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Unformatted text preview: Homework 2 Solutions Spring 2010 Required Problems Chapter 20 5. The efficiency is the work done by the engine, divided by the heat input to the engine from the burning of the gasoline. Both energy terms are expressed in terms of a rate. The gasoline provides the input energy, and the horsepower of the engine represents the output work. ( 29 4 H H 746W 1J s 25hp 1hp 1W 0.21 or 21% kcal 1gal km 1h 4186J 3.0 10 95 gal 38km h 3600s kcal W W t e Q Q t = = = = × 6. ( a ) For the net work done by the engine to be positive, the path must be carried out clockwise. Then the work done by process bc is positive, the work done by process ca is negative, and the work done by process ab is 0. From the shape of the graph, we see that bc ca . W W ( b )The efficiency of the engine is given by Eq. 201a. So we need to find the work done and the heat input. At first glance we might assume that we need to find the pressure, volume, and temperature at the three points on the graph. But as shown here, only the temperatures and the first law of thermodynamics are needed, along with ratios that are obtained from the ideal gas law. ab: ( 29 ( 29 3 ab ab int b a b a 2 ab 0 ; V W P V Q E nC T T nR T T = ∆ = = ∆ = = bc: ( 29 c c c int c b bc bc bc b a a b b b 0 ; ln ln ln V V V T E nC T T Q W nRT nRT nRT V V T ∆ = = = = = = ca: ( 29 ( 29 ( 29 a c c ca a a c a c a a a c a a a 1 1 nRT V T W P V V V V nRT nRT nR T T V V T = = = = = ( 29 ( 29 5 ca a c a c 2 P Q nC T T nR T T = = < P V b a c T b = T c ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 c c b b a c b a c b a b bc ca a a a c c b 3 3 3 input ab bc 2 2 2 b a b b a b b a b a a a 3 2 ln ln ln ln ln ln 423K 423K ln 273K 423K 273K 0.0859 8.59% 423K 423K 273K 423K ln 273K T T T nRT nR T T T T T T T T W W W T T T e T T T Q Q Q nR T T nRT T T T T T T T T T + + + + = = = + + + + + = = + = = = Of course, individual values could have been found for the work and heat on each process, and used in the efficiency equation instead of referring everything to the temperatures. 29. ( a )Use the coefficient of performance and the heat that is to be removed ( 29 L Q to calculate the work done. The heat that is to be removed is the amount of heat released by cooling the water, freezing the water, and cooling the ice. We calculate that heat as a positive quantity. ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 2 L L ideal H L H O liquid fusion ice ice H L L H L L L 5 4 4 COP 0.40kg 4186J kg C 25C 3.33 10 J kg 2100J kg C 17C 42K 273K 17K 3.106 10 J 3.1 10 J Q T W T T mc T mL mc T T T Q T T W T T = = → ∆ + + ∆ = = ° ° + × + ° ° = = × ≈ × g g ( b )Now the compressor power ( 29 W t is given, and is to be related to the rate of removing heat from the freezer, L...
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This note was uploaded on 04/23/2010 for the course PHYSICS ?? taught by Professor Zettle during the Spring '08 term at Berkeley.
 Spring '08
 ZETTLE

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