8-27
8-52 R-134a undergoes an isothermal process in a closed system. The work and heat transfer are to be
determined.
Assumptions 1 The system 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 The energy balance for
this system can be expressed as
E E
in
out
Net energy transfer
by heat, work, and mass
Win Qout
T
'E system
R-134a
240 kPa
T1 =T2 =20qC
Change in internal, kinetic,
potential, etc. energies
'U
m(u 2 u1 )
1
2
s
The initial state properties are
P1
T1
240 kPa ½ u1
¾
20qC
¿ s1
246.74 kJ/kg
1.0134 kJ/kg ˜ K
(Table A - 13)
For this isothermal process, the final state properties are (Table A-11)
T2
x2
20qC ½ u 2
¾
0.20
¿ s2
T1
u f x 2 u fg
78.86 (0.20)(162.16) 111.29 kJ/kg
s f x 2 s fg
0.30063 (0.20)(0.62172)
0.42497 kJ/kg ˜ K
The heat transfer is determined from
q in
T0 ( s 2 s1 )
(293 K )(0.42497 1.0134) kJ/kg ˜ K
172.4 kJ/kg
The negative sign shows that the heat is actually transferred from the system. That is,
q out
172.4 kJ/kg
The work required is determined from the energy balance to be
win
q out (u 2 u1 ) 172.4 kJ/kg (111.29 246.74) kJ/kg
36.95 kJ/kg
8-53 The total heat transfer for the process 1-3 shown in the figure is to be determined.
Analysis For a reversible process, the area under the
process line in T-s diagram is equal to the heat transfer
during that process. Then,
Q1-3
T (qC)
3
2
360
Q1-2 Q2-3
2
³
1
3
³
TdS TdS
2
55
T1 T2
( S 2 S1 ) T2 ( S 3 S 2 )
2
(360 273) (55 273) K
(3 1) kJ/K (360 273 K)( 2 3) kJ/K
2
328 kJ
1
1
2
3
S (kJ/K)
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