4 mpa determine the work output of the turbine per

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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 P1 T1 s 5 MPa 450 C wout ? STEAM TURBINE Rate of change in internal, kinetic, potential, etc., energies · E in · m h1 · Wout · E out · Wout · m (h1 · m h2 h2) · (since Q 0 0, ke pe 0) P2 s2 1.4 MPa s1 FIGURE 7...
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