11.95 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. Solution: IAP 2 Compression: Reversible and adiabatic so constants from Eq.8.33-34 P2-P₁(v₁/v₂)-90(7)¹.4-1372 kPa T₂-T₁(v₁/v₂)-1 283.2 x (7)0.4-616.6 K Combustion: constant volume T T3 T₂+qu/Cvo=616.6+1800/0.717-3127 K P3 P₂T/T2 1372 x 3127/616.6-6958 kPa Efficiency and net work TH 1-T₁/T₂ 1-283.2/616.5=0.541 Wnet NTH XH-0.541 x 1800-973.8 kJ/kg Displacement and Pmeff V₁-RT₁/P₁-(0.287 x 283.2)/90-0.9029 m³/kg V₂ (1/7) v₁0.1290 m³/kg WNET 973.8 Pmeff V₁-V₂ 0.9029-0.129 1258 kPa

Elements Of Electromagnetics
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Author:Sadiku, Matthew N. O.
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11.95
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.
Solution:
IAP
N
Compression: Reversible and adiabatic so constants from Eq.8.33-34
P₂ P₁(v₁/v₂) 90(7)¹.4-1372 kPa
T₂=T₁(v₁/v₂)k-1=283.2 × (7)0.4 = 616.6K
T
Combustion: constant volume
T3 T₂+9H/Cv0 = 616.6+1800/0.717=3127 K
P3 P₂T3/T2= 1372 x 3127/616.6=6958 kPa
Efficiency and net work
TH 1-T₁/T₂ 1-283.2/616.5=0.541
Wnet-THX9H=0.541 x 1800-973.8 kJ/kg
Displacement and Pmeff
V₁-RT₁/P₁-(0.287 x 283.2)/90-0.9029 m³/kg
V₂=(1/7) v₁ = 0.1290 m³/kg
Pmeff
973.8
WNET
V₁-V₂ 0.9029-0.129
1258 kPa
Transcribed Image Text:11.95 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. Solution: IAP N Compression: Reversible and adiabatic so constants from Eq.8.33-34 P₂ P₁(v₁/v₂) 90(7)¹.4-1372 kPa T₂=T₁(v₁/v₂)k-1=283.2 × (7)0.4 = 616.6K T Combustion: constant volume T3 T₂+9H/Cv0 = 616.6+1800/0.717=3127 K P3 P₂T3/T2= 1372 x 3127/616.6=6958 kPa Efficiency and net work TH 1-T₁/T₂ 1-283.2/616.5=0.541 Wnet-THX9H=0.541 x 1800-973.8 kJ/kg Displacement and Pmeff V₁-RT₁/P₁-(0.287 x 283.2)/90-0.9029 m³/kg V₂=(1/7) v₁ = 0.1290 m³/kg Pmeff 973.8 WNET V₁-V₂ 0.9029-0.129 1258 kPa
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