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Unformatted text preview: 1 MAE 301: Engineering Thermodynamics Spring 2010 HOMEWORK #4 ( Solutions ) Problem 1. Problem 4.6 from the book. Solution: No work is done during the process 23 since the area under process line is zero. Then the work done is equal to the area under the process line 12: kJ 300 = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ⋅ − + = − + = = 3 3 1 2 2 1 out , m kPa 1 kJ 1 /kg 0.5)m kg)(1.0 (2 2 )kPa 500 (100 ) ( 2 Area v v m P P W b Problem 2. Problem 4.19 from the book. Solution: The process is quasiequilibrium, and Nitrogen can be considered as an ideal gas. The gas constant for nitrogen is R = 0.2968 kJ/kg.K (Table A2a). The boundary work for this polytropic process can be determined from kJ 89.0 1.4 1 300)K K)(360 kJ/kg kg)(0.2968 (2 1 ) ( 1 1 2 2 1 1 1 2 2 out , − = − − ⋅ = − − = − − = = ∫ n T T mR n P P d P W b V V V The negative sign indicates that work is done on the system (work input). Problem 3. Problem 4.25 from the book. Solution: The properties of air are R = 0.287 kJ/kg.K , k = 1.4 (Table A2a). For the isothermal expansion process: 3 1 1 m 01341 ....
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 Spring '08
 Hassan
 Thermodynamics, Energy, Potential Energy, kPa

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