641_Physics ProblemsTechnical Physics

# 641_Physics ProblemsTechnical Physics - Chapter 22 P22.26...

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Chapter 22 643 P22.26 COP COP Carnot cycle = 0100 . or Q W Q W hh = F H G I K J = F H G I K J 0 100 0 100 1 .. Carnot cycle Carnot efficiency Q W T TT h h hc = F H G I K J = F H G I K J = 293 268 117 . K 293 K K FIG. P22.26 Thus, . joules of energy enter the room by heat for each joule of work done. P22.27 COP Carnot refrig a f == = = T T Q W c c 400 289 00138 . . ∴= W 72 2 . J per 1 J energy removed by heat. P22.28 A Carnot refrigerator runs on minimum power. For it: Q T Q T h h c c = so Qt T T h h c c = . Solving part (b) first: (b) Q t Q t T T h c = F H G I K J = F H G I K J F H G I K J = 800 298 873 10 1 243 6 . MJ h K 273 K Jh h 3 600 s kW bg ej (a) W t Q t Q t =−= × = 800 10 3600 204 6 . . kW ±sh W P22.29 e W Q h 0350 . WQ h = . QWQ =+ QQ ch = 0650 . COP refrigerator = Q W Q Q h 186 . . . *P22.30 To have the same efficiencies as engines, 11 −= T T T T cp hp cr hr the pump and refrigerator must operate
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## This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

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