What is the Navier-Stokes Energy equation?

Question

I'm trying to understand the basics of fluid dynamics and the Navier-Stokes equations by following the short book A Mathematical Introduction to Fluid Dynamics by Chorin and Marsden (I'm an applied mathematician, not a physicist, so please be patient with me. Also I wasn't sure if this question was better suited for MSE or PSE. I chose here.)

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As I understand it, to uniquely determine the three important quantities in a fluid mechanics problem which are $\mathrm u$, $P$ and $\rho$, one needs five equations - one for each component of $\mathrm u$ (which can be written as a single vector equation, reducing our number of equations to 3), and two more for the scalars $P,\rho$. To get these three equations, we need three physical conditions. They are

• Conservation of mass
• Conservation of momentum
• Conservation of energy

I want to derive these three equations in the most general form possible. What I mean by that is our fluid may be

• compressible ($\nabla\cdot\mathrm u\neq0$)
• viscous ($\mu,\lambda\neq0$)
• vortical ($\nabla\times\mathrm u\neq0$)
• Subject to thermodynamic effects

The easiest part is conservation of mass. A short derivation will show that mass conservation may be described mathematically via the continuity equation,

$$\partial_t\rho+\nabla\cdot(\rho~\mathrm u)=0\tag{1}$$

Conservation of momentum is considerably more difficult, but after some work I was able to eventually arrive at

$$\rho\frac{D\mathrm u}{Dt}=-\nabla P+\nabla\cdot \boldsymbol{\tau}+\mathrm{F}\tag{2A}$$

Where,

• $D/Dt=\partial_t +\mathrm u\cdot\nabla$ is the material derivative,
• $\mathrm{F}$ denotes any external forces ($\mathrm{F}=0$ in an isolated system),
• $\boldsymbol{\tau}$ is the deviatoric stress tensor, a $(1,1)$ tensor

defined as $$\boldsymbol{\tau}=\lambda\mathbf{I}~\nabla\cdot\mathrm u+\mu\big(\nabla\mathrm u+(\nabla\mathrm u)^{\mathrm T}\big)$$ Where $\mathbf I$ is the identity, $\mu$ is the dynamic viscosity, and $\lambda$ is the bulk viscosity. Using the identity $\nabla\cdot(\nabla\mathrm u)^{\mathrm T}=\nabla(\nabla\cdot\mathrm u)$ we can expand out this equation to reach

$$\rho(\partial_t\mathrm u+\mathrm u\cdot\nabla\mathrm u)=-\nabla P+(\lambda+\mu)\nabla(\nabla\cdot\mathrm u)+\mu\boldsymbol{\triangle}\mathrm u+\mathrm F\tag{2B}$$

This equation is given on page 33 of the linked text.

However, conservation of energy still eludes me. NASA appears to have a published formula, but it is only given in the special case of Cartesian coordinates, and I would much rather a coordinate-free representation. This page gives a coordinate-free description:

$$\partial_t\left[\rho~\left(e+\frac{|\mathrm u|^2}{2}\right)\right]+\nabla\cdot\left[\rho\mathrm u~\left(e+\frac{|\mathrm u|^2}{2}\right)\right]=\nabla\cdot(k\nabla T-P\mathrm u+\boldsymbol{\tau}\cdot\mathrm u)+\mathrm u\cdot\mathrm F+\mathcal{Q}\tag{3}$$

But they don't offer a derivation, nor do they even explain what the symbols $e,k,\mathcal Q$ mean. My guess(?) is that $k$ is some kind of thermal diffusivity, $e$ is a function of position and time and tells you something about the internal or potential energy, and $\mathcal{Q}$ is some external heat source. Am I right? And in most cases we would know $e(\mathrm r,t)$ and $T(\mathrm r,t)$ as given, and not have to solve for them, correct?

If we assume that the fluid has no thermal or potential energy, the energy can be written as purely kinetic: $$E_{\text{tot}}=\frac{1}{2}\int_{\Omega(t)}\rho(\mathrm r,t)~|\mathrm u(\mathrm r,t)|^2~\mathrm d\mu(\mathrm r)$$ Here $\Omega(t)$ is the region that our fluid occupies at time $t$. Using the transport theorem, one can arrive at $$\frac{\mathrm d E_{\text{tot}}}{\mathrm dt}=\frac{\mathrm d}{\mathrm dt}\int_{\Omega(t)}\rho |\mathrm u|^2\mathrm d\mu=\int_{\Omega(t)}\rho\mathrm u\cdot\frac{D\mathrm u}{Dt}\mathrm d\mu$$ So it would appear that a sufficient constraint for conservation of energy in this case is

$$\rho\mathrm u\cdot\frac{D\mathrm u}{Dt}=0\implies \mathrm u\cdot\big(\partial_t\mathrm u+\mathrm u\cdot\nabla\mathrm u\big)=0\tag{4}$$

Which is a slightly modified form of the well known Burgers' equation.

But what if there is thermal and kinetic energy? What can I do then? Can anyone please help me derive and make sense of equation $(3)$, and reduce it to $(4)$ in a special case?

Thanks so much.

EDIT: Definitions for various uses of the symbol $\nabla$.

