EXAMPLE 8-5 Isentropic Expansion of Steam in a Turbine Steam enters an adiabatic turbine at 5 MPa and 450°C and leaves a

[answerhappygod]

Example 8 5 Isentropic Expansion Of Steam In A Turbine Steam Enters An Adiabatic Turbine At 5 Mpa And 450 C And Leaves A 1 (36.64 KiB) Viewed 27 times

EXAMPLE 8-5 Isentropic Expansion of Steam in a Turbine Steam enters an adiabatic turbine at 5 MPa and 450°C and leaves at a pressure of 1.4 MPa. Determine the work output of the turbine per unit mass of steam if the process is reversible. SOLUTION Steam is expanded in an adiabatic turbine to a specified pressure in a reversible manner. The work output of the turbine is to be determined. Assumptions 1 This is a steady-flow process since there is no change with time at any point and thus Amy 0, AE-0, and AS - 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 We take the turbine as the system (Fig. 8-15). 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 mm, m. The power output of the turbine is determined from the rate form of the energy balance, Jo dE Has mh, +mh (since 0-0, kepen 0) -(₂-₂) The inlet state is completely specified since two properties are given. But only one property (pressure) is given at the final state, and we need one more prop- erty to fix it. The second property comes from the observation that the process is reversible and adiabatic, and thus isentropic. Therefore, 5, 5, and State 1: P₁-5 MPa] 7₁-450°C) P-1.4 MPa] ₁-3317.2 kJ/kg 4-6.8210 kJ/kg-K State 2: h₂-2967.4 kJ/kg 444 5 MPa Applied Thermodynamics 14MP A 5 MP ₁450°C ₂1 Steam Sabine Isentropic expansion P 14 MP Fy=h FIGURE 8-15 Schematic and T-s diagram for Example 8-5 Then, the work output of the turbine per unit mass of the steam becomes Wot-hy-hy-3317.2-2967.4-349.8 kJ/kg 14