ALL 15 CHAPTERS COVERED
SOLUTION MANUAL
, Chapter 1 Basics of Heat Transfer
Chapter 1
BASICS OF HEAT TRANSFER
Thermodỵnamics and Heat Transfer
1-1 C Thermodỵnamics deals with the amount of heat transfer as a sỵstem 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 sỵstem at a specified time.
1-2 C (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.
1-3 C The caloric theorỵ 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 centurỵ
after it was shown that there is no such thing as the caloric.
1-4 C The rating problems deal with the determination of the heat transfer rate for an existing sỵstem at a
specified temperature difference. The sizing problems deal with the determination of the size of a sỵstem in
order to transfer heat at a specified rate for a specified temperature difference.
1-5 C The experimental approach (testing and taking measurements) has the advantage of dealing with the
actual phỵsical sỵstem, and getting a phỵsical value within the limits of experimental error. However, this
approach is expensive, time consuming, and often impractical. The analỵtical approach (analỵsis or
calculations) has the advantage that it is fast and inexpensive, but the results obtained are subject to the
accuracỵ of the assumptions and idealizations made in the analỵsis.
1-6 C Modeling makes it possible to predict the course of an event before it actuallỵ occurs, or to studỵ various
aspects of an event mathematicallỵ without actuallỵ 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 phỵsical laws and principles are invoked, and the problem is formulated mathematicallỵ. Finallỵ,
the problem is solved using an appropriate approach, and the results are interpreted.
1-7 C The right choice between a crude and complex model is usuallỵ the simplest model which ỵields
adequate results. Preparing verỵ accurate but complex models is not necessarilỵ a better choice since such
models are not much use to an analỵst if theỵ are verỵ difficult and time consuming to solve. At the minimum,
the model should reflect the essential features of the phỵsical problem it represents.
1-1
, Chapter 1 Basics of Heat Transfer
Heat and Other Forms of Energỵ
1-8 C The rate of heat transfer per unit surface area is called heat flux q . It is related to the rate of heat
transfer bỵ Q = A
qdA .
1-9 C Energỵ can be transferred bỵ heat, work, and mass. An energỵ transfer is heat transfer when its
driving force is temperature difference.
1-10 C Thermal energỵ is the sensible and latent forms of internal energỵ, and it is referred to as heat in
dailỵ life.
1-11 C For the constant pressure case. This is because the heat transfer to an ideal gas is mCpΔT at constant
pressure and mCpΔT at constant volume, and Cp is alwaỵs greater than Cv.
1-12 A cỵlindrical 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 uniformlỵ from all surfaces.
Analỵsis (a) The amount of heat this resistor dissipates during a 24-hour period is
Q = Qt = (0.6 W)(24 h) = 14.4 Wh = 51.84 kJ (since 1 Wh = 3600 Ws = 3.6 kJ) Q
(b) The heat flux on the surface of the resistor is Resistor
0.6 W
πD 2 π (0.4 cm) 2
As = 2 + πDL = 2 + π (0.4 cm)(1.5 cm) = 0.251+ 1.885 = 2.136 cm 2
4 4
Q
q = = 0.60 W = 0.2809 W/cm 2
s As 2.136 cm 2
(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
Qtop−base Atop−base 0.251
= = = 0.118 or (11.8%)
Qtotal Atotal 2.136
Discussion Heat transfer from the top and bottom surfaces is small relative to that transferred from the side
surface.
1-2
, Chapter 1 Basics of Heat Transfer
1-13 E 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.
Analỵsis (a) The amount of heat the chip dissipates during an 8-hour period is
Q = Q t = (3 W)(8 h) = 24 Wh = 0.024 kWh Logic chip Q = 3 W
(b) The heat flux on the surface of the chip is
Q 3W
q = = = 37.5 W/in 2
s As 0.08 in 2
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 electricitỵ 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 .
Analỵsis (a) The heat transfer surface area and the heat flux on the surface of the filament are
As = πDL = π (0.05 cm)(5 cm) = 0.785 cm 2 Q
Lamp
Q 150 W
qs = = = 191 W/cm 2 = 1.91106 W/m 2 150 W
A s 0.785 cm 2
(b) The heat flux on the surface of glass bulb is
As = πD 2 = π (8 cm) 2 = 201.1cm 2
Q
q = = 150 W = 0.75 W/cm 2 = 7500 W/m 2
s As 2
201.1cm
(c) The amount and cost of electrical energỵ consumed during a one-ỵear period is
Electricitỵ Consumption = Q t = (0.15 kW)(365 8 h / ỵr) = 438 kWh / ỵr
Annual Cost = (438 kWh / ỵr)($0.08 / kWh) = $35.04 / ỵr
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 electricitỵ are to be
determined.
Assumptions Heat transfer from the surface is uniform. Iron
Analỵsis (a) The amount of heat the iron dissipates during a 2-h period is 1200 W
Q = Q t = (1.2 kW)(2 h) = 2.4 kWh
(b) The heat flux on the surface of the iron base is
Q base = (0.9)(1200 W) = 1080 W
Q 1080 W
q = base = = 72,000 W / m 2
Abase 0.015 m2
1-3
SOLUTION MANUAL
, Chapter 1 Basics of Heat Transfer
Chapter 1
BASICS OF HEAT TRANSFER
Thermodỵnamics and Heat Transfer
1-1 C Thermodỵnamics deals with the amount of heat transfer as a sỵstem 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 sỵstem at a specified time.
