Specific Exergy vs Specific Flow Exergy

[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.

 

[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.

 

[Chestermiller]

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.

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?