3-1
Solutions Manual
for
Thermodynamics: An Engineering Approach
Seventh Edition in SI Units
Yunus A. Cengel, Michael A. Boles
McGraw-Hill, 2011
Chapter 3
PROPERTIES OF PURE SUBSTANCES
PROPRIETARY AND CONFIDENTIAL
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3-2
Pure Substances, Phase Change Processes, Property Diagrams
3-1C A vapor that is about to condense is saturated vapor; otherwise it is superheated vapor.
3-2C No.
3-3C The temperature will also increase since the boiling or saturation temperature of a pure substance depends on
pressure.
3-4C Because one cannot be varied while holding the other constant. In other words, when one changes, so does the other
one.
3-5C At critical point the saturated liquid and the saturated vapor states are identical. At triple point the three phases of a
pure substance coexist in equilibrium.
3-6C Yes.
3-7C Case (c) when the pan is covered with a heavy lid. Because the heavier the lid, the greater the pressure in the pan, and
thus the greater the cooking temperature.
3-8C At supercritical pressures, there is no distinct phase change process. The liquid uniformly and gradually expands into
a vapor. At subcritical pressures, there is always a distinct surface between the phases.
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3-3
Property Tables
3-9C A perfectly fitting pot and its lid often stick after cooking as a result of the vacuum created inside as the temperature
and thus the corresponding saturation pressure inside the pan drops. An easy way of removing the lid is to reheat the food.
When the temperature rises to boiling level, the pressure rises to atmospheric value and thus the lid will come right off.
3-10C The molar mass of gasoline (C8H18) is 114 kg/kmol, which is much larger than the molar mass of air that is 29
kg/kmol. Therefore, the gasoline vapor will settle down instead of rising even if it is at a much higher temperature than the
surrounding air. As a result, the warm mixture of air and gasoline on top of an open gasoline will most likely settle down
instead of rising in a cooler environment
3-11C Yes. Otherwise we can create energy by alternately vaporizing and condensing a substance.
3-12C No. Because in the thermodynamic analysis we deal with the changes in properties; and the changes are
independent of the selected reference state.
3-13C The term hfg represents the amount of energy needed to vaporize a unit mass of saturated liquid at a specified
temperature or pressure. It can be determined from hfg = hg - hf .
3-14C Yes. It decreases with increasing pressure and becomes zero at the critical pressure.
3-15C Yes; the higher the temperature the lower the hfg value.
3-16C Quality is the fraction of vapor in a saturated liquid-vapor mixture. It has no meaning in the superheated vapor
region.
3-17C Completely vaporizing 1 kg of saturated liquid at 1 atm pressure since the higher the pressure, the lower the hfg .
3-18C No. Quality is a mass ratio, and it is not identical to the volume ratio.
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3-4
3-19C The compressed liquid can be approximated as a saturated liquid at the given temperature. Thus v T , P ≅ v f @ T .
3-20C Ice can be made by evacuating the air in a water tank. During evacuation, vapor is also thrown out, and thus the
vapor pressure in the tank drops, causing a difference between the vapor pressures at the water surface and in the tank. This
pressure difference is the driving force of vaporization, and forces the liquid to evaporate. But the liquid must absorb the
heat of vaporization before it can vaporize, and it absorbs it from the liquid and the air in the neighborhood, causing the
temperature in the tank to drop. The process continues until water starts freezing. The process can be made more efficient
by insulating the tank well so that the entire heat of vaporization comes essentially from the water.
3-21
Complete the following table for H2 O:
T, °C
P, kPa
v, m3 / kg
Phase description
50
12.35
7.72
Saturated mixture
143.6
400
0.4624
Saturated vapor
250
500
0.4744
Superheated vapor
110
350
0.001051
Compressed liquid
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3-5
3-22
Problem 3-21 is reconsidered. The missing properties of water are to be determined using EES, and the
solution is to be repeated for refrigerant-134a, refrigerant-22, and ammonia.
Analysis The problem is solved using EES, and the solution is given below.
"Given"
T[1]=50 [C]
v[1]=7.72 [m^3/kg]
P[2]=400 [kPa]
x[2]=1
T[3]=250 [C]
P[3]=500 [kPa]
T[4]=110 [C]
P[4]=350 [kPa]
"Analysis"
Fluid$='steam_iapws' "Change the Fluid to R134a, R22 and Ammonia and solve"
P[1]=pressure(Fluid$, T=T[1], v=v[1])
x[1]=quality(Fluid$, T=T[1], v=v[1])
T[2]=temperature(Fluid$, P=P[2], x=x[2])
v[2]=volume(Fluid$, P=P[2], x=x[2])
v[3]=volume(Fluid$, P=P[3], T=T[3])
x[3]=quality(Fluid$, P=P[3], T=T[3])
v[4]=volume(Fluid$, P=P[4], T=T[4])
x[4]=quality(Fluid$, P=P[4], T=T[4])
"x = 100 for superheated vapor and x = -100 for compressed liquid"
SOLUTION for water
T [C]
P [kPa]
x
v [kg/m3]
50.00
12.35
0.641
9
7.72
143.6
1
400.00
1
0.4624
250.0
0
500.00
100
0.4744
110.0
0
350.00
-100
0.00105
1
SOLUTION for R-134a
T [C]
P [kPa]
x
v [kg/m3]
50.00
3.41
100
7.72
8.91
400.00
1
0.0512
250.0
0
500.00
-
-
110.0
0
350.00
100
0.08666
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3-6
SOLUTION for R-22
T [C]
P [kPa]
X
v [kg/m3]
50.00
4.02
100
7.72
-6.56
400.00
1
0.05817
250.0
0
500.00
100
0.09959
110.0
0
350.00
100
0.103
SOLUTION for Ammonia
T [C]
P [kPa]
X
v [kg/m3]
50.00
20.40
100
7.72
-1.89
400.00
1
0.3094
250.0
0
500.00
100
0.5076
110.0
0
350.00
100
0.5269
Steam
700
600
500
T [C]
400
300
8600 kPa
2600 kPa
200
500 kPa
100
45 kPa
0
0.0
1.0
2.0
3.0
4.0
5.0
6.0
7.0
8.0
9.0
10.0
s [kJ/kg-K]
Steam
700
600
500
T [C]
400
300
8600 kPa
2600 kPa
200
500 kPa
100
0
10-4
45 kPa
10-3
10-2
10-1
100
101
102
103
v [m 3/kg]
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3-7
Steam
105
104
250 C
P [kPa]
10
3
170 C
102
110 C
75 C
101
100
10-3
10-2
10-1
100
101
102
3
v [m /kg]
Steam
105
104
250 C
P [kPa]
10
3
170 C
102
110 C
75 C
101
100
0
500
1000
1500
2000
2500
3000
h [kJ/kg]
Steam
4000
8600 kPa
h [kJ/kg]
3500
2600 kPa
3000
500 kPa
2500
45 kPa
2000
1500
1000
500
0
0.0
1.0
2.0
3.0
4.0
5.0
6.0
7.0
8.0
9.0
10.0
s [kJ/kg-K]
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3-8
3-23 Complete the following table for H2 O:
T, °C
P, kPa
h, kJ / kg
x
Phase description
120.21
200
2045.8
0.7
Saturated mixture
140
361.53
1800
0.565
Saturated mixture
177.66
950
752.74
0.0
Saturated liquid
80
500
335.37
---
Compressed liquid
350.0
800
3162.2
---
Superheated vapor
3-24 Complete the following table for Refrigerant-134a:
T, °C
P, kPa
v, m3 / kg
Phase description
-12
320
0.000750
Compressed liquid
30
770.64
0.0065
Saturated mixture
18.73
550
0.03741
Saturated vapor
60
600
0.04139
Superheated vapor
3-25 Complete the following table for water:
*
P, kPa
T, oC
v, m3/kg
h, kJ/kg
200
120.2
0.8858
2706.3
Condition description and quality, if
applicable
x = 1, Saturated vapor
270.3
130
1959.3
x = 0.650, Two-phase mixture
201.8
400
1.5358
3277.0
Superheated vapor
800
30
0.001004*
125.74*
Compressed liquid
450
147.90
-
-
Insufficient information
Approximated as saturated liquid at the given temperature of 30oC
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3-9
3-26 Complete the following table for Refrigerant-134a:
T, °C
21.55
-10
-14
70
44
P, kPa
600
200.74
500
1200
1131
h, kJ /kg
180
162.13
33.40
300.61
272.95
x
0.545
0.6
----1.0
Phase description
Saturated mixture
Saturated mixture
Compressed liquid
Superheated vapor
Saturated vapor
3-27 A piston-cylinder device contains R-134a at a specified state. Heat is transferred to R-134a. The final pressure, the
volume change of the cylinder, and the enthalpy change are to be determined.
Analysis (a) The final pressure is equal to the initial pressure, which is determined from
P2 = P1 = Patm +
mpg
2
πD /4
= 88 kPa +
(12 kg)(9.81 m/s 2 ) ⎛⎜
1 kN
2
⎜
π (0.25 m) /4 ⎝ 1000 kg.m/s 2
⎞
⎟ = 90.4 kPa
⎟
⎠
(b) The specific volume and enthalpy of R-134a at the initial state of 90.4 kPa and -10°C and at the final state of 90.4 kPa
and 15°C are (from EES)
v1 = 0.2302 m3/kg
h1 = 247.76 kJ/kg
v 2 = 0.2544 m3/kg
h2 = 268.16 kJ/kg
The initial and the final volumes and the volume change are
V1 = mv 1 = (0.85 kg)(0.2302 m 3 /kg) = 0.1957 m 3
3
V 2 = mv 2 = (0.85 kg)(0.2544 m /kg) = 0.2162 m
3
R-134a
0.85 kg
-10°C
Q
∆V = 0.2162 − 0.1957 = 0.0205 m 3
(c) The total enthalpy change is determined from
∆H = m(h2 − h1 ) = (0.85 kg)(268.16 − 247.76) kJ/kg = 17.4 kJ/kg
3-28 The temperature of R-134a at a specified state is to be determined.
Analysis Since the specified specific volume is higher than vg for 800 kPa, this is a superheated vapor state. From R-134a
tables,
P = 800 kPa
⎫
⎬ T = 60°C (Table A - 13)
v = 0.030 m 3 /kg ⎭
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3-10
3-29 A rigid container that is filled with water is cooled. The initial temperature and the final pressure are to be determined.
Analysis This is a constant volume process. The specific volume is
v1 = v 2 =
V
m
=
H2O
2 MPa
1 kg
150 L
0.150 m 3
= 0.150 m 3 /kg
1 kg
The initial state is superheated vapor. The temperature is determined to be
P1 = 2 MPa
⎫
⎬ T1 = 395°C
v 1 = 0.150 m 3 /kg ⎭
(Table A - 6)
Q
P
This is a constant volume cooling process (v = V /m = constant). The final
state is saturated mixture and thus the pressure is the saturation pressure at
the final temperature:
1
2
T2 = 40°C
⎫
⎬ P = Psat @ 40°C = 7.385 kPa (Table A - 4)
3
v 2 = v 1 = 0.150 m /kg ⎭ 2
v
3-30 A rigid container that is filled with R-134a is heated. The final temperature and initial pressure are to be determined.
Analysis This is a constant volume process. The specific volume is
v1 = v 2 =
V
m
=
R-134a
-40°C
10 kg
1.348 m3
1.348 m 3
= 0.1348 m 3 /kg
10 kg
The initial state is determined to be a mixture, and thus the pressure is the
saturation pressure at the given temperature
P
P1 = Psat @ - 40°C = 51.25 kPa (Table A - 11)
The final state is superheated vapor and the temperature is determined by
interpolation to be
P2 = 200 kPa
⎫
⎬ T2 = 66.3°C (Table A - 13)
v 2 = 0.1348 m 3 /kg ⎭
2
1
v
3-31 The enthalpy of R-134a at a specified state is to be determined.
Analysis The specific volume is
v=
V
m
=
9 m3
= 0.03 m 3 /kg
300 kg
Inspection of Table A-11 indicates that this is a mixture of liquid and vapor. Using the properties at 10°C line, the quality
and the enthalpy are determined to be
x=
v −v f
v fg
=
(0.03 − 0.0007930) m 3 /kg
(0.049403 − 0.0007930) m 3 /kg
= 0.6008
h = h f + xh fg = 65.43 + (0.6008)(190.73) = 180.02 kJ/kg
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3-11
3-32 The specific volume of R-134a at a specified state is to be determined.
Analysis Since the given temperature is higher than the saturation temperature for 200 kPa, this is a superheated vapor state.
The specific volume is then
P = 200 kPa ⎫
3
⎬ v = 0.11646 m /kg (Table A - 13)
T = 25°C ⎭
3-33 A spring-loaded piston-cylinder device is filled with R-134a. The water now undergoes a process until its volume
increases by 40%. The final temperature and the enthalpy are to be determined.
Analysis From Table A-11, the initial specific volume is
v 1 = v f + x1v fg = 0.0007142 + (0.80)(0.27090 − 0.0007142) = 0.21686 m 3 /kg
and the initial volume will be
V1 = mv 1 = (0.1 kg)(0.21686 m 3 /kg) = 0.021686 m 3
P
With a 40% increase in the volume, the final volume will be
3
V 2 = 1.4V1 = 1.4(0.021686 m ) = 0.03036 m
2
3
1
The distance that the piston moves between the initial and final conditions is
v
4(0.03036 − 0.021686) m 3
∆V
∆V
=
=
= 0.1227 m
A p πD 2 / 4
π (0.30 m) 2
∆x =
As a result of the compression of the spring, the pressure difference between the initial and final states is
∆P =
4(6.6 kN/m)(0.1227 m)
∆F k∆x
k∆x
=
=
=
= 11.46 kN/m 2 = 11.46 kPa
2
2
Ap
A p πD / 4
π (0.30 m)
The initial pressure is
P1 = Psat @ -34°C = 69.56 kPa (Table A - 11)
The final pressure is then
P2 = P1 + ∆P = 69.56 + 11.46 = 81.02 kPa
and the final specific volume is
v2 =
V2
m
=
0.03036 m 3
= 0.003036 m 3 /kg
0.1 kg
At this final state, the temperature and enthalpy are
P2 = 81.02 kPa
⎫ T1 = −30.9°C
⎬
v 2 = 0.003036 m 3 /kg ⎭ h1 = 13.73 kJ/kg
(from EES)
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3-12
3-34 A piston-cylinder device that is filled with water is cooled. The final pressure and volume of the water are to be
determined.
Analysis The initial specific volume is
v1 =
V1
m
=
0.1546 m 3
= 0.1546 m 3 /kg
1 kg
H2O
350°C
1 kg
0.1546 m3
This is a constant-pressure process. The initial state is determined to be superheated
vapor and thus the pressure is determined to be
T1 = 350°C
⎫
⎬ P = P2 = 1.8 MPa (Table A - 6)
v 1 = 0.1546 m 3 /kg ⎭ 1
P
The saturation temperature at 1.8 MPa is 207.11°C. Since the final temperature
is less than this temperature, the final state is compressed liquid. Using the
incompressible liquid approximation,
1
2
v 2 = v f @ 100°C = 0.001043 m 3 /kg (Table A - 4)
v
The final volume is then
V 2 = mv 2 = (1 kg)(0.001043 m 3 /kg) = 0.001043 m 3
3-35 A piston-cylinder device that is filled with R-134a is heated. The final volume is to be determined.
