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Chapter 2 Properties of Fluids
Solutions Manual for
Fluid Mechanics: Fundamentals and Applications Third Edition Yunus A. Çengel & John M. Cimbala McGraw-Hill, 2013
CHAPTER 2 PROPERTIES OF FLUIDS
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Chapter 2 Properties of Fluids Density and Specific Gravity
2-1C 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. Discussion
Mass, number of moles, and molar mass are often confused. Molar mass is also called molecular weight.
2-2C 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.
2-3C Solution
We are to define specific gravity and discuss its relationship to density.
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 H2O = 1000 kg/m3). That is, SG / H2O . When specific gravity is known, density is determined from SG H2O . Discussion
Specific gravity is dimensionless and unitless [it is just a number without dimensions or units].
2-4C Solution
We are to decide if the specific weight is an extensive or intensive property.
Analysis
The original specific weight is
1
W 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 1 V /2
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.
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Chapter 2 Properties of Fluids 2-5C Solution We are to define the state postulate. Analysis The state postulate is expressed as: The state of a simple compressible system is completely specified by two independent, intensive properties. Discussion
An example of an intensive property is temperature.
2-6C 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. Discussion Air and many other gases at room temperature and pressure can be approximated as ideal gases without any significant loss of accuracy.
2-7C 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 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.
2-8 Solution Analysis
The volume and the weight of a fluid are given. Its mass and density are to be determined. Knowing the weight, the mass and the density of the fluid are determined to be
m
W 225 N g 9.80 m/s 2
1 kg m/s 3 1N
23.0 kg
m 23.0 kg 0.957 kg/L V 24 L Discussion Note that mass is an intrinsic property, but weight is not.
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Chapter 2 Properties of Fluids 2-9 Solution Assumptions
The pressure in a container that is filled with air is to be determined. At specified conditions, air behaves as an ideal gas.
Properties
The gas constant of air is R 0.287
Analysis
The definition of the specific volume gives v
kJ kPa m3 kPa m3 0.287 (see also Table A-1). kg K kJ kg K
V 0.100 m 3 0.100 m 3 /kg m 1 kg
Using the ideal gas equation of state, the pressure is
Pv RT
P
RT (0.287 kPa m 3 /kg K)(27 273.15 K) 861 kPa v 0.100 m3 /kg
Discussion
In ideal gas calculations, it saves time to convert the gas constant to appropriate units.
2-10E Solution Assumptions
The volume of a tank that is filled with argon at a specified state is to be determined. At specified conditions, argon behaves as an ideal gas.
Properties Analysis
The gas constant of argon is obtained from Table A-1E, R = 0.2686 psiaft3/lbmR. According to the ideal gas equation of state,
V
mRT (1 lbm)(0.2686 psia ft 3 /lbm R)(100 460 R) 0.7521 ft 3 P 200 psia
Discussion
In ideal gas calculations, it saves time to write the gas constant in appropriate units.
2-11E Solution Assumptions
The specific volume of oxygen at a specified state is to be determined. At specified conditions, oxygen behaves as an ideal gas.
Properties Analysis
The gas constant of oxygen is obtained from Table A-1E, R = 0.3353 psiaft3/lbmR. According to the ideal gas equation of state,
v
Discussion
RT (0.3353 psia ft 3 /lbm R)(80 460 R) 4.53 ft 3 /lbm P 40 psia
In ideal gas calculations, it saves time to write the gas constant in appropriate units.
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Chapter 2 Properties of Fluids 2-12E Solution An automobile tire is under-inflated with air. The amount of air that needs to be added to the tire to raise its pressure to the recommended value is to be determined. Assumptions
1 At specified conditions, air behaves as an ideal gas. 2 The volume of the tire remains constant.
Properties
The gas constant of air is Ru 53.34
Analysis
The initial and final absolute pressures in the tire are
ft lbf 1 psia lbm R 144 lbf/ft 2
P1 = Pg1 + Patm = 20 + 14.6 = 34.6 psia P2 = Pg2 + Patm = 30 + 14.6 = 44.6 psia
psia ft . 0.3794 lbm R 3
Tire 2.60 ft3 90F 20 psia
Treating air as an ideal gas, the initial mass in the tire is PV (34.6 psia)(2.60 ft 3 ) m1 1 0.4416 lbm RT1 (0.3704 psia ft 3 /lbm R)(550 R) Noting that the temperature and the volume of the tire remain constant, the final mass in the tire becomes PV (44.6 psia)(2.60 ft 3 ) m2 2 0.5692 lbm RT2 (0.3704 psia ft 3 /lbm R)(550 R) Thus the amount of air that needs to be added is m m 2 m1 0.5692 0.4416 0.128 lbm Discussion
Notice that absolute rather than gage pressure must be used in calculations with the ideal gas law.
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Chapter 2 Properties of Fluids 2-13 Solution An automobile tire is inflated with air. The pressure rise of air in the tire when the tire is heated and the amount of air that must be bled off to reduce the temperature to the original value are to be determined. Assumptions
1 At specified conditions, air behaves as an ideal gas. 2 The volume of the tire remains constant.
Properties
The gas constant of air is R 0.287
Analysis
Initially, the absolute pressure in the tire is
kJ kPa m3 kPa m3 . 0.287 kg K kJ kg K
P1 Pg Patm 210 100 310 kPa
Treating air as an ideal gas and assuming the volume of the tire to remain constant, the final pressure in the tire is determined from 323K P1V1 P2V 2 T P2 2 P1 (310kPa) 336 kPa T1 T2 T1 298K Tire 25C
Thus the pressure rise is P P2 P1 336 310 26.0 kPa
210 kPa
The amount of air that needs to be bled off to restore pressure to its original value is m1
P1V (310kPa)(0.025m 3 ) 0.0906kg RT1 (0.287kPa m 3 /kg K)(298K)
m2
P2V (310kPa)(0.025m 3 ) 0.0836kg RT2 (0.287 kPa m 3 /kg K)(323K) m m1 m 2 0.0906 0.0836 0.0070 kg
Discussion
Notice that absolute rather than gage pressure must be used in calculations with the ideal gas law.
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Chapter 2 Properties of Fluids 2-14 Solution
A balloon is filled with helium gas. The number of moles and the mass of helium are to be determined.
Assumptions
At specified conditions, helium behaves as an ideal gas.
Properties The molar mass of helium is 4.003 kg/kmol. The temperature of the helium gas is 20oC, which we must convert to absolute temperature for use in the equations: T = 20 + 273.15 = 293.15 K. The universal gas constant is kJ kPa m3 kPa m3 Ru 8.31447 . 8.31447 kmol K kJ kmol K Analysis
The volume of the sphere is
4 4 V r 3 (4.5 m)3 381.704 m3 3 3 Assuming ideal gas behavior, the number of moles of He is determined from
N
PV (200 kPa)(381.704 m3 ) 31.321 kmol 31.3 kmol RuT (8.31447 kPa m3 /kmol K)(293.15 K)
He D=9m 20C 200 kPa
Then the mass of He is determined from
m NM (31.321 kmol)(4.003 kg/kmol) 125.38 kg 125 kg Discussion Although the helium mass may seem large (about the mass of an adult football player!), it is much smaller than that of the air it displaces, and that is why helium balloons rise in the air.
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Chapter 2 Properties of Fluids 2-15 Solution A balloon is filled with helium gas. The effect of the balloon diameter on the mass of helium is to be investigated, and the results are to be tabulated and plotted. Properties The molar mass of helium is 4.003 kg/kmol. The temperature of the helium gas is 20oC, which we must convert to absolute temperature for use in the equations: T = 20 + 273.15 = 293.15 K. The universal gas constant is kJ kPa m3 kPa m3 Ru 8.31447 . 8.31447 kmol K kJ kmol K Analysis The EES Equations window is shown below, followed by the two parametric tables and the plot (we overlaid the two cases to get them to appear on the same plot). P = 100 kPa:
P = 200 kPa:
P = 200 kPa
P = 100 kPa
Discussion
Mass increases with diameter as expected, but not linearly since volume is proportional to D3. 2-8
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Chapter 2 Properties of Fluids
2-16 Solution A cylindrical tank contains methanol at a specified mass and volume. The methanol’s weight, density, and specific gravity and the force needed to accelerate the tank at a specified rate are to be determined. Assumptions
1 The volume of the tank remains constant.
Properties
The density of water is 1000 kg/m3.
Analysis
The methanol’s weight, density, and specific gravity are
The force needed to accelerate the tank at the given rate is
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Chapter 2 Properties of Fluids 2-17 Solution Using the data for the density of R-134a in Table A-4, an expression for the density as a function of temperature in a specified form is to be obtained. Analysis
An Excel sheet gives the following results. Therefore we obtain
Temp
Temp,K
Density
Rel. Error, %
‐20
253
1359
‐1.801766
‐10
263
1327
‐0.2446119
0
273
1295
0.8180695
10
283
1261
1.50943695
20
293
1226
1.71892333
30
303
1188
1.57525253
40
313
1147
1.04219704
50
323
1102
0.16279492
60
333
1053
‐1.1173789
70
343
996.2
‐2.502108
80
353
928.2
‐3.693816
90
363
837.7
‐3.4076638
100
373
651.7
10.0190272
The relative accuracy is quite reasonable except the last data point.
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Chapter 2 Properties of Fluids 2-18E Solution A rigid tank contains slightly pressurized air. The amount of air that needs to be added to the tank to raise its pressure and temperature to the recommended values is to be determined. Assumptions
1 At specified conditions, air behaves as an ideal gas. 2 The volume of the tank remains constant.
Properties
The gas constant of air is Ru 53.34
ft lbf 1 psia lbm R 144 lbf/ft 2
psia ft . The air temperature is 0.3794 lbm R 3
70oF = 70 + 459.67 = 529.67 R Analysis
Treating air as an ideal gas, the initial volume and the final mass in the tank are determined to be
V
m1 RT1 (40 lbm)(0.3704 psia ft 3 /lbm R)(529.67 R) 392.380 ft 3 P1 20 psia
m2
P2V (35 psia)(392.380 ft 3 ) 67.413 lbm RT2 (0.3704 psia ft 3 /lbm R)(550 R)
Air, 40 lbm 20 psia 70F
Thus the amount of air added is
m m2 m1 67.413 40.0 27.413 lbm 27.4 lbm Discussion
As the temperature slowly decreases due to heat transfer, the pressure will also decrease.
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Chapter 2 Properties of Fluids 2-19 Solution A relation for the variation of density with elevation is to be obtained, the density at 7 km elevation is to be calculated, and the mass of the atmosphere using the correlation is to be estimated. Assumptions 1 Atmospheric air behaves as an ideal gas. 2 The earth is perfectly spherical with a radius of 6377 km at sea level, and the thickness of the atmosphere is 25 km. Properties
The density data are given in tabular form as a function of radius and elevation, where r = z + 6377 km:
, kg/m3
1.4 1.2 1 3
1.225 1.112 1.007 0.9093 0.8194 0.7364 0.6601 0.5258 0.4135 0.1948 0.08891 0.04008
, kg/m
z, km 0 1 2 3 4 5 6 8 10 15 20 25
r, km 6377 6378 6379 6380 6381 6382 6383 6385 6387 6392 6397 6402
0.8 0.6 0.4 0.2 0 0
5
10
15
20
25
z, km
Analysis Using EES, (1) Define a trivial function “rho= a+z” in the Equation window, (2) select new parametric table from Tables, and type the data in a two-column table, (3) select Plot and plot the data, and (4) select Plot and click on curve fit to get curve fit window. Then specify 2nd order polynomial and enter/edit equation. The results are:
(z) = a + bz + cz2 = 1.20252 – 0.101674z + 0.0022375z2 for the unit of kg/m3, (or, (z) = (1.20252 – 0.101674z + 0.0022375z2)109 for the unit of kg/km3) where z is the vertical distance from the earth surface at sea level. At z = 7 km, the equation gives = 0.600 kg/m3. (b) The mass of atmosphere is evaluated by integration to be m
V
dV
h
z 0
(a bz cz 2 )4 (r0 z ) 2 dz 4
h
z 0
(a bz cz 2 )(r02 2r0 z z 2 )dz
4 ar02 h r0 (2a br0 )h 2 / 2 (a 2br0 cr02 )h 3 / 3 (b 2cr0 )h 4 / 4 ch 5 / 5 where r0 = 6377 km is the radius of the earth, h = 25 km is the thickness of the atmosphere. Also, a = 1.20252, b = -0.101674, and c = 0.0022375 are the constants in the density function. Substituting and multiplying by the factor 109 to convert the density from units of kg/km3 to kg/m3, the mass of the atmosphere is determined to be approximately
m = 5.091018 kg EES Solution for final result: a = 1.2025166 b = -0.10167 c = 0.0022375 r = 6377 h = 25 m = 4*pi*(a*r^2*h+r*(2*a+b*r)*h^2/2+(a+2*b*r+c*r^2)*h^3/3+(b+2*c*r)*h^4/4+c*h^5/5)*1E+9 Discussion
At 7 km, the density of the air is approximately half of its value at sea level.
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Chapter 2 Properties of Fluids Vapor Pressure and Cavitation
2-20C Solution
We are to define and discuss cavitation.
Analysis In the flow of a liquid, cavitation is the vaporization that may occur at locations where the pressure drops below the vapor pressure. The vapor bubbles collapse as they are swept away from the low pressure regions, generating highly destructive, extremely high-pressure waves. This phenomenon is a common cause for drop in performance and even the erosion of impeller blades. Discussion The word “cavitation” comes from the fact that a vapor bubble or “cavity” appears in the liquid. Not all cavitation is undesirable. It turns out that some underwater vehicles employ “super cavitation” on purpose to reduce drag.
2-21C Solution
We are to discuss whether the boiling temperature of water increases as pressure increases.
Analysis Yes. The saturation temperature of a pure substance depends on pressure; in fact, it increases with pressure. The higher the pressure, the higher the saturation or boiling temperature. Discussion This fact is easily seen by looking at the saturated water property tables. Note that boiling temperature and saturation pressure at a given pressure are equivalent.
2-22C Solution increases.
We are to determine if temperature increases or remains constant when the pressure of a boiling substance
Analysis If the pressure of a substance increases during a boiling process, the temperature also increases since the boiling (or saturation) temperature of a pure substance depends on pressure and increases with it. Discussion We are assuming that the liquid will continue to boil. If the pressure is increased fast enough, boiling may stop until the temperature has time to reach its new (higher) boiling temperature. A pressure cooker uses this principle.
2-23C Solution
We are to define vapor pressure and discuss its relationship to saturation pressure.
Analysis The vapor pressure Pv of a pure substance is defined as the pressure exerted by a vapor in phase equilibrium with its liquid at a given temperature. In general, the pressure of a vapor or gas, whether it exists alone or in a mixture with other gases, is called the partial pressure. During phase change processes between the liquid and vapor phases of a pure substance, the saturation pressure and the vapor pressure are equivalent since the vapor is pure. Discussion Partial pressure is not necessarily equal to vapor pressure. For example, on a dry day (low relative humidity), the partial pressure of water vapor in the air is less than the vapor pressure of water. If, however, the relative humidity is 100%, the partial pressure and the vapor pressure are equal.
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Chapter 2 Properties of Fluids 2-24E Solution
The minimum pressure in a pump is given. It is to be determined if there is a danger of cavitation.
Properties
The vapor pressure of water at 70F is 0.3632 psia.
Analysis To avoid cavitation, the pressure everywhere in the flow should remain above the vapor (or saturation) pressure at the given temperature, which is Pv Psat @ 70 F 0.3632 psia
The minimum pressure in the pump is 0.1 psia, which is less than the vapor pressure. Therefore, there is danger of cavitation in the pump. Discussion Note that the vapor pressure increases with increasing temperature, and the danger of cavitation increases at higher fluid temperatures.
2-25 Solution
The minimum pressure in a pump to avoid cavitation is to be determined.
