Thermodynamics

Chapter 9 | 499P 3 v 3T 3 P 2v 2T 2S P 3 P 2 a T 3T 2ba v 2v 3b(b) The net work output for the cycle is determined either by finding theboundary (PdV) work involved in each process by integration and addingthem or by finding the net heat transfer that is equivalent to the net workdone during the cycle. We take the latter approach. However, first we needto find the internal energy of the air at state 4:Process 3-4 (isentropic expansion of an ideal gas):Process 4-1 (constant-volume heat rejection): 11.7997 MPa2 a 1575.1 K b112 4.345 MPa652.4 Kv r4v r3 v 4v 3 r S v r4 rv r3 18216.1082 48.864 S T 4 795.6 Ku 4 588.74 kJ>kgThus,q out u 1 u 4 S q out u 4 u 1q out 588.74 206.91 381.83 kJ>kgw net q net q in q out 800 381.83 418.17 kJ/kg(c) The thermal efficiency of the cycle is determined from its definition:Under the cold-air-standard assumptions (constant specific heat values atroom temperature), the thermal efficiency would be (Eq. 9–8)which is considerably different from the value obtained above. Therefore,care should be exercised in utilizing the cold-air-standard assumptions.(d ) The mean effective pressure is determined from its definition, Eq. 9–4:whereThus,v 1 RT 1P 1MEP h th w netq inh th,Otto 1 1r k1 1 r 1k 1 182 11.4 0.565 or 56.5%MEP 418.17 kJ>kg 0.523 or 52.3%800 kJ>kgw netv 1 v 2 10.287 kPa # m 3 >kg # K21290 K2100 kPa418.17 kJ>kgw netv 1 v 1 >r 10.832 m 3 >kg2 11 1 82 a 1 kPa # m31 kJw netv 1 11 1>r2 0.832 m 3 >kgb 574 kPaDiscussion Note that a constant pressure of 574 kPa during the powerstroke would produce the same net work output as the entire cycle.