Integrating Eq. (169) over the range of the whole tray volume, entropy generation P on a tray is:
P = jV adV = | ^ (_ —^ dV+\ v_vnt t —^ dV (170)
where the interfacial volume Vmerface = A • Ax, A is the vapor-liquid phase contact area(i.e. bubble area), Ax is the total height of vapor and liquid films and calculated by Ax = D/kc, and kc is the total liquid-phase mass transfer coefficient. In terms of the assumption of thermodynamic interface, the energy consumption is mainly concentrated on the vapor and liquid films. So the second term in the right side of Eq. (170) can be omitted, and we have:
Ignoring temperature gradient on the tray and substituting for JA in Eq. (167),
At the steady-state of nonequilibrium thermodynamics, inter-coefficient LA B and the corresponding chemical potential gradient remain constant, that is, VjuA_R = ^A~H
Ac and Vc = ——. By taking Eq. (168) into Eq. (172), and integrating Eq. (172), Ax p = j^~da-b • (ACA_B)2 . Ar = -¡t—D\_H • (AcA_Rf (173)
where AcA_B =(clA-cB)-(c*4-c"B). cA and c* are mole concentration equilibrium with components A and B in the vapor phase respectively. From Eq. (173), entropy generation rate per unit interfacial volume crv is obtained:
It is concluded from Eq. (173) that
(1) Entropy generation P at nonequilibrium state is always greater than zero. When the system reaches equilibrium, P = 0 because AcA_B = 0.
(2) At the steady-state of nonequilibrium thermodynamics, — = 0 due to constant values of dt v, La_b, Da_b and AcA_B.
At equilibrium state P = 0 and at the linear range of nonequilibrium thermodynamics, the nonequilibrium steady-state is stable and doesn't spontaneously form the regular structure in time and space. The results obtained from this text are consistent with those from classic nonequilibrium thermodynamics, which means that the deducing results are reasonable. Evidently, the above analysis is also suitable for common extractive distillation with tray column. Of course, in this case, only three components of A, B and P excluding the new component S formed by reaction, are concerned.
Furthermore, the results can also be extended to other special distillation processes such as adsorption distillation, catalytic distillation, etc. Let us see the following example of nonequilibrium thermodynamic analysis for adsorption distillation.
Example: In the adsorption distillation column, the tiny solid particles are used as the adsorption agent, and blended with liquid phase on the tray. There exists four components: M (the heavy key component), N (the light key component), P (the carrier for carrying S) and S (adsorption agent) involved , The following reversible chemical reaction may take place on the tray,
It is required to set up the condition that the separation process can occur from the viewpoint of nonequilibrium thermodynamics.
Solution: The similar procedure as reactive extractive distillation can be deduced for this situation, and the results similar as Eqs. (163)- (165) and Eq. (173) will be obtained. The interested reader can try to do it.
The most difference between equilibrium thermodynamics and nonequilibrium thermodynamics is that mass transfer (as well as heat transfer) should be considered in the latter. In the linear range of nonequilibrium thermodynamics, for binary mixtures or for diffusion of dilute species i in a c-component mixture, Fick's law of diffusion postulates a linear dependence of the flux J, with respect to the molar average mixture velocity u and its composition gradient Vx,:
The molar flux Nl with respect to a laboratory-fixed coordinate reference frame is given by
N, =C,u, =C,x,u, = Ji+x,Nl = -C,£>,Vx, +x,N, (177)
If one takes the review that Eq. (175) provides a definition of the effective Fick diffusivity Dejf of component i in a c-component mixture, then this parameter shows a complicated, often unpredictable behaviour because the fundamental driving force for diffusion is the gradient of chemical potential rather than mole fraction or concentration gradient.
For multi-component mass transfer, it is common to use the generalized Maxwell-Stefan equation [76-80], which is divided into implicit and explicit forms. The implicit formulation of Maxwell-Stefan equation is related to the chemical potential gradients, and the molar transfer rate N't' in the liquid phase and N] in the vapor phase are, respectively, x/- (179)
RT' dij h C',K'Lka and v a v c v at1' v atv y, ^ x, Nk - xk N,
RT drj f^i C, Klka where a is the effective interfacial area per unit volume (m2 m"3).
The k\\ and k\k represent the mass transfer coefficients of the i-k pair in the liquid and vapor phases, and how to calculate them will be discussed in what follows.
Undoubtedly, it is very difficult to directly obtain Nf and Nf from Eqs. (179) and (180).
However, the explicit formulation of Maxwell-Stefan equation is more preferred to use.
Moreover, for the convenience, j\', J) , Nj' and N] are determined as mass transfer rates instead of mass transfer fluxes. For the liquid phase, by using (c-l)X(c-l) matrix notation,
and the elements of the matrix [B'lk ] are:
where x, is the molar fraction of /'-component in the liquid phase.
The elements of the (c-1) X (c-1) matrix of thermodynamic factors \Y'ik ] are:
As we know, 8 function (the Kronecker delta) is defined as: 8lk =1 as i-k: otherwise
Comparing to Eq. (181), the thermodynamic factors [T^ ] don't appear in Eq. (185). It is due to the assumption of ideal gas at or below middle pressure in the vapor phase such that in this case [T^ ] = / (unit matrix).
The elements of the matrix [Bfk ] are:
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