# Mechanical Pressure vs. Thermodynamic Pressure

In fluid mechanics I learned about the difference between the mechanical and thermodynamic pressure. From Stoke's hypothesis we equate the mechanical pressure and thermodynamic pressure to each other as $p = p_m = p_t$. However I had one concern.

Let's say that we have a pool water a couple of meters deep. Let's also assume that the entire body of water is of uniform temperature and density (as is the case for many hydrostatics problems). From hydrostatics, the absolute pressure at the very bottom of the pool would be $p = p_0 + {\rho}gh$. However, if I am assuming that the absolute pressure at the bottom is equivalent to thermodynamic pressure (according to Stoke's hypothesis), then the absolute pressure must also be $p = p(\rho,T)$ from the state postulate.

But I assumed uniform temperature and density, so according to thermodynamics, my pressure should be uniform throughout the fluid. But from from a hydrostatics perspective, the pressure should be higher the deeper the fluid. What is going on here?

• I do have one more question however. If thermodynamic pressure and mechanical pressure essentially become decoupled in an incompressible fluid, how is the Stokes hypothesis still valid? – mechanician100 Sep 9 '16 at 6:27

The correct equation is $$\frac{dp}{dh}=\rho(p,T) g$$ So, to be strictly correct, you need to take into account the effect of pressure on density. Then, everything will be consistent. If you consider the material to be incompressible (constant density), the pressure cannot be determined from an equation of state.
It's because of the assumptions you made. As soon as you assume constant density, you have now decoupled thermodynamic and mechanical pressures. The thermodynamic pressure is now constant, and the mechanical pressure is the dynamic pressure, $1/2 \rho u^2$.