ALL CHAPTER 1-18 COVERED
SOLUTIONS MANUAL
,TABLE OF CONTENTS
1) Introduction and Basic Concepts
2) Energy, Energy Transfer, and General Energy Analysis
3) Properties of Pure Substances
4) Energy Analysis of Closed Systems
5) Mass and Energy Analysis of Control Volumes
6) The Second Law of Thermodynamics
7) Entropy
8) Entropy Analysis
9) Exergy
10) Gas Power Cycles
11) Vapor and Combined Power Cycles
12) Refrigeration Cycles
13) Thermodynamic Property Relations
14) Gas Mixtures
15) Gas-Vapor Mixtures and Air-Conditioning
16) Chemical Reactions
17) Chemical and Phase Equilibrium
18) Compressible Flow
,
, 1-2
Thermodynamics
18-1 C Classical thermodynamics is based on experimental observations whereas statistical
thermodynamics is based on the average behavior of large groups of particles.
18-2 C On a downhill road the potential energy of the bicyclist is being converted to kinetic energy, and
thus the bicyclist picks up speed. There is no creation of energy, and thus no violation of the conservation of
energy principle.
18-3 C A car going uphill without the engine running would increase the energy of the car, and thus it would be
a violation of the first law of thermodynamics. Therefore, this cannot happen. Using a level meter (a device with
an air bubble between two marks of a horizontal water tube) it can shown that the road that looks uphill to the
eye is actually downhill.
18-4 C There is no truth to his claim. It violates the second law of thermodynamics.
Mass, Force, and Units
18-5 C Kg-mass is the mass unit in the SI system whereas kg-force is a force unit. 1-kg-force is the force
required to accelerate a 1-kg mass by 9.807 m/s2. In other words, the weight of 1-kg mass at sea level is 1 kg-
force.
18-6 C In this unit, the word light refers to the speed of light. The light-year unit is then the product of a
velocity and time. Hence, this product forms a distance dimension and unit.
18-7 C There is no acceleration, thus the net force is zero in both cases.
18-8 The variation of gravitational acceleration above the sea level is given as a function of altitude. The
height at which the weight of a body will decrease by 0.3% is to be determined.
Analysis The weight of a body at the elevation z can be expressed as z
W mg m(9.807 3.32 10 6 z)
In our case,
W (1 0.3 /100)Ws 0.997Ws 0.997mg s
0.997(m)(9.807)
Substituting,
6
0.997(9.807) (9.807 3.32 10 z) z 8862 m
0
PROPRIETARY MATERIAL. © 2015 McGraw-Hill Education. 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.
, 1-3
18-9 The mass of an object is given. Its weight is to be determined.
Analysis Applying Newton's second law, the weight is determined to be
W mg (200 kg)(9.6 m/s2 ) 1920N
18-10 A plastic tank is filled with water. The weight of the combined system is to be
determined.
Assumptions The density of water is constant throughout.
Properties The density of water is given to be = 1000
kg/m3. Analysis The mass of the water in the tank and the mtank = 3 kg
total mass are
V =0.2 m 3
mw = V =(1000 kg/m3)(0.2 m3) = 200 kg H2O
mtotal = mw + mtank = 200 + 3 = 203 kg
Thus, 1
W mg (203 kg)(9.81 m/s2 ) 1991 N
N
1 kg m/s2
18-11 E The constant-pressure specific heat of air given in a specified unit is to be expressed in various units.
Analysis Using proper unit conversions, the constant-pressure specific heat is determined in various units to be
1 kJ/kg K
cp (1.005 kJ/kg 1.005kJ/kg K
C)
1 kJ/kg C
1000 J 1 kg
cp (1.005 kJ/kg 1.005 J/g C
C)
1 kJ 1000 g
c (1.005 kJ/kg 1
0.240kcal/kg C
p C) kcal
4.1868 kJ
1
cp (1.005 kJ/kg C)
Btu/lbm
0.240Btu/lbm F
F
4.1868 kJ/kg C
PROPRIETARY MATERIAL. © 2015 McGraw-Hill Education. 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.
, 1-4
18-12 A rock is thrown upward with a specified force. The acceleration of the rock is to be determined.
Analysis The weight of the rock is
2 1N
W mg (3 kg)(9.79 m/s 29.37 N
)
1 kg m/s2
Then the net force that acts on the rock is
Fnet F up F down 200 29.37 170.6 N
From the Newton's second law, the acceleration of the rock
becomes Stone
F 170.6 N 1 kg m/s2
a 56.9 m/s2
m 3 kg 1 N
PROPRIETARY MATERIAL. © 2015 McGraw-Hill Education. 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.
, 1-5
18-13 Problem 1-12 is reconsidered. The entire EES solution is to be printed out, including the numerical
results with proper units.
