How is enthalpy change equal to work input/output in an isentropic process?

I'm an engineering student, and I'm fascinated by thermodynamics. I'm taking a module for heat engines, and I was recently looking at an example problem found in Thermodynamics: An Engineering Approach by Cengel and Boles (example 9-8). The author has taken the back work ratio by taking the enthalpy difference between each stage of the compressor and turbine. My question is, I don't see how a change in enthalpy can be equal to work input/output in the isentropic compressor and turbine of a Brayton cycle. Isn't enthalpy only equal to heat added if the process is isobaric? In an isentropic process, the enthalpy change will be equal to the expansion work plus the pressure increase, right? Thanks in advance!

• This is related to the open system (control volume) version for the first law of thermodynamics for a steady flow process. As an engineer, you should be familiar with this. Has this not been covered in your thermodynamics course? – Chet Miller Nov 25 '18 at 16:41
• Also, do you think that enthalpy only applies to isobaric processes? – Chet Miller Nov 25 '18 at 16:46
• Unfortunately not, we've been left to do all the theory by ourselves, the lecturer just does the questions without any explanation whatsoever. I never memorize stuff, I need to understand what's going on if I'm to get through the exams, and I like thermodynamics so I want to learn this anyway :). – Thiraj Wegala Nov 26 '18 at 0:59
• I know enthalpy can be used to find heat input in constant pressure heat addition processes (furnace, intercooler, reheater, etc). But how is enthalpy difference equal to work output in an isentropic process? Why not use internal energy instead, since isentropic processes are adiabatic and the dQ is zero? Mathematical answers will be appreciated, I just want some proof for this. – Thiraj Wegala Nov 26 '18 at 1:08

That said, for a compressor or turbine operating adiabatically at steady state, the open system version of the first law of thermodynamics tells us that: $$\dot{m}\Delta h=W_S$$where $$\dot{m}$$ is the mass flow rate through the compressor or turbine, $$\Delta h$$ is the change in enthalpy per unit mass of the working fluid in passing through the device, and $$W_S$$ is the so-called "shaft work." This is not the total amount of work, but only the part of the work delivered to- or derived from the rotating shaft. Not included in $$W_S$$ is the work required to push working fluid into- and out of the device.