Chapter 1 Basics of Heat Transfer 1-1 Chapter 1 BASICS OF HEAT TRANSFER Notes | EduRev

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: Chapter 1 Basics of Heat Transfer 1-1 Chapter 1 BASICS OF HEAT TRANSFER Notes | EduRev

 Page 1


Chapter 1 Basics of Heat Transfer 
 1-1 
Chapter 1 
BASICS OF HEAT TRANSFER 
 
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 (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. 
 
1-3C 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. 
 
1-4C  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.  
 
1-5C  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.  
 
1-6C 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. 
 
1-7C  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. 
Page 2


Chapter 1 Basics of Heat Transfer 
 1-1 
Chapter 1 
BASICS OF HEAT TRANSFER 
 
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 (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. 
 
1-3C 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. 
 
1-4C  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.  
 
1-5C  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.  
 
1-6C 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. 
 
1-7C  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. 
Chapter 1 Basics of Heat Transfer 
 1-2 
 
Heat and Other Forms of Energy 
 
1-8C  The rate of heat transfer per unit surface area is called heat flux ? q . It is related to the rate of heat 
transfer by 
?
?
A
dA q Q ?
?
. 
 
1-9C  Energy can be transferred by heat, work, and mass. An energy transfer is heat transfer when its 
driving force is temperature difference. 
 
1-10C Thermal energy is the sensible and latent forms of internal energy, and it is referred to as heat in 
daily life. 
 
1-11C  For the constant pressure case. This is because the heat transfer to an ideal gas is mC p ?T at constant 
pressure and mC p ?T at constant volume, and C p is always greater than C v. 
 
1-12 A cylindrical resistor on a circuit board dissipates 0.6 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. 
Analysis (a) The amount of heat this resistor dissipates during a 24-hour period is 
    kJ  51.84 =  Wh 14.4 ? ? ? ?  h) W)(24  6 . 0 ( t Q Q
?
   (since 1 Wh = 3600 Ws = 3.6 kJ) 
(b) The heat flux on the surface of the resistor is 
       
2
2 2
cm 136 . 2 885 . 1 251 . 0 cm) cm)(1.5 4 . 0 (
4
cm) 4 . 0 (
2
4
2 ? ? ? ? ? ? ? ?
?
?
?
DL
D
A
s
 
       
2
 W/cm 0.2809 ? ? ?
2
cm 136 . 2
W 60 . 0
s
s
A
Q
q
?
? 
(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 
 
Q
Q
A
A
top base
total
top base
total
 or  (11.8%)
? ?
? ? ?
0 251
2136
.
.
0.118   
Discussion Heat transfer from the top and bottom surfaces is small relative to that transferred from the side 
surface. 
   
Q
?
 
Resistor 
0.6 W 
Page 3


Chapter 1 Basics of Heat Transfer 
 1-1 
Chapter 1 
BASICS OF HEAT TRANSFER 
 
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 (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. 
 
1-3C 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. 
 
1-4C  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.  
 
1-5C  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.  
 
1-6C 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. 
 
1-7C  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. 
Chapter 1 Basics of Heat Transfer 
 1-2 
 
Heat and Other Forms of Energy 
 
1-8C  The rate of heat transfer per unit surface area is called heat flux ? q . It is related to the rate of heat 
transfer by 
?
?
A
dA q Q ?
?
. 
 
1-9C  Energy can be transferred by heat, work, and mass. An energy transfer is heat transfer when its 
driving force is temperature difference. 
 
1-10C Thermal energy is the sensible and latent forms of internal energy, and it is referred to as heat in 
daily life. 
 
1-11C  For the constant pressure case. This is because the heat transfer to an ideal gas is mC p ?T at constant 
pressure and mC p ?T at constant volume, and C p is always greater than C v. 
 
