ME235Lecture%2015

# ME235Lecture%2015 - Engine Power Cycles & Refrigeration...

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1 Engine Power Cycles & Refrigeration Systems () v cyl cyl cyl cylinders N nt displaceme cyl cylinder nt displaceme crank r V V CR Ratio n Compressio S A N V V N V S A V V V R S = = = = = = = min max min max min max ) ( 2 Engine variables Work done ө Work done: min max / min max known is P(v) until integrate not can / / / : ; : ; Obviously v v w P v v P Pdv w where w m W cyl net meff meff cyl net cyl net cyl net = = 3 2 1 ( ) min max / V V P W meff cyl net = Work done by the whole engine: 2 1 60 2 1 60 1 2 = = = RPM V P RPM w m N W w m N W disp meff cycle complete to needed are rev because net cyl net cyl 48 47 6 &

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2 Idealized Engine Cycle: Otto Cycle 1-2 Isoentropic compression 2-3 Const. ‘v’ heat addition 3-4 Isoentropic expansion 4-1 Const. ‘v’ heat rejection Thermal efficiency (assuming const. sp. heat): ( ) () 1 1 1 ) ( ) ( 1 1 2 3 2 1 4 1 C const. assuming 2 3 1 4 v = = = = T T T T T T T T T C m T T T C m Q Q Q Q Q T e appropriat at v v H L H L H th 3 2 1 η & RT P cons P ga idea isoentrop k 1 2 1 3 1 1 3 4 4 3 4 1 3 2 1 3 1 1 2 1 1 2 : & , since : . T T V V V V T T also V V V V V V V V T T gives Pv const Pv gas ideal c isoentropi k k k k = = = = = = = = = ( ) 2 1 2 3 2 1 4 1 1 1 1 1 T T T T T T T T th = = Thus: Thermal efficiency: : ; 1 1 1 1 3 4 2 1 1 1 1 2 2 1 2 1 2 3 2 1 4 1 V V V V CR the is r where r V V T T T T T T T T T T v k v k th = = = = = = . . . 1 1 : ) important ( 1 2 1 only CR of function a is efficiency e i r T T Thus k v th = = Deviation from ideal Otto cycle: C v (T) Combustion replaces Heat Tran @ Hi-T Open cycle – inlet & exhaust processes. Combustion incomplete Not adiabatic Irreversibilities
3 To approximate an actual spark-ignition engine consider an air-standard Otto cycle that has a heat addition of 1800 kJ/kg of air, a compression ratio of 7, and a pressure and temperature at the beginning of the compression process of 90 kPa, 10 ° C. Assuming constant specific heat, with the value from Table A.5, determine the maximum pressure and temperature of the cycle, the thermal efficiency of the cycle and the mean effective pressure . Compression: Reversible and adiabatic so constant s: Thus, P 2 = P 1 (v 1 /v 2 ) k = 90(7) 1.4 = 1372 kPa T 2 = T 1 (v 1 /v 2 ) k-1 = 283.2 × ( 7) 0.4 = 616.6 K Combustion: constant volume T 3 = T 2 + q H /C V0 = 616.6 + 1800/0.717 = 3127 K P 3 = P 2 T 3 /T 2 = 1372 × 3127 / 616.6 = 6958 kPa Efficiency and net work η TH = 1 - T 1 /T 2 = 1 - 283.2/616.5 = 0.541 w net = η TH × q H = 0.541 × 1800 = 973.8 kJ/kg Displacement and P meff v 1 = RT 1 /P 1 = (0.287 × 283.2)/90 = 0.9029 m 3 /kg v 2 = (1/7) v 1 = 0.1290 m 3 /kg P meff = w NET v 1 -v 2 = 973.8 0.9029 - 0.129 = 1258 kPa A four stroke gasoline engine has a compression ratio of 10:1 with 4 cylinders of total displacement 2.3 L.

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## This note was uploaded on 09/15/2010 for the course ME 235 taught by Professor Borgnakke during the Fall '07 term at University of Michigan.

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ME235Lecture%2015 - Engine Power Cycles & Refrigeration...

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