3
$\begingroup$

My textbook Fundamentals of engineering thermodynamics, Moran and Shapiro states:

The exergy is the maximum theoretical work obtainable for an overall system consisting of a system and the environment as the system comes into equilibrium with the environment (passes to the dead state).

The exergy of a system (which I'll call $\chi$), at a specified state is given by the expression: $$ \chi = (U-U_o)+p_o(V-V_o)-T_o(S-S_o) +KE +PE$$ Where $S_o, U_o$ etc. denote the dead states of a system. An Exergy change is : $$\chi_2 - \chi_1 = (U_2-U_1)+p_o(V_2-V_1)-T_o(S_2-S_1) +(K.E._2-K.E._1) +(P.E._2-P.E._1)-^?T_o\sigma_o$$

Where $\sigma_o$ is the "entropy production" and I'm not sure if it belongs there or not. The equation is clearly related to the Gibbs free energy, but the gibbs free energy is for an isothermal and isobaric process (natural variables are T and P). It seems obvious to me that if a system is passing to the dead state it's temperature and pressure could change, for example: as a cup of coffee comes into equilibrium with the environment it's temperature will eventually be the same as the temperature of the "reservoir".(Note that these equations can also be expressed as rates)

I haven't really been satisfied with the depth of the physics in my textbook and have had trouble finding good conceptual and mathematical explanations of exergy and exergy destruction.

Can someone present a thermodynamic or statistical-mechanical description of exergy and exergy destruction? The intuitive mechanical engineer explanation I've come across is that it is about the economics of energy use (for example: heating your house with a radiator would be a poor use of exergy), but I'd like something more mathematically rigorous if possible. Online references or a generalized example would be great too.

$\endgroup$
2
$\begingroup$

If you call $ \chi $ the exergy (availability) then $ \chi = U + p_o V - T_o S $ where $p_o, T_o$ are the pressure and temperature of the environment (and are assumed to be constant). To find the maximum amount of useful work that can be extracted form the system it is sufficient to analyze reversible processes only so that $ dU=TdS-pdV $ and then the exergy change for reversible process is $$ \chi_2 -\chi_1 = U_2 -U_1 + p_o (V_2 - V_1) - T_o (S_2 - S_1) $$. from which we have $$ d\chi = (T-T_o)dS -(p-p_o)dV $$ Let me quote here Pippard's magnificent book, page 101, in which Pippard calls $A$ the availability, your exergy $\chi$:

"If the change is reversible, - TdS is the heat extracted from the system, which may be most usefully employed by transferring it to an ideal Carnot engine working between the temperatures T and To. The work done by the engine will then be - (T - To) dS. Also if the system be allowed to expand reversibly by an amount dV, an additional pressure (P - Po) must be applied externally, and the useful work of expansion is that performed against the additional pressure; the remaining work done by the system is not useful, being expended in pushing back the atmosphere (or any other passive source of the external pressure Po). Hence (P -Po)d V is the useful work, and the two work terms add to give -dA. Thus the decrease in A is a measure of the maximum useful work available. This interpretation of A leads us to expect that A takes a minimum value when T = To and P = Po, for then the system is in equilibrium with its surroundings and cannot act as a source of useful work."

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.