Analysis The problem is solved using EES, and the solution is given below.
"The weight of the rock is"
W=m*g
m=3 [kg]
g=9.79 [m/s2]
"The force balance on the rock yields the net force acting on the rock as"
F_up=200 [N]
F_net = F_up - F_down
F_down=W
"The acceleration of the rock is determined from Newton's second law."
F_net=m*a
"To Run the program, press F2 or select Solve from the Calculate menu."
SOLUTION
a=56.88 [m/s^2]
F_down=29.37 [N]
F_net=170.6 [N]
F_up=200 [N]
g=9.79 [m/s2]
m=3 [kg]
W=29.37 [N]
m [kg] a [m/s2] 200
1 190.2
2 90.21 160
3 56.88
4 40.21
5 30.21 120
a [m/s ]
2
6 23.54
7 18.78
8 15.21 80
9 12.43
10 10.21
40
0
1 2 3 4 5 6 7 8 9 10
m [kg]
PROPRIETARY MATERIAL. © 2015 McGraw-Hill Education. 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.
, 1-6
18-14 A resistance heater is used to heat water to desired temperature. The amount of electric energy used in
kWh and kJ are to be determined.
Analysis The resistance heater consumes electric energy at a rate of 4 kW or 4 kJ/s. Then the total amount of
electric energy used in 3 hours becomes
Total energy = (Energy per unit time)(Time interval)
= (4 kW)(3 h)
= 12 kWh
Noting that 1 kWh = (1 kJ/s)(3600 s) = 3600 kJ,
Total energy = (12 kWh)(3600 kJ/kWh)
= 43,200 kJ
Discussion Note kW is a unit for power whereas kWh is a unit for energy.
18-15 E An astronaut took his scales with him to space. It is to be determined how much he will weigh on the
spring and beam scales in space.
Analysis (a) A spring scale measures weight, which is the local gravitational force applied on a body:
2 1 lbf
W mg (150 lbm)(5.48 ) 25.5 lbf
2
ft/s 32.2
lbm ft/s
(b) A beam scale compares masses and thus is not affected by the variations in gravitational acceleration. The
beam scale will read what it reads on earth,
W 150 lbf
18-16 A gas tank is being filled with gasoline at a specified flow rate. Based on unit considerations alone, a
relation is to be obtained for the filling time.
Assumptions Gasoline is an incompressible substance and the flow rate is constant.
Analysis The filling time depends on the volume of the tank and the discharge rate of gasoline. Also, we know that the un
of time is „seconds‟. Therefore, the independent quantities should be arranged such that we end up with the unit of secon
Putting the given information into perspective, we have
t [s] V[L], and V [L/s}
It is obvious that the only way to end up with the unit “s” for time is to divide the tank volume by the discharge rate.
Therefore, the desired relation is
V
t
V
Discussion Note that this approach may not work for cases that involve dimensionless (and thus unitless) quantities.
PROPRIETARY MATERIAL. © 2015 McGraw-Hill Education. 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.
, 1-7
18-17 A pool is to be filled with water using a hose. Based on unit considerations, a relation is to be obtained for
the volume of the pool.
Assumptions Water is an incompressible substance and the average flow velocity is constant.
Analysis The pool volume depends on the filling time, the cross-sectional area which depends on hose diameter,
and flow velocity. Also, we know that the unit of volume is m3. Therefore, the independent quantities should be
arranged such that we end up with the unit of seconds. Putting the given information into perspective, we have
V[m3] is a function of t [s], D [m], and V [m/s}
It is obvious that the only way to end up with the unit “m3” for volume is to multiply the quantities t and V with the square of
D. Therefore, the desired relation is
V = CD2Vt
where the constant of proportionality is obtained for a round hose, namely, C =π/4 so that V= ( D2/4)Vt.
Discussion Note that the values of dimensionless constants of proportionality cannot be determined with this approach.
Systems, Properties, State, and Processes
18-18 C The radiator should be analyzed as an open system since mass is crossing the boundaries of the system.
18-19 C The system is taken as the air contained in the piston-cylinder device. This system is a closed or
fixed mass system since no mass enters or leaves it.
18-20 C A can of soft drink should be analyzed as a closed system since no mass is crossing the boundaries of the
system.
18-21 C Intensive properties do not depend on the size (extent) of the system but extensive properties do.
18-22 C If we were to divide the system into smaller portions, the weight of each portion would also be
smaller. Hence, the weight is an extensive property.
18-23 C Yes, because temperature and pressure are two independent properties and the air in an isolated
room is a simple compressible system.
PROPRIETARY MATERIAL. © 2015 McGraw-Hill Education. 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.
