+⋅⋅×=−=
=°−°⋅=−= ∞
ss
ss
TTAQ
TThAQ
σε&
&
and 
Btu/h 9176=+=+= 35515625radconvtotal QQQ &&& 
 
 
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1-103 Temperature of the stainless steel sheet going through an annealing process inside an electrically heated oven is to be 
determined. 
Assumptions 1 Steady operating conditions 
exist. 2 Temperature of the stainless steel 
sheet is uniform. 3 Radiation heat transfer 
between stainless steel sheet and surrounding 
oven surfaces is between a small object and a 
large enclosure. 
Properties The emissivity of the stainless 
steel sheet is given to be 0.40. 
Analysis The amount of heat transfer by 
radiation between the sheet and the 
surrounding oven surfaces is balanced by the 
convection heat transfer between the sheet 
and the ambient air: 
0convrad =− qq && 
0)()( 44surr =−−− ∞TThTT ssεσ 
0K )]273600()[K W/m10(K ])273750)[(K W/m1067.5)(40.0( 2444428 =+−⋅−−+⋅× − ss TT 
Solving the above equation by EES software (Copy the following line and paste on a blank EES screen to verify solution): 
0.40*5.67e-8*((750+273)^4-T_s^4)-10*(T_s-(600+273))=0 
The temperature of the stainless steel sheet is 
C 736 °== K 1009sT 
Discussion Note that the energy balance equation involving radiation heat transfer used for solving the stainless steel sheet 
temperature must be used with absolute temperature. 
 
 
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1-104 The upper surface temperature of a silicon wafer undergoing heat treatment in a vacuum chamber by infrared heater is 
to be determined. 
Assumptions 1 Steady operating conditions exist. 2 Radiation heat transfer between upper wafer surface and surroundings is 
between a small object and a large enclosure. 3 One-dimensional conduction in wafer. 4 The silicon wafer has constant 
properties. 5 No hot spot exists on the wafer. 
Properties The thermal conductivity of silicon at 1000 K is 31.2 W/m · K (Table A-3). 
Analysis The heat transfer through the 
thickness of the wafer by conduction 
is equal to net heat transfer at the 
upper wafer surface: 
 radabscond qqq &&& −=
 )( 4surr
4
,IR
, TTq
L
TT
k us
lss,u −−=
−
εσα & 
 
444
,
428
2
6
K )310)(K W/m1067.5)(70.0(
) W/m200000)(70.0(
m) 10725(
K )1000(
)K W/m2.31(
−⋅×−
=
×
−
⋅
−
−
us
s,u
T
T
 
Copy the following line and paste on a blank EES screen to solve the above equation: 
 31.2*(T_su-1000)/725e-6=0.70*200000-0.70*5.67e-8*(T_su^4-310^4) 
Solving by EES software, the upper surface temperature of silicon wafer is 
K 1002=usT , 
Discussion Excessive temperature difference across the wafer thickness will cause warping in the silicon wafer. 
 
 
 
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Problem Solving Techniques and EES 
 
1-105C Despite the convenience and capability the engineering software packages offer, they are still just tools, and they 
will not replace the traditional engineering courses. They will simply cause a shift in emphasis in the course material from 
mathematics to physics. They are of great value in engineering practice, however, as engineers today rely on software 
packages for solving large and complex problems ina short time, and perform optimization studies efficiently. 
 
 
 
1-106 We are to determine a positive real root of the following equation using EES: 3.5x3 – 10x0.5 – 3x = −4. 
Analysis Using EES software, copy the following lines and paste on a blank EES screen to verify the solution: 
3.5*x^3-10*x^0.5-3*x = -4 
Answer: x = 1.554 
Discussion To obtain the solution in EES, click on the icon that looks like a calculator, or Calculate-Solve. 
 
 
 
1-107 We are to solve a system of 2 equations and 2 unknowns using EES. 
Analysis Using EES software, copy the following lines and paste on a blank EES screen to verify the solution: 
x^3-y^2=10.5 
3*x*y+y=4.6 
Answers: x = 2.215, y = 0.6018 
Discussion To obtain the solution in EES, click on the icon that looks like a calculator, or Calculate-Solve. 
 
 
 
1-108 We are to solve a system of 3 equations with 3 unknowns using EES. 
Analysis Using EES software, copy the following lines and paste on a blank EES screen to verify the solution: 
2*x-y+z=5 
3*x^2+2*y=z+2 
x*y+2*z=8 
Answers: x = 1.141, y = 0.8159, z = 3.535. 
Discussion To obtain the solution in EES, click on the icon that looks like a calculator, or Calculate-Solve. 
 
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1-109 We are to solve a system of 3 equations with 3 unknowns using EES. 
Analysis Using EES software, copy the following lines and paste on a blank EES screen to verify the solution: 
x^2*y-z=1.5 
x-3*y^0.5+x*z=-2 
x+y-z=4.2 
Answers: x = 0.9149, y = 10.95, z = 7.665 
Discussion To obtain the solution in EES, click on the icon that looks like a calculator, or Calculate-Solve. 
 
 
 
1-110 The squares of the number from 1 to 100 in increments of 10 are to be evaluated using the parametric table and 
plot features of EES. 
Analysis The problem is solved using EES, and the solution is given below. 
 
x=1 
y=x^2 
 
 
x y 
1 
10 
20 
30 
40 
50 
60 
70 
80 
90 
100 
1 
100 
400 
900 
1600 
2500 
3600 
4900 
6400 
8100 
10000 
0 20 40 60 80 100
0
2200
4400
6600
8800
11000
x 
y 
 
 
 
 
 
 
 
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Special Topic: Thermal Comfort 
 
1-111C The metabolism refers to the burning of foods such as carbohydrates, fat, and protein in order to perform the 
necessary bodily functions. The metabolic rate for an average man ranges from 108 W while reading, writing, typing, or 
listening to a lecture in a classroom in a seated position to 1250 W at age 20 (730 at age 70) during strenuous exercise. The 
corresponding rates for women are about 30 percent lower. Maximum metabolic rates of trained athletes can exceed 2000 W. 
We are interested in metabolic rate of the occupants of a building when we deal with heating and air conditioning because 
the metabolic rate represents the rate at which a body generates heat and dissipates it to the room. This body heat contributes 
to the heating in winter, but it adds to the cooling load of the building in summer. 
 
 
1-112C The metabolic rate is proportional to the size of the body, and the metabolic rate of women, in general, is lower than 
that of men because of their smaller size. Clothing serves as insulation, and the thicker the clothing, the lower the 
environmental temperature that feels comfortable. 
 
 
1-113C Asymmetric thermal radiation is caused by the cold surfaces of large windows, uninsulated walls, or cold products 
on one side, and the warm surfaces of gas or electric radiant heating panels on the walls or ceiling, solar heated masonry 
walls or ceilings on the other. Asymmetric radiation causes discomfort by exposing different sides of the body to surfaces at 
different temperatures and thus to different rates of heat loss or gain by radiation. A person whose left side is exposed to a 
cold window, for example, will feel like heat is being drained from that side of his or her body. 
 
 
1-114C (a) Draft causes undesired local cooling of the human body by exposing parts of the body to high heat transfer 
coefficients. (b) Direct contact with cold floor surfaces causes localized discomfort in the feet by excessive heat loss by 
conduction, dropping the temperature of the bottom of the feet to uncomfortable levels. 
 
 
1-115C Stratification is the formation of vertical still air layers in a room at difference temperatures, with highest 
temperatures occurring near the ceiling. It is likely to occur at places with high ceilings. It causes discomfort by exposing the 
head and the feet to different temperatures. This effect can be prevented or minimized by using destratification fans (ceiling 
fans running in reverse). 
 
 
1-116C It is necessary to ventilate buildings to provide adequate fresh air and to get rid of excess carbon dioxide, 
contaminants, odors, and humidity. Ventilation increases the energy consumption for heating in winter by replacing the 
warm indoors air by the colder outdoors air. Ventilation also increases the energy consumption for cooling in summer by 
replacing the cold indoors air by the warm outdoors air. It is not a good idea to keep the bathroom fans on all the time since 
they will waste energy by expelling conditioned air (warm in winter and cool in summer) by the unconditioned outdoor air. 
 
 
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1-117 The windows of a house in Atlanta are of double door type with wood frames and metal spacers. The average rate of 
heat loss through the windows in winter is to be determined. 
Assumptions 1 Steady operating conditions exist. 2 Heat losses associated with the infiltration of air through the 
cracks/openings are not considered. 
Q&
Window 
22°C 
11.3°C
Analysis The rate of heat transfer through the window can be determined from 
)( oiwindowoverallavg window, TTAUQ −=& 
where Ti and To are the indoor and outdoor air temperatures, respectively, Uoverall is the U-
factor (the overall heat transfer coefficient) of the window, and Awindow is the window area. 
Substituting, 
 W535=°−°⋅= C)3.1122)(m C)(20 W/m50.2( 22avg window,Q&
Discussion This is the “average” rate of heat transfer through the window in winter in the absence of any infiltration. 
 
 
 
 
1-118 Boiling experiments are conducted by heating water at 1 atm pressure with an electric resistance wire, and measuring 
the power consumed by the wire as well as temperatures. The boiling heat transfer coefficient is to be determined. 
Assumptions 1 Steady operating conditions exist. 2 Heat losses from the water 
container are negligible. 
Water 
100°C 
 
 
Heater 
120°C 
Analysis The heat transfer area of the heater wire is 
 2m 004398.0m) m)(0.70 002.0( === ππDLA
Noting that 4100 W of electric power is consumed when the heater surface 
temperature is 120°C, the boiling heat transfer coefficient is determined from 
Newton’s law of cooling to be )( satTThAQ s −=&
 C W/m46,600 2 °⋅=
°−
=
−
=
C)100)(120m (0.004398
 W4100
 )( 2satTTA
Qh
s
&
 
 
 
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Review Problems 
 
1-119 The power required to maintain the soldering iron tip at 400 °C is to be determined. 
 
Assumptions 1 Steady operating conditions exist since the tip surface and the surroundingair temperatures remain constant. 
2 The thermal properties of the tip and the convection heat transfer coefficient are constant and uniform. 3 The surrounding 
surfaces are at the same temperature as the air. 
Properties The emissivity of the tip is given to be 0.80. 
Analysis The total heat transfer area of the soldering iron tip is 
24
2
2
m 1062.1
)m 02.0)(m 0025.0(4/)m 0025.0(
4/
−×=
+=
+=
ππ
ππ DLDAs
 
The rate of heat transfer by convection is 
 W541
C )20400)(m 1062.1)(C W/m25(
)(
242
tipconv
.
TThAQ s
=
°−×°⋅=
−=
−
∞
&
 
The rate of heat transfer by radiation is 
 W451
K ])27320()273400)[(m 1062.1)(K W/m1067.5)(80.0(
)(
44424428
4
surr
4
tiprad
.
TTAQ s
=
+−+×⋅×=
−=
−−
εσ&
 
Thus, the power required is equal to the total rate of heat transfer from the tip by both convection and radiation: 
 W2.99=+=+= W45.1 W54.1radconvtotal QQQ &&& 
Discussion If the soldering iron tip is highly polished with an emissivity of 0.05, the power required to maintain the tip at 
400 °C will reduce to 1.63 W, or by 45.5%. 
 
 
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1-120 A standing man is subjected to high winds and thus high convection coefficients. The rate of heat loss from this man 
by convection in still air at 20°C, in windy air, and the wind chill temperature are to be determined. 
Assumptions 1 A standing man can be modeled as a 30-cm diameter, 170-cm long vertical cylinder with both the top and 
bottom surfaces insulated. 2 The exposed surface temperature of the person and the convection heat transfer coefficient is 
constant and uniform. 3 Heat loss by radiation is negligible. 
Analysis The heat transfer surface area of the person is 
 As = πDL = π(0.3 m)(1.70 m) = 1.60 m2
The rate of heat loss from this man by convection in still air is 
 Qstill air = hAs∆T = (15 W/m2·°C)(1.60 m2)(34 - 20)°C = 336 W 
In windy air it would be 
 Qwindy air = hAs∆T = (30 W/m2·°C)(1.60 m2)(34 - 20)°C = 672 W 
Windy weather To lose heat at this rate in still air, the air temperature must be 
 672 W = (hAs∆T)still air = (15 W/m²·°C)(1.60 m²)(34 - Teffective)°C 
which gives 
 Teffective = 6°C 
That is, the windy air at 20°C feels as cold as still air at 6°C as a result of the wind-chill effect. 
 
 
 
 
1-121 The backside of the thin metal plate is insulated and the front side is exposed to solar radiation. The surface 
temperature of the plate is to be determined when it stabilizes. 
Assumptions 1 Steady operating conditions exist. 2 Heat transfer through the insulated side of the plate is negligible. 3 The 
heat transfer coefficient is constant and uniform over the plate. 4 Radiation heat transfer is negligible. 
Properties The solar absorptivity of the plate is given to be α = 0.7. 
 . 
Qrad
α = 0.7 
air, 10°C 
550 W/m2
Analysis When the heat loss from the plate by convection equals the solar radiation 
absorbed, the surface temperature of the plate can be determined from 
 
)10(C)W/m (25W/m 0557.0
)(
22
solar
convabsorbedsolar
−°⋅=××
−=
=
ss
oss
TAA
TThAQ
QQ
&
&&
α
Canceling the surface area As and solving for Ts gives 
 C25.4°=sT
 
 
 
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1-122 A room is to be heated by 1 ton of hot water contained in a tank placed in the room. The minimum initial temperature 
of the water is to be determined if it to meet the heating requirements of this room for a 24-h period. 
Assumptions 1 Water is an incompressible substance with constant specific heats. 2 Air is an ideal gas with constant specific 
heats. 3 The energy stored in the container itself is negligible relative to the energy stored in water. 4 The room is 
maintained at 20°C at all times. 5 The hot water is to meet the heating requirements of this room for a 24-h period. 
Properties The specific heat of water at room temperature is c = 4.18 kJ/kg·°C (Table A-9). 
Analysis Heat loss from the room during a 24-h period is 
 Qloss = (10,000 kJ/h)(24 h) = 240,000 kJ 
Taking the contents of the room, including the water, as our system, the energy balance can be written as 
 ( ) ( ) 0airwaterout
energies etc. potential, 
kinetic, internal,in Change
system
mass and work,heat,by 
nsferenergy traNet 
 UUUQEEE outin ∆+∆=∆=−→∆=− 4342143421
 
10,000 kJ/h
or 
water 
20°C 
 -Qout = [mc(T2 - T1)]water
Substituting, 
 -240,000 kJ = (1000 kg)(4.18 kJ/kg·°C)(20 - T1) 
It gives 
 T1 = 77.4°C 
where T1 is the temperature of the water when it is first brought into the room. 
 
 
 
 
1-123 The base surface of a cubical furnace is surrounded by black surfaces at a specified temperature. The net rate of 
radiation heat transfer to the base surface from the top and side surfaces is to be determined. 
Assumptions 1 Steady operating conditions exist. 2 The top and side surfaces of the furnace closely approximate black 
surfaces. 3 The properties of the surfaces are constant. 
Properties The emissivity of the base surface is ε = 0.4. 
Black furnace 
1200 K 
Base, 800 K
Analysis The base surface is completely surrounded by the top and side surfaces. Then 
using the radiation relation for a surface completely surrounded by another large (or 
black) surface, the net rate of radiation heat transfer from the top and side surfaces to 
the base is determined to be 
 
kW 340==
−××=
−=
 W,660339
])K 800(K) 1200)[(K. W/m10)(5.67m 33)(4.0(
)(
4442-82
4
surr
4
basebaserad, TTAQ σε&
 
 
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1-124 Engine valves are to be heated in a heat treatment section. The amount of heat transfer, the average rate of heat 
transfer, the average heat flux, and the number of valves that can be heat treated daily are to be determined. 
Assumptions Constant properties given in the problem can be used. 
Properties The average specific heat and density of valves are given to be cp = 440 J/kg.°C and ρ = 7840 kg/m3. 
Analysis (a) The amount of heat transferred to the valve is simply the change in its internal energy, and is determined from 
 
kJ 26.35=C40)C)(800kJ/kg kg)(0.440 0788.0(
)( 12
°−°⋅=
−=∆= TTmcUQ p
Engine valve 
T1 = 40°C 
T2 = 800°C 
D = 0.8 cm 
L = 10 cm 
(b) The average rate of heat transfer can be determined from 
 W87.8==
×
=
∆
= kW 0878.0
s 605
kJ 35.26
avg t
QQ& 
(c) The average heat flux is determined from 
 24 W/m101.75×====
m) m)(0.1 008.0(2
 W8.87
π2
avgavg
ave πDL
Q
A
Q
q
s
&&
& 
(d) The number of valves that can be heat treated daily is 
 valves 3000=×=
min 5
 valves)min)(25 6010(
 valvesofNumber 
 
 
 
 
1-125 The glass cover of a flat plate solar collector with specified inner and outer surface temperatures is considered. The 
fraction of heat lost from the glass cover by radiation is to be determined. 
Assumptions 1 Steady operating conditions exist since the surface temperatures of the glass remain constant at the specified 
values. 2 Thermal properties of the glass are constant. 
Properties The thermal conductivity of the glass is given to be k = 0.7 W/m⋅°C. 
Analysis Under steady conditions, the rate of heat transfer through the glass by conduction is 
 W 583
m 0.006
C)31(33)m C)(2.5W/m (0.7 2cond =
°−
°⋅=
∆
=
L
TkAQ& 
 A = 2.5 m2
 31°C 33°C
 L=0.6 Air, 15°C 
h=10 W/m2.°C 
Q&
The rate of heat transfer from the glass by convection is 
 W400C)15)(31m C)(2.5W/m (10 22conv =°−°⋅=∆= ThAQ& 
Under steady conditions, the heattransferred through the cover by 
conduction should be transferred from the outer surface by convection and 
radiation. That is, 
 W183400583convcondrad =−=−= QQQ &&&
Then the fraction of heat transferred by radiation becomes 
0.314===
583
183
cond
rad
Q
Q
f
&
&
 (or 31.4%) 
 
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1-126 The range of U-factors for windows are given. The range for the rate of heat loss through the window of a house is to 
be determined. 
Assumptions 1 Steady operating conditions exist. 2 Heat losses associated with the 
infiltration of air through the cracks/openings are not considered. 
20°C 
-8°C
Window
Q&Analysis The rate of heat transfer through the window can be determined from 
 )(windowoverallwindow oi TTAUQ −=& 
where Ti and To are the indoor and outdoor air temperatures, respectively, Uoverall is the U-
factor (the overall heat transfer coefficient) of the window, and Awindow is the window area. 
Substituting, 
Maximum heat loss: W378=°−−×°⋅= C)]8(20)[m 8.1C)(1.2 W/m25.6( 22max window,Q&
Minimum heat loss: W76=°−−×°⋅= C)]8(20)[m 8.1C)(1.2 W/m25.1( 22min window,Q&
Discussion Note that the rate of heat loss through windows of identical size may differ by a factor of 5, depending on how 
the windows are constructed. 
 
 
 
 
1-127 Prob. 1-126 is reconsidered. The rate of heat loss through the window as a function of the U-factor is to be 
plotted. 
Analysis The problem is solved using EES, and the solution is given below. 
 
"GIVEN" 
A=1.2*1.8 [m^2] 
T_1=20 [C] 
T_2=-8 [C] 
U=1.25 [W/m^2-C] 
 
1 2 3 4 5 6 7
50
100
150
200
250
300
350
400
U [W/m2-C]
Q
w
in
do
w
 [
W
]
 
"ANALYSIS" 
Q_dot_window=U*A*(T_1-T_2) 
 
 
U 
[W/m2.C] 
Qwindow 
[W] 
1.25 75.6 
1.75 105.8 
2.25 136.1 
2.75 166.3 
3.25 196.6 
3.75 226.8 
4.25 257 
4.75 287.3 
5.25 317.5 
5.75 347.8 
6.25 378 
 
 
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1-128 An electric heater placed in a room consumes 500 W power when its surfaces are at 120°C. The surface temperature 
when the heater consumes 700 W is to be determined without and with the consideration of radiation. 
Assumptions 1 Steady operating conditions exist. 2 The temperature is uniform over the surface. 
Analysis (a) Neglecting radiation, the convection heat 
transfer coefficient is determined from 
 
( )
CW/m02
C20120)m 25.0(
 W500
)(
2
2
°⋅=
°−
=
−
=
∞TTA
Q
h
s
&
 
The surface temperature when the heater consumes 700 W is 
C160°=
°⋅
+°=+= ∞
)m 25.0(C)W/m02(
 W700C20
22hA
Q
TTs
&
 
eW&
T∞ , h 
qrad
qconv
Tw 
Ts 
A, ε 
(b) Considering radiation, the convection heat transfer coefficient is determined from 
 
[ ]
( )
CW/m58.12
C20120)m 25.0(
K) 283(K) 393()KW/m1067.5)(m 5(0.75)(0.2- W500
)(
)(
2
2
444282
4
surr
4
°⋅=
°−
−⋅×
=
−
−−
=
−
∞TTA
TTAQ
h
s
sσε&
 
Then the surface temperature becomes 
 
( )
[ ]
C152.9°==
−×+−=
−+−=
−
∞
K 9.425
K) 283()1067.5)(5(0.75)(0.2)293)(25.0)(58.12(700
)(
448
4
surr
4
s
ss
ss
T
TT
TTATThAQ σε&
Discussion Neglecting radiation changed Ts by more than 7°C, so assumption is not correct in this case. 
 
 
 
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1-129 An ice skating rink is located in a room is considered. The refrigeration load of the system and the time it takes to melt 
3 mm of ice are to be determined. 
Assumptions 1 Steady operating conditions exist in part (a). 2 The surface is insulated on the back side in part (b). 
Properties The heat of fusion and the density of ice are given to be 333.7 kJ/kg and 920 kg/m3, respectively. 
 
Insulation
Qload Ts = 0°C 
Tair = 20°C 
h = 10 W/m2⋅K 
Tw = 25°C 
Control Volume 
 Qrad Qconv
Refrigerator 
 
 
 
 
 
 
 Ice 
 
Analysis (a) The refrigeration load is determined from 
( )
[ ] W156,300=−××+−×=
−+−=
− 448
4
s
4
airload
273298)1067.5)(1240((0.95))020)(1240)(10(
)( TTATThAQ ws σε& 
(b) The time it takes to melt 3 mm of ice is determined from 
 min 47.2==××== s 2831
J/s 300,156
)J/kg 10)(333.7kg/m m)(920 )(0.003m 12(40 332
loadQ
hLW
t if
&
δρ
 
 
 
 
1-130 The surface temperature of an engine block that generates 50 kW of power output is to be determined. 
Assumptions 1 Steady operating conditions exist. 2 Temperature inside the engine compartment is uniform. 3 Heat transfer 
by radiation is not considered. 
Analysis With a net engine efficiency of 35%, which means 65% of the generated power output are heat loss by convection: 
kW 5.32)35.01)(kW 50()1(outconv =−=−= ηWQ && 
From Newton’s law of cooling, the heat transfer by convection is given as 
)(conv ∞−= TThAQ ss& 
Rearranging, the engine block surface temperature is 
C 841 °=°+
°⋅
×
=+= ∞ C 157)m 950C)( W/m50(
 W105.32
22
3
conv
.
T
hA
Q
T
s
s
&
 
Discussion Due to the complex geometry of the engine block, hot spots are likely to occur with temperatures much higher 
than 841 °C. 
 
 
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Fundamentals of Engineering (FE) Exam Problems 
 
1-131 Which equation below is used to determine the heat flux for conduction? 
(a) 
dx
dTkA− (b) (c) Tk grad − )( 12 TTh − (d) 
4 (e) None of them Tεσ
Answer (b) Tk grad −
 
 
 
1-132 Which equation below is used to determine the heat flux for convection? 
(a) 
dx
dTkA− (b) (c) Tk grad − )( 12 TTh − (d) 
4 (e) None of them Tεσ
Answer (c) )( 12 TTh −
 
 
 
1-133 Which equation below is used to determine the heat flux emitted by thermal radiation from a surface? 
(a) 
dx
dTkA− (b) (c) Tk grad − )( 12 TTh − (d) 
4 (e) None of them Tεσ
Answer (d) 4Tεσ
 
 
 
1-134 Consider two different materials, A and B. The ratio of thermal conductivities is kA/kB = 13, the ratio of the densities is 
ρA/ρB = 0.045, and the ratio of specific heats is cp,A/cp,B = 16.9. The ratio of the thermal diffusivities αA/αB is 
(a) 4882 (b) 17.1 (c) 0.06 (d) 0.1 (e) 0.03 
Answer (b) 17.1 
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES 
screen. 
 
k_A\k_B=13 
rho_A\rho_B=0.045 
c_p_A\c_p_B=16.9 
 
"From the definition of thermal diffusivity, alpha = k/(rho*c-p)" 
alpha_A\alpha_B=k_A\k_B*(1/rho_A\rho_B)*(1/c_p_A\c_p_B) 
 
"Some Wrong Solutions with Common Mistakes" 
W1_alpha_A\alpha_B=k_A\k_B*rho_A\rho_B*(1/c_p_A\c_p_B) "Not inversing density ratio" 
W2_alpha_A\alpha_B=k_A\k_B*(1/rho_A\rho_B)*c_p_A\c_p_B "Not inversing specific heat ratio" 
W3_alpha_A\alpha_B=1/(k_A\k_B)*(1/rho_A\rho_B)*(1/c_p_A\c_p_B) "Inversing conductivity ratio" 
W4_alpha_A\alpha_B=1/(k_A\k_B*(1/rho_A\rho_B)*(1/c_p_A\c_p_B)) "Taking the inverse of result" 
 
 
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1-135 A 2-kW electric resistance heater in a room is turned on and kept on for 50 minutes. The amount of energy transferred 
to the room by the heater is 
(a) 2 kJ (b) 100 kJ (c) 6000 kJ (d) 7200 kJ (e) 12,000 kJ 
Answer (c) 6000 kJ 
 
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the followinglines on a blank EES 
screen. 
 
We= 2 [kJ/s] 
time=50*60 [s] 
We_total=We*time [kJ] 
 
"Wrong Solutions:" 
W1_Etotal=We*time/60 "using minutes instead of s" 
W2_Etotal=We "ignoring time" 
 
 
 
1-136 A hot 16 cm × 16 cm × 16 cm cubical iron block is cooled at an average rate of 80 W. The heat flux is 
(a) 195 W/m2 (b) 521 W/m2 (c) 3125 W/m2 (d) 7100 W/m2 (e) 19,500 W/m2
Answer (b) 521 W/m2
 
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES 
screen. 
 
a=0.16 [m] 
Q_dot=80 [W] 
A_s=6*a^2 
q=Q_dot/A_s 
 
"Some Wrong Solutions with Common Mistakes" 
W1_q=Q_dot/a^2 "Using wrong equation for area" 
W2_q=Q_dot/a^3 "Using volume instead of area" 
 
 
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1-137 A 2-kW electric resistance heater submerged in 30-kg water is turned on and kept on for 10 min. During the process, 
500 kJ of heat is lost from the water. The temperature rise of water is 
(a) 5.6°C (b) 9.6°C (c) 13.6°C (d) 23.3°C (e) 42.5°C 
Answer (a) 5.6°C 
 
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES 
screen. 
 
C=4.18 [kJ/kg-K] 
m=30 [kg] 
Q_loss=500 [kJ] 
time=10*60 [s] 
W_e=2 [kJ/s] 
 
"Applying energy balance E_in-E_out=dE_system gives" 
time*W_e-Q_loss = dU_system 
dU_system=m*C*DELTAT 
 
“Some Wrong Solutions with Common Mistakes:” 
time*W_e = m*C*W1_T "Ignoring heat loss" 
time*W_e+Q_loss = m*C*W2_T "Adding heat loss instead of subtracting" 
time*W_e-Q_loss = m*1.0*W3_T "Using specific heat of air or not using specific heat" 
 
 
 
1-138 Eggs with a mass of 0.15 kg per egg and a specific heat of 3.32 kJ/kg⋅°C are cooled from 32°C to 10°C at a rate of 200 
eggs per minute. The rate of heat removal from the eggs is 
(a) 7.3 kW (b) 53 kW (c) 17 kW (d) 438 kW (e) 37 kW 
Answer (e) 37 kW 
 
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES 
screen. 
 
C=3.32 [kJ/kg-K] 
m_egg=0.15 [kg] 
T1=32 [C] 
T2=10 [C] 
n=200 "eggs/min" 
m=n*m_egg/60 "kg/s" 
"Applying energy balance E_in-E_out=dE_system gives" 
"-E_out = dU_system" 
Qout=m*C*(T1-T2) "kJ/s" 
“Some Wrong Solutions with Common Mistakes:” 
W1_Qout = m*C*T1 "Using T1 only" 
W2_Qout = m_egg*C*(T1-T2) "Using one egg only" 
W3_Qout = m*C*T2 "Using T2 only" 
W4_Qout=m_egg*C*(T1-T2)*60 "Finding kJ/min" 
 
 
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1-139 Steel balls at 140°C with a specific heat of 0.50 kJ/kg⋅°C are quenched in an oil bath to an average temperature of 
85°C at a rate of 35 balls per minute. If the average mass of steel balls is 1.2 kg, the rate of heat transfer from the balls to the 
oil is 
(a) 33 kJ/s (b) 1980 kJ/s (c) 49 kJ/s (d) 30 kJ/s (e) 19 kJ/s 
Answer (e) 19 kJ/s 
 
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES 
screen. 
 
c=0.50 [kJ/kg-K] 
m1=1.2 [kg] 
T1=140 [C] 
T2=85 [C] 
n=35 "balls/min" 
m=n*m1/60 "kg/s" 
"Applying energy balance E_in-E_out=dE_system gives" 
"-E_out = dU_system" 
Qout=m*c*(T1-T2) "kJ/s" 
“Some Wrong Solutions with Common Mistakes:” 
W1_Qout = m*c*T1 "Using T1 only" 
W2_Qout = m1*c*(T1-T2) "Using one egg only" 
W3_Qout = m*c*T2 "Using T2 only" 
W4_Qout=m1*c*(T1-T2)*60 "Finding kJ/min" 
 
 
 
1-140 A cold bottled drink (m = 2.5 kg, cp = 4200 J/kg⋅°C) at 5°C is left on a table in a room. The average temperature 
of the drink is observed to rise to 15°C in 30 minutes. The average rate of heat transfer to the drink is 
(a) 23 W (b) 29 W (c) 58 W (d) 88 W (e) 122 W 
Answer: (c) 58 W 
 
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES 
screen. 
 
c=4200 [J/kg-K] 
m=2.5 [kg] 
T1=5 [C] 
T2=15 [C] 
time = 30*60 [s] 
"Applying energy balance E_in-E_out=dE_system gives" 
Q=m*c*(T2-T1) 
Qave=Q/time 
 
“Some Wrong Solutions with Common Mistakes:” 
W1_Qave = m*c*T1/time "Using T1 only" 
W2_Qave = c*(T2-T1)/time "Not using mass" 
W3_Qave = m*c*T2/time "Using T2 only" 
 
 
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1-141 Water enters a pipe at 20ºC at a rate of 0.50 kg/s and is heated to 60ºC. The rate of heat transfer to the water is 
(a) 20 kW (b) 42 kW (c) 84 kW (d) 126 kW (e) 334 kW 
Answer (c) 84 kW 
 
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES 
screen. 
 
T_in=20 [C] 
T_out=60 [C] 
m_dot=0.50 [kg/s] 
c_p=4.18 [kJ/kg-C] 
Q_dot=m_dot*c_p*(T_out-T_in) 
 
"Some Wrong Solutions with Common Mistakes" 
W1_Q_dot=m_dot*(T_out-T_in) "Not using specific heat" 
W2_Q_dot=c_p*(T_out-T_in) "Not using mass flow rate" 
W3_Q_dot=m_dot*c_p*T_out "Using exit temperature instead of temperature change" 
 
 
 
1-142 Air enters a 12-m-long, 7-cm-diameter pipe at 50ºC at a rate of 0.06 kg/s. The air is cooled at an average rate of 400 W 
per m2 surface area of the pipe. The air temperature at the exit of the pipe is 
(a) 4.3ºC (b) 17.5ºC (c) 32.5ºC (d) 43.4ºC (e) 45.8ºC 
Answer (c) 32.5ºC 
 
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES 
screen. 
 
L=12 [m] 
D=0.07 [m] 
T1=50 [C] 
m_dot=0.06 [kg/s] 
q=400 [W/m^2] 
 
A=pi*D*L 
Q_dot=q*A 
c_p=1007 [J/kg-C] "Table A-15" 
Q_dot=m_dot*c_p*(T1-T2) 
 
"Some Wrong Solutions with Common Mistakes" 
q=m_dot*c_p*(T1-W1_T2) "Using heat flux, q instead of rate of heat transfer, Q_dot" 
Q_dot=m_dot*4180*(T1-W2_T2) "Using specific heat of water" 
Q_dot=m_dot*c_p*W3_T2 "Using exit temperature instead of temperature change" 
 
 
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
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1-143 Heat is lost steadily through a 0.5-cm thick 2 m × 3 m window glass whose thermal conductivity is 0.7 W/m⋅°C. The 
inner and outer surface temperatures of the glass are measured to be 12°C to 9°C. The rate of heat loss by conduction 
through the glass is 
(a) 420 W (b) 5040 W (c) 17,600 W (d) 1256 W (e) 2520 W 
Answer (e) 2520 W 
 
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES 
screen. 
 
A=3*2 [m^2] 
L=0.005 [m] 
T1=12 [C] 
T2=9 [C] 
k=0.7 [W/m-C] 
Q=k*A*(T1-T2)/L 
 
“Some Wrong Solutions with Common Mistakes:” 
W1_Q=k*(T1-T2)/L "Not using area" 
W2_Q=k*2*A*(T1-T2)/L "Using areas of both surfaces" 
W3_Q=k*A*(T1+T2)/L "Adding temperatures instead of subtracting" 
W4_Q=k*A*L*(T1-T2) "Multiplying by thickness instead of dividing by it" 
 
 
 
1-144 The East wall of an electrically heated house is 9 m long, 3 m high, and 0.35 m thick, and it has an effective thermal 
conductivity of 0.7 W/m.°C. If the inner and outer surface temperatures of wall are 15°C and 6°C, the rate of heat loss 
through the wall is 
(a) 486 W (b) 60 W (c) 1134 W (d) 972 W (e) 2085 W 
Answer (a) 486 W 
 
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES 
screen. 
 
A=3*9 [m^2] 
L=0.35 [m] 
k=0.7 [W/m-C] 
T1=15 [C] 
T2=6 [C] 
Q_cond=k*A*(T1-T2)/L 
 
"Wrong Solutions:" 
W1_Q=k*(T1-T2)/L "Not using area" 
W2_Q=k*2*A*(T1-T2)/L "Using areas of both surfaces"W3_Q=k*A*(T1+T2)/L "Adding temperatures instead of subtracting" 
W4_Q=k*A*L*(T1-T2) "Multiplying by thickness instead of dividing by it" 
 
 
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1-145 Steady heat conduction occurs through a 0.3-m thick 9 m by 3 m composite wall at a rate of 1.2 kW. If the inner and 
outer surface temperatures of the wall are 15°C and 7°C, the effective thermal conductivity of the wall is 
(a) 0.61 W/m⋅°C (b) 0.83 W/m⋅°C (c) 1.7 W/m⋅°C (d) 2.2 W/m⋅°C (e) 5.1 W/m⋅°C 
Answer (c) 1.7 W/m⋅°C 
 
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES 
screen. 
 
A=9*3 [m^2] 
L=0.3 [m] 
T1=15 [C] 
T2=7 [C] 
Q=1200 [W] 
Q=k*A*(T1-T2)/L 
 
"Wrong Solutions:" 
Q=W1_k*(T1-T2)/L "Not using area" 
Q=W2_k*2*A*(T1-T2)/L "Using areas of both surfaces" 
Q=W3_k*A*(T1+T2)/L "Adding temperatures instead of subtracting" 
Q=W4_k*A*L*(T1-T2) "Multiplying by thickness instead of dividing by it" 
 
 
 
1-146 Heat is lost through a brick wall (k = 0.72 W/m·ºC), which is 4 m long, 3 m wide, and 25 cm thick at a rate of 500 W. 
If the inner surface of the wall is at 22ºC, the temperature at the midplane of the wall is 
(a) 0ºC (b) 7.5ºC (c) 11.0ºC (d) 14.8ºC (e) 22ºC 
Answer (d) 14.8ºC 
 
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES 
screen. 
 
k=0.72 [W/m-C] 
Length=4 [m] 
Width=3 [m] 
L=0.25 [m] 
Q_dot=500 [W] 
T1=22 [C] 
 
A=Length*Width 
Q_dot=k*A*(T1-T_middle)/(0.5*L) 
 
"Some Wrong Solutions with Common Mistakes" 
Q_dot=k*A*(T1-W1_T_middle)/L "Using L instead of 0.5L" 
W2_T_middle=T1/2 "Just taking the half of the given temperature" 
 
 
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
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1-147 A 10-cm high and 20-cm wide circuit board houses on its surface 100 closely spaced chips, each generating heat at a 
rate of 0.12 W and transferring it by convection and radiation to the surrounding medium at 40°C. Heat transfer from the 
back surface of the board is negligible. If the combined convection and radiation heat transfer coefficient on the surface of 
the board is 22 W/m2⋅°C, the average surface temperature of the chips is 
(a) 41°C (b) 54°C (c) 67°C (d) 76°C (e) 82°C 
Answer (c) 67°C 
 
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES 
screen. 
 
A=0.1*0.2 [m^2] 
Q= 100*0.12 [W] 
Tair=40 [C] 
h=22 [W/m^2-C] 
Q= h*A*(Ts-Tair) 
 
"Wrong Solutions:" 
Q= h*(W1_Ts-Tair) "Not using area" 
Q= h*2*A*(W2_Ts-Tair) "Using both sides of surfaces" 
Q= h*A*(W3_Ts+Tair) "Adding temperatures instead of subtracting" 
Q/100= h*A*(W4_Ts-Tair) "Considering 1 chip only" 
 
 
 
1-148 A 40-cm-long, 0.4-cm-diameter electric resistance wire submerged in water is used to determine the convection heat 
transfer coefficient in water during boiling at 1 atm pressure. The surface temperature of the wire is measured to be 114°C 
when a wattmeter indicates the electric power consumption to be 7.6 kW. The heat transfer coefficient is 
(a) 108 kW/m2⋅°C (b) 13.3 kW/m2⋅°C (c) 68.1 kW/m2⋅°C (d) 0.76 kW/m2⋅°C (e) 256 kW/m2⋅°C 
Answer (a) 108 kW/m2⋅°C 
 
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES 
screen. 
 
