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Unformatted text preview: 6-132 Equations
Set 22: Problem 6.132 Air enters a turbine operaitng at steady state at 6 bar and 1100 K and expands isentropically to a state where the temperature is 700 K. Determine the pressure at the exit and the work per kg of air. Ignore kinetic and potential energy contributions and assume that air obeys the ideal gas law. $Units bar kJ K System: Turbine steady-sate energy balance on turbine q = 0 [kJ/kg] T1 = 1100 [K] P1 = 6 [bar] T2 = 700 [K] a) Using Tables s1 = 3.07732 [kJ/kg K] - R ln s2 = s1 isentropic P2 1 [bar] specific entropy at state 2 - determines P2 P1 1 [bar] specific entropy at state 1 using table A-22 (5) (6) (7) (8) (9) (10) (11) adiabatic entering temperature entering pressure existing temperature (1) (2) (3) (4) s2 = 2.57277 [kJ/kg K] - R ln R= R# MW (Air) gas constant for air energy balance to determin wtable enthalpy values from table A-22 h1 - h2 + q - wtable = 0 h1 = 1161.07 [kJ/kg] h2 = 713.27 [kJ/kg] b) Using EES h1 - h2 + q - wEES = 0 h1 = h (Air, T = T1 ) h2 = h (Air, T = T2 ) energy balance on trubine specific enthalpy at turbine inlet specific enthalpy at turbine exit - but T is not known yet specific entropy at turbine inlet (12) (13) (14) (15) (16) s1 = s (Air, T = T1 , P = P1 ) s2 = s1 isentropic s2 = s (Air, T = T2 , P = P2 ) specific entropy at turbine exit - determines P (17) EES returns essentially the same values as in table A-22. $TabStops 0.2 3.5 in Solution h1 = 1161 [kJ/kg] q = 0 [kJ/kg] s2 = 2.563 [kJ/kg-K] h2 = 713.3 [kJ/kg] R = 0.287 [kJ/kg-K] wEES = 447.7 [kJ/kg] P2 = 1.034 [bar] s1 = 2.563 [kJ/kg-K] wtable = 447.8 [kJ/kg] Arrays
Row 1 2 hi [kJ/kg] 1161 713.6 Pi [bar] 6 1.035 si [kJ/kg-K] 6.566 6.566 Ti [K] 1100 700 ...
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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