I have a hard time to wrap my head around pressure in the Navier-Stokes equation! It may sounds ridiculous but still I cannot understand the true meaning of pressure in the Navier-Stokes equation. Let's do some math to explain my purpose more accurately! Let's start from basics of the physics and in my opinion that would be the first equation in the classical thermodynamics as equation of state. We assume: there is a fluid, which have a equation of state as:
$$\rho = \rho(P,T)$$
Where $\rho$ is the density of the fluid, $P$ is the pressure, and $T$ is the temperature. Let's take a derivative from this equation to have:
$$d\rho = (\frac{\partial \rho}{\partial P})_{T} dP + (\frac{\partial \rho}{\partial T})_{P} dT$$
Let's assume that our fluid is in the thermal equilibrium and its temperature will not change, as a result: $d T = 0$
So, we have:
$$d \rho = (\frac{\partial \rho}{\partial P})_{T} dP$$
I know it's a lot of assumption but again let's assume that density change because of pressure change is not nonlinear and our fluid in fact behaves like a ideal gas. As a result, I call $(\frac{\partial \rho}{\partial P})_{T}$ the inverse square of the speed of sound, which is a constant number, as:
$$(\frac{\partial \rho}{\partial P})_{T} = c_{s}^{-2}$$
So, finally we have:
$$d \rho = c_{s}^{-2} d P$$
Or:
$$\Delta \rho = c_{s}^{-2} \Delta P$$
Or again:
$$(\rho - \rho_{f}) = c_{s}^{-2} (P - P_{0})$$
Where $\rho_{f}$ is the density of the fluid at the rest or reference, which is a tabulated value for each fluid, and $P_{0}$ is the reference pressure.
Now, I would assume my fluid is an incompressible fluid and it means (density is constant and it is really constant!):
$$\rho = \rho_{f}$$
As a result, because, every fluid regardless of its compressibility or incompressibility has a finite speed of sound, I would argue that:
$$P = P_{0}$$
Or in other word, strictly speaking pressure should be equal to the reference pressure.
Now, I proved that for an incompressible fluid as long as density is constant, pressure should also be a constant. So in incompressible Navier-Stokes equation we have:
$$\rho_{f} \frac{\partial \mathbf{u}}{\partial t} + \rho_{f} (\mathbf{u} \cdot \nabla)\mathbf{u} = -\nabla P + \nabla \cdot \tau$$
And I showed that for incompressible fluid, P is just constant, so: $\nabla P = 0$!
As a result, I could simplify the Navier-Stokes equation as:
$$\rho_{f} \frac{\partial \mathbf{u}}{\partial t} + \rho_{f} (\mathbf{u} \cdot \nabla)\mathbf{u} = \nabla \cdot \tau$$
Now let's back to my original question:
Based on these calculations I would say that pressure in the incompressible Navier-Stokes equation is just a dummy variable, which does not have any physical meaning! I appreciate if someone could explain this to me!
Answer
There are two pressures: Thermodynamic pressure $p_\text{thermo}$, and Mechanical pressure $p_\text{mech}$. Thermodynamic pressure, a concept from equilibrium thermodynamics and therefore applicable only to a static fluid, is given by an equation of state: $p_\text{thermo}=f(\rho,T)$, where $\rho$ is fluid density and $T$ its temperature. A moving fluid is not in equilibrium and its $p_\text{thermo}$ is not defined. Mechanical pressure is the isotropic part of the stress tensor and is defined for a moving fluid too; $p_\text{mech}$ appears in the Navier-Stokes equation.
If a static fluid is isothermal and has constant density ($\rho,T$ fixed) then $p_\text{thermo}$ is also fixed. But mechanical pressure given by hydrostatic equation varies with depth in an isothermal constant-density fluid.
Your derivation mixes up the two pressures. The relations are: \begin{align} \mathrm{d}\rho&=c_s^{-2}\,\mathrm{d}p_\text{mech}\\ \mathrm{d}\rho&=\left(\frac{\partial\rho}{\partial T}\right)\,\mathrm{d}T+\left(\frac{\partial\rho}{\partial p_\text{thermo}}\right)\,\mathrm{d}p_\text{thermo}. \end{align} The latter equation from thermodynamics is applicable only to a static fluid. The former is not a thermodynamic equation. An incompressible fluid is defined to be one whose density doesn't depend on its mechanical pressure $p_\text{mech}$; it doesn't say that $p_\text{mech}$ can't vary. Therefore as you approach the limit of an incompressible fluid, $\mathrm{d}\rho\to0$, we must necessarily have $c_s\to\infty$. It is incorrect to say "...every fluid regardless of its compressibility or incompressibility has a finite speed of sound..."; incompressible fluids don't exist, so a priori you wouldn't know what sound speed should be assigned to a hypothetical fluid; to be consistent with the definition of incompressibility however a variation in $p_\text{mech}$ must be allowed which demands that sound speed in an hypothetical incompressible fluid be infinite.
P.S. Here's article1 and article2 that may interest you (NB: both are PDFs).
No comments:
Post a Comment