L=0.4 [m] 
D=0.004 [m] 
A=pi*D*L [m^2] 
We=7.6 [kW] 
Ts=114 [C] 
Tf=100 [C] “Boiling temperature of water at 1 atm" 
We= h*A*(Ts-Tf) 
 
"Wrong Solutions:" 
We= W1_h*(Ts-Tf) "Not using area" 
We= W2_h*(L*pi*D^2/4)*(Ts-Tf) "Using volume instead of area" 
We= W3_h*A*Ts "Using Ts instead of temp difference" 
 
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1-149 Over 90 percent of the energy dissipated by an incandescent light bulb is in the form of heat, not light. What is the 
temperature of a vacuum-enclosed tungsten filament with an exposed surface area of 2.03 cm2 in a 100 W incandescent light 
bulb? The emissivity of tungsten at the anticipated high temperatures is about 0.35. Note that the light bulb consumes 100 W 
of electrical energy, and dissipates all of it by radiation. 
(a) 1870 K (b) 2230 K (c) 2640 K (d) 3120 K (e) 2980 K 
Answer (b) 2230 K 
 
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES 
screen. 
 
e =0.35 
Q=100 [W] 
A=2.03E-4 [m^2] 
Q=e*A*sigma#*T^4 
 
 
 
1-150 Commercial surface coating processes often use infrared lamps to speed the curing of the coating. A 1-mm-thick, 
teflon (k = 0.45 W/m⋅K) coating is applied to a 4 m × 4 m surface using this process. Once the coating reaches steady-state, 
the temperature of its two surfaces are 50oC and 45oC. What is the minimum rate at which power must be supplied to the 
infrared lamps steadily? 
(a) 36 kW (b) 40 kW (c) 44 kW (d) 48 kW (e) 52 kW 
Answer (a) 36 kW 
 
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES 
screen. 
 
k=0.45 [W/m-K] 
A=16 [m^2] 
t=0.001 [m] 
dT=5 [C] 
Q=k*A*dT/t 
 
 
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
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1-151 A 10 cm × 12 cm × 14 cm rectangular prism object made of hardwood (ρ = 721 kg/m3, cp = 1.26 kJ/kg·ºC) is 
cooled from 100ºC to the room temperature of 20ºC in 54 minutes. The approximate heat transfer coefficient during this 
process is 
(a) 0.47 W/m2·ºC (b) 5.5 W/m2·ºC (c) 8 W/m2·ºC (d) 11 W/m2·ºC (e) 17,830 W/m2·ºC 
Answer (d) 11 W/m2·ºC 
 
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES 
screen. 
 
a=0.10 [m] 
b=0.12 [m] 
c=0.14 [m] 
rho=721 [kg/m^3] 
c_p=1260 [J/kg-C] 
T1=100 [C] 
T2=20 [C] 
time=54*60 [s] 
V=a*b*c 
m=rho*V 
Q=m*c_p*(T1-T2) 
Q_dot=Q/time 
T_ave=1/2*(T1+T2) 
T_infinity=T2 
A_s=2*a*b+2*a*c+2*b*c 
Q_dot=h*A_s*(T_ave-T_infinity) 
 
"Some Wrong Solutions with Common Mistakes" 
Q_dot=W1_h*A_s*(T1-T2) "Using T1 instead of T_ave" 
Q_dot=W2_h*(T1-T2) "Not using A" 
Q=W3_h*A_s*(T1-T2) "Using Q instead of Q_dot " 
 
 
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1-152 A 25-cm diameter black ball at 130°C is suspended in air, and is losing heat to the surrounding air at 25°C by 
convection with a heat transfer coefficient of 12 W/m2⋅°C, and by radiation to the surrounding surfaces at 15°C. The total 
rate of heat transfer from the black ball is 
(a) 217 W (b) 247 W (c) 251 W (d) 465 W (e) 2365 W 
Answer: (d) 465 W 
 
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES 
screen. 
 
sigma=5.67E-8 [W/m^2-K^4] 
eps=1 
D=0.25 [m] 
A=pi*D^2 
h_conv=12 [W/m^2-C] 
Ts=130 [C] 
Tf=25 [C] 
Tsurr=15 [C] 
Q_conv=h_conv*A*(Ts-Tf) 
Q_rad=eps*sigma*A*((Ts+273)^4-(Tsurr+273)^4) 
Q_total=Q_conv+Q_rad 
 
"Wrong Solutions:" 
W1_Ql=Q_conv "Ignoring radiation" 
W2_Q=Q_rad "ignoring convection" 
W3_Q=Q_conv+eps*sigma*A*(Ts^4-Tsurr^4) "Using C in radiation calculations" 
W4_Q=Q_total/A "not using area" 
 
 
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1-153 A 3-m2 black surface at 140°C is losing heat to the surrounding air at 35°C by convection with a heat transfer 
coefficient of 16 W/m2⋅°C, and by radiation to the surrounding surfaces at 15°C. The total rate of heat loss from the surface 
is 
(a) 5105 W (b) 2940 W (c) 3779 W (d) 8819 W (e) 5040 W 
Answer (d) 8819 W 
 
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES 
screen. 
 
sigma=5.67E-8 [W/m^2-K^4] 
eps=1 
A=3 [m^2] 
h_conv=16 [W/m^2-C] 
Ts=140 [C] 
Tf=35 [C] 
Tsurr=15 [C] 
Q_conv=h_conv*A*(Ts-Tf) 
Q_rad=eps*sigma*A*((Ts+273)^4-(Tsurr+273)^4) 
Q_total=Q_conv+Q_rad 
 
“Some Wrong Solutions with Common Mistakes:” 
W1_Ql=Q_conv "Ignoring radiation" 
W2_Q=Q_rad "ignoring convection" 
W3_Q=Q_conv+eps*sigma*A*(Ts^4-Tsurr^4) "Using C in radiation calculations" 
W4_Q=Q_total/A "not using area" 
 
 
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1-154 A person’s head can be approximated as a 25-cm diameter sphere at 35°C with an emissivity of 0.95. Heat is lost from 
the head to the surrounding air at 25°C by convection with a heat transfer coefficient of 11 W/m2⋅°C, and by radiation to the 
surrounding surfaces at 10°C. Disregarding the neck, determine the total rate of heat loss from the head. 
(a) 22 W (b) 27 W (c) 49 W (d) 172 W (e) 249 W 
Answer: (c) 49 W 
 
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES 
screen. 
 
sigma=5.67E-8 [W/m^2-K^4] 
eps=0.95 
D=0.25 [m] 
A=pi*D^2 
h_conv=11 [W/m^2-C] 
Ts=35 [C] 
Tf=25 [C] 
Tsurr=10 [C] 
Q_conv=h_conv*A*(Ts-Tf) 
Q_rad=eps*sigma*A*((Ts+273)^4-(Tsurr+273)^4) 
Q_total=Q_conv+Q_rad 
 
"Wrong Solutions:" 
W1_Ql=Q_conv "Ignoring radiation" 
W2_Q=Q_rad "ignoring convection" 
W3_Q=Q_conv+eps*sigma*A*(Ts^4-Tsurr^4) "Using C in radiation calculations" 
W4_Q=Q_total/A "not using area" 
 
 
 
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1-155 A 25-cm-long, 0.4-cm-diameter electric resistance wire is used to determine the convection heat transfer coefficient in 
air at 25°C experimentally. The surface temperature of the wire is measured to be 230°C when the electric power 
consumption is 180 W. If the radiation heat loss from the wire is calculated to be 60 W, the convection heat transfer 
coefficient is 
(a) 186 W/m2⋅°C (b) 280 W/m2⋅°C (c) 373 W/m2⋅°C (d) 585 W/m2⋅°C (e) 620 W/m2⋅°C 
Answer (a) 186 W/m2⋅°C 
 
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES 
screen. 
 
L=0.25 [m] 
D=0.004 [m] 
A=pi*D*L 
We=180 [W] 
Ts=230 [C] 
Tf=25 [C] 
Qrad = 60 
We- Qrad = h*A*(Ts-Tf) 
 
“Some Wrong Solutions with Common Mistakes:” 
We- Qrad = W1_h*(Ts-Tf) "Not using area" 
We- Qrad = W2_h*(L*D)*(Ts-Tf) "Using D*L for area" 
We+ Qrad = W3_h*A*(Ts-Tf) "Adding Q_rad instead of subtracting" 
We= W4_h*A*(Ts-Tf) "Disregarding Q_rad" 
 
 
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1-156 A room is heated by a 1.2 kW electric resistance heater whose wires have a diameter of 4 mm and a total length 
of 3.4 m. The air in the room is at 23ºC and the interior surfaces of the room are at 17ºC. The convection heat transfer 
coefficient on the surface of the wires is 8 W/m2·ºC. If the rates of heat transfer from the wires to the room by 
convection and by radiation are equal, the surface temperature of the wires is 
(a) 3534ºC (b) 1778ºC (c) 1772ºC (d) 98ºC (e) 25ºC 
Answer (b) 1778ºC 
 
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES 
screen. 
 
D=0.004 [m] 
L=3.4 [m] 
W_dot_e=1200 [W] 
T_infinity=23 [C] 
T_surr=17 [C] 
h=8 [W/m^2-C] 
 
A=pi*D*L 
Q_dot_conv=W_dot_e/2 
Q_dot_conv=h*A*(T_s-T_infinity) 
 
"Some Wrong Solutions with Common Mistakes" 
Q_dot_conv=h*A*(W1_T_s-T_surr) "Using T_surr instead of T_infinity" 
Q_dot_conv/1000=h*A*(W2_T_s-T_infinity) "Using kW unit for the rate of heat transfer" 
Q_dot_conv=h*(W3_T_s-T_infinity) "Not using surface area of the wires" 
W_dot_e=h*A*(W4_T_s-T_infinity) "Using total heat transfer" 
 
 
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1-157 A person standing in a room loses heat to the air in the room by convection and to the surrounding surfaces by 
radiation. Both the air in the room and the surrounding surfaces are at 20ºC. The exposed surfaces of the person is 1.5 m2 and 
has an average temperature of 32ºC, and an emissivity of 0.90. If the rates of heat transfer from the person by convection and 
by radiation are equal, the combined heat transfer coefficient is 
(a) 0.008 W/m2·ºC (b) 3.0 W/m2·ºC (c) 5.5 W/m2·ºC (d) 8.3 W/m2·ºC (e) 10.9 W/m2·ºC 
Answer (e) 10.9 W/m2·ºC 
 
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES 
screen. 
 
T_infinity=20 [C] 
T_surr=20 [C] 
T_s=32 [C] 
A=1.5 [m^2] 
epsilon=0.90 
 
sigma=5.67E-8 [W/m^2-K^4] 
Q_dot_rad=epsilon*A*sigma*((T_s+273)^4-(T_surr+273)^4) 
Q_dot_total=2*Q_dot_rad 
Q_dot_total=h_combined*A*(T_s-T_infinity) 
 
"Some Wrong Solutions with Common Mistakes" 
Q_dot_rad=W1_h_combined*A*(T_s-T_infinity) "Using radiation heat transfer instead of total heat transfer" 
Q_dot_rad_1=epsilon*A*sigma*(T_s^4-T_surr^4) "Using C unit for temperature in radiation calculation" 
2*Q_dot_rad_1=W2_h_combined*A*(T_s-T_infinity) 
 
 
 
1-158 While driving down a highway early in the evening, the air flow over an automobile establishes an overall heat 
transfer coefficient of 18 W/m2⋅K. The passenger cabin of this automobile exposes 9 m2 of surface to the moving ambient air. 
On a day when the ambient temperature is 33oC, how much cooling must the air conditioning system supply to maintain a 
temperature of 20oC in the passenger cabin? 
(a) 670 W (b) 1284 W (c) 2106 W (d) 2565 W (e) 3210 W 
Answer (c) 2106 W 
 
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES 
screen. 
 
h=18 [W/m^2-C] 
A=9 [m^2] 
T_1=33 [C] 
T_2=20 [C] 
Q=h*A*(T_2-T_1) 
 
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1-159 On a still clear night, the sky appears to be a blackbody with an equivalent temperature of 250 K. What is the air 
temperature when a strawberry field cools to 0°C and freezes if the heat transfer coefficient between the plants and the air is 
6 W/m2⋅oC because of a light breeze and the plants have an emissivity of 0.9? 
(a) 14oC (b) 7oC (c) 3oC (d) 0oC (e) –3°C 
Answer (a) 14oC 
 
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES 
screen. 
 
e=0.9 
h=6 [W/m^2-K] 
T_1=273 [K] 
T_2=250 [K] 
h*(T-T_1)=e*sigma#*(T_1^4-T_2^4) 
 
 
 
1-160 . . . 1-163 Design and Essay Problems 
 
 
 
 
 
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2-1
 
Solutions Manual 
for 
Heat and Mass Transfer: Fundamentals & Applications 
Fourth Edition 
 Yunus A. Cengel & Afshin J. Ghajar 
McGraw-Hill, 2011 
 
 
 
 
Chapter 2 
HEAT CONDUCTION EQUATION 
 
 
 
 
 
 
 
 
PROPRIETARY AND CONFIDENTIAL 
 
This Manual is the proprietary property of The McGraw-Hill Companies, Inc. (“McGraw-Hill”) and 
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otherwise, without the prior written permission of McGraw-Hill. 
 
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2-2
 
Introduction 
 
2-1C Heat transfer is a vector quantity since it has direction as well as magnitude. Therefore, we must specify both direction 
and magnitude in order to describe heat transfer completely at a point. Temperature, on the other hand, is a scalar quantity. 
 
 
2-2C The heat transfer process from the kitchen air to the refrigerated space is 
transient in nature since the thermal conditions in the kitchen and the 
refrigerator, in general, change with time. However, we would analyze this 
problem as a steady heat transfer problem under the worst anticipated conditions 
such as the lowest thermostat setting for the refrigerated space, and the 
anticipated highest temperature in the kitchen (the so-called design conditions). 
If the compressor is large enough to keep the refrigerated space at the desired 
temperature setting under the presumed worst conditions, then it is large enough 
to do so under all conditions by cycling on and off. Heat transfer into the 
refrigerated space is three-dimensional in nature since heat will be entering 
through all six sides of the refrigerator. However, heat transfer through any wall 
or floor takes place in the direction normal to the surface, and thus it can be 
analyzed as being one-dimensional. Therefore, this problem can be simplified 
greatly by considering the heat transfer to be onedimensional at each of the four 
sides as well as the top and bottom sections, and then by adding the calculated 
values of heat transfer at each surface. 
 
 
2-3C The term steady implies no change with time at any point within the medium while transient implies variation with 
time or time dependence. Therefore, the temperature or heat flux remains unchanged with time during steady heat transfer 
through a medium at any location although both quantities may vary from one location to another. During transient heat 
transfer, the temperature and heat flux may vary with time as well as location. Heat transfer is one-dimensional if it occurs 
primarily in one direction. It is two-dimensional if heat tranfer in the third dimension is negligible. 
 
 
2-4C Heat transfer through the walls, door, and the top and bottom sections of an oven is transient in nature since the thermal 
conditions in the kitchen and the oven, in general, change with time. However, we would analyze this problem as a steady 
heat transfer problem under the worst anticipated conditions such as the highest temperature setting for the oven, and the 
anticipated lowest temperature in the kitchen (the so called “design” conditions). If the heating element of the oven is large 
enough to keep the oven at the desired temperature setting under the presumed worst conditions, then it is large enough to do 
so under all conditions by cycling on and off. 
 Heat transfer from the oven is three-dimensional in nature since heat will be entering through all six sides of the 
oven. However, heat transfer through any wall or floor takes place in the direction normal to the surface, and thus it can be 
analyzed as being one-dimensional. Therefore, this problem can be simplified greatly by considering the heat transfer as 
being one- dimensional at each of the four sides as well as the top and bottom sections, and then by adding the calculated 
values of heat transfers at each surface. 
 
 
2-5C Heat transfer to a potato in an oven can be modeled as one-dimensional since temperature differences (and thus heat 
transfer) will exist in the radial direction only because of symmetry about the center point. This would be a transient heat 
transfer process since the temperature at any point within the potato will change with time during cooking. Also, we would 
use the spherical coordinate system to solve this problem since the entire outer surface of a spherical body can be described 
by a constant value of the radius in spherical coordinates. We would place the origin at the center of the potato. 
 
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2-3
 
2-6C Assuming the egg to be round, heat transfer to an egg in boiling water can be modeled as one-dimensional since 
temperature differences (and thus heat transfer) will primarily exist in the radial direction only because of symmetry about 
the center point. This would be a transient heat transfer process since the temperature at any point within the egg will change 
with time during cooking. Also, we would use the spherical coordinate system to solve this problem since the entire outer 
surface of a spherical body can be described by a constant value of the radius in spherical coordinates. We would place the 
origin at the center of the egg. 
 
 
2-7C Heat transfer to a hot dog can be modeled as two-dimensional since temperature differences (and thus heat transfer) 
will exist in the radial and axial directions (but there will be symmetry about the center line and no heat transfer in the 
azimuthal direction. This would be a transient heat transfer process since the temperature at any point within the hot dog will 
change with time during cooking. Also, we would use the cylindrical coordinate system to solve this problem since a 
cylinder is best described in cylindrical coordinates. Also, we would place the origin somewhere on the center line, possibly 
at the center of the hot dog. Heat transfer in a very long hot dog could be considered to be one-dimensional in preliminary 
calculations. 
 
 
2-8C Heat transfer to a roast beef in an oven would be transient since the temperature at any point within the roast will 
change with time during cooking. Also, by approximating the roast as a spherical object, this heat transfer process can be 
modeled as one-dimensional since temperature differences (and thus heat transfer) will primarily exist in the radial direction 
because of symmetry about the center point. 
 
 
2-9C Heat loss from a hot water tank in a house to the surrounding medium can be considered to be a steady heat transfer 
problem. Also, it can be considered to be two-dimensional since temperature differences (and thus heat transfer) will exist in 
the radial and axial directions (but there will be symmetry about the center line and no heat transfer in the azimuthal 
direction.) 
 
 
2-10C Yes, the heat flux vector at a point P on an isothermalsurface of a medium has to be perpendicular to the surface at 
that point. 
 
 
2-11C Isotropic materials have the same properties in all directions, and we do not need to be concerned about the variation 
of properties with direction for such materials. The properties of anisotropic materials such as the fibrous or composite 
materials, however, may change with direction. 
 
 
2-12C In heat conduction analysis, the conversion of electrical, chemical, or nuclear energy into heat (or thermal) energy in 
solids is called heat generation. 
 
 
2-13C The phrase “thermal energy generation” is equivalent to “heat generation,” and they are used interchangeably. They 
imply the conversion of some other form of energy into thermal energy. The phrase “energy generation,” however, is vague 
since the form of energy generated is not clear. 
 
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2-4
 
2-14C Heat transfer to a canned drink can be modeled as two-dimensional since temperature differences (and thus heat 
transfer) will exist in the radial and axial directions (but there will be symmetry about the center line and no heat transfer in 
the azimuthal direction. This would be a transient heat transfer process since the temperature at any point within the drink 
will change with time during heating. Also, we would use the cylindrical coordinate system to solve this problem since a 
cylinder is best described in cylindrical coordinates. Also, we would place the origin somewhere on the center line, possibly 
at the center of the bottom surface. 
 
 
 
 
2-15 A certain thermopile used for heat flux meters is considered. The minimum heat flux this meter can detect is to be 
determined. 
Assumptions 1 Steady operating conditions exist. 
Properties The thermal conductivity of kapton is given to be 0.345 W/m⋅K. 
Analysis The minimum heat flux can be determined from 
2 W/m17.3=°°⋅=∆=
m 002.0
C1.0)C W/m345.0(
L
tkq& 
 
 
 
 
2-16 The rate of heat generation per unit volume in the uranium rods is given. The total rate of heat generation in each rod is 
to be determined. 
g = 2×108 W/m3Assumptions Heat is generated uniformly in the uranium rods. 
D = 5 cmAnalysis The total rate of heat generation in the rod is determined 
by multiplying the rate of heat generation per unit volume by the 
volume of the rod L = 1 m
 kW 393= W10.933m) 1](4/m) 05.0([) W/m102()4/( 52382genrodgengen ×=×=== ππ LDeeE &&& V
 
 
 
 
2-17 The variation of the absorption of solar energy in a solar pond with depth is given. A relation for the total rate of heat 
generation in a water layer at the top of the pond is to be determined. 
Assumptions Absorption of solar radiation by water is modeled as heat generation. 
Analysis The total rate of heat generation in a water layer of surface area A and thickness L at the top of the pond is 
determined by integration to be 
 
b
)e(1eA bL0
−−
=
− −=
−
=== ∫∫
&
&&&&
LbxL
x
bx
b
eeAAdxeedeE
0
0
0
0gengen )(
V
V 
 
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2-5
 
2-18 The rate of heat generation per unit volume in a stainless steel plate is given. The heat flux on the surface of the plate is 
to be determined. 
Assumptions Heat is generated uniformly in steel plate. 
 e 
 
 L 
Analysis We consider a unit surface area of 1 m2. The total rate of 
heat generation in this section of the plate is 
 W101.5m) )(0.03m 1)( W/m105()( 5236genplategengen ×=×=×== LAeeE &&& V 
Noting that this heat will be dissipated from both sides of the plate, the heat flux on 
either surface of the plate becomes 
2kW/m 75==
×
×
== 2
2
5
plate
gen W/m000,75
m 12
 W105.1
A
E
q
&
& 
 
 
 
 
2-19E The power consumed by the resistance wire of an iron is given. The heat generation and the heat flux are to be 
determined. 
Assumptions Heat is generated uniformly in the resistance wire. 
q = 800 W Analysis An 800 W iron will convert electrical energy into 
heat in the wire at a rate of 800 W. Therefore, the rate of heat 
generation in a resistance wire is simply equal to the power 
rating of a resistance heater. Then the rate of heat generation in 
the wire per unit volume is determined by dividing the total 
rate of heat generation by the volume of the wire to be 
D = 0.08 in
L = 15 in
37 ftBtu/h 106.256 ⋅×=⎟
⎠
⎞
⎜
⎝
⎛===
 W1
Btu/h 412.3
ft) 12/15](4/ft) 12/08.0([
 W800
)4/( 22
gen
wire
gen
gen ππ LD
EE
e
&&
&
V
 
Similarly, heat flux on the outer surface of the wire as a result of this heat generation is determined by dividing the total rate 
of heat generation by the surface area of the wire to be 
 25 ftBtu/h 101.043 ⋅×=⎟
⎠
⎞
⎜
⎝
⎛===
 W1
Btu/h 412.3
ft) 12/15(ft) 12/08.0(
 W800gen
wire
gen
ππDL
E
A
E
q
&&
& 
Discussion Note that heat generation is expressed per unit volume in Btu/h⋅ft3 whereas heat flux is expressed per unit surface 
area in Btu/h⋅ft2. 
 
 
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2-6
 
2-20E Prob. 2-19E is reconsidered. The surface heat flux as a function of wire diameter is to be plotted. 
Analysis The problem is solved using EES, and the solution is given below. 
 
"GIVEN" 
E_dot=800 [W] 
L=15 [in] 
D=0.08 [in] 
 
"ANALYSIS" 
g_dot=E_dot/V_wire*Convert(W, Btu/h) 
V_wire=pi*D^2/4*L*Convert(in^3, ft^3) 
q_dot=E_dot/A_wire*Convert(W, Btu/h) 
A_wire=pi*D*L*Convert(in^2, ft^2) 
 
 
D 
[in] 
q 
[Btu/h.ft2] 
0.02 
0.04 
0.06 
0.08 
0.1 
0.12 
0.14 
0.16 
0.18 
0.2 
0.22 
0.24 
0.26 
0.28 
0.3 
0.32 
0.34 
0.36 
0.38 
0.4 
417069 
208535 
139023 
104267 
83414 
69512 
59581 
52134 
46341 
41707 
37915 
34756 
32082 
29791 
27805 
26067 
24533 
23171 
21951 
20853 
0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4
0
50000
100000
150000
200000
250000
300000
350000
400000
450000
D [in]
q 
 [B
tu
/h
-ft
2 ]
 
 
 
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preparation. If you are a student using this Manual, you are using it without permission. 
2-7
 
Heat Conduction Equation 
 
2-21C The one-dimensional transient heat conduction equation for a plane wall with constant thermal conductivity and heat 
generation is 
t
T
αk
e
x
T
∂
∂
=+
∂
∂ 1gen
2
2 &
. Here T is the temperature, x is the space variable, is the heat generation per unit 
volume, k is the thermal conductivity, α is the thermal diffusivity, and t is the time. 
gene&
 
 
2-22C The one-dimensional transient heat conduction equation for a long cylinder with constant thermal conductivity and 
heat generation is 
t
T
k
e
r
Tr
rr ∂
∂
α
=+⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
∂
∂ 11 gen& . Here T is the temperature, r is the space variable, g is the heat generation per 
unit volume, k is the thermal conductivity, α is the thermal diffusivity, and t is the time. 
 
 
2-23 We consider a thin element of thickness ∆x in a large plane wall (see Fig. 2-12 in the text). The density of the wall is ρ, 
the specific heat is c, and the area of the wall normal to the direction of heat transfer is A. In the absence of any heat 
generation, an energy balance on this thin element of thickness ∆x during a small time interval ∆t can be expressed as 
 
t
E
QQ xxx ∆
∆
=− ∆+
element&& 
where 
 )()(element ttttttttt TTxcATTmcEEE −∆=−=−=∆ ∆+∆+∆+ ρ 
Substituting, 
 
t
TT
xcAQQ tttxxx ∆
−
∆=− ∆+∆+ ρ&& 
Dividing by A∆x gives 
 
t
TT
c
x
QQ
A
tttxxx
∆
−
=
∆
−
− ∆+∆+ ρ
&&1 
Taking the limitas and yields ∆x → 0 ∆t → 0
 
t
Tρc
x
TkA
xA ∂
∂
=⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
∂
∂1 
since from the definition of the derivative and Fourier’s law of heat conduction, 
 ⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
−
∂
∂
=
∂
∂
=
∆
−∆+
→∆ x
TkA
xx
Q
x
QQ xxx
x
&&
0
lim 
Noting that the area A of a plane wall is constant, the one-dimensional transient heat conduction equation in a plane wall with 
constant thermal conductivity k becomes 
 
t
T
αx
T
∂
∂
=
∂
∂ 1
2
2
 
where the property ck ρα /= is the thermal diffusivity of the material. 
 
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2-8
 
2-24 We consider a thin cylindrical shell element of thickness ∆r in a long cylinder (see Fig. 2-14 in the text). The density of 
the cylinder is ρ, the specific heat is c, and the length is L. The area of the cylinder normal to the direction of heat transfer at 
any location is rLA π2= where r is the value of the radius at that location. Note that the heat transfer area A depends on r in 
this case, and thus it varies with location. An energy balance on this thin cylindrical shell element of thickness ∆r during a 
small time interval ∆t can be expressed as 
 
t
E
EQQ rrr ∆
∆
=+− ∆+
element
element
&&& 
where 
 )()(element ttttttttt TTrcATTmcEEE −∆=−=−=∆ ∆+∆+∆+ ρ 
 rAeeE ∆== genelementgenelement &&& V
Substituting, 
 
t
TT
rcArAeQQ tttrrr ∆
−
∆=∆+− ∆+∆+ ρgen&&& 
where rLA π2= . Dividing the equation above by A∆r gives 
 
t
TT
ce
r
QQ
A
tttrrr
∆
−
=+
∆
−
− ∆+∆+ ρgen
1
&
&&
 
Taking the limit as and yields 0→∆r 0→∆t
 
t
Tce
r
TkA
rA ∂
∂
ρ=+⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
∂
∂
gen
1
& 
since, from the definition of the derivative and Fourier’s law of heat conduction, 
 ⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
−
∂
∂
=
∂
∂
=
∆
−∆+
→∆ r
TkA
rr
Q
r
QQ rrr
r
&&
0
lim 
Noting that the heat transfer area in this case is rLA π2= and the thermal conductivity is constant, the one-dimensional 
transient heat conduction equation in a cylinder becomes 
 
t
Te
r
Tr
rr ∂
∂
α
=+⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
∂
∂ 11
gen& 
where ck ρα /= is the thermal diffusivity of the material. 
 
 
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2-9
 
2-25 We consider a thin spherical shell element of thickness ∆r in a sphere (see Fig. 2-16 in the text).. The density of the 
sphere is ρ, the specific heat is c, and the length is L. The area of the sphere normal to the direction of heat transfer at any 
location is where r is the value of the radius at that location. Note that the heat transfer area A depends on r in this 
case, and thus it varies with location. When there is no heat generation, an energy balance on this thin spherical shell 
element of thickness ∆r during a small time interval ∆t can be expressed as 
24 rA π=
 
t
E
QQ rrr ∆
∆
=− ∆+
element&& 
where 
 )()(element ttttttttt TTrcATTmcEEE −∆=−=−=∆ ∆+∆+∆+ ρ 
Substituting, 
 
t
TTrcAQQ tttrrr ∆
−
∆=− ∆+∆+ ρ&& 
where . Dividing the equation above by A∆r gives 24 rA π=
 
t
TT
c
r
QQ
A
tttrrr
∆
−
=
∆
−
− ∆+∆+ ρ
&&1 
Taking the limit as and yields 0→∆r 0→∆t
 
t
Tρc
r
TkA
rA ∂
∂
=⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
∂
∂1 
since, from the definition of the derivative and Fourier’s law of heat conduction, 
 ⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
−
∂
∂
=
∂
∂
=
∆
−∆+
→∆ r
TkA
rr
Q
r
QQ rrr
r
&&
0
lim 
Noting that the heat transfer area in this case is and the thermal conductivity k is constant, the one-dimensional 
transient heat conduction equation in a sphere becomes 
24 rA π=
 
t
T
αr
Tr
rr ∂
∂
=⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
∂
∂ 11 2
2
 
where ck ρα /= is the thermal diffusivity of the material. 
 
 
 
 
2-26 For a medium in which the heat conduction equation is given in its simplest by 
t
T
x
T
∂
∂
α
=
∂
∂ 1
2
2
: 
(a) Heat transfer is transient, (b) it is one-dimensional, (c) there is no heat generation, and (d) the thermal conductivity is 
constant. 
 
 
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2-10
 
2-27 For a medium in which the heat conduction equation is given by 
t
T
y
T
x
T
∂
∂
α
=
∂
∂
+
∂
∂ 1
2
2
2
2
: 
(a) Heat transfer is transient, (b) it is two-dimensional, (c) there is no heat generation, and (d) the thermal conductivity is 
constant. 
 
 
 
 
2-28 For a medium in which the heat conduction equation is given by 01 gen =+⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
∂
∂
+⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
∂
∂ e
z
Tk
zr
Tkr
rr
& : 
(a) Heat transfer is steady, (b) it is two-dimensional, (c) there is heat generation, and (d) the thermal conductivity is variable. 
 
 
 
 
2-29 For a medium in which the heat conduction equation is given in its simplest by 01 gen =+⎟
⎠
⎞
⎜
⎝
⎛ e
dr
dTrk
dr
d
r
& : 
(a) Heat transfer is steady, (b) it is one-dimensional, (c) there is heat generation, and (d) the thermal conductivity is variable. 
 
 
 
 
2-30 For a medium in which the heat conduction equation is given by 
t
T
αr
Tr
rr ∂
∂
=⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
∂
∂ 11 2
2
 
(a) Heat transfer is transient, (b) it is one-dimensional, (c) there is no heat generation, and (d) the thermal conductivity is 
constant. 
 
 
 
 
2-31 For a medium in which the heat conduction equation is given in its simplest by 02
2
=+
dr
dT
dr
Tdr : 
(a) Heat transfer is steady, (b) it is one-dimensional, (c) there is no heat generation, and (d) the thermal conductivity is 
constant. 
 
 
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2-11
 
2-32 We consider a small rectangular element of length ∆x, width ∆y, and height ∆z = 1 (similar to the one in Fig. 2-20). The 
density of the body is ρ and the specific heat is c. Noting that heat conduction is two-dimensional and assuming no heat 
generation, an energy balance on this element during a small time interval ∆t can be expressed as 
⎟
⎟
⎟
⎠
⎞
⎜
⎜
⎜
⎝
⎛
=
⎟
⎟
⎟
⎠
⎞
⎜
⎜
⎜
⎝
⎛
∆+∆
−
⎟
⎟
⎟
⎠
⎞
⎜
⎜
⎜
⎝
⎛
element theof 
content energy the
 of change of Rate 
 and +
at surfaces at the 
conductionheat of Rate
 and at surfaces
 at the conduction
heat of Rate 
yyxxyx
 
or 
t
E
QQQQ yyxxyx ∆
∆
=−−+ ∆+∆+
element&&&& 
Noting that the volume of the element is 1element ×∆∆=∆∆∆= yxzyxV , the change in the energy content of the element can 
be expressed as 
 )()(element ttttttttt TTyxcTTmcEEE −∆∆=−=−=∆ ∆+∆+∆+ ρ 
Substituting, 
t
TT
yxcQQQQ tttyyxxyx ∆
−
∆∆=−−+ ∆+∆+∆+ ρ&&&& 
Dividing by ∆x∆y gives 
 
t
TT
c
y
QQ
xx
QQ
y
tttyyyxxx
∆
−
=
∆
−
∆
−
∆
−
∆
− ∆+∆+∆+ ρ
&&&& 11 
Taking the thermal conductivity k to be constant and noting that the heat transfer surface areas of the element for heat 
conduction in the x and y directions are ,1 and 1 ×∆=×∆= xAyA yx respectively, and taking the limit as 0 and , , →∆∆∆ tyx 
yields 
 
t
T
αy
T
x
T
∂
∂
=
∂
∂
+
∂
∂ 1
2
2
2
2
 
since, from the definition of the derivative and Fourier’s law of heat conduction, 
 
2
2
0
111lim
x
Tk
x
Tk
xx
Tzyk
xzyx
Q
zyx
QQ
zy
xxxx
x ∂
∂
−=⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
∂
∂
−=⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
∆∆−
∂
∂
∆∆
=
∂
∂
∆∆
=
∆
−
∆∆
∆+
→∆
&&
 
 
2
2
0
111lim
y
Tk
y
Tk
yy
Tzxk
yzxy
Q
zxy
QQ
zx
yyyy
y ∂
∂
−=⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
∂
∂
∂
∂
−=⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
∂
∂
∆∆−
∂
∂
∆∆
=
∂
∂
∆∆
=
∆
−
∆∆
∆+
→∆
&&
 
Here the property ck ρα /= is the thermal diffusivity of the material. 
 
 
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2-12
 
2-33 We consider a thin ring shaped volume element of width ∆z and thickness ∆r in a cylinder. The density of the cylinder 
is ρ and the specific heat is c. In general, an energy balance on this ring element during a small time interval ∆t can be 
expressed as 
 
t
E
QQQQ zzzrrr ∆
∆
=−+− ∆+∆+
element)()( &&&& 
r r r+∆r 
∆z 
But the change in the energy content of the element can be expressed as 
 )()2()(element ttttttttt TTzrrcTTmcEEE −∆∆πρ=−=−=∆ ∆+∆+∆+ 
Substituting, 
 
t
TT
zrrcQQQQ tttzzzrrr ∆
−
∆∆=−+− ∆+∆+∆+ )2()()( πρ&&&& 
Dividing the equation above by zrr ∆∆ )2( π gives 
 
t
TT
c
z
QQ
rrr
QQ
zr
tttzzzrrr
∆
−
=
∆
−
∆
−
∆
−
∆
− ∆+∆+∆+ ρ
ππ
&&&&
2
1
2
1 
Noting that the heat transfer surface areas of the element for heat conduction in the r and z directions are 
,2 and 2 rrAzrA zr ∆=∆= ππ respectively, and taking the limit as 0 and , →∆∆∆ tzr yields 
 
t
Tc
z
Tk
z
Tk
rr
Tkr
rr ∂
∂
ρ=⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
∂
∂
+⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
∂φ
∂
∂φ
∂
+⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
∂
∂
2
11 
since, from the definition of the derivative and Fourier’s law of heat conduction, 
⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
∂
∂
−=⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
∆π−
∂
∂
∆π
=
∂
∂
∆π
=
∆
−
∆π
∆+
→∆ r
Tkr
rrr
Tzrk
rzrr
Q
zrr
QQ
zr
rrr
r
1)2(
2
1
2
1
2
1lim
0
&&
 
⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
∂
∂
−=⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
∆π−
∂
∂
∆π
=
∂
∂
∆π
=
∆
−
∆π
∆+
→∆ z
Tk
zz
Trrk
zrrz
Q
rrz
QQ
rr
zzzz
z
)2(
2
1
2
1
2
1lim
0
&&
 
For the case of constant thermal conductivity the equation above reduces to 
 
t
T
z
T
r
Tr
rr ∂
∂
α
=
∂
∂
+⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
∂
∂ 11
2
2
 
where ck ρα /= is the thermal diffusivity of the material. For the case of steady heat conduction with no heat generation it 
reduces to 
 01
2
2
=
∂
∂
+⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
∂
∂
z
T
r
Tr
rr
 
 
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2-13
 
2-34 Consider a thin disk element of thickness ∆z and diameter D in a long cylinder (Fig. P2-34). The density of the cylinder 
is ρ, the specific heat is c, and the area of the cylinder normal to the direction of heat transfer is , which is 
constant. An energy balance on this thin element of thickness ∆z during a small time interval ∆t can be expressed as 
4/2DA π=
⎟
⎟
⎟
⎠
⎞
⎜
⎜
⎜
⎝
⎛
=
⎟
⎟
⎟
⎠
⎞
⎜
⎜
⎜
⎝
⎛
+
⎟
⎟
⎟
⎠
⎞
⎜
⎜
⎜
⎝
⎛
∆
−
⎟
⎟
⎟
⎠
⎞
⎜
⎜
⎜
⎝
⎛
element theof 
content energy the
 of change of Rate 
element the
inside generation
heat of Rate 
 +at surface 
 at the conduction 
heat of Rate 
at surface the
at conduction 
heat of Rate 
zzz
 
or, 
 
t
E
EQQ zzz ∆
∆
=+− ∆+
element
element
&&& 
But the change in the energy content of the element and the rate of heat generation within the element can be expressed as 
 )()(element ttttttttt TTzcATTmcEEE −∆=−=−=∆ ∆+∆+∆+ ρ 
and 
 zAeeE ∆== genelementgenelement &&& V
Substituting, 
 
t
TT
zcAzAeQQ tttzzz ∆
−
∆=∆+− ∆+∆+ ρgen&&& 
Dividing by A∆z gives 
 
t
TT
ce
z
QQ
A
tttzzz
∆
−
=+
∆
−
− ∆+∆+ ρgen
1
&
&&
 
Taking the limit as and yields 0→∆z 0→∆t
 
t
Tce
z
TkA
zA ∂
∂
ρ=+⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
∂
∂
gen
1
& 
since, from the definition of the derivative and Fourier’s law of heat conduction, 
 ⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
−
∂
∂
=
∂
∂
=
∆
−∆+
→∆ z
TkA
zz
Q
z
QQ zzz
z
&&
0
lim 
Noting that the area A and the thermal conductivity k are constant, the one-dimensional transient heat conduction equation in 
the axial direction in a long cylinder becomes 
 
t
T
k
e
z
T
∂
∂
α
=+
∂
∂ 1gen
2
2 &
 
where the property ck ρα /= is the thermal diffusivity of the material. 
 
