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?

• In the equation for de, shouldn't that be a $d\rho$? – Chet Miller Nov 8 '18 at 2:12
• @ChesterMiller Yes You are right, sorry for the typo – drlh Nov 8 '18 at 2:21
• 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. – Chet Miller Nov 8 '18 at 2:36
• 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. – Chet Miller Nov 8 '18 at 3:55
• Perhaps he meant the remark to apply to a constant density fluid. – Deep Nov 8 '18 at 10:03

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.
• This seems like a nice answer; however, the statement in the question was 'if $p=p(\rho)$'. Is that implied, if $\rho=\rho(p)$? – Time4Tea Nov 12 '18 at 18:15
• @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. – hyportnex Nov 12 '18 at 18:28
• 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? – Time4Tea Nov 13 '18 at 14:25
• 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. – hyportnex Nov 13 '18 at 14:35