1-12 A cylindrical resistor on a circuit board dissipates 0.6 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. 
Analysis (a) The amount of heat this resistor dissipates during a 24-hour period is 
    kJ  51.84 =  Wh 14.4 ? ? ? ?  h) W)(24  6 . 0 ( t Q Q
?
   (since 1 Wh = 3600 Ws = 3.6 kJ) 
(b) The heat flux on the surface of the resistor is 
       
2
2 2
cm 136 . 2 885 . 1 251 . 0 cm) cm)(1.5 4 . 0 (
4
cm) 4 . 0 (
2
4
2 ? ? ? ? ? ? ? ?
?
?
?
DL
D
A
s
 
       
2
 W/cm 0.2809 ? ? ?
2
cm 136 . 2
W 60 . 0
s
s
A
Q
q
?
? 
(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 
 
Q
Q
A
A
top base
total
top base
total
 or  (11.8%)
? ?
? ? ?
0 251
2136
.
.
0.118   
Discussion Heat transfer from the top and bottom surfaces is small relative to that transferred from the side 
surface. 
   
Q
?
 
Resistor 
0.6 W 
Chapter 1 Basics of Heat Transfer 
 1-3 
1-13E 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 
 Q Q t ? ? ? ?
?
( )( ) ? 3 8 24 W h Wh 0.024 kWh 
(b) The heat flux on the surface of the chip is 
 
2
 W/in 37.5 ? ? ?
2
in 08 . 0
W 3
s
s
A
Q
q
?
? 
 
1-14 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 
 
2
cm 785 . 0 ) cm 5 )( cm 05 . 0 ( ? ? ? ? ?DL A
s
 
 
2 6
 W/m 10 1.91 ? ? ? ? ?
2
2
W/cm 191
cm 785 . 0
W 150
s
s
A
Q
q
?
? 
(b) The heat flux on the surface of glass bulb is 
 
2 2 2
cm 1 . 201 cm) 8 ( ? ? ? ? ?D A
s
 
 
2
 W/m 7500 ? ? ? ?
2
2
W/cm 75 . 0
cm 1 . 201
W 150
s
s
A
Q
q
?
? 
(c) The amount and cost of electrical energy consumed during a one-year period is 
 
Electricity Consumption kW)(365 8 h / yr) 438 kWh / yr
Annual Cost = (438 kWh / yr)($0.08 kWh)
? ? ? ?
?
?
( .
/
Q t ? 015
$35.04 / yr
 
 
1-15 A 1200 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. 
Analysis (a) The amount of heat the iron dissipates during a 2-h period is 
 Q Q t ? ? ?
?
( . ? 12 kW)(2 h) 2.4 kWh 
(b) The heat flux on the surface of the iron base is 
 
?
( . )( Q
base
 W) = 1080 W ? 0 9 1200 
 ?
?
.
q
Q
A
? ? ?
base
base
2
 W
 m
1080
0 015
72,000 W / m
2
 
(c) The cost of electricity consumed during this period is 
 Cost of electricity = (2.4 kWh) ($0.07 kWh) ? ? / $0.17 
Logic chip 
 W 3 ? Q
?
 
Iron 
1200 W 
Q
?
 Lamp 
150 W 
Page 4


Chapter 1 Basics of Heat Transfer 
 1-1 
Chapter 1 
BASICS OF HEAT TRANSFER 
 
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 (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. 
 
1-3C 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. 
 
1-4C  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.  
 
1-5C  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.  
 
1-6C 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. 
 
1-7C  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. 
Chapter 1 Basics of Heat Transfer 
 1-2 
 
Heat and Other Forms of Energy 
 
1-8C  The rate of heat transfer per unit surface area is called heat flux ? q . It is related to the rate of heat 
transfer by 
?
?
A
dA q Q ?
?
. 
 
1-9C  Energy can be transferred by heat, work, and mass. An energy transfer is heat transfer when its 
driving force is temperature difference. 
 
1-10C Thermal energy is the sensible and latent forms of internal energy, and it is referred to as heat in 
daily life. 
 
1-11C  For the constant pressure case. This is because the heat transfer to an ideal gas is mC p ?T at constant 
pressure and mC p ?T at constant volume, and C p is always greater than C v. 
 
