Thermodynamics Chapter 3 Homework with Solutions
Chapter 3 Homework and Solutions 7th Edition
Quentin McRae: Salt Lake Community College
Thermodynamics Chapter 3 Homework with Solutions
Problem 3.22
Complete this table for H 2 O:
Cengel/Boles, Thermodynamics 7th edition 2011©The McGraw-Hill Companies.
Quentin McRae: Salt Lake Community College
Thermodynamics Chapter 3 Homework with Solutions
Solution 3.22
Complete the following table for H 2 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
Cengel/Boles, Thermodynamics 7th edition 2011©The McGraw-Hill Companies.
Quentin McRae: Salt Lake Community College
Thermodynamics Chapter 3 Homework with Solutions
Problem 3.24E
Complete this table for H 2 O:
Cengel/Boles, Thermodynamics 7th edition 2011©The McGraw-Hill Companies.
Quentin McRae: Salt Lake Community College
Thermodynamics Chapter 3 Homework with Solutions
Solution 3.24E
Complete the following table for H 2 O:
T, F
P, psia
u, Btu / lbm
Phase description
300
67.03
782
Saturated mixture
267.22
40
236.02
Saturated liquid
500
120
1174.4
Superheated vapor
400
400
373.84
Compressed liquid
Cengel/Boles, Thermodynamics 7th edition 2011©The McGraw-Hill Companies.
Quentin McRae: Salt Lake Community College
Thermodynamics Chapter 3 Homework with Solutions
Problem 3.26
Complete this table for H 2 O:
Cengel/Boles, Thermodynamics 7th edition 2011©The McGraw-Hill Companies.
Quentin McRae: Salt Lake Community College
Thermodynamics Chapter 3 Homework with Solutions
Solution 3.26
Complete the following table for H 2 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
Cengel/Boles, Thermodynamics 7th edition 2011©The McGraw-Hill Companies.
Quentin McRae: Salt Lake Community College
Thermodynamics Chapter 3 Homework with Solutions
Problem 3.27
Complete this table for refrigerant-134a:
Cengel/Boles, Thermodynamics 7th edition 2011©The McGraw-Hill Companies.
Quentin McRae: Salt Lake Community College
Thermodynamics Chapter 3 Homework with Solutions
Solution 3.27
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
Cengel/Boles, Thermodynamics 7th edition 2011©The McGraw-Hill Companies.
Quentin McRae: Salt Lake Community College
Thermodynamics Chapter 3 Homework with Solutions
Problem 3.50
Water in a 5-cm-deep pan is observed to boil at 98°C. At what temperature will the water
in a 40-cm-deep pan boil? Assume both pans are full of water.
Cengel/Boles, Thermodynamics 7th edition 2011©The McGraw-Hill Companies.
Quentin McRae: Salt Lake Community College
Thermodynamics Chapter 3 Homework with Solutions
Solution 3.50
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  = 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@98 C  94.39 kPa
(Table A-4)
5 cm
The pressure difference between the bottoms of two pans is

1 kPa
 1000 kg/m  s 2

P   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)
Cengel/Boles, Thermodynamics 7th edition 2011©The McGraw-Hill Companies.
Quentin McRae: Salt Lake Community College
Thermodynamics Chapter 3 Homework with Solutions
40
Problem 3.59
A piston–cylinder device contains 0.8 kg of steam at 300°C and 1 MPa. Steam is cooled
at constant pressure until one-half of the mass condenses.
(a) Show the process on a T-v diagram.
(b) Find the final temperature.
(c) Determine the volume change.
Cengel/Boles, Thermodynamics 7th edition 2011©The McGraw-Hill Companies.
Quentin McRae: Salt Lake Community College
Thermodynamics Chapter 3 Homework with Solutions
Solution 3.59
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)
H O
2
(c) The quality at the final state is specified to be x 2 = 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  C  1
P2  1.0 MPa
x2  0.5
300C
1 MPa
(Table A-6)

 v 2  v f  x2v fg
  0.001127  0.5  (0.19436  0.001127)
 0.09775 m3/kg
T
1
2
Thus,
ΔV  m(v 2  v 1 )  (0.8 kg)(0.09775  0.25799)m 3 /kg  0.1282 m 3
v
Cengel/Boles, Thermodynamics 7th edition 2011©The McGraw-Hill Companies.
Quentin McRae: Salt Lake Community College
Thermodynamics Chapter 3 Homework with Solutions
Problem 3.60
A rigid tank contains water vapor at 250°C and an unknown pressure. When the tank is
cooled to 124°C, the vapor starts condensing. Estimate the initial pressure in the tank.
Cengel/Boles, Thermodynamics 7th edition 2011©The McGraw-Hill Companies.
Quentin McRae: Salt Lake Community College
Thermodynamics Chapter 3 Homework with Solutions
Solution 3.60
Answer: 0.30 MPa
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
H2O
initial specific volume is equal to the final specific volume that is
T 1 = 250C
3
v 1  v 2  v g @124C  0.79270 m /kg (Table A-4)
P1 = ?
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
25
15
1
2
v
Cengel/Boles, Thermodynamics 7th edition 2011©The McGraw-Hill Companies.
Quentin McRae: Salt Lake Community College
Thermodynamics Chapter 3 Homework with Solutions
Problem 3.83
The pressure in an automobile tire depends on the temperature of the air in the tire. When
the air temperature is 25°C, the pressure gage reads 210 kPa. If the volume of the tire is
0.025 m3, determine the pressure rise in the tire when the air temperature in the tire rises
to 50°C. Also, determine the amount of air that must be bled off to restore pressure to its
original value at this temperature. Assume the atmospheric pressure is 100 kPa.
Cengel/Boles, Thermodynamics 7th edition 2011©The McGraw-Hill Companies.
Quentin McRae: Salt Lake Community College
Thermodynamics Chapter 3 Homework with Solutions
Solution 3.83
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). Tire
Analysis Initially, the absolute pressure in the tire is
25C
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
P1V1 P2V2
T
323 K


