4-27
4-37 Saturated vapor water is cooled at constant pressure to a saturated liquid. The heat transferred 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

Net energy transfer
by heat, work, and mass

E system



Change in internal, kinetic,
potential, etc. energies
 q out  wb,out  u  u 2  u1
(since KE = PE = 0)
Water
40 kPa
sat. vap.
 q out  wb,out  (u 2  u1 )
 q out  h2  h1
Q
q out  h1  h2
since u + wb = h during a constant pressure quasi-equilibrium
process. Since water changes from saturated liquid to saturated vapor,
we have
q out  h g  h f  h fg @ 40 kPa  2318.4 kJ/kg (Table A-5)
P
2
1
The specific volumes at the initial and final states are
v 1  v g @ 40 kPa  3.993 m 3 / kg
v
v 2  v f @ 40 kPa  0.001026 m / kg
3
Then the work done is determined from
wb,out 

2
1
 1 kJ 
P dV  P(v 2  v 1 )  (40 kPa)(0.001026  3.9933)m 3 
  159.7 kJ/kg
 1 kPa  m 3 
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