Properties
The vapor pressure of water at 20C is 2.339 kPa.
Analysis To avoid cavitation, the pressure anywhere in the system should not be allowed to drop below the vapor (or saturation) pressure at the given temperature. That is, Pmin [email protected]C 2.339 kPa Therefore, the lowest pressure that can exist in the pump is 2.339 kPa. Discussion Note that the vapor pressure increases with increasing temperature, and thus the risk of cavitation is greater at higher fluid temperatures.
2-26 Solution
The minimum pressure in a piping system to avoid cavitation is to be determined.
Properties
The vapor pressure of water at 30C is 4.246 kPa.
Analysis To avoid cavitation, the pressure anywhere in the flow should not be allowed to drop below the vapor (or saturation) pressure at the given temperature. That is, Pmin [email protected]C 4.246 kPa
Therefore, the pressure should be maintained above 4.246 kPa everywhere in flow. Discussion Note that the vapor pressure increases with increasing temperature, and thus the risk of cavitation is greater at higher fluid temperatures.
2-27 Solution
The minimum pressure in a pump is given. It is to be determined if there is a danger of cavitation.
Properties
The vapor pressure of water at 20C is 2.339 kPa.
Analysis To avoid cavitation, the pressure everywhere in the flow should remain above the vapor (or saturation) pressure at the given temperature, which is
Pv Psat @ 20C 2.339 kPa The minimum pressure in the pump is 2 kPa, which is less than the vapor pressure. Therefore, a there is danger of cavitation in the pump. Discussion Note that the vapor pressure increases with increasing temperature, and thus there is a greater danger of cavitation at higher fluid temperatures. 2-14 PROPRIETARY MATERIAL. © 2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.
Chapter 2 Properties of Fluids
Energy and Specific Heats
2-28C Solution
We are to define and discuss flow energy.
Analysis Flow energy or flow work is the energy needed to push a fluid into or out of a control volume. Fluids at rest do not possess any flow energy. Discussion Flow energy is not a fundamental quantity, like kinetic or potential energy. However, it is a useful concept in fluid mechanics since fluids are often forced into and out of control volumes in practice.
2-29C Solution
We are to compare the energies of flowing and non-flowing fluids.
Analysis A flowing fluid possesses flow energy, which is the energy needed to push a fluid into or out of a control volume, in addition to the forms of energy possessed by a non-flowing fluid. The total energy of a non-flowing fluid consists of internal and potential energies. If the fluid is moving as a rigid body, but not flowing, it may also have kinetic energy (e.g., gasoline in a tank truck moving down the highway at constant speed with no sloshing). The total energy of a flowing fluid consists of internal, kinetic, potential, and flow energies. Discussion
Flow energy is not to be confused with kinetic energy, even though both are zero when the fluid is at rest.
2-30C Solution
We are to discuss the difference between macroscopic and microscopic forms of energy.
Analysis The macroscopic forms of energy are those a system possesses as a whole with respect to some outside reference frame. The microscopic forms of energy, on the other hand, are those related to the molecular structure of a system and the degree of the molecular activity, and are independent of outside reference frames. Discussion
We mostly deal with macroscopic forms of energy in fluid mechanics.
2-31C Solution
We are to define total energy and identify its constituents.
Analysis 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. Discussion All three constituents of total energy (kinetic, potential, and internal) need to be considered in an analysis of a general fluid flow.
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Chapter 2 Properties of Fluids 2-32C Solution
We are to list the forms of energy that contribute to the internal energy of a system.
Analysis 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. Discussion We deal with the flow of a single phase fluid in most problems in this textbook; therefore, latent, chemical, and nuclear energies do not need to be considered.
2-33C Solution
We are to discuss the relationship between heat, internal energy, and thermal energy.
Analysis Thermal energy is the sensible and latent forms of internal energy. It does not include chemical or nuclear forms of energy. In common terminology, thermal energy is referred to as heat. However, like work, heat is not a property, whereas thermal energy is a property. Discussion Technically speaking, “heat” is defined only when there is heat transfer, whereas the energy state of a substance can always be defined, even if no heat transfer is taking place.
2-34C Solution
We are to explain how changes in internal energy can be determined.
Analysis Using specific heat values at the average temperature, the changes in the specific internal energy of ideal gases can be determined from u c v, avg T . For incompressible substances, cp cv c and u c avg T . Discussion
If the fluid can be treated as neither incompressible nor an ideal gas, property tables must be used.
2-35C Solution
We are to explain how changes in enthalpy can be determined.
Analysis Using specific heat values at the average temperature, the changes in specific enthalpy of ideal gases can be determined from h c p ,avg T . For incompressible substances, cp cv c and h u vP c avg T vP . Discussion
If the fluid can be treated as neither incompressible nor an ideal gas, property tables must be used.
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Chapter 2 Properties of Fluids 2-36 Solution determined.
The total energy of saturated water vapor flowing in a pipe at a specified velocity and elevation is to be
Analysis
The total energy of a flowing fluid is given by (Eq. 28)
The enthalpy of the vapor at the specified temperature can be found in any thermo text to be energy is determined as
. Then the total
Note that only 0.047% of the total energy comes from the combination of kinetic and potential energies, which explains why we usually neglect kinetic and potential energies in most flow systems.
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Chapter 2 Properties of Fluids
Compressibility
2-37C Solution
We are to discuss the coefficient of compressibility and the isothermal compressibility.
Analysis The coefficient of compressibility represents the variation of pressure of a fluid with volume or density at constant temperature. Isothermal compressibility is the inverse of the coefficient of compressibility, and it represents the fractional change in volume or density corresponding to a change in pressure. Discussion
The coefficient of compressibility of an ideal gas is equal to its absolute pressure.
2-38C Solution
We are to define the coefficient of volume expansion.
Analysis The coefficient of volume expansion represents the variation of the density of a fluid with temperature at constant pressure. It differs from the coefficient of compressibility in that the latter represents the variation of pressure of a fluid with density at constant temperature. Discussion
The coefficient of volume expansion of an ideal gas is equal to the inverse of its absolute temperature.
2-39C Solution
We are to discuss the sign of the coefficient of compressibility and the coefficient of volume expansion.
Analysis The coefficient of compressibility of a fluid cannot be negative, but the coefficient of volume expansion can be negative (e.g., liquid water below 4C). Discussion
This is the reason that ice floats on water.
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Chapter 2 Properties of Fluids 2-40 Solution Water at a given temperature and pressure is heated to a higher temperature at constant pressure. The change in the density of water is to be determined. Assumptions 1 The coefficient of volume expansion is constant in the given temperature range. 2 An approximate analysis is performed by replacing differential changes in quantities by finite changes. Properties The density of water at 15C and 1 atm pressure is 1 = 999.1 kg/m3. The coefficient of volume expansion at the average temperature of (15+95)/2 = 55C is = 0.484 10-3 K-1. Analysis When differential quantities are replaced by differences and the properties and are assumed to be constant, the change in density in terms of the changes in pressure and temperature is expressed approximately as P T
The change in density due to the change of temperature from 15C to 95C at constant pressure is T (0.484 10 3 K -1 )(999.1 kg/m 3 )(95 15)K 38.7 kg/m 3
Noting that 2 1 , the density of water at 95C and 1 atm is
Discussion
2 1 999.1 (38.7) 960.4 kg/m 3 which is very close to the listed value of 961.5 kg/m3 at 95C in water table in the Appendix. This is mostly due to varying with temperature almost linearly. Note that the density of water decreases while being heated, as expected. This problem can be solved more accurately using differential analysis when functional forms of properties are available.
2-41 Solution The percent increase in the density of an ideal gas is given for a moderate pressure. The percent increase in density of the gas when compressed at a higher pressure is to be determined. Assumptions
The gas behaves an ideal gas.
Analysis
For an ideal gas, P = RT and (P / ) T RT P / , and thus ideal gas P . Therefore, the coefficient
of compressibility of an ideal gas is equal to its absolute pressure, and the coefficient of compressibility of the gas increases with increasing pressure. P P and rearranging Substituting = P into the definition of the coefficient of compressibility v / v / gives
P P
Therefore, the percent increase of density of an ideal gas during isothermal compression is equal to the percent increase in pressure. At 10 atm: At 1000 atm:
P 11 10 10% P 10 P 1001 1000 0.1% P 1000
Therefore, a pressure change of 1 atm causes a density change of 10% at 10 atm and a density change of 1% at 100 atm. Discussion
If temperature were also allowed to change, the relationship would not be so simple.
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Chapter 2 Properties of Fluids 2-42 Solution
Using the definition of the coefficient of volume expansion and the expression ideal gas 1 / T , it is to be
shown that the percent increase in the specific volume of an ideal gas during isobaric expansion is equal to the percent increase in absolute temperature. Assumptions
The gas behaves an ideal gas.
Analysis The coefficient of volume expansion can be expressed as
Noting that ideal gas 1 / T for an ideal gas and rearranging give
v / v 1 v . T v T P
v T v T
Therefore, the percent increase in the specific volume of an ideal gas during isobaric expansion is equal to the percent increase in absolute temperature. Discussion
We must be careful to use absolute temperature (K or R), not relative temperature (oC or oF).
2-43 Solution Water at a given temperature and pressure is compressed to a high pressure isothermally. The increase in the density of water is to be determined. Assumptions 1 The isothermal compressibility is constant in the given pressure range. 2 An approximate analysis is performed by replacing differential changes by finite changes. Properties The density of water at 20C and 1 atm pressure is 1 = 998 kg/m3. The isothermal compressibility of water is given to be = 4.80 10-5 atm-1. Analysis When differential quantities are replaced by differences and the properties and are assumed to be constant, the change in density in terms of the changes in pressure and temperature is expressed approximately as P T
The change in density due to a change of pressure from 1 atm to 400 atm at constant temperature is P (4.80 10 5 atm -1 )(998 kg/m 3 )(400 1)atm 19.2 kg/m 3
Discussion Note that the density of water increases from 998 to 1017.2 kg/m3 while being compressed, as expected. This problem can be solved more accurately using differential analysis when functional forms of properties are available.
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Chapter 2 Properties of Fluids 2-44 Solution determined. Assumptions Analysis
The volume of an ideal gas is reduced by half at constant temperature. The change in pressure is to be The process is isothermal and thus the temperature remains constant. For an ideal gas of fixed mass undergoing an isothermal process, the ideal gas relation reduces to
P2V 2 P1V1 T2 T1
P2V 2 P1V1
P2
V1 V P1 1 P1 2 P1 V2 0.5V1
Therefore, the change in pressure becomes P P2 P1 2 P1 P1 P1
Discussion
Note that at constant temperature, pressure and volume of an ideal gas are inversely proportional.
2-45 Solution Saturated refrigerant-134a at a given temperature is cooled at constant pressure. The change in the density of the refrigerant is to be determined. Assumptions 1 The coefficient of volume expansion is constant in the given temperature range. 2 An approximate analysis is performed by replacing differential changes in quantities by finite changes. Properties The density of saturated liquid R-134a at 10C is 1 =1261 kg/m3. The coefficient of volume expansion at the average temperature of (10+0)/2 = 5C is = 0.00269 K-1. Analysis When differential quantities are replaced by differences and the properties and are assumed to be constant, the change in density in terms of the changes in pressure and temperature is expressed approximately as P T
The change in density due to the change of temperature from 10C to 0C at constant pressure is T (0.00269 K -1 )(1261 kg/m 3 )(0 10)K 33.9 kg/m 3
Discussion
Noting that 2 1 , the density of R-134a at 0C is
2 1 1261 33.9 1294.9 kg/m 3 which is almost identical to the listed value of 1295 kg/m3 at 0C in R-134a table in the Appendix. This is mostly due to varying with temperature almost linearly. Note that the density increases during cooling, as expected.
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Chapter 2 Properties of Fluids 2-46 Solution A water tank completely filled with water can withstand tension caused by a volume expansion of 0.8%. The maximum temperature rise allowed in the tank without jeopardizing safety is to be determined. Assumptions 1 The coefficient of volume expansion is constant. 2 An approximate analysis is performed by replacing differential changes in quantities by finite changes. 3 The effect of pressure is disregarded. Properties
The average volume expansion coefficient is given to be = 0.377 10-3 K-1.
Analysis When differential quantities are replaced by differences and the properties and are assumed to be constant, the change in density in terms of the changes in pressure and temperature is expressed approximately as P T
A volume increase of 0.8% corresponds to a density decrease of 0.8%, which can be expressed as 0.008 . Then the decrease in density due to a temperature rise of T at constant pressure is 0.008 T
Solving for T and substituting, the maximum temperature rise is determined to be T
0.008
0.008 0.377 10 3 K -1
21.2 K 21.2C
Discussion This result is conservative since in reality the increasing pressure will tend to compress the water and increase its density.
2-47 Solution A water tank completely filled with water can withstand tension caused by a volume expansion of 1.5%. The maximum temperature rise allowed in the tank without jeopardizing safety is to be determined. Assumptions 1 The coefficient of volume expansion is constant. 2 An approximate analysis is performed by replacing differential changes in quantities by finite changes. 3 The effect of pressure is disregarded. Properties
The average volume expansion coefficient is given to be = 0.377 10-3 K-1.
Analysis When differential quantities are replaced by differences and the properties and are assumed to be constant, the change in density in terms of the changes in pressure and temperature is expressed approximately as P T
A volume increase of 1.5% corresponds to a density decrease of 1.5%, which can be expressed as 0.015 . Then the decrease in density due to a temperature rise of T at constant pressure is 0.015 T
Solving for T and substituting, the maximum temperature rise is determined to be T
0.015
0.015 0.377 10 3 K -1
39.8 K 39.8C
Discussion This result is conservative since in reality the increasing pressure will tend to compress the water and increase its density. The change in temperature is exactly half of that of the previous problem, as expected.
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Chapter 2 Properties of Fluids 2-48 Solution The density of seawater at the free surface and the bulk modulus of elasticity are given. The density and pressure at a depth of 2500 m are to be determined. Assumptions 1 The temperature and the bulk modulus of elasticity of seawater is constant. 2 The gravitational acceleration remains constant. Properties The density of seawater at free surface where the pressure is given to be 1030 kg/m3, and the bulk modulus of elasticity of seawater is given to be 2.34 109 N/m2. Analysis
The coefficient of compressibility or the bulk modulus of elasticity of fluids is expressed as P
T
or
dP d
(at constant T )
The differential pressure change across a differential fluid height of dz is given as
z=0
dP gdz
z
Combining the two relations above and rearranging,
gdz dz g 2 d d
d
2
gdz
2500 m
Integrating from z = 0 where 0 1030 kg/m 3 to z = z where gives d
0
2
g
z
dz
0
1
0
1
gz
Solving for gives the variation of density with depth as
1
1 / 0 gz /
Substituting into the pressure change relation dP gdz and integrating from z = 0 where P P0 98 kPa to z = z where P = P gives
P
dP
P0
z
0
gdz 1 / 0 gz /
1 P P0 ln 1 0 gz /
which is the desired relation for the variation of pressure in seawater with depth. At z = 2500 m, the values of density and pressure are determined by substitution to be
1 1 /(1030 kg/m ) (9.81 m/s )(2500 m) /( 2.34 10 N/m ) 3
2
9
2
1041 kg/m 3
1 P (98,000 Pa) (2.34 10 9 N/m 2 ) ln 3 2 9 2 1 (1030 kg/m )(9.81 m/s )(2500 m) /( 2.34 10 N/m ) 2.550 10 7 Pa 25.50 MPa
since 1 Pa = 1 N/m2 = 1 kg/ms2 and 1 kPa = 1000 Pa. Discussion Note that if we assumed = o = constant at 1030 kg/m3, the pressure at 2500 m would be P P0 gz = 0.098 + 25.26 = 25.36 MPa. Then the density at 2500 m is estimated to be P (1030)(2340 MPa) 1 (25.26 MPa) 11.1 kg/m 3 and thus = 1041 kg/m3
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Chapter 2 Properties of Fluids 2-49E Solution The coefficient of compressibility of water is given. The pressure increases required to reduce the volume of water by 1 percent and then by 2 percent are to be determined. Assumptions 1 The coefficient of compressibility is constant. 2 The temperature remains constant. Properties The coefficient of compressibility of water is given to be 7×105 psia. Analysis (a) A volume decrease of 1 percent can mathematically be expressed as v V 0.01 v V The coefficient of compressibility is expressed as P P v v /v T
v
Rearranging and substituting, the required pressure increase is determined to be v 5 P (7 10 psia )(0.01) 7,000 psia v (b) Similarly, the required pressure increase for a volume reduction of 2 percent becomes v 5 P (7 10 psia )(0.02) 14,000 psia v Discussion Note that at extremely high pressures are required to compress water to an appreciable amount.
