4-40
4-84 The specific volume of steam is to be determined using the ideal gas relation, the compressibility chart, and the steam
R = 0.4615 kPa·m3/kg·K, Tcr = 647.1 K, Pcr = 22.06 MPa
Analysis (a) From the ideal gas equation of state,
3
⋅⋅
K) K)(723/kgmkPa (0.4615
3
RT
H2O
3.5 MPa
4-44
4-90 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,
2
v
4-45
4-91 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
2
P
(b) From the compressibility chart (EES function for compressibility factor is used)
9791.0
1.64
K 304.2
K 500
0.407
MPa 7.39
MPa 3
1
1
1,
1
=
===
===
Z
T
T
T
P
P
P
cr
R
cr
R
9656.0
1.48
K 304.2
K 450
0.407
MPa 7.39
MPa 3
2
2
2,
2
=
===
===
Z
T
T
T
P
P
P
cr
R
cr
R
Thus,
/kgm 0.06165
3
=
⋅⋅
== kPa) (3000
K) K)(500/kgmkPa 89kg/s)(0.18 (0.9791)(2
3
1
11
1
P
RTmZ
V
3
mkg/ 32.44=
⋅⋅
==
K) K)(500/kgmkPa .1889(0.9791)(0
kPa) (3000
3
11
1
1
RTZ
P
ρ
/kgm 0.05472 3
=
⋅⋅
== kPa) (3000
K) K)(450/kgmkPa 89kg/s)(0.18 (0.9656)(2 3
2
22
2P
RTmZ
V
4-93E Water in a pressure cooker boils at 260°F. The absolute pressure in the pressure cooker is to be determined.
Analysis The absolute pressure in the pressure cooker is the saturation
4-94 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
(a) The volume and mass flow rates may be determined from ideal gas relation as
/sm 0.5543
3
=== kPa 3000
K) 00/kmol.K)(5kPa.m 314kmol/s)(8. 4.0(
3
1
1
P
TRN
u
V
kg/s 17.60=== K) /kg.K)(500kPa.m (0.1889
)/m 3kPa)(0.554 (3000
3
3
1
11
1
s
RT
P
m
V
0.4 kmol/s
4-51
4-102 Superheated refrigerant–134a is cooled at constant pressure until it exists as a compressed liquid. The changes in total
volume and internal energy are to be determined, and the process is to be shown on a T-v diagram.
Analysis The refrigerant is a superheated vapor at the initial state and a compressed liquid at the final state. From Tables A–
13 and A–11,
/kgm 0.019502
kJ/kg 277.23
C07
MPa .21
3
1
1
1
1
=
=
°=
=
u
T
P
v
(b)
3
m 0.187−=−=−=∆ /kgm 0.019502)160kg)(0.0008 (10)( 3
12
vvV
m
4-103 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
T
v
1
4-53
4-106 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,
MPa 22.06 K, 647.1 ,K/kgmkPa 0.4615
crcr
3
==⋅⋅= PTR
Analysis (a) From the ideal gas equation of state,
⋅⋅
K) K)(673/kgmkPa (0.4615
3
RT
57.0
1.48
K) K)(647.1/kgmkPa (0.4615
kPa) 0/kg)(22,06m (0.02
/
040.1
K 647.1
K 673
3
3
crcr
actual
cr
=
=
⋅⋅
==
===
R
R
R
P
PRT
T
T
T
v
v
=°= P
/kgm 0.02
v
4-107 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
/kgm 0.0008580
m 0.03
3
3
1
1
v
V
since
R-134a
4-54
4-108 Problem 4-107 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.
4-55
4-109 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).
4-56
4-112 The table is completed as follows:
P, kPa
T, oC
v
, m3/kg
u, kJ/kg
Condition description and quality, if
applicable
300
250
0.7921
2728.9
Superheated vapor
300
133.52
0.3058
1560.0
x = 0.504, Two–phase mixture
101.42
100
–
–
Insufficient information
3000
180
0.001127*
761.92*
Compressed liquid
* Approximated as saturated liquid at the given temperature of 180oC
4-113 … 4-115 Design and Essay Problems
4-115 It is helium.
