To start with I know thermodynamics deals with processes at equilibrium. Hence the thermodynamic pressure should most likely be the pressure of a fluid at equilibrium.
I'm not sure if a fluid flow (in general unsteady) is in thermodynamic equilibrium (say flow in a channel which has a pressure gradient) and so would the static pressure at a point in the channel be different from the thermodynamic pressure?
What does this entail about the ideal gas law $p = \rho RT$? can it be used for moving flow? What is the pressure in the equation referring to; mechanical or thermodynamic?
EDIT: To clear up any confusion- In a given flow we can measure the pressure at any point, say using a pitot tube to get the stagnation and static pressure. My question is then, is the static pressure we measure (which is by definition an "$F/A$" (force / area) quantity any different from the thermodynamic pressure? The pressure in $P = \rho RT$ must be referring to the thermodynamic pressure, since the equation is derived purely from the laws of thermodynamics. However, in all literature I have encountered, compressible flows use the ideal gas equation to as a link between the incompressible variables ($p, \mathbf{V}$) and the full set of compressible variables ($p, \mathbf{V}, \rho, T $). So it seems the two pressures are equivalent?
Best Answer
The difference has to do with the fact that when you sum the normal stress on each face of a differential fluid element using the Newtonian constitutive law, you get something different from the thermodynamic pressure, which is what you normally think of as "pressure". There's a good explanation of it in Viscous Fluid Flow by Frank White.
So the constitutive law for a fluid (or any continuum) is what connects the stress to the strain. For a Newtonian fluid, the constitutive law is:
$$\tau_{ij} = -p\delta_{ij}+\mu(u_{i,j}+u_{j,i}) + \delta_{ij}\lambda u_{k,k}$$
Where $\mu$ is the dynamic viscosity and $\lambda$ is the bulk viscosity, both properties of the fluid. When you sum this over all the faces of the fluid element, you get:
$$\tau_{ii}=-3(p-u_{i,i}(\frac{2}{3}\mu+\lambda))$$
Divide by -1/3 to get:
$$p_{mech} = p_{therm}-u_{i,i}(\frac{2}{3}\mu+\lambda)$$
The original pressure term was the thermodynamic pressure and I added a subscript to make it a little clearer in the last equation. These two pressures are different by the product of the divergence of the velocity and a term related to the material properties. If you're talking about incompressible flow, then there's no difference at all because the divergence of the velocity is zero. If you're talking about compressible flow, then the difference is still small, but depends on how compressible the fluid is and how big this fluid property term is. Stokes basically assumed away the problem by saying that $$\frac{2}{3}\mu+\lambda=0$$ Which is "Stokes Hypothesis"