P21_031 - WT L / ( T H T L ) = mc ( T f T i ) + mL F . We...

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31. (a) If T H is the temperature of the high-temperature reservoir and T L is the temperature of the low- temperature reservoir, then the maximum efficiency of the engine is ε = T H T L T H = (800 + 40) K (800 + 273) K =0 . 78 . (b) The efficiency is deFned by ε = | W | / | Q H | ,where W is the work done by the engine and Q H is the heat input. W is positive. Over a complete cycle, Q H = W + | Q L | ,where Q L is the heat output, so ε = W/ ( W + | Q L | )and | Q L | = W [(1 ) 1]. Now ε =( T H T L ) /T H ,where T H is the temperature of the high-temperature heat reservoir and T L is the temperature of the low-temperature reservoir. Thus, 1 ε 1= T L T H T L and | Q L | = WT L T H T L . The heat output is used to melt ice at temperature T i = 40 C. The ice must be brought to 0 C, then melted, so | Q L | = mc ( T f T i )+ mL F ,whe re m is the mass of ice melted, T f is the melting temperature (0 C), c is the speciFc heat of ice, and L F
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Unformatted text preview: WT L / ( T H T L ) = mc ( T f T i ) + mL F . We dierentiate with respect to time and replace dW/dt with P , the power output of the engine, and obtain PT L / ( T H T L ) = ( dm/dt )[ c ( T f T i ) + L F ]. Thus, dm dt = PT L T H T L 1 c ( T f T i ) + L F . Now, P = 100 10 6 W, T L = 0 + 273 = 273 K, T H = 800 + 273 = 1073 K, T i = 40 + 273 = 233 K, T f = 0 + 273 = 273 K, c = 2220 J / kg K, and L F = 333 10 3 J / kg, so dm dt = (100 10 6 J / s)(273 K) 1073 K 273 K 1 (2220 J / kg K)(273 K 233 K) + 333 10 3 J / kg = 82 kg / s . We note that the engine is now operated between 0 C and 800 C....
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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