, 1-8
THOSE WERE PREVIEW PAGES
TO DOWNLOAD THE FULL PDF
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SOLUTIONS MANUAL
,TABLE OF CONTENTS
1) Introduction and Basic Concepts
2) Energy, Energy Transfer, and General Energy Analysis
3) Properties of Pure Substances
4) Energy Analysis of Closed Systems
5) Mass and Energy Analysis of Control Volumes
6) The Second Law of Thermodynamics
7) Entropy
8) Entropy Analysis
9) Exergy
10) Gas Power Cycles
11) Vapor and Combined Power Cycles
12) Refrigeration Cycles
13) Thermodynamic Property Relations
14) Gas Mixtures
15) Gas-Vapor Mixtures and Air-Conditioning
16) Chemical Reactions
17) Chemical and Phase Equilibrium
18) Compressible Flow
,
, 1-2
Thermodynamics
18-1 C Classical thermodynamics is based on experimental observations whereas statistical
thermodynamics is based on the average behavior of large groups of particles.
18-2 C On a downhill road the potential energy of the bicyclist is being converted to kinetic energy, and
thus the bicyclist picks up speed. There is no creation of energy, and thus no violation of the conservation of
energy principle.
18-3 C A car going uphill without the engine running would increase the energy of the car, and thus it would be
a violation of the first law of thermodynamics. Therefore, this cannot happen. Using a level meter (a device with
an air bubble between two marks of a horizontal water tube) it can shown that the road that looks uphill to the
eye is actually downhill.
18-4 C There is no truth to his claim. It violates the second law of thermodynamics.
Mass, Force, and Units
18-5 C Kg-mass is the mass unit in the SI system whereas kg-force is a force unit. 1-kg-force is the force
required to accelerate a 1-kg mass by 9.807 m/s2. In other words, the weight of 1-kg mass at sea level is 1 kg-
force.
18-6 C In this unit, the word light refers to the speed of light. The light-year unit is then the product of a
velocity and time. Hence, this product forms a distance dimension and unit.
18-7 C There is no acceleration, thus the net force is zero in both cases.
18-8 The variation of gravitational acceleration above the sea level is given as a function of altitude. The
height at which the weight of a body will decrease by 0.3% is to be determined.
Analysis The weight of a body at the elevation z can be expressed as z
W mg m(9.807 3.32 10 6 z)
In our case,
W (1 0.3 /100)Ws 0.997Ws 0.997mg s
0.997(m)(9.807)
Substituting,
6
0.997(9.807) (9.807 3.32 10 z) z 8862 m
0
PROPRIETARY MATERIAL. © 2015 McGraw-Hill Education. 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.
, 1-3
18-9 The mass of an object is given. Its weight is to be determined.
Analysis Applying Newton's second law, the weight is determined to be
W mg (200 kg)(9.6 m/s2 ) 1920N
18-10 A plastic tank is filled with water. The weight of the combined system is to be
determined.
Assumptions The density of water is constant throughout.
Properties The density of water is given to be = 1000
kg/m3. Analysis The mass of the water in the tank and the mtank = 3 kg
total mass are
V =0.2 m 3
mw = V =(1000 kg/m3)(0.2 m3) = 200 kg H2O
mtotal = mw + mtank = 200 + 3 = 203 kg
Thus, 1
W mg (203 kg)(9.81 m/s2 ) 1991 N
N
1 kg m/s2
18-11 E The constant-pressure specific heat of air given in a specified unit is to be expressed in various units.
Analysis Using proper unit conversions, the constant-pressure specific heat is determined in various units to be
1 kJ/kg K
cp (1.005 kJ/kg 1.005kJ/kg K
C)
1 kJ/kg C
1000 J 1 kg
cp (1.005 kJ/kg 1.005 J/g C
C)
1 kJ 1000 g
c (1.005 kJ/kg 1
0.240kcal/kg C
p C) kcal
4.1868 kJ
1
cp (1.005 kJ/kg C)
Btu/lbm
0.240Btu/lbm F
F
4.1868 kJ/kg C
PROPRIETARY MATERIAL. © 2015 McGraw-Hill Education. 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.
, 1-4
18-12 A rock is thrown upward with a specified force. The acceleration of the rock is to be determined.
Analysis The weight of the rock is
2 1N
W mg (3 kg)(9.79 m/s 29.37 N
)
1 kg m/s2
Then the net force that acts on the rock is
Fnet F up F down 200 29.37 170.6 N
From the Newton's second law, the acceleration of the rock
becomes Stone
F 170.6 N 1 kg m/s2
a 56.9 m/s2
m 3 kg 1 N
PROPRIETARY MATERIAL. © 2015 McGraw-Hill Education. 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.
, 1-5
18-13 Problem 1-12 is reconsidered. The entire EES solution is to be printed out, including the numerical
results with proper units.
Analysis The problem is solved using EES, and the solution is given below.
