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Unformatted text preview: ME320 Thermodynamics Homework set #7 Due: Wednesday 11/05/03 at the start of class 1. For a cycle, is the net work necessarily zero? For what kind of systems will this be the case? Solution: No. This is the case for adiabatic systems only. 2. Can a steadyflow system involve boundary work? Solution: No, there’s no boundary work for a steadyflow process. 3. 479 (Textbook) p 1 = Steam mass flow rate = 12 kg/s 10 MPa T 1 = 450 C p 2 = 10 kPa x 2 = 0.92 V 2 = 50 m/s o V 1 = 80 m/s Solution: This is a steadyflow process since there is no change with time. Po tential energy changes are negligible. The device is adiabatic and thus heat transfer is negligible too. From the steam tables (Table A4 through A6) For P 1 = 10 MPa and T 1 = 450 ◦ C, ν 1 = 0 . 02975 m 3 /kg and h 1 = 3240 . 9 kJ/kg and For P 2 = 10 kPa and x 2 = 0 . 92, h 2 = h f + x 2 h fg = 191 . 83 + 0 . 92 × 2392 . 8 = 2393 . 2 kJ/kg (a) The change in kinetic energy is determined from Δ ke = V 2 2 V 2 1 2 = (50 2 80 2...
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 Spring '08
 Shen
 Energy, 2 kg, 50 m/s, 20 kJ, 90 kg, 80 m/s

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