Consider the Lagrangian energy conservation equation for a viscous, conducting fluid $$ \rho D_t e = -p \nabla\cdot\vec u - \Pi:\nabla\vec u - \nabla\cdot\vec q $$ where $\rho$ is the fluid mass density, $e$ is the specific internal energy, $p$ is the hydrostatic pressure, $\vec u$ is the fluid velocity, $\Pi$ is the viscous stress tensor, and $\vec q$ is the heat flux. The operator $D_t = \partial_t+\vec u\cdot\nabla$ is the Lagrangian derivative. In order to cast this in terms of temperature rather than internal energy, one needs a constitutive relation between $e$ and $T$. I have seen two conflicting presentations.

In one case, an ideal equation of state is assumed, asserting $e = 3kT/2m = c_V T$, where $c_V$ is the ideal gas specific heat at constant volume. One then obtains trivially, $D_t e = c_V D_t T$.

However, in several places I have seen it claimed without reference that instead, the constant-pressure specific heat should appear: $D_t e = c_p D_t T$. An example is seen on the Wikipedia page for thermal diffusivity.

I am having trouble deriving either expression from first principles, that is, thermodynamic relations alone. My difficulty is that $D_t e$ represents the change in energy along a constant-mass trajectory, but it does not seem to constrain the fluid volume or pressure. Nevertheless, here's my best attempt:

For a constant-mass process, I start from the differential form of the 1st Law, $de = Tds - p dv$. This implies $\rho D_t e = \rho T D_t s - \rho p D_t v = \rho T D_t s - p \nabla\cdot\vec u $. This means I can express the energy transport equation in terms of the entropy instead $$ \rho T D_t s = -\Pi: \nabla\vec u - \nabla\cdot\vec q $$ I can now introduce a specific heat $c_? = T (\partial s/\partial T)_?$ via the chain rule to obtain $$ \rho c_? D_t T = -\Pi: \nabla\vec u - \nabla\cdot\vec q $$ where I have marked the underlying thermodynamic process as "?" to represent that it is not specified. So at least following this line of work, I'm at an impasse and still do not know: How should the specific heat appear when expressing energy conservation in terms of the fluid temperature?


1 Answer 1


Use continuity equation $0 = \partial_t \rho + \nabla \cdot (\rho \mathbf{u}) = D_t \rho + \rho \nabla \cdot \mathbf{u}$, to write $\nabla \cdot \mathbf{u} = -\dfrac{1}{\rho} D_t \rho$, and use this expression of the divergence of the velocity in the energy equation to get

$ \mathbb{S}:\mathbb{D} - \nabla \cdot \mathbf{q} = \rho D_t e - \dfrac{P}{\rho} D_t \rho =$
$\qquad \qquad \qquad = \rho\left[ D_t e - \dfrac{P}{\rho^2} D_t \rho \right] = $
$\qquad \qquad \qquad = \rho\left[ D_t e + D_t \left( \dfrac{P}{\rho} \right) - \dfrac{1}{\rho} D_t P \right] = $

$\qquad \qquad \qquad = \rho D_t \left( e + \dfrac{P}{\rho} \right) - D_t P = $
$\qquad \qquad \qquad = \rho D_t h - D_t P$ ,

where I introduced the definition of enthalpy $h = e + \frac{P}{\rho}$, that can be written as $h = c_P T$ for an ideal gas.

If you start from the conservative form of the differential balance

$\partial_t (\rho e) + \nabla \cdot ( \rho e \mathbf{u} ) = - p \nabla \cdot \mathbf{u} + \mathbb{S} : \mathbb{D} - \nabla \cdot \mathbf{q}$,

you can recast it as

$\partial_t (\rho e) + \nabla \cdot \underbrace{\left( \rho \left( e + \dfrac{P}{\rho} \right) \mathbf{u} \right)}_{= \rho h \mathbf{u}} = - \mathbf{u} \cdot \nabla p+ \mathbb{S} : \mathbb{D} - \nabla \cdot \mathbf{q}$

Litte note about your answer. It looks that you used the wrong sign in the dissipation term $\Pi : \nabla \mathbf{u}$ (here I called $\Pi = \mathbb{S}$ and $\mathbb{D}$ the symmetric part of $\nabla \mathbf{u}$).

Local equilibrium in fluid mechanics. Following your derivation of the entropy equation (with the right sign):

$\mathbb{S}:\mathbb{D} - \nabla \cdot \mathbf{q} = \rho T D_t S$

we can write

$\mathbb{S}:\mathbb{D} - \nabla \cdot \mathbf{q} = \rho T D_t S$
$\qquad \qquad \qquad = \rho D_t h - D_t P$
$\qquad \qquad \qquad = \rho D_t e - \dfrac{P}{\rho}D_t \rho$

so that it's quite easy to see that:

  • $T D_t s = D_t h$ for material particles experiencing constant p, so that

    $D_t h = T D_t s = T(\partial_T s)_p D_t T + T(\partial_p s)_T \underbrace{D_t p}_{=0}$$\quad \rightarrow \quad$$D_t h = T(\partial_T s)_p D_t T =: c_p D_t T$

  • $T D_t s = D_t e$ for material particles experiencing constant $\rho$, so that

    $D_t e = T D_t s = T(\partial_T s)_{\rho} D_t T + T(\partial_{\rho} s)_T \underbrace{D_t \rho}_{=0}$$\quad \rightarrow \quad$$D_t e = T(\partial_T s)_{\rho} D_t T =: c_v D_t T$

  • $\begingroup$ I don't see how going to the enthalpy helps. I can arrive at your result from my own by just writing $\rho T ds = \rho dh - dp$. I definitely do not agree that $h = c_p T$. However, if the flow is isobaric, then $D_t p = 0$ and we can use $ D_t h = c_p D_t T$. But that is only for isobaric flow. Is that the best I can do? $\endgroup$
    – Endulum
    Oct 24, 2022 at 18:41
  • $\begingroup$ if you use an ideal gas, you have to believe that $h = c_P T$, or to go back and study ideal gases. And if you transform $e = c_v T$, or even assuming $de = c_v dT$, you're using ideal gas equations, otherwise internal energy is in general a function of 2 independent thermodynamic variables, $e(\rho, s)$ as an example $\endgroup$
    – basics
    Oct 24, 2022 at 19:02
  • $\begingroup$ Moreover, since equations of fluid mechanics assume local thermodynamic equilibrium, you only have to care that the pressure felt by each material particle doesn't change, i.e. $D_t P = 0$, to get the local isobaric condition. About this, I', adding a note to my answer. $\endgroup$
    – basics
    Oct 24, 2022 at 19:06
  • $\begingroup$ I am interested in non-ideal fluids in general. I brought up the ideal gas only to show that in that limit, one can arrive at a temperature equation involving $c_V$. I do agree with your most recent edit that if you restrict the flow to be isochoric/isobaric, then it is clear how $c_V$ or $c_p$ appear respectively. $\endgroup$
    – Endulum
    Oct 24, 2022 at 21:02
  • $\begingroup$ And thanks for catching the sign error on the viscous stress. It's fixed in the original now. $\endgroup$
    – Endulum
    Oct 24, 2022 at 21:04

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