Analysis This is a constant pressure process. The initial specific volume is
v1 =
V
m
=
1.595 m 3
= 0.1595 m 3 /kg
10 kg
R-134a
-26.4°C
10 kg
1.595 m3
The initial state is determined to be a mixture, and thus the pressure is the
saturation pressure at the given temperature
P1 = Psat @ - 26.4°C = 100 kPa (Table A - 12)
The final state is superheated vapor and the specific volume is
P2 = 100 kPa ⎫
3
⎬ v 2 = 0.30138 m /kg (Table A - 13)
T2 = 100°C ⎭
P
1
2
The final volume is then
V 2 = mv 2 = (10 kg)(0.30138 m 3 /kg) = 3.0138 m 3
v
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3-13
3-36 The volume of a container that contains water at a specified state is to be determined.
Analysis The specific volume is determined from steam tables by interpolation to be
P = 100 kPa ⎫
3
⎬ v = 2.4062 m /kg (Table A - 6)
T = 250°C ⎭
Water
3 kg
100 kPa
250°C
The volume of the container is then
V = mv = (3 kg )(2.4062 m 3 /kg) = 7.22 m 3
3-37 A rigid container that is filled with R-134a is heated. The temperature and total enthalpy are to be determined at the
initial and final states.
Analysis This is a constant volume process. The specific volume is
R-134a
300 kPa
10 kg
14 L
0.014 m 3
= 0.0014 m 3 /kg
v1 = v 2 = =
m
10 kg
V
The initial state is determined to be a mixture, and thus the temperature is the
saturation temperature at the given pressure. From Table A-12 by interpolation
P
T1 = Tsat @ 300 kPa = 0.61°C
2
and
x1 =
v1 −v f
v fg
=
(0.0014 − 0.0007736) m 3 /kg
(0.067978 − 0.0007736) m 3 /kg
= 0.009321
h1 = h f + x1 h fg = 52.67 + (0.009321)(198.13) = 54.52 kJ/kg
1
v
The total enthalpy is then
H 1 = mh1 = (10 kg )(54.52 kJ/kg ) = 545.2 kJ
The final state is also saturated mixture. Repeating the calculations at this state,
T2 = Tsat @ 600 kPa = 21.55°C
x2 =
v 2 −v f
v fg
=
(0.0014 − 0.0008199) m 3 /kg
(0.034295 − 0.0008199) m 3 /kg
= 0.01733
h2 = h f + x 2 h fg = 81.51 + (0.01733)(180.90) = 84.64 kJ/kg
H 2 = mh2 = (10 kg )(84.64 kJ/kg ) = 846.4 kJ
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3-14
3-38 A piston-cylinder device that is filled with R-134a is cooled at constant pressure. The final temperature and the change
of total internal energy are to be determined.
Analysis The initial specific volume is
v1 =
V
=
m
12.322 m 3
= 0.12322 m 3 /kg
100 kg
R-134a
200 kPa
100 kg
12.322 m3
The initial state is superheated and the internal energy at this state is
P1 = 200 kPa
⎫
⎬ u1 = 263.08 kJ/kg (Table A - 13)
v 1 = 0.12322 m 3 /kg ⎭
P
The final specific volume is
v2 =
v1
2
=
0.12322 m 3 / kg
= 0.06161 m 3 /kg
2
2
This is a constant pressure process. The final state is determined to be saturated
mixture whose temperature is
1
v
T2 = Tsat @ 200 kPa = −10.09°C (Table A - 12)
The internal energy at the final state is (Table A-12)
x2 =
v 2 −v f
v fg
=
(0.06161 − 0.0007533) m 3 /kg
(0.099867 − 0.0007533) m 3 /kg
= 0.6140
u 2 = u f + x 2 u fg = 38.28 + (0.6140)(186.21) = 152.61 kJ/kg
Hence, the change in the internal energy is
∆u = u 2 − u1 = 152.61 − 263.08 = −110.47 kJ/kg
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3-15
3-39 A piston-cylinder device fitted with stops contains water at a specified state. Now the water is cooled until a final
pressure. The process is to be indicated on the T-v diagram and the change in internal energy is to be determined.
Analysis The process is shown on T-v diagram. The internal
energy at the initial state is
P1 = 300 kPa ⎫
⎬ u1 = 2728.9 kJ/kg (Table A - 6)
T1 = 250°C ⎭
Water
300 kPa
250°C
State 2 is saturated vapor at the initial pressure. Then,
P2 = 300 kPa
⎫
3
⎬ v 2 = 0.6058 m /kg (Table A - 5)
x 2 = 1 (sat. vapor) ⎭
Process 2-3 is a constant-volume process. Thus,
P3 = 100 kPa
⎫
⎬ u = 1163.3 kJ/kg (Table A - 5)
3
v 3 = 0.6058 m /kg ⎭ 3
300 kPa
T
1
250°C
The overall change in internal energy is
∆u = u1 − u 3 = 2728.9 − 1163.3 = 1566 kJ/kg
Q
100 kPa
2
3
v
3-40 The local atmospheric pressure, and thus the boiling temperature, changes with the weather conditions. The change in
the boiling temperature corresponding to a change of 5 mm of mercury in atmospheric pressure is to be determined.
Properties The saturation pressures of water at 95 and 100°C are 84.609 and 101.42 kPa, respectively (Table A-4). One
mm of mercury is equivalent to 1 mmHg = 0.1333 kPa (inner cover page).
Analysis A change of 5 mm of mercury in atmospheric pressure corresponds to
⎛ 0.1333 kPa ⎞
⎟⎟ = 0.6665 kPa
∆P = (5 mmHg)⎜⎜
⎝ 1 mmHg ⎠
P ± 5 mmHg
At about boiling temperature, the change in boiling temperature per 1 kPa change
in pressure is determined using data at 95 and 100°C to be
(100 − 95)°C
∆T
=
= 0.2974 °C/kPa
∆P (101.42 − 84.609) kPa
Then the change in saturation (boiling) temperature corresponding to a change of 0.147 psia becomes
∆Tboiling = (0.2974 °C/kPa) ∆P = (0.2974 °C/kPa)(0.6665 kPa) = 0.20°C
which is very small. Therefore, the effect of variation of atmospheric pressure on the boiling temperature is negligible.
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3-16
3-41 A person cooks a meal in a pot that is covered with a well-fitting lid, and leaves the food to cool to the room
temperature. It is to be determined if the lid will open or the pan will move up together with the lid when the person
attempts to open the pan by lifting the lid up.
Assumptions 1 The local atmospheric pressure is 1 atm = 101.325 kPa. 2 The weight of the lid is small and thus its effect
on the boiling pressure and temperature is negligible. 3 No air has leaked into the pan during cooling.
Properties The saturation pressure of water at 20°C is 2.3392 kPa (Table A-4).
Analysis Noting that the weight of the lid is negligible, the reaction force F on the lid after cooling at the pan-lid interface
can be determined from a force balance on the lid in the vertical direction to be
PA +F = PatmA
or,
F = A( Patm − P) = (πD 2 / 4)( Patm − P)
=
π (0.3 m) 2
P
2.3392 kPa
(101,325 − 2339.2) Pa
4
= 6997 m 2 Pa = 6997 N (since 1 Pa = 1 N/m 2 )
Patm = 1 atm
The weight of the pan and its contents is
W = mg = (8 kg)(9.81 m/s2 ) = 78.5 N
which is much less than the reaction force of 6997 N at the pan-lid interface. Therefore, the pan will move up together with
the lid when the person attempts to open the pan by lifting the lid up. In fact, it looks like the lid will not open even if the
mass of the pan and its contents is several hundred kg.
3-42 Water is boiled at 1 atm pressure in a pan placed on an electric burner. The water level drops by 10 cm in 45 min
during boiling. The rate of heat transfer to the water is to be determined.
Properties The properties of water at 1 atm and thus at a saturation temperature of Tsat = 100°C are hfg = 2256.5 kJ/kg and
vf = 0.001043 m3/kg (Table A-4).
Analysis The rate of evaporation of water is
mevap =
m& evap =
Vevap (πD 2 / 4) L [π (0.25 m)2 / 4](0.10 m)
=
=
= 4.704 kg
vf
vf
0.001043
mevap
∆t
=
H2O
1 atm
4.704 kg
= 0.001742 kg/s
45 × 60 s
Then the rate of heat transfer to water becomes
Q& = m& evap h fg = (0.001742 kg/s)(2256.5 kJ/kg) = 3.93 kW
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3-17
3-43 Water is boiled at a location where the atmospheric pressure is 79.5 kPa in a pan placed on an electric burner. The
water level drops by 10 cm in 45 min during boiling. The rate of heat transfer to the water is to be determined.
Properties The properties of water at 79.5 kPa are Tsat = 93.3°C, hfg = 2273.9 kJ/kg and vf = 0.001038 m3/kg (Table A-5).
Analysis The rate of evaporation of water is
m evap =
m& evap =
V evap
vf
m evap
∆t
=
=
(πD 2 / 4) L
vf
=
H 2O
79.5 kPa
[π (0.25 m) 2 / 4](0.10 m)
= 4.727 kg
0.001038
4.727 kg
= 0.001751 kg/s
45 × 60 s
Then the rate of heat transfer to water becomes
Q& = m& evap h fg = (0.001751 kg/s)(2273.9 kJ/kg) = 3.98 kW
3-44 Saturated steam at Tsat = 40°C condenses on the outer surface of a cooling tube at a rate of 130 kg/h. The rate of heat
transfer from the steam to the cooling water is to be determined.
Assumptions 1 Steady operating conditions exist. 2 The condensate leaves the condenser as a saturated liquid at 30°C.
Properties The properties of water at the saturation temperature of 40°C are hfg = 2406.0 kJ/kg (Table A-4).
Analysis Noting that 2406.0 kJ of heat is released as 1 kg of saturated
vapor at 40°C condenses, the rate of heat transfer from the steam to
the cooling water in the tube is determined directly from
Q& = m& evap h fg
40°C
L = 35 m
D = 3 cm
= (130 kg/h)(2406.0 kJ/kg) = 312,780 kJ/h
= 86.9 kW
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3-18
3-45 The boiling temperature of water in a 5-cm deep pan is given. The boiling temperature in a 40-cm deep pan is to be
determined.
Assumptions Both pans are full of water.
Properties The density of liquid water is approximately ρ = 1000 kg/m3.
Analysis The pressure at the bottom of the 5-cm pan is the saturation
pressure corresponding to the boiling temperature of 98°C:
P = Psat@98o C = 94.39 kPa
(Table A-4)
40 cm
5 cm
The pressure difference between the bottoms of two pans is
⎛
1 kPa
⎜ 1000 kg/m ⋅ s 2
⎝
∆P = ρ g h = (1000 kg/m 3 )(9.807 m/s 2 )(0.35 m)⎜
⎞
⎟ = 3.43 kPa
⎟
⎠
Then the pressure at the bottom of the 40-cm deep pan is
P = 94.39 + 3.43 = 97.82 kPa
Then the boiling temperature becomes
Tboiling = Tsat@97.82
kPa
= 99.0 °C
(Table A-5)
3-46 A vertical piston-cylinder device is filled with water and covered with a 20-kg piston that serves as the lid. The
boiling temperature of water is to be determined.
Analysis The pressure in the cylinder is determined from a force balance on the piston,
PA = PatmA + W
Patm
or,
P = Patm +
mg
A
= (100 kPa) +
= 119.61 kPa
⎞
(20 kg)(9.81 m/s2 ) ⎛
1 kPa
⎜
⎟
2
2⎟
⎜
0.01 m
⎝ 1000 kg/m ⋅ s ⎠
P
W = mg
The boiling temperature is the saturation temperature corresponding to this pressure,
T = Tsat @119.61 kPa = 104.7°C
(Table A-5)
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3-19
3-47 A rigid tank that is filled with saturated liquid-vapor mixture is heated. The temperature at which the liquid in the tank
is completely vaporized is to be determined, and the T-v diagram is to be drawn.
Analysis This is a constant volume process (v = V /m = constant), and
the specific volume is determined to be
v=
V
m
=
H 2O
90°C
1.8 m 3
= 0.12 m 3 /kg
15 kg
When the liquid is completely vaporized the tank will contain
saturated vapor only. Thus,
T
v 2 = v g = 0.12 m 3 /kg
2
The temperature at this point is the temperature that corresponds to
this vg value,
T = Tsat @v
g = 0.12
m 3 /kg
= 202.9°C
1
v
(Table A-4)
3-48
A piston-cylinder device contains a saturated liquid-vapor mixture of water at 800 kPa pressure. The mixture is
heated at constant pressure until the temperature rises to 200°C. The initial temperature, the total mass of water, the final
volume are to be determined, and the P-v diagram is to be drawn.
Analysis (a) Initially two phases coexist in equilibrium, thus we have a saturated liquid-vapor mixture. Then the
temperature in the tank must be the saturation temperature at the specified pressure,
T = Tsat@600 kPa = 158.8°C
(b) The total mass in this case can easily be determined by adding the mass of each phase,
mf =
mg =
Vf
vf
Vg
vg
=
=
0.005 m 3
0.001101 m 3 /kg
0.9 m 3
0.3156 m 3 /kg
= 4.543 kg
P
= 2.852 kg
mt = m f + m g = 4.543 + 2.852 = 7.395 kg
(c) At the final state water is superheated vapor, and its specific volume is
1
2
v
P2 = 600 kPa ⎫
3
⎬ v = 0.3521 m /kg (Table A-6)
T2 = 200 o C ⎭ 2
Then,
V 2 = mt v 2 = (7.395 kg)(0.3521 m 3 /kg) = 2.604 m 3
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3-20
3-49
Problem 3-48 is reconsidered. The effect of pressure on the total mass of water in the tank as the pressure
varies from 0.1 MPa to 1 MPa is to be investigated. The total mass of water is to be plotted against pressure, and results are
to be discussed.
Analysis The problem is solved using EES, and the solution is given below.
P[1]=600 [kPa]
P[2]=P[1]
T[2]=200 [C]
V_f1 = 0.005 [m^3]
V_g1=0.9 [m^3]
spvsat_f1=volume(Steam_iapws, P=P[1],x=0) "sat. liq. specific volume, m^3/kg"
spvsat_g1=volume(Steam_iapws,P=P[1],x=1) "sat. vap. specific volume, m^3/kg"
m_f1=V_f1/spvsat_f1 "sat. liq. mass, kg"
m_g1=V_g1/spvsat_g1 "sat. vap. mass, kg"
m_tot=m_f1+m_g1
V[1]=V_f1+V_g1
spvol[1]=V[1]/m_tot "specific volume1, m^3"
T[1]=temperature(Steam_iapws, P=P[1],v=spvol[1])"C"
"The final volume is calculated from the specific volume at the final T and P"
spvol[2]=volume(Steam_iapws, P=P[2], T=T[2]) "specific volume2, m^3/kg"
V[2]=m_tot*spvol[2]
Steam IAPWS
106
P1 [kPa]
mtot [kg]
100
200
300
400
500
600
700
800
900
1000
5.324
5.731
6.145
6.561
6.978
7.395
7.812
8.23
8.648
9.066
105
P [kPa]
104
200°C
103
2
1
P=600 kPa
102
101
100
10-3
10-2
10-1
100
101
102
3
v [m /kg]
9.5
9
8.5
mtot [kg]
8
7.5
7
6.5
6
5.5
5
100
200
300
400
500
600
700
800
900
1000
P1 [kPa]
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3-21
3-50 Superheated water vapor cools at constant volume until the temperature drops to 120°C. At the final state, the
pressure, the quality, and the enthalpy are to be determined.
