# In a perfect fluid, is incompressibility incompatible with thermodynamic relations?

When we study fluid dynamics we assume that the hypothesis of local equilibrium holds. It means that every point of the fluid is in equilibrium with its immediate surroundings.

If I understand it correctly, this allows us to define intensive thermodynamic parameters in every point, such as:

• internal energy for unit of mass $$\epsilon$$;
• entropy for unit of mass $$s$$;
• pressure $$p$$;
• density $$\rho$$ (actually, we don't need thermodynamics for this one, anyway I list it here because $$\rho$$ enters the thermodynamics relations).

Even better, local equilibrium implies that these quantities are related to each other by equilibrium thermodynamic relations. And in equilibrium we know that all intensive parameters are determined by only two of them. You have to choose two independent parameters, of course.

Anyway, these considerations seem to be in contrast with what happens in a perfect incompressible fluid, as I'll explain now.

Let's consider a steady flow of a perfect incompressible fluid. Since the fluid is perfect, heat conduction doesn't take place, therefore entropy $$s$$ is constant along a streamline. Since the fluid is incompressible, the density $$\rho$$ is a constant too. Therefore we have two intensive parameters that don't change along a streamline. Then, according to ordinary thermodynamics, also all the other intensive parameters should be constant.

But this does not happen for pressure $$p$$. For example, if the flow takes place in a pipe that at some point shrinks, then the speed has to increase and, according to Bernoulli's Theorem, pressure has to drop.

Question: Why did I find this contradiction? Maybe my understanding of local equilibrium is wrong? Or maybe the two assumptions of perfect fluid and incompressibility together are incompatible with normal thermodynamic equilibrium relations? This last option looks plausible to me because these assumptions are both approximations, after all. Maybe together they lead to a sort of "degenerate" thermodynamics, in which pressure can change independently of entropy, energy and density. But I hope that someone has a more precise explanation.

• See replies to this other question: physics.stackexchange.com/questions/427000/… Commented Nov 9, 2019 at 18:49
• @GiorgioP Thank you for the link. Unfortunately, it doesn't dispel my doubts, because the reply to that question focuses on the difference between thermodynamic pressure and mechanical pressure, but I looked it up, and found this expression for the relationship that ties the two pressures: $p_{mech} = p_{therm} - \left( \lambda + \frac{2}{3} \mu \right) \nabla \cdot \vec{v}$, where $\vec{v}$ is the fluid velocity and $\lambda$, $\mu$ are coefficients representing viscosity. Now, if the fluid is perfect there is no viscosity, so $\lambda = \mu = 0$. Commented Nov 10, 2019 at 0:39
• Then in my example the two pressures coincide. I took that formula from equation (4) in article1 linked in the reply. Commented Nov 10, 2019 at 0:43