5-36
5-46 R-134a contained in a spring-loaded piston-cylinder device is cooled until the temperature and
volume drop to specified values. The heat transfer and the work done are to be determined.
Assumptions 1 The cylinder is stationary and thus the kinetic and potential energy changes are zero. 2
There are no work interactions involved other than the boundary work. 3 The thermal energy stored in the
cylinder itself is negligible. 4 The compression or expansion process is quasi-equilibrium.
Analysis We take the contents of the cylinder as the system. This is a closed system since no mass enters
or leaves. The energy balance for this stationary closed system can be expressed as
E E
in
out
'E system
Net energy transfer
by heat, work, and mass
Change in internal, kinetic,
potential, etc. energies
Wb,in Qout
Qout
'U
m(u 2 u1 )
(since KE = PE = 0)
Wb,in m(u 2 u1 )
The initial state properties are (Table A-13)
P1
T1
600 kPa ½ v 1
¾
15qC
¿ u1
P
3
0.055522 m / kg
357.96 kJ/kg
The mass of refrigerant is
m
V1
v1
1
600 kPa
2
0.3 m 3
0.055522 m 3 /kg
5.4033 kg
v
The final specific volume is
v2
V2
m
0.1 m 3
5.4033 kg
0.018507 m 3 /kg
The final state at this specific volume and at -30qC is a saturated mixture. The properties at this state are
(Table A-11)
x2
v 2 v f
0.018507 0.0007203
0.22580 0.0007203
v g v f
u2
u f x 2 u fg
P2
84.43 kPa
0.079024
12.59 (0.079024)(200.52)
28.44 kJ/kg
Since this is a linear process, the work done is equal to the area under the process line 1-2:
Wb,in
Area
P1 P2
(V 1 V 2 )
2
(600 84.43)kPa
§ 1 kJ ·
(0.3 0.1)m 3 ¨
¸
2
© 1 kPa ˜ m 3 ¹
68.44 kJ
Substituting into energy balance equation gives
Qout
Wb,in m(u 2 u1 )
68.44 kJ (5.4033 kg)(28.44 357.96) kJ/kg
1849 kJ
PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
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5-36 volume drop to specified values. The heat transfer and the... There are no work interactions involved other than the boundary...