4-21
4-22
4-48 A vertical piston–cylinder device is filled with water and covered with a 40-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,
kPa 126.15
skg/m 1000
m 0.0150
=
⋅
The boiling temperature is the saturation temperature corresponding to this pressure,
kPa 126.15@sat
4-49 Water is boiled in a pan by supplying electrical heat. The local atmospheric pressure is to be estimated.
Assumptions 75 percent of electricity consumed by the heater is transferred to the water.
The enthalpy of vaporization is determined from
kg 1.19
kJ 2700
boil
Q
Using the data by a trial–error approach in saturation table of water (Table A–5) or using EES as we did, the saturation
which is the local atmospheric pressure.
4-50 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
W = mg
P
4-23
4-51 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,
kPa 600@sat
(b) The total mass in this case can easily be determined by adding the mass of each phase,
kg 7.395=+=+=
===
852.2543.4
/kgm 0.3156
m 0.9
kg 543.4
/kgm 0.001101
m 0.005
3
3
3
3
gft
g
g
f
f
f
mmm
m
v
V
v
V
(c) At the final state water is superheated vapor, and its specific volume is
kPa 006
3
2
=
P
P
v
2
1
4-25
4-53E A rigid tank contains water at a specified pressure. The temperature, total enthalpy, and the mass of each phase are
to be determined.
Analysis (a) The specific volume of the water is
lbm 5
ft 5
3
3
V
At 20 psia,
v
f = 0.01683 ft3/lbm and
v
g = 20.093 ft3/lbm (Table A–12E). Thus the tank contains saturated liquid–vapor
mixture since
v
f <
v
<
v
g , and the temperature must be the saturation temperature at the specified pressure,
F227.92°==
psia 20@sat
TT
(b) The quality of the water and its total enthalpy are determined from
04897.0
01683.00.1
=
−
=
−
=
f
x
vv
(c) The mass of each phase is determined from
lbm 4.755
lbm 0.245
=−=+=
=×==
245.05
504897.0
gtf
tg
mmm
xmm
4-54E A rigid tank contains saturated liquid–vapor mixture of R–134a. The quality and total mass of the refrigerant are to
be determined.
Analysis At 50 psia,
v
f = 0.01252 ft3/lbm and
v
g = 0.94791 ft3/lbm (Table A–12E). The volume occupied by the liquid and
the vapor phases are
33
t
m
H
2
O
5 lbm
4-26
4-55E Superheated water vapor cools at constant volume until the temperature drops to 250°F. 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
F500
psia 180 3
1
1=
=
=
T
P
At 250°F,
v
f = 0.01700 ft3/lbm and
v
g = 13.816 ft3/lbm. Thus at the
final state, the tank will contain saturated liquid–vapor mixture since
v
f <
v
<
v
g , and the final pressure must be the saturation pressure at
the final temperature,
(b) The quality at the final state is determined from
−
−
01700.00433.3
2
f
vv
fgf
T
v
1
H
2
O
180 psia
500°F
4-28
4-57 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 liquid–
vapor mixture, and thus the final temperature must be the saturation
temperature at the final pressure,
12
4-58 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
3
since the vapor starts condensing at 150°C. Then from
Table A–6,
=
°=
3
1
1
/kgm 0.79270
C025 P
T
v
v
H2O
T1= 250°C
T
v
2
1
15
°C
4-29
4-59 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:
v
f = 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
33
11
m 001619.0/kg)m 001157.0kg)( 4.1( ===
vV
m
The volume at the final state is
3
(b) The final state properties are
Water
Q
4-31
4-63 Heat is supplied to a rigid tank 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:
v
f = 0.001157 m3/kg and uf = 850.46 kJ/kg (Table A–4).
11 m 001619.0/kg)m 001157.0kg)( 4.1( ===
which is the 25 percent of total volume. Then, the total volume is determined from
3
25.0
1
(b) Properties after the heat addition process are
m 0.006476
3
3
V
4-64 A piston–cylinder device that is initially filled with water is heated at constant pressure until all the liquid has
vaporized. The mass of water, the final temperature, and the total enthalpy change are to be determined, and the T-
v
diagram
is to be drawn.
C120.21°== kPa 002@sat
TT
v
2
1
4-32
Ideal Gas
4-67C Propane (molar mass = 44.1 kg/kmol) poses a greater fire danger than methane (molar mass = 16 kg/kmol) since
4-68 A rigid tank contains air at a specified state. The gage pressure of the gas in the tank 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 Treating air as an ideal gas, the absolute pressure in the tank is determined from
m 0.4
K) K)(298/kgmkPa kg)(0.287 (5
3
3
⋅⋅
V
mRT
Thus the gage pressure is
atm
g
4-69E 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 = 0.3353 psia⋅ft3/lbm⋅R (Table A–1E).
⋅⋅
R/lbmftpsia 0.3353
3
R
Pg
Air
400 L
25
°
C
4-34
4-73 Problem 4-72 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.
4-35
4-74E An automobile tire is under inflated with air. The amount of air that needs to be added to the tire to raise its pressure
to the recommended value is to be determined.
Assumptions 1 At specified conditions, air behaves as an ideal gas. 2 The volume of the tire remains constant.
4-75 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).
3
m 2.326
V
4-37
Compressibility Factor
4-81E The temperature of R–134a is to be determined using the ideal gas relation, the compressibility chart, and the R–134a
tables.
Properties The gas constant, the critical pressure, and the critical temperature of refrigerant–134a are, from Table A–1E,
Analysis (a) From the ideal gas equation of state,
⋅⋅
R)/lbmftpsia (0.10517
/lbm)ft 84psia)(0.13 (400
3
3
R
P
v
(b) From the compressibility chart (Fig. A–28a),
0.679
psia 588.7
psia 400
===
cr
R
P
P
P
/lbmft 0.1384
=
v