Analysis This is a constant volume process (v = V/m = constant), and the initial specific volume is determined to be
P1 = 1.4 MPa ⎫
3
⎬ v = 0.16356 m /kg
T1 = 250 o C ⎭ 1
(Table A-6)
H2O
1.4 MPa
250°C
At 120°C, vf = 0.001060 m3/kg and vg = 0.89133 m3/kg. Thus at the final state, the
tank will contain saturated liquid-vapor mixture since vf < v < vg , and the final
pressure must be the saturation pressure at the final temperature,
P = Psat@120o C = 198.67 kPa
T
1
(b) The quality at the final state is determined from
x2 =
v 2 −v f
v fg
=
0.16356 − 0.001060
= 0.1825
0.89133 − 0.001060
(c) The enthalpy at the final state is determined from
h = h f + xh fg = 503.81 + 0.1825 × 2202.1 = 905.7 kJ/kg
2
v
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3-22
3-51
Problem 3-50 is reconsidered. The effect of initial pressure on the quality of water at the final state as the
pressure varies from 700 kPa to 2000 kPa is to be investigated. The quality is to be plotted against initial pressure, and the
results are to be discussed.
Analysis The problem is solved using EES, and the solution is given below.
T[1]=250 [C]
P[1]=1400 [kPa]
T[2]=120 [C]
v[ 1]=volume(steam_iapws, T=T[1], P=P[1])
v[2]=v[1]
P[2]=pressure(steam_iapws, T=T[2], v=v[2])
h[2]=enthalpy(steam_iapws, T=T[2], v=v[2])
x[2]=quality(steam_iapws, T=T[2], v=v[2])
x2
0.4
0.3766
0.3282
0.2904
0.2602
0.2355
0.2149
0.1975
0.1825
0.1696
0.1582
0.1482
0.1392
0.1312
0.124
0.35
0.3
x[2]
P1
[kPa]
700
800
900
1000
1100
1200
1300
1400
1500
1600
1700
1800
1900
2000
0.25
0.2
0.15
0.1
800
1000
1200
1400
1600
1800
2000
P[1] [kPa]
SteamIAPWS
700
600
T [°C]
500
400
300
200
1
1400 kPa
198.7 kPa
100
2
0.05 0.1
0
10-3
10-2
10-1
100
0.2
0.5
101
102
3
v [m /kg]
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3-23
3-52 The properties of compressed liquid water at a specified state are to be determined using the compressed liquid tables,
and also by using the saturated liquid approximation, and the results are to be compared.
Analysis Compressed liquid can be approximated as saturated liquid at the given temperature. Then from Table A-4,
T = 80°C ⇒
v ≅ v f @ 80°C = 0.001029 m 3 /kg (0.90% error)
u ≅ u f @ 80°C = 334.97 kJ/kg
h ≅ h f @ 80°C = 335.02 kJ/kg
(1.35% error)
(4.53% error)
From compressed liquid table (Table A-7),
v = 0.00102 m 3 /kg
P = 20 MPa ⎫
u = 330.50 kJ/kg
T = 80°C ⎬⎭
h = 350.90 kJ/kg
The percent errors involved in the saturated liquid approximation are listed above in parentheses.
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3-24
3-53
Problem 3-52 is reconsidered. Using EES, the indicated properties of compressed liquid are to be determined,
and they are to be compared to those obtained using the saturated liquid approximation.
Analysis The problem is solved using EES, and the solution is given below.
"Given"
T=80 [C]
P=20000 [kPa]
"Analysis"
Fluid$='steam_iapws'
"Saturated liquid assumption"
v_f=volume(Fluid$, T=T, x=0)
u_f=intenergy(Fluid$, T=T, x=0)
h_f=enthalpy(Fluid$, T=T, x=0)
"Compressed liquid treatment"
v=volume(Fluid$, T=T, P=P)
u=intenergy(Fluid$, T=T, P=P)
h=enthalpy(Fluid$, T=T, P=P)
"Percentage Errors"
error_v=100*(v_f-v)/v
error_u=100*(u_f-u)/u
error_h=100*(h-h_f)/h
SOLUTION
error_h=4.527
error_u=1.351
error_v=0.8987
Fluid$='steam_iapws'
h=350.90 [kJ/kg]
h_f=335.02 [kJ/kg]
P=20000 [kPa]
T=80 [C]
u=330.50 [kJ/kg]
u_f=334.97 [kJ/kg]
v=0.00102 [m^3/kg]
v_f=0.001029 [m^3/kg]
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3-25
3-54 Superheated steam in a piston-cylinder device is cooled at constant pressure until half of the mass condenses. The final
temperature and the volume change are to be determined, and the process should be shown on a T-v diagram.
Analysis (b) At the final state the cylinder contains saturated liquidvapor mixture, and thus the final temperature must be the saturation
temperature at the final pressure,
T = Tsat@1 MPa = 179.88 °C
(Table A-5)
H2O
300°C
1 MPa
(c) The quality at the final state is specified to be x2 = 0.5. The specific
volumes at the initial and the final states are
P1 = 1.0 MPa ⎫
3
⎬ v = 0.25799 m /kg
T1 = 300 o C ⎭ 1
(Table A-6)
T
1
P2 = 1.0 MPa ⎫
⎬ v 2 = v f + x2v fg
x2 = 0.5
⎭ = 0.001127 + 0.5 × (0.19436 − 0.001127)
= 0.09775 m3/kg
2
Thus,
v
∆V = m(v 2 − v 1 ) = (0.8 kg)(0.0977 5 − 0.25799) m 3 /kg = −0.128 2 m 3
3-55 The water in a rigid tank is cooled until the vapor starts condensing. The initial pressure in the tank is to be
determined.
Analysis This is a constant volume process (v = V /m = constant), and the
initial specific volume is equal to the final specific volume that is
25
3
v 1 = v 2 = v g @124°C = 0.79270 m /kg (Table A-4)
since the vapor starts condensing at 150°C. Then from
Table A-6,
T1 = 250°C
⎫
⎬ P = 0.30 MPa
3
v 1 = 0.79270 m /kg ⎭ 1
T °C
1
H2O
T1= 250°C
P1 = ?
15
2
v
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3-26
3-56 Heat is supplied to a piston-cylinder device that contains water at a specified state. The volume of the tank, the final
temperature and pressure, and the internal energy change of water are to be determined.
Properties The saturated liquid properties of water at 200°C are: vf = 0.001157 m3/kg and uf = 850.46 kJ/kg (Table A-4).
Analysis (a) The cylinder initially contains saturated liquid water. The volume of the cylinder at the initial state is
V 1 = mv 1 = (1.4 kg)(0.001157 m 3 /kg) = 0.001619 m 3
The volume at the final state is
V = 4(0.001619) = 0.006476 m 3
Water
1.4 kg, 200°C
sat. liq.
(b) The final state properties are
v2 =
V
m
=
0.006476 m3
= 0.004626 m3 / kg
1.4 kg
v 2 = 0.004626 m3 / kg ⎫⎪
T2 = 371.3 °C
⎬ P2 = 21,367 kPa
⎪⎭
u2 = 2201.5 kJ/kg
x2 = 1
Q
(Table A-4 or A-5 or EES)
(c) The total internal energy change is determined from
∆U = m(u 2 − u1 ) = (1.4 kg)(2201.5 - 850.46) kJ/kg = 1892 kJ
3-57 The errors involved in using the specific volume and enthalpy of water by the incompressible liquid approximation are
to be determined.
Analysis The state of water is compressed liquid. From the steam tables,
P = 20 MPa ⎫ v = 0.0010679 m 3 /kg
⎬
T = 140°C ⎭ h = 602.07 kJ/kg
(Table A - 7)
Based upon the incompressible liquid approximation,
P = 20 MPa ⎫ v ≅ v f @ 140°C = 0.001080 m 3 /kg
(Table A - 4)
⎬
T = 140°C ⎭ h ≅ h f @ 140°C = 589.16 kJ/kg
The errors involved are
Percent Error (specific volume) =
Percent Error (enthalpy) =
0.0010679 − 0.001080
× 100 = −1.13%
0.0010679
602.07 − 589.16
× 100 = 2.14%
602.07
which are quite acceptable in most engineering calculations.
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3-27
3-58 A piston-cylinder device that is filled with R-134a is heated. The volume change is to be determined.
Analysis The initial specific volume is
P1 = 60 kPa ⎫
3
⎬ v 1 = 0.33608 m /kg (Table A - 13)
T1 = −20°C ⎭
R-134a
60 kPa
-20°C
100 g
and the initial volume is
V 1 = mv 1 = (0.100 kg)(0.33608 m 3 /kg) = 0.033608 m 3
At the final state, we have
P2 = 60 kPa ⎫
3
⎬ v 2 = 0.50410 m /kg (Table A - 13)
T2 = 100°C ⎭
P
V 2 = mv 2 = (0.100 kg)(0.50410 m 3 /kg) = 0.050410 m 3
1
2
The volume change is then
v
∆V = V 2 −V 1 = 0.050410 − 0.033608 = 0.0168 m 3
3-59 A rigid vessel is filled with refrigerant-134a. The total volume and the total internal energy are to be determined.
Properties The properties of R-134a at the given state are (Table A-13).
P = 800 kPa ⎫ u = 327.87 kJ/kg
T = 120 o C ⎬⎭ v = 0.037625 m 3 /kg
Analysis The total volume and internal energy are determined from
V = mv = (2 kg)(0.037625 m 3 /kg) = 0.0753 m 3
R-134a
2 kg
800 kPa
120°C
U = mu = (2 kg)(327.87 kJ/kg) = 655.7 kJ
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3-28
3-60 A rigid vessel contains R-134a at specified temperature. The pressure, total internal energy, and the volume of the
liquid phase are to be determined.
Analysis (a) The specific volume of the refrigerant is
v=
V
m
=
0.5 m 3
= 0.05 m 3 /kg
10 kg
At -20°C, vf = 0.0007362 m3/kg and vg = 0.14729 m3/kg (Table A-11). Thus the tank
contains saturated liquid-vapor mixture since vf < v < vg , and the pressure must be the
saturation pressure at the specified temperature,
R-134a
10 kg
-20°C
P = Psat@ − 20o C = 132.82 kPa
(b) The quality of the refrigerant-134a and its total internal energy are determined from
x=
v −v f
v fg
=
0.05 − 0.0007362
= 0.3361
0.14729 − 0.0007362
u = u f + xu fg = 25.39 + 0.3361× 193.45 = 90.42 kJ/kg
U = mu = (10 kg)(90.42 kJ/kg) = 904.2 kJ
(c) The mass of the liquid phase and its volume are determined from
m f = (1 − x) mt = (1 − 0.3361) × 10 = 6.639 kg
V f = m f v f = (6.639 kg)(0.0007362 m3/kg) = 0.00489 m 3
3-61
The Pessure-Enthalpy diagram of R-134a showing some constant-temperature and constant-entropy lines are
obtained using Property Plot feature of EES.
P [kPa]
1.2
kJ
/kg
1
0.8
0.5
0.3
0.2
104
-K
R134a
105
70°C
103
40°C
10°C
-10°C
102
101
-100
-30°C
0
100
200
300
400
500
h [kJ/kg]
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3-29
3-62 A rigid vessel that contains a saturated liquid-vapor mixture is heated until it reaches the critical state. The mass of the
liquid water and the volume occupied by the liquid at the initial state are to be determined.
Analysis This is a constant volume process (v = V /m = constant) to the critical state, and thus the initial specific volume
will be equal to the final specific volume, which is equal to the critical specific volume of water,
v 1 = v 2 = v cr = 0.003106 m 3 /kg
(last row of Table A-4)
The total mass is
0.3 m 3
V
m= =
= 96.60 kg
v 0.003106 m 3 /kg
At 150°C, vf = 0.001091 m3/kg and vg = 0.39248 m3/kg (Table A4). Then the quality of water at the initial state is
x1 =
v1 −v f
v fg
=
T
cp
H 2O
150°C
0.003106 − 0.001091
= 0.005149
0.39248 − 0.001091
vcr
v
Then the mass of the liquid phase and its volume at the initial state are determined from
m f = (1 − x1 ) mt = (1 − 0.005149)(96.60) = 96.10 kg
V f = m f v f = (96.10 kg)(0.001091 m3/kg) = 0.105 m 3
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3-30
Ideal Gas
3-63C Mass m is simply the amount of matter; molar mass M is the mass of one mole in grams or the mass of one kmol in
kilograms. These two are related to each other by m = NM, where N is the number of moles.
3-64C A gas can be treated as an ideal gas when it is at a high temperature or low pressure relative to its critical
temperature and pressure.
3-65C 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 of the gas.
3-66C Propane (molar mass = 44.1 kg/kmol) poses a greater fire danger than methane (molar mass = 16 kg/kmol) since
propane is heavier than air (molar mass = 29 kg/kmol), and it will settle near the floor. Methane, on the other hand, is
lighter than air and thus it will rise and leak out.
3-67 The specific volume of nitrogen at a specified state is to be determined.
Assumptions At specified conditions, nitrogen behaves as an ideal gas.
Properties The gas constant of nitrogen is R = 0.2968 kJ/kg⋅K (Table A-1).
Analysis According to the ideal gas equation of state,
v=
RT (0.2968 kPa ⋅ m 3 /kg ⋅ K)(227 + 273 K)
=
= 0.495 m 3 /kg
P
300 kPa
3-68 The temperature in a container that is filled with oxygen is to be determined.
Assumptions At specified conditions, oxygen behaves as an ideal gas.
Properties The gas constant of oxygen is R = 0.2598 kPa⋅m3/kg⋅K (Table A-1).
Analysis The definition of the specific volume gives
v=
V
m
=
0.09 m 3
= 0.1 m 3 /kg
0.9 kg
Using the ideal gas equation of state, the temperature is
T=
(600 kPa)(0.1 m 3 /kg)
Pv
=
= 231 K
R 0.2598 kPa ⋅ m 3 /kg ⋅ K
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3-31
3-69 The volume of a container that is filled with helium at a specified state is to be determined.
Assumptions At specified conditions, helium behaves as an ideal gas.
Properties The gas constant of helium is R = 2.0769 kJ/kg⋅K (Table A-1).
Analysis According to the ideal gas equation of state,
V =
mRT (2 kg)(2.0769 kPa ⋅ m 3 /kg ⋅ K)(27 + 273 K)
=
= 4.154 m 3
P
300 kPa
3-70 A balloon is filled with helium gas. The mole number and the mass of helium in the balloon are to be determined.
Assumptions At specified conditions, helium behaves as an ideal gas.
Properties The universal gas constant is Ru = 8.314 kPa.m3/kmol.K. The molar mass of helium is 4.0 kg/kmol (Table A-1).
Analysis The volume of the sphere is
4
3
4
3
V = π r 3 = π (4.5 m) 3 = 381.7 m 3
Assuming ideal gas behavior, the mole numbers of He is determined from
N=
(200 kPa)(381.7 m 3 )
PV
=
= 30.61 kmol
Ru T (8.314 kPa ⋅ m 3 /kmol ⋅ K)(300 K)
He
D=9m
27°C
200 kPa
Then the mass of He can be determined from
m = NM = (30.61 kmol)(4.0 kg/kmol) = 123 kg
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3-32
3-71
Problem 3-70 is to be reconsidered. The effect of the balloon diameter on the mass of helium contained in the
balloon is to be determined for the pressures of (a) 100 kPa and (b) 200 kPa as the diameter varies from 5 m to 15 m. The
mass of helium is to be plotted against the diameter for both cases.
Analysis The problem is solved using EES, and the solution is given below.
