hw12_s13_sol

Then q 750 6892 1734 k t3 3 t2 0718 cv

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Unformatted text preview: v 2 k 1 3008 0.4 689.2 K From the ideal gas law, T 689 .2 P2 rP1 2 895 1746 kPa . T1 300 For an ideal gas with constant Cv, qin = Cv(T3 ‐ T2). Then q 750 689.2 1734 K . T3 3 T2 0.718 Cv From the ideal gas law, T 1734 P3 rP1 3 895 4392 kPa . T1 300 b) Find the net work leaving the engine (wnet,out). From an energy balance over the cycle, wnet,out = qin ‐ qout. We start with the isentropic expansion from 3 to 4: T4 v 3 T3 v 4 k 1 v or T4 T3 3 v 4 k 1 1 1734 8 0 .4 754.7 K . Then qout = Cv(T4 ‐ T1) = 0.718(754.7 ‐ 300) = 326.5 kJ/kg. The work output is wnet,out = qin ‐ qout = 750 ‐ 326.5 = 423.5 kJ/kg c) The thermal efficiency is th,otto w net,out qin 423.5 0.5647 750 d) The mean effective pressure (MEP) is MEP w net,out v1 v 2 w net,out 423.5 534.1 kPa RT1 RT2 300 689.2 0.2870 P1 P2 95 1746 Problem 2 (10 points) An ideal Otto cycle has a compression ratio of 8. At the beginning of the compression process, air is at 95 kPa and 27 C, and 750 kJ/kg of heat is transferred to the air during the constant‐volume heat‐addition process. Taking in...
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