1-12 A cylindrical resistor on a circuit board dissipates 0.6 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. 
Analysis (a) The amount of heat this resistor dissipates during a 24-hour period is 
    kJ  51.84 =  Wh 14.4 ? ? ? ?  h) W)(24  6 . 0 ( t Q Q
?
   (since 1 Wh = 3600 Ws = 3.6 kJ) 
(b) The heat flux on the surface of the resistor is 
       
2
2 2
cm 136 . 2 885 . 1 251 . 0 cm) cm)(1.5 4 . 0 (
4
cm) 4 . 0 (
2
4
2 ? ? ? ? ? ? ? ?
?
?
?
DL
D
A
s
 
       
2
 W/cm 0.2809 ? ? ?
2
cm 136 . 2
W 60 . 0
s
s
A
Q
q
?
? 
(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 
 
Q
Q
A
A
top base
total
top base
total
 or  (11.8%)
? ?
? ? ?
0 251
2136
.
.
0.118   
Discussion Heat transfer from the top and bottom surfaces is small relative to that transferred from the side 
surface. 
   
Q
?
 
Resistor 
0.6 W 
Chapter 1 Basics of Heat Transfer 
 1-3 
1-13E 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 
 Q Q t ? ? ? ?
?
( )( ) ? 3 8 24 W h Wh 0.024 kWh 
(b) The heat flux on the surface of the chip is 
 
2
 W/in 37.5 ? ? ?
2
in 08 . 0
W 3
s
s
A
Q
q
?
? 
 
1-14 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 
 
2
cm 785 . 0 ) cm 5 )( cm 05 . 0 ( ? ? ? ? ?DL A
s
 
 
2 6
 W/m 10 1.91 ? ? ? ? ?
2
2
W/cm 191
cm 785 . 0
W 150
s
s
A
Q
q
?
? 
(b) The heat flux on the surface of glass bulb is 
 
2 2 2
cm 1 . 201 cm) 8 ( ? ? ? ? ?D A
s
 
 
2
 W/m 7500 ? ? ? ?
2
2
W/cm 75 . 0
cm 1 . 201
W 150
s
s
A
Q
q
?
? 
(c) The amount and cost of electrical energy consumed during a one-year period is 
 
Electricity Consumption kW)(365 8 h / yr) 438 kWh / yr
Annual Cost = (438 kWh / yr)($0.08 kWh)
? ? ? ?
?
?
( .
/
Q t ? 015
$35.04 / yr
 
 
1-15 A 1200 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. 
Analysis (a) The amount of heat the iron dissipates during a 2-h period is 
 Q Q t ? ? ?
?
( . ? 12 kW)(2 h) 2.4 kWh 
(b) The heat flux on the surface of the iron base is 
 
?
( . )( Q
base
 W) = 1080 W ? 0 9 1200 
 ?
?
.
q
Q
A
? ? ?
base
base
2
 W
 m
1080
0 015
72,000 W / m
2
 
(c) The cost of electricity consumed during this period is 
 Cost of electricity = (2.4 kWh) ($0.07 kWh) ? ? / $0.17 
Logic chip 
 W 3 ? Q
?
 
Iron 
1200 W 
Q
?
 Lamp 
150 W 
Chapter 1 Basics of Heat Transfer 
 1-4 
1-16  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 
 W 4 . 14 W) 12 . 0 )( 120 ( ? ? Q
?
 
 kWh 0.144 ? ? ? ? h) kW)(10 0144 . 0 ( t Q Q
?
 
(b) The heat flux on the surface of the circuit board is 
 
2
m 03 . 0 ) m 2 . 0 )( m 15 . 0 ( ? ?
s
A 
 
2
 W/m 480 ? ? ?
2
m 03 . 0
W 4 . 14
s
s
A
Q
q
?
? 
 
1-17  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 ? = 2,700 kg/m
3
 and 90 . 0 ?
p
C kJ/kg. ?C. 
Analysis The amount of energy added to the ball is simply the change in its 
internal energy, and is determined from 
 E U mC T T
transfer
? ? ? ? ( )
2 1
 
where 
 m V D ? ? ? ? ?
?
?
?
6 6
2700 015 4 77
3
( )( . . kg / m m) kg
3 3
 
Substituting, 
 E
transfer
 kg)(0.90 kJ / kg. C)(200-80) C = ? ? ? ( . 4 77 515 kJ 
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. 
 