"Given Data"
{D=9 [m]}
T=27 [C]
P=200 [kPa]
R_u=8.314 [kJ/kmol-K]
"Solution"
P*V=N*R_u*(T+273)
V=4*pi*(D/2)^3/3
m=N*MOLARMASS(Helium)
m
[kg]
21.01
38.35
63.31
97.25
141.6
197.6
266.9
350.6
450.2
567.2
600
500
400
m [kg]
D
[m]
5
6.111
7.222
8.333
9.444
10.56
11.67
12.78
13.89
15
P=200 kPa
300
P=200 kPa
200
P=100 kPa
100
0
5
7
9
D [m]
11
13
15
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3-33
3-72 Two rigid tanks connected by a valve to each other contain air at specified conditions. The volume of the second tank
and the final equilibrium pressure when the valve is opened are to be determined.
Assumptions At specified conditions, air behaves as an ideal gas.
Properties The gas constant of air is R = 0.287 kPa.m3/kg.K (Table A-1).
Analysis Let's call the first and the second tanks A and B. Treating air as an ideal gas, the volume of the second tank and
the mass of air in the first tank are determined to be
⎛ m RT ⎞
V B = ⎜⎜ 1 1 ⎟⎟ =
⎝ P1 ⎠ B
(5 kg)(0.287 kPa ⋅ m3/kg ⋅ K)(308 K)
= 2.21 m 3
200 kPa
⎛ PV ⎞
(500 kPa)(1.0 m3 )
= 5.846 kg
m A = ⎜⎜ 1 ⎟⎟ =
3
⎝ RT1 ⎠ A (0.287 kPa ⋅ m /kg ⋅ K)(298 K)
Thus,
A
B
Air
V = 1 m3
T = 25°C
P = 500 kPa
×
Air
m = 5 kg
T = 35°C
P = 200 kPa
V = V A + V B = 1.0 + 2.21 = 3.21 m3
m = mA + mB = 5.846 + 5.0 = 10.846 kg
Then the final equilibrium pressure becomes
P2 =
mRT2
V
=
(10.846 kg)(0.287 kPa ⋅ m3 /kg ⋅ K)(293 K)
3.21 m3
= 284.1 kPa
3-73 An elastic tank contains air at a specified state. The volume is doubled at the same pressure. The initial volume and
the final temperature are to be determined.
Assumptions At specified conditions, air behaves as an ideal gas.
Analysis According to the ideal gas equation of state,
PV = nRu T
(225 kPa)V = (1 kmol)(8.314 kPa ⋅ m 3 /kmol ⋅ K)(18 + 273) K
V = 0.0930 m 3
V 2 T2
T2
=
⎯
⎯→ 2 =
⎯
⎯→ T2 = 582 K = 309°C
V1 T1
(18 + 273) K
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3-34
3-74 An ideal gas in a rigid tank is cooled to a final gage pressure. The final temperature is to be determined.
Assumptions The gas is specified as an ideal gas so that ideal gas relation can be used.
Analysis According to the ideal gas equation of state at constant volume,
m1 = m1
P1V 1 P2V 2
=
T1
T2
Since
Patm = 100 kPa
V1 = V 2
Then,
Ideal gas
1227°C
200 kPa
(gage)
Q
P1 P2
=
T1 T2
T2 = T1
P2
(50 + 100) kPa
= [(1227 + 273) K ]
= 750 K = 477°C
P2
(200 + 100) kPa
3-75 One side of a two-sided tank contains an ideal gas while the other side is evacuated. The partition is removed and the
gas fills the entire tank. The gas is also heated to a final pressure. The final temperature is to be determined.
Assumptions The gas is specified as an ideal gas so that ideal gas relation can be used.
Analysis According to the ideal gas equation of state,
P2 = P1
V 2 = V1 + 2V1 = 3V1
Applying these,
m1 = m1
Ideal
gas
927°C
V1
Evacuated
2V1
Q
P1V 1 P2V 2
=
T1
T2
V1
T1
=
V2
T2
T2 = T1
V2
3V
= T1 1 = 3T1 = 3[927 + 273) K ] = 3600 K = 3327°C
V1
V1
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3-35
3-76 A piston-cylinder device containing argon undergoes an isothermal process. The final pressure is to be determined.
Assumptions At specified conditions, argon behaves as an ideal gas.
Properties The gas constant of argon is R = 0.2081 kJ/kg⋅K (Table A-1).
Analysis Since the temperature remains constant, the ideal gas equation gives
Argon
0.6 kg
0.05 m3
550 kPa
PV
PV
m= 1 1 = 2 2 ⎯
⎯→ P1V1 = P2V 2
RT
RT
which when solved for final pressure becomes
P2 = P1
V1
V
= P1 1 = 0.5P1 = 0.5(550 kPa ) = 275 kPa
V2
2V1
3-77 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 kPa.m3/kg.K (Table A-1).
Analysis Initially, the absolute pressure in the tire is
P1 = Pg + Patm = 210 + 100 = 310kPa
Treating air as an ideal gas and assuming the volume of the tire to
remain constant, the final pressure in the tire can be determined from
Tire
25°C
323 K
P1V1 P2V 2
T
=
⎯
⎯→ P2 = 2 P1 =
(310 kPa) = 336 kPa
T1
T2
T1
298 K
Thus the pressure rise is
∆P = P2 − P1 = 336 − 310 = 26 kPa
The amount of air that needs to be bled off to restore pressure to its original value is
m1 =
m2 =
(310 kPa)(0.025 m3 )
P1V
=
= 0.0906 kg
RT1 (0.287 kPa ⋅ m3/kg ⋅ K)(298 K)
(310 kPa)(0.025 m3 )
P1V
=
= 0.0836 kg
RT2 (0.287 kPa ⋅ m3/kg ⋅ K)(323 K)
∆m = m1 − m2 = 0.0906 − 0.0836 = 0.0070 kg
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3-36
Compressibility Factor
3-78C It represent the deviation from ideal gas behavior. The further away it is from 1, the more the gas deviates from
ideal gas behavior.
3-79C All gases have the same compressibility factor Z at the same reduced temperature and pressure.
3-80C Reduced pressure is the pressure normalized with respect to the critical pressure; and reduced temperature is the
temperature normalized with respect to the critical temperature.
3-81 The specific volume of steam is to be determined using the ideal gas relation, the compressibility chart, and the steam
tables. The errors involved in the first two approaches are also to be determined.
Properties The gas constant, the critical pressure, and the critical temperature of water are, from Table A-1,
R = 0.4615 kPa·m3/kg·K,
Tcr = 647.1 K,
Pcr = 22.06 MPa
Analysis (a) From the ideal gas equation of state,
v=
RT (0.4615 kPa ⋅ m 3 /kg ⋅ K)(623.15 K)
=
= 0.01917 m 3 /kg (67.0% error)
P
15,000 kPa
(b) From the compressibility chart (Fig. A-15),
10 MPa
P
⎫
=
= 0.453 ⎪
Pcr 22.06 MPa
⎪
⎬ Z = 0.65
673 K
T
⎪
TR =
=
= 1.04
⎪⎭
Tcr 647.1 K
PR =
H2O
15 MPa
350°C
Thus,
v = Zv ideal = (0.65)(0.01917 m 3 /kg) = 0.01246 m 3 /kg (8.5% error)
(c) From the superheated steam table (Table A-6),
P = 15 MPa
T = 350°C
} v = 0.01148 m /kg
3
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3-37
3-82
Problem 3-81 is reconsidered. The problem is to be solved using the general compressibility factor feature of
EES (or other) software. The specific volume of water for the three cases at 15 MPa over the temperature range of 350°C to
600°C in 25°C intervals is to be compared, and the % error involved in the ideal gas approximation is to be plotted against
temperature.
Analysis The problem is solved using EES, and the solution is given below.
P=15 [MPa]*Convert(MPa,kPa)
{T_Celsius= 350 [C]}
T=T_Celsius+273 "[K]"
T_critical=T_CRIT(Steam_iapws)
P_critical=P_CRIT(Steam_iapws)
{v=Vol/m}
P_table=P; P_comp=P;P_idealgas=P
T_table=T; T_comp=T;T_idealgas=T
v_table=volume(Steam_iapws,P=P_table,T=T_table) "EES data for steam as a real gas"
{P_table=pressure(Steam_iapws, T=T_table,v=v)}
{T_sat=temperature(Steam_iapws,P=P_table,v=v)}
MM=MOLARMASS(water)
R_u=8.314 [kJ/kmol-K] "Universal gas constant"
R=R_u/MM "[kJ/kg-K], Particular gas constant"
P_idealgas*v_idealgas=R*T_idealgas "Ideal gas equation"
z = COMPRESS(T_comp/T_critical,P_comp/P_critical)
P_comp*v_comp=z*R*T_comp "generalized Compressibility factor"
Error_idealgas=Abs(v_table-v_idealgas)/v_table*Convert(, %)
Error_comp=Abs(v_table-v_comp)/v_table*Convert(, %)
Errorcomp [%]
9.447
2.725
0.4344
0.5995
1.101
1.337
1.428
1.437
1.397
1.329
1.245
Errorideal gas [%]
67.22
43.53
32.21
25.23
20.44
16.92
14.22
12.1
10.39
8.976
7.802
TCelcius [C]
350
375
400
425
450
475
500
525
550
575
600
Specific Volume
Percent Error [%]
70
60
Ideal Gas
50
Compressibility Factor
40
Steam at 15 MPa
30
20
10
0
300
350
400
450
500
550
600
TCe lsius [C]
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3-38
3-83 The specific volume of R-134a is to be determined using the ideal gas relation, the compressibility chart, and the R134a tables. The errors involved in the first two approaches are also to be determined.
Properties The gas constant, the critical pressure, and the critical temperature of refrigerant-134a are, from Table A-1,
R = 0.08149 kPa·m3/kg·K,
Tcr = 374.2 K,
Pcr = 4.059 MPa
Analysis (a) From the ideal gas equation of state,
v=
RT (0.08149 kPa ⋅ m3/kg ⋅ K)(343 K)
=
= 0.03105 m3 /kg
P
900 kPa
(13.3% error )
(b) From the compressibility chart (Fig. A-15),
0.9 MPa
P
⎫
=
= 0.222 ⎪
Pcr 4.059 MPa
⎪
⎬ Z = 0.894
343 K
T
=
= 0.917 ⎪
TR =
⎪⎭
Tcr 374.2 K
R-134a
0.9 MPa
PR =
70°C
Thus,
v = Zv ideal = (0.894)(0.03105 m 3 /kg) = 0.02776 m 3 /kg
(1.3%error)
(c) From the superheated refrigerant table (Table A-13),
}
P = 0.9 MPa v = 0.027413 m3 /kg
T = 70°C
3-84 The specific volume of steam is to be determined using the ideal gas relation, the compressibility chart, and the steam
tables. The errors involved in the first two approaches are also to be determined.
Properties The gas constant, the critical pressure, and the critical temperature of water are, from Table A-1,
R = 0.4615 kPa·m3/kg·K,
Tcr = 647.1 K,
Pcr = 22.06 MPa
Analysis (a) From the ideal gas equation of state,
v=
RT (0.4615 kPa ⋅ m3/kg ⋅ K)(723 K)
=
= 0.09533 m3 /kg
P
3500 kPa
(3.7% error)
(b) From the compressibility chart (Fig. A-15),
3.5 MPa
P
⎫
=
= 0.159 ⎪
Pcr 22.06 MPa
⎪
⎬ Z = 0.961
723 K
T
⎪
=
= 1.12
TR =
⎪⎭
Tcr 647.1 K
PR =
H 2O
3.5 MPa
450°C
Thus,
v = Zv ideal = (0.961)(0.09533 m 3 /kg) = 0.09161 m 3 /kg
(0.4% error)
(c) From the superheated steam table (Table A-6),
P = 3.5 MPa
T = 450°C
} v = 0.09196 m /kg
3
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3-39
3-85 Ethane in a rigid vessel is heated. The final pressure is to be determined using the compressibility chart.
Properties The gas constant, the critical pressure, and the critical temperature of ethane are, from Table A-1,
R = 0.2765 kPa·m3/kg·K,
Tcr = 305.5 K,
Pcr = 4.48 MPa
Analysis From the compressibility chart at the initial state (Fig. A-15),
⎫
⎪
⎪
⎬ Z1 = 0.960
P1
550 kPa
PR1 =
=
= 0.123 ⎪
⎪⎭
Pcr 4480 kPa
TR1 =
T1
313 K
=
= 1.025
Tcr 305.5 K
Ethane
550 kPa
Q
40°C
The specific volume does not change during the process. Then,
v1 = v 2 =
Z 1 RT1 (0.960)(0.2765 kPa ⋅ m 3 /kg ⋅ K)(313 K)
=
= 0.1511 m 3 /kg
P1
550 kPa
At the final state,
⎫
⎪
⎪
⎬ Z 2 = 1.0
3
v 2,actual
0.1511 m /kg
⎪
=
=
=
8
.
01
⎪
RTcr /Pcr (0.2765 kPa ⋅ m 3 /kg ⋅ K)(305.5 K)/(4480 kPa)
⎭
TR 2 =
v R2
T2
553 K
=
= 1.810
Tcr 305.5 K
Thus,
P2 =
Z 2 RT2
v2
=
(1.0)(0.2765 kPa ⋅ m 3 /kg ⋅ K)(553 K)
0.1511 m 3 /kg
= 1012 kPa
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3-40
3-86 Ethylene is heated at constant pressure. The specific volume change of ethylene is to be determined using the
compressibility chart.
Properties The gas constant, the critical pressure, and the critical temperature of ethane are, from Table A-1,
R = 0.2964 kPa·m3/kg·K,
Tcr = 282.4 K,
Pcr = 5.12 MPa
Analysis From the compressibility chart at the initial and final states (Fig. A-15),
⎫
⎪
⎪
⎬ Z1 = 0.56
P
5 MPa
= 1 =
= 0.977 ⎪
⎪⎭
Pcr 5.12 MPa
T R1 =
PR1
T1
293 K
=
= 1.038
Tcr 282.4 K
T2
⎫
473 K
=
= 1.675 ⎪
Tcr 282.4 KR
⎬ Z1 = 0.961
⎪
= PR1 = 0.977
⎭
TR 2 =
PR 2
Ethylene
5 MPa
20°C
Q
The specific volume change is
R
( Z 2 T2 − Z 1T1 )
P
0.2964 kPa ⋅ m 3 /kg ⋅ K
[(0.961)(473 K) − (0.56)(293 K)]
=
5000 kPa
∆v =
= 0.0172 m 3 /kg
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3-41
3-87 Water vapor is heated at constant pressure. The final temperature is to be determined using ideal gas equation, the
compressibility charts, and the steam tables.
Properties The gas constant, the critical pressure, and the critical temperature of water are, from Table A-1,
R = 0.4615 kPa·m3/kg·K,
Tcr = 647.1 K,
Pcr = 22.06 MPa
Analysis (a) From the ideal gas equation,
T2 = T1
v2
= (350 + 273 K )(2) = 1246 K
v1
(b) The pressure of the steam is
Water
350°C
sat. vapor
P1 = P2 = Psat@350°C = 16,529 kPa
From the compressibility chart at the initial state (Fig. A-15),
⎫
⎪
⎪
⎬ Z1 = 0.593, v R1 = 0.75
P1 16.529 MPa
=
=
= 0.749 ⎪
⎪⎭
Pcr
22.06 MPa
T R1 =
PR1
Q
T1
623 K
=
= 0.963
Tcr 647.1 KR
At the final state,
PR 2 = PR1 = 0.749
v R 2 = 2v R1
⎫
⎬ Z 2 = 0.88
= 2(0.75) = 1.50 ⎭
Thus,
T2 =
P2v 2
P v T
16,529 kPa (1.50)(647.1 K)
= 2 R 2 cr =
= 826 K
Z 2 R Z 2 Pcr
0.88
22,060 kPa
(c) From the superheated steam table,
T1 = 350°C ⎫
3
⎬ v 1 = 0.008806 m /kg
x1 = 1
⎭
(Table A-4)
P2 = 16,529 kPa
⎫
⎬ T = 477°C = 750 K
3
v 2 = 2v 1 = 0.01761 m /kg ⎭ 2
(from Table A-6 or EES)
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3-42
3-88 Methane is heated at constant pressure. The final temperature is to be determined using ideal gas equation and the
compressibility charts.