 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 
(310 kPa)(0.025 m3 )
P1V

 0.0906 kg
RT1 (0.287 kPa  m3/kg  K)(298 K)
m2 
(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
Cengel/Boles, Thermodynamics 7th edition 2011©The McGraw-Hill Companies.
Quentin McRae: Salt Lake Community College
Thermodynamics Chapter 3 Homework with Solutions
Problem 3.90
Determine the specific volume of superheated water vapor at 3.5 MPa and 450°C based
on (a) the ideal-gas equation, (b) the generalized compressibility chart, and (c) the steam
tables. Determine the error involved in the first two cases.
Cengel/Boles, Thermodynamics 7th edition 2011©The McGraw-Hill Companies.
Quentin McRae: Salt Lake Community College
Thermodynamics Chapter 3 Homework with Solutions
Solution 3.90
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,
T cr = 647.1 K,
P cr = 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),
P
3.5 MPa


 0.159 
Pcr 22.06 MPa

 Z  0.961
T
723 K

TR 

 1.12

Tcr 647.1 K
PR 
H2O
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
Cengel/Boles, Thermodynamics 7th edition 2011©The McGraw-Hill Companies.
Quentin McRae: Salt Lake Community College
Thermodynamics Chapter 3 Homework with Solutions
Problem 3.99C
What is the physical significance of the two constants that appear in the van der Waals
equation of state? On what basis are they determined?
Cengel/Boles, Thermodynamics 7th edition 2011©The McGraw-Hill Companies.
Quentin McRae: Salt Lake Community College
Thermodynamics Chapter 3 Homework with Solutions
Solution 3.99C
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.
Cengel/Boles, Thermodynamics 7th edition 2011©The McGraw-Hill Companies.
Quentin McRae: Salt Lake Community College
Thermodynamics Chapter 3 Homework with Solutions
Solution 3.103E
Refrigerant-134a at 160 psia has a specific volume of 0.3479 ft3/lbm. Determine the
temperature of the refrigerant based on (a) the ideal-gas equation, (b) the van der Waals
equation, and (c) the refrigerant tables.
Cengel/Boles, Thermodynamics 7th edition 2011©The McGraw-Hill Companies.
Quentin McRae: Salt Lake Community College
Thermodynamics Chapter 3 Homework with Solutions
Solution 3.103E
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-1E)
R = 0.1052 psia·ft3/lbm·R,
T cr = 673.6 R,
P cr = 588. 7 psia
Analysis (a) From the ideal gas equation of state,
T
Pv (160 psia)(0.3479 ft 3 /lbm)

 529 R
R
0.1052 psia  ft 3 /lbm  R
(b) The van der Waals constants for the refrigerant are determined from
a
27 R 2Tcr2 (27)(0.1052 psia  ft 3 /lbm  R) 2 (673.6 R) 2

 3.591 ft 6  psia/lbm 2
64 Pcr
(64)(588.7 psia)
b
RTcr (0.1052 psia  ft 3 /lbm  R)(673.6 R)

 0.0150 ft 3 /lbm
8 Pcr
8  588.7 psia
T
1
1 
a 
3.591 
0.3479  0.0150  600 R
160 
 P  2 v  b  

R
0.1052 
v 
(0.3479) 2 
Then,
(c) From the superheated refrigerant table (Table A-13E),
P  160 psia

T  160F (620 R)
v  0.3479 ft 3 /lbm 
Cengel/Boles, Thermodynamics 7th edition 2011©The McGraw-Hill Companies.
Quentin McRae: Salt Lake Community College
Thermodynamics Chapter 3 Homework with Solutions
Problem 3.123
A 20-m3 tank contains nitrogen at 23°C and 600 kPa. Some nitrogen is allowed to escape
until the pressure in the tank drops to 400 kPa. If the temperature at this point is 20°C,
determine the amount of nitrogen that has escaped.
Cengel/Boles, Thermodynamics 7th edition 2011©The McGraw-Hill Companies.
Quentin McRae: Salt Lake Community College
Thermodynamics Chapter 3 Homework with Solutions
Solution 3.123
Answer: 44.6 kg
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 N 2 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 N 2 that escaped is
m  m1  m 2  136.6  92.0  44.6 kg
N2
600 kPa
23C
20 m3
Cengel/Boles, Thermodynamics 7th edition 2011©The McGraw-Hill Companies.
Quentin McRae: Salt Lake Community College
Thermodynamics Chapter 3 Homework with Solutions