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Solutions Manual for Fluid Mechanics Fundamentals and Applications 4th Edition by Cengel IBSN 1259696537\
Chapter 2 Properties of Fluids
Solutions Manual for
Fluid Mechanics: Fundamentals and Applications
Fourth Edition
Yunus A. Çengel & John M. Cimbala
McGraw-Hill Education, 2018
M
ED
Chapter 2
PROPERTIES OF FLUIDS
ST
U
D
PROPRIETARY AND CONFIDENTIAL
Y
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 professors and instructors for
use in preparing for the classes using the affiliated textbook. No
other use or distribution of this Manual is permitted. This Manual
may not be sold and may not be distributed to or used by any
student or other third party. No part of this Manual may be
reproduced, displayed or distributed in any form or by any means,
electronic or otherwise, without the prior written permission of
McGraw-Hill Education.
2-1
, Chapter 2 Properties of Fluids
Density and Specific Gravity
2-1C
Solution We are to discuss the difference between intensive and extensive properties.
Analysis Intensive properties do not depend on the size (extent) of the system but extensive properties do depend
on the size (extent) of the system.
Discussion An example of an intensive property is temperature. An example of an extensive property is mass.
M
2-2C
ED
Solution We are to discuss the difference between mass and molar mass.
Analysis Mass m is the actual mass in grams or kilograms; molar mass M is the mass per mole in grams/mol or
kg/kmol. These two are related to each other by m = NM, where N is the number of moles.
ST
Discussion Mass, number of moles, and molar mass are often confused. Molar mass is also called molecular weight.
U
2-3C
Solution We are to define specific gravity and discuss its relationship to density.
D
Analysis The specific gravity, or relative density, is defined as the ratio of the density of a substance to the density
of some standard substance at a specified temperature (the standard is water at 4°C, for which ρH2O = 1000 kg/m ). That
3
Y
is, SG = ρ/ρH2O . When specific gravity is known, density is determined from ρ = SG×ρH2O .
Discussion Specific gravity is dimensionless and unitless [it is just a number without dimensions or units].
, Chapter 2 Properties of Fluids
2-4C
Solution We are to decide if the specific weight is an extensive or intensive property.
Analysis The original specific weight is
W
γ =
1
V
If we were to divide the system into two halves, each half weighs W/2 and occupies a volume of V /2. The specific weight
of one of these halves is
W /2
γ= =γ
V /2 1
which is the same as the original specific weight. Hence, specific weight is an intensive property.
Discussion If specific weight were an extensive property, its value for half of the system would be halved.
M
ED
2-5C
Solution We are to discuss the applicability of the ideal gas law.
Analysis A gas can be treated as an ideal gas when it is at a high temperature and/or a low pressure relative to its
critical temperature and pressure.
ST
Discussion Air and many other gases at room temperature and pressure can be approximated as ideal gases without any
significant loss of accuracy.
U
2-6C
D
Solution We are to discuss the difference between R and Ru.
Analysis Ru is the universal gas constant that is the same for all gases, whereas R is the specific gas constant that is
Y
different for different gases. These two are related to each other by R = Ru / M , where M is the molar mass (also called
the molecular weight) of the gas.
Discussion Since molar mass has dimensions of mass per mole, R and Ru do not have the same dimensions or units.
, Chapter 2 Properties of Fluids
2-7
Solution The pressure in a container that is filled with air is to be determined.
Assumptions At specified conditions, air behaves as an ideal gas.
kJ ⎛⎜ kPa ⋅m3 ⎟⎞ kPa ⋅m3
The gas constant of air is R = 0.287 = 0.287
⎟⎟
Properties (see also Table A-1).
kg⋅ K ⎜⎜⎝ kJ j kg⋅K
Analysis The definition of the specific volume gives
V 0.075 m3
v= = = 0.075 m 3/kg
m 1 kg
Using the ideal gas equation of state, the pressure is
RT (0.287 kPa⋅ m3/kg⋅K)(27 +273 K)
Pv = RT → P = = = 1148 kPa
v 0.075 m3/kg
Discussion In ideal gas calculations, it saves time to convert the gas constant to appropriate units.
M
ED
2-8E
Solution The volume of a tank that is filled with argon at a specified state is to be determined.
Assumptions At specified conditions, argon behaves as an ideal gas.
3
Properties The gas constant of argon is obtained from Table A-1E, R = 0.2686 psia⋅ft /lbm⋅R.
ST
Analysis According to the ideal gas equation of state,
mRT (1 lbm)(0.2686 psia ⋅ ft3/lbm⋅R)(100 + 460 R)
V= = = 0.7521 ft 3
P 200 psia
Discussion In ideal gas calculations, it saves time to write the gas constant in appropriate units.
U
D
2-9E
Solution The specific volume of oxygen at a specified state is to be determined.
Y
Assumptions At specified conditions, oxygen behaves as an ideal gas.
3
Properties The gas constant of oxygen is obtained from Table A-1E, R = 0.3353 psia⋅ft /lbm⋅R.
Analysis According to the ideal gas equation of state,
RT (0.3353 psia ⋅ ft3/lbm⋅R)(80 + 460 R)
v= = = 4.53 ft 3 /lbm
P 40 psia
Discussion In ideal gas calculations, it saves time to write the gas constant in appropriate units.
2-4
PROPRIETARY MATERIAL. © 2018 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.