Properties The gas constant, the critical pressure, and the critical temperature of methane are, from Table A-1,
R = 0.5182 kPa·m3/kg·K,
Tcr = 191.1 K,
Pcr = 4.64 MPa
Analysis From the ideal gas equation,
T2 = T1
v2
= (300 K )(1.5) = 450 K
v1
From the compressibility chart at the initial state (Fig. A-15),
PR1
Methane
8 MPa
300 K
⎫
⎪
⎪
⎬ Z1 = 0.88, v R1 = 0.80
P1
8 MPa
=
=
= 1.724 ⎪
⎪⎭
Pcr 4.64 MPa
T R1 =
T1
300 K
=
= 1.570
Tcr 191.1 K
Q
At the final state,
PR 2 = PR1 = 1.724
v R 2 = 1.5v R1
⎫
⎬ Z 2 = 0.975
= 1.5(0.80) = 1.2 ⎭
Thus,
T2 =
P2v 2
P v T
8000 kPa (1.2)(191.1 K)
= 2 R 2 cr =
= 406 K
Z 2 R Z 2 Pcr
0.975
4640 kPa
Of these two results, the accuracy of the second result is limited by the accuracy with which the charts may be read.
Accepting the error associated with reading charts, the second temperature is the more accurate.
3-89 The percent error involved in treating CO2 at a specified state as an ideal gas is to be determined.
Properties The critical pressure, and the critical temperature of CO2 are, from Table A-1,
Tcr = 304.2 K and Pcr = 7.39 MPa
Analysis From the compressibility chart (Fig. A-15),
5 MPa
P
⎫
PR =
=
= 0.677 ⎪
Pcr 7.39 MPa
⎪
⎬ Z = 0.69
298 K
T
TR =
=
= 0.980 ⎪
⎪⎭
Tcr 304.2 K
CO2
5 MPa
25°C
Then the error involved in treating CO2 as an ideal gas is
Error =
v − v ideal
1
1
= 1− = 1−
= −0.44.9 or 44.9%
Z
0.69
v
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3-43
3-90 CO2 gas flows through a pipe. The volume flow rate and the density at the inlet and the volume flow rate at the exit of
the pipe are to be determined.
Properties The gas constant, the critical pressure, and the critical temperature of CO2 are (Table A-1)
R = 0.1889 kPa·m3/kg·K,
Tcr = 304.2 K,
Pcr = 7.39 MPa
Analysis
3 MPa
500 K
2 kg/s
CO2
450 K
(a) From the ideal gas equation of state,
V&1 =
m& RT1 (2 kg/s)(0.1889 kPa ⋅ m 3 /kg ⋅ K)(500 K)
=
= 0.06297 m 3 /kg (2.1% error)
(3000 kPa)
P1
ρ1 =
P1
(3000 kPa)
=
= 31.76 kg/m 3 (2.1% error)
RT1 (0.1889 kPa ⋅ m 3 /kg ⋅ K)(500 K)
V&2 =
m& RT2 (2 kg/s)(0.1889 kPa ⋅ m 3 /kg ⋅ K)(450 K)
=
= 0.05667 m 3 /kg (3.6% error)
(3000 kPa)
P2
(b) From the compressibility chart (EES function for compressibility factor is used)
3 MPa
P1
⎫
=
= 0.407 ⎪
Pcr 7.39 MPa
⎪
⎬ Z1 = 0.9791
500 K
T
= 1.64 ⎪
TR ,1 = 1 =
⎪⎭
Tcr 304.2 K
PR =
3 MPa
P2
⎫
=
= 0.407 ⎪
Pcr 7.39 MPa
⎪
⎬ Z 2 = 0.9656
450 K
T2
=
=
= 1.48 ⎪
⎪⎭
Tcr 304.2 K
PR =
TR , 2
Thus,
V&1 =
Z 1 m& RT1 (0.9791)(2 kg/s)(0.1889 kPa ⋅ m 3 /kg ⋅ K)(500 K)
=
= 0.06165 m 3 /kg
(3000 kPa)
P1
ρ1 =
P1
(3000 kPa)
=
= 32.44 kg/m 3
Z1 RT1 (0.9791)(0.1889 kPa ⋅ m 3 /kg ⋅ K)(500 K)
V&2 =
Z 2 m& RT2 (0.9656)(2 kg/s)(0.1889 kPa ⋅ m 3 /kg ⋅ K)(450 K)
=
= 0.05472 m 3 /kg
(3000 kPa)
P2
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3-44
3-91 The specific volume of nitrogen gas is to be determined using the ideal gas relation and the compressibility chart. The
errors involved in these two approaches are also to be determined.
Properties The gas constant, the critical pressure, and the critical temperature of nitrogen are, from Table A-1,
R = 0.2968 kPa·m3/kg·K,
Tcr = 126.2 K,
Pcr = 3.39 MPa
Analysis (a) From the ideal gas equation of state,
v=
RT (0.2968 kPa ⋅ m 3 /kg ⋅ K)(150 K)
=
= 0.004452 m 3 /kg
P
10,000 kPa
(86.4% error )
(b) From the compressibility chart (Fig. A-15),
10 MPa
P
⎫
=
= 2.95 ⎪
Pcr 3.39 MPa
⎪
⎬ Z = 0.54
150 K
T
TR =
=
= 1.19 ⎪
⎪⎭
Tcr 126.2 K
N2
10 MPa
150 K
PR =
Thus,
v = Zv ideal = (0.54)(0.004452 m 3 /kg) = 0.002404 m 3 /kg
(0.7% error)
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3-45
Other Equations of State
3-92C The constant a represents the increase in pressure as a result of intermolecular forces; the constant b represents the
volume occupied by the molecules. They are determined from the requirement that the critical isotherm has an inflection
point at the critical point.
3-93 Carbon monoxide is heated in a piston-cylinder device. The final volume of the carbon monoxide is to be determined
using the ideal gas equation and the Benedict-Webb-Rubin equation of state.
Properties The gas constant and molar mass of CO are (Table A-1)
R = 0.2968 kPa·m3/kg·K, M = 28.011 kg/kmol
Analysis (a) From the ideal gas equation of state,
mRT2 (0.100 kg)(0.2968 kPa ⋅ m 3 /kg ⋅ K)(473 K)
V2 =
=
= 0.0140 m 3
1000 kPa
P
CO
1000 kPa
200°C
Q
(b) Using the coefficients of Table 3-4 for carbon dioxide and the given data, the
Benedict-Webb-Rubin equation of state for state 2 is
P2 =
1000 =
+
Ru T2
v2
⎛
C ⎞ 1
bR T − a aα
c ⎛
γ ⎞
+ 6 + 3 2 ⎜1 + 2 ⎟ exp(−γ / v 2 )
+ ⎜ B0 Ru T2 − A0 − 02 ⎟ 2 + u 23
⎟v
⎜
v
v
v
v
T
T
⎠
2 ⎠
2 ⎝
⎝
(8.314)(473)
v2
⎛
8.673 × 10 5
+ ⎜⎜ 0.05454 × 8.314 × 473 − 135.9 −
44732
⎝
⎞ 1
0.002632 × 8.314 × 473 − 3.71
⎟
⎟v 2 +
v3
⎠
3.71 × 0.000135 1.054 × 10 5 ⎛ 0.0060 ⎞
+ 3
⎜1 +
⎟ exp(−0.0060 / v 2 )
6
2
2
v
v (473) ⎝
v
⎠
The solution of this equation by an equation solver such as EES gives
v 2 = 3.952 m 3 /kmol
Then,
v2 =
v2
M
=
3.952 m 3 /kmol
= 0.1411 m 3 /kg
28.011 kg/kmol
V 2 = mv 2 = (0.100 kg)(0.1411 m 3 /kg) = 0.0141 m 3
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3-46
3-94 Methane is heated in a rigid container. The final pressure of the methane is to be determined using the ideal gas
equation and the Benedict-Webb-Rubin equation of state.
Analysis (a) From the ideal gas equation of state,
P2 = P1
T2
573 K
= (80 kPa)
= 156.5 kPa
293 K
T1
Methane
80 kPa
20°C
(b) The specific molar volume of the methane is
v1 = v 2 =
Q
Ru T1 (8.314 kPa ⋅ m 3 /kmol ⋅ K)(293 K)
=
= 29.41 m 3 /kmol
P1
80 kPa
Using the coefficients of Table 3-4 for methane and the given data, the Benedict-Webb-Rubin equation of state for state 2
gives
P2 =
=
+
Ru T2
v2
⎛
C ⎞ 1
bR T − a aα
c ⎛
γ ⎞
+ 6 + 3 2 ⎜1 + 2 ⎟ exp(−γ / v 2 )
+ ⎜ B0 Ru T2 − A0 − 02 ⎟ 2 + u 23
⎜
⎟
T2 ⎠ v
v
v
v T2 ⎝ v ⎠
⎝
(8.314)(573) ⎛⎜
2.286 × 10 6
+ 0.04260 × 8.314 × 573 − 187.91 −
⎜
29.41
5732
⎝
5.00 × 1.244 × 10 −4
29.416
= 162.0 kPa
+
⎞ 1
0.003380 × 8.314 × 573 − 5.00
⎟
+
⎟ 29.412
29.413
⎠
2.578 × 10 5 ⎛
0.0060 ⎞
⎜1 +
⎟ exp(−0.0060 / 29.412 )
3
2
29.41 (573) ⎝
29.412 ⎠
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3-47
3-95 Carbon monoxide is heated in a rigid container. The final pressure of the CO is to be determined using the ideal gas
equation and the Benedict-Webb-Rubin equation of state.
Properties The gas constant and molar mass of CO are (Table A-1)
R = 0.2968 kPa·m3/kg·K, M = 28.011 kg/kmol
Analysis (a) From the ideal gas equation of state,
T
700 K
P2 = P1 2 = (101 kPa)
= 240.5 kPa
294 K
T1
CO
101 kPa
Q
21°C
The specific molar volume of the CO in SI units is
v1 = v 2 =
Ru T1 (8.314 kPa ⋅ m 3 /kmol ⋅ K)(294 K)
=
= 24.20 m 3 /kmol
P1
101 kPa
(b) The specific molar volume of the CO in SI units is
v1 = v 2 =
Ru T1 (8.314 kPa ⋅ m 3 /kmol ⋅ K)(294 K)
=
= 24.20 m 3 /kmol
P1
101 kPa
Using the coefficients of Table 3-4 for CO and the given data, the Benedict-Webb-Rubin equation of state for state 2 gives
⎛
C ⎞ 1 bR T − a aα
c ⎛
γ ⎞
+ ⎜⎜ B0 RuT2 − A0 − 02 ⎟⎟ 2 + u 23
+ 6 + 3 2 ⎜1 + 2 ⎟ exp(−γ / v 2 )
v2 ⎝
T2 ⎠ v
v
v
v T2 ⎝ v ⎠
(8.314)(700) ⎛⎜
8.673 ×105 ⎞⎟ 1
0.002632 × 8.314 × 700 − 3.71
+
=
+ 0.05454 × 8.314 × 700 − 135.87 −
2
⎜
⎟ 24.202
24.20
700
24.203
⎝
⎠
P2 =
RuT2
3.71×1.350 ×10− 4
1.054 ×105 ⎛ 0.0060 ⎞
+
1+
exp(−0.0060 / 24.202 )
6
3
2 ⎜
2⎟
24.20
24.20 (700) ⎝ 24.20 ⎠
= 240.8 kPa
+
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3-48
3-96 The temperature of R-134a in a tank at a specified state is to be determined using the ideal gas relation, the van der
Waals equation, and the refrigerant tables.
Properties The gas constant, critical pressure, and critical temperature of R-134a are (Table A-1)
R = 0.08149 kPa·m3/kg·K,
Tcr = 374.2 K,
Pcr = 4059 kPa
Analysis (a) From the ideal gas equation of state,
T=
Pv (1200 kPa)(0.0195 m 3 /kg)
=
= 287.2 K
R
0.08149 kPa ⋅ m 3 /kg ⋅ K
(b) The van der Waals constants for the refrigerant are determined from
a=
27 R 2Tcr2 (27)(0.08149 kPa ⋅ m 3 /kg ⋅ K) 2 (374.2 K) 2
=
= 0.09663 m 6 ⋅ kPa/kg 2
64 Pcr
(64)(4059 kPa)
b=
RTcr (0.08149 kPa ⋅ m 3 /kg ⋅ K)(374.2 K)
=
= 0.000939 m 3 /kg
8 Pcr
8 × 4059 kPa
T=
1⎛
a
⎜P + 2
R⎝
v
Then,
⎛
⎞
1
⎞
⎜1200 + 0.09663 ⎟(0.0195 − 0.000939 ) = 331.2 K
⎟(v − b ) =
2 ⎟
⎜
0.08149
(0.0195) ⎠
⎠
⎝
(c) From the superheated refrigerant table (Table A-13),
P = 1200 kPa
⎫
v = 0.0195 m 3 /kg ⎬⎭
T = 70°C (343 K)
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3-49
3-97
The pressure of nitrogen in a tank at a specified state is to be determined using the ideal gas relation and the
Beattie-Bridgeman equation. The error involved in each case is to be determined.
Properties The gas constant and molar mass of nitrogen are (Table A-1)
R = 0.2968 kPa·m3/kg·K and M = 28.013 kg/kmol
Analysis (a) From the ideal gas equation of state,
P=
RT
v
=
(0.2968 kPa ⋅ m3 /kg ⋅ K)(150 K)
0.041884 m3 /kg
= 1063 kPa (6.3% error )
N2
0.041884 m3/kg
150 K
(b) The constants in the Beattie-Bridgeman equation are
⎛ a⎞
⎛ 0.02617 ⎞
A = Ao ⎜1 − ⎟ = 136.2315⎜1 −
⎟ = 133.193
v
1.1733 ⎠
⎝
⎠
⎝
⎛ b⎞
⎛ − 0.00691 ⎞
B = Bo ⎜1 − ⎟ = 0.05046⎜1 −
⎟ = 0.05076
1.1733 ⎠
⎝ v ⎠
⎝
c = 4.2 × 10 4 m 3 ⋅ K 3 /kmol
since
v = Mv = (28.013 kg/kmol)(0.041884 m 3 /kg) = 1.1733 m 3 /kmol .
Substituting,
P=
4.2 × 10 4 ⎞⎟
RuT ⎛
c ⎞
A 8.314 × 150 ⎛⎜
−
+
−
=
−
1
1
B
(
)
(1.1733 + 0.05076) − 133.1932
v
⎜
⎟
2
3
2
2 ⎜
3⎟
v ⎝ vT ⎠
v
(1.1733) ⎝ 1.1733 × 150 ⎠
(1.1733)
= 1000.4 kPa (negligible error)
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3-50
3-98
Problem 3-97 is reconsidered. Using EES (or other) software, the pressure results of the ideal gas and BeattieBridgeman equations with nitrogen data supplied by EES are to be compared. The temperature is to be plotted versus
specific volume for a pressure of 1000 kPa with respect to the saturated liquid and saturated vapor lines of nitrogen over the
range of 110 K < T < 150 K.
