1-1
Solution Manual For
Thermodynamics an Engineering Approach 6th Edition by Yunus A
Chapter 1-17they illustrate how business concepts are applied in real-world scenarios. Group discussions and practicing sample case studies can aid in refining analytical
skills.________________________________________
Chapter 1
INTRODUCTION AND BASIC CONCEPTS
Thermodynamics
1-1 C Classical thermodynamics is based on experimental observations whereas statistical thermodynamics
is based on the average behavior of large groups of particles.
1-2 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.
1-3 C There is no truth to his claim. It violates the second law of thermodynamics.
1-4 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.
Mass, Force, and Units
1-5 C Pound-mass lbm is the mass unit in English system whereas pound-force lbf is the force unit. One
pound-force is the force required to accelerate a mass of 32.174 lbm by 1 ft/s2. In other words, the weight
of a 1-lbm mass at sea level is 1 lbf.
1-6 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.
1-7 C There is no acceleration, thus the net force is zero in both cases.
, 1-2
1-8 E The weight of a man on earth is given. His weight on the moon is to be determined.
Analysis Applying Newton's second law to the weight force gives
2
W 180 lbf 32.174 lbm ft/s
W mg m 180.4 lbm
g 32.10ft/s2 1 lbf
Mass is invariant and the man will have the same mass on the moon. Then, his weight on the moon will be
2 1lbf
W mg (180.4lbm)(5.47 ft/s 30.7 lbf
)
32.174 lbmft/s
2
1-9 The interior dimensions of a room are given. The mass and weight of the air in the room are to be
determined.
Assumptions The density of air is constant throughout the room.
Properties The density of air is given to be = 1.16 kg/m3.
Analysis The mass of the air in the room is ROOM
AIR
m V (1.16 kg/m3 )(6 6 8 m3 ) 334.1 kg
6X6X8 m3
Thus,
1N
W mg (334.1 kg)(9.81 m/s ) 2
2 3277 N
1 kgm/s
1-10 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 1% is to be determined.
z
Analysis The weight of a body at the elevation z can be expressed as
W mg m(9.807 3.32 106 z)
In our case,
W 0.99Ws 0.99mgs 0.99(m)(9.807)
Substituting, 0
Sea level
0.99(9.81) (9.81 3.32106 z) z
29,539 m
they illustrate how business concepts are applied in real-world scenarios. Group
discussions and practicing sample case studies can aid in refining analytical
skills.________________________________________
, 1-3
1-11 E 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
2 1lbf
W mg (10lbm)(32.0 ft/s 9.95 lbf
)
32.174 lbmft/s
2
1-12 The acceleration of an aircraft is given in g’s. The net upward force acting on a man in the aircraft is
to be determined.
Analysis From the Newton's second law, the force applied is
1N
F ma m(6 g) (90 kg)(6 9.81m/s ) 2
5297 N
2
1 kg
m/s
1-13 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 (5 kg)(9.79 m/s )
48.95 N
1 kg m/s2
Then the net force that acts on the rock is
Fnet Fup Fdown 150 48.95 101.05 N
Stone
From the Newton's second law, the acceleration of the rock becomes
a
F 101.05 N1 kgm/s2 2they illustrate how business concepts are applied in real-world scenarios. Group discussions and practicing sample
20.2 m/s
case studies can aid in refining analytical skills.________________________________________
m 5kg 1N
, 1-4
1-14 EES Problem 1-13 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.
they illustrate how business concepts are applied in real-world scenarios. Group discussions and practicing sample case studies can aid in refining analytical
skills.________________________________________
W=m*g "[N]"
m=5 [kg]
g=9.79 [m/s^2]
"The force balance on the rock yields the net force acting on the rock as"
F_net = F_up - F_down"[N]"
F_up=150 [N]
F_down=W"[N]"
"The acceleration of the rock is determined from Newton's second law."
F_net=a*m
"To Run the program, press F2 or click on the calculator icon from the Calculate menu"
SOLUTION
a=20.21 [m/s^2]
F_down=48.95 [N]
F_net=101.1 [N]
F_up=150 [N]
g=9.79 [m/s^2]
m=5 [kg]
W=48.95 [N]
1-15 Gravitational acceleration g and thus the weight of bodies decreases with increasing elevation. The
percent reduction in the weight of an airplane cruising at 13,000 m is to be determined.
Properties The gravitational acceleration g is given to be 9.807 m/s2 at sea level and 9.767 m/s2 at an
altitude of 13,000 m.
Analysis Weight is proportional to the gravitational acceleration g, and thus the
percent reduction in weight is equivalent to the percent reduction in the
gravitational acceleration, which is determined from
g 9.807 9.767
%Reduction in weight %Reduction in g 100 100 0.41%
g 9.807
Therefore, the airplane and the people in it will weight 0.41% less at 13,000 m
altitude.
Discussion Note that the weight loss at cruising altitudes is negligible.
