five kilograms of water initially at 4000 kPA, 200 degrres are contained in a pi
ID: 1860582 • Letter: F
Question
five kilograms of water initially at 4000 kPA, 200 degrres are contained in a piston-cylinder device. A variable force and appropriate energy transfers are applied to the piston so that the water temperature remains constant as wthe water expands to final volumesof .49m^3 and 1.7606m^3 in two processes. sketch the processes on a T-v diagram relative to the saturation lines and determine the final pressure, the total volume change, and the total internal energy change of the water during these processes.
Explanation / Answer
From steam tables, at 4000 kPa and 200 deg C, we get, v1 = 0.00115 m^3/kg and u1 = 849 kJ/kg and quality = subcooled liquid.
Hence, inital volume = 5*0.00115 = 0.00575 m^3
We have v2 = 0.49/5 = 0.098 m^3/kg
From steam table at T2 = T1 = 200 deg C and v2 = 0.098 m^3/kg, we get P2 = 1550 kPa and quality x2 = 0.768, u2 = 2190 kJ/kg
We have v3 = 1.7606/5 = 0.35212 m^3/kg
From steam table at T3 = T1 = 200 deg C and v3 = 0.35212 m^3/kg, we get P3 = 600 kPa and quality x3 = superheated steam, u3 = 2640 kJ/kg
Total volume change in process 1 is = 0.49 - 0.00575 = 0.48425 m^3
Total volume change in process 2 is = 1.7606 - 0.00575 = 1.75485 m^3
Total internal energy change in process 1 = m*(u2 - u1) = 5*(2190 - 849) = 6705 kJ
Total internal energy change in process 2 = m*(u3 - u1) = 5*(2640 - 849) = 8955 kJ
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