Analysis The problem is solved using EES, and the solution is given below.
Function BeattBridg(T,v,M,R_u)
v_bar=v*M "Conversion from m^3/kg to m^3/kmol"
"The constants for the Beattie-Bridgeman equation of state are found in text"
Ao=136.2315; aa=0.02617; Bo=0.05046; bb=-0.00691; cc=4.20*1E4
B=Bo*(1-bb/v_bar)
A=Ao*(1-aa/v_bar)
"The Beattie-Bridgeman equation of state is"
BeattBridg:=R_u*T/(v_bar**2)*(1-cc/(v_bar*T**3))*(v_bar+B)-A/v_bar**2
End
T=150 [K]
v=0.041884 [m^3/kg]
P_exper=1000 [kPa]
T_table=T; T_BB=T;T_idealgas=T
P_table=PRESSURE(Nitrogen,T=T_table,v=v) "EES data for nitrogen as a real gas"
{T_table=temperature(Nitrogen, P=P_table,v=v)}
M=MOLARMASS(Nitrogen)
R_u=8.314 [kJ/kmol-K] "Universal gas constant"
R=R_u/M "Particular gas constant"
P_idealgas=R*T_idealgas/v "Ideal gas equation"
P_BB=BeattBridg(T_BB,v,M,R_u) "Beattie-Bridgeman equation of state Function"
PBB [kPa]
1000
1000
1000
1000
1000
1000
1000
Ptable [kPa]
1000
1000
1000
1000
1000
1000
1000
160
Pidealgas [kPa]
1000
1000
1000
1000
1000
1000
1000
v [m3/kg]
0.01
0.02
0.025
0.03
0.035
0.04
0.05
TBB [K]
91.23
95.52
105
116.8
130.1
144.4
174.6
Tideal gas [K]
33.69
67.39
84.23
101.1
117.9
134.8
168.5
Ttable [K]
103.8
103.8
106.1
117.2
130.1
144.3
174.5
Nitrogen, T vs v for P=1000 kPa
Ideal Gas
150
140
Beattie-Bridgem an
EES Table Value
T [K]
130
120
110
1000 kPa
100
90
80
70
10 -3
10 -2
10 -1
3
v [m /kg]
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3-51
3-99 Carbon dioxide is compressed in a piston-cylinder device in a polytropic process. The final temperature is to be
determined using the ideal gas and van der Waals equations.
Properties The gas constant, molar mass, critical pressure, and critical temperature of carbon dioxide are (Table A-1)
R = 0.1889 kPa·m3/kg·K,
M = 44.01 kg/kmol,
Tcr = 304.2 K,
Pcr = 7.39 MPa
Analysis (a) The specific volume at the initial state is
v1 =
RT1 (0.1889 kPa ⋅ m 3 /kg ⋅ K)(473 K)
=
= 0.08935 m 3 /kg
P1
1000 kPa
According to process specification,
⎛P
v 2 = v 1 ⎜⎜ 1
⎝ P2
⎞
⎟⎟
⎠
1/ n
⎛ 1000 kPa ⎞
= (0.08935 m 3 /kg)⎜
⎟
⎝ 3000 kPa ⎠
1 / 1 .2
= 0.03577 m 3 /kg
CO2
1 MPa
200°C
The final temperature is then
T2 =
P2v 2 (3000 kPa)(0.03577 m 3 /kg)
=
= 568 K
R
0.1889 kPa ⋅ m 3 /kg ⋅ K
(b) The van der Waals constants for carbon dioxide are determined from
a=
27 R 2 Tcr2 (27)(0.1889 kPa ⋅ m 3 /kg ⋅ K) 2 (304.2 K) 2
=
= 0.1885 m 6 ⋅ kPa/kg 2
64 Pcr
(64)(7390 kPa)
b=
RTcr (0.1889 kPa ⋅ m 3 /kg ⋅ K)(304.2 K)
=
= 0.0009720 m 3 /kg
8Pcr
8 × 7390 kPa
Applying the van der Waals equation to the initial state,
a ⎞
⎛
⎜ P + 2 ⎟(v − b) = RT
v ⎠
⎝
0.1885 ⎞
⎛
⎜1000 +
⎟(v − 0.0009720) = (0.1889)(473)
v2 ⎠
⎝
Solving this equation by trial-error or by EES gives
v 1 = 0.08821 m 3 /kg
According to process specification,
⎛ P1
⎝ P2
v 2 = v 1 ⎜⎜
⎞
⎟⎟
⎠
1/ n
⎛ 1000 kPa ⎞
= (0.08821 m 3 /kg)⎜
⎟
⎝ 3000 kPa ⎠
1 / 1 .2
= 0.03531 m 3 /kg
Applying the van der Waals equation to the final state,
a ⎞
⎛
⎜ P + 2 ⎟(v − b) = RT
v ⎠
⎝
0.1885 ⎞
⎛
⎜ 3000 +
⎟(0.03531 − 0.0009720) = (0.1889)T
0.035312 ⎠
⎝
Solving for the final temperature gives
T2 = 573 K
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3-52
Special Topic: Vapor Pressure and Phase Equilibrium
3-100 A glass of water is left in a room. The vapor pressures at the free surface of the water and in the room far from the
glass are to be determined.
Assumptions The water in the glass is at a uniform temperature.
Properties The saturation pressure of water is 2.339 kPa at 20°C, and 1.706 kPa at 15°C (Table A-4).
Analysis The vapor pressure at the water surface is the saturation pressure of water at the
water temperature,
Pv , water surface = Psat @ Twater = Psat@15°C = 1.706 kPa
H2O
15°C
Noting that the air in the room is not saturated, the vapor pressure in the room far from the
glass is
Pv , air = φPsat @ Tair = φPsat@20°C = (0.4)(2.339 kPa) = 0.936 kPa
3-101 The vapor pressure in the air at the beach when the air temperature is 30°C is claimed to be 5.2 kPa. The validity of
this claim is to be evaluated.
Properties The saturation pressure of water at 30°C is 4.247 kPa (Table A-4).
Analysis The maximum vapor pressure in the air is the saturation pressure of water
at the given temperature, which is
Pv, max = Psat @ Tair = Psat@30°C = 4.247 kPa
30°C
WATER
which is less than the claimed value of 5.2 kPa. Therefore, the claim is false.
3-102 The temperature and relative humidity of air over a swimming pool are given. The water temperature of the
swimming pool when phase equilibrium conditions are established is to be determined.
Assumptions The temperature and relative humidity of air over the pool remain constant.
Properties The saturation pressure of water at 20°C is 2.339 kPa (Table A-4).
Analysis The vapor pressure of air over the swimming pool is
Patm, 20°C
Pv , air = φPsat @ Tair = φPsat@20°C = (0.4)(2.339 kPa) = 0.9357 kPa
Phase equilibrium will be established when the vapor pressure at the water surface
equals the vapor pressure of air far from the surface. Therefore,
POOL
Pv, water surface = Pv, air = 0.9357 kPa
and
Twater = Tsat @ Pv = Tsat @ 0.9357
kPa
= 6.0°C
Discussion Note that the water temperature drops to 6.0°C in an environment at 20°C when phase equilibrium is
established.
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3-53
3-103 A person buys a supposedly cold drink in a hot and humid summer day, yet no condensation occurs on the drink. The
claim that the temperature of the drink is below 10°C is to be evaluated.
Properties The saturation pressure of water at 35°C is 5.629 kPa (Table A-4).
Analysis The vapor pressure of air is
Pv , air = φPsat @ Tair = φPsat@35°C = (0.7)(5.629 kPa) = 3.940 kPa
35°C
70%
The saturation temperature corresponding to this pressure (called the dew-point
temperature) is
Tsat = Tsat @ Pv = Tsat@3.940 kPa = 28.7 °C
That is, the vapor in the air will condense at temperatures below 28.7°C. Noting that no condensation is observed on the
can, the claim that the drink is at 10°C is false.
3-104 A thermos bottle half-filled with water is left open to air in a room at a specified temperature and pressure. The
temperature of water when phase equilibrium is established is to be determined.
Assumptions The temperature and relative humidity of air over the bottle remain constant.
Properties The saturation pressure of water at 20°C is 2.3392 kPa (Table A-4).
Analysis The vapor pressure of air in the room is
Pv, air = φPsat @ Tair = φPsat@20°C = (0.35)(2.3392 kPa) = 0.8187 kPa
Phase equilibrium will be established when the vapor pressure at the water surface
equals the vapor pressure of air far from the surface. Therefore,
Pv, water surface = Pv, air = 0.8187 kPa
Thermos
bottle
20°C
35%
and
Twater = Tsat @ Pv = Tsat @ 0.8187
kPa
= 4.1°C
Discussion Note that the water temperature drops to 4.1°C in an environment at 20°C when phase equilibrium is
established.
3-105 Two rooms are identical except that they are maintained at different temperatures and relative humidities. The room
that contains more moisture is to be determined.
Properties The saturation pressure of water is 2.339 kPa at 20°C, and 3.17 kPa at 25°C (Table A-4).
Analysis The vapor pressures in the two rooms are
Room 1:
Pv1 = φ1 Psat @ T1 = φ1 Psat@25°C = (0.4)(3.17 kPa) = 1.27 kPa
Room 2:
Pv 2 = φ 2 Psat @ T2 = φ 2 Psat@20°C = (0.55)(2.339 kPa) = 1.29 kPa
Therefore, room 1 at 30°C and 40% relative humidity contains more moisture.
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3-54
Review Problems
3-106 Nitrogen gas in a rigid tank is heated to a final gage pressure. The final temperature is to be determined.
Assumptions At specified conditions, nitrogen behaves as an ideal gas.
Analysis According to the ideal gas equation of state at constant volume,
m1 = m1
P1V1 P2V 2
=
T1
T2
Patm = 100 kPa
Since V1 = V 2
P1 P2
=
T1 T2
T2 = T1
Nitrogen gas
227°C
100 kPa
(gage)
Q
P2
(250 + 100) kPa
= [(227 + 273) K ]
= 875 K = 602°C
P2
(100 + 100) kPa
3-107 Carbon dioxide flows through a pipe at a given state. The volume and mass flow rates and the density of CO2 at the
given state and the volume flow rate at the exit of the pipe are to be determined.
Analysis
3 MPa
500 K
0.4 kmol/s
CO2
450 K
(a) The volume and mass flow rates may be determined from ideal gas relation as
V&1 =
N& Ru T1 (0.4 kmol/s)(8.314 kPa.m 3 /kmol.K)(500 K)
=
= 0.5543 m 3 /s
P
3000 kPa
m& 1 =
P1V&1
(3000 kPa)(0.5543 m3 / s)
=
= 17.60 kg/s
RT1 (0.1889 kPa.m3 /kg.K)(500 K)
The density is
ρ1 =
m& 1
(17.60 kg/s)
=
= 31.76 kg/m 3
&
V1 (0.5543 m 3 /s)
(b) The volume flow rate at the exit is
V&2 =
N& Ru T2 (0.4 kmol/s)(8.314 kPa.m 3 /kmol.K)(450 K)
=
= 0.4988 m 3 /s
P
3000 kPa
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3-55
3-108 The cylinder conditions before the heat addition process is specified. The pressure after the heat addition process is
to be determined.
Assumptions 1 The contents of cylinder are approximated by the air properties.
2 Air is an ideal gas.
Analysis The final pressure may be determined from the ideal gas relation
P2 =
Combustion
chamber
1.4 MPa
450°C
T2
⎛ 1600 + 273 K ⎞
P1 = ⎜
⎟(1400 kPa) = 3627 kPa
T1
⎝ 450 + 273 K ⎠
3-109 The cylinder conditions before the heat addition process is specified. The temperature after the heat addition process
is to be determined.
Assumptions 1 The contents of cylinder is approximated by the air
properties. 2 Air is an ideal gas.
Combustion
chamber
950 K
75 cm3
Analysis The ratio of the initial to the final mass is
AF
22
22
m1
=
=
=
m2 AF + 1 22 + 1 23
The final temperature may be determined from ideal gas relation
T2 =
3
m1 V 2
⎛ 22 ⎞⎛ 150 cm
T1 = ⎜ ⎟⎜
m 2 V1
⎝ 23 ⎠⎜⎝ 75 cm 3
⎞
⎟(950 K) = 1817 K
⎟
⎠
3-110 A rigid container that is filled with R-13a is heated. The initial pressure and the final temperature are to be
determined.
Analysis The initial specific volume is 0.090 m3/kg. Using this with
the initial temperature reveals that the initial state is a mixture. The
initial pressure is then the saturation pressure,
T1 = −40°C
R-134a
-40°C
1 kg
0.090 m3
⎫
⎬ P1 = Psat @ -40°C = 51.25 kPa (Table A - 11)
v 1 = 0.090 m 3 /kg ⎭
This is a constant volume cooling process (v = V /m = constant). The final
state is superheated vapor and the final temperature is then
P2 = 280 kPa
⎫
⎬ T = 50°C (Table A - 13)
3
v 2 = v 1 = 0.090 m /kg ⎭ 2
P
2
1
v
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3-56
3-111 The volume of chamber 1 of the two-piston cylinder shown in the figure is to be determined.
Assumptions At specified conditions, helium behaves as an ideal gas.
Properties The gas constant of helium is R = 2.0769 kJ/kg⋅K (Table A-1).
P2A2
Analysis Since the water vapor in chamber 2 is condensing, the pressure in
this chamber is the saturation pressure,
P2 = Psat @ 200°C = 1555 kPa
(Table A-4)
Summing the forces acting on the piston in the vertical direction gives
P1 = P2
⎛D
A2
= P2 ⎜⎜ 2
A1
⎝ D1
2
2
⎞
⎛ 4⎞
⎟⎟ = (1555 kPa)⎜ ⎟ = 248.8 kPa
⎝ 10 ⎠
⎠
According to the ideal gas equation of state,
V1 =
P1A1
3
mRT (1 kg)(2.0769 kPa ⋅ m /kg ⋅ K)(200 + 273 K)
=
= 3.95 m 3
P1
248.8 kPa
3-112 The volume of chamber 1 of the two-piston cylinder shown in the figure is to be determined.
Assumptions At specified conditions, air behaves as an ideal gas.
Properties The gas constant of air is R = 0.287 kPa⋅m3/kg⋅K (Table A-1).
Analysis Since R-134a in chamber 2 is condensing, the pressure in this chamber is the saturation pressure,
P2 = Psat @ 48°C = 1253.6 kPa
(Table A-11)
Summing the forces acting on the piston in the vertical direction gives
F2
F2 + F3 = F1
P2 A2 + P3 ( A1 − A2 ) = P1 A1
F3
which when solved for P1 gives
P1 = P2
⎛ A ⎞
A2
+ P3 ⎜⎜1 − 2 ⎟⎟
A1
A1 ⎠
⎝
since the areas of the piston faces are given by A = πD 2 / 4 the above equation
becomes
⎛D
P1 = P2 ⎜⎜ 2
⎝ D1
2
⎡ ⎛D
⎞
⎟⎟ + P3 ⎢1 − ⎜⎜ 2
⎢ ⎝ D1
⎠
⎣
⎞
⎟⎟
⎠
F1
2⎤
⎥
⎥
⎦
2
⎡ ⎛ 5 ⎞2 ⎤
⎛5⎞
= (1253.6 kPa)⎜ ⎟ + (200 kPa) ⎢1 − ⎜ ⎟ ⎥
⎝8⎠
⎢⎣ ⎝ 8 ⎠ ⎥⎦
= 611.6 kPa
According to the ideal gas equation of state,
V1 =
mRT (0.2 kg)(0.287 kPa ⋅ m 3 /kg ⋅ K)(48 + 273 K)
=
= 0.0301 m 3
P1
611.6 kPa
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3-57
3-113 The difference in the volume of chamber 1 for two cases of pressure in chamber 3 is to be determined.
