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In this post the fact that the mass continuity equation in a mixture of gases has no diffusion term, i.e., $$\frac{\partial\rho}{\partial t}+\nabla\cdot(\rho\vec{v})=0$$ has been discussed. Nevertheless, I'm struggling to come up with a proof starting from the convection-diffusion-reaction equation for each species. It seems to me that the starting point should be $$\frac{\partial n_{j}}{\partial t}+\nabla\cdot(n_{j}\vec{v}_{j}-D_{j}\nabla n_{j})=S_{j}$$ where $\vec{v}_{j}$ as opposed to $\vec v$ denotes the convective velocity vector of species $j$. I can define $n_{j}=\omega_{j}n$ where $\omega_{j}$ denotes the partial fraction of the fluid/gas made by species $j$ and $n$ denotes total denisty such that $$\sum_{j}\frac{\partial n_{j}}{\partial t}=\sum_{j}\frac{\partial}{\partial t} (\omega_{j}n)=\sum_{j}\omega_{j}\frac{\partial}{\partial t} n+\sum_{j}n\frac{\partial}{\partial t} \omega_{j}=\frac{\partial n}{\partial t}\sum_{j}\omega_{j}+n\frac{\partial}{\partial t}\sum_{j} \omega_{j}=\frac{\partial n}{\partial t}$$ I can go ahead with the same analysis for the convective term by defining $$\vec v=\frac{\sum_{j}n_{j}\vec{v}_{j}}{n}$$ and since mass is not produced $\sum_{j}S_{j}=0$ for the source term but the cancellation of the diffusion term still eludes me. I would appreciate any help with this final part. I'm curious if Maxwell-Stefan relation is necessary here.

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  • It seems to me the diffusion term should not be present in your first equation. The diffusion term is supposed to represent diffusional transport relative to the mass average velocity v. – Chet Miller May 21 '22 at 11:00
  • @ChetMiller Agreed. Apparently, in kinetic theory diffusion is defined based on the relative (average) velocity of species with respect to one another. For example, in a binary mixture with bulk velocity $\vec v$, $\vec{v}{1}-\vec{v}{2}=\vec{v}{1}-\vec{v}+\vec{v}-\vec{v}{2}=(\vec{v}{1}-\vec{v})-(\vec{v}{2}-\vec{v})$. Still, the $D_{j}\nabla n_{j}$ term seems to be a result of this definition so that the equation for species is consistent with bulk mass continuity in Navier-Stokes/Euler equations. See https://en.wikipedia.org/wiki/Diffusion#Diffusion_coefficient_in_kinetic_theory_of_gases – Newbie May 21 '22 at 13:14
  • Since this is not the actual answer to the question I will leave it as a comment: Chapter 4 of Joel H. Ferziger, H. G. Kaper - Mathematical theory of transport processes in gases covers the answer to my question. – Newbie May 21 '22 at 14:35
  • I stand by what I said. The term in parenthesis should read $vn_j-nD_j\nabla \omega_j$ – Chet Miller May 22 '22 at 00:26
  • @ChetMiller Shouldn't the diffusion be $n_{j}D_{j}\nabla\omega_{j}$, i.e. $n\rightarrow n_{j}$ in your comment? – Newbie May 22 '22 at 04:04
  • No it shouldn't. Where did you get that idea from? – Chet Miller May 22 '22 at 11:01
  • @ChetMiller Lam, Phys. Fluids 18, 073101 (2006), writes the component conservation equation as $\partial\rho_{i}/\partial t+\nabla\cdot(\rho_{i}\vec{V})+\nabla\cdot(\rho_{i}\vec{v}{i})=S{i}$ where $\vec V$ is the bulk velocity and $\vec{v}{i}=\vec{V}{i}-\vec{V}$ is the diffusion velocity and it has a gradient component $\vec{v}{i}=-\sum{j}D_{ij}\nabla\omega_{j}$. Assuming $D_{ij}=D_{i}\delta_{ij}$ and dividing the equation by $m_{i}$ ($\rho_{i}=m_{i}n_{i}$), it seems to me that the diffusion term will have $n_{i}$ instead of $n$. – Newbie May 22 '22 at 11:32
  • I think it makes sense to have $n_{i}$ since $D_{ij}\nabla\omega_{j}$ will be the velocity accounting for the reaction of component $i$ to the inhomogeneity of component $j=1,\ldots,N$ ($\nabla\omega_{j}$). The flux of component $i$'s reaction becomes $n_{i}D_{ij}\nabla\omega_{j}$. – Newbie May 22 '22 at 11:59
  • This makes no sense to me, and is not even correct when j is a trace species. – Chet Miller May 22 '22 at 12:18
  • @ChetMiller Let us continue this discussion in chat. – Newbie May 22 '22 at 12:27
  • @ChetMiller If $i$ is a trace species, $\nabla(\omega_{i})=\nabla(\frac{n_{i}}{n})=\frac{\nabla(n_{i})n-\nabla(n)n_{i}}{n^{2}}\simeq \frac{1}{n}\nabla(n_{i})$. Assuming $D_{ij}$ is diagonal, then we have for the diffusion velocity $v_{i}=-D_{ii}\frac{1}{n}\nabla(n_{i})$ so that the diffusion flux is $n_{i}v_{i}=-D_{ii}\frac{n_{i}}{n}\nabla(n_{i})=-D_{ii}\omega_{i}\nabla(n_{i})=-D'{i}\nabla(n{i})$. Does this sound ok? – Newbie May 22 '22 at 15:20
  • No. See Transport Phenomena by Bird et al . – Chet Miller May 22 '22 at 20:54
  • @ChetMiller Bird equation (24.2-3): $\vec{j}{\alpha}=\rho{\alpha}\sum_{\beta}D_{\alpha\beta}\vec{d}{\beta}-D^{T}{\alpha}\nabla\ln T$ – Newbie May 22 '22 at 21:43
  • It's the same form in equation (6.3-32) of Mathematical Theory of Transport Processes in Gases by Ferziger and Kaper, and Lam, Phys. Fluids 18, 073101 (2006), equation (1a). The term related to density included in $\vec{d}{j}$ is $\nabla(\omega{j})$. Therefore, assuming $D_{\alpha\beta}=D_{\alpha}\delta_{\alpha\beta}$ the diffusion current becomes $\vec{j}{\alpha}=\rho{\alpha}D_{\alpha}\nabla(\omega_{\alpha})$. – Newbie May 22 '22 at 21:53
  • In Chapter 17, Bird et al give the mass flux of species A in a binary mixture as $\rho_Av-\rho D_{AB}\nabla{\omega_A}$ (Eqns. 17.8-1 and 17.7-3). – Chet Miller May 24 '22 at 09:14

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