 
 
2-35 For a medium in which the heat conduction equation is given by 
t
TT
rr
Tr
rr ∂
∂
α
=
∂φ
∂
θ
+⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
∂
∂ 1
sin
11
2
2
22
2
2
 
(a) Heat transfer is transient, (b) it is two-dimensional, (c) there is no heat generation, and (d) the thermal conductivity is 
constant. 
 
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
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2-14
 
Boundary and Initial Conditions; Formulation of Heat Conduction Problems 
 
2-36C The mathematical expressions of the thermal conditions at the boundaries are called the boundary conditions. To 
describe a heat transfer problem completely, two boundary conditions must be given for each direction of the coordinate 
system along which heat transfer is significant. Therefore, we need to specify four boundary conditions for two-dimensional 
problems. 
 
 
2-37C The mathematical expression for the temperature distribution of the medium initially is called the initial condition. 
We need only one initial condition for a heat conduction problem regardless of the dimension since the conduction equation 
is first order in time (it involves the first derivative of temperature with respect to time). Therefore, we need only 1 initial 
condition for a two-dimensional problem. 
 
 
2-38C A heat transfer problem that is symmetric about a plane, line, or point is said to have thermal symmetry about that 
plane, line, or point. The thermal symmetry boundary condition is a mathematical expression of this thermal symmetry. It is 
equivalent to insulation or zero heat flux boundary condition, and is expressed at a point x0 as 0/),( 0 =∂∂ xtxT . 
 
 
2-39C The boundary condition at a perfectly insulated surface (at x = 0, for example) can be expressed as 
0),0(or 0),0( =
∂
∂
=
∂
∂
−
x
tT
x
tTk
 
which indicates zero heat flux.
 
 
 
2-40C Yes, the temperature profile in a medium must be perpendicular to an insulated surface since the slope 0/ =∂∂ xT at 
that surface. 
 
 
2-41C We try to avoid the radiation boundary condition in heat transfer analysis because it is a non-linear expression that 
causes mathematical difficulties while solving the problem; often making it impossible to obtain analytical solutions. 
 
 
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preparation. If you are a student using this Manual, you are using it without permission. 
2-15
 
2-42 Heat conduction through the bottom section of an aluminum pan that is used to cook stew on top of an electric range is 
considered (Fig. P2-48). Assuming variable thermal conductivity and one-dimensional heat transfer, the mathematical 
formulation (the differential equation and the boundary conditions) of this heat conduction problem is to be obtained for 
steady operation. 
Assumptions 1 Heat transfer is given to be steady and one-dimensional. 2 Thermal conductivity is given to be variable. 3 
There is no heat generation in the medium. 4 The top surface at x = L is subjected to specified temperature and the bottom 
surface at x = 0 is subjected to uniform heat flux. 
Analysis The heat flux at the bottom of the pan is 
 2
22
gen
s
 W/m831,31
4/m) 18.0(
 W)900(90.0
4/
=
×
===
ππD
E
A
Q
q ss
&&
& 
Then the differential equation and the boundary conditions for this heat conduction problem can be expressed as 
 0=⎟
⎠
⎞
⎜
⎝
⎛
dx
dTk
dx
d 
 
C108)( 
 W/m831,31)0( 2
°==
==−
L
s
TLT
q
dx
dTk &
 
 
 
 
2-43 A spherical container of inner radius , outer radius , and thermal conductivity k is 
given. The boundary condition on the inner surface of the container for steady one-dimensional 
conduction is to be expressed for the following cases: 
r1 r2
(a) Specified temperature of 50°C: C50)( 1 °=rT
(b) Specified heat flux of 45 W/m2 towards the center: 21 W/m45
)(
=
dr
rdT
k 
(c)Convection to a medium at with a heat transfer coefficient of h: ∞T ])([
)(
1
1
∞−= TrThdr
rdT
k 
r2r1
 
 
 
2-44 Heat is generated in a long wire of radius ro covered with a plastic insulation layer at a constant rate of . The heat 
flux boundary condition at the interface (radius r
gene&
o) in terms of the heat generated is to be expressed. The total heat generated 
in the wire and the heat flux at the interface are 
 
2)2(
)(
)(
gen
2
gengen
2
genwiregengen
o
o
os
s
o
re
Lr
Lre
A
E
A
Q
q
LreeE
&&&&
&
&&&
====
==
π
π
πV
 D egen 
L
Assuming steady one-dimensional conduction in the radial direction, the heat flux boundary condition can be expressed as 
 
2
)( gen oo re
dr
rdT
k
&
=− 
 
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2-16
 
2-45 A long pipe of inner radius r1, outer radius r2, and thermal conductivity 
k is considered. The outer surface of the pipe is subjected to convection to a 
medium at with a heat transfer coefficient of h. Assuming steady one-
dimensional conduction in the radial direction, the convection boundary 
condition on the outer surface of the pipe can be expressed as 
∞T
r2r1
h, T∞
 ])([)( 22 ∞−=− TrThdr
rdTk 
 
 
 
 
2-46 A spherical shell of inner radius r1, outer radius r2, and thermal 
conductivity k is considered. The outer surface of the shell is subjected to 
radiation to surrounding surfaces at . Assuming no convection and 
steady one-dimensional conduction in the radial direction, the radiation 
boundary condition on the outer surface of the shell can be expressed as 
surrT
k 
Tsurr
r2r1
ε 
 [ ]4surr422 )()( TrTdr
rdT
k −=− εσ 
 
 
 
 
2-47 A spherical container consists of two spherical layers A and B that are at 
perfect contact. The radius of the interface is ro. Assuming transient one-
dimensional conduction in the radial direction, the boundary conditions at the 
interface can be expressed as 
ro
 ),(),( trTtrT oBoA =
and 
r
trT
k
r
trT
k oBB
oA
A ∂
∂
−=
∂
∂
−
),(),(
 
 
 
 
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preparation. If you are a student using this Manual, you are using it without permission. 
2-17
 
2-48 Heat conduction through the bottom section of a steel pan that is used to boil water on top of an electric range is 
considered. Assuming constant thermal conductivity and one-dimensional heat transfer, the mathematical formulation (the 
differential equation and the boundary conditions) of this heat conduction problem is to be obtained for steady operation. 
Assumptions 1 Heat transfer is given to be steady and one-dimensional. 2 Thermal conductivity is given to be constant. 3 
There is no heat generation in the medium. 4 The top surface at x = L is subjected to convection and the bottom surface at x = 
0 is subjected to uniform heat flux. 
Analysis The heat flux at the bottom of the pan is 
 2
22
gen W/m820,33
4/m) 20.0(
 W)1250(85.0
4/
=
×
===
ππD
E
A
Q
q
s
s
s
&&
& 
Then the differential equation and the boundary conditions for this heat conduction problem can be expressed as 
 02
2
=
dx
Td 
 
])([
)(
 W/m280,33)0( 2
∞−=−
==−
TLTh
dx
LdT
k
q
dx
dTk s&
 
 
 
 
 
2-49E A 2-kW resistance heater wire is used for space heating. Assuming constant thermal conductivity and one-
dimensional heat transfer, the mathematical formulation (the differential equation and the boundary conditions) of this heat 
conduction problem is to be obtained for steady operation. 
Assumptions 1 Heat transfer is given to be steady and one-dimensional. 2 Thermal conductivity is given to be constant. 3 
Heat is generated uniformly in the wire. 
2 kW 
Analysis The heat flux at the surface of the wire is 
D = 0.12 in
 2gen
s
 W/in.7353
in) in)(15 06.0(2
 W2000
2
====
ππ Lr
E
A
Q
q
o
s
s
&&
& L = 15 in
Noting that there is thermal symmetry about the center line and there is uniform heat flux at the outer surface, the differential 
equation and the boundary conditions for this heat conduction problem can be expressed as 
 01 gen =+⎟
⎠
⎞
⎜
⎝
⎛
k
e
dr
dTr
dr
d
r
&
 
 
2 W/in.7353
)(
0)0(
==−
=
s
o q
dr
rdT
k
dr
dT
&
 
 
 
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2-18
 
2-50 The outer surface of the East wall of a house exchanges heat with both convection and radiation., while the interior 
surface is subjected to convection only. Assuming the heat transfer through the wall to be steady and one-dimensional, the 
mathematical formulation (the differential equation and the boundary and initial conditions) of this heat conduction problem 
is to be obtained. 
Assumptions 1 Heat transfer is given to be steady and one-dimensional. 2 Thermal 
conductivity is given to be constant. 3 There is no heat generation in the medium. 
4 The outer surface at x = L is subjected to convection and radiation while the inner 
surface at x = 0 is subjected to convection only. 
x 
T∞2
h2
L 
Tsky
T∞1
h1
Analysis Expressing all the temperatures in Kelvin, the differential equation and the 
boundary conditions for this heat conduction problem can be expressed as 
 02
2
=
dx
Td 
)]0([)0( 11 TThdx
dTk −=− ∞ 
[ ]4sky4221 )(])([)( TLTTLThdx
LdTk −+−=− ∞ σε 
 
 
 
 
2-51 A spherical metal ball that is heated in an oven to a temperature of Ti throughout is dropped into a large body of water 
at T∞ where it is cooled by convection. Assuming constant thermal conductivity and transient one-dimensional heat transfer, 
the mathematical formulation (the differential equation and the boundary and initial conditions) of this heat conduction 
problem is to be obtained. 
Assumptions 1 Heat transfer is given to be transient and one-dimensional. 2 Thermal conductivity is given to be constant. 3 
There is no heat generation in the medium. 4 The outer surface at r = r0 is subjected to convection. 
Analysis Noting that there is thermal symmetry about the midpoint and convection at the outer surface, the differential 
equation and the boundary conditions for this heat conduction problem can be expressed as 
 
t
T
r
Tr
rr ∂
∂
α
=⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
∂
∂ 11 2
2
 
k r2
Ti
T∞
h 
 
i
o
o
TrT
TrTh
r
trT
k
r
tT
=
−=
∂
∂
−
=
∂
∂
∞
)0,( 
])([
),(
0
),0(
 
 
 
 
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2-19
 
2-52 A spherical metal ball that is heated in an oven to a temperature of Ti throughout is allowed to cool in ambient air at T∞ 
by convection and radiation. Assuming constant thermal conductivity and transient one-dimensional heat transfer, the 
mathematical formulation (the differential equation and the boundary and initial conditions) of this heat conduction problem 
is to be obtained. 
Assumptions 1 Heat transfer is given to be transient and one-dimensional. 2 Thermal conductivity is given to be variable. 3 
There is no heat generation in the medium. 4 The outer surface at r = ro is subjected to convection and radiation. 
Analysis Noting that there is thermal symmetry about the midpoint and convection and radiation at the outer surface and 
expressing all temperatures in Rankine, the differential equation and the boundary conditions for this heat conduction 
problem can be expressed as 
k r2
T∞
h 
Tsurr
ε 
Ti
 
t
Tc
r
Tkr
rr ∂
∂
ρ=⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
∂
∂ 2
2
1 
 
i
oo
o
TrT
TrTTrTh
r
trT
k
rtT
=
−εσ+−=
∂
∂
−
=
∂
∂
∞
)0,( 
])([])([
),(
0
),0(
 
4
surr
4 
 
 
 
 
2-53 Water flows through a pipe whose outer surface is wrapped with a thin electric heater that consumes 400 W per m 
length of the pipe. The exposed surface of the heater is heavily insulated so that the entire heat generated in the heater is 
transferred to the pipe. Heat is transferred from the inner surface of the pipe to the water by convection. Assuming constant 
thermal conductivity and one-dimensional heat transfer, the mathematical formulation (the differential equation and the 
boundary conditions) of the heat conduction in the pipe is to be obtained for steady operation. 
Assumptions 1 Heat transfer is given to be steady and one-dimensional. 2 Thermal conductivity is given to be constant. 3 
There is no heat generation in the medium. 4 The outer surface at r = r2 is subjected to uniform heat flux and the inner 
surface at r = r1 is subjected to convection. 
Analysis The heat flux at the outer surface of the pipe is 
r1 r2
Q = 400 W
h 
T∞
 2
2s
 W/m4.979
m) cm)(1 065.0(2
 W400
2
====
ππ Lr
Q
A
Q
q sss
&&
& 
Noting that there is thermal symmetry about the center line and there is 
uniform heat flux at the outer surface, the differential equation and the 
boundary conditions for this heat conduction problem can be expressed as 
 0=⎟
⎠
⎞
⎜
⎝
⎛
dr
dTr
dr
d 
 
22
1
 W/m6.734
)(
]90)([85])([
)(
==
−=−= ∞
s
ii
q
dr
rdT
k
rTTrTh
dr
rdT
k
&
 
 
 
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preparation. If you are a student using this Manual, you are using it without permission. 
2-20
 
Solution of Steady One-Dimensional Heat Conduction Problems 
 
2-54C Yes, this claim is reasonable since no heat is entering the cylinder and thus there can be no heat transfer from the 
cylinder in steady operation. This condition will be satisfied only when there are no temperature differences within the 
cylinder and the outer surface temperature of the cylinder is the equal to the temperature of the surrounding medium. 
 
 
2-55C Yes, the temperature in a plane wall with constant thermal conductivity and no heat generation will vary linearly 
during steady one-dimensional heat conduction even when the wall loses heat by radiation from its surfaces. This is because 
the steady heat conduction equation in a plane wall is = 0 whose solution is regardless of the 
boundary conditions. The solution function represents a straight line whose slope is C
22 / dxTd 21)( CxCxT +=
1. 
 
 
2-56C Yes, in the case of constant thermal conductivity and no heat generation, the temperature in a solid cylindrical rod 
whose ends are maintained at constant but different temperatures while the side surface is perfectly insulated will vary 
linearly during steady one-dimensional heat conduction. This is because the steady heat conduction equation in this case is 
= 0 whose solution is 22 / dxTd 21)( CxCxT += which represents a straight line whose slope is C1. 
 
 
2-57C Yes, this claim is reasonable since in the absence of any heat generation the rate of heat transfer through a plain wall 
in steady operation must be constant. But the value of this constant must be zero since one side of the wall is perfectly 
insulated. Therefore, there can be no temperature difference between different parts of the wall; that is, the temperature in a 
plane wall must be uniform in steady operation. 
 
 
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2-21
 
2-58 A 20-mm thick draw batch furnace front is subjected to uniform 
heat flux on the inside surface, while the outside surface is subjected 
to convection and radiation heat transfer. The inside surface 
temperature of the furnace front is to be determined. 
Assumptions 1 Heat conduction is steady. 2 One dimensional heat 
conduction across the furnace front thickness. 3 Thermal properties 
are constant. 4 Inside and outside surface temperatures are constant. 
Properties Emissivity and thermal conductivity are given to be 0.30 
and 25 W/m · K, respectively 
Analysis The uniform heat flux subjected on the inside 
surface is equal to the sum of heat fluxes transferred by 
convection and radiation on the outside surface: 
)()( 4surr
4
0 TTTThq LL −+−= ∞ εσ& 
 
444428
22
K ])27320()[K W/m1067.5)(30.0(
K )]27320()[K W/m10( W/m5000
+−⋅×+
+−⋅=
−
L
L
T
T
Copy the following line and paste on a blank EES screen to solve the above equation: 
5000=10*(T_L-(20+273))+0.30*5.67e-8*(T_L^4-(20+273)^4) 
Solving by EES software, the outside surface temperature of the furnace front is 
 K 594=LT
For steady heat conduction, the Fourier’s law of heat conduction can be expressed as 
 
dx
dTkq −=0& 
Knowing that the heat flux and thermal conductivity are constant, integrating the differential equation once with respect to x 
yields 
 1
0)( Cx
k
q
xT +−=
&
 
Applying the boundary condition gives 
 :Lx = 1
0)( CL
k
q
TLT L +−==
&
 → LTLk
q
C += 01
&
 
Substituting into the general solution, the variation of temperature in the furnace front is determined to be 1C
 LTxLk
q
xT +−= )()( 0
&
 
The inside surface temperature of the furnace front is 
 K 598=+
⋅
=+== K 594m) 020.0(
K W/m25
 W/m5000)0(
2
0
0 LTLk
q
TT
&
 
Discussion By insulating the furnace front, heat loss from the outer surface can be reduced. 
 
 
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preparation. If you are a student using this Manual, you are using it without permission. 
2-22
 
2-59 A large plane wall is subjected to specified heat flux and temperature on the left surface and no conditions on the right 
surface. The mathematical formulation, the variation of temperature in the plate, and the right surface temperature are to be 
determined for steady one-dimensional heat transfer. 
Assumptions 1 Heat conduction is steady and one-dimensional since the wall is large relative to its thickness, and the thermal 
conditions on both sides of the wall are uniform. 2 Thermal conductivity is constant. 3 There is no heat generation in the 
wall. 
Properties The thermal conductivity is given to be k =2.5 W/m⋅°C. 
L=0.3 m 
q=700 W/m2
T1=80°C 
k 
Analysis (a) Taking the direction normal to the surface of the wall to 
be the x direction with x = 0 at the left surface, the mathematical 
formulation of this problem can be expressed as 
 02
2
=
dx
Td 
and 20 W/m700
)0(
==− q
dx
dTk & 
 C80)0( 1 °== TT
x 
 (b) Integrating the differential equation twice with respect to x yields 
 1Cdx
dT
= 
 21)( CxCxT +=
where C1 and C2 are arbitrary constants. Applying the boundary conditions give 
Heat flux at x = 0: 
k
q
CqkC 0101 
&
& −=→=− 
Temperature at x = 0: 12121 0)0( TCTCCT =→=+×= 
Substituting C1 and C2 into the general solution, the variation of temperature is determined to be 
 80280C80
C W/m5.2
 W/m700)(
2
1
0 +−=°+
°⋅
−=+−= xxTx
k
q
xT
&
 
 (c) The temperature at x = L (the right surface of the wall) is 
 C-4°=+×−= 80m) 3.0(280)(LT
Note that the right surface temperature is lower as expected. 
 
 
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2-23
 
2-60 A large plane wall is subjected to specified heat flux and temperature on the left surface and no conditions on the right 
surface. The mathematical formulation, the variation of temperaturein the plate, and the right surface temperature are to be 
determined for steady one-dimensional heat transfer. 
Assumptions 1 Heat conduction is steady and one-dimensional since the wall is large relative to its thickness, and the thermal 
conditions on both sides of the wall are uniform. 2 Thermal conductivity is constant. 3 There is no heat generation in the 
wall. 
Properties The thermal conductivity is given to be k =2.5 W/m⋅°C. 
Analysis (a) Taking the direction normal to the surface of the wall to 
be the x direction with x = 0 at the left surface, the mathematical 
formulation of this problem can be expressed as 
L=0.3 m 
q=1050 W/m2
T1=90°C 
k 
02
2
=
dx
Td 
and 
2
0 W/m1050
)0(
==− q
dx
dTk & 
C90)0( 1 °== TT x 
 (b) Integrating the differential equation twice with respect to x yields 
 1Cdx
dT
= 
 21)( CxCxT +=
where C1 and C2 are arbitrary constants. Applying the boundary conditions give 
Heat flux at x = 0: 
k
q
CqkC 0101 
&
& −=→=− 
Temperature at x = 0: 12121 0)0( TCTCCT =→=+×= 
Substituting C1 and C2 into the general solution, the variation of temperature is determined to be 
 90420C90
C W/m5.2
 W/m1050)(
2
1
0 +−=°+
°⋅
−=+−= xxTx
k
q
xT
&
 
 (c) The temperature at x = L (the right surface of the wall) is 
 C-36°=+×−= 90m) 3.0(420)(LT
Note that the right surface temperature is lower as expected. 
 
 
 
PROPRIETARY MATERIAL. © 2011 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. 
2-24
 
2-61 A large plane wall is subjected to specified temperature on the left surface and convection on the right surface. The 
mathematical formulation, the variation of temperature, and the rate of heat transfer are to be determined for steady one-
dimensional heat transfer. 
Assumptions 1 Heat conduction is steady and one-dimensional. 2 Thermal conductivity is constant. 3 There is no heat 
generation. 
Properties The thermal conductivity is given to be k = 1.8 W/m⋅°C. 
Analysis (a) Taking the direction normal to the surface of the wall to be the x direction with x = 0 at the left surface, the 
mathematical formulation of this problem can be expressed as 
0
2
2
=
dx
Td 
x 
T∞ =25°C 
h=24 W/m2.°C
L=0.4 m 
T1=90°C 
A=30 m2
k and 
C90)0( 1 °== TT 
])([)( ∞−=− TLThdx
LdTk 
(b) Integrating the differential equation twice with respect to x yields 
1Cdx
dT
= 
21)( CxCxT += 
where C and C1 2 are arbitrary constants. Applying the boundary conditions give 
x = 0: 1221 0)0( TCCCT =→+×=
x = L: 
hLk
TTh
C
hLk
TCh
CTCLChkC
+
−
−=→
+
−
−=→−+=− ∞∞∞
)(
 
)(
 ])[( 11
2
1211 
Substituting into the general solution, the variation of temperature is determined to be 21 and CC
 
x
x
Tx
hLk
TTh
xT
3.9090
C90
m) 4.0)(C W/m24()C W/m8.1(
C)2590)(C W/m24(
)(
)(
2
2
1
1
−=
°+
°⋅+°⋅
°−°⋅
−=
+
+
−
−= ∞
 
 (c) The rate of heat conduction through the wall is 
 
 W7389=
°⋅+°⋅
°−°⋅
°⋅=
+
−
=−=−= ∞
m) 4.0)(C W/m24()C W/m8.1(
C)2590)(C W/m24()m 30)(C W/m8.1(
)(
2
2
2
1
1wall hLk
TTh
kAkAC
dx
dTkAQ&
 
Note that under steady conditions the rate of heat conduction through a plain wall is constant. 
 
 
PROPRIETARY MATERIAL. © 2011 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. 
2-25
 
2-62 The top and bottom surfaces of a solid cylindrical rod are maintained at constant temperatures of 20°C and 95°C while 
the side surface is perfectly insulated. The rate of heat transfer through the rod is to be determined for the cases of copper, 
steel, and granite rod. 
Assumptions 1 Heat conduction is steady and one-dimensional. 2 Thermal conductivity is constant. 3 There is no heat 
generation. 
Properties The thermal conductivities are given to be k = 380 W/m⋅°C for copper, k = 18 W/m⋅°C for steel, and k = 1.2 
W/m⋅°C for granite. 
Analysis Noting that the heat transfer area (the area normal to 
the direction of heat transfer) is constant, the rate of heat 
transfer along the rod is determined from 
Insulated
T1=25°C
D = 0.05 m T2=95°C L
TT
kAQ 21
−
=& 
where L = 0.15 m and the heat transfer area A is 
L=0.15 m2322 m 10964.14/m) 05.0(4/ −×=== ππDA 
Then the heat transfer rate for each case is determined as follows: 
(a) Copper: W373.1=°−×°⋅=
−
= −
m 0.15
C20)(95
)m 10C)(1.964 W/m380( 2321
L
TT
kAQ& 
(b) Steel: W17.7=°−×°⋅=
−
= −
m 0.15
C20)(95)m 10C)(1.964 W/m18( 2321
L
TT
kAQ& 
(c) Granite: W1.2=°−×°⋅=
−
= −
m 0.15
C20)(95
)m 10C)(1.964 W/m2.1( 2321
L
TT
kAQ& 
Discussion: The steady rate of heat conduction can differ by orders of magnitude, depending on the thermal conductivity of 
the material. 
 
 
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preparation. If you are a student using this Manual, you are using it without permission. 
2-26
 
2-63 Prob. 2-62 is reconsidered. The rate of heat transfer as a function of the thermal conductivity of the rod is to be 
plotted. 
Analysis The problem is solved using EES, and the solution is given below. 
 
"GIVEN" 
L=0.15 [m] 
D=0.05 [m] 
T_1=20 [C] 
T_2=95 [C] 
k=1.2 [W/m-C] 
 
"ANALYSIS" 
A=pi*D^2/4 
Q_dot=k*A*(T_2-T_1)/L 
 
 
k 
[W/m.C] 
Q 
[W] 
1 0.9817 
22 21.6 
43 42.22 
64 62.83 
85 83.45 
106 104.1 
127 124.7 
148 145.3 
169 165.9 
190 186.5 
211 207.1 
232 227.8 
253 248.4 
274 269 
295 289.6 
316 310.2 
337 330.8 
358 351.5 
379 372.1 
400 392.7 
0 50 100 150 200 250 300 350 400
0
50
100
150
200
250
300
350
400
k [W/m-C]
Q
 [
W
]
 
 
 
 
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preparation. If you are a student using this Manual, you are using it without permission. 
2-27
 
2-64 The base plate of a household iron is subjected to specified heat flux on the left surface and to specified temperature on 
the right surface. The mathematical formulation, the variation of temperature in the plate, and the inner surface temperature 
are to be determined for steady one-dimensional heat transfer. 
Assumptions 1 Heat conduction is steady and one-dimensional since the surface area of the base plate is large relative to its 
thickness, and the thermal conditions on both sides of the plate are uniform. 2 Thermal conductivity is constant. 3 There is 
no heat generation in the plate. 4 Heat loss through the upper part of the iron is negligible. 
Properties The thermal conductivity is given to be k = 60 W/m⋅°C. 
Analysis (a) Noting that the upper part of the iron is well insulated and thus the entire heat generated in the resistance wires 
is transferred to the base plate, the heat flux through the inner surface is determined to be 
 2
24
base
0
0 W/m000,50
m 10160
 W800
=
×
==
−A
Q
q
&
& 
x 
T2 =112°C
L=0.6 cm 
Q =800 W 
A=160 cm2
k 
Taking the direction normal to the surface of the wall to be the x 
direction with x = 0 at the left surface, the mathematical formulation 
of this problem can be expressed as 
02
2
=
dx
Td 
and 20 W/m000,50
)0(
==− q
dx
dT
k & 
C112)( 2 °== TLT 
 (b) Integrating the differential equation twice with respect to x yields 
1Cdx
dT
= 
21)( CxCxT += 
where C1 and C2 are arbitrary constants. Applying the boundary conditions give 
x = 0: 
k
q
CqkC 0101 
&
& −=→=− 
x = L: 
k
Lq
TCLCTCTCLCLT 022122221 )(
&
+=→−=→=+= 
Substituting into the general solution, the variation of temperature is determined to be 21 and CC
 
112)006.0(3.833
C112
C W/m60
m)006.0)( W/m000,50(
)(
)(
2
2
00
2
0
+−=°+
°⋅
−
=
+
−
=++−=
x
x
T
k
xLq
k
Lq
Tx
k
q
xT
&&&
 
 (c) The temperature at x = 0 (the inner surface of the plate) is 
 C117°=+−= 112)0006.0(3.833)0(T
Note that the inner surface temperature is higher than the exposed surface temperature, as expected. 
 
PROPRIETARY MATERIAL. © 2011 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. 
2-28
 
2-65 The base plate of a household iron is subjected to specified heat flux on the left surface and to specified temperature on 
the right surface. The mathematical formulation, the variation of temperature in the plate, and the inner surface temperature 
are to be determined for steady one-dimensional heat transfer. 
Assumptions 1 Heat conduction is steady and one-dimensional since the surface area of the base plate is large relative to its 
thickness, and the thermal conditions on both sides of the plate are uniform. 2 Thermal conductivity is constant. 3 There is 
no heat generation in the plate. 4 Heat loss through the upper part of the iron is negligible. 
Properties The thermal conductivity is given to be k = 60 W/m⋅°C. 
Analysis (a) Noting that the upper part of the iron is well insulated and thus 
the entire heat generated in the resistance wires is transferred to the base 
plate, the heat flux through the inner surface is determined to be 
 2
24
base
0
0 W/m000,75
m 10160
 W1200
=
×
==
−A
Q
q
&
& 
Taking the direction normal to the surface of the wall to be the x direction with 
x = 0 at the left surface, the mathematical formulation of this problem can be 
expressed as 
x 
T2 =112°C
L=0.6 cm
Q=1200 W 
A=160 cm2
k 
02
2
=
dx
Td 
and 20 W/m000,75
)0(
==− q
dx
dT
k & 
C112)( 2 °== TLT 
 (b) Integrating the differential equation twice with respect to x yields 
1Cdx
dT
= 
21)( CxCxT += 
where C1 and C2 are arbitrary constants. Applying the boundary conditions give 
x = 0: 
k
q
CqkC 0101 
&
& −=→=− 
x = L: 
k
Lq
TCLCTCTCLCLT 022122221 )(
&
+=→−=→=+= 
Substituting C into the general solution, the variation of temperature is determined to be C21 and 
 
112)006.0(1250
C112
C W/m60
m)006.0)( W/m000,75(
)(
)(
2
2
00
2
0
+−=
°+
°⋅
−
=
+
−
=++−=
x
x
T
k
xLq
k
Lq
Tx
k
q
xT
&&&
 
(c) The temperature at x = 0 (the inner surface of the plate) is 
 C119.5°=+−= 112)0006.0(1250)0(T
Note that the inner surface temperature is higher than the exposed surface temperature, as expected. 
 
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preparation. If you are a student using this Manual, you are using it without permission. 
2-29
 
2-66 Prob. 2-64 is reconsidered. The temperature as a function of the distance is to be plotted. 
Analysis The problem is solved using EES, and the solution is given below. 
 
"GIVEN" 
Q_dot=800 [W] 
L=0.006 [m] 
A_base=160E-4 [m^2] 
k=60 [W/m-C] 
T_2=112 [C] 
 
"ANALYSIS" 
q_dot_0=Q_dot/A_base 
T=q_dot_0*(L-x)/k+T_2 "Variation of temperature" 
"x is the parameter to be varied" 
 
 
x 
[m] 
T 
[C] 
0 
0.0006 
0.0012 
0.0018 
0.0024 
0.003 
0.0036 
0.0042 
0.0048 
0.0054 
0.006 
117 
116.5 
116 
115.5 
115 
114.5 
114 
113.5 
113 
112.5 
112 
0 0.001 0.002 0.003 0.004 0.005 0.006
112
113
114
115
116
117
x [m]
T 
 [C
]
 
 
 
 
 
PROPRIETARY MATERIAL. © 2011 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. 
2-30
 
2-67 Chilled water flows in a pipe that is well insulated from outside. The mathematical formulation and the variation of 
temperature in the pipe are to be determined for steady one-dimensional heat transfer. 
Assumptions 1 Heat conduction is steady and one-dimensional since the pipe is long relative to its thickness, and there is 
thermal symmetry about the center line. 2 Thermal conductivity is constant. 3 There is no heat generation in the pipe. 
Analysis (a) Noting that heat transfer is one-dimensional in the radial r direction, the mathematical formulation of this 
problem can be expressed as 
r2
r1
 Insulated 
0=⎟
⎠
⎞
⎜
⎝
⎛
dr
dTr
dr
d 
and )]([
)(
1
1 rTTh
dr
rdT
k f −=− Water Tf
 0
)( 2 =
dr
rdT
 L
(b) Integrating the differential equation once with respect to r gives 
1Cdr
dTr = 
Dividing both sides of the equation above by r to bring it to a readily integrable form and then integrating, 
r
C
dr
dT 1= 
21 ln)( CrCrT += 
where C1 and C2 are arbitrary constants. Applying the boundary conditions give 
r = r2: 00 1
2
1 =→= C
r
C
 
r = r1: 
ff
f
TCCTh
CrCTh
r
C
k
=→−=
+−=−
22
211
1
1
)(0
)]ln([
 
Substituting C1 and C2 into the general solution, the variation of temperature is determined to be 
 fTrT =)(
This result is not surprising since steady operating conditions exist. 
 
 
PROPRIETARY MATERIAL. © 2011 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. 
2-31
 
2-68 The convection heat transfer coefficient between the surface of a pipe carrying superheated vapor and the surrounding 
air is to be determined. 
Assumptions 1 Heat conduction is steady and one-dimensional and there is thermal symmetry about the centerline. 2 
Thermal properties are constant. 3 There is no heat generation in the pipe. 4 Heat transfer by radiation is negligible. 
Properties The constant pressure specific heat of vapor is given to be 2190 J/kg · °C and the pipe thermal conductivity is 17 
W/m · °C. 
 
Analysis The inner and outer radii of the pipe are 
 m 025.02/m 05.01 ==r
 m 031.0m 006.0m 025.02 =+=r
The rate of heat loss from the vapor in the pipe can be determined from 
 W4599C )7(C)J/kg 2190)(kg/s 3.0()( outinloss =°°⋅=−= TTcmQ p&&
For steady one-dimensional heat conduction in cylindrical coordinates, the heat conduction equation can be expressed as 
 0=⎟
⎠
⎞
⎜
⎝
⎛
dr
dTr
dr
d 
and 
Lr
Q
A
Q
dr
rdT
k
1
lossloss1
 2
)(
π
&&
==− (heat flux at the inner pipe surface) 
 (inner pipe surface temperature) C 120)( 1 °=rT
Integrating the differential equation once with respect to r gives 
 
r
C
dr
dT 1= 
Integrating with respect to r again gives 
 21 ln)( CrCrT +=
where and are arbitrary constants. Applying the boundary conditions gives 1C 2C
 :1rr =
1
1
1
loss1
 2
1)(
r
C
Lr
Q
kdr
rdT
=−=
π
&
 →
kL
Q
C loss1 2
1 &
π
−= 
 :1rr = 21loss1 ln2
1)( Cr
kL
Q
rT +−=
&
π
 → )(ln
2
1
11
loss
2 rTrkL
Q
C +=
&
π
 
Substituting and into the general solution, the variation of temperature is determined to be 1C 2C
 
PROPRIETARY MATERIAL. © 2011 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. 
2-32
 
 
)()/ln(
2
1
)(ln
2
1ln
2
1)(
11
loss
11
lossloss
rTrr
kL
Q
rTr
kL
Q
r
kL
Q
rT
+−=
++−=
&
&&
π
ππ 
The outer pipe surface temperature is 
 
C 1.119
C 120
025.0
031.0ln
)m 10)(C W/m17(
 W4599
2
1
)()/ln(
2
1)( 112
loss
2
°=
°+⎟
⎠
⎞
⎜
⎝
⎛
°⋅
−=
+−=
π
π
rTrr
kL
Q
rT
&
 
From Newton’s law of cooling, the rate of heat loss at the outer pipe surface by convection is 
 [ ]∞−= TrTLrhQ )() 2( 22loss π&
Rearranging and the convection heat transfer coefficient is determined to be 
 C W/m25.1 2 °⋅=
°−
=
−
=
∞ C )251.119)(m 10)(m 031.0(2
 W4599
])([ 2 22
loss
ππ TrTLr
Q
h
&
 
Discussion If the pipe wall is thicker, the temperature difference between the inner and outerpipe surfaces will be greater. If 
the pipe has very high thermal conductivity or the pipe wall thickness is very small, then the temperature difference between 
the inner and outer pipe surfaces may be negligible. 
 
 
 
PROPRIETARY MATERIAL. © 2011 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. 
2-33
 
2-69 A subsea pipeline is transporting liquid hydrocarbon. The temperature variation in the pipeline wall, the inner surface 
temperature of the pipeline, the mathematical expression for the rate of heat loss from the liquid hydrocarbon, and the heat 
flux through the outer pipeline surface are to be determined. 
Assumptions 1 Heat conduction is steady and one-dimensional and there is thermal symmetry about the centerline. 2 
Thermal properties are constant. 3 There is no heat generation in the pipeline. 
Properties The pipeline thermal conductivity is given to be 60 W/m · °C. 
 
Analysis The inner and outer radii of the pipeline are 
 m 25.02/m 5.01 ==r
 m 258.0m 008.0m 25.02 =+=r
(a) For steady one-dimensional heat conduction in cylindrical coordinates, the heat conduction equation can be expressed as 
 0=⎟
⎠
⎞
⎜
⎝
⎛
dr
dTr
dr
d 
and )]([
)(
111
1 rTTh
dr
rdT
k , −=− ∞ (convection at the inner pipeline surface) 
 ])([
)(
2,22
2
∞−=− TrThdr
rdT
k (convection at the outer pipeline surface) 
Integrating the differential equation once with respect to r gives 
 
r
C
dr
dT 1= 
Integrating with respect to r again gives 
 21 ln)( CrCrT +=
where and are arbitrary constants. Applying the boundary conditions gives 1C 2C
 :1rr = )ln( 21111
1
11 CrCTh
r
C
k
dr
)dT(r
k , −−=−=− ∞ 
 :2rr = )ln(
)(
2,2212
2
12
∞−+=−=− TCrChr
C
k
dr
rdT
k 
1C and can be expressed explicitly as 2C
 
)/()/ln()/( 221211
2,1
1 hrkrrhrk
TT
C ,
++
−
−= ∞∞ 
 ⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−
++
−
−= ∞∞∞ 1
11221211
2,1
12 ln)/()/ln()/(
r
hr
k
hrkrrhrk
TT
TC ,, 
Substituting and into the general solution, the variation of temperature is determined to be 1C 2C
 
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preparation. If you are a student using this Manual, you are using it without permission. 
2-34
 
 11
11221211
2,1 )/ln(
)/()/ln()/(
)( ,
, Trr
hr
k
hrkrrhrk
TT
rT ∞
∞∞ +⎥
⎦
⎤
⎢
⎣
⎡
+
++
−
−= 
(b) The inner surface temperature of the pipeline is 
 
C 45.5 °=
°+
°⋅
°⋅
+⎟
⎠
⎞
⎜
⎝
⎛+
°⋅
°⋅
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
°⋅
°⋅
°−
−=
+⎥
⎦
⎤
⎢
⎣
⎡
+
++
−
−= ∞
∞∞
C 70
)C W/m150)(m 258.0(
C W/m60
25.0
258.0ln
)C W/m250)(m 25.0(
C W/m60
)C W/m250)(m 25.0(
C W/m60C )570(
)/ln(
)/()/ln()/(
)(
22
2
111
11221211
2,1
1 ,
, Trr
hr
k
hrkrrhrk
TT
rT
 
(c) The mathematical expression for the rate of heat loss through the pipeline can be determined from Fourier’s law to be 
 
22
12
11
2,1
1
2
2
loss
 2
1
2
)/ln(
 2
1
2
)(
) 2(
LhrLk
rr
Lhr
TT
LkC
dr
rdT
Lrk
dr
dTkAQ
,
πππ
ππ
++
−
=
−=−=
−=
∞∞
&
 
(d) Again from Fourier’s law, the heat flux through the outer pipeline surface is 
 
2 W/m5947=
⎟
⎠
⎞
⎜
⎝
⎛ °⋅
°⋅
°⋅
+⎟
⎠
⎞
⎜
⎝
⎛+
°⋅
°⋅
°−
=
++
−
=
−=−=−=
∞∞
m 258.0
C W/m60
)C W/m150)(m 258.0(
C W/m60
25.0
258.0ln
)C W/m250)(m 25.0(
C W/m60
C )570(
)/()/ln()/(
)(
22
2221211
2,1
2
12
2
r
k
hrkrrhrk
TT
r
C
k
dr
rdT
k
dr
dTkq
,
&
 
Discussion Knowledge of the inner pipeline surface temperature can be used to control wax deposition blockages in the 
pipeline. 
 