1-18  The body temperature of a man rises from 37°C to 39°C during strenuous exercise. The resulting 
increase in the thermal energy content of the body is to be determined. 
Assumptions The body temperature changes uniformly. 
Properties The average specific heat of the human body is given to be 3.6 
kJ/kg.°C. 
Analysis The change in the sensible internal energy content of the body as a  
result of the body temperature rising 2 ?C during strenuous exercise is 
 ?U = mC ?T = (70 kg)(3.6 kJ/kg. ?C)(2 ?C) = 504 kJ 
Chips,  
0.12 W 
15 cm 
20 cm 
Q
?
 
Metal 
ball 
E 
Page 5


Chapter 1 Basics of Heat Transfer 
 1-1 
Chapter 1 
BASICS OF HEAT TRANSFER 
 
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 (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. 
 
1-3C 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. 
 
1-4C  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.  
 
1-5C  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.  
 
1-6C 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. 
 
1-7C  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. 
Chapter 1 Basics of Heat Transfer 
 1-2 
 
Heat and Other Forms of Energy 
 
1-8C  The rate of heat transfer per unit surface area is called heat flux ? q . It is related to the rate of heat 
transfer by 
?
?
A
dA q Q ?
?
. 
 
1-9C  Energy can be transferred by heat, work, and mass. An energy transfer is heat transfer when its 
driving force is temperature difference. 
 
1-10C Thermal energy is the sensible and latent forms of internal energy, and it is referred to as heat in 
daily life. 
 
1-11C  For the constant pressure case. This is because the heat transfer to an ideal gas is mC p ?T at constant 
pressure and mC p ?T at constant volume, and C p is always greater than C v. 
 
1-12 A cylindrical resistor on a circuit board dissipates 0.6 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. 
Analysis (a) The amount of heat this resistor dissipates during a 24-hour period is 
    kJ  51.84 =  Wh 14.4 ? ? ? ?  h) W)(24  6 . 0 ( t Q Q
?
   (since 1 Wh = 3600 Ws = 3.6 kJ) 
(b) The heat flux on the surface of the resistor is 
       
2
2 2
cm 136 . 2 885 . 1 251 . 0 cm) cm)(1.5 4 . 0 (
4
cm) 4 . 0 (
2
4
2 ? ? ? ? ? ? ? ?
?
?
?
DL
D
A
s
 
       
2
 W/cm 0.2809 ? ? ?
2
cm 136 . 2
W 60 . 0
s
s
A
Q
q
?
? 
(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 
 
Q
Q
A
A
top base
total
top base
total
 or  (11.8%)
? ?
? ? ?
0 251
2136
.
.
0.118   
Discussion Heat transfer from the top and bottom surfaces is small relative to that transferred from the side 
surface. 
   
Q
?
 
Resistor 
0.6 W 
Chapter 1 Basics of Heat Transfer 
 1-3 
1-13E 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 
 Q Q t ? ? ? ?
?
( )( ) ? 3 8 24 W h Wh 0.024 kWh 
(b) The heat flux on the surface of the chip is 
 
2
 W/in 37.5 ? ? ?
2
in 08 . 0
W 3
s
s
A
Q
q
?
? 
 
1-14 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 
 
2
cm 785 . 0 ) cm 5 )( cm 05 . 0 ( ? ? ? ? ?DL A
s
 
 
2 6
 W/m 10 1.91 ? ? ? ? ?
2
2
W/cm 191
cm 785 . 0
W 150
s
s
A
Q
q
?
? 
(b) The heat flux on the surface of glass bulb is 
 
2 2 2
cm 1 . 201 cm) 8 ( ? ? ? ? ?D A
s
 
 
2
 W/m 7500 ? ? ? ?
2
2
W/cm 75 . 0
cm 1 . 201
W 150
s
s
A
Q
q
?
? 
(c) The amount and cost of electrical energy consumed during a one-year period is 
 