Solution Manual For
Thermodynamics an Engineering Approach 6th Edition by Yunus A
Chapter 1-17they illustrate how business concepts are applied in real-world scenarios. Group discussions and practicing sample case studies can aid in refining analytical
skills.________________________________________
Chapter 1
INTRODUCTION AND BASIC CONCEPTS
Thermodynamics
1-1 C Classical thermodynamics is based on experimental observations whereas statistical thermodynamics
is based on the average behavior of large groups of particles.
1-2 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.
1-3 C There is no truth to his claim. It violates the second law of thermodynamics.
1-4 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.
Mass, Force, and Units
1-5 C Pound-mass lbm is the mass unit in English system whereas pound-force lbf is the force unit. One
pound-force is the force required to accelerate a mass of 32.174 lbm by 1 ft/s2. In other words, the weight
of a 1-lbm mass at sea level is 1 lbf.
1-6 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.
1-7 C There is no acceleration, thus the net force is zero in both cases.
, 1-2
1-8 E The weight of a man on earth is given. His weight on the moon is to be determined.
Analysis Applying Newton's second law to the weight force gives
2
W 180 lbf 32.174 lbm ft/s
W mg m 180.4 lbm
g 32.10ft/s2 1 lbf
Mass is invariant and the man will have the same mass on the moon. Then, his weight on the moon will be
2 1lbf
W mg (180.4lbm)(5.47 ft/s 30.7 lbf
)
32.174 lbmft/s
2
1-9 The interior dimensions of a room are given. The mass and weight of the air in the room are to be
determined.
Assumptions The density of air is constant throughout the room.
Properties The density of air is given to be = 1.16 kg/m3.
Analysis The mass of the air in the room is ROOM
AIR
m V (1.16 kg/m3 )(6 6 8 m3 ) 334.1 kg
6X6X8 m3
Thus,
1N
W mg (334.1 kg)(9.81 m/s ) 2
2 3277 N
1 kgm/s
1-10 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 1% is to be determined.
z
Analysis The weight of a body at the elevation z can be expressed as
W mg m(9.807 3.32 106 z)
In our case,
W 0.99Ws 0.99mgs 0.99(m)(9.807)
Substituting, 0
Sea level
0.99(9.81) (9.81 3.32106 z) z
29,539 m
they illustrate how business concepts are applied in real-world scenarios. Group
discussions and practicing sample case studies can aid in refining analytical
skills.________________________________________
, 1-3
1-11 E 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
2 1lbf
W mg (10lbm)(32.0 ft/s 9.95 lbf
)
32.174 lbmft/s
2
1-12 The acceleration of an aircraft is given in g’s. The net upward force acting on a man in the aircraft is
to be determined.
Analysis From the Newton's second law, the force applied is
1N
F ma m(6 g) (90 kg)(6 9.81m/s ) 2
5297 N
2
1 kg
m/s
1-13 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 (5 kg)(9.79 m/s )
48.95 N
1 kg m/s2
Then the net force that acts on the rock is
Fnet Fup Fdown 150 48.95 101.05 N
Stone
From the Newton's second law, the acceleration of the rock becomes
a
F 101.05 N1 kgm/s2 2they illustrate how business concepts are applied in real-world scenarios. Group discussions and practicing sample
20.2 m/s
case studies can aid in refining analytical skills.________________________________________
m 5kg 1N
, 1-4
1-14 EES Problem 1-13 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.
they illustrate how business concepts are applied in real-world scenarios. Group discussions and practicing sample case studies can aid in refining analytical
skills.________________________________________
W=m*g "[N]"
m=5 [kg]
g=9.79 [m/s^2]
"The force balance on the rock yields the net force acting on the rock as"
F_net = F_up - F_down"[N]"
F_up=150 [N]
F_down=W"[N]"
"The acceleration of the rock is determined from Newton's second law."
F_net=a*m
"To Run the program, press F2 or click on the calculator icon from the Calculate menu"
SOLUTION
a=20.21 [m/s^2]
F_down=48.95 [N]
F_net=101.1 [N]
F_up=150 [N]
g=9.79 [m/s^2]
m=5 [kg]
W=48.95 [N]
1-15 Gravitational acceleration g and thus the weight of bodies decreases with increasing elevation. The
percent reduction in the weight of an airplane cruising at 13,000 m is to be determined.
Properties The gravitational acceleration g is given to be 9.807 m/s2 at sea level and 9.767 m/s2 at an
altitude of 13,000 m.
Analysis Weight is proportional to the gravitational acceleration g, and thus the
percent reduction in weight is equivalent to the percent reduction in the
gravitational acceleration, which is determined from
g 9.807 9.767
%Reduction in weight %Reduction in g 100 100 0.41%
g 9.807
Therefore, the airplane and the people in it will weight 0.41% less at 13,000 m
altitude.
Discussion Note that the weight loss at cruising altitudes is negligible.
