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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?

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    $\begingroup$ In the equation for de, shouldn't that be a $d\rho$? $\endgroup$ – Chet Miller Nov 8 '18 at 2:12
  • $\begingroup$ @ChesterMiller Yes You are right, sorry for the typo $\endgroup$ – drlh Nov 8 '18 at 2:21
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    $\begingroup$ What the author says about p being a function of $\rho$ only makes no sense to me either, because, even for an ideal gas, the temperature must also be varying in tandem for an isentropic deformation. $\endgroup$ – Chet Miller Nov 8 '18 at 2:36
  • $\begingroup$ Maybe it means that, since dw is an exact differential, it follows that at constant S, the term $dp/\rho$ must be an exact differential. $\endgroup$ – Chet Miller Nov 8 '18 at 3:55
  • $\begingroup$ Perhaps he meant the remark to apply to a constant density fluid. $\endgroup$ – Deep Nov 8 '18 at 10:03
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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.

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  • $\begingroup$ This seems like a nice answer; however, the statement in the question was 'if $p=p(\rho)$'. Is that implied, if $\rho=\rho(p)$? $\endgroup$ – Time4Tea Nov 12 '18 at 18:15
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    $\begingroup$ @Time4Tea thermodynamically $p=p(\rho)$ is the same as $\rho = \rho (p)$, that is neither variable depends on $s$; obviously they are the same assumptions whenever either is invertible, but even if it is not invertible globally it should be equivalent over finite monotonic segments of the variables. $\endgroup$ – hyportnex Nov 12 '18 at 18:28
  • $\begingroup$ Ok, I see. So, it's the fact that $\rho=\rho (p)$ that implies that the enthalpy can be split into separate functions, $f(p)$ and $g(s)$? So, could another way of looking at it be that $T=T(s)$ (since $T(s,p)=g'(s)$), which means that entropy could change if heat were added or removed, but as you say, not due to the mechanical state of the flow? $\endgroup$ – Time4Tea Nov 13 '18 at 14:25
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    $\begingroup$ Yes, that is my interpretation, too. I would also make the distinction that $\rho = \rho (p)$ being equivalent to $p=p(\rho)$ is a mathematical "detail" regarding continuous functions without apparent physical significance but that we can write the enthalpy as $g(s)+f(p)$ as equal consequence of either assuming $\rho = \rho (p)$ or $T=T(s)$ is physically interesting in that both imply the uncoupled, ie., separate thermal and mechanical nature of the flow. $\endgroup$ – hyportnex Nov 13 '18 at 14:35

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