Thermodynamics – Does Isentropic Flow Imply Homentropic Flow in Euler’s Equations?

entropyfluid dynamicsthermodynamics

In the derivation of the Euler's equation of motion for ideal fluids by Landau and Lifshitz, I am able to follow the derivation of the following equation
$$ \frac{Dv}{Dt} = \frac{\partial v}{\partial t} + (\textbf{v} \cdot \nabla) \textbf{v} = – \frac{\nabla p}{\rho} + \textbf{g}$$
This is assuming there are no viscous forces present. Furthermore if we assume there is no heat input to the system, then the process is fully isentropic, and this is expressed via the equation:
$$ \frac{Ds}{Dt} = \frac{\partial s}{\partial t} + \textbf{v} \cdot \nabla s = 0$$
This is clear to me. However, introducing thermodynamic quantities and potentials into the game is what has me confused. Namely, when we introduce the specific volume $V$, specific entropy $s$, specific enthalpy $w$, specific internal energy $u$, we promote them to fields (i.e. functions of $\textbf{x}$ and $t$ ) and then assume that they locally obey the usual thermodynamic equations:
$$ dw = T ds + V dp $$
$$ du = T ds – p dV $$
The notation of the differential $d$ sort of confuses me. What I think the above equation implies is three seperate equations, e.g. for the enthalpy equation is given by:
$$ \frac{Dw}{Dt} = T \frac{Ds}{Dt} + V \frac{Dp}{Dt}$$
$$ \frac{\partial w}{\partial t} = T \frac{\partial s}{\partial t} + V \frac{\partial p}{\partial t}$$
$$ \nabla w = T \nabla s + V \nabla p$$

and I believe this is so because the third equation above appears on page 9 in Landau and Lifshitz. Now Landau and Lifshitz makes the claim that isentropic condition $\frac{Ds}{Dt} =0$ implies that $\nabla w = – \frac{1}{\rho} \nabla p$. But in order for this to be true, per above we need $\nabla s = 0$, which is not the same thing as $\frac{Ds}{Dt} =0$.
We need the equation $\nabla w = V \nabla p = \frac{1}{\rho} \nabla p $ in order to write as Euler's equation as
$$ \frac{Dv}{Dt} = \frac{\partial v}{\partial t} + (\textbf{v} \cdot \nabla) \textbf{v} = – \nabla w+ \textbf{g}$$
So I guess my main questions are : where is the logic wrong here? Is my interpretation of the equation with the differentials wrong? How does $\frac{Ds}{Dt} = 0$ imply that $\nabla w = V \nabla p$?

Best Answer

You're right with your interpretation of the differentials, from differantials to differential operators acting on fields.

You're right that in continuum mechanics, we are promoting thermodynamic variables to fields, since we often assume local thermodynamic equilibrium, i.e. time to reach thermodynamic equilibrium is much faster than characteristic time of fludi dynamics processes.

The last step you're missing, and Landau is not telling you explicitly is that

  1. $\dfrac{D s }{D t} = 0$ and
  2. uniform conditions at far field, $s(\mathbf{r}_{\infty, inflow}) = \overline{s}$, and
  3. no shock occurring in the domain of interest (since shocks produce entropy, even though you can't approach them with differential equations, if you model them as discontinuities in a inviscid model),

imply that entropy is uniform across the whole domain of interest $s(\mathbf{r}) = \overline{s}$.

You can get a "qualitative proof" of the statement above thinking at the meaning of the material derivative, i.e. the rare of change of the quantity involved as it's seen by a material particle. If every material particle entering the domain of interest has initial entropy $\overline{s}$ (assumption 2.), and there is no entropy production (assumption 3., no shock), all the fluid particles in the domain have the same entropy, and thus the space derivatives are zero as well.

Related Question