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8-60 Steam is expanded in an isentropic turbine. The work produced is to be determined.
Assumptions 1 This is a steady-flow process since there is no change with time. 2 The process is isentropic
Analysis There is one inlet and two exits. We take the turbine as the system, which is a control volume
since mass crosses the boundary. The energy balance for this steady-flow system can be expressed in the
rate form as
E E
in
out
Rate of net energy transfer
by heat, work, and mass
E in
m 1 h1
W
out
0
4 MPa
5 kg/s
Rate of change in internal, kinetic,
potential, etc. energies
Steam
turbine
E out
m 2 h2 m 3 h3 W out
700 kPa
m 1 h1 m 2 h2 m 3 h3
50 kPa
100°C
From a mass balance,
m 2
0.05m 1
(0.05)(5 kg/s)
0.25 kg/s
m 3
0.95m 1
(0.95)(5 kg/s)
4.75 kg/s
T
Noting that the expansion process is isentropic, the
enthalpies at three states are determined as follows:
P3
T3
50 kPa ½ h3
¾
100qC ¿ s 3
P1
4 MPa
s1
s3
P2
700 kPa
s2
s3
2682.4 kJ/kg
7.6953 kJ/kg  K
1
4 MPa
0.7 MPa
50 kPa
(Table A - 6)
½
¾ h1
7.6953 kJ/kg  K ¿
3979.3 kJ/kg (Table A - 6)
½
¾ h2
7.6953 kJ/kg  K ¿
3309.1 kJ/kg
2
3
s
(Table A - 6)
Substituting,
W out
m 1 h1 m 2 h2 m 3 h3
(5 kg/s)(3979.3 kJ/kg) (0.25 kg/s)(3309.1 kJ/kg) (4.75 kg/s)(2682.4 kJ/kg)
6328 kW
PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
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