 
 
PROPRIETARY MATERIAL. © 2011 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. 
2-35
 
2-70E A steam pipe is subjected to convection on the inner surface and to specified temperature on the outer surface. The 
mathematical formulation, the variation of temperature in the pipe, and the rate of heat loss are to be determined for steady 
one-dimensional heat transfer. 
Assumptions 1 Heat conduction is steady and one-dimensional since the pipe is long relative to its thickness, and there is 
thermal symmetry about the center line. 2 Thermal conductivity is constant. 3 There is no heat generation in the pipe. 
Properties The thermal conductivity is given to be k = 7.2 Btu/h⋅ft⋅°F. 
Analysis (a) Noting that heat transfer is one-dimensional in the radial r direction, the mathematical formulation of this 
problem can be expressed as 
0=⎟
⎠
⎞
⎜
⎝
⎛
dr
dTr
dr
d T =175°F 
Steam 
300°F 
h=12.5
and )]([
)(
1
1 rTTh
dr
rdT
k −=− ∞ 
F175)( 22 °== TrT 
L = 30 ft(b) Integrating the differential equation once with respect to r gives 
 1Cdr
dTr = 
Dividing both sides of the equation above by r to bring it to a readily integrable form and then integrating, 
 
r
C
dr
dT 1= 
 21 ln)( CrCrT +=
where C1 and C2 are arbitrary constants. Applying the boundary conditions give 
r = r1: )]ln([ 211
1
1 CrCTh
r
C
k +−=− ∞ 
r = r2: 22212 ln)( TCrCrT =+=
Solving for C1 and C2 simultaneously gives 
 2
11
2
2
22122
11
2
2
1 ln
ln
ln and 
ln
r
hr
k
r
r
TT
TrCTC
hr
k
r
r
TT
C
+
−
−=−=
+
−
= ∞∞ 
Substituting C1 and C2 into the general solution, the variation of temperature is determined to be 
 
F175
in 4.2
ln36.34F175
in 4.2
ln
)ft 12/2)(FftBtu/h 5.12(
FftBtu/h 2.7
2
4.2ln
F)300175(
ln
ln
)ln(lnlnln)(
2
2
2
11
2
2
2212121
°+−=°+
°⋅⋅
°⋅⋅
+
°−
=
+
+
−
=+−=−+= ∞
rr
T
r
r
hr
k
r
r
TT
TrrCrCTrCrT
 
 (c) The rate of heat conduction through the pipe is 
Btu/h 46,630=
°⋅⋅
°⋅⋅
+
°−
°⋅⋅−=
+
−
−=−=−= ∞
)ft 12/2)(FftBtu/h 5.12(
FftBtu/h 2.7
2
4.2ln
F)300175(F)ftBtu/h 2.7ft)( 30(2
ln
2)2(
2
11
2
21
π
ππ
hr
k
r
r
TT
Lk
r
C
rLk
dr
dTkAQ&
 
 
PROPRIETARY MATERIAL. © 2011 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. 
2-36
 
2-71 A spherical container is subjected to specified temperature on the inner surface and convection on the outer surface. The 
mathematical formulation, the variation of temperature, and the rate of heat transfer are to be determined for steady one-
dimensional heat transfer. 
Assumptions 1 Heat conduction is steady and one-dimensional since there is no change with time and there is thermal 
symmetry about the midpoint. 2 Thermal conductivity is constant. 3 There is no heat generation. 
Properties The thermal conductivity is given to be k = 30 W/m⋅°C. 
Analysis (a) Noting that heat transfer is one-dimensional in the radial r direction, the mathematical formulation of this 
problem can be expressed as 
02 =⎟
⎠
⎞
⎜
⎝
⎛
dr
dTr
dr
d k 
r2
T1
T∞
h 
r1and C0)( 11 °== TrT
])([)( 22 ∞−=− TrThdr
rdTk 
(b) Integrating the differential equation once with respect to r gives 
 1
2 C
dr
dTr = 
Dividing both sides of the equation above by r to bring it to a readily integrable form and then integrating, 
 
2
1
r
C
dr
dT
= 
 21)( Cr
CrT +−= 
where C1 and C2 are arbitrary constants. Applying the boundary conditions give 
r = r1: 12
1
1
1 )( TCr
C
rT =+−= 
r = r2: ⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−+−=− ∞TCr
Ch
r
Ck 2
2
1
2
2
1 
Solving for C1 and C2 simultaneously gives 
 
1
2
21
2
1
1
1
1
12
21
2
12
1
1
 and 
1
)(
r
r
hr
k
r
r
TT
T
r
C
TC
hr
k
r
r
TTr
C
−−
−
+=+=
−−
−
= ∞∞ 
Substituting C1 and C2 into the general solution, the variation of temperature is determined to be 
 
)/1.205.1(63.29C01.2
2
1.2
)m 1.2)(C W/m18(
C W/m30
2
1.21
C)250(
1
11)(
2
1
2
1
2
21
2
1
1
1
1
1
1
1
1
r
r
T
r
r
r
r
hr
k
r
r
TT
T
rr
C
r
C
T
r
C
rT
−=°+⎟
⎠
⎞
⎜
⎝
⎛ −
°⋅
°⋅−−
°−
=
+⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−
−−
−
=+⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−=++−= ∞
 
(c) The rate of heat conduction through the wall is 
 
 W23,460=
°⋅
°⋅
−−
°−
°⋅−=
−−
−
−=−=−=−= ∞
)m 1.2)(C W/m18(
C W/m30
2
1.21
C)250(m) 1.2()C W/m30(4
1
)(44)4(
2
21
2
12
12
12
π
πππ
hr
k
r
r
TTrkkC
r
Crk
dr
dTkAQ&
 
 
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preparation. If you are a student using this Manual, you are using it without permission. 
2-37
 
2-72E A large plate is subjected to convection, radiation, and specified temperature on the top surface and no conditions on 
the bottom surface. The mathematical formulation, the variation of temperature in the plate, and the bottom surface 
temperature are to be determined for steady one-dimensional heat transfer. 
Assumptions 1 Heat conduction is steady and one-dimensional since the plate is large relative to its thickness, and the 
thermal conditions on both sides of the plate are uniform. 2 Thermal conductivity is constant. 3 There is no heat generation 
in the plate. 
Properties The thermal conductivity and emissivity are given to be 
k =7.2 Btu/h⋅ft⋅°F and ε = 0.7. 
x 80°F 
ε 
L
T∞
h 
Tsky
Analysis (a) Taking the direction normal to the surface of the plate to be 
the x direction with x = 0 at the bottom surface, and the mathematical 
formulation of this problem can be expressed as 
02
2
=
dx
Td 
and ])460[(][])([])([)( 4sky
4
22
4
sky
4 TTTThTLTTLTh
dx
LdT
k −++−=−+−=− ∞∞ εσεσ 
F80)( 2 °== TLT 
 (b) Integrating the differential equation twice with respect to x yields 
 1C
dx
dT
= 
 21)( CxCxT +=
where C1 and C2 are arbitrary constants. Applying the boundary conditions give 
Convection at x = L: 
kTTTThC
TTTThkC
/]})460[(][{ 
])460[(][ 
4
sky
4
221
4
sky
4
221
−++−−=→
−++−=−
∞
∞
εσ
εσ
Temperature at x = L: LCTCTCLCLT 122221 )( −=→=+×= 
Substituting C1 and C2 into the general solution, the variation of temperature is determined to be 
)3/1(3.1180
ft )12/4(
FftBtu/h 2.7
]R) 480()R 540)[(RftBtu/h 100.7(0.1714+F)9080)(FftBtu/h 12(F80
)(
])460[(][
)()()(
44428-2
4
sky
4
22
212121
x
x
xL
k
TTTTh
TCxLTLCTxCxT
−−=
−
°⋅⋅
−⋅⋅×°−°⋅⋅
+°=
−
−++−
+=−−=−+= ∞
εσ
 
(c) The temperature at x = 0 (the bottom surface of the plate) is 
 F76.2°=−×−= )03/1(3.1180)0(T
 
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2-38
 
2-73E A large plate is subjected to convection and specified temperature on the top surface and no conditions on the bottom 
surface. The mathematical formulation, the variation of temperature in the plate, and the bottom surface temperature are to 
be determined for steady one-dimensional heat transfer. 
Assumptions 1 Heat conduction is steady and one-dimensional since the plate is large relative to its thickness, and the 
thermal conditions on both sides of the plate are uniform. 2 Thermal conductivity is constant. 3 There is no heat generation 
in the plate. 
Properties The thermal conductivity is given to be k =7.2 Btu/h⋅ft⋅°F. 
Analysis (a) Taking the direction normal to the surface of the plate to be the x direction with x = 0 at the bottom surface, the 
mathematical formulation of this problem can be expressed as 
0
2
2
=
dx
Td x 80°F 
L
T∞
h 
and )(])([)( 2 ∞∞ −=−=− TThTLThdx
LdTk 
F80)( 2 °== TLT 
 (b) Integrating the differential equation twice with respect to x yields 
 1Cdx
dT
= 
 21)( CxCxT +=
where C1 and C2 are arbitrary constants. Applying the boundary conditions give 
Convection at x = L: kTThCTThkC /)( )( 2121 ∞∞ −−=→−=− 
Temperature at x = L: LCTCTCLCLT 122221 )( −=→=+×= 
Substituting C1 and C2 into the general solution, the variation of temperature is determined to be 
 
)3/1(7.1680
ft )12/4(
FftBtu/h 2.7
F)9080)(FftBtu/h 12(F80
)(
)(
)()()(
2
2
212121
x
x
xL
k
TTh
TCxLTLCTxCxT
−−=
−
°⋅⋅
°−°⋅⋅
+°=
−
−
+=−−=−+= ∞
 
 (c) The temperature at x = 0 (the bottom surface of the plate) is 
 F74.4°=−×−= )03/1(7.1680)0(T
 
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2-39
 
2-74 A spherical container is subjected to uniform heat flux on the outer surface and specified temperature on the inner 
surface. The mathematical formulation, the variation of temperature in the pipe, and the outer surface temperature, and the 
maximum rate of hot water supply are to be determined for steady one-dimensional heat transfer. 
Assumptions 1 Heat conduction is steady and one-dimensional since there is no change with time and there is thermal 
symmetry about the mid point. 2 Thermal conductivity is constant. 3 There is no heat generation in the container. 
Properties The thermal conductivity is given to be k = 1.5 W/m⋅°C. The specific heat of water at the average temperature of 
(100+20)/2 = 60°C is 4.185 kJ/kg⋅°C (Table A-9). 
Analysis (a) Noting that the 90% of the 800 W generated by the strip heater is transferred to the container, the heat flux 
through the outer surface is determined to be 
 222
22
 W/m8.340
m) (0.414
 W80090.0
4
=
×
===
ππr
Q
A
Q
q sss
&&
& 
Noting that heat transfer is one-dimensional in the radial r direction and heat flux is in the negative r direction, the 
mathematical formulation of this problem can be expressed as 
02 =⎟
⎠
⎞
⎜
⎝
⎛
dr
dTr
dr
d 
and C120)( 11 °== TrT
sqdr
rdT
k &=
)( 2 
(b) Integrating the differential equation once with respect to r gives 
r1 r2
T1k 
r
Heater
Insulation
 1
2 C
dr
dTr = 
Dividing both sides of the equation above by r2 and then integrating, 
 
2
1
r
C
dr
dT
= 
 2
1)( C
r
C
rT +−= 
where C1 and C2 are arbitrary constants. Applying the boundary conditions give 
r = r2: k
rq
Cq
r
C
k ss
2
2
12
2
1 
&
& =→= 
r = r1: 
1
2
2
1
1
1
122
1
1
11 )( kr
rq
T
r
C
TCC
r
C
TrT s
&
+=+=→+−== 
Substituting C1 and C2 into the general solution, the variation of temperature is determined to be 
⎟
⎠
⎞
⎜
⎝
⎛ −+=
°⋅
⎟
⎠
⎞
⎜
⎝
⎛ −+°=
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−+=⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−+=++−=+−=
rr
k
rq
rr
TC
rr
T
r
C
T
r
C
C
r
C
rT s
15.219.38120
C W/m5.1
m) 41.0)( W/m8.340(1
m 40.0
1C120
1111)(
22
2
2
1
11
1
1
1
1
1
1
2
1 &
 
(c) The outer surface temperature is determined by direct substitution to be 
Outer surface (r = r2): C122.3°=⎟
⎠
⎞
⎜
⎝
⎛ −+=⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−+=
41.0
15.219.3812015.219.38120)(
2
2 r
rT 
Noting that the maximum rate of heat supply to the water is W,720= W 8009.0 × water can be heated from 20 to 100°C at a 
rate of 
 kg/h 7.74=kg/s 002151.0
C20)C)(100kJ/kg (4.185
kJ/s 720.0 =
°−°⋅
=
∆
=→∆=
Tc
QmTcmQ
p
p
&
&&& 
 
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2-40
 
2-75 Prob. 2-74 is reconsidered. The temperature as a function of the radius is to be plotted. 
Analysis The problem is solved using EES, and the solution is given below. 
 
"GIVEN" 
r_1=0.40 [m] 
r_2=0.41 [m] 
k=1.5 [W/m-C] 
T_1=120 [C] 
Q_dot=800 [W] 
f_loss=0.10 
 
"ANALYSIS" 
q_dot_s=((1-f_loss)*Q_dot)/A 
A=4*pi*r_2^2 
T=T_1+(1/r_1-1/r)*(q_dot_s*r_2^2)/k "Variation of temperature" 
 
 
r 
[m] 
T 
[C] 
0.4 
0.4011 
0.4022 
0.4033 
0.4044 
0.4056 
0.4067 
0.4078 
0.4089 
0.41 
120 
120.3 
120.5 
120.8 
121 
121.3 
121.6 
121.8 
122.1 
122.3 
0.4 0.402 0.404 0.406 0.408 0.41
120
120.5
121
121.5
122
122.5
r [m]
T 
 [C
]PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
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2-41
 
Heat Generation in a Solid 
 
2-76C The cylinder will have a higher center temperature since the cylinder has less surface area to lose heat from per unit 
volume than the sphere. 
 
 
2-77C Heat generation in a solid is simply conversion of some form of energy into sensible heat energy. Some examples of 
heat generations are resistance heating in wires, exothermic chemical reactions in a solid, and nuclear reactions in nuclear 
fuel rods. 
 
 
2-78C The rate of heat generation inside an iron becomes equal to the rate of heat loss from the iron when steady operating 
conditions are reached and the temperature of the iron stabilizes. 
 
 
2-79C No, it is not possible since the highest temperature in the plate will occur at its center, and heat cannot flow “uphill.” 
 
 
2-80C No. Heat generation in a solid is simply the conversion of some form of energy into sensible heat energy. For example 
resistance heating in wires is conversion of electrical energy to heat. 
 
 
 
 
2-81 Heat is generated uniformly in a large brass plate. One side of the plate is insulated while the other side is subjected to 
convection. The location and values of the highest and the lowest temperatures in the plate are to be determined. 
Assumptions 1 Heat transfer is steady since there is no indication of any change with time. 2 Heat transfer is one-
dimensional since the plate is large relative to its thickness, and there is thermal symmetry about the center plane 3 Thermal 
conductivity is constant. 4 Heat generation is uniform. 
Properties The thermal conductivity is given to be k =111 W/m⋅°C. 
Analysis This insulated plate whose thickness is L is equivalent to one-half of 
an uninsulated plate whose thickness is 2L since the midplane of the 
uninsulated plate can be treated as insulated surface. The highest temperature 
will occur at the insulated surface while the lowest temperature will occur at 
the surface which is exposed to the environment. Note that L in the following 
relations is the full thickness of the given plate since the insulated side 
represents the center surface of a plate whose thickness is doubled. The 
desired values are determined directly from 
Insulated 
k 
egen
 L=5 cm 
T∞ =25°C 
h=44 W/m2.°C
C252.3°=
°⋅
×
+°=+= ∞
C W/m44
m) 05.0)( W/m102(C25
2
35
gen
h
Le
TTs
&
 
C254.6°=
°⋅
×
+°=+=
C) W/m111(2
m) 05.0)( W/m102(C3.252
2
2352
gen
k
Le
TT so
&
 
 
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preparation. If you are a student using this Manual, you are using it without permission. 
2-42
 
2-82 Prob. 2-81 is reconsidered. The effect of the heat transfer coefficient on the highest and lowest temperatures in 
the plate is to be investigated. 
Analysis The problem is solved using EES, and the solution is given below. 
 
"GIVEN" 
L=0.05 [m] 
k=111 [W/m-C] 
g_dot=2E5 [W/m^3] 
T_infinity=25 [C] 
h=44 [W/m^2-C] 
 
"ANALYSIS" 
T_min=T_infinity+(g_dot*L)/h 
T_max=T_min+(g_dot*L^2)/(2*k) 
 
 
h 
[W/m2.C] 
Tmin 
[C] 
Tmax 
[C] 
20 525 527.3 
25 425 427.3 
30 358.3 360.6 
35 310.7 313 
40 275 277.3 
45 247.2 249.5 
50 225 227.3 
55 206.8 209.1 
60 191.7 193.9 
65 178.8 181.1 
70 167.9 170.1 
75 158.3 160.6 
80 150 152.3 
85 142.6 144.9 
90 136.1 138.4 
95 130.3 132.5 
100 125 127.3 
20 30 40 50 60 70 80 90 100
100
150
200
250
300
350
400
450
500
550
h [W/m2-C]
T m
in
 [
C
]
 
 
20 30 40 50 60 70 80 90 100
100
150
200
250
300
350
400
450
500
550
h [W/m2-C]
T m
ax
 [
C
]
 
 
 
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preparation. If you are a student using this Manual, you are using it without permission. 
2-43
 
2-83 A cylindrical nuclear fuel rod is cooled by water flowing through its encased concentric tube. The average temperature 
of the cooling water is to be determined. 
Assumptions 1 Heat conduction is steady and one-dimensional. 2 Thermal properties are constant. 3 Heat generation in the 
fuel rod is uniform. 
Properties The thermal conductivity is given to be 30 W/m · °C. 
 
Analysis The rate of heat transfer by convection at the fuel rod surface is equal to that of the concentric tube surface: 
 )()( tube,2,2rod,1,1 ssss TTAhTTAh −=− ∞∞
 ))( 2())( 2( tube,22rod,11 ss TTLrhTTLrh −=− ∞∞ ππ 
 ∞∞ +−= TTTrh
rh
T ss )( tube,
11
22
rod, (a) 
The average temperature of the cooling water can be determined by applying Eq. 2-68: 
 
1
1gen
rod, 2h
re
TTs
&
+= ∞ (b) 
Substituting Eq. (a) into Eq. (b) and solving for the average temperature of the cooling water gives 
 
1
1gen
tube,
11
22
2
)(
h
re
TTTT
rh
rh
s
&
+=+− ∞∞∞ 
 
C 71.25 °=
°+
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
°⋅
×
=
+=∞
C 40
)C W/m2000(2
)m 005.0)( W/m1050(
m 010.0
m 005.0
2
2
36
tube,
2
1gen
2
1
sTh
re
r
r
T
&
 
Discussion The given information is not sufficient for one to determine the fuel rod surface temperature. The convection heat 
transfer coefficient for the fuel rod surface (h1) or the centerline temperature of the fuel rod (T0) is needed to determine the 
fuel rod surface temperature. 
 
 
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preparation. If you are a student using this Manual, you are using it without permission. 
2-44
 
2-84 A spherical communication satellite orbiting in space absorbs solar radiation while losing heat to deep space by thermal 
radiation. The heat generation rate and the surface temperature of the satellite are to be determined. 
Assumptions 1 Heat transfer is steady and one-dimensional. 2 Heat generation is uniform. 3 Thermal properties are constant. 
Properties The properties of the satellite are given to be ε = 0.75, α = 0.10, and k = 5 W/m · K. 
Analysis For steady one-dimensional heat conduction in sphere, the differential equation is 
 01 gen22 =+⎟⎠
⎞
⎜
⎝
⎛
k
e
dr
dTr
dr
d
r
&
 
and (midpoint temperature of the satellite) K 273)0( 0 == TT
 0)0( =
dr
dT (thermal symmetry about the midpoint) 
Multiply both sides of the differential equation by 2r and rearranging gives 
 2gen2 r
k
e
dr
dTr
dr
d &
−=⎟
⎠
⎞
⎜
⎝
⎛ 
Integrating with respect to r gives 
 1
3
gen2
3
Cr
k
e
dr
dTr +−=
&
 (a) 
Applying the boundary condition at the midpoint (thermal symmetry about the midpoint), 
 :0=r 1
gen 0)0(0 C
k
e
dr
dT
+×−=×
&
 → 01 =C 
Dividing both sides of Eq. (a) by 2r and integrating, 
 r
k
e
dr
dT
3
gen&−= 
and 2
2gen
6
)( Cr
k
e
rT +−=
&
 (b) 
Applying the boundary condition at the midpoint (midpoint temperature of the satellite), 
 :0=r 2
gen
0 06
C
k
e
T +×−=
&
 → 02 TC = 
Substituting into Eq. (b), the variation of temperature is determined to be 2C
 0
2gen
6
)( Tr
k
e
rT +−=
&
 
At the satellite surface ( ), the temperature is orr =
 0
2gen
6
Tr
k
e
T os +−=
&
 (c) 
Also, the rate of heat transfer at the surface of the satellite can be expressed as 
 solar
4
space
43
gen )( 3
4 qATTAre sssso && αεσπ −−=⎟
⎠
⎞
⎜
⎝
⎛ where 0space =T 
The surface temperature of the satellite can be explicitly expressed as 
 
4/1
solargen
4/1
solargen
3 3/ 
3
41
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛ +
=⎥
⎦
⎤
⎢
⎣
⎡
⎟
⎠
⎞
⎜
⎝
⎛ +=
εσ
α
απ
εσ
qre
qAer
A
T sosso
s
s
&&
&& (d) 
 
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2-45
 
Substituting Eq. (c) into Eq. (d) 
 0
2gen
4/1
solargen6
3/
Tr
k
eqre
o
so +−=⎟⎟
⎠
⎞
⎜⎜
⎝
⎛ + &&&
εσ
α
 
 K 273
)K W/m5(6
)m 25.1(
)K W/m1067.5)(75.0(
) W/m1000)(10.0(3/)m 25.1( 2gen
4/1
428
2
gen +
⋅
−=
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
⋅×
+
−
ee &&
 
Copy the following line and paste on a blank EES screen to solve the above equation: 
((e_gen*1.25/3+0.10*1000)/(0.75*5.67e-8))^(1/4)=-e_gen*1.25^2/(6*5)+273 
Solving by EES software, the heat generation rate is 
3 W/m233=gene& 
Using Eq. (c), the surface temperature of the satellite is determined to be 
 K 261=+
⋅
−= K 273)m 25.1(
)K W/m5(6
) W/m233( 2
3
sT 
Discussion The surface temperature of the satellite in space is well below freezing point of water. 
 
 
 
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preparation. If you are a student using this Manual, you are using it without permission. 
2-46
 
2-85 A 2-kW resistance heater wire with a specified surface temperature is used to boil water. The center temperature of the 
wire is to be determined. 
Assumptions 1 Heat transfer is steady since there is no change with time. 2 Heat transfer 
is one-dimensional since there is thermal symmetry about the center line and no change in 
the axial direction. 3 Thermal conductivity is constant. 4 Heat generation in the heater is 
uniform. 
230°C
D 
Properties The thermal conductivity is given to be k = 20 W/m⋅°C. 
Analysis The resistance heater converts electric energy into heat at a rate of 2 kW. The 
rate of heat generation per unit volume of the wire is 
r
 38
22
gen
wire
gen
gen W/m10768.1
m) (0.9m) 002.0(
 W2000
×====
ππ Lr
EE
e
o
&&
&
V
 
The center temperature of the wire is then determined from Eq. 2-71 to be 
 C238.8°=
°
×
+°=+=
C) W/m.20(4
m) 002.0)( W/m10768.1(C230
4
2382
gen
k
re
TT oso
&
 
 
 
 
 
2-86 Heat is generated in a long solid cylinder with a specified surface temperature. The variation of temperature in the 
cylinder is given by 
 s
o
o T
r
r
k
re
rT +
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−=
22
gen 1)(
&
 
80°C 
k 
D
(a) Heat conduction is steady since there is no time t variable involved. 
(b) Heat conduction is a one-dimensional. 
(c) Using Eq. (1), the heat flux on the surface of the cylinder at r = ro 
is determined from its definition to be r
 
2 W/cm280=cm) 4)( W/cm35(22
2
2)(
3
gen2
2
gen
2
2
gen
0
==
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−−=
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−−=−=
=
o
o
oo
rro
oo
s
re
r
r
k
re
k
r
r
k
re
k
dr
rdT
kq
&
&
&
&
 
 
 
PROPRIETARY MATERIAL. © 2011 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. 
2-47
 
2-87 Prob. 2-86 is reconsidered. The temperature as a function of the radius is to be plotted. 
Analysis The problem is solved using EES, and the solution is given below. 
 
"GIVEN" 
r_0=0.04 [m] 
k=25 [W/m-C] 
g_dot_0=35E+6 [W/m^3] 
T_s=80 [C] 
 
"ANALYSIS" 
T=(g_dot_0*r_0^2)/k*(1-(r/r_0)^2)+T_s "Variation of temperature" 
 
 
 
r [m] T [C] 
0 2320 
0.004444 2292 
0.008889 2209 
0.01333 2071 
0.01778 1878 
0.02222 1629 
0.02667 1324 
0.03111 964.9 
0.03556 550.1 
0.04 80 
0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04
0
500
1000
1500
2000
2500
r [m]
T 
 [C
]
 
 
 
 
 
 
 
 
PROPRIETARY MATERIAL. © 2011 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. 
2-48
 
2-88 Heat is generated in a large plane wall whose one side is insulated while the other side is subjected to convection. The 
mathematical formulation, the variation of temperature in the wall, the relation for the surface temperature, and the relation 
for the maximum temperature rise in the plate are to be determined for steady one-dimensional heat transfer. 
Assumptions 1 Heat transfer is steady since there is no indication of any change with time. 2 Heat transfer is one-
dimensional since the wall is large relative to its thickness. 3 Thermal conductivity is constant. 4 Heat generation is uniform. 
Analysis (a) Noting that heat transfer is steady and one-dimensional in x direction, the mathematical formulation of this 
problem can be expressed as 
0gen
2
2
=+
k
e
dx
Td & 
T∞ 
h 
k 
egen
Insulated
 L x 
and 0)0( =
dx
dT (insulated surface at x = 0) 
 ])([)( ∞−=− TLThdx
LdTk 
(b) Rearranging the differential equation and integrating, 
 1
gengen
2
2
 Cx
k
e
dx
dT
k
e
dx
Td
+−=→−=
&&
 
Integrating one more time, 
 21
2
gen
2
)( CxC
k
xe
xT ++
−
=
&
 (1) 
Applying the boundary conditions: 
B.C. at x = 0: 0 0)0(0
)0(
11
gen =→=+
−
→= CC
k
e
dx
dT &
 
B. C. at x = L: 
∞∞
∞
++=→+−
−
=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−+
−
=⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛ −
−
hT
k
Leh
LeCChT
k
Leh
Le
TC
k
Le
hL
k
e
k
22
2
2
gen
gen22
2
gen
gen
2
2
gengen
&
&
&
&
&&
 
Dividing by h: ∞++= Tk
Le
h
Le
C
2
2
gengen
2
&&
 
Substituting the C1 and relations into Eq. (1) and rearranging give C2
 ∞∞ ++−=+++
−
= T
h
Le
xL
k
e
T
k
Le
h
Le
k
xe
xT gen22gen
2
gengen
2
gen )(
222
)(
&&&&&
 
which is the desired solution for the temperature distribution in the wall as a function of x. 
(c) The temperatures at two surfaces and the temperature difference between these surfaces are 
 
k
Le
LTTT
T
h
Le
LT
T
h
Le
k
Le
T
2
)()0(
)(
2
)0(
2
gen
max
gen
gen
2
gen
&
&
&&
=−=∆
+=
++=
∞
∞
 
Discussion These relations are obtained without using differential equations in the text (see Eqs. 2-67 and 2-73). 
 
PROPRIETARY MATERIAL. © 2011 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. 
2-49
 
2-89 A long homogeneous resistance heater wire with specified convection conditions at the surface is used to boil water. 
The mathematical formulation, the variation of temperature in the wire, and the temperature at the centerline of the wire are 
to be determined. 
Assumptions 1 Heat transfer is steady since there is no indication of any change with time. 2 Heat transfer is one-
dimensional since there is thermal symmetry about the center line and no change in the axial direction. 3 Thermal 
conductivity is constant. 4 Heat generation in the wire is uniform. 
Properties The thermal conductivity is given to be k = 15.2 W/m⋅K. 
Heater 
r 
0 
ro 
Analysis Noting that heat transfer is steady and one-dimensional in 
the radial r direction, the mathematical formulation of this problem 
can be expressed as 
T∞
h Water 
01 gen =+⎟
⎠
⎞
⎜
⎝
⎛
k
e
dr
dTr
dr
d
r
&
 
and ])([)( ∞−=− TrThdr
rdTk oo (convection at the outer surface) 
0)0( =
dr
dT (thermal symmetry about the centerline) 
Multiplying both sides of the differential equation by r and rearranging gives 
 r
k
e
dr
dTr
dr
d gen&−=⎟
⎠
⎞
⎜
⎝
⎛ 
Integrating with respect to r gives 
 1
2
gen
2
Cr
k
e
dr
dTr +−=
&
 (a) 
It is convenient at this point to apply the second boundary condition since it is related to the first derivative of the 
temperature by replacing all occurrences of r and dT/dr in the equation above by zero. It yields 
B.C. at r = 0: 0 0
2
)0(
0 11
gen =→+×−=× CC
k
e
dr
dT &
 
Dividing both sides of Eq. (a) by r to bring it to a readily integrable form and integrating, 
 r
k
e
dr
dT
2
gen&−= 
and 2
2gen
4
)( Cr
k
e
rT +−=
&
 (b) 
Applying the second boundary condition at orr = , 
B. C. at : orr =
2gengen
22
2gengen
42
 
42 o
o
o
o r
k
e
h
re
TCTCr
k
e
h
k
re
k
&&&&
++=→⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−+−= ∞∞ 
Substituting this relation into Eq. (b) and rearranging give 2C
 
h
re
rr
k
e
TrToo 2
)(
4
)( gen22gen
&&
+−+= ∞ 
which is the desired solution for the temperature distribution in the wire as a function of r. Then the temperature at the center 
line (r = 0) is determined by substituting the known quantities to be 
 
C125°=
⋅×
×
+
⋅×
×
°=
++= ∞
K) W/m3200(2
)m 006.0)( W/m10(16.4
K) W/m2.15(4
m) 006.0)( W/m10(16.4+C100
24
)0(
2
36236
gen2gen
h
re
r
k
e
TT oo
&&
 
Thus the centerline temperature will be 25°C above the temperature of the surface of the wire. 
 
PROPRIETARY MATERIAL. © 2011 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. 
2-50
 
2-90 Prob. 2-89 is reconsidered. The temperature at the centerline of the wire as a function of the heat generation is to 
be plotted. 
Analysis The problem is solved using EES, and the solution is given below. 
 
"GIVEN" 
r_0=0.006 [m] 
k=15.2 [W/m-K] 
e_dot=16.4 [W/cm^3] 
T_infinity=100 [C] 
h=3200 [W/m^2-K] 
 
"ANALYSIS" 
T_0=T_infinity+e_dot*Convert(W/cm^3,W/m^3)/(4*k)*(r_0^2-r^2) +e_dot*Convert(W/cm^3,W/m^3)*r_0/(2*h) 
"Variation of temperature" 
r=0 "for centerline temperature" 
 
 
e 
[W/cm3] 
T0 
[F] 
10 
20 
30 
40 
50 
60 
70 
80 
90 
100 
115.3 
130.6 
145.9 
161.2 
176.5 
191.8 
207.1 
222.4 
237.7 
253 
10 20 30 40 50 60 70 80 90 100
100
120
140
160
180
200
220
240
260
e [W/cm3]
T 0
 [
C]
 
 
 
 
 
 
PROPRIETARY MATERIAL. © 2011 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. 
2-51
 
2-91 A nuclear fuel rod with a specified surface temperature is used as the fuel in a nuclear reactor. The center temperature 
of the rod is to be determined. 
Assumptions 1 Heat transfer is steady since there is no indication of any 
change with time. 2 Heat transfer is one-dimensional since there is thermal 
symmetry about the center line and no change in the axial direction. 
3 Thermal conductivity is constant. 4 Heat generation in the rod is uniform. 
Properties The thermal conductivity is given to be k = 29.5 W/m⋅°C. 
Analysis The center temperature of the rod is determined from 
C228°=
°
×
+°=+=
C) W/m.5.29(4
m) 005.0)( W/m104(
C220
4
2372
gen
k
re
TT oso
&
 
 
 
 
 
2-92 Both sides of a large stainless steel plate in which heat is generated uniformly are exposed to convection with the 
environment. The location and values of the highest and the lowest temperatures in the plate are to be determined. 
Assumptions 1 Heat transfer is steady since there is no indication of any change with time. 2 Heat transfer is one-
dimensional since the plate is large relative to its thickness, and there is thermal symmetry about the center plane 3 Thermal 
conductivity is constant. 4 Heat generation is uniform. 
Properties The thermal conductivity is given to be k =15.1 W/m⋅°C. 
Analysis The lowest temperature will occur at surfaces of plate 
while the highest temperature will occur at the midplane. Their 
values are determined directly from 
C155°=
°⋅
×
+°=+= ∞
C W/m60
m) 015.0)( W/m105(C30
2
35
gen
h
Le
TTs
&
 C158.7°=
°⋅
×
+°=+=
C) W/m1.15(2
m) 015.0)( W/m105(
C155
2
2352
gen
k
Le
TT so
&
 
 egen
220°C
Uranium rod
T∞ =30°C 
h=60 W/m2⋅°C 
k 
egen
2L=3 cm 
T∞ =30°C 
h=60 W/m2.°C
 
 
PROPRIETARY MATERIAL. © 2011 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. 
2-52
 
2-93 A long resistance heater wire is subjected to convection at its outer surface. The surface temperature of the wire is to be 
determined using the applicable relations directly and by solving the applicable differential equation. 
Assumptions 1 Heat transfer is steady since there is no indication of any change with time. 2 Heat transfer is one-
dimensional since there is thermal symmetry about the center line and no change in the axial direction. 3 Thermal 
conductivity is constant. 4 Heat generation in the wire is uniform. 
Properties The thermal conductivity is given to be k = 15.1 W/m⋅°C. 
 k 
egen
T∞
h 
0 ro
Analysis (a) The heat generation per unit volume of the wire is 
38
22
gen
wire
gen
gen W/m10592.1m) (6m) 001.0(
 W3000
×====
ππ Lr
EE
e
o
&&
&
V
 
T∞
h The surface temperature of the wire is then (Eq. 2-68) 
C475°=
°⋅
×
+°=+= ∞
C) W/m175(2
m) 001.0)( W/m10592.1(C20
2 2
38
gen
h
re
TT os
&
 
r 
(b) The mathematical formulation of this problem can be expressed as 
 01 gen =+⎟
⎠
⎞
⎜
⎝
⎛
k
e
dr
dTr
dr
d
r
&
 
and ])([)( ∞−=− TrThdr
rdTk oo (convection at the outer surface) 
0
)0(
=
dr
dT (thermal symmetry about the centerline) 
Multiplying both sides of the differential equation by r and integrating gives 
 r
k
e
dr
dTr
dr
d gen&−=⎟
⎠
⎞
⎜
⎝
⎛ → 1
2
gen
2
Cr
k
e
dr
dTr +−=
&
 (a) 
Applying the boundary condition at the center line, 
B.C. at r = 0: 0 0
2
)0(
0 11
gen =→+×−=× CC
k
e
dr
dT &
 
Dividing both sides of Eq. (a) by r to bring it to a readily integrable form and integrating, 
 r
k
e
dr
dT
2
gen&−= → 2
2gen
4
)( Cr
k
e
rT +−=
&
 (b) 
Applying the boundary condition at , orr =
B. C. at : orr =
2gengen
22
2gengen
42
 
42 o
o
o
o r
k
e
h
re
TCTCr
k
e
h
k
re
k
&&&&
++=→⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−+−=− ∞∞ 
Substituting this C relation into Eq. (b) and rearranging give 2
 
h
re
rr
k
e
TrT oo 2
)(
4
)( gen22gen
&&
+−+= ∞ 
which is the temperature distribution in the wire as a function of r. Then the temperature of the wire at the surface (r = ro ) is 
determined by substituting the known quantities to be 
C475°=
°⋅
×
+°=+=+−+= ∞∞
C) W/m175(2
m) 001.0)( W/m10592.1(C20
22
)(
4
)( 2
38
gen0gen22gen
0 h
re
T
h
re
rr
k
e
TrT ooo
&&&
 
Note that both approaches give the same result. 
 
