Spring 2006
1
ME 201
Thermodynamics
Homework #7 Solutions
1. Refrigerant 134a as saturated vapor at 0.5 MPa is isentropically compressed by a
compressor in a refrigeration plant to 1.2 MPa. Determine the enthalpy change for
the process and the final fluid phase.
Solution:
Substance Type: Compressible (R134a)
Problem Type: Process(Isentropic)
State 1
State 2
T
1
=
15.71
°
C
T
2
=
49.3
°
C
P
1
= 0.5 MPa
P
2
= 1.2 MPa
h
1
=
259.30 kJ/kg
h
2
=
277.4 kJ/kg
s
1
=
0.9240 kJ/(kg
⋅
K)
s
2
=
0.9240 kJ/(kg
⋅
K)
phase: sat.vap.
phase:
sup.vap.
Italicized
values are from R134a tables,
bold
values are calculated.
At state 1 we know the pressure and that we have saturated liquid, so that the state is
fixed. Going to the saturation pressure table, Table A12, we find
kJ/kg
259.30
=
h
K),
kJ/(kg
40
2
9
0.
=
s
C,
5.71
1
=
T
1
1
1
⋅
°
At state 2 we know the pressure and that we have an isentropic process or
K)
kJ/(kg
.9240
0
=
s
=
s
1
2
⋅
To determine the fluid phase, we go to the saturation pressure table at 1.2 MPa and
find
K)
kJ/(kg
0.9130
=
s
and
K)
kJ/(kg
0.4244
=
s
g
f
⋅
⋅
Since s
2
> s
g
, we have a superheated vapor at state 2. Going to the superheat tables,
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '06
 SOMERTON
 Thermodynamics, 0.215 m, 0.9130 kJ, 1.51 kJ

Click to edit the document details