- #1

leafjerky

- 43

- 7

## Homework Statement

Okay, so these are usually pretty easy for me to understand, but this one doesn't make sense.

10 kg of water in a piston cylinder arrangement exists as saturated liquid/vapor at 100 kPa, with a quality of 50%. It is now heated so the volume triples. The mass of the piston is such that a cylinder pressure of 200 kPa will float it. Find the final temperature and the heat transfer in the process.

## Homework Equations

v = V/m

v - specific volume

V - volume

m - mass

y = y

_{f}+xy

_{fg}

## The Attempt at a Solution

State 1:[/B]

m

_{1}= m

_{2}= m = 10kg

x

_{1 }= .5 ---> 2-phase mixture

P

_{1}= 100 kPa = 1 bar

v

_{1}= .0010432 m

^{3}/kg + (.5)(1.694 - .0010432)

v

_{1}= .8475 m

^{3}/kg

**State 2:**

v

_{2}= 3*v1 = 2.5425 m

^{3}/kg

P

_{2}= 200 kPa --> pressure is constant from this point on

Solutions I have found online say that the final temp is in the 800's. How is that possible? I went into my steam tables (Fundamentals of Engineering Thermodynamics, Moran) and I couldn't figure out the state. I assume it is a superheated vapour. Is this right? If so, I go to the tables and the temp could probably be interpolated between 1.5 bar and 3 bar, but even then it's not close to 800. I believe the answer is 827 or 829 degrees C. Thanks for any help.