2-50E Solution
We are to estimate the energy required to heat up the water in a hot-water tank.
Assumptions 1 There are no losses. 2 The pressure in the tank remains constant at 1 atm. 3 An approximate analysis is performed by replacing differential changes in quantities by finite changes. Properties The specific heat of water is approximated as a constant, whose value is 0.999 Btu/lbmR at the average temperature of (60 + 110)/2 = 85oF. In fact, c remains constant at 0.999 Btu/lbmR (to three digits) from 60oF to 110oF. For this same temperature range, the density varies from 62.36 lbm/ft3 at 60oF to 61.86 lbm/ft3 at 110oF. We approximate the density as constant, whose value is 62.17 lbm/ft3 at the average temperature of 85oF. Analysis
For a constant pressure process, u cavg T . Since this is energy per unit mass, we must multiply by the
total mass of the water in the tank, i.e., U mcavg T V cavg T . Thus, 35.315 ft 3 31,135 Btu 31,100 Btu U Vcavg T (62.17 lbm/ft 3 )(75 gal)(0.999 Btu/lbm R)[(110 - 60)R] 264.17 gal
where we note temperature differences are identical in oF and R. Discussion We give the final answer to 3 significant digits. The actual energy required will be greater than this, due to heat transfer losses and other inefficiencies in the hot-water heating system.
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Chapter 2 Properties of Fluids 2-51 Solution
We are to prove that the coefficient of volume expansion for an ideal gas is equal to 1/T.
Assumptions
1 Temperature and pressure are in the range that the gas can be approximated as an ideal gas.
Analysis
The ideal gas law is P RT , which we re-write as
P 1 . By definition, . Thus, T P RT
substitution and differentiation yields
ideal gas
P 1 RT T
1 P 1 1/T RT 2 T P
where both pressure and the gas constant R are treated as constants in the differentiation. Discussion The coefficient of volume expansion of an ideal gas is not constant, but rather decreases with temperature. However, for small temperature differences, is often approximated as a constant with little loss of accuracy.
2-52 Solution The coefficient of compressibility of nitrogen gas is to be estimated using Van der Waals equation of state. The result is to be compared to ideal gas and experimental values. Assumptions
1 Nitrogen gas obeys the Van der Waals equation of state.
Analysis
From the definition we have
since
The gas constant of nitrogen is
(Table A1). Substituting given data we obtain
For the ideal gas behavior, the coefficient of compressibility is equal to the pressure (Eq. 215). Therefore we get
whichis in error by
compared to experimentally measured pressure.
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Chapter 2 Properties of Fluids 2-53 Solution The water contained in a piston-cylinder device is compressed isothermally. The energy needed is to be determined. Assumptions 1 The coefficient of compressibility of water remains unchanged during the compression. Analysis We take the water in the cylinder as the system. The energy needed to compress water is equal to the work done on the system, and can be expressed as
From the definition of coefficient of compressibility we have
Rearranging we obtain
which can be integrated from the initial state to any state as follows:
from which we obtain
Substituting in Eq. 1 we have
or
In terms of finite changes, the fractional change due to change in pressure can be expressed approximately as (Eq. 323)
or at 20 . Realizing that 10 kg water where is the isothermal compressibility of water, which is the final volume of water is determined to be occupies initially a volume of Then the work done on the water is
from which we obtain since
.
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Chapter 2 Properties of Fluids 2-54 Solution The water contained in a piston-cylinder device is compressed isothermally and the pressure increases linearly. The energy needed is to be determined. Assumptions
1 The pressure increases linearly.
Analysis We take the water in the cylinder as the system. The energy needed to compress water is equal to the work done on the system, and can be expressed as
For a linear pressure increase we take
In terms of finite changes, the fractional change due to change in pressure can be expressed approximately as (Eq. 323)
or
at 20 . Realizing that 10 kg water where is the isothermal compressibility of water, which is the final volume of water is determined to be occupies initially a volume of
Therefore the work expression becomes
or
Thus, we conclude that linear pressure increase approximation does not work well since it gives almost ten times larger work.
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Chapter 2 Properties of Fluids
Speed of Sound
2-55C Solution
We are to define and discuss sound and how it is generated and how it travels.
Analysis Sound is an infinitesimally small pressure wave. It is generated by a small disturbance in a medium. It travels by wave propagation. Sound waves cannot travel in a vacuum. Discussion
Electromagnetic waves, like light and radio waves, can travel in a vacuum, but sound cannot.
2-56C Solution
We are to discuss whether sound travels faster in warm or cool air.
Analysis
Sound travels faster in warm (higher temperature) air since c kRT .
Discussion On the microscopic scale, we can imagine the air molecules moving around at higher speed in warmer air, leading to higher propagation of disturbances.
2-57C Solution
We are to compare the speed of sound in air, helium, and argon.
Analysis Sound travels fastest in helium, since c kRT and helium has the highest kR value. It is about 0.40 for air, 0.35 for argon, and 3.46 for helium. Discussion temperature.
2-58C Solution
We are assuming, of course, that these gases behave as ideal gases – a good approximation at room
We are to compare the speed of sound in air at two different pressures, but the same temperature.
Analysis Air at specified conditions will behave like an ideal gas, and the speed of sound in an ideal gas depends on temperature only. Therefore, the speed of sound is the same in both mediums. Discussion
If the temperature were different, however, the speed of sound would be different.
2-59C Solution
We are to examine whether the Mach number remains constant in constant-velocity flow.
Analysis In general, no, because the Mach number also depends on the speed of sound in gas, which depends on the temperature of the gas. The Mach number remains constant only if the temperature and the velocity are constant. Discussion It turns out that the speed of sound is not a strong function of pressure. In fact, it is not a function of pressure at all for an ideal gas.
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Chapter 2 Properties of Fluids 2-60C Solution
We are to state whether the propagation of sound waves is an isentropic process.
Analysis Yes, the propagation of sound waves is nearly isentropic. Because the amplitude of an ordinary sound wave is very small, and it does not cause any significant change in temperature and pressure. Discussion
No process is truly isentropic, but the increase of entropy due to sound propagation is negligibly small.
2-61C Solution
We are to discuss sonic velocity – specifically, whether it is constant or it changes.
Analysis The sonic speed in a medium depends on the properties of the medium, and it changes as the properties of the medium change. Discussion
The most common example is the change in speed of sound due to temperature change.
2-62 Solution
The Mach number of a passenger plane for specified limiting operating conditions is to be determined.
Assumptions Properties Analysis
Air is an ideal gas with constant specific heats at room temperature. The gas constant of air is R = 0.287 kJ/kg·K. Its specific heat ratio at room temperature is k = 1.4. From the speed of sound relation
1000 m 2 / s 2 c kRT (1.4)(0.287 kJ/kg K)(-60 273 K) 1 kJ/kg
293 m/s
Thus, the Mach number corresponding to the maximum cruising speed of the plane is Ma
Discussion same result.
V max (945 / 3.6) m/s 0.897 c 293 m/s
Note that this is a subsonic flight since Ma < 1. Also, using a k value at -60C would give practically the
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Chapter 2 Properties of Fluids 2-63 Solution Carbon dioxide flows through a nozzle. The inlet temperature and velocity and the exit temperature of CO2 are specified. The Mach number is to be determined at the inlet and exit of the nozzle. Assumptions 1 CO2 is an ideal gas with constant specific heats at room temperature. 2 This is a steady-flow process. Properties The gas constant of carbon dioxide is R = 0.1889 kJ/kg·K. Its constant pressure specific heat and specific heat ratio at room temperature are cp = 0.8439 kJ/kgK and k = 1.288. Analysis (a) At the inlet 1000 m 2 / s 2 c1 k1 RT1 (1.288)(0.1889 kJ/kg K)(1200 K) 1 kJ/kg
540.3 m/s
Thus, Ma 1
V1 50 m/s 0.0925 c1 540.3 m/s
1200 K 50 m/s
Carbon dioxide
400 K
(b) At the exit, 1000 m 2 / s 2 c 2 k 2 RT2 (1.288)(0.1889 kJ/kg K)(400 K) 1 kJ/kg
312.0 m/s
The nozzle exit velocity is determined from the steady-flow energy balance relation, 0 h2 h1
V 2 2 V1 2 2
0 (0.8439 kJ/kg K)(400 1200 K)
0 c p (T2 T1 ) V 2 2 (50 m/s) 2 2
V 2 2 V1 2 2
1 kJ/kg V 2 1163 m/s 1000 m 2 / s 2
Thus, Ma 2
V 2 1163 m/s 3.73 c2 312 m/s
Discussion The specific heats and their ratio k change with temperature, and the accuracy of the results can be improved by accounting for this variation. Using EES (or another property database):
At 1200 K: cp = 1.278 kJ/kgK, k = 1.173
c1 = 516 m/s,
V1 = 50 m/s,
Ma1 = 0.0969
At 400 K: cp = 0.9383 kJ/kgK, k = 1.252 c2 = 308 m/s, V2 = 1356 m/s, Ma2 = 4.41 Therefore, the constant specific heat assumption results in an error of 4.5% at the inlet and 15.5% at the exit in the Mach number, which are significant.
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Chapter 2 Properties of Fluids 2-64 Solution Nitrogen flows through a heat exchanger. The inlet temperature, pressure, and velocity and the exit pressure and velocity are specified. The Mach number is to be determined at the inlet and exit of the heat exchanger. Assumptions 1 N2 is an ideal gas. 2 This is a steady-flow process. 3 The potential energy change is negligible. Properties The gas constant of N2 is R = 0.2968 kJ/kg·K. Its constant pressure specific heat and specific heat ratio at room temperature are cp = 1.040 kJ/kgK and k = 1.4. 1000 m 2 / s 2 c1 k1 RT1 (1.400)(0.2968 kJ/kg K)(283 K) 1 kJ/kg
Analysis
342.9 m/s
Thus,
120kJ/kg
V 100 m/s 0.292 Ma 1 1 c1 342.9 m/s
150 kPa 10C 100 m/s
From the energy balance on the heat exchanger, qin c p (T2 T1 )
Nitrogen
100 kPa 200 m/s
V2 2 V12 2
120 kJ/kg (1.040 kJ/kg.C)(T2 10C)
(200 m/s) 2 (100 m/s) 2 1 kJ/kg 2 1000 m 2 / s 2
It yields T2 = 111C = 384 K 1000 m 2 / s 2 c 2 k 2 RT2 (1.4 )(0.2968 kJ/kg K)(384 K) 1 kJ/kg
399 m/s
Thus, Ma 2
V 2 200 m/s 0.501 c 2 399 m/s
Discussion The specific heats and their ratio k change with temperature, and the accuracy of the results can be improved by accounting for this variation. Using EES (or another property database):
At 10C : cp = 1.038 kJ/kgK, k = 1.400
c1 = 343 m/s,
V1 = 100 m/s,
Ma1 = 0.292
At 111C cp = 1.041 kJ/kgK, k = 1.399 c2 = 399 m/s, V2 = 200 m/s, Ma2 = 0.501 Therefore, the constant specific heat assumption results in an error of 4.5% at the inlet and 15.5% at the exit in the Mach number, which are almost identical to the values obtained assuming constant specific heats.
2-65 Solution
The speed of sound in refrigerant-134a at a specified state is to be determined.
Assumptions Properties Analysis
R-134a is an ideal gas with constant specific heats at room temperature. The gas constant of R-134a is R = 0.08149 kJ/kg·K. Its specific heat ratio at room temperature is k = 1.108. From the ideal-gas speed of sound relation,
1000 m 2 / s 2 c kRT (1.108)(0.08149 kJ/kg K)(60 273 K) 1 kJ/kg
Discussion
173 m/s
Note that the speed of sound is independent of pressure for ideal gases.
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Chapter 2 Properties of Fluids 2-66 Solution The Mach number of an aircraft and the speed of sound in air are to be determined at two specified temperatures. Assumptions Air is an ideal gas with constant specific heats at room temperature. Properties The gas constant of air is R = 0.287 kJ/kg·K. Its specific heat ratio at room temperature is k = 1.4. Analysis From the definitions of the speed of sound and the Mach number, (a) At 300 K, 1000 m 2 / s 2 c kRT (1.4)(0.287 kJ/kg K)(300 K) 1 kJ/kg V 330 m/s and 0.951 Ma c 347 m/s (b) At 800 K,
347 m/s
1000 m 2 / s 2 567 m/s c kRT (1.4)(0.287 kJ/kg K)(800 K) 1 kJ/kg V 330 m/s 0.582 and Ma c 567 m/s Discussion Note that a constant Mach number does not necessarily indicate constant speed. The Mach number of a rocket, for example, will be increasing even when it ascends at constant speed. Also, the specific heat ratio k changes with temperature.
2-67E Solution Steam flows through a device at a specified state and velocity. The Mach number of steam is to be determined assuming ideal gas behavior. Assumptions Properties Analysis
Steam is an ideal gas with constant specific heats. The gas constant of steam is R = 0.1102 Btu/lbm·R. Its specific heat ratio is given to be k = 1.3. From the ideal-gas speed of sound relation,
25,037 ft 2 / s 2 c kRT (1.3)(0.1102 Btu/lbm R)(1160 R) 1 Btu/lbm
2040 ft/s
Thus, Ma
V 900 ft/s 0.441 c 2040 ft/s
Using property data from steam tables and not assuming ideal gas behavior, it can be shown that the Mach Discussion number in steam at the specified state is 0.446, which is sufficiently close to the ideal-gas value of 0.441. Therefore, the ideal gas approximation is a reasonable one in this case.
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Chapter 2 Properties of Fluids 2-68E
Solution Problem 2-67e is reconsidered. The variation of Mach number with temperature as the temperature changes between 350 and 700F is to be investigated, and the results are to be plotted. Analysis
The EES Equations window is printed below, along with the tabulated and plotted results.
T=Temperature+460 R=0.1102 V=900 k=1.3 c=SQRT(k*R*T*25037) Ma=V/c Mach number Ma 0.528 0.520 0.512 0.505 0.498 0.491 0.485 0.479 0.473 0.467 0.462 0.456 0.451 0.446 0.441
0.54
0.52
0.5
Ma
Temperature, T, F 350 375 400 425 450 475 500 525 550 575 600 625 650 675 700
0.48
0.46
0.44 350
400
450
500
550
600
650
700
Temperature, °F Discussion
Note that for a specified flow speed, the Mach number decreases with increasing temperature, as expected.
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Chapter 2 Properties of Fluids 2-69E Solution The inlet state and the exit pressure of air are given for an isentropic expansion process. The ratio of the initial to the final speed of sound is to be determined. Assumptions
Air is an ideal gas with constant specific heats at room temperature.
Properties The properties of air are R = 0.06855 Btu/lbm·R and k = 1.4. The specific heat ratio k varies with temperature, but in our case this change is very small and can be disregarded. Analysis
The final temperature of air is determined from the isentropic relation of ideal gases, P T2 T1 2 P1
( k 1) / k
60 (659.7 R) 170
(1.4 1) / 1.4
489.9 R
Treating k as a constant, the ratio of the initial to the final speed of sound can be expressed as Ratio
Discussion
c2 c1
k1RT1 k2 RT2
T1 T2
659.7 1.16 489.9
Note that the speed of sound is proportional to the square root of thermodynamic temperature.