Electricity Consumption kW)(365 8 h / yr) 438 kWh / yr
Annual Cost = (438 kWh / yr)($0.08 kWh)
? ? ? ?
?
?
( .
/
Q t ? 015
$35.04 / yr
 
 
1-15 A 1200 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. 
Analysis (a) The amount of heat the iron dissipates during a 2-h period is 
 Q Q t ? ? ?
?
( . ? 12 kW)(2 h) 2.4 kWh 
(b) The heat flux on the surface of the iron base is 
 
?
( . )( Q
base
 W) = 1080 W ? 0 9 1200 
 ?
?
.
q
Q
A
? ? ?
base
base
2
 W
 m
1080
0 015
72,000 W / m
2
 
(c) The cost of electricity consumed during this period is 
 Cost of electricity = (2.4 kWh) ($0.07 kWh) ? ? / $0.17 
Logic chip 
 W 3 ? Q
?
 
Iron 
1200 W 
Q
?
 Lamp 
150 W 
Chapter 1 Basics of Heat Transfer 
 1-4 
1-16  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 
 W 4 . 14 W) 12 . 0 )( 120 ( ? ? Q
?
 
 kWh 0.144 ? ? ? ? h) kW)(10 0144 . 0 ( t Q Q
?
 
(b) The heat flux on the surface of the circuit board is 
 
2
m 03 . 0 ) m 2 . 0 )( m 15 . 0 ( ? ?
s
A 
 
2
 W/m 480 ? ? ?
2
m 03 . 0
W 4 . 14
s
s
A
Q
q
?
? 
 
1-17  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 ? = 2,700 kg/m
3
 and 90 . 0 ?
p
C kJ/kg. ?C. 
Analysis The amount of energy added to the ball is simply the change in its 
internal energy, and is determined from 
 E U mC T T
transfer
? ? ? ? ( )
2 1
 
where 
 m V D ? ? ? ? ?
?
?
?
6 6
2700 015 4 77
3
( )( . . kg / m m) kg
3 3
 
Substituting, 
 E
transfer
 kg)(0.90 kJ / kg. C)(200-80) C = ? ? ? ( . 4 77 515 kJ 
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. 
 
1-18  The body temperature of a man rises from 37°C to 39°C during strenuous exercise. The resulting 
increase in the thermal energy content of the body is to be determined. 
Assumptions The body temperature changes uniformly. 
Properties The average specific heat of the human body is given to be 3.6 
kJ/kg.°C. 
Analysis The change in the sensible internal energy content of the body as a  
result of the body temperature rising 2 ?C during strenuous exercise is 
 ?U = mC ?T = (70 kg)(3.6 kJ/kg. ?C)(2 ?C) = 504 kJ 
Chips,  
0.12 W 
15 cm 
20 cm 
Q
?
 
Metal 
ball 
E 
Chapter 1 Basics of Heat Transfer 
 1-5 
1-19 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 C
p
 = 1.007 kJ/kg. ?C (Table A-15). 
Analysis The volume of the air in the house is 
 V ? ? ? ( ( floor space)(height) m )(3 m) m
2 3
200 600 
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 314 , 11
K) 273.15 + /kg.K)(5 kPa.m 287 . 0 (
day) / m 600 kPa)(16.8 6 . 89 (
) ACH (
3
3
house air
air
?
?
?
?
? ?
o
o
o
o
RT
V P
RT
V P
m
?
?
 
 Noting that outdoor air enters at 5 ?C and leaves at 22 ?C, the energy loss of this house per day is  
 
kWh/day 53.8 = kJ/day 681 , 193 C ) 5 C)(22 kJ/kg. .007 kg/day)(1 314 , 11 (
) (
outdoors indoors p air infilt
? ? ? ? ?
? ? T T C m Q ?
?
 
At a unit cost of $0.082/kWh, the cost of this electrical energy lost by infiltration is 
     $4.41/day ? ? ) $0.082/kWh kWh/day)( 8 . 53 ( energy) of cost t used)(Uni (Energy = Cost Enegy 
5 ?C 
0.7 ACH 
22 ?C 
AIR 
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