Question #219904
In an air standard Otto cyclc the maximum and minimum tcmperatures arc 1400
1
Expert's answer
2021-07-26T05:57:02-0400

Compression ratio

Q23=Cv(T3T2)800=0.718(1673T2)T2=16738000.718=558.8KT2T1=(v1v2)γ1v1v2=(558.8288)10.4=5.243Q_{2-3}=C_v(T_3-T_2)\\ 800=0.718(1673-T_2)\\ T_2= 1673-\frac{800}{0.718}=558.8 K\\ \frac{T_2}{T_1}=(\frac{v_1}{v_2})^{\gamma-1}\\ \frac{v_1}{v_2}=(\frac{558.8}{288})^{\frac{1}{0.4}}= 5.243


Thermal Efficiency

η=11rcr1=115.2430.4=48.45%η= 1-\frac{1}{r_c^{r-1}}=1-\frac{1}{5.243^{0.4}}=48.45 \%


The ratio of maximum to minimum pressures

T2T1=(P2P1)γ1γP2P1=(558.8288)1.40.4=10.174.............1P3P2=(T3T2)=1673558.8=2.993............2Eqn 1 Eqn 2P3P1=10.1742.993=30.45\frac{T_2}{T_1}=(\frac{P_2}{P_1})^{\frac{\gamma-1}{\gamma}}\\ \frac{P_2}{P_1}=(\frac{558.8}{288})^{\frac{1.4}{0.4}}=10.174.............1\\ \frac{P_3}{P_2}=(\frac{T_3}{T_2})=\frac{1673}{558.8}=2.993............2\\ \therefore Eqn \space 1 \space *Eqn \space 2\\ \frac{P_3}{P_1}=10.174*2.993=30.45


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