SPSolution33-35

# SPSolution33-35 - Naik, Sameer V. Solution for Special...

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Naik, Sameer V. Fall 2009 1000 Solution for Special Problem 33 Given: - Two stage air compressor with intercooler for minimum work input requirement Find: - Heat rejected (kJ/kg) in the intercooler - Isentropic efficiency (%) of the second compressor stage - Total entropy generation (kJ/kg-K) within the system System: Assumptions: - Steady state - Negligible changes in kinetic and potential energy - Air behaves as ideal gas - No heat transfer (adiabatic) for both compressors - No mechanical interaction for the intercooler - Heat exchange from the intercooler occurs at the surrounding temperature Basic Equation(s): For minimum compressor work, the process must be reversible and adiabatic in both compressors isentropic 22 11 r r isentropic p P p P    and 44 33 r r isentropic p P p P  1/2 23 1 4 PP P P  Isentropic efficiency for the compressors: 21 , s comp I hh and 43 , s comp II

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Naik, Sameer V. Fall 2009 1000 Energy balance for the intercooler:    22 23 2 3 3 2 2 intercooler intercooler intercooler intercooler dE QWm h h g z z q h h dt       Entropy balance for the compressors: ,, 00 2 12 , 21 1 ln comp I j comp I comp I air j j dS Q P ms s s s s s R dt T m P      4 34 , 43 3 ln comp II j comp II comp II air j j dS Q P dt T m P Entropy balance for the intercooler: 32 3 2 ln j intercooler intercooler intercooler intercooler j js u r r intercooler air surr Q dS q dt T m T qP ssR TP    Solution: (a) For minimum input work requirement in both compressor stages, we get: 1240.95 kPa PP  Using Table A-22, at T 1 = 300 K, we get: 1 kJ 300.19 kg h , 0 1 kJ 1.70203 kg-K s and 1 1.386 r p 2 1 1240.95 kPa 1.386 17.2 100 kPa rr P pp P Interpolating in Table A-22, we get: 2 609.1 K s T and 2 kJ 616.58 kg s h Using isentropic efficiency for the first compressor stage: 2 616.58 300.19 0.9 300.19 h 2 kJ 651.75 kg h . Interpolating in Table A-22, we get: 2 642.4 K T and 0 2 kJ 2.48112 kg-K s Using Table A-22, at T 3 = 340 K, we get: 3 kJ 340.42 kg h , 0 3 kJ 1.8279 kg-K s and 3 2.149 r p Considering the energy balance for the intercooler, the heat rejected by the intercooler is: intercooler qh h i.e. kJ 311.3 kg intercooler q ; negative sign indicates heat loss from the intercooler to its surrounding.
Naik, Sameer V. Fall 2009 1000 (b) 4 43 3 15,400 kPa 2.149 26.68 1240.95 kPa rr P pp P  Interpolating in Table A-22, we get: 4 685.8 K s T and 4 kJ 698.03 kg s h Using Table A-22, at T 4 = 750 K, we get: 4 kJ 767.29 kg

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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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SPSolution33-35 - Naik, Sameer V. Solution for Special...

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