PROPRIETARY MATERIAL. © 2011 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. 
2-53
 
2-94E Heat is generated uniformly in a resistance heater wire. The temperature difference between the center and the surface 
of the wire is to be determined. 
Assumptions 1 Heat transfer is steady since there is no change with time. 2 Heat 
transfer is one-dimensional since there is thermal symmetry about the center line 
and no change in the axial direction. 3 Thermal conductivity is constant. 4 Heat 
generation in the heater is uniform. 
Heater 
r 
0 
ro 
Ts
Properties The thermal conductivity is given to be k = 5.8 Btu/h⋅ft⋅°F. 
Analysis The resistance heater converts electric energy into heat at a rate 
of 3 kW. The rate of heat generation per unit length of the wire is 
 38
22
wire
gen
gen ftBtu/h 10933.2
ft) (1ft) 12/04.0(
Btu/h) 14.34123(
⋅×=
×
===
ππ Lr
EE
e
o
gen
&&
&
V
 
Then the temperature difference between the centerline and the surface becomes 
 F140.5°=
°⋅⋅
⋅×
==∆
F)ftBtu/h 8.5(4
ft) 12/04.0)(ftBtu/h 10933.2(
4
2382
gen
max k
re
T o
&
 
 
 
 
 
2-95E Heat is generated uniformly in a resistance heater wire. The temperature difference between the center and the surface 
of the wire is to be determined. 
Assumptions 1 Heat transfer is steady since there is no change with time. 2 Heat 
transfer is one-dimensional since there is thermal symmetry about the center line 
and no change in the axial direction. 3 Thermal conductivity is constant. 4 Heat 
generation in the heater is uniform. 
Properties The thermal conductivity is given to be k = 4.5 Btu/h⋅ft⋅°F. 
Analysis The resistance heater converts electric energy into heat at a rate 
of 3 kW. The rate of heat generation per unit volume of the wire is 
 38
22gen
wire
gen
gen ftBtu/h 10933.2
ft) (1ft) 12/04.0(
Btu/h) 14.34123(
⋅×=
×
===
ππ Lr
EE
e
o
&&
&
V
 Heater 
r 
ro 
0 
Ts
Then the temperature difference between the centerline and the surface becomes 
 F181.0°=
°⋅⋅
⋅×
==∆
F)ftBtu/h 5.4(4
ft) 12/04.0)(ftBtu/h 10933.2(
4
2382
gen
max k
re
T o
&
 
 
PROPRIETARY MATERIAL. © 2011 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. 
2-54
 
2-96 Heat is generated uniformly in a spherical radioactive material with specified surface temperature. The mathematical 
formulation, the variation of temperature in the sphere, and the center temperature are to be determined for steady one-
dimensional heat transfer. 
Assumptions 1 Heat transfer is steady since there is no indication of any changes with time. 2 Heat transfer is one-
dimensional since there is thermal symmetry about the mid point. 3 Thermal conductivity is constant. 4 Heat generation is 
uniform. 
Properties The thermal conductivity is given to be k = 15 W/m⋅°C. 
Analysis (a) Noting that heat transfer is steady and one-dimensional in the radial 
r direction, the mathematical formulation of this problem can be expressed as 
k 
egen
Ts=110°C
r
0 ro
 constant with 01 gen
gen2
2
==+⎟
⎠
⎞
⎜
⎝
⎛ e
k
e
dr
dTr
dr
d
r
&
&
 
and 110°C (specified surface temperature) == so TrT )(
0)0( =
dr
dT (thermal symmetry about the mid point) 
(b) Multiplying both sides of the differential equation by r2 and rearranging gives 
 2gen2 r
k
e
dr
dTr
dr
d &
−=⎟
⎠
⎞
⎜
⎝
⎛ 
Integrating with respect to r gives 
 1
3
gen2
3
Cr
k
e
dr
dTr +−=
&
 (a) 
Applying the boundary condition at the mid point, 
B.C. at r = 0: 0 0
3
)0(
0 11
gen =→+×−=× CC
k
e
dr
dT & 
Dividing both sides of Eq. (a) by r2 to bring it to a readily integrable form and integrating, 
 r
k
e
dr
dT
3
gen&−= 
and 2
2gen
6
)( Cr
k
e
rT +−=
&
 (b) 
Applying the other boundary condition at r r= 0 , 
B. C. at : orr =
2gen
22
2gen
6
 
6 osos
r
k
e
TCCr
k
e
T
&&
+=→+−= 
Substituting this relation into Eq. (b) and rearranging give 2C
 )(
6
)( 22gen rr
k
e
TrT os −+=
&
 
which is the desired solution for the temperature distribution in the wire as a function of r. 
(c) The temperature at the center of the sphere (r = 0) is determined by substituting the known quantities to be 
C999°=
°⋅×
×
°=+=−+=
C)m W/ 15(6
)m 04.0)( W/m10(5+C110
6
)0(
6
)0(
2372
gen22gen
k
re
Tr
k
e
TT osos
&&
 
Thus the temperature at center will be 999°C above the temperature of the outer surface of the sphere. 
 
PROPRIETARY MATERIAL. © 2011 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. 
2-55
 
2-97 Prob. 2-96 is reconsidered. The temperature as a function of the radius is to be plotted. Also, the center 
temperature of the sphere as a function of the thermal conductivity is to be plotted. 
Analysis The problem is solved using EES, and the solution is given below. 
 
"GIVEN" 
r_0=0.04 [m] 
g_dot=5E7 [W/m^3] 
T_s=110 [C] 
k=15 [W/m-C] 
r=0 [m] 
 
"ANALYSIS" 
T=T_s+g_dot/(6*k)*(r_0^2-r^2) "Temperature distribution as a function of r" 
T_0=T_s+g_dot/(6*k)*r_0^2 "Temperature at the center (r=0)" 
 
 
 
r 
[m] 
T 
[C] 
0 
0.002105 
0.004211 
0.006316 
0.008421 
0.01053 
0.01263 
0.01474 
0.01684 
0.01895 
0.02105 
0.02316 
0.02526 
0.02737 
0.02947 
0.03158 
0.03368 
0.03579 
0.03789 
0.04 
998.9 
996.4 
989 
976.7 
959.5 
937.3 
910.2 
878.2 
841.3 
799.4 
752.7 
701 
644.3 
582.8 
516.3 
444.9 
368.5 
287.3 
201.1 
110 
0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04
100
200
300
400
500
600
700
800
900
1000
r [m]
T 
 [C
]
 
 
PROPRIETARY MATERIAL. © 2011 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. 
2-56
 
 
k 
[W/m.C] 
T0 
[C] 
10 
30.53 
51.05 
71.58 
92.11 
112.6 
133.2 
153.7 
174.2 
194.7 
215.3 
235.8 
256.3 
276.8 
297.4 
317.9 
338.4 
358.9 
379.5 
400 
1443 
546.8 
371.2 
296.3 
254.8 
228.4 
210.1 
196.8 
186.5 
178.5 
171.9 
166.5 
162 
158.2 
154.8 
151.9 
149.4 
147.1 
145.1 
143.3 
0 50 100 150 200 250 300 350 400
0
200
400
600
800
1000
1200
1400
1600
k [W/m-C]
T 0
 [
C]
 
 
 
PROPRIETARY MATERIAL. © 2011 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. 
2-57
 
2-98 A long homogeneous resistance heater wire with specified surface temperature is used to heat the air. The temperature 
of the wire 3.5 mm from the center is to be determined in steady operation. 
Assumptions 1 Heat transfer is steady since there is no indication of any change with time. 2 Heat transfer is one-
dimensional since there is thermal symmetry about the center line and no change in the axial direction. 3 Thermal 
conductivity is constant. 4 Heat generation in the wire is uniform. 
Properties The thermal conductivity is given to be k = 6 W/m⋅°C. 
Analysis Noting that heat transfer is steady and one-dimensional in the radial r 
direction, the mathematical formulation of this problem can be expressed as 
 gene&
180°C
Resistance wire 
ro
r
01 gen =+⎟
⎠
⎞
⎜
⎝
⎛
k
e
dr
dTr
dr
d
r
&
 
and 180°C (specified surface temperature) == so TrT )(
0
)0(
=
dr
dT (thermal symmetry about the centerline) 
Multiplying both sides of the differential equation by r and rearranging gives 
 r
k
e
dr
dTr
dr
d gen&−=⎟
⎠
⎞
⎜
⎝
⎛ 
Integrating with respect to r gives 
 1
2
gen
2
Cr
k
e
dr
dTr +−=
&
 (a) 
It is convenient at this point to apply the boundary condition at the center since it is related to the first derivative of the 
temperature. It yields 
B.C. at r = 0: 0 0
2
)0(
0 11
gen =→+×−=× CC
k
e
dr
dT & 
Dividing both sides of Eq. (a) by r to bring it to a readily integrable form and integrating, 
 r
k
e
dr
dT
2
gen&−= 
and 2
2gen
4
)( Cr
k
e
rT +−=
&
 (b) 
Applying the other boundary condition at orr = , 
B. C. at : orr =
2gen
22
2gen
4
 
4 osos
r
k
e
TCCr
k
e
T
&&
+=→+−= 
Substituting this C relation into Eq. (b) and rearranging give 2
 )(
4
)( 22gen rr
k
e
TrT os −+=
&
 
which is the desired solution for the temperature distribution in the wire as a function of r. The temperature 3.5 mm from the 
center line (r = 0.0035 m) is determined by substituting the known quantities to be 
C207°=−
°⋅×
×
°=−+= ])m 0035.0(m) 005.0[(
C)m W/ 6(4
 W/m105+C180)(
4
)m 0035.0( 22
37
22gen rr
k
e
TT os
&
 
Thus the temperature at that location will be about 20°C above the temperature of the outer surface of the wire. 
 
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2-58
 
2-99 Heat is generated in a large plane wall whose one side is insulated while the other side is maintained at a specified 
temperature. The mathematical formulation, the variation of temperature in the wall, and the temperature of the insulated 
surface are to be determined for steady one-dimensional heat transfer. 
Assumptions 1 Heat transfer is steady since there is no indication of any change with time. 2 Heat transfer is one-
dimensional since the wall is large relative to its thickness, and there is thermal symmetry about the center plane. 3 Thermal 
conductivity is constant. 4 Heat generation varies with location in the x direction.Properties The thermal conductivity is given to be k = 30 W/m⋅°C. 
Analysis (a) Noting that heat transfer is steady and one-dimensional in x 
direction, the mathematical formulation of this problem can be expressed as 
T2 =30°C
k 
gene& 
Insulated 
 L x
0
)(gen
2
2
=+
k
xe
dx
Td & 
where and = 8×10Lxeee /5.00gen
−= && 0e&
6 W/m3
and 0)0( =
dx
dT (insulated surface at x = 0) 
 30°C (specified surface temperature) == 2)( TLT
(b) Rearranging the differential equation and integrating, 
 1
/5.00
1
/5.0
0/5.00
2
2 2
 
/5.0
 Ce
k
Le
dx
dTC
L
e
k
e
dx
dTe
k
e
dx
Td LxLxLx +=→+
−
−=→−= −
−
− &&& 
Integrating one more time, 
 
4
)( 
/5.0
2
)( 21
/5.0
2
0
21
/5.0
0 CxCe
k
Le
xTCxC
L
e
k
Le
xT Lx
Lx
++−=→++
−
= −
− &&
 (1) 
Applying the boundary conditions: 
B.C. at x = 0: 
k
Le
CC
k
Le
Ce
k
Le
dx
dT L 0
11
0
1
/05.00 2 
2
0 
2)0( &&&
−=→+=→+= ×− 
B. C. at x = L: 
k
Le
e
k
Le
TCCLCe
k
Le
TLT LL
2
05.0
2
0
2221
/5.0
2
0
2
24
 
4
)(
&&&
++=→++−== −− 
Substituting the C1 and C2 relations into Eq. (1) and rearranging give 
 )]/1(2)(4[)( /5.05.0
2
0
2 Lxeek
Le
TxT Lx −+−+= −−
&
 
which is the desired solution for the temperature distribution in the wall as a function of x. 
(c) The temperature at the insulate surface (x = 0) is determined by substituting the known quantities to be 
 
C314°=−+−
°⋅
×
+°=
−+−+=
−
−
)]02()1(4[
C) W/m30(
m) 05.0)( W/m10(8
C30
)]/02()(4[)0(
5.0
236
05.0
2
0
2
e
Lee
k
Le
TT
&
 
Therefore, there is a temperature difference of almost 300°C between the two sides of the plate. 
 
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preparation. If you are a student using this Manual, you are using it without permission. 
2-59
 
2-100 Prob. 2-99 is reconsidered. The heat generation as a function of the distance is to be plotted. 
Analysis The problem is solved using EES, and the solution is given below. 
 
"GIVEN" 
L=0.05 [m] 
T_s=30 [C] 
k=30 [W/m-C] 
e_dot_0=8E6 [W/m^3] 
 
"ANALYSIS" 
e_dot=e_dot_0*exp((-0.5*x)/L) "Heat generation as a function of x" 
"x is the parameter to be varied" 
 
 
x 
[m] 
e 
[W/m3] 
0 8.000E+06 
0.005 7.610E+06 
0.01 7.239E+06 
0.015 6.886E+06 
0.02 6.550E+06 
0.025 6.230E+06 
0.03 5.927E+06 
0.035 5.638E+06 
0.04 5.363E+06 
0.045 5.101E+06 
0.05 4.852E+06 
0 0.01 0.02 0.03 0.04 0.05
4.500x106
5.000x106
5.500x106
6.000x106
6.500x106
7.000x106
7.500x106
8.000x106
x [m]
e 
 [W
/m
3 ]
 
 
 
 
 
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2-60
 
Variable Thermal Conductivity, k(T) 
 
2-101C No, the temperature variation in a plain wall will not be linear when the thermal conductivity varies with 
temperature. 
 
 
2-102C The thermal conductivity of a medium, in general, varies with temperature. 
 
 
2-103C During steady one-dimensional heat conduction in a plane wall in which the thermal conductivity varies linearly, the 
error involved in heat transfer calculation by assuming constant thermal conductivity at the average temperature is (a) none. 
 
 
2-104C During steady one-dimensional heat conduction in a plane wall, long cylinder, and sphere with constant thermal 
conductivity and no heat generation, the temperature in only the plane wall will vary linearly. 
 
 
2-105C Yes, when the thermal conductivity of a medium varies linearly with temperature, the average thermal conductivity 
is always equivalent to the conductivity value at the average temperature. 
 
 
 
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preparation. If you are a student using this Manual, you are using it without permission. 
2-61
 
2-106 A silicon wafer with variable thermal conductivity is subjected to uniform heat flux at the lower surface. The 
maximum allowable heat flux such that the temperature difference across the wafer thickness does not exceed 2 °C is to be 
determined. 
Assumptions 1 Heat conduction is steady and one-dimensional. 
2 There is no heat generation. 3 Thermal conductivity varies 
with temperature. 
Properties The thermal conductivity is given to be k(T) = (a + 
bT + cT2) W/m · K. 
Analysis For steady heat transfer, the Fourier’s law of heat 
conduction can be expressed as 
 
dx
dTcTbTa
dx
dTTkq )()( 2++−=−=& 
Separating variable and integrating from 0=x where 1)0( TT = to Lx = where 2)( TLT = , we obtain 
 dTcTbTadxq
T
T
L
∫∫ ++−=
2
1
)( 2
0
&
Performing the integration gives 
 ⎥⎦
⎤
⎢⎣
⎡ −+−+−−= )(
3
)(
2
)( 31
3
2
2
1
2
212 TT
cTTbTTaLq& 
The maximum allowable heat flux such that the temperature difference across the wafer thickness does not exceeding 2 °C is 
 
25 W/m101.35×=
×
⎥⎦
⎤
⎢⎣
⎡ −+−−−
−=
− )m 10925(
 W/m)600598(
3
00111.0)600598(
2
29.1)600598(437
6
3322
q& 
Discussion For heat flux less than 135 kW/m2, the temperature difference across the silicon wafer thickness will be 
maintained below 2 °C. 
 
 
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2-62
 
2-107 A plate with variable conductivity is subjected to specified 
temperatures on both sides. The rate of heat transfer through the plate is to be 
determined. 
Assumptions 1 Heat transfer is given to be steady and one-dimensional. 2 
Thermal conductivity varies quadratically. 3 There is no heat generation. 
Properties The thermal conductivity is given to be . )1()( 20 TkTk β+=
Analysis When the variation of thermal conductivity with temperature k(T) is 
known, the average value of the thermal conductivity in the temperature 
range between can be determined from 21 and TT
 
( ) ( )
( )⎥⎦
⎤
⎢⎣
⎡ +++=
−
⎥⎦
⎤
⎢⎣
⎡ −+−
=
−
⎟
⎠
⎞
⎜
⎝
⎛ +
=
−
+
=
−
=
∫∫
2
121
2
20
12
3
1
3
2120
12
3
0
12
2
0
12
avg
3
1
33
)1()(
2
1
2
1
2
1
TTTTk
TT
TTTTk
TT
TTk
TT
dTTk
TT
dTTk
k
T
T
T
T
T
T
β
ββ
β
 
T2
k(T)
T1
 L x
This relation is based on the requirement that the rate of heat transfer through a medium with constant average thermal 
conductivity kavg equals the rate of heat transfer through the same medium with variable conductivity k(T). Then the rate of 
heat conduction through the plate can be determined to be 
 ( )
L
TT
ATTTTk
L
TT
AkQ 212121
2
20
21
avg 3
1
−
⎥⎦
⎤
⎢⎣
⎡ +++=
−
=
β& 
Discussion We would obtain the same result if we substituted the given k(T) relation into the second part of Eq. 2-76, and 
performed the indicated integration. 
 
 
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2-63
 
2-108 A cylindrical shell with variable conductivity is subjected to specified temperatures on both sides. The variation of 
temperature and the rate of heat transfer through the shell are to be determined. 
Assumptions 1 Heat transfer is given to be steady and one-dimensional. 2 Thermal conductivity varies linearly. 3 There is 
no heat generation. 
Properties The thermal conductivity is given to be )1()( 0 TkTk β+= . 
r2
k(T)
T1
r1
r 
T2
Analysis (a) The rate of heat transfer through the shell is expressed as 
 
)/ln(
2
12
21
avgcylinder rr
TT
LkQ
−
= π& 
where L is the length of the cylinder, r1 is the inner radius, and 
r2 is the outer radius, and 
 ⎟
⎠
⎞
⎜
⎝
⎛ ++==
2
1)( 120avgavg
TTkTkk β 
is the average thermal conductivity. 
(b) To determine thetemperature distribution in the shell, we begin with the Fourier’s law of heat conduction expressed as 
 
dr
dTATkQ )(−=& 
where the rate of conduction heat transfer is constant and the heat conduction area A = 2πrL is variable. Separating the 
variables in the above equation and integrating from r = r
Q&
1 where 11 )( TrT = to any r where , we get TrT =)(
 ∫∫ −=
T
T
r
r
dTTkL
r
drQ
11
)(2π& 
Substituting )1()( 0 TkTk β+= and performing the integrations gives 
 ]2/)()[(2ln 21
2
10
1
TTTTLk
r
rQ −+−−= βπ& 
Substituting the expression from part (a) and rearranging give &Q
 02)(
)/ln(
)/ln(22
1
2
121
12
1
0
avg2 =−−−++ TTTT
rr
rr
k
k
TT
βββ
 
which is a quadratic equation in the unknown temperature T. Using the quadratic formula, the temperature distribution T(r) 
in the cylindrical shell is determined to be 
 1
2
121
12
1
0
avg
2
2)(
)/ln(
)/ln(211)( TTTT
rr
rr
k
k
rT
ββββ
++−−±−= 
Discussion The proper sign of the square root term (+ or -) is determined from the requirement that the temperature at any 
point within the medium must remain between . 21 and TT
 
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2-64
 
2-109 A spherical shell with variable conductivity is subjected to specified temperatures on both sides. The variation of 
temperature and the rate of heat transfer through the shell are to be determined. 
Assumptions 1 Heat transfer is given to be steady and one-dimensional. 2 Thermal conductivity varies linearly. 3 There is 
no heat generation. 
Properties The thermal conductivity is given to be )1()( 0 TkTk β+= . 
Analysis (a) The rate of heat transfer through the shell is expressed as 
k(T) 
r2
T1
T2
r 
r1
 
12
21
21avgsphere 4 rr
TT
rrkQ
−
−
= π& 
where r1 is the inner radius, r2 is the outer radius, and 
 ⎟
⎠
⎞
⎜
⎝
⎛ ++==
2
1)( 120avgavg
TTkTkk β 
is the average thermal conductivity. 
(b) To determine the temperature distribution in the shell, we begin with the Fourier’s law of heat conduction expressed as 
 
dr
dTATkQ )(−=& 
where the rate of conduction heat transfer is constant and the heat conduction area A = 4πrQ& 2 is variable. Separating the 
variables in the above equation and integrating from r = r1 where 11 )( TrT = to any r where , we get TrT =)(
 ∫∫ −=
T
T
r
r
dTTk
r
drQ
11
)(4
2
π& 
Substituting )1()( 0 TkTk β+= and performing the integrations gives 
 ]2/)()[(411 21
2
10
1
TTTTk
rr
Q −+−−=⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
− βπ& 
Substituting the expression from part (a) and rearranging give Q&
 02)(
)(
)(22
1
2
121
12
12
0
avg2 =−−−
−
−
++ TTTT
rrr
rrr
k
k
TT
βββ
 
which is a quadratic equation in the unknown temperature T. Using the quadratic formula, the temperature distribution T(r) 
in the cylindrical shell is determined to be 
 1
2
121
12
12
0
avg
2
2)(
)(
)(211)( TTTT
rrr
rrr
k
k
rT
ββββ
++−
−
−
−±−= 
Discussion The proper sign of the square root term (+ or -) is determined from the requirement that the temperature at any 
point within the medium must remain between . 21 and TT
 
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2-65
 
2-110 A plate with variable conductivity is subjected to specified temperatures on both sides. The rate of heat transfer 
through the plate is to be determined. 
Assumptions 1 Heat transfer is given to be steady and one-dimensional. 2 Thermal conductivity varies linearly. 3 There is 
no heat generation. 
Properties The thermal conductivity is given to be )1()( 0 TkTk β+= . 
 L 
T1 
k(T) 
Analysis The average thermal conductivity of the medium in this case is simply 
the conductivity value at the average temperature since the thermal conductivity 
varies linearly with temperature, and is determined to be 
 
K W/m66.24
2
K 350)+(500)K 10(8.7+1K) W/m18(
2
1)(
1-4-
12
0avgave
⋅=
⎟
⎠
⎞
⎜
⎝
⎛ ×⋅=
⎟
⎠
⎞
⎜
⎝
⎛ +
+==
TT
kTkk β
 
T2
Then the rate of heat conduction through the plate becomes 
 kW 22.2==−×⋅=
−
= W,19022
m15.0
0)K35(500m) 0.6 m K)(1.5 W/m66.24(21avg L
TT
AkQ& 
Discussion We would obtain the same result if we substituted the given k(T) relation into the second part of Eq, 2-76, and 
performed the indicated integration. 
 
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2-66
 
2-111 Prob. 2-110 is reconsidered. The rate of heat conduction through the plate as a function of the temperature of 
the hot side of the plate is to be plotted. 
Analysis The problem is solved using EES, and the solution is given below. 
 
"GIVEN" 
A=1.5*0.6 [m^2] 
L=0.15 [m] 
T_1=500 [K] 
T_2=350 [K] 
k_0=18 [W/m-K] 
beta=8.7E-4 [1/K] 
 
"ANALYSIS" 
k=k_0*(1+beta*T) 
T=1/2*(T_1+T_2) 
Q_dot=k*A*(T_1-T_2)/L 
 
 
T1 
[W] 
Q 
[W] 
400 
425 
450 
475 
500 
525 
550 
575 
600 
625 
650 
675 
700 
7162 
10831 
14558 
18345 
22190 
26094 
30056 
34078 
38158 
42297 
46494 
50750 
55065 400 450 500 550 600 650 700
0
10000
20000
30000
40000
50000
60000
T1 [K]
Q
 [
W
]
 
 
 
 
 
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2-67
 
Special Topic: Review of Differential equations 
 
2-112C We utilize appropriate simplifying assumptions when deriving differential equations to obtain an equation that we 
can deal with and solve. 
 
 
2-113C A variable is a quantity which may assume various values during a study. A variable whose value can be changed 
arbitrarily is called an independent variable (or argument). A variable whose value depends on the value of other variables 
and thus cannot be varied independently is called a dependent variable (or a function). 
 
 
2-114C A differential equation may involve more than one dependent or independent variable. For example, the equation 
t
txT
k
e
x
txT
∂
∂
α
=+
∂
∂ ),(1),( gen
2
2 &
 has one dependent (T) and 2 independent variables (x and t). the equation 
t
txW
t
txT
x
txW
x
txT
∂
∂
α
+
∂
∂
α
=
∂
∂
+
∂
∂ ),(1),(1),(),(
2
2
 has 2 dependent (T and W) and 2 independent variables (x and t). 
 
 
2-115C Geometrically, the derivative of a function y(x) at a point represents the slope of the tangent line to the graph of the 
function at that point. The derivative of a function that depends on two or more independent variables with respect to one 
variable while holding the other variables constant is called the partial derivative. Ordinary and partial derivatives are 
equivalent for functions that depend on a single independent variable. 
 
 
2-116C The order of a derivative represents the number of times a function is differentiated, whereas the degree of a 
derivative represents how many times a derivative is multiplied by itself. For example, is the third order derivative of y, 
whereas is the third degree of the first derivative of y. 
y ′′′
3)( y ′
 
 
2-117C For a function , the partial derivative ),( yxf xf ∂∂ / will be equal to the ordinary derivative when f does not 
depend on y or this dependence is negligible. 
dxdf /
 
 
2-118C For a function , the derivative does not have to be a function of x. The derivative will be a constant 
when the f is a linear function of x. 
)(xf dxdf /
 
 
2-119C Integration is the inverse of derivation. Derivation increases the order of a derivative by one, integration reduces it 
by one. 
 
 
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2-68
 
2-120C A differential equation involves derivatives, an algebraic equation does not. 
 
 
2-121C A differential equation that involves only ordinary derivatives is called an ordinary differential equation, and a 
differential equation that involves partial derivatives is called a partial differential equation. 
 
 
2-122C The order of a differential equation is the order of the highest order derivative in the equation. 
 
 
2-123C A differential equation is said to be linear if the dependent variable and all of its derivatives are of the first degree, 
and their coefficients depend on the independent variable only. In other words, a differential equation is linear if it can be 
written in a form which does not involve (1) any powers of the dependent variable or its derivatives such as or , (2) 
any products of the dependent variable or its derivatives such as 
3y 2)(y′
yy ′ or yy ′′′′ , and (3) any other nonlinear functions of the 
dependent variable such as or . Otherwise, it is nonlinear. sin y ye
 
 
2-124C A linear homogeneous differential equation of order n is expressed in the most general form as 
 0)()( )( 1
)1(
1
)( =+′+++ −
− yxfyxfyxfy nn
nn L
Each term in a linear homogeneous equation contains the dependent variable or one of its derivatives after the equation is 
cleared of any common factors. The equation is linear and homogeneous since each term is linear in y, and 
contains the dependent variable or one of its derivatives. 
04 2 =−′′ yxy
 
 
2-125C A differential equation is said to have constant coefficients if the coefficients of all the terms which involve the 
dependent variable or its derivatives are constants. If, after cleared of any common factors, any of the terms with the 
dependent variable or its derivatives involve the independent variable as a coefficient, that equation is said to have variable 
coefficients The equation has variable coefficients whereas the equation has constant 
coefficients. 
04 2 =−′′ yxy 04 =−′′ yy
 
 
2-126C A linear differential equation that involves a single term with the derivatives can be solved by direct integration. 
 
 
2-127C The general solution of a 3rd order linear and homogeneous differential equation will involve 3 arbitrary constants. 
 
 
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2-69
 
Review Problems 
 
2-128 In a quenching process, steel ball bearings at a given instant have a rate of temperature decrease of 50 K/s. The rate of 
heat loss is to be determined. 
Assumptions 1 Heat conduction is one-dimensional. 2 There is no heat generation. 3 Thermal properties are constant. 
Properties The properties of the steel ball bearings are given to be c = 500 J/kg · K, k = 60 W/m · K, and ρ = 7900 kg/m3. 
Analysis The thermal diffusivity on the steel ball bearing is 
 /sm 1019.15
)KJ/kg 500)(kg/m 7900(
K W/m60 26
3
−×=
⋅
⋅
==
c
k
ρ
α 
The given rate of temperature decrease can be expressed as 
 K/s 50
)(
−=
dt
rdT o 
For one-dimensional transient heat conduction in a sphere with no heat generation, the differential equation is 
 
t
T
r
Tr
rr ∂
∂
=⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
∂
∂
α
11 2
2 
Substituting the thermal diffusivity and the rate of temperature decrease, the differential equation can be written as 
 
/sm 1019.15
K/s 501
26
2
2 −×
−
=⎟
⎠
⎞
⎜
⎝
⎛
dr
dTr
dr
d
r
 
Multiply both sides of the differential equation by 2r and rearranging gives 
 226
2
/sm 1019.15
K/s 50 r
dr
dTr
dr
d
−×
−
=⎟
⎠
⎞
⎜
⎝
⎛ 
Integrating with respect to r gives 
 1
3
26
2
3/sm 1019.15
K/s 50 Cr
dr
dTr +⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
×
−
=
−
 (a) 
Applying the boundary condition at the midpoint (thermal symmetry about the midpoint), 
 :0=r 126 3
0
/sm 1019.15
K/s 50)0(0 C
dr
dT
+⎟
⎠
⎞
⎜
⎝
⎛
×
−
=×
−
 → 01 =C 
Dividing both sides of Eq. (a) by 2r gives 
 ⎟
⎠
⎞
⎜
⎝
⎛
×
−
=
− 3/sm 1019.15
K/s 50
26
r
dr
dT 
The rate of heat loss through the steel ball bearing surface can be determined from Fourier’s law to be 
 
kW 12.9=
⎟
⎠
⎞
⎜
⎝
⎛
×
⋅=
⎟
⎠
⎞
⎜
⎝
⎛
×
=−=
−=
−
−
3
m 025.0
/sm 1019.15
K/s 50)m 025.0)(4)(K W/m60(
3/sm 1019.15
K/s 50) 4(
)(
) 4(
26
2
26
22
loss
π
ππ oo
o
o
r
rk
dr
rdT
rk
dr
dTkAQ&
 
Discussion The rate of heat loss through the steel ball bearing surface determined here is for the given instant when the rate 
of temperature decrease is 50 K/s. 
 
PROPRIETARY MATERIAL. © 2011 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. 
2-70
 
2-129 A spherical reactor of 5-cm diameter operating at steady condition has its heat generation suddenly set to 9 MW/m3. 
The time rate of temperature change in the reactor is to be determined. 
Assumptions 1 Heat conduction is one-dimensional. 2 Heat generation is uniform. 3 Thermal properties are constant. 
Properties The properties of the reactor are given to be c = 200 J/kg·°C, k = 40 W/m·°C, and ρ = 9000 kg/m3. 
Analysis The thermal diffusivity of the reactor is 
 /sm 1022.22
)CJ/kg 200)(kg/m 9000(
C W/m40 26
3
−×=
°⋅
°⋅
==
c
k
ρ
α 
For one-dimensional transient heat conduction in a sphere with heat generation, the differential equation is 
 
t
T
k
e
r
Tr
rr ∂
∂
=+⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
∂
∂
α
11 gen2
2
&
 or ⎥
⎦
⎤
⎢
⎣
⎡
+⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
∂
∂
=
∂
∂
k
e
r
Tr
rrt
T gen2
2
1 &α 
At the instant when the heat generation of reactor is suddenly set to 90 MW/m3 (t = 0), the temperature variation can be 
expressed by the given T(r) = a – br2, hence 
 
[ ]
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
+−=⎥
⎦
⎤
⎢
⎣
⎡
+−=
⎭
⎬
⎫
⎩
⎨
⎧
+−
∂
∂
=
⎭
⎬
⎫
⎩
⎨
⎧
+⎥⎦
⎤
⎢⎣
⎡ −
∂
∂
∂
∂
=
∂
∂
k
e
b
k
e
br
r
k
e
brr
rrk
e
bra
r
r
rrt
T
gengen2
2
gen2
2
gen22
2
6)6(1
)2(1)(1
&&
&&
αα
αα
 
The time rate of temperature change in the reactor when the heat generation suddenly set to 9 MW/m3 is determined to be 
 
C/s 61.7 °−=
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
°⋅
×
+°×−×=⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
+−=
∂
∂ −
C W/m40
 W/m109)C/m 105(6)/sm 1022.22(6
36
2526gen
k
e
b
t
T &α 
Discussion Since the time rate of temperature change is a negative value, this indicates that the heat generation of reactor is 
suddenly decreased to 9 MW/m3. 
 
 
 
PROPRIETARY MATERIAL. © 2011 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. 
2-71
 
2-130 A small hot metal object is allowed to cool in an environment by convection. The differential equation that describes 
the variation of temperature of the ball with time is to be derived. 
Assumptions 1 The temperature of the metal object changes uniformly with time during cooling so that T = T(t). 2 The 
density, specific heat, and thermal conductivity of the body are constant. 3 There is no heat generation. 
Analysis Consider a body of arbitrary shape of mass m, volume V, surface area A, density ρ, and specific heat cp initially at a 
uniform temperature Ti. At time t = 0, the body is placed into a medium at temperature , and heat transfer takes place 
between the body and its environment with a heat transfer coefficient h. 
∞T
A
m, c, Ti
T=T(t) 
h 
T∞
 During a differential time interval dt, the temperature of the body rises by a 
differential amount dT. Noting that the temperature changes with time only, an energy 
balance of the solid for the time interval dt can be expressed as 
 ⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
=⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
dtdt duringbody theof 
energy in the decreaseThe
 during 
body thefromfer Heat trans
or )()( dTmcdtTThA ps −=− ∞
Noting that Vρ=m and since )( ∞−= TTddT =∞T constant, the equation above can be rearranged as 
 dt
c
hA
TT
TTd
p
s
Vρ
−=
−
−
∞
∞ )( 
which is the desired differential equation. 
 
 
 
 
2-131 A long rectangular bar is initially at a uniform temperature of Ti. The surfaces of the bar at x = 0 and y = 0 are 
insulated while heat is lost from the other two surfaces by convection. The mathematical formulation of this heat conduction 
problem is to be expressed for transient two-dimensional heat transfer with no heat generation. 
Assumptions 1 Heat transfer is transient and two-dimensional. 2 Thermal conductivity is constant. 3 There is no heat 
generation. 
Analysis The differential equation and the boundary conditions for this heat conduction problem can be expressed as 
 
t
T
y
T
x
T
∂
∂
α
=
∂
∂
+
∂
∂ 1
2
2
2
2
 h, T∞
h, T∞
b
a 
 
0),,0(
0),0,(
=
∂
∂
=
∂
∂
y
tyT
x
txT
 
 
]),,([),,(
]),,([),,(
∞
∞
−=
∂
∂
−
−=
∂
∂
−
TtbxTh
x
tbxTk
TtyaTh
y
tyaTk
 Insulated
iTyxT =)0,,( 
 
PROPRIETARY MATERIAL. © 2011 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. 
2-72
 
2-132 Heat is generated at a constant rate in a short cylinder. Heat is lost from the cylindrical surface at r = ro by convection 
to the surrounding medium at temperature with a heat transfer coefficient of h. The bottom surface of the cylinder at r = 
0 is insulated, the top surface at z = H is subjected to uniform heat flux , and the cylindrical surface at r = r
∞T
hq& o is subjected 
to convection. The mathematical formulation of this problem is to be expressed for steady two-dimensional heat transfer. 
Assumptions 1 Heat transfer is given to be steady and two-dimensional. 2 Thermal conductivity is constant. 3 Heat is 
generated uniformly. 
Analysis The differential equation and the boundary conditions for this heat conduction problem can be expressed as 
01 gen
2
2
=+
∂
∂
+⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
∂
∂
k
e
z
T
r
Tr
rr
&
 
ro
z 
egen
qH
Hqz
HrT
k
z
rT
&=
∂
∂
=
∂
∂
),(
0)0,(
 h 
T∞
]),([
),(
0
),0(
∞−=∂
∂
−
=
∂
∂
TzrTh
r
zrT
k
r
zT
o
o
 
 
 
 
2-133E The concrete slab roof of a house is subjected to specified temperature at the bottom surface and convection and 
radiation at the top surface. The temperature of the top surface of the roof and the rate of heat transfer are to be determined 
when steady operating conditions are reached. 
Assumptions 1 Steady operating conditions are reached. 2 Heat transfer is one-dimensional since the roof area is large 
relative to its thickness, and the thermal conditions on both sides of the roof are uniform. 3 Thermal properties are constant. 
4 There is no heat generation in the wall. 
Properties The thermal conductivity and emissivity are given to be k = 1.1 Btu/h⋅ft⋅°F and ε = 0.8. 
Analysis In steady operation, heat conduction through the roof must be equal to net heat transfer from the outer surface. 
Therefore, taking the outer surface temperature of the roof to be T2 (in °F), 
 ])460[()( 4sky
4
22
21 TTATThA
L
TT
kA −++−=
−
∞ σε x T∞
h 
T1
Tsky
LCanceling the area A and substituting the known quantities, 
444
2
428
2
22
R]310)460)[(RftBtu/h 101714.0(8.0 
F)50)(FftBtu/h 2.3(
ft 0.8
F)62(
)FftBtu/h 1.1(
−+⋅⋅×+
°−°⋅⋅=
°−
°⋅⋅
− T
T
T
 
Using an equation solver (or the trial and error method), the outer surface temperature is determined to be 
 T2 = 38°F 
Then the rate of heat transfer through the roof becomes 
 Btu/h 28,875=°−×°⋅⋅=
−
=
ft 0.8
F)3862(
)ft 3525)(FftBtu/h 1.1( 221
L
TT
kAQ& 
Discussion The positive sign indicates that the direction of heat transfer is from the inside to the outside. Therefore, the 
house is losing heat as expected. 
 
