Question about Isentropic flow

Question

I'm reading a book "A Mathematical Introduction to Fluid Mechanics" by Alexandre J. Chorin that says that the flow is isentropic if

$$ \nabla w = \frac{1}{\rho}\nabla p $$

where $w$ is an enthalpy, $\rho$ is density and $p$ is a pressure.

Then Chorin recalls the following two facts from thermodynamics

$$ dw = Tds + \frac{1}{\rho}dp $$ and $$ de = Tds + \frac{p}{\rho^2}d\rho $$ where $e$ is the internal energy, $s$ entropy and $T$ temperature.

But I do not understand the next claim of the book:

if $p$ is a function of $\rho$ only, then the flow is clearly isentropic.

Why is this true?

Answer 1

The underlying assumption in local equilibrium theories is that any thermodynamic variable locally depends only on two parameters, here they are $s$ and $p$. So the enthalpy is taken to be $w=w(s,p)$ and also $dw=\frac{\partial w}{\partial s}ds + \frac{\partial w}{\partial p} dp=Tds+\frac {1}{\rho}dp$ at every point; here $T=T(s,p)$ and $\rho = \rho (s,p)$.

If one also assumes that $\rho = \rho (p)$ independent of entropy then one must also have both $\frac{\partial w}{\partial s} = T$ and $\frac{\partial w}{\partial p} =\frac {1}{\rho}$. So integrating the latter we get $w(s,p)=f(p)+g(s)$ and then $T(s,p)=g'(s)$ and thus the enthalpy change is really the sum of two independent exact differentials, one in the $s$ and the other in $p$ (or $\rho$) so the two are completely unrelated $dw=g'(s)ds + f'(p)dp = dg(s) + df(p)$ and one can change without affecting the other.

It may be a misnomer to say that it is isentropic, i.e., $ds=0$, as a consequence of $\rho$ being dependent on $p$ alone; the important point is that entropy change itself has no effect on the fluid flow and specifically on its mechanical state.