Solutions Manual for Fluid Mechanics Fundamentals and Applications 4th Edition by Cengel IBSN 1259696537\
Chapter 2 Properties of Fluids
Solutions Manual for
Fluid Mechanics: Fundamentals and Applications
Fourth Edition
Yunus A. Çengel & John M. Cimbala
McGraw-Hill Education, 2018
M
ED
Chapter 2
PROPERTIES OF FLUIDS
ST
U
D
PROPRIETARY AND CONFIDENTIAL
Y
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 professors and instructors for
use in preparing for the classes using the affiliated textbook. No
other use or distribution of this Manual is permitted. This Manual
may not be sold and may not be distributed to or used by any
student or other third party. No part of this Manual may be
reproduced, displayed or distributed in any form or by any means,
electronic or otherwise, without the prior written permission of
McGraw-Hill Education.
2-1
, Chapter 2 Properties of Fluids
Density and Specific Gravity
2-1C
Solution We are to discuss the difference between intensive and extensive properties.
Analysis Intensive properties do not depend on the size (extent) of the system but extensive properties do depend
on the size (extent) of the system.
Discussion An example of an intensive property is temperature. An example of an extensive property is mass.
M
2-2C
ED
Solution We are to discuss the difference between mass and molar mass.
Analysis Mass m is the actual mass in grams or kilograms; molar mass M is the mass per mole in grams/mol or
kg/kmol. These two are related to each other by m = NM, where N is the number of moles.
ST
Discussion Mass, number of moles, and molar mass are often confused. Molar mass is also called molecular weight.
U
2-3C
Solution We are to define specific gravity and discuss its relationship to density.
D
Analysis The specific gravity, or relative density, is defined as the ratio of the density of a substance to the density
of some standard substance at a specified temperature (the standard is water at 4°C, for which ρH2O = 1000 kg/m ). That
3
Y
is, SG = ρ/ρH2O . When specific gravity is known, density is determined from ρ = SG×ρH2O .
Discussion Specific gravity is dimensionless and unitless [it is just a number without dimensions or units].
, Chapter 2 Properties of Fluids
2-4C
Solution We are to decide if the specific weight is an extensive or intensive property.
Analysis The original specific weight is
W
γ =
1
V
If we were to divide the system into two halves, each half weighs W/2 and occupies a volume of V /2. The specific weight
of one of these halves is
W /2
γ= =γ
V /2 1
which is the same as the original specific weight. Hence, specific weight is an intensive property.
Discussion If specific weight were an extensive property, its value for half of the system would be halved.
M
ED
2-5C
Solution We are to discuss the applicability of the ideal gas law.
Analysis A gas can be treated as an ideal gas when it is at a high temperature and/or a low pressure relative to its
critical temperature and pressure.
ST
Discussion Air and many other gases at room temperature and pressure can be approximated as ideal gases without any
significant loss of accuracy.
U
2-6C
D
Solution We are to discuss the difference between R and Ru.
Analysis Ru is the universal gas constant that is the same for all gases, whereas R is the specific gas constant that is
Y
different for different gases. These two are related to each other by R = Ru / M , where M is the molar mass (also called
the molecular weight) of the gas.
Discussion Since molar mass has dimensions of mass per mole, R and Ru do not have the same dimensions or units.
, Chapter 2 Properties of Fluids
2-7
Solution The pressure in a container that is filled with air is to be determined.
Assumptions At specified conditions, air behaves as an ideal gas.
kJ ⎛⎜ kPa ⋅m3 ⎟⎞ kPa ⋅m3
The gas constant of air is R = 0.287 = 0.287
⎟⎟
Properties (see also Table A-1).
kg⋅ K ⎜⎜⎝ kJ j kg⋅K
Analysis The definition of the specific volume gives
V 0.075 m3
v= = = 0.075 m 3/kg
m 1 kg
Using the ideal gas equation of state, the pressure is
RT (0.287 kPa⋅ m3/kg⋅K)(27 +273 K)
Pv = RT → P = = = 1148 kPa
v 0.075 m3/kg
Discussion In ideal gas calculations, it saves time to convert the gas constant to appropriate units.
M
ED
2-8E
Solution The volume of a tank that is filled with argon at a specified state is to be determined.
Assumptions At specified conditions, argon behaves as an ideal gas.
3
Properties The gas constant of argon is obtained from Table A-1E, R = 0.2686 psia⋅ft /lbm⋅R.
ST
Analysis According to the ideal gas equation of state,
mRT (1 lbm)(0.2686 psia ⋅ ft3/lbm⋅R)(100 + 460 R)
V= = = 0.7521 ft 3
P 200 psia
Discussion In ideal gas calculations, it saves time to write the gas constant in appropriate units.
U
D
2-9E
Solution The specific volume of oxygen at a specified state is to be determined.
Y
Assumptions At specified conditions, oxygen behaves as an ideal gas.
3
Properties The gas constant of oxygen is obtained from Table A-1E, R = 0.3353 psia⋅ft /lbm⋅R.
Analysis According to the ideal gas equation of state,
RT (0.3353 psia ⋅ ft3/lbm⋅R)(80 + 460 R)
v= = = 4.53 ft 3 /lbm
P 40 psia
Discussion In ideal gas calculations, it saves time to write the gas constant in appropriate units.
2-4
PROPRIETARY MATERIAL. © 2018 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.