If $\mathbf{T}$ is an $(r,s)$ tensor, then I define the components of $\nabla \mathbf{T}$ in a coordinate system with Christoffel symbols $\Gamma^i_{jk}$ as (Using Einstein summation)

$$( \nabla \mathbf{T})^{i_{1} \dotsc i_{r}}{}_{j_{1} \dotsc j_{s} \ k} =\begin{matrix} \partial _{k} T^{i_{1} \dotsc i_{r}}{}_{j_{1} \dotsc j_{s}}\\ +\Gamma _{lk}^{i_{1}} T^{l\ i_{2} \dotsc i_{r}}{}_{j_{1} \dotsc j_{s}} +\cdots +\Gamma _{lk}^{i_{r}} T^{i_{1} \dotsc i_{r-1} \ l}{}_{j_{1} \dotsc j_{s}}\\ -\Gamma _{j_{1} k}^{l} T^{i_{1} \dotsc i_{r}}{}_{l\ j_{2} \dotsc j_{s}} -\cdots -\Gamma _{j_{s} k}^{l} T^{i_{1} \dotsc i_{r}}{}_{j_{1} \dotsc j_{s-1} \ l} \end{matrix}$$

I define the components the divergence of a $(1,s)$ tensor $\mathbf{T}$ as

$$(\nabla\cdot\mathbf{T})_{j_1\dots j_s}= (\nabla \mathbf{T})^i{}_{j_1\dots j_s~i}$$

And lastly I define the Laplacian of an $(r,s)$ tensor as

$$(\boldsymbol{\triangle}\mathbf{T})^{i_1\dots i_r}{}_{j_1\dots j_s}=g^{kl}(\nabla(\nabla\mathbf{T}))^{i_1\dots i_r}{}_{j_1\dots j_s~kl}$$

Where $g^{ij}$ is the $i,j$th component of the inverse metric.

Answer 1

The kinetic energy equation for stationary flow is just Bernoulli: $$ ({\bf v}\cdot\nabla) \left( \frac 12 |{\bf v}|^2+ h\right)=0 $$ where $h$ is the enthalpy $E+PV$ per unit mass.

For the thermal energy I'm going to ignore viscosity and heat flow, but the general idea should be obtained from the following:

For a non-relativistic fluid we can write the equations of fluid flow as $$ \left(\frac{\partial}{\partial t}+{\bf v}\cdot \nabla\right)\rho= -\rho\,{\rm div\,} {\bf v}\nonumber\\ \left(\frac{\partial}{\partial t}+{\bf v}\cdot \nabla\right) {\bf v}= -\frac{1}{\rho} \nabla p\nonumber\\ \left(\frac{\partial}{\partial t}+{\bf v}\cdot \nabla\right)E=-\frac{p}{\rho} {\rm div\,}{\bf v} .\nonumber $$ In deriving the last (energy) equation we have assumed that there is no heat flow so only thing changing $E$, the energy per unit mass, is the work $-pdV=-pd(1/\rho)$. It is therefore not surprising that we can combine the first and last of these equations to get $$ \frac 1 T \left(\frac{\partial}{\partial t}+{\bf v}\cdot \nabla\right) E+ \frac{p}{T}\left(\frac{\partial}{\partial t}+{\bf v}\cdot \nabla\right)\frac{1}{\rho}=0. $$ Since $$ TdS= dE+pd\left(\frac{1}{\rho}\right), $$ this last is $$ \left(\frac{\partial}{\partial t}+{\bf v}\cdot \nabla\right) S=0. $$ Combining this convective constancy of the entropy with mass conservation, we have $$ \frac{\partial \rho S}{\partial t}+ {\rm div\,} (\rho S {\bf v})=0. $$ Equivalently, this is $$ \frac{\partial s}{\partial t}+ {\rm div\,}(s {\bf v})=0, $$ where $s= \rho S=S/V$.

With $\varepsilon= \rho E= E/V$, the internal energy (non) conservation equation can be written as $$ \frac{\partial \varepsilon }{\partial t}+ {\rm div\,} \varepsilon {\bf v}+ P\,{\rm div\,}{\bf v}=0. $$ Thus $$ T\left(\frac{\partial s}{\partial t}+ {\rm div\,}s {\bf v}\right) =T\left(\frac{\partial s}{\partial t}+ {\rm div\,}s {\bf v}\right)+\mu \left(\frac{\partial \rho}{\partial t}+ {\rm div\,} \rho {\bf v}\right) -\left( \frac{\partial \varepsilon}{\partial t}+ {\rm div\,} \varepsilon {\bf v} +P\,{\rm div\,}{\bf v}\right)\nonumber \\ =\left(T \frac{\partial s}{\partial t}+\mu \frac{\partial \rho}{\partial t}-\frac{\partial \varepsilon}{\partial t}\right)+ {\bf v}\cdot \left(T \nabla s +\mu \nabla \rho-\nabla \varepsilon \right) \nonumber\\ \qquad +\left(Ts+\mu \rho -\varepsilon-P\right){\rm div\,} {\bf v}.\nonumber $$ The last RHS is zero because the first two terms compose the convective derivative of $ Tds+\mu d\rho -d\varepsilon$, which is zero, and also $TS+\mu N-E-PV$ is identically zero.

Thermo relations: We have that the Gibbs free energy is $E-TS+pV = \mu N$, so, with the definitions $$ \epsilon = E/V,\quad n=N/V,\quad s=S/V, $$ we can write the first law $$ dE=TdS-pdV +\mu dN $$ as $$ d\epsilon = \frac 1 V (TdS-pdV+\mu dN) - E \frac{dV}{V^2}\nonumber\\ = \frac 1 V (TdS+\mu dN) - \frac {dV}{V^2}(TS+\mu N)\nonumber\\ = Tds +\mu dn\nonumber $$