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Thermodynamics an Engineering Approach 8th edition Solutions Manual

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Thermodynamics an Engineering Approach 8th edition Solutions Manual

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  • May 28, 2024
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2-1

Solutions Manual for
Thermodynamics: An Engineering Approach
8th Edition
Yunus A. Cengel, Michael A. Boles
McGraw-Hill, 2015




Chapter 2
ENERGY, ENERGY TRANSFER, AND
GENERAL ENERGY ANALYSIS




PROPRIETARY AND CONFIDENTIAL


This Manual is the proprietary property of McGraw-Hill Education and 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 Education: This Manual is being provided only to authorized
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be distributed to or used by any student or other third party. No part of this Manual may be
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without the prior written permission of McGraw-Hill Education.




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.

, 2-2
Forms of Energy


2-1C The sum of all forms of the energy a system possesses is called total energy. In the absence of magnetic, electrical and
surface tension effects, the total energy of a system consists of the kinetic, potential, and internal energies.




2-2C The internal energy of a system is made up of sensible, latent, chemical and nuclear energies. The sensible internal
energy is due to translational, rotational, and vibrational effects.




2-3C Thermal energy is the sensible and latent forms of internal energy, and it is referred to as heat in daily life.




2-4C The mechanical energy is the form of energy that can be converted to mechanical work completely and directly by a
mechanical device such as a propeller. It differs from thermal energy in that thermal energy cannot be converted to work
directly and completely. The forms of mechanical energy of a fluid stream are kinetic, potential, and flow energies.




2-5C Hydrogen is also a fuel, since it can be burned, but it is not an energy source since there are no hydrogen reserves in
the world. Hydrogen can be obtained from water by using another energy source, such as solar or nuclear energy, and then
the hydrogen obtained can be used as a fuel to power cars or generators. Therefore, it is more proper to view hydrogen is an
energy carrier than an energy source.




2-6C In electric heaters, electrical energy is converted to sensible internal energy.




2-7C The forms of energy involved are electrical energy and sensible internal energy. Electrical energy is converted to
sensible internal energy, which is transferred to the water as heat.




2-8E The total kinetic energy of an object is given is to be determined.
Analysis The total kinetic energy of the object is given by

V2 (50 ft/s ) 2  1 Btu/lbm 
KE  m  (10 lbm)    0.499 Btu  0.50 Btu
 25,037 ft 2 /s 2 
2 2  




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.

, 2-3
2-9E The total potential energy of an object is to be determined.
Analysis Substituting the given data into the potential energy expression gives
 1 Btu/lbm 
PE  mgz  (200 lbm)(32.2 ft/s 2 )(10 ft)  2 2
  2.57 Btu

 25,037 ft /s 




2-10 A person with his suitcase goes up to the 10th floor in an elevator. The part of the energy of the elevator stored in the
suitcase is to be determined.
Assumptions 1 The vibrational effects in the elevator are negligible.
Analysis The energy stored in the suitcase is stored in the form of potential energy, which is mgz. Therefore,
 1 kJ/kg 
Esuitcase  PE  mgz  (30 kg)(9.81 m/s 2 )(35 m)   10.3 kJ
 1000 m 2 /s 2 
Therefore, the suitcase on 10th floor has 10.3 kJ more energy compared to an identical suitcase on the lobby level.
Discussion Noting that 1 kWh = 3600 kJ, the energy transferred to the suitcase is 10.3/3600 = 0.0029 kWh, which is very
small.




2-11 A hydraulic turbine-generator is to generate electricity from the water of a large reservoir. The power generation
potential is to be determined.
Assumptions 1 The elevation of the reservoir remains constant.
2 The mechanical energy of water at the turbine exit is
negligible.
Analysis The total mechanical energy water in a reservoir
120 m
possesses is equivalent to the potential energy of water at the
free surface, and it can be converted to work entirely.
Therefore, the power potential of water is its potential energy,
which is gz per unit mass, and m gz for a given mass flow rate.
Turbine Generator
 1 kJ/kg 
emech  pe  gz  (9.81 m/s )(120 m)
2
  1.177 kJ/kg
 1000 m 2 /s 2 
Then the power generation potential becomes
 1 kW 
W max  E mech  m emech  (2400 kg/s)(1.177 kJ/kg)   2825 kW
 1 kJ/s 
Therefore, the reservoir has the potential to generate 2825 kW of power.
Discussion This problem can also be solved by considering a point at the turbine inlet, and using flow energy instead of
potential energy. It would give the same result since the flow energy at the turbine inlet is equal to the potential energy at the
free surface of the reservoir.




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.

, 2-4
2-12 Wind is blowing steadily at a certain velocity. The mechanical energy of air per unit mass and the power generation
potential are to be determined.
Assumptions The wind is blowing steadily at a constant
uniform velocity.
Wind
Properties The density of air is given to be  = 1.25 kg/m3. Wind turbine
Analysis Kinetic energy is the only form of mechanical
10 m/s
energy the wind possesses, and it can be converted to work 60 m
entirely. Therefore, the power potential of the wind is its
kinetic energy, which is V2/2 per unit mass, and m V for
a given mass flow rate:

V 2 (10 m/s) 2  1 kJ/kg 
e mech  ke      0.050 kJ/kg
2 2  1000 m 2 /s 2 

D2  (60 m)2
  VA  V
m  (1.25 kg/m3 )(10 m/s)  35,340 kg/s
4 4

W max  E mech  m
 emech  (35,340 kg/s)(0.050 kJ/kg)  1770 kW

Therefore, 1770 kW of actual power can be generated by this wind turbine at the stated conditions.
Discussion The power generation of a wind turbine is proportional to the cube of the wind velocity, and thus the power
generation will change strongly with the wind conditions.




2-13 A water jet strikes the buckets located on the perimeter of a wheel at a specified velocity and flow rate. The power
generation potential of this system is to be determined.
Assumptions Water jet flows steadily at the specified speed and flow rate.
Analysis Kinetic energy is the only form of harvestable mechanical
energy the water jet possesses, and it can be converted to work entirely.
Therefore, the power potential of the water jet is its kinetic energy,
which is V2/2 per unit mass, and m V for a given mass flow rate:

V 2 (60 m/s)2  1 kJ/kg 
emech  ke    2 2
 1.8 kJ/kg Shaft
2 2  1000 m /s 

Wmax  E mech  m emech Nozzle
 1 kW 
 (120 kg/s)(1.8 kJ/kg)   216 kW
 1 kJ/s 
Vj
Therefore, 216 kW of power can be generated by this water jet at the
stated conditions.
Discussion An actual hydroelectric turbine (such as the Pelton wheel) can convert over 90% of this potential to actual
electric power.




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.

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