Is the isentropic work equal to the reversible work?

[Andres13_X]

Hello, I am working in a proyect and studying the exergy topic, a doubt came. Hope someone could help me to understand better the concept.

Imagine we have a turbine with the inlet and outlet temperatures and pressures and it is known it is bad isolated and so it releases 200kW of heat, and we want to figure out the maximum outlet work it can produce. To find the maximum work, I would have some ideas:

1) We consider the process is isentropic and so we set s1 = s2, and with the outlet pressure we find the isentropic entahpy. Then we perfome the energy balance to find the maximum work, considering the heat transferred.
2) same than the previous, but we ignore the heat that is comming out.
3) Perfom an exergy analysys to find the reversible work.

I am hesisating which approach to follow since for all of them I got different answers, in the next order: W2>W1>W3. In the thermodynamic book of Cengel, to find the maximum outlet work, they use an exergy balance, taking the exergy destroyed equal to 0 and ignoring the heat transferred. But I don't know why this answer is different than the method 1 or 2 I said. That's why my doubt is if reversible work is the same than isentropic work. Thank you!

 

[chicopee]

The isentropic expansion process is ideal work. At the outlet, entropy will have increase and to find the entropy of the outlet steam, you have to know the outlet pressure and the outlet enthalpy. The actual work will then be the difference in enthalpy between the inlet and outlet. The efficiency will then be a ration between the enthalpy difference during the actual work to the difference in enthalpy during the ideal work. See the attachment for clarification.

 

[chicopee]

One more thing, You'll either will need steam tables or a Mollier diagram to figure out the enthalpy and entropy values at the inlet and outlet of the steam turbine.

 

[chicopee]

Rereading your OP I failed to properly answer your question. The answer to your question in the title is YES because you have to look at the definition of entropy being dS=> dQ*/T whereby dQ* is a partial integral(since I can not make the greek symbol delta). When dS=0 then we have an adiabatic condition in the process and in my attachment of my first response such process would going from point 1 to 2 and that represent ideal reversible work.
To get your head around Clausius' work on entropy takes some doing at least it was for me when I first got into the realm entropy but with a little of perseverance you'll appreciate Clausius' insight in this subject and I personally can only marvel at his work.