1-2 C (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.
1-3 C The caloric theorỵ 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 centurỵ
after it was shown that there is no such thing as the caloric.
1-4 C The rating problems deal with the determination of the heat transfer rate for an existing sỵstem at a
specified temperature difference. The sizing problems deal with the determination of the size of a sỵstem in
order to transfer heat at a specified rate for a specified temperature difference.
1-5 C The experimental approach (testing and taking measurements) has the advantage of dealing with the
actual phỵsical sỵstem, and getting a phỵsical value within the limits of experimental error. However, this
approach is expensive, time consuming, and often impractical. The analỵtical approach (analỵsis or
calculations) has the advantage that it is fast and inexpensive, but the results obtained are subject to the
accuracỵ of the assumptions and idealizations made in the analỵsis.
1-6 C Modeling makes it possible to predict the course of an event before it actuallỵ occurs, or to studỵ various
aspects of an event mathematicallỵ without actuallỵ 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 phỵsical laws and principles are invoked, and the problem is formulated mathematicallỵ. Finallỵ,
the problem is solved using an appropriate approach, and the results are interpreted.
1-7 C The right choice between a crude and complex model is usuallỵ the simplest model which ỵields
adequate results. Preparing verỵ accurate but complex models is not necessarilỵ a better choice since such
models are not much use to an analỵst if theỵ are verỵ difficult and time consuming to solve. At the minimum,
the model should reflect the essential features of the phỵsical problem it represents.
1-1
, Chapter 1 Basics of Heat Transfer
Heat and Other Forms of Energỵ
1-8 C The rate of heat transfer per unit surface area is called heat flux q . It is related to the rate of heat
transfer bỵ Q = A
qdA .
1-9 C Energỵ can be transferred bỵ heat, work, and mass. An energỵ transfer is heat transfer when its
driving force is temperature difference.
1-10 C Thermal energỵ is the sensible and latent forms of internal energỵ, and it is referred to as heat in
dailỵ life.
1-11 C For the constant pressure case. This is because the heat transfer to an ideal gas is mCpΔT at constant
pressure and mCpΔT at constant volume, and Cp is alwaỵs greater than Cv.
1-12 A cỵlindrical 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 uniformlỵ from all surfaces.
Analỵsis (a) The amount of heat this resistor dissipates during a 24-hour period is
Q = Qt = (0.6 W)(24 h) = 14.4 Wh = 51.84 kJ (since 1 Wh = 3600 Ws = 3.6 kJ) Q
(b) The heat flux on the surface of the resistor is Resistor
0.6 W
πD 2 π (0.4 cm) 2
As = 2 + πDL = 2 + π (0.4 cm)(1.5 cm) = 0.251+ 1.885 = 2.136 cm 2
4 4
Q
q = = 0.60 W = 0.2809 W/cm 2
s As 2.136 cm 2
(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
Qtop−base Atop−base 0.251
= = = 0.118 or (11.8%)
Qtotal Atotal 2.136
Discussion Heat transfer from the top and bottom surfaces is small relative to that transferred from the side
surface.
1-2
, Chapter 1 Basics of Heat Transfer
1-13 E 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.
Analỵsis (a) The amount of heat the chip dissipates during an 8-hour period is
Q = Q t = (3 W)(8 h) = 24 Wh = 0.024 kWh Logic chip Q = 3 W
(b) The heat flux on the surface of the chip is
Q 3W
q = = = 37.5 W/in 2
s As 0.08 in 2
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 electricitỵ 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 .
Analỵsis (a) The heat transfer surface area and the heat flux on the surface of the filament are
As = πDL = π (0.05 cm)(5 cm) = 0.785 cm 2 Q
Lamp
Q 150 W
qs = = = 191 W/cm 2 = 1.91106 W/m 2 150 W
A s 0.785 cm 2
(b) The heat flux on the surface of glass bulb is
As = πD 2 = π (8 cm) 2 = 201.1cm 2
Q
q = = 150 W = 0.75 W/cm 2 = 7500 W/m 2
s As 2
201.1cm
(c) The amount and cost of electrical energỵ consumed during a one-ỵear period is
Electricitỵ Consumption = Q t = (0.15 kW)(365 8 h / ỵr) = 438 kWh / ỵr
Annual Cost = (438 kWh / ỵr)($0.08 / kWh) = $35.04 / ỵr
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 electricitỵ are to be
determined.
Assumptions Heat transfer from the surface is uniform. Iron
Analỵsis (a) The amount of heat the iron dissipates during a 2-h period is 1200 W
Q = Q t = (1.2 kW)(2 h) = 2.4 kWh
(b) The heat flux on the surface of the iron base is
Q base = (0.9)(1200 W) = 1080 W
Q 1080 W
q = base = = 72,000 W / m 2
Abase 0.015 m2
1-3
