An ideal Otto cycle has a compression ratio of 7. At the beginning of the compression process, P1 = 90 kPa, T1 = 27°C and V1 = 0.004 m³. The maximum cycle temperature is 1137°C. For each repetition of the cycle, calculate the heat rejection and the net work production. Also, calculate the thermal efficiency and mean effective pressure for this cycle. Use constant specific heats at room temperature. The properties of air at room temperature are Cp = 1.005kJ / kgK Cv = 0.718kJ / k gK R = 0.287kJ / k gK and k = 1.4 The heat rejection is kJ. kJ. The net work production is The thermal efficiency is The mean effective pressure kPa.
For air,
R = 0.287 KJ/kgK CV = 0.718 KJ/kgK K = 1.4 P1 = 90 kPa T1 = 27°C = 300 K V1 = 0.004 m^3
Compression ratio
r =7 r = V1/V2 = 7
m = P1V1/RT1
m = (90 kPa × 0.004m^3)/(0.287 KJ/kgK × 300 K)
m = 0.00418 kg
Process 1-2 isentropic,
(P2/P1) = (V1/V2)^ɣ = (r)^ɣ
(P2/P1) = (7)^1.4
P2 = 90 × (7)^1.4 = 1372.08 kPa
T2 = T1 (V1/V2)^ɣ-1
T2 = 300 × (7)^1.4-1
T2 = 653.37 K
Consider state 3,
P2V2/T2 = P3V3/T3
P3 = (T3/T3) × P2
P3 = (1400/653.37) × 1372.08
P3 = 2940 kPa
Process 3-4,
T4 = (T3)/(V4/V3)^ɣ-1
T4 = 1400/(7)^1.4-1
T4 = 642.8 K
Calculate the heat input to the cycle,
Qin = m×Cv(T3-T1)
Qin = 0.00418 × 0.718 (1400-653.37)
Qin = 2.24 KJ
Calculate the heat rejected from the cycle,
Qout = m×Cv(T4-T1)
Qout = 0.00418×0.718(642.8-300)
Qout = 1.028 KJ
Net Work = Qin -Qout
Net Work = (2.24 – 1.028) KJ
Net Work = 1.211 KJ
ηthermal = (Net Work/Qin) × 100
ηthermal = (1.24/2.24)×100
ηthermal = 54.08 %
Pmean = (Net Work)/V1(1-1/ɣ)
Pmean= (1.211)/0.04(1-1/1.4)
Pmean = 353 KPa