I'm trying to understand a derivation from "The Thermodynamics of Linear Fluids and Fluid Mixtures," by Miloslav Pekař and Ivan Samohýl (2014). The derivation produces an expression for the entropy generation in a multi-component continuum when viscosity, diffusion, heat transfer, and chemical reactions are all present.

Both the book and I agree that the entropy generation can be written as: $$ r T \dot{s}_\text{gen} = r T \sum_\alpha X^\alpha \frac{\text{D}^\alpha}{\text{D}t}s^\alpha + T \sum_\alpha \dot{\Omega}^\alpha s^\alpha + \vec{\nabla}\cdot \vec{\dot{q}}^\text{cond} - \frac{\vec{\dot{q}}^\text{cond}}{T}\cdot \vec{\nabla}T - r \dot{q}^\text{rad,in} $$ Where

  • $r$ is the density (rho didn't show up properly in my preview)
  • $\alpha$ is the species label
  • $X^\alpha$ is the mass fraction of species $\alpha$
  • $\dot{\Omega}^\alpha$ is the rate of production of $\alpha$ by chemical reactions [units density per time]
  • $\frac{\text{D}^\alpha}{\text{D}t}$ is a variation of the material derivative where the species velocity $\vec{v}^\alpha$ is used instead of the bulk velocity: $\frac{\text{D}^\alpha}{\text{D}t} \phi = \frac{\partial}{\partial T} \phi + \vec{v}^\alpha\cdot \vec{\nabla}\phi$

(In the book, this is the combination of Eqns 4.84 and 4.85, although the notation they use is slightly different)

The book and I also agree that the approach to simplifying this equation is to re-express the first term on the right-hand side, but our choices for rearrangement are different.

The Problem

The book's expression for the entropy generation can be arranged as \begin{align*} r T \dot{s}_\text{gen} =& - \sum_\alpha \dot{\Omega}^\alpha g^\alpha - \overbrace{\sum_\alpha \left( \frac{\partial r^\alpha f^\alpha}{\partial T} + r^\alpha s^\alpha \right) \frac{\partial T}{\partial t}}^A \\ &+ \underbrace{\sum_\alpha \sum_\beta \left[ g^\beta \delta^{\alpha \beta} + \frac{\partial r^\alpha f^\alpha}{\partial r^\beta} \right] \vec{v}^\alpha \cdot \vec{\nabla} r^\beta }_B + \underbrace{\sum_\alpha \left( r^\alpha \sum_\beta \frac{\partial r^\beta f^\beta}{\partial r^\alpha} - r^\alpha f^\alpha \right) \vec{\nabla} \cdot \vec{v}^\alpha}_C \end{align*} (where $g^\alpha$ is the Gibbs Free Energy of species $\alpha$) while I conclude that it is $$ r T \dot{s}_\text{gen} = - \sum_\alpha \dot{\Omega}^\alpha g^\alpha + P \sum_\alpha \vec{\nabla} \cdot \vec{v}^\alpha $$ I can convince myself that $A = 0$ and that $C$ reduces to my second term. I can see that $B = 0$ if the thermodynamic properties of $\alpha$ are independent of the density of all other species $\beta$, but I don't believe that this is true in the general case where there may be inter-species interaction energy.

This suggests to me that there is some discrepancy between our approaches.

My Question

  • Is there some way of proving that $B = 0$? I believe that this would require a proof that the bracketed factor is zero, since the velocity and density gradient are independently variable
  • Assuming that my approach (below) is incorrect, where is the error? Is there a simple algebra mistake somewhere, or a fundamental problem with the idea of NOT involving the Helmholtz Free Energy as in the book's approach (also below)?

My Approach

I choose to apply the Gibbs Relation: \begin{equation} \text{d} u = T \text{d}s + P \text{d}v + \sum_\alpha \mu^\alpha \text{d}X^\alpha \end{equation} With the substitution \begin{align} \phi &= \sum_\alpha \phi^\alpha X^\alpha &\text{where }\quad\phi=(u,s,v) \\ \text{d} \phi &= \sum_\alpha \left( \phi^\alpha \text{d} X^\alpha + X^\alpha \text{d}\phi^\alpha \right) \end{align} and the identity $\mu^\alpha = u^\alpha + Pv^\alpha - Ts^\alpha$ allowing the Gibbs Expression to be rewritten as \begin{align} \sum_\alpha X^\alpha \text{d} u^\alpha &= T\sum_\alpha X^\alpha \text{d}s^\alpha + P\sum_\alpha X^\alpha \text{d}v^\alpha \end{align} Which implies that \begin{align} r T\sum_\alpha X^\alpha \frac{\text{D}^\alpha}{\text{D}t} s^\alpha = r \sum_\alpha X^\alpha \frac{\text{D}^\alpha}{\text{D}t} u^\alpha + r P \sum_\alpha X^\alpha \frac{\text{D}^\alpha}{\text{D}t} v^\alpha \end{align} ...which is convenient because the two expressions on the right-hand side can be found from conservation of energy and conservation of species respectively.