PROPRIETARY MATERIAL. © 2011 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. 
2-73
 
2-134 A steam pipe is subjected to convection on both the inner and outer surfaces. The mathematical formulation of the 
problem and expressions for the variation of temperature in the pipe and on the outer surface temperature are to be obtained 
for steady one-dimensional heat transfer. 
Assumptions 1 Heat conduction is steady and one-dimensional since the pipe is long relative to its thickness, and there is 
thermal symmetry about the center line. 2 Thermal conductivity is constant. 3 There is no heat generation in the pipe. 
Analysis (a) Noting that heat transfer is steady and one-dimensional in the radial 
r direction, the mathematical formulation of this problem can be expressed as 
r2
To
ho 
Ti
hi 
r1
r 
0=⎟
⎠
⎞
⎜
⎝
⎛
dr
dTr
dr
d 
and )]([)( 11 rTThdr
rdTk ii −=− 
])([)( 22 oo TrThdr
rdTk −=− 
(b) Integrating the differential equation once with respect to r gives 
 1Cdr
dTr = 
Dividing both sides of the equation above by r to bring it to a readily integrable form and then integrating, 
 
r
C
dr
dT 1= 
 21 ln)( CrCrT +=
where C1 and C2 are arbitrary constants. Applying the boundary conditions give 
r = r1: )]ln([ 211
1
1 CrCTh
r
Ck ii +−=− 
r = r2: ])ln[( 221
2
1
oo TCrChr
Ck −+=− 
Solving for C1 and C2 simultaneously gives 
 ⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−
++
−
−=⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−−=
++
−
=
1
1
211
2
0
1
112
211
2
0
1 ln
ln
ln and 
ln rh
kr
rh
k
rh
k
r
r
TTT
rh
krCTC
rh
k
rh
k
r
r
TTC
i
oi
i
i
i
i
oi
i 
Substituting into the general solution and simplifying, we get the variation of temperature to be 21 and CC
 
211
2
11
1
111
ln
ln
)(lnln)(
rh
k
rh
k
r
r
rh
k
r
r
T
rh
krCTrCrT
oi
i
i
i
i
++
+
+=−−+= 
 (c) The outer surface temperature is determined by simply replacing r in the relation above by r2. We get 
 
211
2
11
2
2
ln
ln
)(
rh
k
rh
k
r
r
rh
k
r
r
TrT
oi
i
i
++
+
+= 
 
PROPRIETARY MATERIAL. © 2011 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. 
2-74
 
2-135 A spherical liquid nitrogen container is subjected to specified temperature on the inner surface and convection on the 
outer surface. The mathematical formulation, the variation of temperature, and the rate of evaporation of nitrogen are to be 
determined for steady one-dimensional heat transfer. 
Assumptions 1 Heat conduction is steady and one-dimensional since there is no change with time and there is thermal 
symmetry about the midpoint. 2 Thermal conductivity is constant. 3 There is no heat generation. 
Properties The thermal conductivity of the tank is given to be k = 12 W/m⋅°C. Also, hfg = 198 kJ/kg for nitrogen. 
Analysis (a) Noting that heat transfer is one-dimensional in the radial r direction, the mathematical formulation of this 
problem can be expressed as 
02 =⎟
⎠
⎞
⎜
⎝
⎛
dr
dTr
dr
d 
N2
-196°C r r2
r1
h 
T∞ 
and C196)( 11 °−== TrT
])([)( 22 ∞−=− TrThdr
rdTk 
(b) Integrating the differential equation once with respect to r gives 
 1
2 C
dr
dTr = 
Dividing both sides of the equation above by r to bring it to a readily integrable form and then integrating, 
2
1
r
C
dr
dT
= → 21)( Cr
C
rT +−= 
where C1 and C2 are arbitrary constants. Applying the boundary conditions give 
r = r1: 12
1
1
1 )( TCr
C
rT =+−= 
r = r2: ⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−+−=− ∞TCr
Ch
r
Ck 2
2
1
2
2
1 
Solving for C1 and C2 simultaneously gives 
 
1
2
21
2
1
1
1
1
12
21
2
12
1
1
 and 
1
)(
r
r
hr
k
r
r
TT
T
r
C
TC
hr
k
r
r
TTr
C
−−
−
+=+=
−−
−
= ∞∞ 
Substituting C1 and C2 into the general solution,the variation of temperature is determined to be 
196)/1.205.1(1013C)196(1.2
2
1.2
)m 1.2)(C W/m35(
C W/m12
2
1.21
C)20196(
1
11)(
2
1
2
1
2
21
2
1
1
1
1
1
1
1
1
−−=°−+⎟
⎠
⎞
⎜
⎝
⎛ −
°⋅
°⋅
−−
°−−
=
+⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−
−−
−
=+⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−=++−= ∞
r
r
T
r
r
r
r
hr
k
r
r
TT
T
rr
C
r
C
T
r
C
rT
 
(c) The rate of heat transfer through the wall and the rate of evaporation of nitrogen are determined from 
negative) since tank the(to W ,710320
)m 1.2)(C W/m35(
C W/m12
2
1.21
C)20196(m) 1.2()C W/m12(4
1
)(
44)4(
2
21
2
12
12
12
 −=
°⋅
°⋅
−−
°−−
°⋅−=
−−
−
−=−=−=−= ∞
π
πππ
hr
k
r
r
TTr
kkC
r
C
rk
dx
dTkAQ&
 
kg/s 1.62===
J/kg 000,198
J/s 700,320
fgh
Qm
&
& 
 
PROPRIETARY MATERIAL. © 2011 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. 
2-75
 
2-136 A spherical liquid oxygen container is subjected to specified temperature on the inner surface and convection on the 
outer surface. The mathematical formulation, the variation of temperature, and the rate of evaporation of oxygen are to be 
determined for steady one-dimensional heat transfer. 
Assumptions 1 Heat conduction is steady and one-dimensional since there is no change with time and there is thermal 
symmetry about the midpoint. 2 Thermal conductivity is constant. 3 There is no heat generation. 
Properties The thermal conductivity of the tank is given to be k = 12 W/m⋅°C. Also, hfg = 213 kJ/kg for oxygen. 
Analysis (a) Noting that heat transfer is one-dimensional in the radial r direction, the mathematical formulation of this 
problem can be expressed as 
02 =⎟
⎠
⎞
⎜
⎝
⎛
dr
dTr
dr
d 
O2
-183°C r r2
r1
h 
T∞ 
and C183)( 11 °−== TrT
])([)( 22 ∞−=− TrThdr
rdTk 
(b) Integrating the differential equation once with respect to r gives 
 1
2 C
dr
dTr = 
Dividing both sides of the equation above by r to bring it to a readily integrable form and then integrating, 
 
2
1
r
C
dr
dT
= → 21)( Cr
C
rT +−= 
 where C1 and C2 are arbitrary constants. Applying the boundary conditions give 
r = r1: 12
1
1
1 )( TCr
C
rT =+−= 
r = r2: ⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−+−=− ∞TCr
Ch
r
Ck 2
2
1
2
2
1 
Solving for C1 and C2 simultaneously gives 
 
1
2
21
2
1
1
1
1
12
21
2
12
1
1
 and 
1
)(
r
r
hr
k
r
r
TT
T
r
C
TC
hr
k
r
r
TTr
C
−−
−
+=+=
−−
−
= ∞∞ 
Substituting C1 and C2 into the general solution, the variation of temperature is determined to be 
 
183)/1.205.1(9.951C)183(1.2
2
1.2
)m 1.2)(C W/m35(
C W/m12
2
1.21
C)20183(
1
11)(
2
1
2
1
2
21
2
1
1
1
1
1
1
1
1
−−=°−+⎟
⎠
⎞
⎜
⎝
⎛ −
°⋅
°⋅
−−
°−−
=
+⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−
−−
−
=+⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−=++−= ∞
r
r
T
r
r
r
r
hr
k
r
r
TT
T
rr
C
r
C
T
r
C
rT
 
(c) The rate of heat transfer through the wall and the rate of evaporation of oxygen are determined from 
negative) since tank the(to W400,301
)m 1.2)(C W/m35(
C W/m12
2
1.21
C)20183(m) 1.2()C W/m12(4
1
)(
44)4(
2
21
2
12
12
12
 −=
°⋅
°⋅
−−
°−−
°⋅−=
−−
−
−=−=−=−= ∞
π
πππ
hr
k
r
r
TTr
kkC
r
C
rk
dx
dTkAQ&
 
kg/s 1.42===
J/kg 000,213
J/s 400,301
fgh
Qm
&
& 
 
PROPRIETARY MATERIAL. © 2011 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. 
2-76
 
2-137 A large plane wall is subjected to convection, radiation, and specified temperature on the right surface and no 
conditions on the left surface. The mathematical formulation, the variation of temperature in the wall, and the left surface 
temperature are to be determined for steady one-dimensional heat transfer. 
Assumptions 1 Heat conduction is steady and one-dimensional since the wall is large relative to its thickness, and the thermal 
conditions on both sides of the wall are uniform. 2 Thermal conductivity is constant. 3 There is no heat generation in the 
wall. 
Properties The thermal conductivity and emissivity are given to be k = 8.4 W/m⋅°C and ε = 0.7. 
Analysis (a) Taking the direction normal to the surface of the wall to be the x direction with x = 0 at the left surface, and the 
mathematical formulation of this problem can be expressed as 
02
2
=
dx
Td 
and ])273[(][])([])([)( 4surr
4
22
4
surr
4 TTTThTLTTLTh
dx
LdTk −++−=−+−=− ∞∞ εσεσ 
C45)( 2 °== TLT 
45°C
ε 
Tsurr
 L 
h 
T∞
x
(b) Integrating the differential equation twice with respect to x yields 
 1Cdx
dT
= 
 21)( CxCxT +=
where C1 and C2 are arbitrary constants. Applying the boundary conditions give 
Convection at x = L 
kTTTThC
TTTThkC
/]})273[(][{ 
])273[(][ 
4
surr
4
221
4
surr
4
221
−+εσ+−−=→
−+εσ+−=−
∞
∞
Temperature at x = L: LCTCTCLCLT 122221 )( −=→=+×= 
Substituting C1 and C2 into the general solution, the variation of temperature is determined to be 
)4.0(23.4845
m )4.0(
C W/m4.8
]K) 290()K 318)[(K W/m100.7(5.67+C)2545)(C W/m14(C45
)(])273[(][)()()(
444282
4
surr
4
22
212121
x
x
xL
k
TTTThTCxLTLCTxCxT
−+=
−
°⋅
−⋅×°−°⋅
+°=
−
−+εσ+−
+=−−=−+=
−
∞
 
 (c) The temperature at x = 0 (the left surface of the wall) is 
 C64.3°=−+= )04.0(23.4845)0(T
 
PROPRIETARY MATERIAL. © 2011 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. 
2-77
 
2-138 The base plate of an iron is subjected to specified heat flux on the left surface and convection and radiation on the 
right surface. The mathematical formulation, and an expression for the outer surface temperature and its value are to be 
determined for steady one-dimensional heat transfer. 
Assumptions 1 Heat conduction is steady and one-dimensional. 2 Thermal conductivity is constant. 3 There is no heat 
generation. 4 Heat loss through the upper part of the iron is negligible. 
Properties The thermal conductivity and emissivity are given to be k = 18 W/m⋅°C and ε = 0.7. 
Analysis (a) Noting that the upper part of the iron is well insulated and thus the entire 
heat generated in the resistance wires is transferred to the base plate, the heat flux 
through the inner surface is determined to be 
ε
q 
Tsurr
 L
h 
T∞
x
 224
base
0
0 W/m0000,80m 10150
 W1200
=
×
==
−A
Q
q
&
& 
Taking the direction normal to the surface of the wall to be the x direction with x = 0 at 
the left surface, the mathematical formulation of this problem can be expressed as 
0
2
2
=
dx
Td 
and 20 W/m000,80
)0(
==− q
dx
dTk & 
])273[(][])([])([)( 4surr
4
22
4
surr
4 TTTThTLTTLTh
dx
LdTk −++−=−+−=− ∞∞ εσεσ
(b) Integrating the differential equation twice with respect to x yields 
 1Cdx
dT
= 
 21)( CxCxT +=
where C1 and C2 are arbitrary constants. Applying the boundary conditions give 
x = 0: 
k
q
CqkC 0101 
&
& −=→=− 
x = L: ])273[(][ 4surr
4
221 TTTThkC −++−=− ∞ εσ
Eliminating the constant C1 from the two relations above gives the following expression for the outer surface temperature T2, 
 0
4
surr
4
22 ])273[()( qTTTTh &=−++− ∞ εσ
(c) Substituting the known quantities into the implicit relation above gives 
 2442
428
2
2 W/m000,80]295)273)[(K W/m1067.5(7.0)26)(C W/m30( =−+⋅×+−°⋅ − TT
Using an equation solver (or a trial and error approach), the outer surface temperature is determined from the relation above 
to be 
 T2 = 819°C 
 
 
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2-78
 
2-139 The base plate of an iron is subjected to specified heat flux on the left surface and convectionand radiation on the 
right surface. The mathematical formulation, and an expression for the outer surface temperature and its value are to be 
determined for steady one-dimensional heat transfer. 
Assumptions 1 Heat conduction is steady and one-dimensional. 2 Thermal conductivity is constant. 3 There is no heat 
generation. 4 Heat loss through the upper part of the iron is negligible. 
Properties The thermal conductivity and emissivity are given to be k = 18 W/m⋅°C and ε = 0.7. 
Analysis (a) Noting that the upper part of the iron is well insulated and thus the entire 
heat generated in the resistance wires is transferred to the base plate, the heat flux 
through the inner surface is determined to be 
ε
q 
Tsurr
 L
h 
T∞
x
 2
24
base
0
0 W/m000,100
m 10150
 W1500
=
×
==
−A
Q
q
&
& 
Taking the direction normal to the surface of the wall to be the x direction with x = 0 at 
the left surface, the mathematical formulation of this problem can be expressed as 
0
2
2
=
dx
Td 
and 20 W/m000,100
)0(
==− q
dx
dT
k & 
])273[(][])([])([)( 4surr
4
22
4
surr
4 TTTThTLTTLTh
dx
LdTk −++−=−+−=− ∞∞ εσεσ
(b) Integrating the differential equation twice with respect to x yields
 
 1Cdx
dT
= 
 21)( CxCxT +=
where C1 and C2 are arbitrary constants. Applying the boundary conditions give 
x = 0: 
k
q
CqkC 0101 
&
& −=→=− 
x = L: ])273[(][ 4surr
4
221 TTTThkC −++−=− ∞ εσ
Eliminating the constant C1 from the two relations above gives the following expression for the outer surface temperature T2, 
 0
4
surr
4
22 ])273[()( qTTTTh &=−++− ∞ εσ
 (c) Substituting the known quantities into the implicit relation above gives 
 2442
428
2
2 W/m000,100]295)273)[(K W/m1067.5(7.0)26)(C W/m30( =−+⋅×+−°⋅ − TT
Using an equation solver (or a trial and error approach), the outer surface temperature is determined from the relation above 
to be 
 T2 = 896°C 
 
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2-79
 
2-140E A large plane wall is subjected to a specified temperature on the left (inner) surface and solar radiation and heat loss 
by radiation to space on the right (outer) surface. The temperature of the right surface of the wall and the rate of heat transfer 
are to be determined when steady operating conditions are reached. 
Assumptions 1 Steady operating conditions are reached. 2 Heat transfer is one-
dimensional since the wall is large relative to its thickness, and the thermal conditions 
on both sides of the wall are uniform. 3 Thermal properties are constant. 4 There is no 
heat generation in the wall. 
qsolar520 R 
T2
 L x
Properties The properties of the plate are given to be k = 1.2 Btu/h⋅ft⋅°F and 
ε = 0.80, and 60.0=sα . 
Analysis In steady operation, heat conduction through the wall must be equal 
to net heat transfer from the outer surface. Therefore, taking the outer surface 
temperature of the plate to be T2 (absolute, in R), 
 solar
4
2
21 qATA
L
TTkA ssss &αεσ −=
− 
Canceling the area A and substituting the known quantities, 
)ftBtu/h 300(60.0)RftBtu/h 101714.0(8.0
ft 0.8
R) 520(
)FftBtu/h 2.1( 242
4282 ⋅−⋅⋅×=
−
°⋅⋅ − T
T
 
Solving for T2 gives the outer surface temperature to be 
T2 = 553.9 R 
Then the rate of heat transfer through the wall becomes 
 2ftBtu/h 50.9 ⋅−=−°⋅⋅=−=
ft 0.8
R )9.553520(
)FftBtu/h 2.1(21
L
TT
kq& (per unit area) 
Discussion The negative sign indicates that the direction of heat transfer is from the outside to the inside. Therefore, the 
structure is gaining heat. 
 
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2-80
 
2-141E A large plane wall is subjected to a specified temperature on the left (inner) surface and heat loss by radiation to 
space on the right (outer) surface. The temperature of the right surface of the wall and the rate of heat transfer are to be 
determined when steady operating conditions are reached. 
Assumptions 1 Steady operating conditions are reached. 2 Heat transfer is one-dimensional since the wall is large relative to 
its thickness, and the thermal conditions on both sides of the wall are uniform. 3 Thermal properties are constant. 4 There is 
no heat generation in the wall. 
Properties The properties of the plate are given to be k = 1.2 Btu/h⋅ft⋅°F and ε = 0.80. 
Analysis In steady operation, heat conduction through the wall must be equal to net heat 
transfer from the outer surface. Therefore, taking the outer surface temperature of the 
plate to be T2 (absolute, in R), 
520 R 
T2
 L x
 42
21 TA
L
TT
kA ss εσ=
−
 
Canceling the area A and substituting the known quantities, 
4
2
4282 )RftBtu/h 101714.0(8.0
ft 0.5
R) 520(
)FftBtu/h 2.1( T
T
⋅⋅×=
−
°⋅⋅ − 
Solving for T2 gives the outer surface temperature to be T2 = 487.7 R 
Then the rate of heat transfer through the wall becomes 
2ftBtu/h 77.5 ⋅=
−
°⋅⋅=
−
=
ft 0.5
R )7.487520(
)FftBtu/h 2.1(21
L
TT
kq& (per unit area) 
Discussion The positive sign indicates that the direction of heat transfer is from the inside to the outside. Therefore, the 
structure is losing heat as expected. 
 
 
PROPRIETARY MATERIAL. © 2011 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. 
2-81
 
2-142 The surface and interface temperatures of a resistance wire covered with a plastic layer are to be determined. 
Assumptions 1 Heat transfer is steady since there is no change with time. 2 Heat transfer is one-dimensional since this two-
layer heat transfer problem possesses symmetry about the center line and involves no change in the axial direction, and thus 
T = T(r) . 3 Thermal conductivities are constant. 4 Heat generation in the wire is uniform. 
Properties It is given that and C W/m18wire °⋅=k C W/m8.1plastic °⋅=k . 
Analysis Letting denote the unknown interface temperature, the mathematical formulation of the heat transfer problem in 
the wire can be expressed as 
IT
01 gen =+⎟
⎠
⎞
⎜
⎝
⎛
k
e
dr
dTr
dr
d
r
&
 
r2
T∞
h
egen
r1
r 
with and ITrT =)( 1 0
)0(
=
dr
dT 
Multiplying both sides of the differential equation by r, 
rearranging, and integrating give 
 r
k
e
dr
dTr
dr
d gen&−=⎟
⎠
⎞
⎜
⎝
⎛ → 1
2
gen
2
Cr
k
e
dr
dTr +−=
&
 (a) 
Applying the boundary condition at the center (r = 0) gives 
B.C. at r = 0: 0 0
2
)0(
0 11
gen =→+×−=× CC
k
e
dr
dT &
 
Dividing both sides of Eq. (a) by r to bring it to a readily integrable form and integrating, 
 r
k
e
dr
dT
2
gen&−= → 2
2gen
4
)( Cr
k
e
rT +−=
&
 (b) 
Applying the other boundary condition at 1rr = , 
B. C. at : 1rr =
2
1
gen
22
2
1
gen
4
 
4
r
k
e
TCCr
k
e
T II
&&
+=→+−= 
Substituting this relation into Eq. (b) and rearranging give 2C
 )(
4
)( 221
wire
gen
wire rrk
e
TrT I −+=
&
 (c) 
Plastic layer The mathematical formulation of heat transfer problem in the plastic can be expressed as 
0=⎟
⎠
⎞
⎜
⎝
⎛
dr
dTr
dr
d 
with and ITrT =)( 1 ])([
)(
2
2
∞−=− TrThdr
rdTk 
The solution of the differential equation is determined by integration to be 
 1Cdr
dTr = → 
r
C
dr
dT 1= → 21 ln)( CrCrT += 
where C1 and C2 are arbitrary constants. Applying the boundary conditions give 
r = r1: 112211 ln ln rCTCTCrC II −=→=+ 
r = r2: ])ln[( 221
2
1
∞−+=− TCrChr
Ck → 
21
2
1
ln
hr
k
r
r
TTC I
+
−
= ∞ 
 
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2-82
 
Substituting C1 and C2 into the general solution, the variation of temperature in plastic is determined to be 
 
1
2
plastic
1
2
111plastic ln
ln
lnln)(
r
r
hr
k
r
r
TT
TrCTrCrT III
+
−
+=−+= ∞ 
We have already utilized the first interface condition by setting the wire and plastic layer temperatures equal to at the 
interface . The interface temperature is determined from the second interface condition that the heat flux in the wire 
and the plastic layer at must be the same: 
IT
1rr = IT
1rr =
 
1
2
plastic
1
2
plastic
1gen1plastic
plastic
1wire
wire
1
ln
2
 
)()(
r
hr
k
r
r
TT
k
re
dr
rdT
k
dr
rdT
k I
+
−
−=→−=− ∞
&
 
Solving for and substituting the given values, the interface temperature is determined to be IT
C255.6°°+⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
°⋅
°⋅
+
°⋅
×
=
+⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
+= ∞
=C25
m) C)(0.007 W/m(14
C W/m1.8
m 0.003
m 007.0ln
C) W/m2(1.8
m) 003.0)( W/m108.4(
ln
2
2
236
2
plastic
1
2
plastic
2
1gen T
hr
k
r
r
k
re
TI
&
 
Knowing the interface temperature, the temperature at the center line (r = 0) is obtained by substituting the known quantities 
into Eq. (c), 
 C256.2°=
°⋅×
×
°=+=
C) W/m(184
m) 003.0)( W/m108.4(+C6.255
4
)0(
236
wire
2
1gen
wire k
re
TT I
&
 
Thus the temperature of the centerline will be slightly above the interface temperature. 
 
PROPRIETARY MATERIAL. © 2011 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. 
2-83
 
2-143 A cylindrical shell with variable conductivity is subjected to 
specified temperatures on both sides. The rate of heat transfer through 
the shell is to be determined. 
r2
k(T)
T1
r1
r 
T2
Assumptions 1 Heat transfer is given to be steady and one-
dimensional. 2 Thermal conductivity varies quadratically. 3 There is 
no heat generation. 
Properties The thermal conductivity is given to be 
. )1()( 20 TkTk β+=
Analysis When the variation of thermal conductivity with temperature 
k(T) is known, the average value of the thermal conductivity in the 
temperature range between is determined from 21 and TT
 
( ) ( )
( )⎥⎦
⎤
⎢⎣
⎡ +++=
−
⎥⎦
⎤
⎢⎣
⎡ −+−
=
−
⎟
⎠
⎞
⎜
⎝
⎛ +
=
−
+
=
−
=
∫∫
2
121
2
20
12
3
1
3
2120
12
3
0
12
2
0
12
avg
3
1
33
)1()(
2
1
2
1
2
1
TTTTk
TT
TTTTk
TT
TTk
TT
dTTk
TT
dTTk
k
T
T
T
T
T
T
β
ββ
β
 
This relation is based on the requirement that the rate of heat transfer through a medium with constant average thermal 
conductivity equals the rate of heat transfer through the same medium with variable conductivity k(T). avgk
Then the rate of heat conduction through the cylindrical shell can be determined from Eq. 2-77 to be 
 ( )
)/ln(3
12
)/ln(
2
12
212
121
2
20
12
21
avgcylinder rr
TT
LTTTTk
rr
TT
LkQ
−
⎥⎦
⎤
⎢⎣
⎡ +++=
−
=
βππ& 
Discussion We would obtain the same result if we substituted the given k(T) relation into the second part of Eq. 2-77, and 
performed the indicated integration. 
 
 
 
 
2-144 Heat is generated uniformly in a cylindrical uranium fuel rod. The temperature difference between the center and the 
surface of the fuel rod is to be determined. 
Assumptions 1 Heat transfer is steady since there is no indication of any change with time. 2 Heat transfer is one-
dimensional since there is thermal symmetry about the center line and no change in the axial direction. 3 Thermal 
conductivity is constant. 4 Heat generation is uniform. 
Properties The thermal conductivity of uranium at room temperature 
is k = 27.6 W/m⋅°C (Table A-3). Ts
Analysis The temperature difference between the center and the 
surface of the fuel rods is determined from De
 C9.1°=
°
×
==−
C) W/m.6.27(4
m) 005.0)( W/m104(
4
2372
gen
k
re
TT oso
&
 
 
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2-84
 
2-145 A large plane wall is subjected to convection on the inner and outer surfaces. The mathematical formulation, the 
variation of temperature, and the temperatures at the inner and outer surfaces to be determined for steady one-dimensional 
heat transfer. 
Assumptions 1 Heat conduction is steady and one-dimensional. 2 Thermal conductivity is constant. 3 There is no heat 
generation. 
Properties The thermal conductivity is given to be k = 0.77 W/m⋅°C. 
Analysis (a) Taking the direction normal to the surface of the wall to be the x direction with x = 0 at the inner surface, the 
mathematical formulation of this problem can be expressed as 
02
2
=
dx
Td 
h2
T∞2
 L 
h1
T∞1
k and 
dx
dT
kTTh
)0(
)]0([ 11 −=−∞ 
])([)( 22 ∞−=− TLThdx
LdTk 
(b) Integrating the differential equation twice with respect to x yields 
1Cdx
dT
= 
21)( CxCxT += 
where C1 and C2 are arbitrary constants. Applying the boundary conditions give 
x = 0: 12111 )]0([ kCCCTh −=+×−∞
x = L: ])[( 22121 ∞−+=− TCLChkC
Substituting the given values, these equations can be written as 
 12 77.0)22(8 CC −=−
 )82.0)(12(77.0 211 −+=− CCC
Solving these equations simultaneously give 
 26.18 84.38 21 =−= CC
Substituting into the general solution, the variation of temperature is determined to be 21 and CC
 xxT 84.3826.18)( −=
 (c) The temperatures at the inner and outer surfaces are 
 
C10.5
C18.3
°=×−=
°=×−=
2.084.3826.18)(
084.3826.18)0(
LT
T
 
 
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2-85
 
2-146 A hollow pipe is subjected to specified temperatures at the inner and outer surfaces. There is also heat generation in 
the pipe. The variation of temperature in the pipe and the center surface temperature of the pipe are to be determined for 
steady one-dimensional heat transfer. 
Assumptions 1 Heat conduction is steady and one-dimensional since the pipe is long relative to its thickness, and there is 
thermal symmetry about the centerline. 2 Thermal conductivity is constant. 
Properties The thermal conductivity is given to be k = 14 W/m⋅°C. 
Analysis The rate of heat generation is determined from 
 [ ]
3
222
1
2
2
gen W/m750,26
4/)m 17(m) 3.0(m) 4.0(
 W000,25
4/)(
=
−
=
−
==
ππ LDD
WWe
&&
&
V
 
Noting that heat transfer is one-dimensional in the radial r direction, the mathematical formulation of this problem can be 
expressed as 
01 gen =+⎟
⎠
⎞
⎜
⎝
⎛
k
e
dr
dTr
dr
d
r
&
 
r2
egen
T1
r1
r 
T2
and C60)( 11 °== TrT
C80)( 22 °== TrT 
Rearranging the differential equation 
 0gen =
−
=⎟
⎠
⎞
⎜
⎝
⎛
k
re
dr
dTr
dr
d & 
and then integrating once with respect to r, 
 1
2
gen
2
C
k
re
dr
dTr +
−
=
&
 
Rearranging the differential equation again 
 
r
C
k
re
dr
dT 1gen
2
+
−
=
&
 
and finally integrating again with respect to r, we obtain 
 21
2
gen ln
4
)( CrC
k
re
rT ++
−
=
&
 
where C1 and C2 are arbitrary constants. Applying the boundary conditions give 
r = r1: 211
2
1gen
1 ln4
)( CrC
k
re
rT ++
−
=
&
 
r = r2: 221
2
2gen
2 ln4
)( CrC
k
re
rT ++
−
=
&
 
Substituting the given values, these equations can be written as 
 21
2
)15.0ln(
)14(4
)15.0)(750,26(60 CC ++−= 
 21
2
)20.0ln(
)14(4
)20.0)(750,26(80 CC ++−= 
Solving for simultaneously gives 21 and CC
 8.257 58.98 21 == CC
Substituting into the general solution, the variation of temperature is determined to be 21 and CC
 rrrrrT ln58.987.4778.2578.257ln58.98
)14(4
750,26)(2
2
+−=++
−
= 
The temperature at the center surface of the pipe is determined by setting radius r to be 17.5 cm, which is the average of the 
inner radius and outer radius. 
 C71.3°=+−= )175.0ln(58.98)175.0(7.4778.257)( 2rT
 
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preparation. If you are a student using this Manual, you are using it without permission. 
2-86
 
2-147 Heat is generated in a plane wall. Heat is supplied from one side which is insulated while the other side is subjected to 
convection with water. The convection coefficient, the variation of temperature in the wall, and the location and the value of 
the maximum temperature in the wall are to be determined. 
Assumptions 1 Heat transfer is steady since there is no indication of any change with time. 2 Heat transfer is one-
dimensional since the wall is large relative to its thickness. 3 Thermal conductivity is constant. 4 Heat generation is uniform. 
Analysis (a) Noting that the heat flux and the heat generated will be 
transferred to the water, the heat transfer coefficient is determined from 
the Newton’s law of cooling to be Tsk 
x 
 T∞ , h 
Heater 
Insulation 
sq&
gene&
L 
C W/m400 2 °⋅=
°−
+
=
−
+
=
∞
C40)(90
m) )(0.04 W/m(10) W/m(16,000 352
gen
TT
Leq
h
s
s &&
 
(b) The variation of temperature in the wall is in the form of 
T(x) = ax2+bx+c. First, the coefficient a is determined as 
follows 
 
k
e
dT
Tdke
dx
Tdk gen
2
2
gen2
2
0
&
& −=→=+ 
cbxx
k
e
Tbx
k
e
dx
dT
++−=+−= 2gengen
2
 and 
&&
 → 2
35
gen C/m2500
)C W/m20(2
 W/m10
2
°−=
°⋅
=−=
k
e
a
&
Applying the first boundary condition: 
 x = 0, T(0) = Ts → c = Ts = 90ºC 
As the second boundary condition, we can use either 
 s
Lx
q
dx
dTk −=−
=
→ sqbk
Le
k =⎟⎟
⎠
⎞
⎜
⎜
⎝
⎛
+− gen
&
→ ( ) ( ) C/m100004.01016000
20
11 5
gen °=×+=+= Leqk
b s & 
or )(
0
∞
=
−−=− TTh
dx
dTk s
x
 
k(a×0+b) = h(Ts -T∞) → C/m1000)4090(20
400
°=−=b 
Substituting the coefficients, the variation of temperature becomes 
 9010002500)( 2 ++−= xxxT
(c) The x-coordinate of Tmax is xvertex= -b/(2a) = 1000/(2×2500) = 0.2 m = 20 cm. This is outside of the wall boundary, to the 
left, so Tmax is at the left surface of the wall. Its value is determined to be 
 C126°=++−=++−== 90)04.0(1000)04.0(25009010002500)( 22max LLLTT
The direction of qs(L) (in the negative x direction) indicates that at x = L the temperature increases in the positive x direction. 
If a is negative, the T plot is like in Fig. 1, which shows Tmax at x=L. If a is positive, the T plot could only be like in Fig. 2, 
which is incompatible with the direction of heat transfer at the surface in contact with the water. So, temperature distribution 
can only be like in Fig. 1, where Tmax is at x=L, and this was determined without using numerical values for a, b, or c. 
 
Slope 
Fig. 1 
qs(L) 
qs(0) 
Here, heat transfer 
and slope are 
incompatible 
Slope Fig. 2 qs(L) qs(0) 
 
 
 
 
 
 
This part could also be answered to without any information about the nature of the T(x) function, using qualitative 
arguments only. At steady state, heat cannot go from right to left at any location. There is no way out through the left surface 
because of the adiabatic insulation, so it would accumulate somewhere, contradicting the steady state assumption. Therefore, 
the temperature must continually decrease from left to right, and Tmax is at x = L. 
 
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2-148 Heat is generated in a plane wall. The temperature distribution in the wall is given. The surface temperature, the heat 
generation rate, the surface heat fluxes, and the relationship between these heat fluxes, the heat generation rate, and the 
geometry of the wall are to be determined. 
Assumptions 1 Heat transfer is steady since there is no indication of any change with time. 2 Heat transfer is one-
dimensional since the wall is large relative to its thickness. 3 Thermal conductivity is constant. 4 Heat generation is uniform. 
Analysis (a) The variation of temperature is symmetric about x = 0. The surface temperature is 
C67.5°=°×−°=−=−== 2242 )m 025.0)(C/m102(C80)()( bLaLTLTTs 
The plot of temperatures across the wall thickness is given below. 
 
-0.025 -0.015 -0.005 0.005 0.015 0.025
66
68
70
72
74
76
78
80
82
x [m]
T 
[C
]
 
T∞ 
h 
72ºC k 
 
gene& 
 L -L
72ºC
x
 
(b) The volumetric rate of heat generation is 
 35 W/m103.2 ×=°×°⋅=−−=⎯→⎯=+ )C/m102(C) W/m8(2)2(0 24gengen2
2
bkee
dx
Tdk && 
(c) The heat fluxes at the two surfaces are 
 
[ ] 2
2
 W/m8000
 W/m8000
−=°×°⋅−=−−−=−=−
=°×°⋅=−−=−=
m) 025.0)(C/m102(C) W/m8(2)(2()(
m) 025.0)(C/m102(C) W/m8(2)2()(
24
24
Lbk
dx
dTkLq
bLk
dx
dTkLq
L
s
L
s
&
&
 
(d) The relationship between these fluxes, the heat generation rate and the geometry of the wall is 
 
[ ]
[ ]
LeLqLq
LWHeWHLqLq
eALqLq
EE
ss
ss
ss
gen
gen
gen
genout
2)()(
)2()()(
)()(
&&&
&&&
&&&
&&
=−+
=−+
=−+
=
V
Discussion Note that in this relation the absolute values of heat fluxes should be used. Substituting numerical values gives 
1000 W/m2 on both sides of the equation, and thus verifying the relationship. 
 
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2-88
 
2-149 Steady one-dimensional heat conduction takes place in a long slab. It is to be shown that the heat flux in steady 
operation is given by ⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
+
=
wTT
TT
W
kq
*
0
**
ln& . Also, the heat flux is to be calculated for a given set of parameters. 
Assumptions 1 Heat transfer is steady. 2 Heat transfer is one-dimensional since the wall is large relative to its thickness. 
Analysis The derivation is given as follows 
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
+
=
−=⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
+
−−=+
−=
+
+
−
=−=
∫∫
w
w
T
T
WT
T
TT
TT
W
kq
k
Wq
TT
TT
W
k
qTT
dx
k
q
TT
dT
dx
dT
TT
k
dx
dTkq
w
w
*
0
**
*
0
*
*
*
*
0
**
*
*
ln
ln
)0()ln(
or 
)(
0
0
&
&
&
&
&
 
The heat flux for the given values is 
 25 W/m101.42×−=⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−
−×
=⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
+
=
K)4001000(
K)6001000(ln
m 0.2
 W/m107ln
4
*
0
**
wTT
TT
W
kq& 
 
 
 
 
2-150 A spherical ball in which heat is generated uniformly is exposed to iced-water. The temperatures at the center and at 
the surface of the ball are to be determined. 
Assumptions 1 Heat transfer is steady since there is no indication of any change with time. 2 Heat transfer is one-
dimensional., and there is thermal symmetry about the center point. 3 Thermal conductivity is constant. 4 Heat generation is 
uniform. 
Properties The thermal conductivity is given to be k = 45 W/m⋅°C. 
D 
gene&
h 
T∞
Analysis The temperatures at the center and at the surface of the ball 
are determined directly from 
 C140°=
°
×
+°=+= ∞
C). W/m1200(3
m) 12.0)( W/m102.4(C0
3 2
36
gen
h
re
TT os
&
 C364°=
°
×
+°=+=
C) W/m.45(6
m) 12.0)( W/m102.4(C140
6
2362
gen
0 k
re
TT os
&
 
 
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2-151 A 10-m tall exhaust stack discharging exhaust gases at a rate of 1.2 kg/s is subjected to solar radiation and convection 
at the outer surface. The variation of temperature in the exhaust stack and the inner surface temperature of the exhaust stack 
are to be determined. 
Assumptions 1 Heat conduction is steady and one-dimensional andthere is thermal symmetry about the centerline. 2 
Thermal properties are constant. 3 There is no heat generation in the pipe. 
Properties The constant pressure specific heat of exhaust gases is given to be 1600 J/kg · °C and the pipe thermal 
conductivity is 40 W/m · K. Both the emissivity and solar absorptivity of the exhaust stack outer surface are 0.9. 
 
Analysis The outer and inner radii of the pipe are 
 m 5.02/m 12 ==r
 m 4.0m 1.0m 5.01 =−=r 
The outer surface area of the exhaust stack is 
 222, m 42.31)m 10)(m 5.0(2 2 === ππ LrAs
The rate of heat loss from the exhaust gases in the exhaust stack can be determined from 
 W57600C )30(C)J/kg 1600)(kg/s 2.1()( outinloss =°°⋅=−= TTcmQ p&&
The heat loss on the outer surface of the exhaust stack by radiation and convection can be expressed as 
 solar
4
surr
4
22
2,
loss ])([ ])([ qTrTTrTh
A
Q
s
s
&
&
αεσ −−+−= ∞ 
 
) W/m150)(9.0(K ])27327()()[K W/m1067.5)(9.0(
K )]27327()()[K W/m8(
m 42.31
 W57600
2444
2
428
2
2
2
−+−⋅×+
+−⋅=
− rT
rT
 
Copy the following line and paste on a blank EES screen to solve the above equation: 
57600/31.42=8*(T_r2-(27+273))+0.9*5.67e-8*(T_r2^4-(27+273)^4)-0.9*150 
Solving by EES software, the outside surface temperature of the furnace front is 
 K 7.412)( 2 =rT
(a) For steady one-dimensional heat conduction in cylindrical coordinates, the heat conduction equation can be expressed as 
 0=⎟
⎠
⎞
⎜
⎝
⎛
dr
dTr
dr
d 
and 
Lr
Q
A
Q
dr
rdT
k
s 1
loss
1,
loss1
 2
)(
π
&&
==− (heat flux at the inner exhaust stack surface) 
 (outer exhaust stack surface temperature) K 7.412)( 2 =rT
Integrating the differential equation once with respect to r gives 
 
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2-90
 
 
r
C
dr
dT 1= 
Integrating with respect to r again gives 
 21 ln)( CrCrT +=
where and are arbitrary constants. Applying the boundary conditions gives 1C 2C
 :1rr =
1
1
1
loss1
 2
1)(
r
C
Lr
Q
kdr
rdT
=−=
π
&
 →
kL
Q
C loss1 2
1 &
π
−= 
 :2rr = 22
loss
2 ln2
1)( Cr
kL
Q
rT +−=
&
π
 → )(ln
2
1
22
loss
2 rTrkL
Q
C +=
&
π
 
Substituting and into the general solution, the variation of temperature is determined to be 1C 2C
 
)()/ln(
2
1
)(ln
2
1ln
2
1)(
22
loss
22
lossloss
rTrr
kL
Q
rTr
kL
Q
r
kL
Q
rT
+−=
++−=
&
&&
π
ππ 
(b) The inner surface temperature of the exhaust stack is 
 
K 418==
+⎟
⎠
⎞
⎜
⎝
⎛
⋅
−=
+−=
K7417
K 7.412
5.0
4.0ln
)m 10)(K W/m40(
 W57600
2
1
)()/ln(
2
1)( 221
loss
1
 .
rTrr
kL
Q
rT
π
π
&
 
Discussion There is a temperature drop of 5 °C from the inner to the outer surface of the exhaust stack. 
 
