# What is the Navier-Stokes Energy equation?

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.)

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.

• Your guesses were mostly right. e is the internal energy per unit mass, k is the thermal conductivity and Q is an external source of heat. The function $e(\rho, T)$ is usually know, as well as the equation of state $T(\rho,P)$ Aug 8, 2021 at 22:39
• The main idea for the NASA energy equation is basically application of the first law of thermodynamics, and its use is typically to calculate the temperature distribution. The e in the equation is the internal energy per unit mass, and is a function of temperature and pressure. k is the thermal conductivity of the fluid, F is the body force per unit volume, and q Aug 9, 2021 at 3:10
• I am probably at a level of expertise now where I can answer this question myself. I will post something if I find the time. Feb 1 at 20:46

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 $$d\epsilon = \frac 1 V (TdS-pdV+\mu dN) - E \frac{dV}{V^2}\nonumber\\ = \frac 1 V (TdS+\mu dN) - \frac {dV}{V^2}(TS+\mu N)\nonumber\\ = Tds +\mu dn\nonumber$$

• I'm still a bit confused... What is the difference between $p,P$? And what are $V,N$? How do I link these equations with (1),(2) in my question to be able to arrive at a unique solution for $\mathrm u,\rho,P$ given suitable initial and boundary conditions? Thanks for the answer. Aug 8, 2021 at 23:19
• @K.defaoite See en.wikipedia.org/wiki/Gibbs_free_energy for a definition of the notation. Aug 8, 2021 at 23:39
• Sorry! Copying from various notes caused a mess. There is no difference in the $p$'s $p=P$, but $N$ is total number of particles, $V$ is volume of a unit mass so mass density os $\rho= 1/V$. The rest is standard thermodynamics. My first two equations are your 1 and 2 and the my third is the energy equation for the internal energy that I think you are asking for. As to solving them, that is a vast subject. Aug 8, 2021 at 23:39