Adiabatic expansion of a saturated vapor mix: Δx, ΔS

[james_a]

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=(v_avg-v_f)/(v_g-v_f)
u_avg=u_f+xu_fg
s_avg=s_f+xs_fg

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 , U_{2, then} u₂ and use the equation u_avg=u_f+xu_fg to find the final value for x, the quality.

relevant values from the steam tables:
@ 200C:
u_f=850.65 kJ/kg
u_fg=1744.7 kJ/kg
v_g=0.13736 m³/kg
v_f=0.001157 m³/kg

@ 50C:
u_f=209.32 kJ/kg
u_fg=2234.2 kJ/kg

x=0.5=(v_avg-v_f)/(v_g-v_f)
v_avg=x(v_g-v_f)+v_f=0.06926 m³/kg
V/v_avg=m=0.7219 kg

u_avg=u_f+xu_fg=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
u_avg=u_f+xu_fg
(u_avg-u_f)/u_fg=(1030.42-209.32)/2231.2=0.368=X₂

b)
On to entropy:
s_avg=s_f+xs_fg
Δs=s_{avg 2}+s_{avg 1}
ΔS=mΔs

from the steam tables:
@ 200C
s_f=2.3309 kJ/kg*K
s_fg=4.1014
@ 50C
s_f=0.7038 kJ/kg*K
s_fg=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
s_f1-0.5s_fg1=s_f2-xs_fg2 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?

 

[Chestermiller]

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

 

[james_a]

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.

 

[Chestermiller]

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

 

[james_a]

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_g@200C=0.12721 m³/kg, recalculating v_avg@200C and m using the same method as shown above, I came out with:
v_avg@200C = 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 s_f1-0.5s_fg1=s_f2-xs_fg2
Q=0
-W=ΔU
ΔU=mΔu=m((u_f2+xu_fg2)-(u_f1+xu_fg1))=-308.92kJ
W=309kJ

 

[Chestermiller]

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_g@200C=0.12721 m³/kg, recalculating v_avg@200C and m using the same method as shown above, I came out with:
v_avg@200C = 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 s_f1-0.5s_fg1=s_f2-xs_fg2
Q=0
-W=ΔU
ΔU=mΔu=m((u_f2+xu_fg2)-(u_f1+xu_fg1))=-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

 

[james_a]

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.