Borgnakke and Sonntag
8.178
A cylinder fitted with a piston contains 0.5 kg of R-134a at 60 ° C, with a quality of 50 percent. The R-134a now expands in an internally reversible polytropic process to ambient temperature, 20 ° C at which point the quality is 100 percent.
Any heat transfer is with a constant-temperature source, which is at 60 ° C. Find the polytropic exponent n and show that this process satisfies the second law of thermodynamics.
Solution:
C.V.: R-134a, Internally Reversible, Polytropic Expansion: PVn = Const.
Cont.Eq.: m
2
= m
1
= m ; Energy Eq.: m(u
2
− u
1
) =
1
Q
2
−
1
W
2
Entropy Eq.: m(s
2
− s
1
) = ∫ dQ/T +
1
S
2 gen
State 1: T1 = 60 o C, x1 = 0.5, Table B.5.1: P
1
= P g
= 1681.8 kPa, v
1
= vf + x
1 v fg
= 0.000951 + 0.5
× 0.010511 = 0.006207 m3/kg s
1
= sf + x
1 s fg
= 1.2857 + 0.5
× 0.4182 = 1.4948 kJ/kg K, u
1
= uf + x
1 u fg
= 286.19 + 0.5
× 121.66 = 347.1 kJ/kg
State 2: T
2
= 20 o
C, x
2
= 1.0, P
2
= P g
= 572.8 kPa, Table B.5.1 v
2
= v g
= 0.03606 m3/kg, s
2
= s g
= 1.7183 kJ/kg-K u
2
= u g
= 389.19 kJ/kg
Process: PVn = Const. =>
P
P
1
2
=
 v v
2
1 n
=> n = ln
P
1
P
2
/
ln v
2 v
1
= 0.6122
1
W
2
= ∫ PdV =
P
2
V
2
- P
1
1-n
V
1
= × 0.03606 - 1681.8 × 0.006207)/(1 - 0.6122) = 13.2 kJ
2nd Law for C.V.: R-134a plus wall out to source:
∆ Snet = m(s
2
− s
1
) −
Q
H
T
H
; Check ∆ Snet > 0
Q
H
=
1
Q
2
= m(u
2
− u
1
) +
1
W
2
= 34.2 kJ
∆ Snet = 0.5(1.7183 - 1.4948) - 34.2/333.15 = 0.0092 kJ/K,
∆ Snet > 0 Process Satisfies 2nd Law
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