SPSolution36-38 - Naik, Sameer V. Solution for Special...

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Naik, Sameer V. Fall 2009 1000 Solution for Special Problem 36 Given: - Ideal reheat Rankine cycle Find: - Thermal efficiency (%) of the cycle - Mass flow rate of steam (kg/s) - Rate of heat transfer (MW) in the condenser - T-s diagram System: Condenser Boiler Pump HP Turbine W pump Q in P 1 = 30 bar T 1 = 500 C P 4 = 0.1 bar P 5 = 0.1 bar saturated liquid P 6 = 30 bar Q out P 2 = 5 bar LP Turbine P 3 = 5 bar T 3 = 500 C Assumptions: - Steady state - Negligible changes in KE and PE in all devices - Negligible pressure drop in boiler and condenser - No mechanical work interaction for boiler and condenser - Isentropic process in turbine and pump Basic Equation(s):  f gf yy x y y  Energy balance for the boiler: 22 16 2 boiler s boiler boiler steam s s dE QWm h h g z z dt        
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Naik, Sameer V. Fall 2009 1000 Energy balance for the high-pressure turbine:  22 12 2 HPturbine s HPturbine HPturbine steam s s dE QWm h h g z z dt      Energy balance for the reheat section of the boiler: 23 2 3 2 reheat s reheat reheat steam s s dE QWmh h g z z dt    Energy balance for the low-pressure turbine: 34 2 LPturbine s LPturbine LPturbine steam s s dE h h g z z dt Energy balance for the condenser: 45 2 condenser s condenser condenser steam s s dE h h g z z dt  Energy balance for the pump: 56 2 pump s pump pump steam s s dE h h g z z dt Isentropic pump work: 4 34 3 3 4 3 pump s wv d P K E P E v P P h h     6 5 16 32 ss s net thermal in s s hh h h w qh h h h    6 5 net net steam net s s s WW m wh h  Solution: State 1: Using Table A-3, at P 1 = 30 bar, T sat = 233.9 C T 1 = 500 C > T sat SHV Using Table A-4, we get: h 1 = 3456.5 kJ/kg, s 1 = 7.2338 kJ/kg-K State 2: For the isentropic process through the high-pressure turbine: s 2s = s 1 = 7.2338 kJ/kg-K and P 2 = 5 bar SHV at the high-pressure turbine exit after isentropic process Using Table A-4, we get: T 2s = 240 C and h 2s = 2939.9 kJ/kg State 3: Using Table A-3, at P 3 = 5 bar, T sat = 151.86 C T 3 = 500 C > T sat SHV Using Table A-4, we get: h 3 = 3483.9 kJ/kg, s 3 = 8.0873 kJ/kg-K
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This note was uploaded on 05/15/2010 for the course ME 200 taught by Professor Gal during the Spring '08 term at Purdue University-West Lafayette.

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SPSolution36-38 - Naik, Sameer V. Solution for Special...

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