Question #219905

In an air standard Otto cycle the maximum and minimum tcmperatures arc 1400°C and 15°C. The heat supplied per kg of air is 800 kJ. Calculate the compression ratio and the thermal eficiency. Calculale also the. ratio of maximum to minimum pressures in the cycle.


Expert's answer

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


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

LATEST TUTORIALS
APPROVED BY CLIENTS