Adiabatic expansion of a saturated vapor mix: Δx, ΔS

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Homework Statement



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]

Homework Equations



Q-W=ΔU
x=(vavg-vf)/(vg-vf)
uavg=uf+xufg
savg=sf+xsfg

The Attempt at a Solution



A)[/B]
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:
@ 200C:
uf=850.65 kJ/kg
ufg=1744.7 kJ/kg
vg=0.13736 m3/kg
vf=0.001157 m3/kg

@ 50C:
uf=209.32 kJ/kg
ufg=2234.2 kJ/kg

x=0.5=(vavg-vf)/(vg-vf)
vavg=x(vg-vf)+vf=0.06926 m3/kg
V/vavg=m=0.7219 kg

uavg=uf+xufg=850.65+0.5(1744.7)=1723 kJ/kg

U1=mu1=0.7219*1723=1243.86 kJ

U2-U1=ΔU=-500kJ
ΔU+U1=U2=-500+1243.86=743.86kJ

U2/m=u2=1030.42 kJ/kg
uavg=uf+xufg
(uavg-uf)/ufg=(1030.42-209.32)/2231.2=0.368=X2

b)
On to entropy:

savg=sf+xsfg
Δs=savg 2+savg 1
ΔS=mΔs

from the steam tables:
@ 200C
sf=2.3309 kJ/kg*K
sfg=4.1014
@ 50C
sf=0.7038 kJ/kg*K
sfg=7.3725

ΔS=mΔs=0.7219((0.7038+0.368*7.3725)-(2.3309+0.5(4.1014)))=-0.69

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?
 
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Answers and Replies

  • #2
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4,678
I've checked over your calculations, and they look correct to me. Certainly, in line with the Clausius inequality, ΔS should be ≥ 0 for an adiabatic process on a closed system. So there must be something wrong with the data in the problem statement, either the 500 kJ work or the 50 C final temperature.

Chet
 
  • #3
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I've checked over your calculations, and they look correct to me. Certainly, in line with the Clausius inequality, ΔS should be ≥ 0 for an adiabatic process on a closed system. So there must be something wrong with the data in the problem statement, either the 500 kJ work or the 50 C final temperature.

Chet
Thanks for looking it over! I was really wondering if I was missing something big... or losing my mind.
 
  • #4
21,163
4,678
Thanks for looking it over! I was really wondering if I was missing something big... or losing my mind.
I think it would be interesting to figure out how much work would be done if the process were adiabatic and reversible (with the final temperature still 50 C) to compare with the 500 kJ.

Incidentally, I checked my steam tables for the specific volume of the vapor at 200 C, and it showed 0.127 rather than 0.137. Maybe one of our tables has a typo.

Chet
 
  • #5
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I think it would be interesting to figure out how much work would be done if the process were adiabatic and reversible (with the final temperature still 50 C) to compare with the 500 kJ.

Incidentally, I checked my steam tables for the specific volume of the vapor at 200 C, and it showed 0.127 rather than 0.137. Maybe one of our tables has a typo.

Chet

Strange. I double checked and the steam table in my physical textbook - which I was using last night - states 0.137, while my digital copy of the textbook, which is a few editions newer states 0.127.

That is an interesting question. I think I know how to go about it.

Going with v[email protected]=0.12721 m3/kg, recalculating v[email protected] and m using the same method as shown above, I came out with:
v[email protected] = 0.06418 m3/kg
m=0.779 kg

Now, if the expansion is reversible and adiabatic, Δs=0 and Δx=0, as was discovered when solving for final quality in sf1-0.5sfg1=sf2-xsfg2
Q=0
-W=ΔU
ΔU=mΔu=m((uf2+xufg2)-(uf1+xufg1))=-308.92kJ
W=309kJ
 
  • #6
21,163
4,678
Strange. I double checked and the steam table in my physical textbook - which I was using last night - states 0.137, while my digital copy of the textbook, which is a few editions newer states 0.127.

That is an interesting question. I think I know how to go about it.

Going with v[email protected]=0.12721 m3/kg, recalculating v[email protected] and m using the same method as shown above, I came out with:
v[email protected] = 0.06418 m3/kg
m=0.779 kg

Now, if the expansion is reversible and adiabatic, Δs=0 and Δx=0, as was discovered when solving for final quality in sf1-0.5sfg1=sf2-xsfg2
Q=0
-W=ΔU
ΔU=mΔu=m((uf2+xufg2)-(uf1+xufg1))=-308.92kJ
W=309kJ
For this particular situation, Δx for an adiabatic reversible expansion to 50 C just happened to come out close to zero. But that would not generally be the case.

You showed that, for the reversible adiabatic expansion, the amount of work was 309, rather than 500. For an irreversible adiabatic expansion, it would be less than 309.

Chet
 
  • #7
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For this particular situation, Δx for an adiabatic reversible expansion to 50 C just happened to come out close to zero. But that would not generally be the case.

You showed that, for the reversible adiabatic expansion, the amount of work was 309, rather than 500. For an irreversible adiabatic expansion, it would be less than 309.

Chet
Thank you for clarifying and pointing that out. I would have went along after today assuming Δx was always nearly 0 for reversible adiabatic processes.
 

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