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Chapter 3
Properties of Pure Substances
359
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 A4,
T = 100°C
⇒
vv
≅=
fC
m
kg
error
u
u
kJ kg
error
hh
k
Jk
g
e
r
r
o
r
@
@
@
./
(
.
(
.
(
.
100
3
100
100
0 001044
0 76%
41894
101%
419 04
2 61%
o
o
o
)
)
)
From compressed liquid table (Table A7),
kJ/kg
430.28
kJ/kg
414.74
/kg
m
0.0010361
C
100
MPa
15
3
=
=
=
⎭
⎬
⎫
=
=
h
u
v
T
P
o
The percent errors involved in the saturated liquid approximation are listed above in parentheses.
360
Problem 359 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.
T = 100
"[C]"
P = 15000
"[kPa]"
v = VOLUME(Steam,T=T,P=P)
"[m^3/kg]"
u = INTENERGY(Steam,T=T,P=P)
"[kJ/kg]"
h = ENTHALPY(Steam,T=T,P=P)
"[kJ/kg]"
v_app = VOLUME(Steam,T=T,x=0)
"[m^3/kg]"
u_app = INTENERGY(Steam,T=T,x=0)
"[kJ/kg]"
h_app_1 = ENTHALPY(Steam,T=T,x=0)
"[kJ/kg]"
h_app_2 = ENTHALPY(Steam,T=T,x=0)+v_app*(Ppressure(Steam,T=T,x=0))
"[kJ/kg]"
SOLUTION
Variables in Main
h=430.3 [kJ/kg]
h_app_1=419.1 [kJ/kg]
h_app_2=434.6 [kJ/kg]
P=15000 [kPa]
T=100 [C]
u=414.7 [kJ/kg]
u_app=419 [kJ/kg]
v=0.001036 [m^3/kg]
v_app=0.001043 [m^3/kg]
315
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View Full Document Chapter 3
Properties of Pure Substances
361E
A rigid tank contains saturated liquidvapor mixture of R134a. The quality and total mass of the
refrigerant are to be determined.
Analysis
At 30 psia,
v
f
= 0.01209 ft
3
/lbm and
v
g
= 1.5408 ft
3
/lbm.
The volume occupied by the liquid and
the vapor phases are
VV
fg
==
1.5 ft
and
13.5 ft
33
R134a
15 ft
3
30 psia
Thus the mass of each phase is
m
V
m
V
f
f
f
g
g
g
=
=
v
v
1.5 ft
0.01209 ft / lbm
124.1lbm
13.5 ft
1.5408 ft / lbm
8.76 lbm
3
3
3
3
Then the total mass and the quality of the refrigerant are
mmm
x
m
m
tfg
g
t
=+
=
+=
=
124.1 8.76
8.76
132.86
132.86 lbm
.
0 0659
316
Chapter 3
Properties of Pure Substances
362
Superheated steam in a pistoncylinder 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
Tv
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,
H
2
O
300
°
C
1 MPa
C
179.91
o
=
=
MPa
[email protected]
T
T
(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
/kg
m
0.2579
C
300
MPa
1.0
3
1
1
1
=
⎭
⎬
⎫
=
=
v
T
P
o
T
v
2
1
/kg
m
.0978
0
)
001127
.
0
19444
.
0
(
5
.
0
001127
.
0
5
.
0
MPa
1.0
3
2
2
2
2
=
−
×
+
=
+
=
⎭
⎬
⎫
=
=
fg
f
v
x
v
v
x
P
Thus,
Δ
Vm
k
g
m k
g
=−
=
−
=
−
()
(
.
)
(
.
.
)
/
vv
21
3
08
00978 02579
0.128m
3
363
The water in a rigid tank is cooled until the vapor starts condensing. The initial pressure in the tank is
to be determined.
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This note was uploaded on 04/14/2008 for the course ME 320 taught by Professor Dr.kinne during the Spring '07 term at University of Texas at Austin.
 Spring '07
 Dr.Kinne

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