# Specific Exergy vs Specific Flow Exergy

1. May 22, 2013

### Suvat99

I'm having some difficulty understanding exactly what the difference between the definitions of these values are. As I understand it, in terms of solving given problems, you have to use the equation for specific exergy (NOT flow) when solving a problem that isn't a "flow" problem (ie: otto cycle, diesel cycle). However, I don't really understand the difference conceptually.

My understanding is: exergy of a state refers to the maximum theoretical energy that can be extracted from it usefully in a given environment, where
e = (u-u0) + p0(v - v0) - T0(s-s0) + (V^2)/2 + gz
Meaning what you can get from it is its internal energy above the normal, the "potential" energy it has from displacing the atmosphere due to expansion, a loss based on its entropy, and its KE + GPE.

For flow exergy, the equation is much the same as above, only the (u-u0)+p0(v-v0) is replaced with (h-h0). I thought these were the same thing...? When, mathematically, would these two formulae give different answers?

I've been given an explanation based on a piston engine having to push against the atmosphere as stuff in it changes, while (say) a rankine cycle is all self contained and never acts to displace the atmosphere. This seems like it's the right concept to explore, but I don't see how to apply it to understanding two equations which mathematically look identical to me.

2. Apr 27, 2017

### ScareCrow271828

Flow exergy just means that exergy is flowing in or out. It might just be more relevant to substitute h for u+pv because that could be more relevant for an exergy flow analysis. Not sure, they are mathematically equivalent though, its just a matter of definition.

Last edited: Apr 27, 2017
3. Apr 27, 2017

### Staff: Mentor

Are you familiar with the comparison between the closed system version of the first law and the open system (control volume) version of the first law?