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656_Physics ProblemsTechnical Physics

# 656_Physics ProblemsTechnical Physics - 658 Heat Engines...

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658 Heat Engines, Entropy, and the Second Law of Thermodynamics P22.65 At point A , PV nRT i i i = and n = 1 00 . mol At point B , 3 PV nRT i i B = so T T B i = 3 At point C , 3 2 P V nRT i i C b gb g = and T T C i = 6 At point D , P V nRT i i D 2 b g = so T T D i = 2 The heat for each step in the cycle is found using C R V = 3 2 and C R P = 5 2 : Q nC T T nRT Q nC T T nRT Q nC T T nRT Q nC T T nRT AB V i i i BC P i i i CD V i i i DA P i i i = = = = = = − = = − 3 3 6 3 7 50 2 6 6 2 2 50 b g b g b g b g . . FIG. P22.65 (a) Therefore, Q Q Q Q nRT h AB BC i entering = = + = 10 5 . (b) Q Q Q Q nRT c CD DA i leaving = = + = 8 50 . (c) Actual efficiency, e Q Q Q h c h = = 0 190 . (d) Carnot efficiency, e T T T T c c h i i = = = 1 1 6 0 833 . *P22.66 S dQ T nC dT T nC T dT nC T nC T T nC T T i f P i f P i f P T T P f i P f i i f = = = = = = F H G I K J z z z 1 ln ln ln ln d i S nC PV nR nR PV nC P f i P = F H G I K J = ln ln3 *P22.67 (a) The ideal gas at constant temperature keeps constant internal energy. As it puts out energy by work in expanding it must take in an equal amount of energy by heat. Thus its entropy
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