# Deriving the heat equation from compressible Navier Stokes equations

Starting from the compressible Navier-Stokes equations, I want to derive the standard form the instationary heat equation.

The energy equation in general form can be written as

\begin{align} \frac{\partial E}{\partial t}+\nabla\cdot \left(H\vec{v}\right)&=\nabla\cdot\left(\mathbf{\sigma}\cdot\vec{v}-\vec{q}\right) \end{align}

or in a more specific form

\begin{align} \frac{\partial \left(\rho \epsilon + \frac{1}{2} \rho |\vec{v}|^2\right)}{\partial t} + \nabla \cdot \left(\left[ \rho \epsilon + \frac{1}{2} \rho |\vec{v}|^2 + p \right] \vec{v} \right) = \dots \\ \dots \nabla \cdot \Bigg\langle \left( \eta \left[ \left( \nabla\otimes\vec{v} \right)^\top+\nabla\otimes\vec{v} \right]-\frac{2}{3}\eta \left(\nabla\cdot\vec{v}\right)\mathbf{1}\right)\cdot\vec{v}+\lambda \nabla T \Bigg\rangle. \end{align}

Using the caloric ideal EoS

\begin{align} \epsilon=c_v T, \quad h=c_p T,\quad h=\epsilon + p/ \rho \end{align}

we get

\begin{align} \frac{\partial \left(\rho c_{v} T + \frac{1}{2} \rho |\vec{v}|^2\right)}{\partial t} + \nabla \cdot \left(\left[ \rho c_{p} T + \frac{1}{2} \rho |\vec{v}|^2 \right] \vec{v} \right) = \dots \\ \dots \nabla \cdot \Bigg\langle \left( \eta \left[ \left( \nabla\otimes\vec{v} \right)^\top+\nabla\otimes\vec{v} \right]-\frac{2}{3}\eta \left(\nabla\cdot\vec{v}\right)\mathbf{1}\right)\cdot\vec{v}+\lambda \nabla T \Bigg\rangle. \end{align}

Now assuming

1. Zero velocity, $$\vec{v}=0$$,
2. Zero viscosity, $$\eta=0$$,

I end up with the following heat equation

\begin{align} \frac{\partial (\rho c_v T)}{\partial t} = \left( \lambda T_{x} \right)_{x} + \left( \lambda T_{y} \right)_{y} + \left( \lambda T_{z} \right)_{z}, \end{align}

and not

\begin{align} \frac{\partial (\rho c_p T)}{\partial t} = \left( \lambda T_{x} \right)_{x} + \left( \lambda T_{y} \right)_{y} + \left( \lambda T_{z} \right)_{z}. \end{align}

as given in most literature, see Wikipedia.

Any hints are appreciated, thanks!

• What happened to the internal energy in the terms on the left side of the equation? Commented Jun 23, 2023 at 11:39
• @ChetMiller I used the caloric ideal EoS and a universal thermodynamic relation: $\epsilon=c_v T, \quad h=c_p T,\quad h=\epsilon + p/ \rho$, see also my edit in the question. Commented Jun 23, 2023 at 11:46
• So you're assuming an ideal gas, right? Isn't the density a function of t? If so, isn't v not equal to zero? Commented Jun 23, 2023 at 11:54
• For an arbitrary material (solid, liquid, or gas), Cp is defined as$$C_p=\left(\frac{\partial h}{\partial T}\right)_P$$So it is far from meaningless. And, $$C_v=\left(\frac{\partial u}{\partial T}\right)_V$$ Commented Jun 23, 2023 at 13:48
• See Transport Phenomena by Bird, et al, Chapter 11 for correct mathematical handling of these quantities. Commented Jun 23, 2023 at 15:11

Let me quote Landau & Lifshitz: Fluid Mechanics, $$\S 50$$, page 188, $$3^{rd}$$ed:
From which it follows the use of $$c_p$$