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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 Pv 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 A1, 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) Pv diagram is
P 1 2 v (c) For a polytropic process: For ideal gas
. . 513.58 6203 / , 25753 / (d) First law: From table A19, So the total heat transfer is: 2. 513.58 2
. 1735.5 1.74 [30points] A massloaded piston/cylinder, as shown in the figure, containing R134a 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 R134a 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 R134a 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 highpressure 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 twostage 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. Twostate 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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This note was uploaded on 04/20/2008 for the course ME 235 taught by Professor Borgnakke during the Winter '07 term at University of Michigan.
 Winter '07
 Borgnakke

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