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Chapter 2 Properties of Fluids 2-70 Solution The inlet state and the exit pressure of air are given for an isentropic expansion process. The ratio of the initial to the final speed of sound is to be determined. Assumptions
Air is an ideal gas with constant specific heats at room temperature.
Properties The properties of air are R = 0.287 kJ/kg·K and k = 1.4. The specific heat ratio k varies with temperature, but in our case this change is very small and can be disregarded. Analysis
The final temperature of air is determined from the isentropic relation of ideal gases, P T2 T1 2 P1
( k 1) / k
0.4 MPa (350.2 K) 2.2 MPa
(1.4 1) / 1.4
215.2 K
Treating k as a constant, the ratio of the initial to the final speed of sound can be expressed as Ratio
Discussion
c2 c1
k1 RT1
k 2 RT2
T1
T2
350.2
1.28
215.2
Note that the speed of sound is proportional to the square root of thermodynamic temperature.
2-71 Solution The inlet state and the exit pressure of helium are given for an isentropic expansion process. The ratio of the initial to the final speed of sound is to be determined. Assumptions
Helium is an ideal gas with constant specific heats at room temperature.
Properties
The properties of helium are R = 2.0769 kJ/kg·K and k = 1.667.
Analysis
The final temperature of helium is determined from the isentropic relation of ideal gases, P T2 T1 2 P1
( k 1) / k
0.4 (350.2 K) 2.2
(1.667 1) / 1.667
177.0 K
The ratio of the initial to the final speed of sound can be expressed as Ratio
Discussion
c2 c1
k1 RT1
k 2 RT2
T1 T2
350.2
1.41
177.0
Note that the speed of sound is proportional to the square root of thermodynamic temperature.
2-72 Solution The expression for the speed of sound for an ideal gas is to be obtained using the isentropic process equation and the definition of the speed of sound. Analysis
The isentropic relation Pvk = A where A is a constant can also be expressed as k
1 P A A k v
Substituting it into the relation for the speed of sound, ( A ) k P c 2 s
kA k 1 k ( A k ) / k ( P / ) kRT s
since for an ideal gas P = RT or RT = P/. Therefore, c kRT , which is the desired relation. Discussion
Notice that pressure has dropped out; the speed of sound in an ideal gas is not a function of pressure.
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Chapter 2 Properties of Fluids
Viscosity
2-73C Solution
We are to define and discuss viscosity.
Analysis Viscosity is a measure of the “stickiness” or “resistance to deformation” of a fluid. It is due to the internal frictional force that develops between different layers of fluids as they are forced to move relative to each other. Viscosity is caused by the cohesive forces between the molecules in liquids, and by the molecular collisions in gases. In general, liquids have higher dynamic viscosities than gases. Discussion The ratio of viscosity to density often appears in the equations of fluid mechanics, and is defined as the kinematic viscosity, = /.
2-74C Solution
We are to discuss Newtonian fluids.
Analysis Fluids whose shear stress is linearly proportional to the velocity gradient (shear strain) are called Newtonian fluids. Most common fluids such as water, air, gasoline, and oils are Newtonian fluids. Discussion
In the differential analysis of fluid flow, only Newtonian fluids are considered in this textbook.
2-75C Solution
We are to discuss how kinematic viscosity varies with temperature in liquids and gases.
Analysis (a) For liquids, the kinematic viscosity decreases with temperature. (b) For gases, the kinematic viscosity increases with temperature. Discussion
You can easily verify this by looking at the appendices.
2-76C Solution
We are to discuss how dynamic viscosity varies with temperature in liquids and gases.
Analysis (a) The dynamic viscosity of liquids decreases with temperature. (b) The dynamic viscosity of gases increases with temperature. Discussion A good way to remember this is that a car engine is much harder to start in the winter because the oil in the engine has a higher viscosity at low temperatures.
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Chapter 2 Properties of Fluids 2-77C Solution We are to compare the settling speed of balls dropped in water and oil; namely, we are to determine which will reach the bottom of the container first. Analysis When two identical small glass balls are dropped into two identical containers, one filled with water and the other with oil, the ball dropped in water will reach the bottom of the container first because of the much lower viscosity of water relative to oil. Discussion Oil is very viscous, with typical values of viscosity approximately 800 times greater than that of water at room temperature.
2-78E Solution The torque and the rpm of a double cylinder viscometer are given. The viscosity of the fluid is to be determined. Assumptions 1 The inner cylinder is completely submerged in the fluid. 2 The viscous effects on the two ends of the inner cylinder are negligible. 3 The fluid is Newtonian.
R
Analysis Substituting the given values, the viscosity of the fluid is determined to be l = 0.035 in T (1.2 lbf ft)(0.035/12 ft) 4 2 fluid 2.72 10 lbf s/ft 4 2 R 3 n L 4 2 (3 / 12 ft) 3 (250 / 60 s -1 )(5 ft) Discussion This is the viscosity value at temperature that existed during the experiment. Viscosity is a strong function of temperature, and the values can be significantly different at different temperatures.
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Chapter 2 Properties of Fluids 2-79 Solution A block is moved at constant velocity on an inclined surface. The force that needs to be applied in the horizontal direction when the block is dry, and the percent reduction in the required force when an oil film is applied on the surface are to be determined. Assumptions 1 The inclined surface is plane (perfectly flat, although tilted). 2 The friction coefficient and the oil film thickness are uniform. 3 The weight of the oil layer is negligible.
The absolute viscosity of oil is given to be = 0.012 Pas = 0.012 Ns/m2.
Properties
Analysis (a) The velocity of the block is constant, and thus its acceleration and the net force acting on it are zero. A free body diagram of the block is given. Then the force balance gives F1 Fx 0 : F1 F f cos 20 FN 1 sin 20 0 (1)
F
y
0:
FN 1 cos 20 F f sin 20 W 0
Friction force: F f fFN 1
(2)
V= 0.8 m/s Ff 200
y
FN1 200 W = 150 N
20
(3)
0
x
Substituting Eq. (3) into Eq. (2) and solving for FN1 gives W 150 N FN 1 177.0 N cos 20 f sin 20 cos 20 0.27 sin 20 Then from Eq. (1): F1 F f cos 20 FN 1 sin 20 (0.27 177 N) cos 20 (177 N) sin 20 105.5 N (b) In this case, the friction force is replaced by the shear force applied on the bottom surface of the block due to the oil. Because of the no-slip condition, the oil film sticks to the inclined surface at the bottom and the lower surface of the block at the top. Then the shear force is expressed as
Fshear w As
V As h 0.012 N s/m 2 0.5 0.2 m 2
V= 0.8 m/s 50 cm F2 20
0.8 m/s 4 10-4 m
0.4 mm
0
Fshear = wAs FN2
W = 150 N
2 .4 N Replacing the friction force by the shear force in part (a),
F F
x
0:
F2 Fshear cos 20 FN 2 sin 20 0
(4)
y
0:
FN 2 cos 20 Fshear sin 20 W 0
(5)
Eq. (5) gives FN 2 ( Fshear sin 20 W ) / cos 20 [(2.4 N ) sin 20 (150 N )] / cos 20 160.5 N Substituting into Eq. (4), the required horizontal force is determined to be F2 Fshear cos 20 FN 2 sin 20 (2.4 N) cos 20 (160.5 N) sin 20 57.2 N Then, our final result is expressed as F F2 105.5 57.2 100% 100% 45.8% Percentage reduction in required force = 1 F1 105.5 Discussion Note that the force required to push the block on the inclined surface reduces significantly by oiling the surface.
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Chapter 2 Properties of Fluids 2-80 Solution The velocity profile of a fluid flowing though a circular pipe is given. The friction drag force exerted on the pipe by the fluid in the flow direction per unit length of the pipe is to be determined. Assumptions
The viscosity of the fluid is constant.
Analysis
The wall shear stress is determined from its definition to be
nu max d rn nr n 1 1 u max n n dr R r R R R rR rR Note that the quantity du /dr is negative in pipe flow, and the negative sign is added to the w relation for pipes to make shear stress in the positive (flow) direction a positive quantity. (Or, du /dr = - du /dy since y = R – r). Then the friction drag force exerted by the fluid on the inner surface of the pipe becomes nu max F w Aw (2R ) L 2nu max L R Therefore, the drag force per unit length of the pipe is
w
du dr
u max
u(r) = umax(1-rn/Rn)
R r 0
umax
F / L 2nu max .
Discussion
Note that the drag force acting on the pipe in this case is independent of the pipe diameter.
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Chapter 2 Properties of Fluids 2-81 Solution A thin flat plate is pulled horizontally through an oil layer sandwiched between two plates, one stationary and the other moving at a constant velocity. The location in oil where the velocity is zero and the force that needs to be applied on the plate are to be determined. Assumptions
1 The thickness of the plate is negligible. 2 The velocity profile in each oil layer is linear.
Properties
The absolute viscosity of oil is given to be = 0.027 Pas = 0.027 Ns/m2.
Analysis (a) The velocity profile in each oil layer relative to the fixed wall is as shown in the figure below. The point of zero velocity is indicated by point A, and its distance from the lower plate is determined from geometric considerations (the similarity of the two triangles in the lower oil layer) to be 2.6 y A 3 yA 0.3
yA = 0.23636 mm Fixed wall
h1=1 mm
V = 3 m/s
h2=2.6 mm y
F
A
yA
Vw= 0.3 m/s Moving wall
(b) The magnitudes of shear forces acting on the upper and lower surfaces of the plate are Fshear, upper w, upper As As
du V 0 3 m/s As (0.027 N s/m 2 )(0.3 0.3 m 2 ) 7.29 N dy h1 1.0 10 -3 m
V Vw du [3 (0.3)] m/s As (0.027 N s/m 2 )(0.3 0.3 m 2 ) 3.08 N h2 dy 2.6 10 -3 m Noting that both shear forces are in the opposite direction of motion of the plate, the force F is determined from a force balance on the plate to be Fshear, lower w, lower As As
F Fshear, upper Fshear, lower 7.29 3.08 10.4 N
Discussion motion.
Note that wall shear is a friction force between a solid and a liquid, and it acts in the opposite direction of
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Chapter 2 Properties of Fluids 2-82 Solution We are to determine the torque required to rotate the inner cylinder of two Inner cylinder concentric cylinders, with the inner cylinder rotating and the outer cylinder stationary. We V are also to explain what happens when the gap gets bigger. Assumptions 1 The fluid is incompressible and Newtonian. 2 End effects (top and bottom) are negligible. 3 The gap is very small so that wall curvature effects are negligible. 4 The gap is so small that the velocity profile in the gap is linear.
h
y
u
Outer cylinder
Analysis (a) We assume a linear velocity profile between the two walls as sketched – the inner wall is moving at speed V = iRi and the outer wall is stationary. The thickness of the gap is h, and we let y be the distance from the outer wall into the fluid (towards the inner wall). Thus,
u V
y du V and h dy h
where
h Ro - Ri and V i Ri Since shear stress has dimensions of force/area, the clockwise (mathematically negative) tangential force acting along the surface of the inner cylinder by the fluid is
F A
i Ri V 2 Ri L 2 Ri L h Ro Ri
But the torque is the tangential force times the moment arm Ri. Also, we are asked for the torque required to turn the inner cylinder. This applied torque is counterclockwise (mathematically positive). Thus,
T FRi
2 Li Ri 3 2 Li Ri 3 Ro Ri h
(b) The above is only an approximation because we assumed a linear velocity profile. As long as the gap is very small, and therefore the wall curvature effects are negligible, this approximation should be very good. Another way to think about this is that when the gap is very small compared to the cylinder radii, a magnified view of the flow in the gap appears similar to flow between two infinite walls (Couette flow). However, as the gap increases, the curvature effects are no longer negligible, and the linear velocity profile is not expected to be a valid approximation. We do not expect the velocity to remain linear as the gap increases. Discussion It is possible to solve for the exact velocity profile for this problem, and therefore the torque can be found analytically, but this has to wait until the differential analysis chapter.
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Chapter 2 Properties of Fluids 2-83 Solution A clutch system is used to transmit torque through an oil film between two identical disks. For specified rotational speeds, the transmitted torque is to be determined. Assumptions
1 The thickness of the oil film is uniform. 2 The rotational speeds of the disks remain constant.
Properties
The absolute viscosity of oil is given to be = 0.38 Ns/m2.
Driving shaft
Driven shaft
30 cm
2 mm
SAE 30W oil
Analysis The disks are rotting in the same direction at different angular speeds of 1 and of 2 . Therefore, we can assume one of the disks to be stationary and the other to be rotating at an angular speed of 1 2 . The velocity gradient anywhere in the oil of film thickness h is V /h where V (1 2 )r is the tangential velocity. Then the wall shear stress anywhere on the surface of the faster disk at a distance r from the axis of rotation can be expressed as ( 2 )r du V 1 w dr h h h Then the shear force acting on a differential area dA on the surface and the torque generation associated with it can be expressed as 2r ( 2 )r dF w dA 1 (2r )dr r 1 h dT rdF
(1 2 )r 2 2 (1 2 ) 3 (2r )dr r dr h h
Integrating, T
2 (1 2 ) h
D/2
r 0
r 3 dr
2 (1 2 ) r 4 h 4
D/2
(1 2 ) D 4
r 0
Noting that 2 n , the relative angular speed is
32h
1 min
1 2 2 n1 n2 2 rad/rev 1450 1398 rev/min 5.445 rad/s , 60 s Substituting, the torque transmitted is determined to be T
(0.38 N s/m 2 )(5.445 /s)(0.30 m) 4 32(0.002 m)
0.82 N m
Discussion Note that the torque transmitted is proportional to the fourth power of disk diameter, and is inversely proportional to the thickness of the oil film.
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Chapter 2 Properties of Fluids 2-84 Solution We are to investigate the effect of oil film thickness on the transmitted torque. Analysis The previous problem is reconsidered. Using EES software, the effect of oil film thickness on the torque transmitted is investigated. Film thickness varied from 0.1 mm to 10 mm, and the results are tabulated and (1 2 ) D 4 . The EES Equations window is printed below, followed by the plotted. The relation used is T 32h tabulated and plotted results. mu=0.38 n1=1450 "rpm" w1=2*pi*n1/60 "rad/s" n2=1398 "rpm" w2=2*pi*n2/60 "rad/s" D=0.3 "m" Tq=pi*mu*(w1-w2)*(D^4)/(32*h) Film thickness h, mm 0.1 0.2 0.4 0.6 0.8 1 2 4 6 8 10
Torque transmitted T, Nm 16.46 8.23 4.11 2.74 2.06 1.65 0.82 0.41 0.27 0.21 0.16
20
Tq, Nm
16
12
8
4
0 0
2
4
6
8
10
h, mm Conclusion Torque transmitted is inversely proportional to oil film thickness, and the film thickness should be as small as possible to maximize the transmitted torque. Discussion value of h.
To obtain the solution in EES, we set up a parametric table, specify h, and let EES calculate T for each
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Chapter 2 Properties of Fluids 2-85 Solution The viscosities of carbon dioxide at two temperatures are given. The constants of Sutherland correlation for carbon dioxide are to be determined and the viscosity of carbon dioxide at a specified temperature is to be predicted and compared to the value in table A-10. Analysis
where
Sutherland correlation is given by Eq. 232 as
is the absolute temperature. Substituting the given values we have
which is a nonlinear system of two algebraic equations. Using EES or any other computer code, one finds the following result:
Using these values the Sutherland correlation becomes
Therefore the viscosity at 100
is found to be
The agreement is perfect and within approximately 0.1%.