"The weight of the rock is"
W=m*g
m=3 [kg]
g=9.79 [m/s2]
"The force balance on the rock yields the net force acting on the rock as"
F_up=200 [N]
F_net = F_up - F_down
F_down=W
"The acceleration of the rock is determined from Newton's second law."
F_net=m*a
"To Run the program, press F2 or select Solve from the Calculate menu."
SOLUTION
a=56.88 [m/s^2]
F_down=29.37 [N]
F_net=170.6 [N]
F_up=200 [N]
g=9.79 [m/s2]
m=3 [kg]
W=29.37 [N]
m [kg] a [m/s2] 200
1 190.2
2 90.21 160
3 56.88
4 40.21
5 30.21 120
a [m/s ]
2
6 23.54
7 18.78
8 15.21 80
9 12.43
10 10.21
40
0
1 2 3 4 5 6 7 8 9 10
m [kg]
PROPRIETARY MATERIAL. © 2015 McGraw-Hill Education. 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.
, 1-6
18-14 A resistance heater is used to heat water to desired temperature. The amount of electric energy used in
kWh and kJ are to be determined.
Analysis The resistance heater consumes electric energy at a rate of 4 kW or 4 kJ/s. Then the total amount of
electric energy used in 3 hours becomes
Total energy = (Energy per unit time)(Time interval)
= (4 kW)(3 h)
= 12 kWh
Noting that 1 kWh = (1 kJ/s)(3600 s) = 3600 kJ,
Total energy = (12 kWh)(3600 kJ/kWh)
= 43,200 kJ
Discussion Note kW is a unit for power whereas kWh is a unit for energy.
18-15 E An astronaut took his scales with him to space. It is to be determined how much he will weigh on the
spring and beam scales in space.
Analysis (a) A spring scale measures weight, which is the local gravitational force applied on a body:
2 1 lbf
W mg (150 lbm)(5.48 ) 25.5 lbf
2
ft/s 32.2
lbm ft/s
(b) A beam scale compares masses and thus is not affected by the variations in gravitational acceleration. The
beam scale will read what it reads on earth,
W 150 lbf
18-16 A gas tank is being filled with gasoline at a specified flow rate. Based on unit considerations alone, a
relation is to be obtained for the filling time.
Assumptions Gasoline is an incompressible substance and the flow rate is constant.
Analysis The filling time depends on the volume of the tank and the discharge rate of gasoline. Also, we know that the un
of time is „seconds‟. Therefore, the independent quantities should be arranged such that we end up with the unit of secon
Putting the given information into perspective, we have
t [s] V[L], and V [L/s}
It is obvious that the only way to end up with the unit “s” for time is to divide the tank volume by the discharge rate.
Therefore, the desired relation is
V
t
V
Discussion Note that this approach may not work for cases that involve dimensionless (and thus unitless) quantities.
PROPRIETARY MATERIAL. © 2015 McGraw-Hill Education. 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.
, 1-7
18-17 A pool is to be filled with water using a hose. Based on unit considerations, a relation is to be obtained for
the volume of the pool.
Assumptions Water is an incompressible substance and the average flow velocity is constant.
Analysis The pool volume depends on the filling time, the cross-sectional area which depends on hose diameter,
and flow velocity. Also, we know that the unit of volume is m3. Therefore, the independent quantities should be
arranged such that we end up with the unit of seconds. Putting the given information into perspective, we have
V[m3] is a function of t [s], D [m], and V [m/s}
It is obvious that the only way to end up with the unit “m3” for volume is to multiply the quantities t and V with the square of
D. Therefore, the desired relation is
V = CD2Vt
where the constant of proportionality is obtained for a round hose, namely, C =π/4 so that V= ( D2/4)Vt.
Discussion Note that the values of dimensionless constants of proportionality cannot be determined with this approach.
Systems, Properties, State, and Processes
18-18 C The radiator should be analyzed as an open system since mass is crossing the boundaries of the system.
18-19 C The system is taken as the air contained in the piston-cylinder device. This system is a closed or
fixed mass system since no mass enters or leaves it.
18-20 C A can of soft drink should be analyzed as a closed system since no mass is crossing the boundaries of the
system.
18-21 C Intensive properties do not depend on the size (extent) of the system but extensive properties do.
18-22 C If we were to divide the system into smaller portions, the weight of each portion would also be
smaller. Hence, the weight is an extensive property.
18-23 C Yes, because temperature and pressure are two independent properties and the air in an isolated
room is a simple compressible system.
PROPRIETARY MATERIAL. © 2015 McGraw-Hill Education. 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.
, 1-8
THOSE WERE PREVIEW PAGES
TO DOWNLOAD THE FULL PDF
CLICK ON THE L.I.N.K
ON THE NEXT PAGE