 
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Fundamentals of Engineering (FE) Exam Problems 
 
2-152 The heat conduction equation in a medium is given in its simplest form as 01 gen =+⎟
⎠
⎞
⎜
⎝
⎛ e
dr
dTrk
dr
d
r
& Select the wrong 
statement below. 
(a) the medium is of cylindrical shape. 
(b) the thermal conductivity of the medium is constant. 
(c) heat transfer through the medium is steady. 
(d) there is heat generation within the medium. 
(e) heat conduction through the medium is one-dimensional. 
Answer (b) thermal conductivity of the medium is constant 
 
 
 
2-153 Heat is generated in a long 0.3-cm-diameter cylindrical electric heater at a rate of 180 W/cm3. The heat flux at the 
surface of the heater in steady operation is 
(a) 12.7 W/cm2 (b) 13.5 W/cm2 (c) 64.7 W/cm2 (d) 180 W/cm2 (e) 191 W/cm2 
Answer (b) 13.5 W/cm2
 
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES 
screen. 
 
"Consider a 1-cm long heater:" 
L=1 [cm] 
e=180 [W/cm^3] 
D=0.3 [cm] 
V=pi*(D^2/4)*L 
A=pi*D*L "[cm^2]” 
Egen=e*V "[W]" 
Qflux=Egen/A "[W/cm^2]" 
 
“Some Wrong Solutions with Common Mistakes:” 
W1=Egen "Ignoring area effect and using the total" 
W2=e/A "Threating g as total generation rate" 
W3=e “ignoring volume and area effects” 
 
 
 
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2-154 Heat is generated in a 10-cm-diameter spherical radioactive material whose thermal conductivity is 25 W/m.°C 
uniformly at a rate of 15 W/cm3. If the surface temperature of the material is measured to be 120°C, the center temperature of 
the material during steady operation is 
(a) 160°C (b) 205°C (c) 280°C (d) 370°C (e) 495°C 
Answer (d) 370°C 
 
D=0.10 
Ts=120 
k=25 
e_gen=15E+6 
T=Ts+e_gen*(D/2)^2/(6*k) 
 
“Some Wrong Solutions with Common Mistakes:” 
W1_T= e_gen*(D/2)^2/(6*k) "Not using Ts" 
W2_T= Ts+e_gen*(D/2)^2/(4*k) "Using the relation for cylinder" 
W3_T= Ts+e_gen*(D/2)^2/(2*k) "Using the relation for slab" 
 
 
 
2-155 Consider a medium in which the heat conduction equation is given in its simplest form as 
 
t
T
r
Tr
rr ∂
∂
α
=⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
∂
∂ 11 2
2
 
(a) Is heat transfer steady or transient? 
(b) Is heat transfer one-, two-, or three-dimensional? 
(c) Is there heat generation in the medium? 
(d) Is the thermal conductivity of the medium constant or variable? 
(e) Is the medium a plane wall, a cylinder, or a sphere? 
(f) Is this differential equation for heat conduction linear or nonlinear? 
Answers: (a) transient, (b) one-dimensional, (c) no, (d) constant, (e) sphere, (f) linear 
 
 
 
2-156 An apple of radius R is losing heat steadily and uniformly from its outer surface to the ambient air at temperature T∞ 
with a convection coefficient of h, and to the surrounding surfaces at temperature Tsurr (all temperatures are absolute 
temperatures). Also, heat is generated within the apple uniformly at a rate of per unit volume. If Tgene& s denotes the outer 
surface temperature, the boundary condition at the outer surface of the apple can be expressed as 
(a) )()( 4surr
4 TTTTh
dr
dTk ss
Rr
−+−=− ∞
=
εσ (b) genss
Rr
eTTTTh
dr
dTk &+−+−=− ∞
=
)()( 4surr
4εσ 
(c) )()( 4surr
4 TTTTh
dr
dTk ss
Rr
−+−= ∞
=
εσ (d) genss
Rr
e
R
RTTTTh
dr
dTk &
2
3
4
surr
4
4
3/4)()(
π
πεσ +−+−= ∞
=
 
(e) None of them 
Answer: (a) )()( 4surr
4 TTTTh
dr
dTk ss
Rr
−+−=− ∞
=
εσ 
Note: Heat generation in the medium has no effect on boundary conditions. 
 
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2-157 A furnace of spherical shape is losing heat steadily and uniformly from its outer surface of radius R to the ambient air 
at temperature T∞ with a convection coefficient of h, and to the surrounding surfaces at temperature Tsurr (all temperatures are 
absolute temperatures). If To denotes the outer surface temperature, the boundary condition at the outer surface of the furnace 
can be expressed as 
 (a) )()( 4surr
4 TTTTh
dr
dTk oo
Rr
−+−=− ∞
=
εσ (b) )()( 4surr
4 TTTTh
dr
dTk oo
Rr
−−−=− ∞
=
εσ 
 
(c) )()( 4surr
4 TTTTh
dr
dTk oo
Rr
−+−= ∞
=
εσ (d) )()( 4surr
4 TTTTh
dr
dTk oo
Rr
−−−= ∞
=
εσ 
 
(e) )()()4( 4surr
42 TTTTh
dr
dTRk oo
Rr
−+−= ∞
=
εσπ 
 
Answer (a) )()( 4surr
4 TTTTh
dr
dTk oo
Rr
−+−=− ∞
=
εσ 
 
 
 
2-158 A plane wall of thickness L is subjected to convection at both surfaces with ambient temperature T∞1 and heat transfer 
coefficient h1 at inner surface, and corresponding T∞2 and h2 values at the outer surface. Taking thepositive direction of x to 
be from the inner surface to the outer surface, the correct expression for the convection boundary condition is 
(a) [ ]))0()0( 11 ∞−= TThdx
dTk (b) [ ]))()( 22 ∞−= TLThdx
LdTk 
(c) [ ]))0( 211 ∞∞ −=− TThdx
dTk (d) [ ))( 212 ∞∞ −=− TThdx
LdTk ] (e) None of them 
Answer (a) [ ]))0()0( 11 ∞−= TThdx
dT
k 
 
 
 
2-159 Consider steady one-dimensional heat conduction through a plane wall, a cylindrical shell, and a spherical shell of 
uniform thickness with constant thermophysical properties and no thermal energy generation. The geometry in which the 
variation of temperature in the direction of heat transfer be linear is 
(a) plane wall (b) cylindrical shell (c) spherical shell (d) all of them (e) none of them 
Answer (a) plane wall 
 
 
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2-160 Consider a large plane wall of thickness L, thermal conductivity k, and surface area A. The left surface of the wall is 
exposed to the ambient air at T∞ with a heat transfer coefficient of h while the right surface is insulated. The variation of 
temperature in the wall for steady one-dimensional heat conduction with no heat generation is 
(a) ∞
−
= T
k
xLhxT )()( (b) ∞+
= T
Lxh
kxT
)5.0(
)( (c) ∞⎟
⎠
⎞
⎜
⎝
⎛ −= T
k
xhxT 1)( (d) ∞−= TxLxT )()(
(e) ∞= TxT )(
Answer (e) ∞= TxT )(
 
 
 
2-161 The variation of temperature in a plane wall is determined to be T(x)=52x+25 where x is in m and T is in °C. If the 
temperature at one surface is 38ºC, the thickness of the wall is 
(a) 0.10 m (b) 0.20 m (c) 0.25 m (d) 0.40 m (e) 0.50 m 
Answer (c) 0.25 m 
 
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES 
screen. 
 
38=52*L+25 
 
 
 
2-162 The variation of temperature in a plane wall is determined to be T(x)=110 - 60x where x is in m and T is in °C. If the 
thickness of the wall is 0.75 m, the temperature difference between the inner and outer surfaces of the wall is 
(a) 30ºC (b) 45ºC (c) 60ºC (d) 75ºC (e) 84ºC 
Answer (b) 45ºC 
 
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES 
screen. 
 
T1=110 [C] 
L=0.75 
T2=110-60*L 
DELTAT=T1-T2 
 
 
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2-163 The temperatures at the inner and outer surfaces of a 15-cm-thick plane wall are measured to be 40ºC and 28ºC, 
respectively. The expression for steady, one-dimensional variation of temperature in the wall is 
(a) (b) 4028)( += xxT 2840)( +−= xxT (c) 2840)( += xxT 
(d) (e) 4080)( +−= xxT 8040)( −= xxT 
Answer (d) 4080)( +−= xxT
 
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES 
screen. 
 
T1=40 [C] 
T2=28 [C] 
L=0.15 [m] 
 
"T(x)=C1x+C2" 
C2=T1 
T2=C1*L+T1 
 
 
 
2-164 Heat is generated in a 3-cm-diameter spherical radioactive material uniformly at a rate of 15 W/cm3. Heat is dissipated 
to the surrounding medium at 25°C with a heat transfer coefficient of 120 W/m2⋅°C. The surface temperature of the material 
in steady operation is 
(a) 56°C (b) 84°C (c) 494°C (d) 650°C (e) 108°C 
Answer (d) 650°C 
 
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES 
screen. 
 
h=120 [W/m^2-C] 
e=15 [W/cm^3] 
Tinf=25 [C] 
D=3 [cm] 
V=pi*D^3/6 "[cm^3]" 
A=pi*D^2/10000 "[m^2]" 
Egen=e*V "[W]" 
Qgen=h*A*(Ts-Tinf) 
 
 
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2-96
 
2-165 Which one of the followings is the correct expression for one-dimensional, steady-state, constant thermal conductivity 
heat conduction equation for a cylinder with heat generation? 
(a) 
t
Tce
r
Trk
rr ∂
∂
=+⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
∂
∂ ρgen
1
& (b) 
t
T
k
e
r
Tr
rr ∂
∂
=+⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
∂
∂
α
11 gen& (c) 
t
T
r
Tr
rr ∂
∂
α
=⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
∂
∂ 11 
(d) 01 gen =+⎟
⎠
⎞
⎜
⎝
⎛
k
e
dr
dTr
dr
d
r
&
 (e) 0=⎟
⎠
⎞
⎜
⎝
⎛
dr
dTr
dr
d 
Answer (d) 01 gen =+⎟
⎠
⎞
⎜
⎝
⎛
k
e
dr
dTr
dr
d
r
&
 
 
 
 
2-166 A solar heat flux is incident on a sidewalk whose thermal conductivity is k, solar absorptivity is αsq& s and convective 
heat transfer coefficient is h. Taking the positive x direction to be towards the sky and disregarding radiation exchange with 
the surroundings surfaces, the correct boundary condition for this sidewalk surface is 
(a) ss qdx
dTk &α=− (b) )( ∞−=− TThdx
dTk (c) ss qTThdx
dTk &α−−=− ∞ )( 
(d) ss qTTh &α=− ∞ )( (e) None of them 
Answer (c) ss qTThdx
dTk &α−−=− ∞ )( 
 
 
 
2-167 Hot water flows through a PVC (k = 0.092 W/m⋅K) pipe whose inner diameter is 2 cm and outer diameter is 2.5 cm. 
The temperature of the interior surface of this pipe is 50oC and the temperature of the exterior surface is 20oC. The rate of 
heat transfer per unit of pipe length is 
(a) 77.7 W/m (b) 89.5 W/m (c) 98.0 W/m (d) 112 W/m (e) 168 W/m 
Answer (a) 77.7 W/m 
 
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES 
screen. 
 
do=2.5 [cm] 
di=2.0 [cm] 
k=0.092 [W/m-C] 
T2=50 [C] 
T1=20 [C] 
Q=2*pi*k*(T2-T1)/LN(do/di) 
 
 
 
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2-168 The thermal conductivity of a solid depends upon the solid’s temperature as k = aT + b where a and b are constants. 
The temperature in a planar layer of this solid as it conducts heat is given by 
(a) aT + b = x + C2 (b) aT + b = C1x2 + C2 (c) aT2 + bT = C1x + C2 
(d) aT2 + bT = C1x2 + C2 (e) None of them 
Answer (c) aT2 + bT = C1x + C2
 
 
 
2-169 Harvested grains, like wheat, undergo a volumetric exothermic reaction while they are being stored. This heat 
generation causes these grains to spoil or even start fires if not controlled properly. Wheat (k = 0.5 W/m⋅K) is stored on the 
ground (effectively an adiabatic surface) in 5-m thick layers. Air at 22°C contacts the upper surface of this layer of wheat 
with h = 3 W/m2⋅K. The temperature distribution inside this layer is given by 
 
2
0
1 ⎟
⎠
⎞
⎜
⎝
⎛−=
−
−
L
x
TT
TT
s
s 
where Ts is the upper surface temperature, T0 is the lower surface temperature, x is measured upwards from the ground, and L 
is the thickness of the layer. When the temperature of the upper surface is 24oC, what is the temperature of the wheat next to 
the ground? 
(a) 42oC (b) 54oC (c) 58oC (d) 63oC (e) 76°C 
Answer (b) 54oC 
 
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES 
screen. 
 
k=0.5 [W/m-K] 
h=3 [W/m2-K] 
L=5[m] 
Ts=24 [C] 
Ta=22 [C] 
To=(h*L/(2*k))*(Ts-Ta)+Ts 
 
 
 
2-170 The conduction equation boundary condition for an adiabatic surface with direction n being normal to the surface is 
(a) T = 0 (b) dT/dn = 0 (c) d2T/dn2 = 0 (d) d3T/dn3 = 0 (e) -kdT/dn = 1 
Answer (b) dT/dn = 0 
 
 
 
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2-171 Heat is generated uniformly in a 4-cm-diameter,12-cm-long solid bar (k = 2.4 W/m⋅ºC). The temperatures at the 
center and at the surface of the bar are measured to be 210ºC and 45ºC, respectively. The rate of heat generation within the 
bar is 
(a) 597 W (b) 760 W b) 826 W (c) 928 W (d) 1020 W 
Answer (a) 597 W 
 
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES 
screen. 
 
D=0.04 [m] 
L=0.12 [m] 
k=2.4 [W/m-C] 
T0=210 [C] 
T_s=45 [C] 
T0-T_s=(e*(D/2)^2)/(4*k) 
V=pi*D^2/4*L 
E_dot_gen=e*V 
 
"Some Wrong Solutions with Common Mistakes" 
W1_V=pi*D*L "Using surface area equation for volume" 
W1_E_dot_gen=e*W1_V 
T0=(W2_e*(D/2)^2)/(4*k) "Using center temperature instead of temperature difference" 
W2_Q_dot_gen=W2_e*V 
W3_Q_dot_gen=e "Using heat generation per unit volume instead of total heat generation as the result" 
 
 
 
 
2-172 .... 2-174 Design and Essay Problems 
 
 
 
 
 
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3-1
 
Solutions Manual 
for 
Heat and Mass Transfer: Fundamentals & Applications 
Fourth Edition 
 Yunus A. Cengel & Afshin J. Ghajar 
McGraw-Hill, 2011 
 
 
 
 
Chapter 3 
STEADY HEAT CONDUCTION 
 
 
 
 
 
 
 
 
PROPRIETARY AND CONFIDENTIAL 
 
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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 
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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. 
 
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3-2
 
&
re, 
mperature. 
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 Convection heat transfer through the wall is expressed as )( ∞−= TThA ss . 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. Therefo
at the outer surface, the temperature will be closer to the surrounding air te
Q
 
 
3-4C 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. 
 
 
3-5C (a) If the lateral surfaces of the rod are insulated, the heat transfer surface area of the cylindrical rod is the bottom or 
the top surface area of the rod, . (b) If the top and the bottom surfaces of the rod are insulated, the heat transfer 
area of the rod is the lateral surface area of the rod, 
4/2DAs π=
DLA π= . 
 
 
3-6C The thermal resistance of a medium represents the resistance of that medium against heat transfer. 
 
 
3-7C 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. 
 
 
3-8C 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 . )/(1 hARconv =
 
 
3-9C The convection and the radiation resistances at a surface are parallel since both the convection and radiation heat 
transfers occur simultaneously. 
 
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preparation. If you are a student using this Manual, you are using it without permission. 
3-3
 
r coefficient is eh
ce will be
3-10C For a surface of A at which the convection and radiation heat transfer coefficients are hconv rad and , the single 
equivalent heat transfe radconvqv hh += when the medium and the surrounding surfaces are at the same 
temperature. Then the equivalent thermal resistan
h
 
 )/(1 AhR eqveqv = . 
 
 
3-11C The thermal resistance network associated with a five-layer composite wall involves five single-layer resistances 
connected in series. 
 
 
3-12C Once the rate of heat transfer is known, the temperature drop across any layer can be determined by multiplying 
heat transfer rate by the thermal resistance across that layer, 
Q&
layerlayer RQT &=∆ 
 
 
3-13C The temperature of each surface in this case can be determined from 
 
)(/)(
)(/)(
22222222
11111111
∞−∞∞−∞
−∞∞−∞∞
+=⎯→⎯−=
−=⎯→⎯−=
ssss
ssss
RQTTRTTQ
RQTTRTTQ
&&
&&
 
where is the thermal resistance between the environment iR −∞ ∞ and surface i. 
 
 
3-14C Yes, it is. 
 window, and thus the heat 
sfer rate will be smaller relative to the one which consists of a single 8 mm thick glass sheet. 
nsfer and slow down the heat gain of the drink wrapped in a 
lanket. Therefore, the drink left on a table will warm up faster. 
 
 
 
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
tran
 
 
3-16C The blanket will introduce additional resistance to heat tra
b
 
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preparation. If you are a student using this Manual, you are using it without permission. 
3-4
 
A 
3-17 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 
significant temperature gradients will exist in the direction from the indoors to the 
outdoors. 3 Thermal conductivity is constant. 
Wall 
5°C
Q&
L= 0.25 m
Properties The thermal conductivity is given to be k = 0.8 W/m⋅°C. 
Analysis The surface area of the wall and the rate of heat loss through the wall 
are 14°C 
 = 2m 18m) 6(m) 3( =×
 W518=°−°⋅=−=
m25.0
C)514()m C)(18 W/m8.0( 221
L
TT
kAQ& 
 
 
 
 
3-18 Heat is transferred steadily to the boiling water in an aluminum pan. The inner surface temperature of the bottom of the 
pan is given. The boiling heat transfer coefficient and the outer surface temperature of the bottomof the pan are to be 
determined. 
Assumptions 1 Steady operating conditions exist. 2 Heat transfer is one-dimensional since the thickness of the bottom of the 
pan is small relative to its diameter. 3 The thermal conductivity of the pan is constant. 
Properties The thermal conductivity of the aluminum pan is given to be k = 237 W/m⋅°C. 
Analysis (a) The boiling heat transfer coefficient is 
2
22
m 0491.0
4
m) 25.0(
4
===
ππDAs 
95°C 
 108°C 
800 W 
C. W/m1254 2 °=
°−− ∞ C)95108)(m 0491.0()( 2TTA ss
==
−= ∞
 W800
)(
Qh
TThAQ ss
&
&
 
The ou r surface temperature of the bottom of the pan is 
 
0.5 cm
(b) te
C108.3°=
°⋅
°=+=
−
=
)m C)(0.0491 W/m237(
m) 005.0 W)(800(+C108
21,,
,,
kA
LQTT
L
TT
kAQ
innersouters
innersouters
&
&
 
 
 
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preparation. If you are a student using this Manual, you are using it without permission. 
3-5
 
3-19 The two surfaces of a window are maintained at specified temperatures. The rate of heat loss through the window and 
the inner surface temperature are to be determined. 
Assumptions 1 Heat transfer through the window is steady since the surface temperatures remain constant at the specified 
values. 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 is constant. 4 Heat transfer by radiation is negligible. 
Properties The thermal conductivity of the glass is given to be k = 0.78 W/m⋅°C. 
Analysis The area of the window and the individual resistances are 
2m 6.3m) 4.2(m) 5.1( =×=A 
L 
Glass 
T1
Q&
C/W 04103.001111.000214.002778.0
C/W 01111.0
)m 6.3(C). W/m25( 222
2,o °=
°
=== conv Ah
RR 11
C/W 002140
)m 6.3(C) W/m.78.0(
m 006.0
C/W 02778.011
2,1,
2
1
glass
°=++=
++=
°=
°
==
°====
convglassconvtal RRRR
Ak
LR
RR
he steady rate of heat transfer through window glass is then 
 
)m 6.3(C). W/m10( 221
1,i
°
conv Ah
.
to
RglassRi Ro
T T∞1 T∞2
 W707=
°
°−−
=
−
= ∞∞
C/W 04103.0
C)]5(24[21
totalR
TT
Q& 
The inner surface temperature of the window glass can be determined from 
 C4.4°=°−°=−=⎯→⎯−= ∞∞ C/W) 8 W)(0.0277707(C241,11
1,
11
conv
conv
RQTT
R
TT
Q && 
 
 
 
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3-6
 
3-20 A double-pane window consists of two layers of glass separated by a stagnant air space. For specified indoors and 
outdoors temperatures, the rate of heat loss through the window and the inner surface temperature of the window are to be 
determined. 
Assumptions 1 Heat transfer through the window is steady since the indoor and 
outdoor temperatures remain constant at the specified values. 2 Heat transfer is 
one-dimensional since any significant temperature gradients will exist in the 
direction from the indoors to the outdoors. 3 Thermal conductivities of the glass 
and air are constant. 4 Heat transfer by radiation is negligible. 
Air 
Properties The thermal conductivity of the glass and air are given to be kglass = 
0.78 W/m⋅°C and kair = 0.026 W/m⋅°C. 
Analysis The area of the window and the individual resistances 
are 
 2m 6.3m) 4.2(m) 5.1( =×=A
 
C/W 16924.0
01111.012821.0)00107.0(202778.02
C/W 01111.0
)m 6.3(C). W/m25(
11
C/W 12821.0
)m 6.3(C) W/m.026.0( 22
2
2 °=
°
=== air Ak
RR m 012.0
C/W 00107.0
) 6.3(C) W/m.78.0(
m 003.0
C/W 02778.011
2,211,
o
2o2
2
2,o
2
1
1
glass31
1,i
°=
+++=+++=
====
°=
°
====
°====
convconvtotal
conv
conv
RRRRR
Ah
RR
L
Ak
L
RRR
R
 
The steady rate of heat transfer through window glass then becomes 
 
R1 R2 R3Ri Ro
T∞1 T∞2R
)m 6.3(C). W/m10( 221 °Ah
m
 W154=
°
°−−
=
−
= ∞∞
C/W16924.0
C)]5(21[21
totalR
TT
Q& 
The inner surface temperature of the window glass can be determined from 
 C16.7°°−°=−=⎯→⎯
−
= ∞
∞ =C/W)8 W)(0.0277154(C211,11
1,
11
conv
conv
RQTT
R
TT
Q && 
 
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3-7
 
3-21 A double-pane window consists of two layers of glass separated by an evacuated space. For specified indoors and 
outdoors temperatures, the rate of heat loss through the window and the inner surface temperature of the window are to be 
determined. 
Assumptions 1 Heat transfer through the window is steady since the indoor and outdoor temperatures remain constant at the 
specified values. 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 glass is constant. 4 Heat transfer by radiation is negligible. 
Properties The thermal conductivity of the glass is given to be kglass = 0.78 W/m⋅°C. 
Analysis Heat cannot be conducted through an evacuated space since the 
thermal conductivity of vacuum is zero (no medium to conduct heat) and thus 
its thermal resistance is zero. Therefore, if radiation is disregarded, the heat 
transfer through the window will be zero. Then the answer of this problem is 
zero since the problem states to disregard radiation. 
Vacuum 
Discussion In reality, heat will be transferred between the glasses by 
radiation. We do not know the inner surface temperatures of windows. In 
order to determine radiation heat resistance we assume them to be 5°C and 
15°C, respectively, and take the emissivity to be 1. Then individual 
resistances are 
 2m 6.3m) 4.2(m) 5.1( =×=A
 
C/W 09505.0
01111.005402.0)00107.0(202778.02
C/W 01111.0
)m 6.3(C). W/m25(
11
C/W 05402.0
]278288][278288)[m 6.3().K W/m1067.5(1 3222428
===
°=
++× −
RR
K
1 
1
C/W 00107.0
)m 6.3(C) W/m.78.0(
m 003.0
C/W 02778.011
2,11,
o
2o2
2
2,o
2
1
1
glass31
1,i
°=
+++=+++=
=
=
°=
°
====
°====
convradconvtotal
conv
conv
RRRRR
Ah
Ak
L
RRR
R
 
The steady rate of heat transfer through window glass then becomes 
 
R1 Rrad R3Ri Ro
T∞1 T∞2R
)m 6.3(C). W/m10( 221 °Ah
))(( 22 ++
=
surrssurrs
rad
TTTTA
R
εσ
 W274=
°
°−−
=
−
= ∞∞
C/W09505.0
C)]5(21[21
totalR
TT
Q& 
The inner surface temperature of the window glass can be determined from 
 C13.4°=°−°=−=⎯→⎯
−
= ∞
∞ C/W)8 W)(0.0277274(C211,11
1,
11
conv
conv
RQTT
R
TT
Q && 
Similarly, the inner surface temperatures of the glasses are calculated to be 13.1 and -1.7°C (we had assumed them to be 15 
and 5°C when determining the radiation resistance). We can improve the result obtained by reevaluating the radiation 
resistance and repeating the calculations. 
 
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3-8
 
3-22 Prob. 3-20 is reconsidered. The rate of heat transfer through the window as a function of the width of air space is 
to be plotted. 
Analysis The problem is solved using EES, and the solution is given below. 
 
"GIVEN" 
A=1.5*2.4 [m^2] 
L_glass=3 [mm] 
k_glass=0.78 [W/m-C] 
L_air=12 [mm] 
T_infinity_1=21 [C] 
T_infinity_2=-5 [C] 
h_1=10 [W/m^2-C] 
h_2=25 [W/m^2-C] 
 
"PROPERTIES" 
k_air=conductivity(Air,T=25) 
 
"ANALYSIS" 
R_conv_1=1/(h_1*A) 
R_glass=(L_glass*Convert(mm, m))/(k_glass*A) 
R_air=(L_air*Convert(mm, m))/(k_air*A) 
R_conv_2=1/(h_2*A) 
R_total=R_conv_1+2*R_glass+R_air+R_conv_2 
Q_dot=(T_infinity_1-T_infinity_2)/R_total 
 
 
Lair 
[mm] 
Q 
[W] 
2 
4 
6 
8 
10 
12 
14 
16 
18 
20 
414 
307.4 
244.5 
202.9 
173.4 
151.4 
134.4 
120.8 
109.7 
100.5 
2 4 6 8 10 12 14 16 18 20
100
150
200250
300
350
400
450
Lair [mm]
Q
 [
W
]
 
 
 
 
 
 
 
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3-9
 
A 
3-23E 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 
 = 2ft 1530)935950(2 =×+× Wall 
T2
Q& 
L
The rate of heat loss during the daytime is 
 Btu/h 6120
ft 1
F)4555(
)ft F)(1530ftBtu/h 40.0( 221day =
°−
°⋅⋅=
−
=
L
TT
kAQ& 
The rate of heat loss during nighttime is 
T1
Btu/h 240,12
ft 1
C)3555(
)ft F)(1530ftBtu/h 40.0( 2
21
night L 
=
°−
°⋅⋅=
−
=
TT
kAQ&
 
The amount of heat loss from the house that night will be 
 
Btu 232,560=
+=+=∆=⎯→⎯= 1410 nightday QQtQQ
Q
Q &&&&
∆
 
)Btu/h 240,12(h) 14()Btu/h 6120(h) 10(
t 
hen the cost of this heat loss for that day becomes T
$6.13== kWh)/09.0)($kWh 3412/560,232(Cost 
 
 
 
 
3-24 A cylindrical resistor on a circuit board dissipates 0.15 W of power steadily in
heat dissipated in 24 h, the surface heat flux, and the surface temperature of the resist
 a specified environment. The amount of 
or are to be determined. 
ssumptions 1 Steady operating conditions exist. 2 Heat is transferred uniformly from all surfaces of the resistor. 
pates during a 24-hour period is 
h) W)(2415.0(tQQ & 
A
Analysis (a) The amount of heat this resistor dissi
Resistor 
0.15 W 
Q& 
 Wh3.6==∆ =
(b) The heat flux on the surface of the resistor is 
2
22
m 000127.0m m)(0.012 003.0(
4
m) 003.0(2
4
2 +=+= ππDLDAs ) =
ππ 
 2 W/m1179===
2m 000127.0
 W15.0
sA
Q
q
&
& 
(c) The surface temperature of the resistor can be determined from 
 C166°=
°⋅
+°=+=⎯→⎯−= ∞∞ )m 7C)(0.00012 W/m(9
 W15.0C35)( 22
s
sss hA
QTTTThAQ
&
& 
 
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3-10
 
3-25 A very thin transparent heating element is attached to the inner surface of an automobile window for defogging 
purposes, the inside surface temperature of the window is to be determined. 
Assumptions 1 Steady operating conditions exist. 2 Heat transfer through the window is one-dimensional. 3 Thermal 
properties are constant. 4 Heat transfer by radiation is negligible. 5 Thermal resistance of the thin heating element is 
negligible. 
Properties Thermal conductivity of the window is given 
to be k = 1.2 W/m · °C. 
Analysis The thermal resistances are 
 
Ah
R
i
i
1
= 
Ah
R
o
o
1
= and 
kA
LR =win 
From energy balance and using the thermal resistance 
concept, the following equation is expressed: 
 
o
o
h
i
i −∞,
RR
TT
Aq
R
TT
+
−
=+ ∞
win
,11 & 
or 
)/(1)/()/(1
,11,
AhkAL
TT
Aq
Ah
TT
o
o
h
i
i
+
−
=+
− ∞∞ & 
o
o
h
i
i
hkL
TT
q
h
TT
/1//1
,11,
+
−
=+
− ∞∞ & 
 
C) W/m100/1(C) W/m2.1/m 005.0(C W/m15/1 2 +°⋅°⋅
)C 5( W/m1300C 22 121 °−−=+−° TT 2 °⋅
 equation: 
(2 -T_1)/(1/15)+1300=(T_1-(-5))/(0.005/1.2+1/100) 
Discussion In actuality, the ambient temperature and the convective heat transfer coefficient outside the automobile vary 
with weather conditions and the automobile speed. To maintain the inner surface temperature of the window, it is necessary 
to vary the heat flux to the heating element according to the outside condition. 
Copy the following line and paste on a blank EES screen to solve the above
 2
Solving by EES software, the inside surface temperature of the window is 
 C 14.9 °=1T 
 
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3-11
 
3-26 A process of bonding a transparent film onto a solid plate is taking place inside a heated chamber. The temperatures 
inside the heated chamber and on the transparent film surface are to be determined. 
Assumptions 1 Steady operating conditions exist. 2 Heat transfer is one-dimensional. 3 Thermal properties are constant. 4 
Heat transfer by radiation is negligible. 5 Thermal contact resistance is negligible. 
Properties The thermal conductivities of the 
transparent film and the solid plate are given to 
be 0.05 W/m · °C and 1.2 W/m · °C, 
respectively. 
Analysis The thermal resistances are 
hA
R 1conv = 
Ak
L
R
f
f
f = 
and 
Aks
s
Using the thermal resistance conc
LR s= 
ept, the 
llowing e uation is expressed: fo q
 
s
b
f
b
R
TT
RR
TT 2
conv
−
=
+
−∞ 
R
inside the cham
earranging and solving for the temperature 
ber yields 
( ) b
f
f 
sss kLR ⎝/
b
bf
b T
k
L
h
TTTRRTTT +⎟
⎟
⎠
⎞
⎜
⎜
⎛
+
−
=++
−
=
12
conv
2 ∞
C 127 °=°+⎟
⎠
⎞
⎜
⎝
⎛
°⋅
+
°⋅°⋅
°−
=∞ C 70C W/m05.0
m 001.0
C W/m70
1
C
 
 W/m2.1/m 013.0
C )5270(
2T 
he surfac temperature of the transparent film is T e 
s
b
f
b
R
TT
R
TT 21 −=− 
 b
f
f
ss
b
bf
s
b T
k
L
kL
TTTR
R
TTT +⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛−
=+
−
=
/
22
1 
C 103 °=°+⎟
⎠
⎞
⎜
⎝
⎛
°⋅°⋅
°−
= C 70
C W/m05.0
m 001.0
C W/m2.1/m 013.0
C )5270(
1T 
Discussion If a thicker transparent film were to be bonded on the solid plate, then the inside temperature of the heated 
chamber would have to be higher to maintain the temperature of the bond at 70 °C. 
 
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3-12
 
3-27 A power transistor dissipates 0.15 W of power steadily in a specified environment. The amount of heat dissipated in 24 
h, the surface heat flux, and the surface temperature of the resistor are to be determined. 
Assumptions 1 Steady operating conditions exist. 2 Heat is transferred uniformly from all surfaces of the transistor. 
Analysis (a) The amount of heat this transistor dissipates during a 
24-hour period is 
Air, 
30°C kWh 0.0036===∆= Wh6.3h) W)(2415.0(tQQ & 
(b) The heat flux on the surface of the transistor is 
2
2
2
m 0001021.0m) m)(0.004 005.0(
4
m) 005.0(
2
4
2
=+=
+=
π
π
ππ DLDAs
 
Power 
Transistor 
0.15 W 
2 W/m1469=== 2m 0001021.0sA
q& W15.0Q
&
(c) The surface temperature of the transistor can be determined from 
C111.6°=
°⋅
+°=+=⎯→⎯−= ∞∞ )m 21C)(0.00010 W/m(18
 W15.0C30)( 22
s
sss hA
QTTTThAQ
&
& 
 
 
 
 
3-28 A circuit board houses 100 chips, each dissipating 0.06 W. The surface heat flux, the surface temperature of the chips, 
and the thermal resistance between the surface of the board and the cooling medium are to be determined. 
 1 Steady operating conditions exist. 2 Assumptions
transferred uniformly from the entire front surface. 
Heat transfer from the back surface of the board is negligible. 2 Heat is 
nalysis (a The heat flux on the surface of the circuit board is A )
Chips 
Ts
Q& 
T∞ 
2m 0216.0m) m)(0.18 12.0( ==sA 
2 W278===
2m 0216.0sA
q& /m× W)06.0100(Q
&
 
e sur ce temperature of the chips is 
 
 
(b) Th fa
−= ∞ )( ss TThAQ&
 
C67.8°=°⋅
×
°=+= ∞
)m 0216.0)(C W/m10(
 W)06.0100(+C40
22
s
s hA
QTT
& 
(c) The thermal resistance is 
 C/W4.63°=
°⋅
==
)m 0216.0(C) W/m10(
11
22
s
conv hA
R 
 
PROPRIETARY MATERIAL. © 2011 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. 
3-13
 
 
3-29 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. 
Qconv
Tskin
 
Qrad
Properties The thermal conductivity of the tissue near the skin is given to 
be k = 0.3 W/m⋅°C. 
Analysis The skin temperature can be determined directly from 
C35.5°=
°⋅
−°=−=
−
=
)m C)(1.7 W/m3.0(
m) 005.0 W)(150(
C37
21
1
kA
LQ
TT
L
TT
kAQ
skin
skin
&
&
 
 
 
 
 
3-30 A double-pane window is considered. The rate of heat loss through the window and the temperature difference across 
 
Properties The thermal conductivities of glass and air are given to be 0.78 W/m⋅K and 0.025 W/m⋅K, respectively. 
Analysis (a) The rate of heat transfer through the window is determined to be 
the largest thermal resistance are to be determined. 
Assumptions 1 Steady operating conditions exist. 2 Heat transfer coefficients are constant. 
[ ]
[ ] W210=
++++
°×
°⋅°⋅°⋅°⋅°⋅ C W/m20C W/m78.0C W/m025.0C W/m78.0C W/m40
2
22
=
++++
°×
=
++++
∆
=
05.0000513.02.0000513.0025.0
C(-20)-20)m 5.11(
1m 004.0m 005.0m 004.01
C(-20)-20)m 5.11(
11
2
og
g
a
a
g
g
i hk
L
k
L
k
L
h
TAQ&
 
b) Noting that the largest resistance is through the air gap, the temperature difference across the air gap is determined from 
 
(
C28°=
×°⋅
===∆
)m 5.11(C) W/m025.0(
m 005.0 W)210(
2Ak
L
QRQT
a
a
aa
&& 
 
 
 
PROPRIETARY MATERIAL. © 2011 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. 
3-14
 
3-31E A wall is constructed of two layers of sheetrock with fiberglass insulation in between. The thermal resistance of the 
wall and its R-value of insulation are to be determined. 
Assumptions 1 Heat transfer through the wall is one-dimensional. 2 Thermal conductivities are constant. 
Properties The thermal conductivities are given to be 
ksheetrock = 0.10 Btu/h⋅ft⋅°F and kinsulation = 0.020 Btu/h⋅ft⋅°F. 
L1 L2 L3 
R1 R2 R3
 
Analysis (a) The surface area of the wall is not given and thus we consider a unit 
surface area (A = 1 ft2). Then the R-value of insulation of the wall becomes 
equivalent to its thermal resistance, which is determined from. 
F.h/Btu.ft 30.17 2 °=+×=+=
°=
°
===
°=
°
====
17.29500.022
F.h/Btu.ft 17.29
F)Btu/h.ft. 020.0(
ft 12/7
F.h/Btu.ft 500.0
F)Btu/h.ft. 10.0(
ft 12/6.0
21
2
2
2
2
2
1
1
31
RRR
k
L
RR
k
L
RRR
total
fiberglass
sheetrock
 
(b) Therefore, this is approximately a R-30 wall in English units. 
 