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Solutions Manual for
Fluid Mechanics: Fundamentals and Applications Third Edition Yunus A. Çengel & John M. Cimbala McGraw-Hill, 2013
CHAPTER 2 PROPERTIES OF FLUIDS
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Chapter 2 Properties of Fluids Density and Specific Gravity
2-1C 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. Discussion
Mass, number of moles, and molar mass are often confused. Molar mass is also called molecular weight.
2-2C 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.
2-3C Solution
We are to define specific gravity and discuss its relationship to density.
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 H2O = 1000 kg/m3). That is, SG / H2O . When specific gravity is known, density is determined from SG H2O . Discussion
Specific gravity is dimensionless and unitless [it is just a number without dimensions or units].
2-4C Solution
We are to decide if the specific weight is an extensive or intensive property.
Analysis
The original specific weight is
1
W 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 1 V /2
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.
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Chapter 2 Properties of Fluids 2-5C Solution We are to define the state postulate. Analysis The state postulate is expressed as: The state of a simple compressible system is completely specified by two independent, intensive properties. Discussion
An example of an intensive property is temperature.
2-6C 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. Discussion Air and many other gases at room temperature and pressure can be approximated as ideal gases without any significant loss of accuracy.
2-7C 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 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.
2-8 Solution Analysis
The volume and the weight of a fluid are given. Its mass and density are to be determined. Knowing the weight, the mass and the density of the fluid are determined to be
m
W 225 N g 9.80 m/s 2
1 kg m/s 3 1N
23.0 kg
m 23.0 kg 0.957 kg/L V 24 L Discussion Note that mass is an intrinsic property, but weight is not.
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Chapter 2 Properties of Fluids 2-9 Solution Assumptions
The pressure in a container that is filled with air is to be determined. At specified conditions, air behaves as an ideal gas.
Properties
The gas constant of air is R 0.287
Analysis
The definition of the specific volume gives v
kJ kPa m3 kPa m3 0.287 (see also Table A-1). kg K kJ kg K
V 0.100 m 3 0.100 m 3 /kg m 1 kg
Using the ideal gas equation of state, the pressure is
Pv RT
P
RT (0.287 kPa m 3 /kg K)(27 273.15 K) 861 kPa v 0.100 m3 /kg
Discussion
In ideal gas calculations, it saves time to convert the gas constant to appropriate units.
2-10E Solution Assumptions
The volume of a tank that is filled with argon at a specified state is to be determined. At specified conditions, argon behaves as an ideal gas.
Properties Analysis
The gas constant of argon is obtained from Table A-1E, R = 0.2686 psiaft3/lbmR. According to the ideal gas equation of state,
V
mRT (1 lbm)(0.2686 psia ft 3 /lbm R)(100 460 R) 0.7521 ft 3 P 200 psia
Discussion
In ideal gas calculations, it saves time to write the gas constant in appropriate units.
2-11E Solution Assumptions
The specific volume of oxygen at a specified state is to be determined. At specified conditions, oxygen behaves as an ideal gas.
Properties Analysis
The gas constant of oxygen is obtained from Table A-1E, R = 0.3353 psiaft3/lbmR. According to the ideal gas equation of state,
v
Discussion
RT (0.3353 psia ft 3 /lbm R)(80 460 R) 4.53 ft 3 /lbm P 40 psia
In ideal gas calculations, it saves time to write the gas constant in appropriate units.
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Chapter 2 Properties of Fluids 2-12E Solution An automobile tire is under-inflated with air. The amount of air that needs to be added to the tire to raise its pressure to the recommended value is to be determined. Assumptions
1 At specified conditions, air behaves as an ideal gas. 2 The volume of the tire remains constant.
Properties
The gas constant of air is Ru 53.34
Analysis
The initial and final absolute pressures in the tire are
ft lbf 1 psia lbm R 144 lbf/ft 2
P1 = Pg1 + Patm = 20 + 14.6 = 34.6 psia P2 = Pg2 + Patm = 30 + 14.6 = 44.6 psia
psia ft . 0.3794 lbm R 3
Tire 2.60 ft3 90F 20 psia
Treating air as an ideal gas, the initial mass in the tire is PV (34.6 psia)(2.60 ft 3 ) m1 1 0.4416 lbm RT1 (0.3704 psia ft 3 /lbm R)(550 R) Noting that the temperature and the volume of the tire remain constant, the final mass in the tire becomes PV (44.6 psia)(2.60 ft 3 ) m2 2 0.5692 lbm RT2 (0.3704 psia ft 3 /lbm R)(550 R) Thus the amount of air that needs to be added is m m 2 m1 0.5692 0.4416 0.128 lbm Discussion
Notice that absolute rather than gage pressure must be used in calculations with the ideal gas law.
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Chapter 2 Properties of Fluids 2-13 Solution An automobile tire is inflated with air. The pressure rise of air in the tire when the tire is heated and the amount of air that must be bled off to reduce the temperature to the original value are to be determined. Assumptions
1 At specified conditions, air behaves as an ideal gas. 2 The volume of the tire remains constant.
Properties
The gas constant of air is R 0.287
Analysis
Initially, the absolute pressure in the tire is
kJ kPa m3 kPa m3 . 0.287 kg K kJ kg K
P1 Pg Patm 210 100 310 kPa
Treating air as an ideal gas and assuming the volume of the tire to remain constant, the final pressure in the tire is determined from 323K P1V1 P2V 2 T P2 2 P1 (310kPa) 336 kPa T1 T2 T1 298K Tire 25C
Thus the pressure rise is P P2 P1 336 310 26.0 kPa
210 kPa
The amount of air that needs to be bled off to restore pressure to its original value is m1
P1V (310kPa)(0.025m 3 ) 0.0906kg RT1 (0.287kPa m 3 /kg K)(298K)
m2
P2V (310kPa)(0.025m 3 ) 0.0836kg RT2 (0.287 kPa m 3 /kg K)(323K) m m1 m 2 0.0906 0.0836 0.0070 kg
Discussion
Notice that absolute rather than gage pressure must be used in calculations with the ideal gas law.
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Chapter 2 Properties of Fluids 2-14 Solution
A balloon is filled with helium gas. The number of moles and the mass of helium are to be determined.
Assumptions
At specified conditions, helium behaves as an ideal gas.
Properties The molar mass of helium is 4.003 kg/kmol. The temperature of the helium gas is 20oC, which we must convert to absolute temperature for use in the equations: T = 20 + 273.15 = 293.15 K. The universal gas constant is kJ kPa m3 kPa m3 Ru 8.31447 . 8.31447 kmol K kJ kmol K Analysis
The volume of the sphere is
4 4 V r 3 (4.5 m)3 381.704 m3 3 3 Assuming ideal gas behavior, the number of moles of He is determined from
N
PV (200 kPa)(381.704 m3 ) 31.321 kmol 31.3 kmol RuT (8.31447 kPa m3 /kmol K)(293.15 K)
He D=9m 20C 200 kPa
Then the mass of He is determined from
m NM (31.321 kmol)(4.003 kg/kmol) 125.38 kg 125 kg Discussion Although the helium mass may seem large (about the mass of an adult football player!), it is much smaller than that of the air it displaces, and that is why helium balloons rise in the air.
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Chapter 2 Properties of Fluids 2-15 Solution A balloon is filled with helium gas. The effect of the balloon diameter on the mass of helium is to be investigated, and the results are to be tabulated and plotted. Properties The molar mass of helium is 4.003 kg/kmol. The temperature of the helium gas is 20oC, which we must convert to absolute temperature for use in the equations: T = 20 + 273.15 = 293.15 K. The universal gas constant is kJ kPa m3 kPa m3 Ru 8.31447 . 8.31447 kmol K kJ kmol K Analysis The EES Equations window is shown below, followed by the two parametric tables and the plot (we overlaid the two cases to get them to appear on the same plot). P = 100 kPa:
P = 200 kPa:
P = 200 kPa
P = 100 kPa
Discussion
Mass increases with diameter as expected, but not linearly since volume is proportional to D3. 2-8
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Chapter 2 Properties of Fluids
2-16 Solution A cylindrical tank contains methanol at a specified mass and volume. The methanol’s weight, density, and specific gravity and the force needed to accelerate the tank at a specified rate are to be determined. Assumptions
1 The volume of the tank remains constant.
Properties
The density of water is 1000 kg/m3.
Analysis
The methanol’s weight, density, and specific gravity are
The force needed to accelerate the tank at the given rate is
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Chapter 2 Properties of Fluids 2-17 Solution Using the data for the density of R-134a in Table A-4, an expression for the density as a function of temperature in a specified form is to be obtained. Analysis
An Excel sheet gives the following results. Therefore we obtain
Temp
Temp,K
Density
Rel. Error, %
‐20
253
1359
‐1.801766
‐10
263
1327
‐0.2446119
0
273
1295
0.8180695
10
283
1261
1.50943695
20
293
1226
1.71892333
30
303
1188
1.57525253
40
313
1147
1.04219704
50
323
1102
0.16279492
60
333
1053
‐1.1173789
70
343
996.2
‐2.502108
80
353
928.2
‐3.693816
90
363
837.7
‐3.4076638
100
373
651.7
10.0190272
The relative accuracy is quite reasonable except the last data point.
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Chapter 2 Properties of Fluids 2-18E Solution A rigid tank contains slightly pressurized air. The amount of air that needs to be added to the tank to raise its pressure and temperature to the recommended values is to be determined. Assumptions
1 At specified conditions, air behaves as an ideal gas. 2 The volume of the tank remains constant.
Properties
The gas constant of air is Ru 53.34
ft lbf 1 psia lbm R 144 lbf/ft 2
psia ft . The air temperature is 0.3794 lbm R 3
70oF = 70 + 459.67 = 529.67 R Analysis
Treating air as an ideal gas, the initial volume and the final mass in the tank are determined to be
V
m1 RT1 (40 lbm)(0.3704 psia ft 3 /lbm R)(529.67 R) 392.380 ft 3 P1 20 psia
m2
P2V (35 psia)(392.380 ft 3 ) 67.413 lbm RT2 (0.3704 psia ft 3 /lbm R)(550 R)
Air, 40 lbm 20 psia 70F
Thus the amount of air added is
m m2 m1 67.413 40.0 27.413 lbm 27.4 lbm Discussion
As the temperature slowly decreases due to heat transfer, the pressure will also decrease.
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Chapter 2 Properties of Fluids 2-19 Solution A relation for the variation of density with elevation is to be obtained, the density at 7 km elevation is to be calculated, and the mass of the atmosphere using the correlation is to be estimated. Assumptions 1 Atmospheric air behaves as an ideal gas. 2 The earth is perfectly spherical with a radius of 6377 km at sea level, and the thickness of the atmosphere is 25 km. Properties
The density data are given in tabular form as a function of radius and elevation, where r = z + 6377 km:
, kg/m3
1.4 1.2 1 3
1.225 1.112 1.007 0.9093 0.8194 0.7364 0.6601 0.5258 0.4135 0.1948 0.08891 0.04008
, kg/m
z, km 0 1 2 3 4 5 6 8 10 15 20 25
r, km 6377 6378 6379 6380 6381 6382 6383 6385 6387 6392 6397 6402
0.8 0.6 0.4 0.2 0 0
5
10
15
20
25
z, km
Analysis Using EES, (1) Define a trivial function “rho= a+z” in the Equation window, (2) select new parametric table from Tables, and type the data in a two-column table, (3) select Plot and plot the data, and (4) select Plot and click on curve fit to get curve fit window. Then specify 2nd order polynomial and enter/edit equation. The results are:
(z) = a + bz + cz2 = 1.20252 – 0.101674z + 0.0022375z2 for the unit of kg/m3, (or, (z) = (1.20252 – 0.101674z + 0.0022375z2)109 for the unit of kg/km3) where z is the vertical distance from the earth surface at sea level. At z = 7 km, the equation gives = 0.600 kg/m3. (b) The mass of atmosphere is evaluated by integration to be m
V
dV
h
z 0
(a bz cz 2 )4 (r0 z ) 2 dz 4
h
z 0
(a bz cz 2 )(r02 2r0 z z 2 )dz
4 ar02 h r0 (2a br0 )h 2 / 2 (a 2br0 cr02 )h 3 / 3 (b 2cr0 )h 4 / 4 ch 5 / 5 where r0 = 6377 km is the radius of the earth, h = 25 km is the thickness of the atmosphere. Also, a = 1.20252, b = -0.101674, and c = 0.0022375 are the constants in the density function. Substituting and multiplying by the factor 109 to convert the density from units of kg/km3 to kg/m3, the mass of the atmosphere is determined to be approximately
m = 5.091018 kg EES Solution for final result: a = 1.2025166 b = -0.10167 c = 0.0022375 r = 6377 h = 25 m = 4*pi*(a*r^2*h+r*(2*a+b*r)*h^2/2+(a+2*b*r+c*r^2)*h^3/3+(b+2*c*r)*h^4/4+c*h^5/5)*1E+9 Discussion
At 7 km, the density of the air is approximately half of its value at sea level.
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Chapter 2 Properties of Fluids Vapor Pressure and Cavitation
2-20C Solution
We are to define and discuss cavitation.
Analysis In the flow of a liquid, cavitation is the vaporization that may occur at locations where the pressure drops below the vapor pressure. The vapor bubbles collapse as they are swept away from the low pressure regions, generating highly destructive, extremely high-pressure waves. This phenomenon is a common cause for drop in performance and even the erosion of impeller blades. Discussion The word “cavitation” comes from the fact that a vapor bubble or “cavity” appears in the liquid. Not all cavitation is undesirable. It turns out that some underwater vehicles employ “super cavitation” on purpose to reduce drag.
2-21C Solution
We are to discuss whether the boiling temperature of water increases as pressure increases.
Analysis Yes. The saturation temperature of a pure substance depends on pressure; in fact, it increases with pressure. The higher the pressure, the higher the saturation or boiling temperature. Discussion This fact is easily seen by looking at the saturated water property tables. Note that boiling temperature and saturation pressure at a given pressure are equivalent.
2-22C Solution increases.
We are to determine if temperature increases or remains constant when the pressure of a boiling substance
Analysis If the pressure of a substance increases during a boiling process, the temperature also increases since the boiling (or saturation) temperature of a pure substance depends on pressure and increases with it. Discussion We are assuming that the liquid will continue to boil. If the pressure is increased fast enough, boiling may stop until the temperature has time to reach its new (higher) boiling temperature. A pressure cooker uses this principle.
2-23C Solution
We are to define vapor pressure and discuss its relationship to saturation pressure.
Analysis The vapor pressure Pv of a pure substance is defined as the pressure exerted by a vapor in phase equilibrium with its liquid at a given temperature. In general, the pressure of a vapor or gas, whether it exists alone or in a mixture with other gases, is called the partial pressure. During phase change processes between the liquid and vapor phases of a pure substance, the saturation pressure and the vapor pressure are equivalent since the vapor is pure. Discussion Partial pressure is not necessarily equal to vapor pressure. For example, on a dry day (low relative humidity), the partial pressure of water vapor in the air is less than the vapor pressure of water. If, however, the relative humidity is 100%, the partial pressure and the vapor pressure are equal.
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Chapter 2 Properties of Fluids 2-24E Solution
The minimum pressure in a pump is given. It is to be determined if there is a danger of cavitation.
Properties
The vapor pressure of water at 70F is 0.3632 psia.
Analysis To avoid cavitation, the pressure everywhere in the flow should remain above the vapor (or saturation) pressure at the given temperature, which is Pv Psat @ 70 F 0.3632 psia
The minimum pressure in the pump is 0.1 psia, which is less than the vapor pressure. Therefore, there is danger of cavitation in the pump. Discussion Note that the vapor pressure increases with increasing temperature, and the danger of cavitation increases at higher fluid temperatures.
2-25 Solution
The minimum pressure in a pump to avoid cavitation is to be determined.
Properties
The vapor pressure of water at 20C is 2.339 kPa.