The Book's Approach

The book chooses to express things in terms of the Helmholz Free Energy $f^\alpha=u^\alpha-Ts^\alpha$, which they refer to as the partial free energy. They begin with a different form of the Gibbs Equation: $$ \text{d} f = - s \text{d}T - P \text{d} s + \sum_\alpha \mu^\alpha \text{d}X^\alpha $$ and apply the same $\phi = \sum_\alpha X^\alpha \phi^\alpha$ substitution as I did in combination with the definition of Helmholtz Free Energy to recover \begin{align} T \sum_\alpha X^\alpha \text{d} s^\alpha &= - \left(\sum_\alpha X^\alpha s^\alpha\right) \text{d}T + \sum_\alpha X^\alpha \text{d} u^\alpha - \sum_\alpha X^\alpha \text{d} f^\alpha \end{align} which implies that \begin{align} r T\sum_\alpha X^\alpha \frac{\text{D}^\alpha}{\text{D}t} s^\alpha = r \sum_\alpha X^\alpha \frac{\text{D}^\alpha}{\text{D}t} u^\alpha - r \sum_\alpha X^\alpha \frac{\text{D}^\alpha}{\text{D}t} f^\alpha - r \sum_\alpha X^\alpha s^\alpha \frac{\text{D}^\alpha}{\text{D}t} T \end{align} I consider this to be inconvenient as it introduces two time derivatives which are not known: the rate of change of temperature, and the rate of change of Helmholtz Free Energy. To try and remedy this, they combine the derivatives of $f^\alpha$ with a source term $\sum_\alpha \dot{\Omega}^\alpha f^\alpha$ (which appears elsewhere in the entropy generation equation) to give derivatives of $r^\alpha f^\alpha = F^\alpha$. They then seem to rely on writing the Helmholtz Free Energy as a function of its natural variables: $$ F^\alpha = f(T,v^\beta) = f(T,1/r^\beta) = f (T,r^\beta) $$ and then expanding derivatives as \begin{align*} \text{d}F^\alpha &= \left(\frac{\partial F^\alpha}{\partial T}\right)\text{d} T + \sum_\beta \left(\frac{\partial F^\alpha}{\partial r^\beta}\right)\text{d} r^\beta \\ \frac{\partial F^\alpha}{\partial \phi} &= \left(\frac{\partial F^\alpha}{\partial T}\right)\frac{\partial T}{\partial \phi} + \sum_\beta \left( \frac{\partial F^\alpha}{\partial r^\beta}\right) \frac{\partial r^\beta}{\partial \phi} \end{align*} to recover the expression listed earlier

Additional thoughts

Based on the form of the equations, it occurs to me that the discrepancy may be related to the difference between $\frac{\text{D}^\alpha}{\text{D}t} r^\alpha$ and $\frac{\text{D}^\beta}{\text{D}t} r^\alpha$: \begin{align} \frac{\text{D}^\beta}{\text{D}t} r^\alpha &= \frac{\partial}{\partial t} r^\alpha + \vec{v}^\beta \cdot \vec{\nabla} r^\alpha \\ &= \left( \frac{\partial}{\partial t} r^\alpha+ \vec{v}^\alpha \cdot \vec{\nabla} r^\alpha\right)+ \vec{v}^\beta \cdot \vec{\nabla} r^\alpha - \vec{v}^\alpha \cdot \vec{\nabla} r^\alpha \\ &= \frac{\text{D}^\alpha}{\text{D}t} r^\alpha + \vec{v}^\beta \cdot \vec{\nabla} r^\alpha - \vec{v}^\alpha \cdot \vec{\nabla} r^\alpha \\ &= \frac{\text{D}^\alpha}{\text{D}t} r^\alpha + \left(1 - \delta^{\alpha \beta} \right) \vec{v}^\beta \cdot \vec{\nabla} r^\alpha \end{align} ...however, I can't see where I would have accidentally used one of these where the other was more appropriate.


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