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midterm1_solution_for%20publish

midterm1_solution_for%20publish - ME235 01 Thermodynamics...

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Unformatted text preview: ME235 01 Thermodynamics Winter 2008 Prof. Angela Violi Midterm 1 February 18, 2008 1. [30 points] 2kg Oxygen goes through a polytropic process, the final volume is 2 times of the initial volume, and the temperature increases from 25 to a final state of 1100 . Without assuming constant specific heat, (a) Find for the polytropic process 0.1886 (10pts) (b) Show the P-v diagram for the process (5pts) (c) Find the total work 513.58 (7pts) (d) Find the amount of heat added 1.74 (8pts) Solution: Oxygen can be assumed as ideal gas From Table A-1, 31.999 / (a) 5, 298 , 1100 ideal gas law gives So , 0.2598 / Polytropic process gives So the polytropic process index n is 1 1 0.1886 (b) P-v diagram is P 1 2 v (c) For a polytropic process: For ideal gas . . 513.58 6203 / , 25753 / (d) First law: From table A-19, So the total heat transfer is: 2. 513.58 2 . 1735.5 1.74 [30points] A mass-loaded piston/cylinder, as shown in the figure, containing R-134a at 8.9 , with a quality of 60% and volume of 100L, while at the stops V = 150L. The system is connected through a valve to a line flowing with saturated liquid R-134a at 0C, the valve is now open until a final inside pressure of 800 kPa is reached, at which point T = 400K. (a) Find the mass of R-134a that enters 0.67 (15pts) (b) Find the work in the process 20 (5pts) (c) Find the heat transfer in the process 762.5 (10pts) Solution: (a) State 1: 8.9 , x=60%, 100 0.1 So 400 0.000797 0.6 0.051201 . . / 0.000797 0.031 / 3.2258 400 127 0.0385 / 400 , the piston is at the stops at the final state, so 3.8961 0.67 0.7009 State 2: 800 , Linear interpolation: Since 800 150 0.15 . . / So the mass flow in is 3.8961 3.2258 0.6703 (Note: if averaged, 0.0382 / , and 3.9267kg, thus (b) Only when the pressure is constant is there work involved 400 0.15 0.1 20 (c) Transient process energy equation: ) State 1: 63.62 0.6 171.45 166.49 / State 2: Linear interpolation 337.341 / Inlet: 51.86 / @ So 3.8961 337.341 0.6703 51.86 762.4891 762.5 3. 3.2258 166.49 20 [40points] A proposal is made to use a geothermal supply of hot water to operate a steam turbine, as shown in the Figure. The high-pressure water at 1.5 MPa, 180C, is throttled into a flash evaporator chamber, which forms liquid and vapor at a lower pressure of 800 kPa. The liquid is discarded while the saturated vapor feeds a two-stage adiabatic turbine, 10% of the entering vapor mass is bled from the turbine at 300kPa and 150C, the remaining steam leaves the turbine at 10 kPa, 90% quality. If the total turbine shaft work is 1 MW (a) Find the mass flow rate that enters the turbine 2.6181 / (16pts) (b) Find the required mass flow rate of hot geothermal water 127.1 / (18pts) (c) Find the quality after the separation of phases through flash evaporator 0.0206 (6pts) 1 Flash Evaporator 2 Sat. vap. High Pressure Turbine Low Pressure Turbine 5 Sat. liq. out. 3 4 Solution: (a) CV. Two-state turbine cont eq.: energy eq.: 10% @ 90% , 2768.3 kJ/kg 2761.2 kJ/kg 191.81 0.9 2392.1 2344.7 kJ/kg , 1 0.1 , 0.9 , . . 2.6181 / (b) CV. Throttling valve + flash evaporator cont eq.: energy eq.: @ 763.05 / @ @ . . . . . . . 720.87 / 2768.3 kJ/kg 127.0834 / 127.1 / (c) the quality after separation of phases is 0.0206 Or alternatively, (b) CV. Throttling valve + flash evaporator cont eq.: energy eq.: 763.05 / @ realizing that So the energy equation then becomes 1 @ @ @ @ @ 2047.5 kJ/kg 0.0206 / 127.1 / . . . . . Then the mass flow rate of hot geothermal water is 127.0944 ...
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