Question #322906

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.


1
Expert's answer
2022-04-04T09:03:57-0400

Solution;

m˙=1lb/s\dot m=1lb/s

r=5.5r=5.5

k=1.32k=1.32

P1=15psiaP_1=15psia

T1=130°F=460+130=590°RT_1=130°F=460+130=590°R

T3=5000°RT_3=5000°R

State 1:

By the Ideal gas equation;

V1=mRT1P1V_1=\frac{mRT_1}{P_1}

V1=1×53.54×59015×144=14.62ft3/sV_1=\frac{1×53.54×590}{15×144}=14.62ft^3/s

State 2:

From the relation;

P1V1k=P2V2kP_1V_1^k=P_2V_2^k

P2=P1(V1V2)kP_2=P_1(\frac{V_1}{V_2})^k

P2=15(5.5)1.32=142.35psiaP_2=15(5.5)^{1.32}=142.35psia

Also;

T2=T1(V1V2)k1T_2=T_1(\frac{V_1}{V_2})^{k-1}

T2=590(5.5)1.321=1018.05°RT_2=590(5.5)^{1.32-1}=1018.05°R

And;

V2˙=V˙1r=14.625.5=2.66ft3/s\dot{V_2}=\frac{\dot V_1}{r}=\frac{14.62}{5.5}=2.66ft^3/s

State 3:

Constant volume heat addition;

V˙3=V˙2=2.66ft3/s\dot V_3=\dot V_2=2.66ft^3/s

T3=5000°RT_3=5000°R

P3=T3(P2T2)P_3=T_3(\frac{P_2}{T_2})

P3=5000(142.351018.05=699.13psiaP_3=5000(\frac{142.35}{1018.05}=699.13psia

State 4:

Isentropic expansion;

Hence;

T4=T3(V3V4)k1T_4=T_3(\frac{V_3}{V_4})^{k-1}

T4=5000(15.5)1.321T_4=5000(\frac{1}{5.5})^{1.32-1}

T4=2897.70°RT_4=2897.70°R

Ans: T4=2897.70°RT_4=2897.70°R

Since process 4-1 is constant volume process;

P4=T4(P1T1)P_4=T_4(\frac{P_1}{T_1})

P4=2897.7(15590)=73.65psiaP_4=2897.7(\frac{15}{590})=73.65psia

Ans: P4=73.65psiaP_4=73.65psia

(c) Heat rejected;

We know that;

Cv=Rk1=53.54(1.320.32)×778C_v=\frac{R}{k-1}=\frac{53.54}{(1.32-0.32)×778}

Cv=0.2151Btu/lb°RC_v=0.2151Btu/lb°R

Qr=m˙Cv(T1T4)Q_r=\dot{m}C_v(T_1-T_4)

Qr=1×0.2151(5902897.7)Q_r=1×0.2151(590-2897.7)

Qr=496.39Btu/sQ_r=-496.39Btu/s

(d)Thermal.efficiency;

η=WQa\eta=\frac{W}{Q_a}

Heat added;

Qa=mCv(T3T2)Q_a=mC_v(T_3-T_2)

Qa=1×0.2151(50001018.05)=856.52Btu/sQ_a=1×0.2151(5000-1018.05)=856.52Btu/s

Work;

W=QaQr=856.52496.39=360.13Btu/sW=Q_a-Q_r=856.52-496.39=360.13Btu/s

Hence;

η=360.13856.52=0.4205\eta=\frac{360.13}{856.52}=0.4205

Ans: 42.05%

(e) Work net in horsepower,hp;

W=360.13×6042.4=509.62hpW=\frac{360.13×60}{42.4}=509.62hp

Ans:509.62hp



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