4 mpa determine the work output of the turbine per

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

This document was uploaded on 11/28/2012.

Ask a homework question - tutors are online