Analysis To avoid cavitation, the pressure anywhere in the system should not be allowed to drop below the vapor (or saturation) pressure at the given temperature. That is, Pmin [email protected]C 2.339 kPa Therefore, the lowest pressure that can exist in the pump is 2.339 kPa. Discussion Note that the vapor pressure increases with increasing temperature, and thus the risk of cavitation is greater at higher fluid temperatures.
2-26 Solution
The minimum pressure in a piping system to avoid cavitation is to be determined.
Properties
The vapor pressure of water at 30C is 4.246 kPa.
Analysis To avoid cavitation, the pressure anywhere in the flow should not be allowed to drop below the vapor (or saturation) pressure at the given temperature. That is, Pmin [email protected]C 4.246 kPa
Therefore, the pressure should be maintained above 4.246 kPa everywhere in flow. Discussion Note that the vapor pressure increases with increasing temperature, and thus the risk of cavitation is greater at higher fluid temperatures.
2-27 Solution
The minimum pressure in a pump is given. It is to be determined if there is a danger of cavitation.
Properties
The vapor pressure of water at 20C is 2.339 kPa.
Analysis To avoid cavitation, the pressure everywhere in the flow should remain above the vapor (or saturation) pressure at the given temperature, which is
Pv Psat @ 20C 2.339 kPa The minimum pressure in the pump is 2 kPa, which is less than the vapor pressure. Therefore, a there is danger of cavitation in the pump. Discussion Note that the vapor pressure increases with increasing temperature, and thus there is a greater danger of cavitation at higher fluid temperatures. 2-14 PROPRIETARY MATERIAL. © 2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.
Chapter 2 Properties of Fluids
Energy and Specific Heats
2-28C Solution
We are to define and discuss flow energy.
Analysis Flow energy or flow work is the energy needed to push a fluid into or out of a control volume. Fluids at rest do not possess any flow energy. Discussion Flow energy is not a fundamental quantity, like kinetic or potential energy. However, it is a useful concept in fluid mechanics since fluids are often forced into and out of control volumes in practice.
2-29C Solution
We are to compare the energies of flowing and non-flowing fluids.
Analysis A flowing fluid possesses flow energy, which is the energy needed to push a fluid into or out of a control volume, in addition to the forms of energy possessed by a non-flowing fluid. The total energy of a non-flowing fluid consists of internal and potential energies. If the fluid is moving as a rigid body, but not flowing, it may also have kinetic energy (e.g., gasoline in a tank truck moving down the highway at constant speed with no sloshing). The total energy of a flowing fluid consists of internal, kinetic, potential, and flow energies. Discussion
Flow energy is not to be confused with kinetic energy, even though both are zero when the fluid is at rest.
2-30C Solution
We are to discuss the difference between macroscopic and microscopic forms of energy.
Analysis The macroscopic forms of energy are those a system possesses as a whole with respect to some outside reference frame. The microscopic forms of energy, on the other hand, are those related to the molecular structure of a system and the degree of the molecular activity, and are independent of outside reference frames. Discussion
We mostly deal with macroscopic forms of energy in fluid mechanics.
2-31C Solution
We are to define total energy and identify its constituents.
Analysis 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. Discussion All three constituents of total energy (kinetic, potential, and internal) need to be considered in an analysis of a general fluid flow.
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Chapter 2 Properties of Fluids 2-32C Solution
We are to list the forms of energy that contribute to the internal energy of a system.
Analysis 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. Discussion We deal with the flow of a single phase fluid in most problems in this textbook; therefore, latent, chemical, and nuclear energies do not need to be considered.
2-33C Solution
We are to discuss the relationship between heat, internal energy, and thermal energy.
Analysis Thermal energy is the sensible and latent forms of internal energy. It does not include chemical or nuclear forms of energy. In common terminology, thermal energy is referred to as heat. However, like work, heat is not a property, whereas thermal energy is a property. Discussion Technically speaking, “heat” is defined only when there is heat transfer, whereas the energy state of a substance can always be defined, even if no heat transfer is taking place.
2-34C Solution
We are to explain how changes in internal energy can be determined.
Analysis Using specific heat values at the average temperature, the changes in the specific internal energy of ideal gases can be determined from u c v, avg T . For incompressible substances, cp cv c and u c avg T . Discussion
If the fluid can be treated as neither incompressible nor an ideal gas, property tables must be used.
2-35C Solution
We are to explain how changes in enthalpy can be determined.
Analysis Using specific heat values at the average temperature, the changes in specific enthalpy of ideal gases can be determined from h c p ,avg T . For incompressible substances, cp cv c and h u vP c avg T vP . Discussion
If the fluid can be treated as neither incompressible nor an ideal gas, property tables must be used.
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Chapter 2 Properties of Fluids 2-36 Solution determined.
The total energy of saturated water vapor flowing in a pipe at a specified velocity and elevation is to be
Analysis
The total energy of a flowing fluid is given by (Eq. 28)
The enthalpy of the vapor at the specified temperature can be found in any thermo text to be energy is determined as
. Then the total
Note that only 0.047% of the total energy comes from the combination of kinetic and potential energies, which explains why we usually neglect kinetic and potential energies in most flow systems.
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Chapter 2 Properties of Fluids
Compressibility
2-37C Solution
We are to discuss the coefficient of compressibility and the isothermal compressibility.
Analysis The coefficient of compressibility represents the variation of pressure of a fluid with volume or density at constant temperature. Isothermal compressibility is the inverse of the coefficient of compressibility, and it represents the fractional change in volume or density corresponding to a change in pressure. Discussion
The coefficient of compressibility of an ideal gas is equal to its absolute pressure.
2-38C Solution
We are to define the coefficient of volume expansion.
Analysis The coefficient of volume expansion represents the variation of the density of a fluid with temperature at constant pressure. It differs from the coefficient of compressibility in that the latter represents the variation of pressure of a fluid with density at constant temperature. Discussion
The coefficient of volume expansion of an ideal gas is equal to the inverse of its absolute temperature.
2-39C Solution
We are to discuss the sign of the coefficient of compressibility and the coefficient of volume expansion.
Analysis The coefficient of compressibility of a fluid cannot be negative, but the coefficient of volume expansion can be negative (e.g., liquid water below 4C). Discussion
This is the reason that ice floats on water.
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Chapter 2 Properties of Fluids 2-40 Solution Water at a given temperature and pressure is heated to a higher temperature at constant pressure. The change in the density of water is to be determined. Assumptions 1 The coefficient of volume expansion is constant in the given temperature range. 2 An approximate analysis is performed by replacing differential changes in quantities by finite changes. Properties The density of water at 15C and 1 atm pressure is 1 = 999.1 kg/m3. The coefficient of volume expansion at the average temperature of (15+95)/2 = 55C is = 0.484 10-3 K-1. Analysis When differential quantities are replaced by differences and the properties and are assumed to be constant, the change in density in terms of the changes in pressure and temperature is expressed approximately as P T
The change in density due to the change of temperature from 15C to 95C at constant pressure is T (0.484 10 3 K -1 )(999.1 kg/m 3 )(95 15)K 38.7 kg/m 3
Noting that 2 1 , the density of water at 95C and 1 atm is
Discussion
2 1 999.1 (38.7) 960.4 kg/m 3 which is very close to the listed value of 961.5 kg/m3 at 95C in water table in the Appendix. This is mostly due to varying with temperature almost linearly. Note that the density of water decreases while being heated, as expected. This problem can be solved more accurately using differential analysis when functional forms of properties are available.
2-41 Solution The percent increase in the density of an ideal gas is given for a moderate pressure. The percent increase in density of the gas when compressed at a higher pressure is to be determined. Assumptions
The gas behaves an ideal gas.
Analysis
For an ideal gas, P = RT and (P / ) T RT P / , and thus ideal gas P . Therefore, the coefficient
of compressibility of an ideal gas is equal to its absolute pressure, and the coefficient of compressibility of the gas increases with increasing pressure. P P and rearranging Substituting = P into the definition of the coefficient of compressibility v / v / gives
P P
Therefore, the percent increase of density of an ideal gas during isothermal compression is equal to the percent increase in pressure. At 10 atm: At 1000 atm:
P 11 10 10% P 10 P 1001 1000 0.1% P 1000
Therefore, a pressure change of 1 atm causes a density change of 10% at 10 atm and a density change of 1% at 100 atm. Discussion
If temperature were also allowed to change, the relationship would not be so simple.
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Chapter 2 Properties of Fluids 2-42 Solution
Using the definition of the coefficient of volume expansion and the expression ideal gas 1 / T , it is to be
shown that the percent increase in the specific volume of an ideal gas during isobaric expansion is equal to the percent increase in absolute temperature. Assumptions
The gas behaves an ideal gas.
Analysis The coefficient of volume expansion can be expressed as
Noting that ideal gas 1 / T for an ideal gas and rearranging give
v / v 1 v . T v T P
v T v T
Therefore, the percent increase in the specific volume of an ideal gas during isobaric expansion is equal to the percent increase in absolute temperature. Discussion
We must be careful to use absolute temperature (K or R), not relative temperature (oC or oF).
2-43 Solution Water at a given temperature and pressure is compressed to a high pressure isothermally. The increase in the density of water is to be determined. Assumptions 1 The isothermal compressibility is constant in the given pressure range. 2 An approximate analysis is performed by replacing differential changes by finite changes. Properties The density of water at 20C and 1 atm pressure is 1 = 998 kg/m3. The isothermal compressibility of water is given to be = 4.80 10-5 atm-1. Analysis When differential quantities are replaced by differences and the properties and are assumed to be constant, the change in density in terms of the changes in pressure and temperature is expressed approximately as P T
The change in density due to a change of pressure from 1 atm to 400 atm at constant temperature is P (4.80 10 5 atm -1 )(998 kg/m 3 )(400 1)atm 19.2 kg/m 3
Discussion Note that the density of water increases from 998 to 1017.2 kg/m3 while being compressed, as expected. This problem can be solved more accurately using differential analysis when functional forms of properties are available.
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Chapter 2 Properties of Fluids 2-44 Solution determined. Assumptions Analysis
The volume of an ideal gas is reduced by half at constant temperature. The change in pressure is to be The process is isothermal and thus the temperature remains constant. For an ideal gas of fixed mass undergoing an isothermal process, the ideal gas relation reduces to
P2V 2 P1V1 T2 T1
P2V 2 P1V1
P2
V1 V P1 1 P1 2 P1 V2 0.5V1
Therefore, the change in pressure becomes P P2 P1 2 P1 P1 P1
Discussion
Note that at constant temperature, pressure and volume of an ideal gas are inversely proportional.
2-45 Solution Saturated refrigerant-134a at a given temperature is cooled at constant pressure. The change in the density of the refrigerant is to be determined. Assumptions 1 The coefficient of volume expansion is constant in the given temperature range. 2 An approximate analysis is performed by replacing differential changes in quantities by finite changes. Properties The density of saturated liquid R-134a at 10C is 1 =1261 kg/m3. The coefficient of volume expansion at the average temperature of (10+0)/2 = 5C is = 0.00269 K-1. Analysis When differential quantities are replaced by differences and the properties and are assumed to be constant, the change in density in terms of the changes in pressure and temperature is expressed approximately as P T
The change in density due to the change of temperature from 10C to 0C at constant pressure is T (0.00269 K -1 )(1261 kg/m 3 )(0 10)K 33.9 kg/m 3
Discussion
Noting that 2 1 , the density of R-134a at 0C is
2 1 1261 33.9 1294.9 kg/m 3 which is almost identical to the listed value of 1295 kg/m3 at 0C in R-134a table in the Appendix. This is mostly due to varying with temperature almost linearly. Note that the density increases during cooling, as expected.
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Chapter 2 Properties of Fluids 2-46 Solution A water tank completely filled with water can withstand tension caused by a volume expansion of 0.8%. The maximum temperature rise allowed in the tank without jeopardizing safety is to be determined. Assumptions 1 The coefficient of volume expansion is constant. 2 An approximate analysis is performed by replacing differential changes in quantities by finite changes. 3 The effect of pressure is disregarded. Properties
The average volume expansion coefficient is given to be = 0.377 10-3 K-1.
Analysis When differential quantities are replaced by differences and the properties and are assumed to be constant, the change in density in terms of the changes in pressure and temperature is expressed approximately as P T
A volume increase of 0.8% corresponds to a density decrease of 0.8%, which can be expressed as 0.008 . Then the decrease in density due to a temperature rise of T at constant pressure is 0.008 T
Solving for T and substituting, the maximum temperature rise is determined to be T
0.008
0.008 0.377 10 3 K -1
21.2 K 21.2C
Discussion This result is conservative since in reality the increasing pressure will tend to compress the water and increase its density.
2-47 Solution A water tank completely filled with water can withstand tension caused by a volume expansion of 1.5%. The maximum temperature rise allowed in the tank without jeopardizing safety is to be determined. Assumptions 1 The coefficient of volume expansion is constant. 2 An approximate analysis is performed by replacing differential changes in quantities by finite changes. 3 The effect of pressure is disregarded. Properties
The average volume expansion coefficient is given to be = 0.377 10-3 K-1.
Analysis When differential quantities are replaced by differences and the properties and are assumed to be constant, the change in density in terms of the changes in pressure and temperature is expressed approximately as P T
A volume increase of 1.5% corresponds to a density decrease of 1.5%, which can be expressed as 0.015 . Then the decrease in density due to a temperature rise of T at constant pressure is 0.015 T
Solving for T and substituting, the maximum temperature rise is determined to be T
0.015
0.015 0.377 10 3 K -1
39.8 K 39.8C
Discussion This result is conservative since in reality the increasing pressure will tend to compress the water and increase its density. The change in temperature is exactly half of that of the previous problem, as expected.
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Chapter 2 Properties of Fluids 2-48 Solution The density of seawater at the free surface and the bulk modulus of elasticity are given. The density and pressure at a depth of 2500 m are to be determined. Assumptions 1 The temperature and the bulk modulus of elasticity of seawater is constant. 2 The gravitational acceleration remains constant. Properties The density of seawater at free surface where the pressure is given to be 1030 kg/m3, and the bulk modulus of elasticity of seawater is given to be 2.34 109 N/m2. Analysis
The coefficient of compressibility or the bulk modulus of elasticity of fluids is expressed as P
T
or
dP d
(at constant T )
The differential pressure change across a differential fluid height of dz is given as
z=0
dP gdz
z
Combining the two relations above and rearranging,
gdz dz g 2 d d
d
2
gdz
2500 m
Integrating from z = 0 where 0 1030 kg/m 3 to z = z where gives d
0
2
g
z
dz
0
1
0
1
gz
Solving for gives the variation of density with depth as
1
1 / 0 gz /
Substituting into the pressure change relation dP gdz and integrating from z = 0 where P P0 98 kPa to z = z where P = P gives
P
dP
P0
z
0
gdz 1 / 0 gz /
1 P P0 ln 1 0 gz /
which is the desired relation for the variation of pressure in seawater with depth. At z = 2500 m, the values of density and pressure are determined by substitution to be
1 1 /(1030 kg/m ) (9.81 m/s )(2500 m) /( 2.34 10 N/m ) 3
2
9
2
1041 kg/m 3
1 P (98,000 Pa) (2.34 10 9 N/m 2 ) ln 3 2 9 2 1 (1030 kg/m )(9.81 m/s )(2500 m) /( 2.34 10 N/m ) 2.550 10 7 Pa 25.50 MPa
since 1 Pa = 1 N/m2 = 1 kg/ms2 and 1 kPa = 1000 Pa. Discussion Note that if we assumed = o = constant at 1030 kg/m3, the pressure at 2500 m would be P P0 gz = 0.098 + 25.26 = 25.36 MPa. Then the density at 2500 m is estimated to be P (1030)(2340 MPa) 1 (25.26 MPa) 11.1 kg/m 3 and thus = 1041 kg/m3
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Chapter 2 Properties of Fluids 2-49E Solution The coefficient of compressibility of water is given. The pressure increases required to reduce the volume of water by 1 percent and then by 2 percent are to be determined. Assumptions 1 The coefficient of compressibility is constant. 2 The temperature remains constant. Properties The coefficient of compressibility of water is given to be 7×105 psia. Analysis (a) A volume decrease of 1 percent can mathematically be expressed as v V 0.01 v V The coefficient of compressibility is expressed as P P v v /v T
v
Rearranging and substituting, the required pressure increase is determined to be v 5 P (7 10 psia )(0.01) 7,000 psia v (b) Similarly, the required pressure increase for a volume reduction of 2 percent becomes v 5 P (7 10 psia )(0.02) 14,000 psia v Discussion Note that at extremely high pressures are required to compress water to an appreciable amount.