Assumptions At specified conditions, air behaves as an ideal gas.
Properties The gas constant of air is R = 0.287 kPa⋅m3/kg⋅K (Table A-1).
Analysis Since R-134a in chamber 2 is condensing, the pressure in this chamber is the saturation pressure,
P2 = Psat @ 48°C = 1253.6 kPa
(Table A-11)
Summing the forces acting on the piston in the vertical direction gives
F2 + F3 = F1
F2
P2 A2 + P3 ( A1 − A2 ) = P1 A1
F3
which when solved for P1 gives
P1 = P2
⎛ A ⎞
A2
+ P3 ⎜⎜1 − 2 ⎟⎟
A1
A1 ⎠
⎝
since the areas of the piston faces are given by A = πD 2 / 4 the above equation
becomes
⎛D
P1 = P2 ⎜⎜ 2
⎝ D1
⎡ ⎛D
⎞
⎟⎟ + P3 ⎢1 − ⎜⎜ 2
⎢ ⎝ D1
⎠
⎣
2
F1
2⎤
⎞
⎟⎟ ⎥
⎠ ⎥⎦
2
⎡ ⎛ 5 ⎞2 ⎤
⎛5⎞
= (1253.6 kPa)⎜ ⎟ + (400 kPa) ⎢1 − ⎜ ⎟ ⎥
⎝8⎠
⎣⎢ ⎝ 8 ⎠ ⎦⎥
= 733.4 kPa
According to the ideal gas equation of state,
V1 =
mRT (0.2 kg)(0.287 kPa ⋅ m 3 /kg ⋅ K)(48 + 273 K)
=
= 0.0251 m 3
733.4 kPa
P1
For a chamber 3 pressure of 200 kPa, the volume of chamber 1 was determined to be 0.0301 m3. Then the change in the
volume of chamber 1 is
∆V = V 2 − V1 = 0.0301 − 0.0251 = 0.005 m 3
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3-58
3-114 Ethane is heated at constant pressure. The final temperature is to be determined using ideal gas equation and the
compressibility charts.
Properties The gas constant, the critical pressure, and the critical temperature of ethane are, from Table A-1,
R = 0.2765 kPa·m3/kg·K,
Tcr = 305.5 K,
Pcr = 4.48 MPa
Analysis From the ideal gas equation,
T2 = T1
v2
= (373 K )(1.6) = 596.8 K
v1
From the compressibility chart at the initial state (Fig. A-15),
⎫
⎪
⎪
⎬ Z1 = 0.61, v R1 = 0.35
P1
10 MPa
=
=
= 2.232 ⎪
⎪⎭
Pcr 4.48 MPa
T R1 =
PR1
T1
373 K
=
= 1.221
Tcr 305.5 K
Ethane
10 MPa
100°C
Q
At the final state,
PR 2 = PR1 = 2.232
v R 2 = 1.6v R1
⎫
⎬ Z 2 = 0.83
= 1.6(0.35) = 0.56 ⎭
Thus,
T2 =
P2v 2 P2 v R 2 Tcr 10,000 kPa (0.56)(305.5 K)
=
=
= 460 K
Z 2 R Z 2 Pcr
0.83
4480 kPa
Of these two results, the accuracy of the second result is limited by the accuracy with which the charts may be read.
Accepting the error associated with reading charts, the second temperature is the more accurate.
3-115 A large tank contains nitrogen at a specified temperature and pressure. Now some nitrogen is allowed to escape, and
the temperature and pressure of nitrogen drop to new values. The amount of nitrogen that has escaped is to be determined.
Properties The gas constant for nitrogen is 0.2968 kPa·m3/kg·K (Table A-1).
Analysis Treating N2 as an ideal gas, the initial and the final masses in the tank are determined to be
m1 =
P1V
(600 kPa)(20 m 3 )
=
= 136.6 kg
RT1 (0.2968kPa ⋅ m 3 /kg ⋅ K)(296 K)
m2 =
P2V
(400 kPa)(20 m 3 )
=
= 92.0 kg
RT2 (0.2968 kPa ⋅ m 3 /kg ⋅ K)(293 K)
Thus the amount of N2 that escaped is
∆m = m1 − m 2 = 136.6 − 92.0 = 44.6 kg
N2
600 kPa
23°C
20 m3
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3-59
3-116 The rigid tank contains saturated liquid-vapor mixture of water. The mixture is heated until it exists in a single
phase. For a given tank volume, it is to be determined if the final phase is a liquid or a vapor.
Analysis This is a constant volume process (v = V /m = constant), and thus the final specific volume will be equal to the
initial specific volume,
v 2 = v1
H2O
The critical specific volume of water is 0.003106 m3/kg. Thus if the final specific
volume is smaller than this value, the water will exist as a liquid, otherwise as a vapor.
V = 4 L ⎯⎯→v =
V
m
V = 400 L ⎯⎯→v =
=
V
m
V=4L
m = 2 kg
T = 50°C
0.004 m3
= 0.002 m3/kg < v cr Thus, liquid.
2 kg
=
0.4 m3
= 0.2 m3/kg > v cr . Thus, vapor.
2 kg
3-117 Two rigid tanks that contain hydrogen at two different states are connected to each other. Now a valve is opened, and
the two gases are allowed to mix while achieving thermal equilibrium with the surroundings. The final pressure in the tanks
is to be determined.
Properties The gas constant for hydrogen is 4.124 kPa·m3/kg·K (Table A-1).
Analysis Let's call the first and the second tanks A and B. Treating H2 as an ideal gas, the total volume and the total mass of
H2 are
A
V = V A + V B = 0.5 + 0.5 = 1.0 m 3
⎛ PV ⎞
(400 kPa)(0.5 m 3 )
= 0.1655 kg
m A = ⎜⎜ 1 ⎟⎟ =
3
⎝ RT1 ⎠ A (4.124 kPa ⋅ m /kg ⋅ K)(293 K)
⎛ PV ⎞
(150 kPa)(0.5 m 3 )
m B = ⎜⎜ 1 ⎟⎟ =
= 0.0563 kg
3
⎝ RT1 ⎠ B (4.124 kPa ⋅ m /kg ⋅ K)(323 K)
m = m A + m B = 0.1655 + 0.0563 = 0.2218 kg
B
H2
V = 0.5 m3
T=20°C
P=400 kPa
×
H2
V = 0.5 m3
T=50°C
P=150 kPa
Then the final pressure can be determined from
P=
mRT2
V
=
(0.2218 kg)(4.124 kPa ⋅ m 3 /kg ⋅ K)(288 K)
1.0 m 3
= 264 kPa
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3-60
3-118
Problem 3-117 is reconsidered. The effect of the surroundings temperature on the final equilibrium pressure
in the tanks is to be investigated. The final pressure in the tanks is to be plotted versus the surroundings temperature, and
the results are to be discussed.
Analysis The problem is solved using EES, and the solution is given below.
"Given Data"
V_A=0.5 [m^3]
T_A=20 [C]
P_A=400 [kPa]
V_B=0.5 [m^3]
T_B=50 [C]
P_B=150 [kPa]
{T_2=15 [C]}
"Solution"
R=R_u/MOLARMASS(H2)
R_u=8.314 [kJ/kmol-K]
V_total=V_A+V_B
m_total=m_A+m_B
P_A*V_A=m_A*R*(T_A+273)
P_B*V_B=m_B*R*(T_B+273)
P_2*V_total=m_total*R*(T_2+273)
T2
[C]
-10
-5
0
5
10
15
20
25
30
280
270
P2 [kPa]
P2
[kPa]
240.6
245.2
249.7
254.3
258.9
263.5
268
272.6
277.2
260
250
240
-10
-5
0
5
10
15
20
25
30
T2 [C]
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3-61
3-119 The pressure in an automobile tire increases during a trip while its volume remains constant. The percent increase in
the absolute temperature of the air in the tire is to be determined.
Assumptions 1 The volume of the tire remains constant. 2 Air is an ideal gas.
TIRE
200 kPa
0.035 m3
Properties The local atmospheric pressure is 90 kPa.
Analysis The absolute pressures in the tire before and after the trip are
P1 � Pgage,1 � Patm � 200 � 90 � 290 kPa
P2 � Pgage,2 � Patm � 220 � 90 � 310 kPa
Noting that air is an ideal gas and the volume is constant, the ratio of absolute
temperatures after and before the trip are
P1V 1 P2V 2
T
P
310 kPa
= 1.069
�
� 2 � 2 =
T1
T2
T1
P1 290 kPa
Therefore, the absolute temperature of air in the tire will increase by 6.9% during this trip.
3-120 The temperature of steam in a tank at a specified state is to be determined using the ideal gas relation, the
generalized chart, and the steam tables.
Properties The gas constant, the critical pressure, and the critical temperature of water are, from Table A-1,
R � 0.4615 kPa � m3/kg � K ,
Tcr � 647.1 K,
Pcr � 22.06 MPa
Analysis (a) From the ideal gas equation of state,
P�
RT
v
�
(0.4615 kPa � m3/kg � K)(673 K)
� 15,529 kPa
0.02 m3/kg
H 2O
0.02 m3/kg
400�C
(b) From the compressibility chart (Fig. A-15a),
�
�
�
� PR � 0.57
v actual
(0.02 m3/kg)(22,060 kPa)
�
� 1.48 �
vR �
�
RTcr / Pcr (0.4615 kPa � m3/kg � K)(647.1 K)
�
TR �
T
673 K
�
� 1.040
Tcr 647.1 K
Thus,
P � PR Pcr � 0.57 � 22,060 � 12,574 kPa
(c) From the superheated steam table,
T � 400�C
� P � 12,576
12,515 kPa
v � 0.02 m 3 /kg ��
(from EES)
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3-62
3-121 One section of a tank is filled with saturated liquid R-134a while the other side is evacuated. The partition is
removed, and the temperature and pressure in the tank are measured. The volume of the tank is to be determined.
Analysis The mass of the refrigerant contained in the tank is
m=
V1
0.03 m 3
=
= 33.58 kg
v 1 0.0008934 m 3 /kg
since
v 1 = v f @1.4 MPa = 0.0008934 m 3 /kg
R-134a
P=1.2 MPa
V =0.03 m3
Evacuated
At the final state (Table A-13),
P2 = 400 kPa ⎫
3
⎬ v 2 = 0.05680 m /kg
⎭
T2 = 30°C
Thus,
V tank = V 2 = mv 2 = (33.58 kg)(0.05680 m 3 /kg) = 1.91 m 3
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3-63
3-122
Problem 3-121 is reconsidered. The effect of the initial pressure of refrigerant-134 on the volume of the tank
is to be investigated as the initial pressure varies from 0.5 MPa to 1.5 MPa. The volume of the tank is to be plotted versus
the initial pressure, and the results are to be discussed.
Analysis The problem is solved using EES, and the solution is given below.
"Given Data"
x_1=0.0
Vol_1=0.03 [m^3]
P_1=1200 [kPa]
T_2=30 [C]
P_2=400 [kPa]
"Solution"
v_1=volume(R134a,P=P_1,x=x_1)
Vol_1=m*v_1
v_2=volume(R134a,P=P_2,T=T_2)
Vol_2=m*v_2
m
[kg]
37.23
36.59
36.01
35.47
34.96
34.48
34.02
33.58
33.15
32.73
32.32
2.15
2.1
2.05
3
Vol2
[m3]
2.114
2.078
2.045
2.015
1.986
1.958
1.932
1.907
1.883
1.859
1.836
Vol2 [m ]
P1
[kPa]
500
600
700
800
900
1000
1100
1200
1300
1400
1500
2
1.95
1.9
1.85
1.8
500
700
900
1100
1300
1500
P1 [kPa]
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3-123 A propane tank contains 5 L of liquid propane at the ambient temperature. Now a leak develops at the top of the tank
and propane starts to leak out. The temperature of propane when the pressure drops to 1 atm and the amount of heat
transferred to the tank by the time the entire propane in the tank is vaporized are to be determined.
Properties The properties of propane at 1 atm are Tsat = -42.1°C, ρ = 581 kg / m 3 , and hfg = 427.8 kJ/kg (Table A-3).
Analysis The temperature of propane when the pressure drops
to 1 atm is simply the saturation pressure at that temperature,
Propane
5L
T = Tsat @1 atm = −42.1° C
20°C
The initial mass of liquid propane is
m = ρV = (581 kg/m 3 )(0.005 m3 ) = 2.905 kg
The amount of heat absorbed is simply the total heat of vaporization,
Leak
Qabsorbed = mh fg = (2.905 kg)(427.8 kJ / kg) = 1243 kJ
3-124 An isobutane tank contains 5 L of liquid isobutane at the ambient temperature. Now a leak develops at the top of the
tank and isobutane starts to leak out. The temperature of isobutane when the pressure drops to 1 atm and the amount of
heat transferred to the tank by the time the entire isobutane in the tank is vaporized are to be determined.
Properties The properties of isobutane at 1 atm are Tsat = -11.7°C, ρ = 593.8 kg / m 3 , and hfg = 367.1 kJ/kg (Table A-3).
Analysis The temperature of isobutane when the pressure drops to 1 atm is simply the saturation pressure at that
temperature,
T = Tsat @1 atm = −11.7° C
Isobutane
5L
The initial mass of liquid isobutane is
20°C
m = ρV = (593.8 kg/m 3 )(0.005 m 3 ) = 2.969kg
The amount of heat absorbed is simply the total heat of vaporization,
Leak
Qabsorbed = mh fg = (2.969 kg)(367.1 kJ / kg) = 1090 kJ
3-125 A tank contains helium at a specified state. Heat is transferred to helium until it reaches a specified temperature. The
final gage pressure of the helium is to be determined.
Assumptions 1 Helium is an ideal gas.
Properties The local atmospheric pressure is given to be 100 kPa.
Analysis Noting that the specific volume of helium in the tank remains
constant, from ideal gas relation, we have
P2 = P1
T2
(300 + 273)K
= (110 + 100 kPa)
= 343.8 kPa
T1
(77 + 273)K
Then the gage pressure becomes
Helium
77ºC
110 kPa
gage
Q
Pgage,2 = P2 − Patm = 343.8 − 100 = 244 kPa
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3-65
3-126 The first eight virial coefficients of a Benedict-Webb-Rubin gas are to be obtained.
Analysis The Benedict-Webb-Rubin equation of state is given by
P=
C ⎞ 1 bR T − a aα
c ⎛
γ ⎞
⎛
+ ⎜ B0 RuT − A0 − 02 ⎟ 2 + u 3 + 6 + 3 2 ⎜1 + 2 ⎟ exp(−γ / v 2 )
T
T
v
v
v
v
v
⎝
⎠
⎝
⎠
RuT
v
Expanding the last term in a series gives
exp(−γ / v 2 ) = 1 −
1 γ2 1 γ3
γ
+
−
+ ....
