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Unformatted text preview: utput of the turbine per unit mass of steam if
the process is reversible. 1.4 MPa Isentropic
expansion 2 SOLUTION We take the turbine as the system (Fig. 7–15). This is a control volume since mass crosses the system boundary during the process. We note that
there is only one inlet and one exit, and thus m 1 m 2 m .
Assumptions 1 This is a steady-flow process since there is no change with time
at any point and thus mCV 0, ECV 0, and SCV 0. 2 The process is reversible. 3 Kinetic and potential energies are negligible. 4 The turbine is adiabatic and thus there is no heat transfer.
Analysis The power output of the turbine is determined from the rate form of
the energy balance, ·
E in E out
14243 · →0 (steady) E system
1442443 Rate of net energy transfer
by heat, work, and mass s2 = s1
T1 s 5 MPa
wout ? STEAM
TURBINE Rate of change in internal, kinetic,
potential, etc., energies ·
m (h1 ·
(since Q 0 0, ke pe 0) P2
s2 1.4 MPa
s1 FIGURE 7...
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- Spring '09