Per the Fundamental Thermodynamic Relation, I know that the chemical potential of $i$ represents the energy which would added to a system if a particle of $i$ were added with all other system properties kept constant. By this definition, the instantaneous energy flux due to diffusion of $i$ should be

$$ \vec{j}_i \mu_i $$

where $\vec{j}_i$ is the mass flux of species $i$. This means the instantaneous energy flux due to all diffusion should be

$$ \sum_i \vec{j}_i \mu_i $$

I've found a source [https://e-reports-ext.llnl.gov/pdf/367674.pdf Equation 12] which suggests that the energy flux due to diffusion is

$$\sum_i \vec{j}_i h_i$$

This is consistent with the familiar expression for the energy flux due to bulk mass transport, but not with my first expression. I know that $\mu$ and $h$ are connected by

$$\mu_i = g_i = h_i - Ts_i $$

but I don't see how I can make the $T s_i$ term cancel out to extract the second expression from the first. It seems that that would require that

$$\sum_i \vec{j}_i s_i = 0 $$

and that this cannot be true in general. This apparently puts the "general" formula from my source at odds with my interpretation of the "general" Fundamental Thermdynamic Relation. What source or assumption is incorrect?

Note: I realize that the source I linked to cites several other sources. I will consult them when I have a chance to get to the library, but I'm concerned that they will also introduce the formula as an assertion rather than a derivation and thus not answer my question.

Note: I've considered the possibility that the arbitrariness of reference states plays a role here. I've concluded that it does not: $\mu$, $h$, and $s$ are only arbitrary up to an additive constant and $\sum_i \vec{j}_i = 0$, so summing the arbitrary constant over all species always returns 0.

  • $\begingroup$ Minor comment to the post (v1): Please consider to mention explicitly author, title, etc. of link, so it is possible to reconstruct link in case of link rot. $\endgroup$
    – Qmechanic
    Commented Dec 8, 2014 at 0:35

2 Answers 2


This question [ Chemical potential in Thermodynamics ] turns out have most of the answer. It wasn't tagged with the 'chemical-potential' label so I didn't see it when I was asking. Anyway, I'll re-state it as it applies to this question: chemical potential is the energy added per particle if entropy and volume are kept constant. The real Fundamental Equation is

$$ du = T ds - PdV + \sum_i \mu_i dy_i $$

For a constant-volume process with diffusive flux $\vec{j}$, the mass crossing the boundary causes a change in $y_i$ and in $s$ - the latter because it carries entropy with it. The overall energy change due to diffusion is then

\begin{align} du &= T ds - PdV + \sum_i \mu_i dy_i \\ \frac{du}{dt}&= T \frac{ds}{dt} - P \frac{dv}{dt}+\sum_i \mu_i \frac{dy_i}{dt} \\ \frac{du}{dt}&= T \sum_i \vec{j}_i s_i - 0 + \sum_i \mu_i \vec{j}_i \\ &= T \sum_i \vec{j}_i s_i + \sum_i \left( h_i -Ts_i \right) \vec{j}_i \\ &= T \sum_i \vec{j}_i s_i + \sum_i \vec{j}_i h_i - T\sum_i s_i \vec{j}_i \\ &= \sum_i \vec{j}_i h_i \end{align}

If volume were changing, we could also consider the impact of this change on the system's energy, but we would consider this additional effect to be the result of mechanical work, not diffusion. We could then consider the system's energy change to be the sum of the diffusive flux (above) and the boundary work.

  • $\begingroup$ God work giving an excellent answer to your own question! This is more or less what I would have posted if you hadn't. $\endgroup$
    – N. Virgo
    Commented Dec 8, 2014 at 1:34

Take an area element $\mathrm{d}S$ perpendicular to the $x$-axis. How much energy is transported through this in unit time? Well, the flow rate is $v_x\mathrm{d}S$, and the energy density is $\frac{1}{2}\rho v^2 + \rho \varepsilon$, where $\varepsilon$ is internal energy. So multiplying these we have the energy transportation rate through $\mathrm{d}S$, $\left(\frac{1}{2}\rho v^2 + \rho \varepsilon\right)v_x\mathrm{d}S$, and finally we get the convective energy flux as we look at all directions (and have $\mathrm{d}S$ be the unit area): $$\left(\frac{1}{2}\rho v^2 + \rho \varepsilon\right)\mathbf{v}$$

That's not all. What can also happen is that the fluid on the minus side of $\mathrm{d}S$ exerts a force $\pi_x\mathrm{d}S$ on the fluid on the plus side. Now this does work: $\pi_x\mathrm{d}S \cdot \mathbf{v} = [\pi \cdot \mathbf{v}]_x \mathrm{d}S$. Generalizing to include $y$ and $z$ directions (and unit area), too, we have $$[\pi \cdot \mathbf{v}]$$

If we include the transportation of heat too, we have the combined energy flux vector $$\mathbf{e} = \left(\frac{1}{2}\rho v^2 + \rho \varepsilon\right)\mathbf{v} + [\pi \cdot \mathbf{v}] + \mathbf{q}$$

Note that $[\pi \cdot \mathbf{v}] = p\mathbf{v} + [\tau \cdot \mathbf{v}]$, and $\rho\varepsilon\mathbf{v} + p\mathbf{v} = \rho\left(\varepsilon + \frac{p}{\rho}\right)\mathbf{v} = \rho h \mathbf{v}$, i.e. $$\mathbf{e} = \left(\frac{1}{2}\rho v^2 + \rho h\right)\mathbf{v} + [\tau \cdot \mathbf{v}] + \mathbf{q}$$

This is to say that it is indeed enthalpy that enters the picture, $\rho \mathbf{v} h = \mathbf{j} h$ being the flux relevant to your question.

  • 1
    $\begingroup$ This is the flux due to bulk motion, which I agree should involve enthalpy. I'm asking about the flux due to diffusion though. $\endgroup$ Commented Dec 7, 2014 at 23:24

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