Solution HW Ch 9

Solution HW Ch 9 - 1 ME 113 THERMODYNAMICS Spring 2008 Ch 9...

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Unformatted text preview: 1 ME 113: THERMODYNAMICS Spring 2008 Ch 9 HW Solution 2 9-38 An ideal Otto cycle with air as the working fluid has a compression ratio of 9.5. The highest pressure and temperature in the cycle, the amount of heat transferred, the thermal efficiency, and the mean effective pressure are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are c p = 1.005 kJ/kg·K, c v = 0.718 kJ/kg·K, R = 0.287 kJ/kg·K, and k = 1.4 (Table A-2). Analysis ( a ) Process 1-2: isentropic compression. ( ) ( ) ( ) ( ) kPa 2338 kPa 100 K 308 K 757.9 9.5 K 757.9 9.5 K 308 1 1 2 2 1 2 1 1 1 2 2 2 0.4 1 2 1 1 2 = = = → = = = = − P T T P T P T P T T k v v v v v v Process 3-4: isentropic expansion. ( ) ( ) K 1969 = = = − 0.4 1 3 4 4 3 9.5 K 800 k T T v v Process 2-3: v = constant heat addition. ( ) kPa 6072 = = = → = kPa 2338 K 757.9 K 1969 2 2 3 3 2 2 2 3 3 3 P T T P T P T P v v ( b ) ( ) ( ) ( ) ( ) kg 10 788 . 6 K 308 K /kg m kPa 0.287 m 0.0006 kPa 100 4 3 3 1 1 1 − × = ⋅ ⋅ = = RT P m V ( ) ( ) ( )( )( ) kJ 0.590 = − ⋅ × = − = − = − K 757.9 1969 K kJ/kg 0.718 kg 10 6.788 4 2 3 2 3 in T T mc u u m Q v (c) Process 4-1: v = constant heat rejection. ( ) ( )( )( ) kJ 0.2 40 K 308 800 K kJ/kg 0.718 kg 10 6.788 ) ( 4 1 4 1 4 out = − ⋅ × − = − = − = − T T mc u u m Q v kJ 0.350 240 . 590 ....
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This homework help was uploaded on 04/12/2008 for the course ME 113 taught by Professor Usaxena during the Spring '07 term at San Jose State.

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Solution HW Ch 9 - 1 ME 113 THERMODYNAMICS Spring 2008 Ch 9...

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