640_Physics ProblemsTechnical Physics

640_Physics ProblemsTechnical Physics - 642 Heat Engines,...

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642 Heat Engines, Entropy, and the Second Law of Thermodynamics For the isothermal process CD : ∆∆ En C T V int == 0 and QW n R T V V D C =− = F H G I K J =⋅ F H G I K J ln . ln . . . 234 549 15 0 24 0 502 mol 8.314 J mol K K kJ bg a f . Finally, for the adiabatic process DA : Q = 0 C T T VA D int mol 3 2 J mol K K kJ = L N M O Q P −= + b g af 2 34 8 314 720 549 4 98 .. . and WQE =− + = + int kJ kJ 04 9 8 4 9 8 . Process Q (kJ) W (kJ) E int (kJ) AB +6.58 –6.58 0 BC 0 –4.98 –4.98 –5.02 +5.02 0 0 +4.98 +4.98 ABCDA +1.56 –1.56 0 The work done by the engine is the negative of the work input. The output work W eng is given by the work column in the table with all signs reversed. (c) e W Q W Q h ABCD eng kJ 6.58 kJ 156 0237 . . or 23 7% . e T T c c h =− =− = 11 549 720 . or 23 7% . P22.23 COP refrig a f = T T c 270 30 0 900
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