```8-142
8-189 An insulated rigid tank is connected to a piston-cylinder device with zero clearance that is
maintained at constant pressure. A valve is opened, and some steam in the tank is allowed to flow into the
cylinder. The final temperatures in the tank and the cylinder are to be determined.
Assumptions 1 Both the tank and cylinder are well-insulated and thus heat transfer is negligible. 2 The
water that remains in the tank underwent a reversible adiabatic process. 3 The thermal energy stored in the
tank and cylinder themselves is negligible. 4 The system is stationary and thus kinetic and potential energy
changes are negligible.
Analysis (a) The steam in tank A undergoes a reversible, adiabatic process, and thus s2 = s1. From the
steam tables (Tables A-4 through A-6),
v 1 v g @500 kPa 0.37483 m 3 /kg
500 kPa ½
¾ u1 u g @500 kPa 2560.7 kJ/kg
sat.vapor
¿s
s g @500 kPa 6.8207 kJ/kg  K
1
P1
150 kPa ½
°
s 2 s1
¾
sat.mixture °¿ v 2, A
P2
T2, A Tsat @150 kPa 111.35 qC
s 2, A s f
6.8207 1.4337
x 2, A
v f x 2, Av fg
s fg
0.9305
0.001053 (0.9305)(1.1594 0.001053) 1.0789 m 3 /kg
u f x 2, A u fg
u 2, A
5.7894
466.97 (0.9305)(2052.3 kJ/kg )
2376.6 kJ/kg
The initial and the final masses in tank A are
m1, A
VA
v1, A
m2, B
m1, A m2, A
Thus,
0.4 m3
and m2, A
1.067 kg
0.37483 m3/kg
VA
v 2, A
0.4 m3
1.0789 m3/kg
0.371 kg
1.067 0.371 0.696 kg
(b) The boundary work done during this process is
Wb,out
³
2
1
P dV
PB V 2, B 0
PB m 2, Bv 2, B
Taking the contents of both the tank and the cylinder
to be the system, the energy balance for this closed
system can be expressed as
E Eout
in
Net energy transfer
by heat, work, and mass
Wb, out
'Esystem
150 kPa
Change in internal, kinetic,
potential, etc. energies
'U
'U A 'U B
Wb,out 'U A 'U B
or,
Sat.
vapor
500 kPa
0.4 m3
PB m2, Bv 2, B m2u2 m1u1 A m2u2 B
m2, B h2, B m2u2 m1u1 A
0
0
0
Thus,
h 2, B
m1u1 m 2 u 2 A
1.0672560.7 0.3712376.6
m 2, B
0.696
2658.8 kJ/kg
At 150 kPa, hf = 467.13 and hg = 2693.1 kJ/kg. Thus at the final state, the cylinder will contain a saturated
liquid-vapor mixture since hf < h2 < hg. Therefore,
T2, B
Tsat @150 kPa
111.35qC
PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
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