Exergy far from thermodynamic equilibrium

Up to this point we have implicitly assumed that a new non-equilibrium state is located not far from thermodynamic equilibrium. Moreover, the transition to the new state was quasi-stationary, so that the system was at a dynamic equilibrium with the environment. In this case, the infinitesimal change of exergy, d(Ex), was very close to the difference of exergies between the current and initial states, iSAmaxl = S(Ex). However, generally speaking it is possible for a system to be moved far from thermodynamic equilibrium where this closeness is not present anymore. How can exergy be calculated in this case? Here we shall follow the method used in the work (Svirezhev, 2001b).

Let the transition be chemical, i.e. it occurs in the field of chemical potentials when work is done against them by means of the change in the number of particles. We assume that the infinitesimal change of exergy described by Eq. (3.3) takes place in an infinitesimal time dt, so that the environment is not able to change. As is evident from equality dN0 = -dN, = -(dN,/dt)dt, if derivative (dNi/dt) is finite then for small dt the value of dNi0 can be also small. This condition is sufficient for relinquishing the previous assumption about the "great" environment, in particular, considering that several Ni0 are small. Thus, we can fully give up the above-mentioned local time condition; to do this we must replace differentials in Eq. (3.3) by the corresponding time derivatives:

Keeping in mind the definition of chemical potentials Mi = Mi(0) + RT ln N,, i = 1,..., n where Ni can be considered as molar concentrations of corresponding chemical substances, and R is the gas constant, we re-write Eq. (4.1) as (T = T0)

There is one more argument on behalf of this replacement. Since the order of change for the chemical potential mf with respect to Nf is logarithmic, should the number Nf change, the value of the chemical potential changes significantly more slowly.

By integrating both sides of Eq. (4.2) with respect to time and taking into account that Ex(t0) = 0 we get p ™ PT f Vi Ni(i) dNi a. J?T Ex(t) = RT0 y_ lfl ——~ dt = RT0

Since the gas constant is measured in J/mol K, the number of particles has to be measured in moles. Then exergy is measured in energy units (joules). If we wish—as before—to use the number of particles we must write Boltzmann's constant k instead of R. As a rule, later on we shall omit the factors RT0, and kT0, remembering them only when it is necessary.

One can see that Ex(t) > 0 for any Ni > 0, except Ni = N0, i = 1,..., n, when Ex ; 0. Also, the value of exergy is determined only for the current and initial states (for this reason the latter will be termed a reference state) and does not depend on characteristics of this transition.

If IdNil << NO, i.e. we do not go far from thermodynamic equilibrium, then Eq. (4.2) is written as

Was this article helpful?

0 0

Post a comment