Problem Statement:
44
An insulated pistoncylinder device contains
0.050
0.8 MPa.
The refigerant is now allowed
to expand in a reversible manner until to pressure drops to
0.4 MPa.
Determine:
a) The final temperature in the cylinder
b) The work done by the refrigerant
Schematic:
Assumptions/Approximations:
The kinetic and potential energy changes are negligible
The cylinder is wellinsulated and thus heat transfer is negligible
The thermal energy stored in the cylinder itself is negligible
The process is stated to be reversible
Physical Laws:
Properties/Sources:
State
P (kPa)
u (kJ/kg)
s (kJ/kg*K)
1
800.000
0.02562
246.79
0.918350
2
8.91
400.000
232.91
63.62
235.07
0.918350
0.247610
0.926900
Calculation:
1.95 kg
Reasoning/Verification/Discussion:
98.74%
None needed
232.91 kJ/kg
27.08
27.08
kJ
m
3
of saturated Refrigerant134a vapor at
m = V/v
1
u
2
= u
f
+ x
2
*(u
g
 u
f
)
x
2
= (s
2
 s
f
)/(s
g
 s
f
)
W
OUT
= m*(u
2
 u
1
)
T (
o
C)
v (m
3
/kg)
v
f
(m
3
/kg)
v
g
(m
3
/kg)
u
f
(kJ/kg)
u
g
(kJ/kg)
s
f
(kJ/kg*K)s
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 Spring '09
 n/a
 Thermodynamics, Energy, Heat, 1.95 kg, 0.30 kJ, 0.180 kJ

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