2
2! v 4 3! v 6
v
Substituting this into the Benedict-Webb-Rubin equation of state and rearranging the first terms gives
P=
Ru T
v
+
R u TB 0 − A0 − C 0 / T 2
v
2
+
bR u T − a
v
3
+
c (1 + γ )
v 5T 2
+
aα
v
6
−
cγ (1 + γ )
v 7T 2
+
1 cγ 2 (1 + γ )
2! v 9 T 2
The virial equation of state is
P=
Ru T
v
+
a (T )
v
2
+
b (T )
v
3
+
c (T )
v
4
d (T )
+
v
5
+
e (T )
v
6
+
f (T )
v
7
+
g (T )
v
8
+
h (T )
v
9
...
Comparing the Benedict-Webb-Rubin equation of state to the virial equation of state, the virial coefficients are
a (T ) = R u TB 0 − A0 − C 0 / T 2
b (T ) = bR u T − a
c (T ) = 0
d (T ) = c (1 + γ ) / T 2
e (T ) = a α
f (T ) = cγ (1 + γ ) / T 2
g (T ) = 0
h (T ) =
1 c γ 2 (1 + γ )
2!
T2
3-127 The table is completed as follows:
*
P, kPa
T, oC
v, m3/kg
u, kJ/kg
300
250
0.7921
2728.9
300
133.52
0.3058
101.42
100
-
-
Insufficient information
3000
180
0.001127*
761.92*
Compressed liquid
1560.0
Condition description and quality, if
applicable
Superheated vapor
x = 0.504, Two-phase mixture
Approximated as saturated liquid at the given temperature of 180oC
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3-128 The table is completed as follows:
*
P, kPa
T, oC
v, m3/kg
u, kJ/kg
200
120.2
0.8858
2529.1
Condition description and quality, if
applicable
Saturated vapor
232.23
125
0.5010
1831.0
x = 0.650, Two-phase mixture
7829
400
0.0352
1000
30
0.001004*
125.73*
Compressed liquid
120.90
105
-
-
Insufficient information
2967.2
Superheated vapor
Approximated as saturated liquid at the given temperature of 30oC
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3-129 Water at a specified state is contained in a tank. It is now cooled. The process will be indicated on the P-v and T- v
diagrams.
Analysis The properties at the initial and final states are
P1 = 300 kPa ⎫
3
⎬ v 1 = 0.7964 m /kg (Table A - 6)
T1 = 250°C ⎭
P2 = 150 kPa
⎫
⎬ T = 111.35°C (Table A - 5)
v 2 = v 1 = 0.7964 m 3 /kg ⎭ 2
Using Property Plot feature of EES, and by adding state points we obtain following diagrams.
Steam IAPWS
106
105
P [kPa]
104
103
1
300 kPa
102
250°C
150 kPa
111,4°C
2
101
100
10-4
10-3
10-2
10-1
v [m3/kg]
100
101
102
Steam IAPWS
700
600
T [°C]
500
400
300
200
100
0
10-3
250°C
1
300 kPa
111.35°C
150 kPa
10-2
10-1
2
100
101
102
v [m3/kg]
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3-130 Water at a specified state is contained in a piston-cylinder device fitted with stops. Water is now heated until a final
pressure. The process will be indicated on the P-v and T- v diagrams.
Analysis The properties at the three states are
P1 = 300 kPa
⎫
⎬ T1 = 133.5°C (Table A - 5)
v 1 = 0.5 m 3 /kg ⎭
P2 = 300 kPa ⎫
3
⎬ v 2 = 0.6058 m /kg, T2 = 133.5°C (Table A - 5)
x 2 = 1 (sat. vap.) ⎭
Q
Water
300 kPa
0.5 m3/kg
P2 = 600 kPa
⎫
⎬ T = 517.8°C (Table A - 6)
3
v 3 = 0.6058 m /kg ⎭ 2
Using Property Plot feature of EES, and by adding state points we obtain following diagrams.
Steam IAPWS
106
105
P [kPa]
104
517.8°C
103
600 kPa
102
300 kPa
3
158.8°C
133.5°C
1
2
101
100
10-4
10-3
10-2
10-1
0.5
100
101
102
v [m3/kg]
Steam IAPWS
700
600 kPa
600
300 kPa
517.8°C
3
T [°C]
500
400
300
200
158.8°C
100
133.5°C
0
10-3
1
10-2
10-1
0.5
2
100
101
102
v [m3/kg]
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3-131 Argon contained in a piston-cylinder device at a given state undergoes a polytropic process. The final temperature is
to be determined using the ideal gas relation and the Beattie-Bridgeman equation.
Analysis (a) The polytropic relations for an ideal gas give
⎛P
T2 = T1 ⎜⎜ 2
⎝ P1
⎞
⎟⎟
⎠
n −1 / n
⎛ 13,790 kPa ⎞
= (149 + 273 K)⎜
⎟
⎝ 6895 kPa ⎠
0 . 6 / 1 .6
= 547 K
(b) The constants in the Beattie-Bridgeman equation are expressed as
0.02328 ⎞
a⎞
⎛
⎛
A = Ao ⎜1 − ⎟ = 130.7802⎜1 −
⎟
v
v
⎝
⎠
⎝
⎠
0
b
⎛
⎞
⎛
⎞
B = Bo ⎜1 − ⎟ = 0.03931⎜1 − ⎟
⎝ v ⎠
⎝ v ⎠
Argon
6895 kPa
149°C
c = 5.99 × 10 4 m 3 ⋅ K 3 /kmol
Substituting these coefficients into the Beattie-Bridgeman equation and using data (P = 6895 kPa, T = 422.2 K, Ru = 8.314
kJ/kmol·K)
P=
Ru T ⎛
c ⎞
A
⎜1 −
⎟(v + B ) − 2
2
3
v ⎝ vT ⎠
v
and solving using an equation solver such as EES gives
v = 0.5120 m 3 /kmol = 8.201 ft 3 /lbmol
From the polytropic equation
⎛ P1
⎝ P2
v 2 = v 1 ⎜⎜
⎞
⎟⎟
⎠
1/ n
⎛1⎞
= (0.5120 m 3 /kmol)⎜ ⎟
⎝2⎠
1 / 1 .6
= 0.3319 m 3 /kmol
Substituting this value into the Beattie-Bridgeman equation and using data (P = 13,790 kPa and Ru = 8.314 kJ/kmol·K),
P=
Ru T ⎛
c ⎞
A
⎜1 −
⎟(v + B ) − 2
2
3
v ⎝ vT ⎠
v
and solving using an equation solver such as EES gives
T2 = 532 K
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Fundamentals of Engineering (FE) Exam Problems
3-132 A rigid tank contains 2 kg of an ideal gas at 4 atm and 40°C. Now a valve is opened, and half of mass of the gas is
allowed to escape. If the final pressure in the tank is 2.2 atm, the final temperature in the tank is
(a) 71°C
(b) 44°C
(c) -100°C
(d) 20°C
(e) 172°C
Answer (a) 71°C
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES
screen. (Similar problems and their solutions can be obtained easily by modifying numerical values).
"When R=constant and V= constant, P1/P2=m1*T1/m2*T2"
m1=2 "kg"
P1=4 "atm"
P2=2.2 "atm"
T1=40+273 "K"
m2=0.5*m1 "kg"
P1/P2=m1*T1/(m2*T2)
T2_C=T2-273 "C"
"Some Wrong Solutions with Common Mistakes:"
P1/P2=m1*(T1-273)/(m2*W1_T2) "Using C instead of K"
P1/P2=m1*T1/(m1*(W2_T2+273)) "Disregarding the decrease in mass"
P1/P2=m1*T1/(m1*W3_T2) "Disregarding the decrease in mass, and not converting to deg. C"
W4_T2=(T1-273)/2 "Taking T2 to be half of T1 since half of the mass is discharged"
3-133 The pressure of an automobile tire is measured to be 190 kPa (gage) before a trip and 215 kPa (gage) after the trip at
a location where the atmospheric pressure is 95 kPa. If the temperature of air in the tire before the trip is 25°C, the air
temperature after the trip is
(a) 51.1°C
(b) 64.2°C
(c) 27.2°C
(d) 28.3°C
(e) 25.0°C
Answer (a) 51.1°C
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES
screen. (Similar problems and their solutions can be obtained easily by modifying numerical values).
"When R, V, and m are constant, P1/P2=T1/T2"
Patm=95
P1=190+Patm "kPa"
P2=215+Patm "kPa"
T1=25+273 "K"
P1/P2=T1/T2
T2_C=T2-273 "C"
"Some Wrong Solutions with Common Mistakes:"
P1/P2=(T1-273)/W1_T2 "Using C instead of K"
(P1-Patm)/(P2-Patm)=T1/(W2_T2+273) "Using gage pressure instead of absolute pressure"
(P1-Patm)/(P2-Patm)=(T1-273)/W3_T2 "Making both of the mistakes above"
W4_T2=T1-273 "Assuming the temperature to remain constant"
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3
3-134 A 300-m rigid tank is filled with saturated liquid-vapor mixture of water at 200 kPa. If 25% of the mass is
liquid and the 75% of the mass is vapor, the total mass in the tank is
(a) 451 kg
(b) 556 kg
(c) 300 kg
(d) 331 kg
(e) 195 kg
Answer (a) 451 kg
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES
screen. (Similar problems and their solutions can be obtained easily by modifying numerical values).
V_tank=300 "m3"
P1=200 "kPa"
x=0.75
v_f=VOLUME(Steam_IAPWS, x=0,P=P1)
v_g=VOLUME(Steam_IAPWS, x=1,P=P1)
v=v_f+x*(v_g-v_f)
m=V_tank/v "kg"
"Some Wrong Solutions with Common Mistakes:"
R=0.4615 "kJ/kg.K"
T=TEMPERATURE(Steam_IAPWS,x=0,P=P1)
P1*V_tank=W1_m*R*(T+273) "Treating steam as ideal gas"
P1*V_tank=W2_m*R*T "Treating steam as ideal gas and using deg.C"
W3_m=V_tank "Taking the density to be 1 kg/m^3"
3-135 Water is boiled at 1 atm pressure in a coffee maker equipped with an immersion-type electric heating element. The
coffee maker initially contains 1 kg of water. Once boiling started, it is observed that half of the water in the coffee maker
evaporated in 10 minutes. If the heat loss from the coffee maker is negligible, the power rating of the heating element is
(a) 3.8 kW
(b) 2.2 kW
(c) 1.9 kW
(d) 1.6 kW
(e) 0.8 kW
Answer (c) 1.9 kW
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES
screen. (Similar problems and their solutions can be obtained easily by modifying numerical values).
m_1=1 "kg"
P=101.325 "kPa"
time=10*60 "s"
m_evap=0.5*m_1
Power*time=m_evap*h_fg "kJ"
h_f=ENTHALPY(Steam_IAPWS, x=0,P=P)
h_g=ENTHALPY(Steam_IAPWS, x=1,P=P)
h_fg=h_g-h_f
"Some Wrong Solutions with Common Mistakes:"
W1_Power*time=m_evap*h_g "Using h_g"
W2_Power*time/60=m_evap*h_g "Using minutes instead of seconds for time"
W3_Power=2*Power "Assuming all the water evaporates"
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3
3-136 A 1-m rigid tank contains 10 kg of water (in any phase or phases) at 160°C. The pressure in the tank is
(a) 738 kPa
(b) 618 kPa
(c) 370 kPa
(d) 2000 kPa
(e) 1618 kPa
Answer (b) 618 kPa
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES
screen. (Similar problems and their solutions can be obtained easily by modifying numerical values).
V_tank=1 "m^3"
m=10 "kg"
v=V_tank/m
T=160 "C"
P=PRESSURE(Steam_IAPWS,v=v,T=T)
"Some Wrong Solutions with Common Mistakes:"
R=0.4615 "kJ/kg.K"
W1_P*V_tank=m*R*(T+273) "Treating steam as ideal gas"
W2_P*V_tank=m*R*T "Treating steam as ideal gas and using deg.C"
3-137 Water is boiling at 1 atm pressure in a stainless steel pan on an electric range. It is observed that 2 kg of liquid
water evaporates in 30 minutes. The rate of heat transfer to the water is
(a) 2.51 kW
(b) 2.32 kW
(c) 2.97 kW
(d) 0.47 kW
(e) 3.12 kW
Answer (a) 2.51 kW
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES
screen. (Similar problems and their solutions can be obtained easily by modifying numerical values).
m_evap=2 "kg"
P=101.325 "kPa"
time=30*60 "s"
Q*time=m_evap*h_fg "kJ"
h_f=ENTHALPY(Steam_IAPWS, x=0,P=P)
h_g=ENTHALPY(Steam_IAPWS, x=1,P=P)
h_fg=h_g-h_f
"Some Wrong Solutions with Common Mistakes:"
W1_Q*time=m_evap*h_g "Using h_g"
W2_Q*time/60=m_evap*h_g "Using minutes instead of seconds for time"
W3_Q*time=m_evap*h_f "Using h_f"
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3-138 Water is boiled in a pan on a stove at sea level. During 10 min of boiling, its is observed that 200 g of water has
evaporated. Then the rate of heat transfer to the water is
(a) 0.84 kJ/min
(b) 45.1 kJ/min
(c) 41.8 kJ/min
(d) 53.5 kJ/min
(e) 225.7 kJ/min
Answer (b) 45.1 kJ/min
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES
screen. (Similar problems and their solutions can be obtained easily by modifying numerical values).
m_evap=0.2 "kg"
P=101.325 "kPa"
time=10 "min"
Q*time=m_evap*h_fg "kJ"
h_f=ENTHALPY(Steam_IAPWS, x=0,P=P)
h_g=ENTHALPY(Steam_IAPWS, x=1,P=P)
h_fg=h_g-h_f
"Some Wrong Solutions with Common Mistakes:"
W1_Q*time=m_evap*h_g "Using h_g"
W2_Q*time*60=m_evap*h_g "Using seconds instead of minutes for time"
W3_Q*time=m_evap*h_f "Using h_f"
3-139 A rigid 3-m3 rigid vessel contains steam at 4 MPa and 500°C. The mass of the steam is
(a) 3 kg
(b) 9 kg
(c) 26 kg
(d) 35 kg
(e) 52 kg
Answer (d) 35 kg
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES
screen. (Similar problems and their solutions can be obtained easily by modifying numerical values).
V=3 "m^3"
m=V/v1 "m^3/kg"
P1=4000 "kPa"
T1=500 "C"
v1=VOLUME(Steam_IAPWS,T=T1,P=P1)
"Some Wrong Solutions with Common Mistakes:"
R=0.4615 "kJ/kg.K"
P1*V=W1_m*R*(T1+273) "Treating steam as ideal gas"
P1*V=W2_m*R*T1 "Treating steam as ideal gas and using deg.C"
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3-140 Consider a sealed can that is filled with refrigerant-134a. The contents of the can are at the room temperature of
25°C. Now a leak developes, and the pressure in the can drops to the local atmospheric pressure of 90 kPa. The
temperature of the refrigerant in the can is expected to drop to (rounded to the nearest integer)
(a) 0°C
(b) -29°C
(c) -16°C
(d) 5°C
(e) 25°C
Answer (b) -29°C
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES
screen. (Similar problems and their solutions can be obtained easily by modifying numerical values).
T1=25 "C"
P2=90 "kPa"
T2=TEMPERATURE(R134a,x=0,P=P2)
"Some Wrong Solutions with Common Mistakes:"
W1_T2=T1 "Assuming temperature remains constant"
3-141 … 3-142 Design and Essay Problems
KJ
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preparation. If you are a student using this Manual, you are using it without permission.