2-50E Solution
We are to estimate the energy required to heat up the water in a hot-water tank.
Assumptions 1 There are no losses. 2 The pressure in the tank remains constant at 1 atm. 3 An approximate analysis is performed by replacing differential changes in quantities by finite changes. Properties The specific heat of water is approximated as a constant, whose value is 0.999 Btu/lbmR at the average temperature of (60 + 110)/2 = 85oF. In fact, c remains constant at 0.999 Btu/lbmR (to three digits) from 60oF to 110oF. For this same temperature range, the density varies from 62.36 lbm/ft3 at 60oF to 61.86 lbm/ft3 at 110oF. We approximate the density as constant, whose value is 62.17 lbm/ft3 at the average temperature of 85oF. Analysis
For a constant pressure process, u cavg T . Since this is energy per unit mass, we must multiply by the
total mass of the water in the tank, i.e., U mcavg T V cavg T . Thus, 35.315 ft 3 31,135 Btu 31,100 Btu U Vcavg T (62.17 lbm/ft 3 )(75 gal)(0.999 Btu/lbm R)[(110 - 60)R] 264.17 gal
where we note temperature differences are identical in oF and R. Discussion We give the final answer to 3 significant digits. The actual energy required will be greater than this, due to heat transfer losses and other inefficiencies in the hot-water heating system.
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Chapter 2 Properties of Fluids 2-51 Solution
We are to prove that the coefficient of volume expansion for an ideal gas is equal to 1/T.
Assumptions
1 Temperature and pressure are in the range that the gas can be approximated as an ideal gas.
Analysis
The ideal gas law is P RT , which we re-write as
P 1 . By definition, . Thus, T P RT
substitution and differentiation yields
ideal gas
P 1 RT T
1 P 1 1/T RT 2 T P
where both pressure and the gas constant R are treated as constants in the differentiation. Discussion The coefficient of volume expansion of an ideal gas is not constant, but rather decreases with temperature. However, for small temperature differences, is often approximated as a constant with little loss of accuracy.
2-52 Solution The coefficient of compressibility of nitrogen gas is to be estimated using Van der Waals equation of state. The result is to be compared to ideal gas and experimental values. Assumptions
1 Nitrogen gas obeys the Van der Waals equation of state.
Analysis
From the definition we have
since
The gas constant of nitrogen is
(Table A1). Substituting given data we obtain
For the ideal gas behavior, the coefficient of compressibility is equal to the pressure (Eq. 215). Therefore we get
whichis in error by
compared to experimentally measured pressure.
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Chapter 2 Properties of Fluids 2-53 Solution The water contained in a piston-cylinder device is compressed isothermally. The energy needed is to be determined. Assumptions 1 The coefficient of compressibility of water remains unchanged during the compression. Analysis We take the water in the cylinder as the system. The energy needed to compress water is equal to the work done on the system, and can be expressed as
From the definition of coefficient of compressibility we have
Rearranging we obtain
which can be integrated from the initial state to any state as follows:
from which we obtain
Substituting in Eq. 1 we have
or
In terms of finite changes, the fractional change due to change in pressure can be expressed approximately as (Eq. 323)
or at 20 . Realizing that 10 kg water where is the isothermal compressibility of water, which is the final volume of water is determined to be occupies initially a volume of Then the work done on the water is
from which we obtain since
.
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Chapter 2 Properties of Fluids 2-54 Solution The water contained in a piston-cylinder device is compressed isothermally and the pressure increases linearly. The energy needed is to be determined. Assumptions
1 The pressure increases linearly.
Analysis We take the water in the cylinder as the system. The energy needed to compress water is equal to the work done on the system, and can be expressed as
For a linear pressure increase we take
In terms of finite changes, the fractional change due to change in pressure can be expressed approximately as (Eq. 323)
or
at 20 . Realizing that 10 kg water where is the isothermal compressibility of water, which is the final volume of water is determined to be occupies initially a volume of
Therefore the work expression becomes
or
Thus, we conclude that linear pressure increase approximation does not work well since it gives almost ten times larger work.
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Chapter 2 Properties of Fluids
Speed of Sound
2-55C Solution
We are to define and discuss sound and how it is generated and how it travels.
Analysis Sound is an infinitesimally small pressure wave. It is generated by a small disturbance in a medium. It travels by wave propagation. Sound waves cannot travel in a vacuum. Discussion
Electromagnetic waves, like light and radio waves, can travel in a vacuum, but sound cannot.
2-56C Solution
We are to discuss whether sound travels faster in warm or cool air.
Analysis
Sound travels faster in warm (higher temperature) air since c kRT .
Discussion On the microscopic scale, we can imagine the air molecules moving around at higher speed in warmer air, leading to higher propagation of disturbances.
2-57C Solution
We are to compare the speed of sound in air, helium, and argon.
Analysis Sound travels fastest in helium, since c kRT and helium has the highest kR value. It is about 0.40 for air, 0.35 for argon, and 3.46 for helium. Discussion temperature.
2-58C Solution
We are assuming, of course, that these gases behave as ideal gases – a good approximation at room
We are to compare the speed of sound in air at two different pressures, but the same temperature.
Analysis Air at specified conditions will behave like an ideal gas, and the speed of sound in an ideal gas depends on temperature only. Therefore, the speed of sound is the same in both mediums. Discussion
If the temperature were different, however, the speed of sound would be different.
2-59C Solution
We are to examine whether the Mach number remains constant in constant-velocity flow.
Analysis In general, no, because the Mach number also depends on the speed of sound in gas, which depends on the temperature of the gas. The Mach number remains constant only if the temperature and the velocity are constant. Discussion It turns out that the speed of sound is not a strong function of pressure. In fact, it is not a function of pressure at all for an ideal gas.
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Chapter 2 Properties of Fluids 2-60C Solution
We are to state whether the propagation of sound waves is an isentropic process.
Analysis Yes, the propagation of sound waves is nearly isentropic. Because the amplitude of an ordinary sound wave is very small, and it does not cause any significant change in temperature and pressure. Discussion
No process is truly isentropic, but the increase of entropy due to sound propagation is negligibly small.
2-61C Solution
We are to discuss sonic velocity – specifically, whether it is constant or it changes.
Analysis The sonic speed in a medium depends on the properties of the medium, and it changes as the properties of the medium change. Discussion
The most common example is the change in speed of sound due to temperature change.
2-62 Solution
The Mach number of a passenger plane for specified limiting operating conditions is to be determined.
Assumptions Properties Analysis
Air is an ideal gas with constant specific heats at room temperature. The gas constant of air is R = 0.287 kJ/kg·K. Its specific heat ratio at room temperature is k = 1.4. From the speed of sound relation
1000 m 2 / s 2 c kRT (1.4)(0.287 kJ/kg K)(-60 273 K) 1 kJ/kg
293 m/s
Thus, the Mach number corresponding to the maximum cruising speed of the plane is Ma
Discussion same result.
V max (945 / 3.6) m/s 0.897 c 293 m/s
Note that this is a subsonic flight since Ma < 1. Also, using a k value at -60C would give practically the
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Chapter 2 Properties of Fluids 2-63 Solution Carbon dioxide flows through a nozzle. The inlet temperature and velocity and the exit temperature of CO2 are specified. The Mach number is to be determined at the inlet and exit of the nozzle. Assumptions 1 CO2 is an ideal gas with constant specific heats at room temperature. 2 This is a steady-flow process. Properties The gas constant of carbon dioxide is R = 0.1889 kJ/kg·K. Its constant pressure specific heat and specific heat ratio at room temperature are cp = 0.8439 kJ/kgK and k = 1.288. Analysis (a) At the inlet 1000 m 2 / s 2 c1 k1 RT1 (1.288)(0.1889 kJ/kg K)(1200 K) 1 kJ/kg
540.3 m/s
Thus, Ma 1
V1 50 m/s 0.0925 c1 540.3 m/s
1200 K 50 m/s
Carbon dioxide
400 K
(b) At the exit, 1000 m 2 / s 2 c 2 k 2 RT2 (1.288)(0.1889 kJ/kg K)(400 K) 1 kJ/kg
312.0 m/s
The nozzle exit velocity is determined from the steady-flow energy balance relation, 0 h2 h1
V 2 2 V1 2 2
0 (0.8439 kJ/kg K)(400 1200 K)
0 c p (T2 T1 ) V 2 2 (50 m/s) 2 2
V 2 2 V1 2 2
1 kJ/kg V 2 1163 m/s 1000 m 2 / s 2
Thus, Ma 2
V 2 1163 m/s 3.73 c2 312 m/s
Discussion The specific heats and their ratio k change with temperature, and the accuracy of the results can be improved by accounting for this variation. Using EES (or another property database):
At 1200 K: cp = 1.278 kJ/kgK, k = 1.173
c1 = 516 m/s,
V1 = 50 m/s,
Ma1 = 0.0969
At 400 K: cp = 0.9383 kJ/kgK, k = 1.252 c2 = 308 m/s, V2 = 1356 m/s, Ma2 = 4.41 Therefore, the constant specific heat assumption results in an error of 4.5% at the inlet and 15.5% at the exit in the Mach number, which are significant.
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Chapter 2 Properties of Fluids 2-64 Solution Nitrogen flows through a heat exchanger. The inlet temperature, pressure, and velocity and the exit pressure and velocity are specified. The Mach number is to be determined at the inlet and exit of the heat exchanger. Assumptions 1 N2 is an ideal gas. 2 This is a steady-flow process. 3 The potential energy change is negligible. Properties The gas constant of N2 is R = 0.2968 kJ/kg·K. Its constant pressure specific heat and specific heat ratio at room temperature are cp = 1.040 kJ/kgK and k = 1.4. 1000 m 2 / s 2 c1 k1 RT1 (1.400)(0.2968 kJ/kg K)(283 K) 1 kJ/kg
Analysis
342.9 m/s
Thus,
120kJ/kg
V 100 m/s 0.292 Ma 1 1 c1 342.9 m/s
150 kPa 10C 100 m/s
From the energy balance on the heat exchanger, qin c p (T2 T1 )
Nitrogen
100 kPa 200 m/s
V2 2 V12 2
120 kJ/kg (1.040 kJ/kg.C)(T2 10C)
(200 m/s) 2 (100 m/s) 2 1 kJ/kg 2 1000 m 2 / s 2
It yields T2 = 111C = 384 K 1000 m 2 / s 2 c 2 k 2 RT2 (1.4 )(0.2968 kJ/kg K)(384 K) 1 kJ/kg
399 m/s
Thus, Ma 2
V 2 200 m/s 0.501 c 2 399 m/s
Discussion The specific heats and their ratio k change with temperature, and the accuracy of the results can be improved by accounting for this variation. Using EES (or another property database):
At 10C : cp = 1.038 kJ/kgK, k = 1.400
c1 = 343 m/s,
V1 = 100 m/s,
Ma1 = 0.292
At 111C cp = 1.041 kJ/kgK, k = 1.399 c2 = 399 m/s, V2 = 200 m/s, Ma2 = 0.501 Therefore, the constant specific heat assumption results in an error of 4.5% at the inlet and 15.5% at the exit in the Mach number, which are almost identical to the values obtained assuming constant specific heats.
2-65 Solution
The speed of sound in refrigerant-134a at a specified state is to be determined.
Assumptions Properties Analysis
R-134a is an ideal gas with constant specific heats at room temperature. The gas constant of R-134a is R = 0.08149 kJ/kg·K. Its specific heat ratio at room temperature is k = 1.108. From the ideal-gas speed of sound relation,
1000 m 2 / s 2 c kRT (1.108)(0.08149 kJ/kg K)(60 273 K) 1 kJ/kg
Discussion
173 m/s
Note that the speed of sound is independent of pressure for ideal gases.
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Chapter 2 Properties of Fluids 2-66 Solution The Mach number of an aircraft and the speed of sound in air are to be determined at two specified temperatures. Assumptions Air is an ideal gas with constant specific heats at room temperature. Properties The gas constant of air is R = 0.287 kJ/kg·K. Its specific heat ratio at room temperature is k = 1.4. Analysis From the definitions of the speed of sound and the Mach number, (a) At 300 K, 1000 m 2 / s 2 c kRT (1.4)(0.287 kJ/kg K)(300 K) 1 kJ/kg V 330 m/s and 0.951 Ma c 347 m/s (b) At 800 K,
347 m/s
1000 m 2 / s 2 567 m/s c kRT (1.4)(0.287 kJ/kg K)(800 K) 1 kJ/kg V 330 m/s 0.582 and Ma c 567 m/s Discussion Note that a constant Mach number does not necessarily indicate constant speed. The Mach number of a rocket, for example, will be increasing even when it ascends at constant speed. Also, the specific heat ratio k changes with temperature.
2-67E Solution Steam flows through a device at a specified state and velocity. The Mach number of steam is to be determined assuming ideal gas behavior. Assumptions Properties Analysis
Steam is an ideal gas with constant specific heats. The gas constant of steam is R = 0.1102 Btu/lbm·R. Its specific heat ratio is given to be k = 1.3. From the ideal-gas speed of sound relation,
25,037 ft 2 / s 2 c kRT (1.3)(0.1102 Btu/lbm R)(1160 R) 1 Btu/lbm
2040 ft/s
Thus, Ma
V 900 ft/s 0.441 c 2040 ft/s
Using property data from steam tables and not assuming ideal gas behavior, it can be shown that the Mach Discussion number in steam at the specified state is 0.446, which is sufficiently close to the ideal-gas value of 0.441. Therefore, the ideal gas approximation is a reasonable one in this case.
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Chapter 2 Properties of Fluids 2-68E
Solution Problem 2-67e is reconsidered. The variation of Mach number with temperature as the temperature changes between 350 and 700F is to be investigated, and the results are to be plotted. Analysis
The EES Equations window is printed below, along with the tabulated and plotted results.
T=Temperature+460 R=0.1102 V=900 k=1.3 c=SQRT(k*R*T*25037) Ma=V/c Mach number Ma 0.528 0.520 0.512 0.505 0.498 0.491 0.485 0.479 0.473 0.467 0.462 0.456 0.451 0.446 0.441
0.54
0.52
0.5
Ma
Temperature, T, F 350 375 400 425 450 475 500 525 550 575 600 625 650 675 700
0.48
0.46
0.44 350
400
450
500
550
600
650
700
Temperature, °F Discussion
Note that for a specified flow speed, the Mach number decreases with increasing temperature, as expected.
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Chapter 2 Properties of Fluids 2-69E Solution The inlet state and the exit pressure of air are given for an isentropic expansion process. The ratio of the initial to the final speed of sound is to be determined. Assumptions
Air is an ideal gas with constant specific heats at room temperature.
Properties The properties of air are R = 0.06855 Btu/lbm·R and k = 1.4. The specific heat ratio k varies with temperature, but in our case this change is very small and can be disregarded. Analysis
The final temperature of air is determined from the isentropic relation of ideal gases, P T2 T1 2 P1
( k 1) / k
60 (659.7 R) 170
(1.4 1) / 1.4
489.9 R
Treating k as a constant, the ratio of the initial to the final speed of sound can be expressed as Ratio
Discussion
c2 c1
k1RT1 k2 RT2
T1 T2
659.7 1.16 489.9
Note that the speed of sound is proportional to the square root of thermodynamic temperature.
