<p>1) The compression ratio of an air-standard Otto cycle is 9.5. The air is ini

Question

<p>1) The compression ratio of an air-standard Otto cycle is 9.5. The air is initially at 100kPa, 35^oC prior to the isentropic compression process. 353.25 kJ/K of heat is ejected out during the constant-volume heat ejection process. using the specific heat values at the room temperature, determine<br />a)The temperature after the isentropic compression<br />b) The temperature before constant-volume heat-ejection process<br />c) The highest pressure in the cycle<br />d) he net work (kJ/kg)<br />e) The thermal efficiency</p>

Explanation / Answer

(a) T3 is the highest temp in the cycle: Isentropic expansion 3-4: T3/T4 = (V4/V3)^(k-1) = (V1/V2)^(k-1) = rk^(k-1) T3 = 800*9.5^0.4 = 1968.7 K P3/P4 = (V4/V3)^k = (V1/V2)^k = rk^k and P4 = P1(T4/T1) constant volume (isometric) P4 = 100*(800/308) = 259.7 kPa P3 = 259.7*9.5^1.4 = 6072.3 kPa (b) Qr = mCv(T4-T1) and m = P1V1/RT1 = 100*0.0006/(0.287*308) = 0.000679 kg Cv = R/(k-1) = 0.287/0.4 = 0.7175 kJ/kg-K Qr = 0.000679*0.7175*(800-308) = 0.24 kJ (c) e = 1 - 1/rk^(k-1) = 1 - (1/9.5^0.4) = 0.594 (59.4%) (d) mep = W / Vd = (Qa - Qr) / Vd and Qa = mCv(T3-T2) and T2 = T1*rk^(k-1) = 308*9.5^0.4 = 757.9 K Qa = 0.000679*0.7175*(1968.7 - 757.9) = 0.59 kJ mep = (0.59 - 0.24) / 0.0006*(1 - 1/9.5) = 652 kPa

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