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## Homework Statement

A saturated water vapor mix in a 0.05m

^{3}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 u

_{1}and the total mass, I can find U

_{1}, U

_{2, then }u

_{2}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 m

^{3}/kg

v

_{f}=0.001157 m

^{3}/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 m

^{3}/kg

V/v

_{avg}=m=0.7219 kg

u

_{avg}=u

_{f}+xu

_{fg}=850.65+0.5(1744.7)=1723 kJ/kg

U

_{1}=mu

_{1}=0.7219*1723=1243.86 kJ

U

_{2}-U

_{1}=ΔU=-500kJ

ΔU+U

_{1}=U

_{2}=-500+1243.86=743.86kJ

U

_{2}/m=u

_{2}=1030.42 kJ/kg

u

_{avg}=u

_{f}+xu

_{fg}

(u

_{avg}-u

_{f})/u

_{fg}=(1030.42-209.32)/2231.2=

**0.368=X**

b)

On to entropy:

_{2}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

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?

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?

Last edited: