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**My Question**- Steam at 200 psi and 600 F flows through a turbine operating adiabatically and exits at atmospheric

pressure. For every kilogram of steam flowing through the turbine there are 150 Btu of shaft work

delivered.

(a) What is the final condition of the exit steam?

(b) What is the maximum amount of work that could be obtained per kg of steam for adiabatic operation between these two pressures?

**Relevant Equations**

1 psi = 6894.76 Pa

T2^k / P2^(k-1) = T1^k / P1^(k-1)

(T2/T1)^k = (P2/P1)^(k-1)

W = n R (T1 - T2) + m L

**My Approach**

(a) By final condition, do we mean phase (solid/liquid/gas)?

P1 = 200 psi = 200 x 6894.76 Pa = 1378952 Pa

P2 = atmospheric pressure = 101325 Pa

T1 = 600 F = (5/9) * (600 - 32) C = 315.56 C or 588.56 K

T2^k / P2^(k-1) = T1^k / P1^(k-1)

(T2/T1)^k = (P2/P1)^(k-1)

For steam, k = 1.3

(T2/588.56)^1.3 = (101325/1378952)^0.3

(T2/588.56)^1.3 = 0.45693

T2/588.56 = 0.54745

T2 = 322.2 K = 49.2 C

This shows that the the steam will convert into liquid. But the problem says that steam exits. This means whole of steam has not condensed into liquid.

*But I am not sure how to find how much of steam will condense.*

(b) Mass of steam = 1 kg = 1000 g

Number of moles of steam n = 1000/18

R = 8.314 in SI units

Latent heat of vaporization of water L = 540 cal/g = 540 * 4.186 J/g

Final pressure is 1 atm. Steam condenses at 100 C (=373 K) at 1 atm

Therefore T2 = 373 K

W = n R (T1 - T2) + m L

W = (1000/18) mol * 8.314 J.mol^-1.K^-1 * (588.56 - 373) + (1000 g) * (540 cal/g) * (4.186 J/cal)

W = 2.36 x 10^6 J or 2.36 MJ

*But I am not sure if I have solved it correctly. If yes, then is it possible to make the solution shorter?*

Note: This is not for my homework. I finished student life some years back. But sometimes I read some things out of my own hobby and collect problems from different sources. I was reading this topic and got stuck in this problem.

So I need help. Thank you.