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## Homework Statement

There are expanded 0.90kg/s of steam at constant pressure from 3MPa and 70% quality to a final state. If the process is nonflow for which W = 121.3kJ/s, find (a) The final temperature, (b) Q, (c) the available part of Q for a sink temperature of to = 27 Celsius

- Answers : a) 282 celsius, b) 966 kJ/s, c) 414 kJ/s

## Homework Equations

Wnf = P(V2-V1)

Q = H2 - H1

but theres mass, so

Wnf = mP(V2-V1)

Q = m(H2 - H1)

## The Attempt at a Solution

I used the steam table for Sat. Pressure. And I got V1 = Vf1 + x(Vfg1) = 47.04095 x 10^-3 m^3/kg

And so,

Wnf = mP(V2-V1)

121.3 = (0.90)(3MPa)(V2 - 47.04095)

So i Got 91.96687593 x10^-3 m^3/kg

Checking it in table 3 from the steam table the temperature is way off the mark of the desired answer.

Is there another way around it? Please help me. The answers are written above ... below the problem...