Question

An Otto cycle operates on 1lb/s of air from 15 psia and 130 F at the beginning of

compression. The temperature at the end of combustion is 5,000 R; compression ratio is 5.5. for

hot-air standard k=1.32. find (a) T4, OR; (b)P4, psia; (c) Qr, BTU/s; (d) thermal efficiency, and (e)

work net in horsepower.

Answer 1

Solution;

$\dot m=1lb/s$

$r=5.5$

$k=1.32$

$P_1=15psia$

$T_1=130°F=460+130=590°R$

$T_3=5000°R$

State 1:

By the Ideal gas equation;

$V_1=\frac{mRT_1}{P_1}$

$V_1=\frac{1×53.54×590}{15×144}=14.62ft^3/s$

State 2:

From the relation;

$P_1V_1^k=P_2V_2^k$

$P_2=P_1(\frac{V_1}{V_2})^k$

$P_2=15(5.5)^{1.32}=142.35psia$

Also;

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

$T_2=590(5.5)^{1.32-1}=1018.05°R$

And;

$\dot{V_2}=\frac{\dot V_1}{r}=\frac{14.62}{5.5}=2.66ft^3/s$

State 3:

Constant volume heat addition;

$\dot V_3=\dot V_2=2.66ft^3/s$

$T_3=5000°R$

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

$P_3=5000(\frac{142.35}{1018.05}=699.13psia$

State 4:

Isentropic expansion;

Hence;

$T_4=T_3(\frac{V_3}{V_4})^{k-1}$

$T_4=5000(\frac{1}{5.5})^{1.32-1}$

$T_4=2897.70°R$

Ans: $T_4=2897.70°R$

Since process 4-1 is constant volume process;

$P_4=T_4(\frac{P_1}{T_1})$

$P_4=2897.7(\frac{15}{590})=73.65psia$

Ans: $P_4=73.65psia$

(c) Heat rejected;

We know that;

$C_v=\frac{R}{k-1}=\frac{53.54}{(1.32-0.32)×778}$

$C_v=0.2151Btu/lb°R$

$Q_r=\dot{m}C_v(T_1-T_4)$

$Q_r=1×0.2151(590-2897.7)$

$Q_r=-496.39Btu/s$

(d)Thermal.efficiency;

$\eta=\frac{W}{Q_a}$

Heat added;

$Q_a=mC_v(T_3-T_2)$

$Q_a=1×0.2151(5000-1018.05)=856.52Btu/s$

Work;

$W=Q_a-Q_r=856.52-496.39=360.13Btu/s$

Hence;

$\eta=\frac{360.13}{856.52}=0.4205$

Ans: 42.05%

(e) Work net in horsepower,hp;

$W=\frac{360.13×60}{42.4}=509.62hp$

Ans:509.62hp