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Chapter 2 Properties of Fluids 2-70 Solution The inlet state and the exit pressure of air are given for an isentropic expansion process. The ratio of the initial to the final speed of sound is to be determined. Assumptions
Air is an ideal gas with constant specific heats at room temperature.
Properties The properties of air are R = 0.287 kJ/kg·K and k = 1.4. The specific heat ratio k varies with temperature, but in our case this change is very small and can be disregarded. Analysis
The final temperature of air is determined from the isentropic relation of ideal gases, P T2 T1 2 P1
( k 1) / k
0.4 MPa (350.2 K) 2.2 MPa
(1.4 1) / 1.4
215.2 K
Treating k as a constant, the ratio of the initial to the final speed of sound can be expressed as Ratio
Discussion
c2 c1
k1 RT1
k 2 RT2
T1
T2
350.2
1.28
215.2
Note that the speed of sound is proportional to the square root of thermodynamic temperature.
2-71 Solution The inlet state and the exit pressure of helium are given for an isentropic expansion process. The ratio of the initial to the final speed of sound is to be determined. Assumptions
Helium is an ideal gas with constant specific heats at room temperature.
Properties
The properties of helium are R = 2.0769 kJ/kg·K and k = 1.667.
Analysis
The final temperature of helium is determined from the isentropic relation of ideal gases, P T2 T1 2 P1
( k 1) / k
0.4 (350.2 K) 2.2
(1.667 1) / 1.667
177.0 K
The ratio of the initial to the final speed of sound can be expressed as Ratio
Discussion
c2 c1
k1 RT1
k 2 RT2
T1 T2
350.2
1.41
177.0
Note that the speed of sound is proportional to the square root of thermodynamic temperature.
2-72 Solution The expression for the speed of sound for an ideal gas is to be obtained using the isentropic process equation and the definition of the speed of sound. Analysis
The isentropic relation Pvk = A where A is a constant can also be expressed as k
1 P A A k v
Substituting it into the relation for the speed of sound, ( A ) k P c 2 s
kA k 1 k ( A k ) / k ( P / ) kRT s
since for an ideal gas P = RT or RT = P/. Therefore, c kRT , which is the desired relation. Discussion
Notice that pressure has dropped out; the speed of sound in an ideal gas is not a function of pressure.
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Chapter 2 Properties of Fluids
Viscosity
2-73C Solution
We are to define and discuss viscosity.
Analysis Viscosity is a measure of the “stickiness” or “resistance to deformation” of a fluid. It is due to the internal frictional force that develops between different layers of fluids as they are forced to move relative to each other. Viscosity is caused by the cohesive forces between the molecules in liquids, and by the molecular collisions in gases. In general, liquids have higher dynamic viscosities than gases. Discussion The ratio of viscosity to density often appears in the equations of fluid mechanics, and is defined as the kinematic viscosity, = /.
2-74C Solution
We are to discuss Newtonian fluids.
Analysis Fluids whose shear stress is linearly proportional to the velocity gradient (shear strain) are called Newtonian fluids. Most common fluids such as water, air, gasoline, and oils are Newtonian fluids. Discussion
In the differential analysis of fluid flow, only Newtonian fluids are considered in this textbook.
2-75C Solution
We are to discuss how kinematic viscosity varies with temperature in liquids and gases.
Analysis (a) For liquids, the kinematic viscosity decreases with temperature. (b) For gases, the kinematic viscosity increases with temperature. Discussion
You can easily verify this by looking at the appendices.
2-76C Solution
We are to discuss how dynamic viscosity varies with temperature in liquids and gases.
Analysis (a) The dynamic viscosity of liquids decreases with temperature. (b) The dynamic viscosity of gases increases with temperature. Discussion A good way to remember this is that a car engine is much harder to start in the winter because the oil in the engine has a higher viscosity at low temperatures.
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Chapter 2 Properties of Fluids 2-77C Solution We are to compare the settling speed of balls dropped in water and oil; namely, we are to determine which will reach the bottom of the container first. Analysis When two identical small glass balls are dropped into two identical containers, one filled with water and the other with oil, the ball dropped in water will reach the bottom of the container first because of the much lower viscosity of water relative to oil. Discussion Oil is very viscous, with typical values of viscosity approximately 800 times greater than that of water at room temperature.
2-78E Solution The torque and the rpm of a double cylinder viscometer are given. The viscosity of the fluid is to be determined. Assumptions 1 The inner cylinder is completely submerged in the fluid. 2 The viscous effects on the two ends of the inner cylinder are negligible. 3 The fluid is Newtonian.
R
Analysis Substituting the given values, the viscosity of the fluid is determined to be l = 0.035 in T (1.2 lbf ft)(0.035/12 ft) 4 2 fluid 2.72 10 lbf s/ft 4 2 R 3 n L 4 2 (3 / 12 ft) 3 (250 / 60 s -1 )(5 ft) Discussion This is the viscosity value at temperature that existed during the experiment. Viscosity is a strong function of temperature, and the values can be significantly different at different temperatures.
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Chapter 2 Properties of Fluids 2-79 Solution A block is moved at constant velocity on an inclined surface. The force that needs to be applied in the horizontal direction when the block is dry, and the percent reduction in the required force when an oil film is applied on the surface are to be determined. Assumptions 1 The inclined surface is plane (perfectly flat, although tilted). 2 The friction coefficient and the oil film thickness are uniform. 3 The weight of the oil layer is negligible.
The absolute viscosity of oil is given to be = 0.012 Pas = 0.012 Ns/m2.
Properties
Analysis (a) The velocity of the block is constant, and thus its acceleration and the net force acting on it are zero. A free body diagram of the block is given. Then the force balance gives F1 Fx 0 : F1 F f cos 20 FN 1 sin 20 0 (1)
F
y
0:
FN 1 cos 20 F f sin 20 W 0
Friction force: F f fFN 1
(2)
V= 0.8 m/s Ff 200
y
FN1 200 W = 150 N
20
(3)
0
x
Substituting Eq. (3) into Eq. (2) and solving for FN1 gives W 150 N FN 1 177.0 N cos 20 f sin 20 cos 20 0.27 sin 20 Then from Eq. (1): F1 F f cos 20 FN 1 sin 20 (0.27 177 N) cos 20 (177 N) sin 20 105.5 N (b) In this case, the friction force is replaced by the shear force applied on the bottom surface of the block due to the oil. Because of the no-slip condition, the oil film sticks to the inclined surface at the bottom and the lower surface of the block at the top. Then the shear force is expressed as
Fshear w As
V As h 0.012 N s/m 2 0.5 0.2 m 2
V= 0.8 m/s 50 cm F2 20
0.8 m/s 4 10-4 m
0.4 mm
0
Fshear = wAs FN2
W = 150 N
2 .4 N Replacing the friction force by the shear force in part (a),
F F
x
0:
F2 Fshear cos 20 FN 2 sin 20 0
(4)
y
0:
FN 2 cos 20 Fshear sin 20 W 0
(5)
Eq. (5) gives FN 2 ( Fshear sin 20 W ) / cos 20 [(2.4 N ) sin 20 (150 N )] / cos 20 160.5 N Substituting into Eq. (4), the required horizontal force is determined to be F2 Fshear cos 20 FN 2 sin 20 (2.4 N) cos 20 (160.5 N) sin 20 57.2 N Then, our final result is expressed as F F2 105.5 57.2 100% 100% 45.8% Percentage reduction in required force = 1 F1 105.5 Discussion Note that the force required to push the block on the inclined surface reduces significantly by oiling the surface.
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Chapter 2 Properties of Fluids 2-80 Solution The velocity profile of a fluid flowing though a circular pipe is given. The friction drag force exerted on the pipe by the fluid in the flow direction per unit length of the pipe is to be determined. Assumptions
The viscosity of the fluid is constant.
Analysis
The wall shear stress is determined from its definition to be
nu max d rn nr n 1 1 u max n n dr R r R R R rR rR Note that the quantity du /dr is negative in pipe flow, and the negative sign is added to the w relation for pipes to make shear stress in the positive (flow) direction a positive quantity. (Or, du /dr = - du /dy since y = R – r). Then the friction drag force exerted by the fluid on the inner surface of the pipe becomes nu max F w Aw (2R ) L 2nu max L R Therefore, the drag force per unit length of the pipe is
w
du dr
u max
u(r) = umax(1-rn/Rn)
R r 0
umax
F / L 2nu max .
Discussion
Note that the drag force acting on the pipe in this case is independent of the pipe diameter.
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Chapter 2 Properties of Fluids 2-81 Solution A thin flat plate is pulled horizontally through an oil layer sandwiched between two plates, one stationary and the other moving at a constant velocity. The location in oil where the velocity is zero and the force that needs to be applied on the plate are to be determined. Assumptions
1 The thickness of the plate is negligible. 2 The velocity profile in each oil layer is linear.
Properties
The absolute viscosity of oil is given to be = 0.027 Pas = 0.027 Ns/m2.
Analysis (a) The velocity profile in each oil layer relative to the fixed wall is as shown in the figure below. The point of zero velocity is indicated by point A, and its distance from the lower plate is determined from geometric considerations (the similarity of the two triangles in the lower oil layer) to be 2.6 y A 3 yA 0.3
yA = 0.23636 mm Fixed wall
h1=1 mm
V = 3 m/s
h2=2.6 mm y
F
A
yA
Vw= 0.3 m/s Moving wall
(b) The magnitudes of shear forces acting on the upper and lower surfaces of the plate are Fshear, upper w, upper As As
du V 0 3 m/s As (0.027 N s/m 2 )(0.3 0.3 m 2 ) 7.29 N dy h1 1.0 10 -3 m
V Vw du [3 (0.3)] m/s As (0.027 N s/m 2 )(0.3 0.3 m 2 ) 3.08 N h2 dy 2.6 10 -3 m Noting that both shear forces are in the opposite direction of motion of the plate, the force F is determined from a force balance on the plate to be Fshear, lower w, lower As As
F Fshear, upper Fshear, lower 7.29 3.08 10.4 N
Discussion motion.
Note that wall shear is a friction force between a solid and a liquid, and it acts in the opposite direction of
2-40 PROPRIETARY MATERIAL. © 2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.
Chapter 2 Properties of Fluids 2-82 Solution We are to determine the torque required to rotate the inner cylinder of two Inner cylinder concentric cylinders, with the inner cylinder rotating and the outer cylinder stationary. We V are also to explain what happens when the gap gets bigger. Assumptions 1 The fluid is incompressible and Newtonian. 2 End effects (top and bottom) are negligible. 3 The gap is very small so that wall curvature effects are negligible. 4 The gap is so small that the velocity profile in the gap is linear.
h
y
u
Outer cylinder
Analysis (a) We assume a linear velocity profile between the two walls as sketched – the inner wall is moving at speed V = iRi and the outer wall is stationary. The thickness of the gap is h, and we let y be the distance from the outer wall into the fluid (towards the inner wall). Thus,
u V
y du V and h dy h
where
h Ro - Ri and V i Ri Since shear stress has dimensions of force/area, the clockwise (mathematically negative) tangential force acting along the surface of the inner cylinder by the fluid is
F A
i Ri V 2 Ri L 2 Ri L h Ro Ri
But the torque is the tangential force times the moment arm Ri. Also, we are asked for the torque required to turn the inner cylinder. This applied torque is counterclockwise (mathematically positive). Thus,
T FRi
2 Li Ri 3 2 Li Ri 3 Ro Ri h
(b) The above is only an approximation because we assumed a linear velocity profile. As long as the gap is very small, and therefore the wall curvature effects are negligible, this approximation should be very good. Another way to think about this is that when the gap is very small compared to the cylinder radii, a magnified view of the flow in the gap appears similar to flow between two infinite walls (Couette flow). However, as the gap increases, the curvature effects are no longer negligible, and the linear velocity profile is not expected to be a valid approximation. We do not expect the velocity to remain linear as the gap increases. Discussion It is possible to solve for the exact velocity profile for this problem, and therefore the torque can be found analytically, but this has to wait until the differential analysis chapter.
2-41 PROPRIETARY MATERIAL. © 2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.
Chapter 2 Properties of Fluids 2-83 Solution A clutch system is used to transmit torque through an oil film between two identical disks. For specified rotational speeds, the transmitted torque is to be determined. Assumptions
1 The thickness of the oil film is uniform. 2 The rotational speeds of the disks remain constant.
Properties
The absolute viscosity of oil is given to be = 0.38 Ns/m2.
Driving shaft
Driven shaft
30 cm
2 mm
SAE 30W oil
Analysis The disks are rotting in the same direction at different angular speeds of 1 and of 2 . Therefore, we can assume one of the disks to be stationary and the other to be rotating at an angular speed of 1 2 . The velocity gradient anywhere in the oil of film thickness h is V /h where V (1 2 )r is the tangential velocity. Then the wall shear stress anywhere on the surface of the faster disk at a distance r from the axis of rotation can be expressed as ( 2 )r du V 1 w dr h h h Then the shear force acting on a differential area dA on the surface and the torque generation associated with it can be expressed as 2r ( 2 )r dF w dA 1 (2r )dr r 1 h dT rdF
(1 2 )r 2 2 (1 2 ) 3 (2r )dr r dr h h
Integrating, T
2 (1 2 ) h
D/2
r 0
r 3 dr
2 (1 2 ) r 4 h 4
D/2
(1 2 ) D 4
r 0
Noting that 2 n , the relative angular speed is
32h
1 min
1 2 2 n1 n2 2 rad/rev 1450 1398 rev/min 5.445 rad/s , 60 s Substituting, the torque transmitted is determined to be T
(0.38 N s/m 2 )(5.445 /s)(0.30 m) 4 32(0.002 m)
0.82 N m
Discussion Note that the torque transmitted is proportional to the fourth power of disk diameter, and is inversely proportional to the thickness of the oil film.
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Chapter 2 Properties of Fluids 2-84 Solution We are to investigate the effect of oil film thickness on the transmitted torque. Analysis The previous problem is reconsidered. Using EES software, the effect of oil film thickness on the torque transmitted is investigated. Film thickness varied from 0.1 mm to 10 mm, and the results are tabulated and (1 2 ) D 4 . The EES Equations window is printed below, followed by the plotted. The relation used is T 32h tabulated and plotted results. mu=0.38 n1=1450 "rpm" w1=2*pi*n1/60 "rad/s" n2=1398 "rpm" w2=2*pi*n2/60 "rad/s" D=0.3 "m" Tq=pi*mu*(w1-w2)*(D^4)/(32*h) Film thickness h, mm 0.1 0.2 0.4 0.6 0.8 1 2 4 6 8 10
Torque transmitted T, Nm 16.46 8.23 4.11 2.74 2.06 1.65 0.82 0.41 0.27 0.21 0.16
20
Tq, Nm
16
12
8
4
0 0
2
4
6
8
10
h, mm Conclusion Torque transmitted is inversely proportional to oil film thickness, and the film thickness should be as small as possible to maximize the transmitted torque. Discussion value of h.
To obtain the solution in EES, we set up a parametric table, specify h, and let EES calculate T for each
2-43 PROPRIETARY MATERIAL. © 2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.
Chapter 2 Properties of Fluids 2-85 Solution The viscosities of carbon dioxide at two temperatures are given. The constants of Sutherland correlation for carbon dioxide are to be determined and the viscosity of carbon dioxide at a specified temperature is to be predicted and compared to the value in table A-10. Analysis
where
Sutherland correlation is given by Eq. 232 as
is the absolute temperature. Substituting the given values we have
which is a nonlinear system of two algebraic equations. Using EES or any other computer code, one finds the following result:
Using these values the Sutherland correlation becomes
Therefore the viscosity at 100
is found to be
The agreement is perfect and within approximately 0.1%.
2-44 PROPRIETARY MATERIAL. © 2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.
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