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!

  • 1
    $\begingroup$ What happened to the internal energy in the terms on the left side of the equation? $\endgroup$ Commented Jun 23, 2023 at 11:39
  • $\begingroup$ @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. $\endgroup$
    – ConvexHull
    Commented Jun 23, 2023 at 11:46
  • $\begingroup$ 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? $\endgroup$ Commented Jun 23, 2023 at 11:54
  • 1
    $\begingroup$ 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, $$C_v=\left(\frac{\partial u}{\partial T}\right)_V$$ $\endgroup$ Commented Jun 23, 2023 at 13:48
  • 2
    $\begingroup$ See Transport Phenomena by Bird, et al, Chapter 11 for correct mathematical handling of these quantities. $\endgroup$ Commented Jun 23, 2023 at 15:11

1 Answer 1


Let me quote Landau & Lifshitz: Fluid Mechanics, $\S 50$, page 188, $3^{rd}$ed:

If the fluid velocity is small compared with the velocity of sound, the pressure variations occurring as a result of the motion are so small that the variation in the density (and in the other thermodynamic quantities) caused by them may be neglected. However, a non-uniformly heated fluid is still not completely incompressible in the sense used previously. The reason is that the density varies with the temperature; this variation cannot in general be neglected, and therefore, even at small velocities, the density of a non-uniformly heated fluid cannot be supposed constant. In determining the derivatives of thermodynamic quantities in this case, it is therefore necessary to suppose the pressure constant, and not the density.

From which it follows the use of $c_p$

  • $\begingroup$ Thank you for the answer. I appreciate it. $\endgroup$
    – ConvexHull
    Commented Jun 24, 2023 at 7:21

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