Steam at 20 bars is in the saturated vapor state (call this state 1) and contain

Question

Steam at 20 bars is in the saturated vapor state (call this state 1) and contained in a pistoncylinder device with a volume of 0.03 m
3
. Assume the steam is cooled at constant volume
(i.e. the piston is held fixed in place) until the temperature reaches 200 C (call this state 2).
Then the steam is expanded isothermally until its volume is twice the initial value (state 3).
(a) Determine the pressures at state 2 and 3. ans. 15.5 bar, ~10 bar
(b) Determine the change in specific internal energy, u, for each of the two processes. ans. -
389 kJ/kg, ~21.6 kJ/kg.

Explanation / Answer

PV = nr T considering 1 mole of gas so , 20/T1 = p2 / T2 473 * 20 = T1*P2 T1 = 610k so p2 = 15.5 bar... p1 v1 =p3 v3 20 *0.03 = 2 *0.03 p3 p3 = 10 bar...

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