A saturated water vapor mix in a 0.05m3 piston-cylinder at 200C with a quality of 0.5 expands adiabatically, producing 500kJ work with a final temperature of 50C. Find:
a) final quality
b) change in entropy[/B]
The Attempt at a Solution
For an adiabatic process Q=0, so -W=ΔU=-500kJ
If I can used the starting parameters to find u1 and the total mass, I can find U1 , U2, then u2 and use the equation uavg=uf+xufg to find the final value for x, the quality.
relevant values from the steam tables:
On to entropy:
Δs=savg 2+savg 1
from the steam tables:
So this was the first red flag for me. I thought ΔS is always supposed to be greater than 0. Then the second realization hit me, isn't ΔS 0 for an adiabatic process? But then, if ΔS=0, I solved for x
sf1-0.5sfg1=sf2-xsfg2 and came out with x=0.499, so essentially no Δs means no Δx. So is final x 0.368 or 0.5?
I've checked over and over my calculations in part a. I've quadruple checked to make sure the values were correctly copied from the tables. I've redone the calculations to make sure I copied the values correctly into my calculator. Is some part of the process in part a incorrect, do the equations I used not apply in this situation for some reason?