# 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? Commented Nov 25, 2018 at 16:41
• Also, do you think that enthalpy only applies to isobaric processes? Commented Nov 25, 2018 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 :). Commented Nov 26, 2018 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. Commented Nov 26, 2018 at 1:08

It requires too extensive a derivation if you are unfamiliar with the open system version of the first law of thermodynamics for a system operating at steady state. So your first step is to go back to your textbook and get an understanding of this derivation. You said that you want to understand the fundamentals and don't want to memorize stuff. Perfect. For an engineer, understanding the relationship between the closed system version of the first law and the open system version (and how to apply the latter) is vital (in my judgment), particularly when dealing with power cycles such as in your present problem.

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.

• I think I've found the answer to my problem just as you've given it. I was comparing the Brayton cycle with the Otto and Diesel cycles, and I seem to have overlooked this "pushing energy" because the former cycle is of course having a flow rate. That said, I will definitely go through the text again. Thank you. Commented Nov 27, 2018 at 0:15

Enthalpy $$H$$ is defined as $$U + PV$$ where $$U$$ is internal energy, $$P$$ is pressure, and $$V$$ is volume. The specific enthalpy (per unit mass) is $$h = u + Pv$$ where $$u$$ is the internal energy per unit mass and $$v$$ is the specific volume (inverse density).

Enthalpy is of special interest for an open thermodynamic system, defined as one where mass can enter or leave the system. In addition to changes in energy within the system due to heat and work added to (taken from) the system, the energy within the system can also change due to mass entering (exiting) the system. (Heat and work do not consider mass transfer.)

The energy added to (or taken from) the system due to mass flowing into (or out of) the open system, is accounted for by the enthalpy entering (or exiting) the system. A mass $$\Delta m$$ entering (exiting) the system adds (removes) energy $$h \Delta m$$ to (from) the system. The $$u \Delta m$$ part of $$h \Delta m$$ is the internal energy of $$\Delta m$$. The other part of $$h \Delta m$$ is $$pv\Delta m$$, sometimes called the flow energy, and it is the work done by (against) the surroundings to push $$\Delta m$$ into (out of) the system.

For an open system where changes in kinetic energy and changes in gravitational potential energy are small, $$Q - W + H_{in} = \Delta U - H_{out}$$, where $$Q$$ is the heat added to the system, $$W$$ is work done by the system, $$\Delta U$$ is the change in the internal energy in the system, and $$H_{in} \enspace and \enspace H_{out}$$ are the total enthalpy entering and exiting the system, respectively. For a closed system (no mass transfer into or out of the system), $$Q - W = \Delta U$$. Note the consideration of the energy associated with mass entering (exiting) the open system using enthalpy.

Specific applications such as for a compressor or for a turbine, are applications of $$Q - W + H_{in} = \Delta U - H_{out}$$ with the appropriate conditions applied.

I find the definitions and developments in an old textbook, Themodynamics by Obert, to be especially good for gaining a physical understanding of basic thermodynamics.