Question #183628

2. 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 12 m 3 , the pressure is 95 kPaa and temperature is 38 oC.

During constant – volume heating, 370 kJ are added per cycle. Compute T 3 , P 3 , T 4 , Qa, Qr, Wnet,

thermal Efficiency, and mean effective pressure.


1
Expert's answer
2021-04-23T08:25:26-0400

Given :

γ=1.34\gamma = 1.34

rk=5r_k = 5

V1=12m3V_1= 12 m^3 V2=V1rk=125=2.4m3V_2 = \frac {V_1}{r_k} = \frac {12}{5}=2.4 m^3

P1=95kPaP_1=95kPa

T1=380C=3110KT_1 = 38^0C = 311^0K

Qs=370kJQ_s = 370 kJ

η=11rkγ1=1151.341=0.4214=42.14\eta=1-\frac{1}{r_k^{\gamma- 1}} = 1-\frac{1}{5^{1.34- 1}} = 0.4214 = 42.14 %


for T2=T1(V1V2)γ1=311×50.34=537.530KT_2= T_1({\frac{V_1}{V_2}})^{\gamma -1} = 311 \times 5^{0.34} = 537.53^0K


Qs=mCv(T3T2)Q_s = mC_v(T_3-T_2)


370=1×0.718×(T3537.53)370 = 1\times 0.718\times(T_3-537.53)


T3=1052.850CT_3 =1052.85 ^0C


T3T4=(rk)γ1\frac{T_3}{T_4}=(r_k)^{\gamma-1}


T4=1052.8550.34=609.1390KT_4 =\frac{1052.85}{5^{0.34}}= 609.139^0K


P2=P1×(rk)γ1=95(5)0.34=164.2kPaP_2 = P_1\times (r_k)^{\gamma-1} = 95(5)^0.34=164.2 kPa


P3=T3T2×P2=1052.85537.53×164.2=321.61kPaP_3 = \frac{T_3}{T_2}\times P_2 = \frac{1052.85}{537.53}\times 164.2 =321.61kPa


Qr=mCv(T4T1)=1×0.718×(609.139311)=214.06KJQ_r = mC_v(T_4-T_1)= 1\times 0.718\times (609.139-311) =214.06 KJ


WNet=QsQr=370214.06=155.94kJkgW_{Net}=Q_s-Q_r = 370 - 214.06 =155.94 \frac{kJ}{kg}


MEP=WnetV1V2=155.94122.4=16.24kPaMEP = \frac{W_{net}}{V_1-V_2} = \frac {155.94}{12-2.4} =16.24kPa



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