Unformatted text preview: 6-142 Equations
Problem 6-142 Steam at 5 MPa and 600 C enters an insulated turbine operating at steady-state and exits as saturated vapor at 50 kPa. Kinetic and potential energy effects are negligible. Determine: a) the work developed by the turbine b) the isentropic turbine efficiency System: turbine q=0 adiabatic operation (1) (2) (3) (4) (5) (6) saturated vapor (7) (8) (9) (10) (11) (12) P1 = 5000 [kPa] T1 = 600 [C] h1 = h (Steam, T = T1 , P = P1 ) s1 = s (Steam, T = T1 , P = P1 ) P2 = 50 [kPa] h2 = h (Steam, P = P2 , x = 1) h1 - h2 + q = w sid,2 = s1 energy balance determines the work per unit mass ideal turbine is isentropic hid,2 = h (Steam, P = P2 , s = sid,2 ) h1 - hid,2 + q = wid w = wid eta energy balance on ideal turbine definition of isentropic turbine efficiency $TabStops 0.2 3.5 in Solution = 0.8959 q = 0 [kJ/kg] w = 1021 [kJ/kg] wid = 1140 [kJ/kg] Arrays
Row 1 2 hi [kJ/kg] 3666 2645 hid,i [kJ/kg] 2527 Pi [kPa] 5000 50 si [kJ/kg-K] 7.259 sid,i [kJ/kg-K] 7.259 Ti [C] 600 ...
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This note was uploaded on 07/29/2008 for the course ME 361 taught by Professor Martin during the Spring '07 term at University of Wisconsin.
- Spring '07