How to find maximum work done in an adiabatic process?

[visharad]

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.

 

[rock.freak667]

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

Using your steam tables you can get the enthalpy h₁ @ 200 psi, 600F and you are given the shaft work. Applying the steady state equation you will see that

w=h₁-h₂ so you can solve for h₂.

and you know that h₂=h_f+x₂h_fg (these are at atmospheric pressure as stated).

The maximum work will require what of all the steam? (i.e. what should x be)

 

[RTW69]

Use steam tables as mentioned or better yet a Mollier Diagram (visual diagram of the steam tables) to locate the initial conditions which are clearly super heated steam.You should be able to determine h1 and s1 knowing T1 and p1. From first law -W(turbine)= Mass(Dot)(h2-h1). Calculate new h2. Assuming adiabatic, reversible s1=s2 We don't know if the final condition is saturated of superheated. Knowing s1=s2 and h2 on your Mollier diagram you can see that you are still in the super heated region and you can determine the final T2 and p2. The maximum work output of the turbine will occur when you can further condense the steam to a saturated gas at p2. Find h2 at p2 when steam is a saturated gas and calculate turbine output -max.

 

[RTW69]

After reconsidering my answer to part B it appears that the maximum work produced by the turbine occurs at the greatest enthalpy difference possible between these pressures. This would occur at the highest temperature the turbine could withstand at 200 psia and the comparable temperature at p2, assuming s1=s2.

 

[rwisz]

To find the maximum work done in an adiabatic process, we can use the first law of thermodynamics which states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system. In an adiabatic process, there is no heat transfer, so the change in internal energy is equal to the work done. Therefore, the maximum work done in an adiabatic process is equal to the change in internal energy.

In this problem, we are given the amount of work done per kg of steam (150 Btu/kg) and the initial and final pressures. We can use the ideal gas law to find the initial and final temperatures of the steam.

P1V1/T1 = P2V2/T2

Since the process is adiabatic, the volume remains constant, so we can cancel out V1 and V2. Rearranging the equation, we get:

T2 = (P2/P1) * T1

Substituting the values, we get:

T2 = (101325 Pa / 1378952 Pa) * 588.56 K = 43.29 K

This shows that the steam will condense into liquid at the final pressure. To find the maximum work done, we need to calculate the change in internal energy of the system. We can use the specific heat capacity of water to find the change in internal energy.

q = m * c * ΔT

Where q is the heat added, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. We know that q is equal to the work done, so we can write:

W = m * c * ΔT

Substituting the values, we get:

W = (1000 g) * (4.186 J/g.K) * (373 K - 588.56 K) = -2.37 MJ

Since the change in internal energy is equal to the work done, the maximum work done per kg of steam is equal to 2.37 MJ. This is the theoretical maximum and in real-life scenarios, the actual work done may be less due to factors such as friction and inefficiencies in the system.

In summary, to find the maximum work done in an adiabatic process, we need to calculate the change in internal energy of the system. This can be done by