The validity of constitutive diffusive fluxes

In transport phenomena the diffusive fluxes for mass, energy and momentum are the constitutive laws: $$\boldsymbol{j}_c=-D\boldsymbol{\nabla}c \quad \boldsymbol{j}_T=-k\boldsymbol{\nabla}T \quad \boldsymbol{\tau}_{\boldsymbol{v}}=-\mu\boldsymbol{\nabla}\boldsymbol{v}$$ with $c$ the mass concentration, $T$ the temperature, $\boldsymbol{v}$ the velocity. The coefficients are the mass diffusion coefficient $D$, the thermal conductivity $k$ and the dynamic viscosity $\mu$.

Typically it is useful to view the diffusive flux in terms of the gradients of concentrations of mass, energy and momentum. For the mass diffusive flux this is already the case as $c$ is the mass concentration, the result is that the units for $D$ are $[m^2/s]$, typical units for diffusion coefficients.

A quick dimensional analysis of the other fluxes show that these aren't in terms of energy and momentum concentration and $k$ and $\mu$ aren't diffusion coefficients, i.e. $k=[W/mK]$ and $\mu=[Ns/ m^2]$. We can proceed to rewrite the fluxes in terms of energy and momentum concentrations:

$$\boldsymbol{j}_T=-\frac{k}{\rho c_p}\boldsymbol{\nabla}\rho c_pT=-\alpha\boldsymbol{\nabla}\epsilon\quad \boldsymbol{\tau}_{\boldsymbol{v}}=-\frac{\mu}{\rho}\boldsymbol{\nabla}\rho\boldsymbol{v}=-\nu\boldsymbol{\nabla}\boldsymbol{p}$$

Here the energy concentration $\epsilon=[J/m^3]$ and momentum concentration $\boldsymbol{p}=[\left(kgm/s\right)/m^3]$, with thermal diffusivity $\alpha=[m^2/s]$ and kinematic viscosity $\nu=[m^2/s]$ which show these are the respective diffusion coefficients for energy and momentum concentration.

The above analysis can only be done under the assumption of incompressibility and this is where my question originates:

Why are the constitutive laws for diffusive fluxes not defined in terms of mass, energy and momentum concentration?

Is it simply because the laws were formulated under the assumption of steady-state and incompressibility? What if incompressibility is not valid, are the laws the invalid?

As practical example of an issue which then arises: for a compressible medium should we still write the advection-diffusion heat equation as: $$\partial_t\rho c_p T + \boldsymbol{\nabla}\cdot \rho c_p \boldsymbol{u} T = k\nabla^2 T$$ or would it be of the following form: $$\partial_t\rho c_p T + \boldsymbol{\nabla}\cdot \rho c_p \boldsymbol{u} T = \alpha\nabla^2 \rho c_p T$$

• I don't recall seeing anyone use that form of the adv-diff equation, it is always without consideration of $\rho$ that I've seen it employed (including my own research involving it). – Kyle Kanos Dec 17 '15 at 16:33
• @KyleKanos: There was a typo in the equation, also made it more specific to temperature. I have also only seen it without $\rho$ considered, however in deriving the adv-diff equation from the Boltzmann equation i discovered i get the second equation... which got me to thinking why $\rho$ is never considered. – nluigi Dec 17 '15 at 17:07

After doing some investigating, I came across the paper Study of an Advection-Reaction-Diffusion equation in a compressible flow field by Federico Bianco, Sergio Chibbaro, Roger Prud'homme (arXiv link). In this paper that give the advection-reaction-diffusion equation as, $$\frac{\partial \rho\phi}{\partial t}+\nabla\cdot\left(\rho\phi\mathbf u\right)=D\nabla\cdot\left(\rho \nabla\phi\right)+s\tag{1}$$ where $\phi$ is a chemical species (in this particular case) and $D$ was assumed to be spatially constant. They derived this formula from the continuity equation (on page 2).

They state,

[This equation differs] from $$\frac{\partial \theta}{\partial t}+\nabla\cdot\left(\theta\mathbf u\right)=D\nabla^2\theta+F(\theta)\tag{6}$$ which is often find in literature to describe the advection, diffusion and reaction of a scalar in a compressible flow. This equation is typical in the study of population dynamics and the scalar $\theta(x,t)$, is the concentration of a population. However, this model is not correct for the concentration of the combustion products. In fact, in equation (6), if $\nabla\cdot\mathbf u\neq0$, the concentration $\theta$ can take values greater then one because is not a fractional parameter.

So it seems that if you are to include the density in the advection-diffusion equation (i.e., $s=0$ case of the above ARD equation), then you should use (1) above, rather than your case (which differs as $\nabla^2(\rho T)\to\nabla\cdot\rho\nabla T$).

• Thanks for your time, it makes sense so i will look into the paper you provide. – nluigi Dec 19 '15 at 16:29
• @nluigi: No problem, glad I could help. I couldn't actually get the thought out of my head while at work yesterday, so I sat down this morning and started digging around. – Kyle Kanos Dec 19 '15 at 17:29

This is an old question, but I think the answer is not quite right.

I think the bottom-line is that the constitutive relation for the energy and momentum current are unique, and the relation for particle diffusion should be generalized.

Because of Galilean invariance the momentum current can only depend on gradients of the velocity, not on the velocity itself. The unique tensor structure (ignoring bulk viscosity) is therefore $${\boldsymbol \tau}_{\boldsymbol{v}}=-\mu\boldsymbol{\nabla}\boldsymbol{v}$$ and we cannot have terms of the form $$(\mu/\rho)\boldsymbol{\nabla}(\rho\boldsymbol{v})$$.

Similarly, the energy current must be proportional to gradients of $$T$$, $$\boldsymbol{j}_T=-k\boldsymbol{\nabla}T .$$ Gradients of $$\rho$$ or $$P$$ violate the second law of thermodynamics (these arguments are explained in more advanced text books on fluid dynamics, see chapter 49 of Landau and Lifshitz, or modern text books on condensed matter physics, see section 8.4 in Chaikin and Lubensky).

The particle (concentration) current is more complicated, because there is no reason not to have both terms involving gradients of concentration (or chemical potential), and gradients of temperature. Also, once there is a concentration (i.e. the fluid has two components) then the heat current can have a term proportional to the gradient of concentration. This gives $${\boldsymbol j} = -D[{\boldsymbol \nabla} c +(k_T/T){\boldsymbol \nabla} T) + (k_P/P){\boldsymbol \nabla} P] ,$$ where $$D,k_T,k_P$$ are independent diffusion constants (which can be functions of the thermodynamic variables), see chapter 58 of Landau and Lifshitz.