 
 
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preparation. If you are a student using this Manual, you are using it without permission. 
3-15
 
 
3-32 The roof of a house with a gas furnace consists of a concrete that is losing heat to the outdoors by radiation and 
convection. The rate of heat transfer through the roof and the money lost through the roof that night during a 14 hour period 
are to be determined. 
Assumptions 1 Steady operating conditions exist. 2 The emissivity 
and thermal conductivity of the roof are constant. Q&
Tin=20°C 
Tsky = 100 K
 
Tair =10°C Properties The thermal conductivity of the concrete is given to be 
k = 2 W/m⋅°C. The emissivity of both surfaces of the roof is given 
to be 0.9. 
L=15 cm 
Analysis When the surrounding surface temperature is different than the ambient 
temperature, the thermal resistances network approach becomes cumbersome in 
problems that involve radiation. Therefore, we will use a different but intuitive 
approach. 
 In steady operation, heat transfer from the room to the roof (by convection 
and radiation) must be equal to the heat transfer from the roof to the surroundings 
(by convection and radiation), that must be equal to the heat transfer through the 
roof by conduction. That is, 
 rad+conv gs,surroundin toroofcond roof,rad+conv roof, toroom QQQQ &&&& === 
Taking the inner and outer surface temperatures of the roof to be Ts,in and Ts,out , respectively, the quantities above can be 
expressed as 
[ ]4,44282
,
224
,
4
, rad+conv roof, toroom
K) 273(K) 27320()K W/m1067.5)(m 300)(9.0( 
C))(20m C)(300 W/m5()()(
+−+⋅×+
°−°⋅=−+−=
−
ins
insinsroominsroomi
T
TTTATTAhQ σε&
 
m 15.0
)m 300)(C W/m2( ,,2,,cond roof,
outsinsoutsins TT
L
TT
kAQ
−
°⋅=
−
=& 
[ ]44,4282m 300)(9.0( +
,
2244
,, rad+conv surr, toroof
K) 100(K) 273()K W/m1067.5)(
C)10)(m C)(300 W/m12()()(
−+⋅×
°−°⋅=−+−=
−
outs
outssurroutssurroutso
T
TTTATTAhQ σε&
 
lving the quations above simultaneously gives 
he total amount of natural gas consumption during a 14-hour period is 
So e
 C1.2and , , out,, °−=°== sins TTQ C7.3 W37,440& 
T
 therms36.22
kJ 1080.080.080.0 ⎜
⎜
⎝
===Q totalgas 5,500
 therm1)s 360014)(kJ/s 440.37(
=⎟⎟
⎠
⎞⎛×∆tQQ & 
inally, the money lost through the roof during that period is 
 
F
$26.8== )therm/20.1$ therms)(36.22(lostMoney 
 
PROPRIETARY MATERIAL. © 2011 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. 
3-16
 
3-33 An exposed hot surface of an industrial natural gas furnace is to be insulated to reduce the heat loss through that section 
of the wall by 90 percent. The thickness of the insulation that needs to be used is to be determined. Also, the length of time it 
will take for the insulation to pay for itself from the energy it saves will be determined. 
Assumptions 1 Heat transfer through the wall is steady and one-dimensional. 2 Thermal conductivities are constant. 3 The 
furnace operates continuously. 4 The given heat transfer coefficient accounts for the radiation effects. 
Properties The thermal conductivity of the glass wool insulation is given to be k = 0.038 W/m⋅°C. 
Analysis The rate of heat transfer without insulation is 
Insulation 
L 
Ts
Rinsulation Ro
 2m 3m) m)(1.5 2( ==A
 W2340C)32110)(m 3(C) W/m10()( 22 =°−°⋅=−= ∞TThAQ s& 
In order to reduce heat loss by 90%, the new heat transfer rate and 
thermal resistance must be T∞
 
C/W 333.0C)32110( °=°−=∆=⎯→⎯∆= TRTQ total&
 
 W234
 W234 W234010.0 =×
QR
Q
total
&
&
and in order to have this thermal resistance, the thickness of insulation must be 
 
=
cm 3.4==
°°⋅
m 034.0
)m C)(3 W/m.038.0()m C)(3 W/m10( 222
L
°=+=
+=+=
C/W 333.01
1
conv
L
kA
L
hA
RRR insulationtotal
 
oting tha eat is saved at a rate of 0.9×2340 = 2106 W and the furnace operates continuously and thus 365×24 = 8760 h 
 furnace efficiency is 78%, the amount of natural gas saved per year is 
 
N t h
per year, and that the
 therms1.807
kJ 105,500
 therm1
h 1
s 3600
0.78
h) kJ/s)(8760 106.2(SavedEnergy =⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛=
∆
=
Efficiency
tQsaved& 
he money aved is 
gy == 
he insulation will pay for its cost of $250 in 
 
T s
Ener(savedMoney= year)(per 8.887$)1.10/therm therms)($1.807(energy) oft Saved)(Cos
T
yr 0.282===
$887.8/yr
$250
savedMoney 
spent Money periodPayback 
which is equal to 3.4 months. 
 
 
PROPRIETARY MATERIAL. © 2011 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. 
3-17
 
∞TThAQ s& 
3-34 An exposed hot surface of an industrial natural gas furnace is to be insulated to reduce the heat loss through that section 
of the wall by 90 percent. The thickness of the insulation that needs to be used is to be determined. Also, the length of time it 
will take for the insulation to pay for itself from the energy it saves will be determined. 
Assumptions 1 Heat transfer through the wall is steady and one-dimensional. 2 Thermal conductivities are constant. 3 The 
furnace operates continuously. 4 The given heat transfer coefficients accounts for the radiation effects. 
 Properties The thermal conductivity of the expanded perlite insulation is given to be k = 0.052 W/m⋅°C. 
Analysis The rate of heat transfer without insulation is 
Insulation 
L 
Ts
Rinsulation Ro
 2m 3m) m)(1.5 2( ==A
 = W2340C)32110)(m 3(C) W/m10()( 22 =°−°⋅=−
In order to reduce heat loss by 90%, the new heat transfer rate and thermal 
resistance must be T∞
 
C/W 333.0C)32110( °=°−=∆=⎯→⎯∆= TRTQ total&
 
 W234
 W234 W234010.0 =×
QR
Q
total
&
&
and in order to have this thermal resistance, the thickness of insulation must be 
 
=
cm 4.7==
°⋅°⋅
m 047.0
)m C)(3 W/m052.0()m C)(3 W/m10( 222
L
°=+=
+=+=
C/W 333.01
1
conv
L
kA
L
hA
RRR insulationtotal
 
oting tha eat is saved at a rate of 0.9×2340 = 2106 W and the furnace operates continuously and thus 365×24 = 8760 h 
 furnace efficiency is 78%, the amount of natural gas saved per year is 
 
N t h
per year, and that the
 therms1.807
kJ 105,500
 therm1
h 1
s 3600
0.78
h) kJ/s)(8760 106.2(SavedEnergy =⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛=
∆
=
Efficiency
tQsaved& 
oney
gy 
The m saved is 
Ener(savedMoney year)(per 8.887$)1.10/therm therms)($1.807(energy) oft Saved)(Cos= == 
he insulation will pay for its cost of $250 in 
 
T
yr 0.282===
$887.8/yr
$250
savedMoney 
spent Money periodPayback 
which is equal to 3.4 months. 
 
 
PROPRIETARY MATERIAL. © 2011 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. 
3-18
 
3-35 Prob. 3-33 is reconsidered. The effect of thermal conductivity on the required insulation thickness is to be 
investigated. 
Analysis The problem is solved using EES, and the solution is given below. 
 
"GIVEN" 
A=2*1.5 [m^2] 
T_s=110 [C] 
T_infinity=32 [C] 
h=10 [W/m^2-C] 
k_ins=0.038 [W/m-C] 
f_reduce=0.90 
 
"ANALYSIS" 
Q_dot_old=h*A*(T_s-T_infinity) 
Q_dot_new=(1-f_reduce)*Q_dot_old 
Q_dot_new=(T_s-T_infinity)/R_total 
R_total=R_conv+R_ins 
R_conv=1/(h*A) 
R_ins=(L_ins*Convert(cm, m))/(k_ins*A) "L_ins is in cm" 
 
 
kins 
[W/m.C] 
Lins 
[cm] 
0.02 
0.025 
0.03 
0.035 
0.04 
0.045 
0.05 
0.055 
0.06 
0.065 
0.07 
0.075 
0.08 
1.8 
2.25 
2.7 
3.15 
3.6 
4.05 
4.5 
4.95 
5.4 
5.85 
6.3 
6.75 
7.2 0.02 0.03 0.04 0.05 0.06 0.07 0.08
1
2
3
4
5
6
7
8
kins [W/m-C]
L i
ns
 [
cm
]
 
 
 
 
 
PROPRIETARY MATERIAL. © 2011 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. 
3-19
 
 
3-36 Two of the walls of a house have no windows while the other two walls have single- or double-pane windows. The 
average rate of heat transfer through each wall, and the amount of money this household will save per heating season by 
converting the single pane windows to double pane windows are to be determined. 
Assumptions 1 Heat transfer through the window is steady since the indoor and outdoor temperatures remain constant at the 
specified values. 2 Heat transfer is one-dimensional since any significant temperature gradients will exist in the direction 
from the indoors to the outdoors. 3 Thermal conductivities of the glass and air are constant. 4 Heat transfer by radiation is 
disregarded. 
Properties The thermal conductivities are given to be k = 0.026 W/m⋅°C for air, and 0.78 W/m⋅°C for glass. 
Analysis The rate of heat transfer through each wall can be determined by applying thermal resistance network. The 
convection resistances at the inner and outer surfaces are common in all cases. 
Walls without windows: 
Wall
C/W 06271.0001389.005775.0003571.0
C/W001389.0
)m 410(C) W/m18( 2
==
o
o Ah
R 11
C/W 05775.0
)m 410(
C/Wm 31.2
C/W 003571.0
)m 410(C) W/m7(
11
walltal
2
2
2
wall
wall
22
°=++=++=
°=
×°⋅
°=
×
°⋅
=
−
==
°=
×°⋅
==
oi
i
i
RRRR
A
valueR
kA
L
R
Ah
 
R
Q&
L
 
to
Then 
RwallRi Ro
 W255.1=
°
°−
=
−
= ∞∞
C/W06271.0
C)824(21
totalR
TT
Q& 
Wall with single pane windows: 
C/W 003063.0000694.0000583.0001786.0
C/W 000694.0
)m 420(C) W/m18(
11
C/W 000583.0
002968.0
15
033382.0
11511
C/W 002968.0
m)8.12.1)(C W/m78.0(
m 005.0
C/W 033382.0
m )8.12.1(5)420(
C/Wm 31.2
C/W 001786.0
)m 420(C) W/m7(
11
eqvtotal
22
o
glasswalleqv
2o2
glass
glass
2
2
wall
wall
22
°=++=++=
°=
×°⋅
==
=→+=+=
°=
×⋅
==
°=
×−×
°⋅
=
−
==
°=
×°⋅
==
oi
o
o
eqv
i
i
RRRR
Ah
R
R
RRR
kA
L
R
A
valueR
kA
L
R
Ah
R
 
Ri Rwall Ro
Rglass
Then 
 W5224=
°
°−
=
−
= ∞∞
C/W003063.0
C)824(
total
21
R
TT
Q& 
 
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preparation. If you are a student using this Manual, you are using it without permission. 
3-20
 
4th wall with double pane windows: 
 
Rwall
Rglass Rair Rglass
Ri 
 
 
Ro 
 
 
C/W 023197.0000694.0020717.0001786.0
C/W 020717.0
27303.0
1511511
eqv
°=⎯→⎯+=+= R
R 033382.0
C/W 27303.0267094.0002968.022
C/W 002968.0
m)12.1)(C W/m78.0(
m 005.0
C/W 033382.0
m)8.12.1(5)420(
C/Wm 31.2
eqvtotal
eqv
windowwall
airglass
22
glass
glass
2
2
wall
wall
°=++=++=
°=+×=+=
°=
×°⋅
==
°=
×−×
°⋅
=
−
==
oi RRRR
RR
RRR
kA
L
R
A
valueR
kA
L
R
8.
C/W 267094.0
m)8.12.1)(C W/m026.0(
m 015.0
2o2
air
air °=
×⋅
==
kA
L
R 
window
Then 
 W690=
°
°−
=
−
= ∞∞
C/W023197.0
C)824(
total
21
R
TT
Q& 
The rate of heat transfer which will be saved if the single pane windows are converted to double pane windows is 
 
The amount of energy and money saved during a 7-month long heating season by switching from single pane to double pane 
windows become 
 
 Money savings = (Energy saved)(Unit cost of energy) = (22,851 kWh)($0.08/kWh) = $1828 
 
 W45346905224
pane
double
pane
singlesave =−=−= QQQ &&& 
kWh 22,851=h) 2430kW)(7 534.4( ××=∆= tQQ savesave & 
 
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preparation. If you are a student using this Manual, you are using it without permission. 
3-21
 
,− outsroomo TTAhQ& 
room and the refrigerated space can be expressed as 
3-37 The wall of a refrigerator is constructed of fiberglass insulation sandwiched between two layers of sheet metal. The 
minimum thickness of insulation that needs to be used in the wall in order to avoid condensation on the outer surfaces is to 
be determined. 
Assumptions 1 Heat transfer through the refrigerator walls is steady since the temperatures of the food compartment and the 
kitchen air remain constant at the specified values. 2 Heat transfer is one-dimensional. 3 Thermal conductivities are constant. 
4 Heat transfer coefficients account for the radiation effects. 
Properties The thermal conductivities are given tobe k = 15.1 W/m⋅°C for sheet metal and 0.035 W/m⋅°C for fiberglass 
insulation. 
Analysis The minimum thickness of insulation can be determined by assuming 
the outer surface temperature of the refrigerator to be 20°C. In steady operation, 
the rate of heat transfer through the refrigerator wall is constant, and thus heat 
transfer between the room and the refrigerated space is equal to the heat transfer 
between the room and the outer surface of the refrigerator. Considering a unit 
surface area, 
1 mm L 1 mm
insulation 
)(=
 W36=C)2024)(m 1(C) W/m9( 22 °−°⋅=
Using the thermal resistance network, heat transfer between the 
Ri R1 Rins R3 Ro
iinsulationmetalo
refrigroom
total
refrigroom
hk
L
k
L
h
TT
AQ
R
TT
Q
121
/ 
 
+⎟
⎠
⎞
⎜
⎝
⎛+⎟
⎠
⎞
⎜
⎝
⎛+
−
=
−
=
&
&
 
Troom Trefrig
S g,ubstitutin 
 
C W/m4
1
C W/m035.0C W/m15.1
m 001.02
C W/m9
1
+
C)224( W/m36 2 °−=
2222 °⋅
+
°⋅
+
°⋅
×
°⋅
L
 
Solv ing for L, the minimum thickness of insulation is determined to be 
 L = 0.00875 m = 0.875 cm 
 
 
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preparation. If you are a student using this Manual, you are using it without permission. 
3-22
 
3-38 Prob. 3-37 is reconsidered. The effects of the thermal conductivities of the insulation material and the sheet 
metal on the thickness of the insulation is to be investigated. 
Analysis The problem is solved using EES, and the solution is given below. 
 
"GIVEN" 
k_ins=0.035 [W/m-C] 
L_metal=0.001 [m] 
k_metal=15.1 [W/m-C] 
T_refrig=2 [C] 
T_kitchen=24 [C] 
h_i=4 [W/m^2-C] 
h_o=9 [W/m^2-C] 
T_s_out=20 [C] 
 
"ANALYSIS" 
A=1 [m^2] “a unit surface area is considered" 
Q_dot=h_o*A*(T_kitchen-T_s_out) 
Q_dot=(T_kitchen-T_refrig)/R_total 
R_total=R_conv_i+2*R_metal+R_ins+R_conv_o 
R_conv_i=1/(h_i*A) 
R_metal=L_metal/(k_metal*A) 
R_ins=(L_ins*Convert(cm, m))/(k_ins*A) "L_ins is in cm" 
R_conv_o=1/(h_o*A) 
 
 
0.02 0.03 0.04 0.05 0.06 0.07 0.08
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2
kins [W/m-C]
L i
ns
 [
cm
]
kins 
[W/m.C] 
Lins 
[cm] 
0.02 
0.025 
0.03 
0.035 
0.04 
0.045 
0.05 
0.055 
0.06 
0.065 
0.07 
0.075 
0.08 
0.4997 
0.6247 
0.7496 
0.8745 
0.9995 
1.124 
1.249 
1.374 
1.499 
1.624 
1.749 
1.874 
1.999 
 
 
 
PROPRIETARY MATERIAL. © 2011 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. 
3-23
 
 
kmetal 
[W/m.C] 
Lins 
[cm] 
10 
30.53 
51.05 
71.58 
92.11 
112.6 
133.2 
153.7 
174.2 
194.7 
215.3 
235.8 
256.3 
276.8 
297.4 
317.9 
338.4 
358.9 
379.5 
400 
0.8743 
0.8748 
0.8749 
0.8749 
0.8749 
0.8749 
0.8749 
0.875 
0.875 
0.875 
0.875 
0.875 
0.875 
0.875 
0.875 
0.875 
0.875 
0.875 
0.875 
0.875 
0 50 100 150 200 250 300 350 400
0.8743
0.8744
0.8745
0.8746
0.8747
0.8748
0.8749
0.875
kmetal [W/m-C]
L i
ns
 [
cm
]
 
 
 
PROPRIETARY MATERIAL. © 2011 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. 
3-24
 
3-39 Heat is to be conducted along a circuit board with a copper layer on one side. The percentages of heat conduction 
along the copper and epoxy layers as well as the effective thermal conductivity of the board are to be determined. 
Assumptions 1 Steady operating conditions exist. 2 Heat transfer is one-dimensional 
since heat transfer from the side surfaces is disregarded 3 Thermal conductivities are 
constant. Copper 
Epoxy 
Ts tepoxytcopper
Properties The thermal conductivities are given to be k = 386 W/m⋅°C for copper and 
0.26 W/m⋅°C for epoxy layers. 
Analysis We take the length in the direction of heat transfer to be L and the width of the 
board to be w. Then heat conduction along this two-layer board can be expressed as 
 
[ ]
L
T
TkATkAQQQ ⎟⎞⎜⎛
∆
+⎟
⎞
⎜
⎛ ∆=+
epoxycopper
&&&
eat conduction along an “equivalent” board of thickness t = tcopper + tepoxy and thermal 
conductivity k can be expressed as 
wktkt
LL
∆
+=
⎠⎝⎠⎝ epoxycopper
epoxycopper
)()(
 
=
H
eff
Q 
L
Twttk
L
TkA⎜
⎝
= Q ∆+=⎟
⎠
⎞⎛ ∆ )( epoxycoppereff
board
& 
 
Setting the two relations above equal to each other and solving for the effective conductivity gives 
epoxycopper
epoxycopper
epoxycopperepoxycopper
)()(
)()()(
tt
ktkt
kktktttk effeff +
+
=⎯→⎯+=+ 
Note that heat conduction is proportional to kt. Substituting, the fractions of heat onducted along the copper and epoxy 
layers as well as the effective thermal conductivity of the board are determined to be 
 c
99.2%
0.8%
====
====
°=+=+=
°=°⋅=
°=°⋅=
992.0
038912.0
0386.0
)(
)(
008.0
038912.0
000312.0
)(
)(
C W/038912.0000312.00386.0)()()(
C W/000312.0m) C)(0.0012 W/m26.0()(
C W/0386.0m) C)(0.0001 W/m386()(
total
copper
copper
total
epoxy
epoxyf
epoxycoppertotal
epoxy
copper
kt
kt
f
kt
kt
ktktkt
kt
kt
 
and 
C W/m.29.9 °=
+
°×+×
=
m )0012.00001.0(
C W/)0012.026.00001.0386(
effk 
 
 
PROPRIETARY MATERIAL. © 2011 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. 
3-25
 
3-40E A thin copper plate is sandwiched between two layers of epoxy boards. The effective thermal conductivity of the 
board along its 9 in long side and the fraction of the heat conducted through copper along that side are to be determined. 
Assumptions 1 Steady operating conditions exist. 2 Heat transfer is one-
dimensional since heat transfer from the side surfaces are disregarded 3 
Thermal conductivities are constant. Copper 
Properties The thermal conductivities are given to be k = 223 Btu/h⋅ft⋅°F for 
copper and 0.15 Btu/h⋅ft⋅°F for epoxy layers. 
Analysis We take the length in the direction of heat transfer to be L and the 
width of the board to be w. Then heat conduction along this two-layer plate 
can be expressed as (we treat the two layers of epoxy as a single layer that is 
twice as thick) 
[ ]
L
Twktkt
L
TkA
L
TkA
QQQ
∆
+=⎟
⎠
⎞
⎜
⎝
⎛ ∆+⎟
⎠
⎞
⎜
⎝
⎛ ∆=
+=
epoxycopper
epoxycopper
epoxycopper
)()(
&&&
 
Epoxy 
Ts
½ tepoxytcopper½ tepoxy
Epoxy 
Heat conduction along an “equivalent” plate of thick ness t = tcopper + tepoxy 
and thermal conductivity keff can be expressed as Q 
 
LL ⎠⎝
epoxycoppereff
board
TwttkTkAQ ∆+=⎟⎞⎜⎛
∆ )(& 
Setting the two relations above equal to each other and solving for the effective conductivity gives 
 
=
epoxycopper
epoxycopperepoxycopper tteffeff +
Note that heat conduction is proportional to kt. Substituting, the 
epoxycopper )()()()()(
ktkt
kktktttk
+
=⎯→⎯+=+ 
fraction of heat conducted along the copper layer and the 
effective th rmal conductivity of the plate are determined to be 
and 
e
FBtu/h. 93292.000375.09292.0)()()(
FBtu/h. 00375.0ft) F)(0.15/12Btu/h.ft. 15.0(2)(
FBtu/h. 9292.0ft) F)(0.05/12Btu/h.ft. 223()(
epoxycoppertotal
epoxy
copper
°=+=+=
°=°=
°=°=
ktktkt
kt
kt
 
F.Btu/h.ft 32.0 2 °=
+
°
=
+
+
=
ft )]12/15.0(2)12/05.0[(
FBtu/h. 93292.0
t
)()(
epoxycopper
epoxycopper
t
ktkt
keff
 
99.6%==== 996.0
93292.0
9292.0
)(
)(
total
copper
copper kt
kt
f 
 
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3-26
 
3-41 Warm air blowing over the inner surface of an automobile windshield is used for defrosting ice accumulated on the 
outer surface. The convection heat transfer coefficient for the warm air blowing over the inner surface of the windshield 
necessary to cause the accumulated ice to begin melting is tobe determined. 
Assumptions 1 Steady operating conditions exist. 2 Heat transfer through the windshield is one-dimensional. 3 Thermal 
properties are constant. 4 Heat transfer by radiation is negligible. 5 The automobile is operating at 1 atm. 
Properties Thermal conductivity of the windshield is given to be k = 1.4 W/m · °C. 
Analysis The thermal resistances are 
 
Ah
R
i
i
1
= 
Ah
R
o
o
1
= 
and 
kAwin
From energy balance and using the thermal res
LR = 
istance 
oncept, th following equation is expressed: c e 
i
i
o
o
RR
TT
R
TT
+
−
=
− ∞∞
win
,11, 
 win
1,o∞
or 
,1 RR
TT
TT
R o
i −
−
−
= ∞ i
k
LTT i −⎟
⎞
⎜
⎛−
= ∞
11 ,1 
hTTh oo ⎟⎠⎜⎝−∞ 1,
ection 
 
i
For the ice to begin melting, the outer surface temperature of the windshield ( 1T ) should be at least 0 °C. The conv
heat transfer coefficient for the warm air is 
C W/m112 2 °⋅=
⎥
⎦
⎤
⎢
⎣
⎡
°⋅
−⎟
⎠
⎞
⎜
⎝
⎛
°⋅°−−
°−
=
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
−⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−
−
=
−
−1
∞
∞
1
2
1,
,1
C W/m4.1
m 005.0
C W/m200
1
C )010(
C )250(
1
k
L
hTT
TT
h
oo
i
i
 
Discussion In practical situations, the ambient temperature and the convective heat transfer coefficient outside the 
automobile vary with weather conditions and the automobile speed. Therefore the convection heat transfer coefficient of the 
warm air necessary to melt the ice should be varied as well. This is done by adjusting the warm air flow rate and temperature. 
 
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3-27
 
3-42 The thermal contact conductance for an aluminum plate attached on a copper plate, that is heated electrically, is to be 
determined. 
Assumptions 1 Steady operating conditions exist. 2 Heat transfer is one-dimensional. 3 Thermal properties are constant. 4 
Heat transfer by radiation is negligible. 
Properties The thermal conductivity of the aluminum plate is given to be 235 W/m · °C. 
Analysis The thermal resistances are 
 
kA
LR =cond 
and 
hA
From energy balance and using the thermal res
R 1= 
istance 
concept, the following equation is expressed: 
 
conv
convcond
1
elec / RRAR
TT
Aq
c ++
−
= ∞& 
or 
)/(1)/(/
1
elec hAkALAR
TT
Aq
++
−
= ∞& 
c
Rearranging the equation and solving for the contact resistance yields 
 
C/Wm102586
CW/m67
1
C W/m235
025.0C )20100(
2 −
°−
=
m 
 W/m5300
1
25
2
elec
1
°⋅×=
°⋅
−
°⋅
−−
−
=
−
∞
 .
hk
L
q
TT
Rc &
 
herma contact conductance is 
Discussion By comparing the value of the thermal contact conductance with the values listed in Table 3-2, the surface 
conditions of the plates appear to be milled. 
The t l
 C W/m16000 2 °⋅== cc Rh /1 
 
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preparation. If you are a student using this Manual, you are using it without permission. 
3-28
 
Thermal Contact Resistance 
 
3-43C The resistance that an interface offers to heat transfer per unit interface area is called thermal contact resistance, . 
The inverse of thermal contact resistance is called the thermal contact conductance. 
cR
 
 
3-44C The thermal contact resistance will be greater for rough surfaces because an interface with rough surfaces will contain 
more air gaps whose thermal conductivity is low. 
 
 
3-45C An interface acts like a very thin layer of insulation, and thus the thermal contact resistance has significance only for 
highly conducting materials like metals. Therefore, the thermal contact resistance can be ignored for two layers of insulation 
pressed against each other. 
 
 
3-46C An interface acts like a very thin layer of insulation, and thus the thermal contact resistance is significant for highly 
conducting materials like metals. Therefore, the thermal contact resistance must be considered for two layers of metals 
pressed against each other. 
 
 
3-47C Heat transfer through the voids at an interface is by conduction and radiation. Evacuating the interface eliminates heat 
transfer by conduction, and thus increases the thermal contact resistance. 
 
 
3-48C Thermal contact resistance can be minimized by (1) applying a thermally conducting liquid on the surfaces before 
they are pressed against each other, (2) by replacing the air at the interface by a better conducting gas such as helium or 
hydrogen, (3) by increasing the interface pressure, and (4) by inserting a soft metallic foil such as tin, silver, copper, nickel, 
or aluminum between the two surfaces. 
 
 
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3-29
 
3-49 The thickness of copper plate whose thermal resistance is equal to the thermal contact resistance is to be determined. 
Properties The thermal conductivity of copper is k = 386 W/m⋅°C. 
Analysis Noting that thermal contact resistance is the inverse of thermal contact conductance, the thermal contact resistance 
is determined to be 
 C/W.m 10143.7
C. W/m000,14
11 25
2
c
For a unit surface area, the therm
°×=
°
== −
h
R 
al resistance of a flat plate is defined as 
c
k
R = where L is the thickness oL f the plate and k is 
e thermal conductivity. Setting he equivalent thickness is determined from t e relation above to be 
 copper. Note that 
e thermal contact resistance in this case is greater than the sum of the thermal resistances of both plates. 
ror involved in the total thermal resistance of the 
ting conditions exist. 2 Heat transfer is one-dimensional since the plate is large. 3 Thermal 
 
ers is given to be hc = 6000 W/m2⋅°C. 
rmal resistances of different layers for unit surface 
a of 1 m
th h,cRR = t
 cm 2.76==°⋅×°⋅=== − m 0276.0C/W)m 10C)(7.143 W/m386( 25ckRkRL 
Therefore, the interface between the two plates offers as much resistance to heat transfer as a 2.76 cm thick
th
 
 
 
 
3-50 A thin copper plate is sandwiched between two epoxy boards. The er
plate if the thermal contact conductances are ignored is to be determined. 
Assumptions 1 Steady opera
conductivities are constant. 
Properties The thermal conductivities are given to be k = 386 W/m⋅°C for copper plates and k = 0.26 W/m⋅°C for epoxy
boards. The contact conductance at the interface of copper-epoxy lay
Analysis The the
are 2 are 
Epoxy 
7 mm 
Epoxy
7 mm 
Copper 
plate 
Q&
C° /W 00017.0
)m C)(1 W/m6000(
11
22
c
contact =
°⋅
==
cAh
R 
C/W 106.2
)m C)(1
 
 W/(386kA m
m 001.0 6
2ate
°×=
°⋅
== −
LR 
 
pl
C/W 02692.0
)m C)(1 W/m(0.26
m 007.0
2epoxy °=°⋅
==
kA
LR 
The total thermal resistance is 
22 epoxyplatecontacttotal ++= RRRR
C/W 05419.002692.02106.200017.02 6 °=×+×+×= −
 
nvolved in the total thermal resistance of the 
ntact resistances are ignored is determined to 
be 
Then the percent error i
plate if the thermal co
Rcontact Rcontact
Repoxy
T1 T2
Repoxy Rplate
0.63%=××=×= 100
05419.0
00017.02100
2
Error%
total
contact
R
R
 
which is negligible. 
 
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3-30
 
3-51 Two cylindrical aluminum bars with ground surfaces are pressed against each other in an insulation sleeve. For 
specified top and bottom surface temperatures, the rate of heat transfer along the cylinders and the temperature drop at the 
interface are to be determined. 
Assumptions 1 Steady operating conditions exist. 2 Heat transfer is 
one-dimensional in the axial direction since the lateral surfaces of 
both cylindersare well-insulated. 3 Thermal conductivities are 
constant. 
Bar Bar
Interface
Properties The thermal conductivity of aluminum bars is given to be 
k = 176 W/m⋅°C. The contact conductance at the interface of 
aluminum-aluminum plates for the case of ground surfaces and of 
20 atm ≈ 2 MPa pressure is hc = 11,400 W/m2⋅°C (Table 3-2). 
Ri Rglass RoAnalysis (a) The thermal resistance network in this case consists of two 
conduction resistance and the contact resistance, and they are 
determined to be 
T1 T2
C/W 0447.0
/4]m) (0.05C)[ W/m400,11(
11
22
c
contact °=
°⋅
==
πcAh
R
C/W 4341.0
/4]m) (0.05C)[ W/m(176
m 15.0
plate ==
LR
2
°=
°⋅ πkA
 
hen the rate of heat transfer is determined to be T
 W142.4=
°
 
×++ )4341.020447.0(2 barcontacttotal RRR
°−
=
∆
=
∆
=
C)20150(TTQ&
C/W 
 
e rate of heat transfer through the bars is 142.4 W. 
) The temperature drop at the interface is determined to be 
 
 
Therefore, th
(b
C6.4°=°==∆ C/W) W)(0.04474.142(contactinterface RQT & 
 
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3-31
 
Generalized Thermal Resistance Networks 
 
3-52C Two approaches used in development of the thermal resistance network in the x-direction for multi-dimensional 
problems are (1) to assume any plane wall normal to the x-axis to be isothermal and (2) to assume any plane parallel to the x-
axis to be adiabatic. 
 
 
3-53C The thermal resistance network approach will give adequate results for multi-dimensional heat transfer problems if 
heat transfer occurs predominantly in one direction. 
 
 
3-54C Parallel resistances indicate simultaneous heat transfer (such as convection and radiation on a surface). Series 
resistances indicate sequential heat transfer (such as two homogeneous layers of a wall). 
 
 
 
 
3-55 A typical section of a building wall is considered. The average heat flux through the wall is to be determined. 
Assumptions 1 Steady operating conditions exist. 
Properties The thermal conductivities are given to be k23b = 50 W/m⋅K, k23a = 0.03 W/m⋅K, k12 = 0.5 W/m⋅K, k34 = 1.0 
W/m⋅K. 
Analysis We consider 1 m2 of wall area. The thermal resistances are 
C/Wm 1.0
C) W/m0.1(
m 1.0
C/Wm 1032.1
0.005)C)(0.6 W/m50(
m 005.0m) 08.0(
)(
C/Wm 645.2
0.005)C)(0.6 W/m03.0(
m 6.0m) 08.0(
)(
C/Wm 02.0
C) W/m5.0(
m 01.0
2
34
34
34
25
23b
b
2323
2
23a
a
2323
2
12
12
12
°⋅=
°⋅
==
°⋅×=
+°⋅
=
+
=
°⋅=
+°⋅
=
+
=
°⋅=
°⋅
==
−
k
t
R
LLk
L
tR
LLk
L
tR
k
t
R
ba
b
ba
a
 
The total thermal resistance and the rate of heat transfer are 
C/Wm 120.01.0
1032.1645.2
1032.1645.202.0 2
5
5
34
2323
2323
12total
°⋅=+⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
×+
×
+=
+⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
+
+=
−
−
R
RR
RR
RR
ba
ba
 
2 W/m125=
⋅
°−
=
−
=
C/Wm 0.120
C)2035(
2
total
14
R
TT
q& 
 
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preparation. If you are a student using this Manual, you are using it without permission. 
3-32
 
 
3-56 A wall consists of horizontal bricks separated by plaster layers. There are also plaster layers on each side of the wall, 
and a rigid foam on the inner side of the wall. The rate of heat transfer through the wall is to be determined. 
Assumptions 1 Heat transfer is steady since there is no indication of change with time. 2 Heat transfer through the wall is 
one-dimensional. 3 Thermal conductivities are constant. 4 Heat transfer by radiation is disregarded. 
Properties The thermal conductivities are given to be k = 0.72 W/m⋅°C for bricks, k = 0.22 W/m⋅°C for plaster layers, and k 
= 0.026 W/m⋅°C for the rigid foam. 
Analysis We consider 1 m deep and 0.28 m high portion of wall which is representative of the entire wall. The thermal 
resistance network and individual resistances are 
R7R6
R5
R4
R3
 
R2
 
R1Ri
T∞1
 
 
T∞2 
 
C/W 737.4179.0804.0)325.0(2747.2357.02
C/W 804.0
45.45
1
833.0
1
45.45
11111
C/W 179.0
)m 128.0(C) W/m20(
11
C/W 833.0
)m 125.0(C) W/m72.0(
m 15.0
C/W45.45
)m 1015.0(C) W/m22.0(
m 15.0
C/W 325.0
)m 128.0(C) W/m22.0(
m 02.0
====
LRRR
C/W 747.2
)m 128.0(C) W/m026.0(
C/W 357.0
)m 1
11
21
543
2
2
2,o
24
253
262
21
221,
°=++++=++++=
°=⎯→⎯++=++=
°=
×°⋅
===
°=
×°⋅
===
°=
×°⋅
====
°=
×°⋅
°=
×°⋅
===
°=
×
===
omiditotal
mid
mid
conv
brick
ocenter
plaster
side
plaster
foam
convi
RRRRRR
R
RRRR
Ah
RR
kA
LRR
Ah
LRRR
kA
kA
RR
RR
 
The steady rate of heat transfer through the wall per is 
 
m 02.0
28.0(C) W/m10(1 °⋅
L
Ah
2m 28.0
 W49.5
C/W737.4
C)]4(22[( 21 =
°
°−−
=
−
= ∞∞
totalR
TT
Q& 
Then steady rate of heat transfer through the entire wall becomes 
 W470=×= 2
2
m 28.0
m)64( W)49.5(totalQ& 
 
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3-33
 
3-57 Prob. 3-56 is reconsidered. The rate of heat transfer through the wall as a function of the thickness of the rigid 
nalysis The problem is solved using EES, and the solution is given below. 
] 
] 
-C] 
 
 [m^2] 
 
ter/(k_plaster*A_3) 
side+R_mid+R_conv_2 
ity_2)/R_total 
_dot_total=Q_dot*A/A_1 
 
] 
foam is to be plotted. 
A
 
"GIVEN" 
A=4*6 [m^2] 
L_brick=0.15 [m] 
L_plaster_center=0.15 [m
L_plaster_side=0.02 [m
"L_foam=2 [cm]" 
k_brick=0.72 [W/m-C] 
k_plaster=0.22 [W/m-C] 
k_foam=0.026 [W/m
T_infinity_1=22 [C] 
T_infinity_2=-4 [C] 
h_1=10 [W/m^2-C] 
h_2=20 [W/m^2-C]
 
A_1=0.28*1 [m^2] 
A_2=0.25*1 [m^2] 
A_3=0.015*1
 
"ANALYSIS" 
R_conv_1=1/(h_1*A_1) 
R_foam=(L_foam*Convert(cm, m))/(k_foam*A_1) "L_foam is in cm"
R_plaster_side=L_plaster_side/(k_plaster*A_1) 
R_plaster_center=L_plaster_cen
R_brick=L_brick/(k_brick*A_2) 
R_conv_2=1/(h_2*A_1) 
1/R_mid=2*1/R_plaster_center+1/R_brick 
R_total=R_conv_1+R_foam+2*R_plaster_
Q_dot=(T_infinity_1-T_infin
Q
 
 
Lfo
[cm
am Q total
[W] 
1 
2 
3 
4 
5 
6 
7 
8 
9 
10 141.7 
662.8 
470.5 
364.8 
297.8 
251.6 
217.8 
192 
171.7 
155.3 
1 2 3 4 5 6 7 8 9 10
100
200
300
400
500
600
700
Q
to
ta
l 
[W
]
 
 
 Lfoam [cm]
 
 
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3-34
 
3-58 A wall is constructed of two layers of sheetrock spaced by 5 cm × 16 cm wood studs. The space between the studs is 
filled with fiberglass insulation. The thermal resistance of the wall and the rate of heat transfer through the wall are to be 
determined. 
Assumptions 1 Heat transfer is steady since there is no indication of change with time. 2 Heat transfer through the wall is 
one-dimensional. 3 Thermal conductivities are constant. 4 Heat transfer coefficients account for the radiation heat transfer. 
Properties The thermal conductivities are given to be k = 0.17 W/m⋅°C for sheetrock, k = 0.11 W/m⋅°C for wood studs, and 
k = 0.034 W/m⋅°C for fiberglass insulation. 
Analysis (a) The representative surface area is . The thermal resistance network and the individual 
thermal resistances are 
2m 65.065.01 =×=A
 
 
 
 
R5R4
R3
R2R1Ri
T∞1 T∞2
 W40.4
C/W 588.6
C)]9(20[
section) m 0.65m 1 a(for 045.0090.0178.6090.0185.0
C/W 178.6
843.7
1
091.29
1111
C/W 045.0
)m 65.0(C) W/m34(
11
C/W 843.7
)m 60.0(C) W/m034.0(
m 16.0
C/W 091.29
)m 05.0(C) W/m11.0(
m 16.0
C/W 090.0
)m 65.0(C) W/m17.0(
m 01.0
C/W 185.011
)m 65.0(C) W/m3.8( 22 °⋅i Ah
21
41
32
2o2
23
22
241
=
°
°−−
=
−
=
×°=++++=
++++=
°=⎯→⎯+=+=
°=
⋅
==
°=
°⋅
===
°=
°⋅
===
°=
°⋅
====
°===
∞∞
total
omiditotal
mid
mid
o
o
fiberglass
stud
sheetrock
i
R
TT
Q
RRRRRR