Question

An ideal Otto engine, operating on the hot-air standard with k=1.34, has a

compression ratio of 5. At the beginning of compression the volume is 6ft3

, the

pressure is 15 psia and temperature is 100F. during constant – volume heating , 500

Btu are added per cycle. Compute Wnet, Qr, thermal Efficiency, and mean

effective pressure.

Answer 1

Solution;

Given;

k=1.34

$V_1=6ft^3$

$P_1=15psia$

$r=5$

$T_1=100F=560°R$

$Q_a=500Btu$

From the relation;

$\frac{T_2}{T_1}=(\frac{V_1}{V_2})^{k-1}=r^{k-1}$

$T_2=560×5^{0.34}=968°R$

The mass can be obtained as;

$m=\frac{PV}{RT}$

$m=\frac{15×144×6}{53.34×560}=0.4339lb$

We can find Cv as;

$C_v=\frac{R}{k-1}=\frac{53.34}{0.34}=156.88ft/lb°R$

Heat added;

$Q_a=mC_v\Delta T$

$Q_a=mC_v(T_3-T_2)$

From which we obtain $T_3$ as;

$T_3=\frac{Q_a}{mC_v}+T_2$

$T_3=\frac{500×778}{0.4339×156.88}+968$

$T_3=6682.69°R$

$P_2=P_1r^k$

$P_2=15×5^{1.34}=129.63psia$

$P_3=P_2(\frac {T_3}{T_2})=129.63(\frac{6682.69}{968})$

$P_3=894.93psia$

$T_4=T_3(\frac {1}{r})^{k-1}$

$T_4=6682.69(\frac15)^{0.34}$

$T_4=3866.47°R$

$Q_r=mC_v(T_1-T_4)$

$Q_r=0.4339×156.88(560-6682.69)$

$Q_r=535.70Btu$

Efficiency;

$\eta=1-\frac{1}{r^{k-1}}$

$\eta=1-\frac{1}{5^{0.34}}=0.4214$

$W_{net}=\eta×Q_a$

$W_{net}=0.4214×500=210Btu$

Mean effective pressure;

$P_m=\frac{W_net}{V_1-V_2}$

$V_2=\frac{V_1}{r}=\frac{6}{5}=1.2ft^3$

$P_m=\frac{210×778}{(6-1.2)×144}$

$P_m=236.37psia$