SPSolution39-41

SPSolution39-41 - Naik, Sameer V. Solution for Special...

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Naik, Sameer V. Fall 2009 1000 Solution for Special Problem 39 Given: - Air-standard Otto cycle with known compression ratio Find: - Temperature (K) and pressure (MPa) after heat addition - Temperature (K) and pressure (MPa) after isentropic expansion - Thermal efficiency (%) of the cycle - Mean effective pressure (MPa) of the cycle System: Assumptions: - Air behaves as ideal gas - Negligible leakage in piston-cylinder device, i.e. closed system (mass constant) - Negligible changes in KE and PE in the piston-cylinder device - Isentropic compression and expansion Basic Equation(s): air PV mR T Constant volume heat addition: 23 23 23 23 3 2 qw u k ep equ u    Constant volume heat rejection: 41 41 41 41 1 4 u k u For isentropic compression: 22 11 1 r r vv r  and r r p P p P
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Naik, Sameer V. Fall 2009 1000 For isentropic expansion: 44 33 r r vv r  and r r p P p P Thermal efficiency:    32 41 23 41 23 3 2 net net in out thermal in in in uu uu qq wqq q q q u u   Mean effective pressure: 12 net W MEP VV Solution: Using Table A-22, at T 1 = 295 K, we get: 1 kJ 210.49 kg u , 1 647.9 r v , and 1 1.3068 r p For isentropic compression process, we have: 22 1 211 11 647.9 72 9 r rrr vVv vvv vV r Interpolating in Table A-22, we get: 2 692 K T , 2 27.592 r p , and 2 kJ 506.03 kg u 2 21 1 27.592 95 kPa 2005.85kPa 1.3068 rr pP p PP p  Mass of air in the cylinder:   63 3 1 95 kPa 2800 10 m 3.14 10 kg kJ 8.314 kmol-K 295 K kg 28.97 kmol air air PV m RT  For constant volume heat addition process, we have: 23 23 3 2 air air Qm qm u u 23 3 kJ 4.3 kJ kJ 506.03 1875.45 kg 3.14 10 kg kg air Q uu m  Interpolating in Table A-22, we get: 3 2203 K T , 3 2 r v , and 3 3157.56 r p For constant volume heat addition process, we have: 3 3 3 23 2 2 2203 K 2005.85 kPa 6385.7 kPa 692 K air air air PV P T mP RT RT P T  For isentropic expansion process, we have: 433 3 921 8 rrrr r v Interpolating in Table A-22, we get: 4 1117.7 K T , 4 178.25 r p , and 4 kJ 860.78 kg u 4 43 3 178.25 6385.7kPa 360.5 kPa 3157.56 p p
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Naik, Sameer V. Fall 2009 1000 For constant volume heat rejection process, we have:  41 41 1 4 2.04 kJ air air Qm qm u u 
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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.

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SPSolution39-41 - Naik, Sameer V. Solution for Special...

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