Pascal's law (fluids): Derivable from fundamental laws?

[greypilgrim]

Hi.

Pascal's law states that static pressure in a confined incompressible fluid without gravity is the same everywhere. Is this law derivable from more fundamental laws? Some thoughts:

• Is Pascal's law part of the definition of the liquid state?
  
• If the liquid operates between two hydraulic cylinders of different diameters $A_1,A_2$, Pascal's law says $\frac{F_1}{A_1}=\frac{F_2}{A_2}$. Incompressibility means $V_1=A_1x_1=A_2x_2=V_2$, combining those equations leads to $W_1=F_1x_1=F_2x_2=W_2$, which is conservation of (mechanical) energy. Is Pascal's law equivalent to conservation of energy?

 

[jbriggs444]

Hi.

Pascal's law states that static pressure in a confined incompressible fluid without gravity is the same everywhere. Is this law derivable from more fundamental laws? Some thoughts:

• Is Pascal's law part of the definition of the liquid state?
  
• If the liquid operates between two hydraulic cylinders of different diameters $A_1,A_2$, Pascal's law says $\frac{F_1}{A_1}=\frac{F_2}{A_2}$. Incompressibility means $V_1=A_1x_1=A_2x_2=V_2$, combining those equations leads to $W_1=F_1x_1=F_2x_2=W_2$, which is conservation of (mechanical) energy. Is Pascal's law equivalent to conservation of energy?

I'd say conservation of momentum and transitivity. F=ma. If you look at the force balance on a small parcel of motionless fluid in the absence of shear stress, the pressure in the one side has to match the pressure out the other. If you arrange a bunch of small parcels back to back, transitivity assures you that the pressure is the same throughout.

 

[coquelicot]

Quoting Wikipedia:
" One can finally condense the whole source in one term, arriving to the incompressible Navier–Stokes equation with conservative external field:
$$\frac{\partial \mathbf{u}}{\partial t} + (\mathbf{u} \cdot \nabla) \mathbf{u} - \nu \, \nabla^2 \mathbf{u} = - \nabla h.$$
The incompressible Navier–Stokes equations with conservative external field is the fundamental equation of hydraulics."

So, Pacal law can be deduced from this equation, like most of the other things.

 

[greypilgrim]

Quoting Wikipedia:
" One can finally condense the whole source in one term, arriving to the incompressible Navier–Stokes equation with conservative external field:

∂u∂t+(u⋅∇)u−ν∇2u=−∇h.​

\frac{\partial \mathbf{u}}{\partial t} + (\mathbf{u} \cdot \nabla) \mathbf{u} - \nu \, \nabla^2 \mathbf{u} = - \nabla h.
The incompressible Navier–Stokes equations with conservative external field is the fundamental equation of hydraulics."

So, Pacal law can be deduced from this equation, like most of the other things.

In a static situation we have $\mathbf{u}=0$ everywhere and at all times and without external fields $h$ is zero as well, so the equation becomes $0=0$. Also, I don't think Pascal's law is true in non-static conditions.

 

[coquelicot]

Does not Pascal law follow immediately from the definition of a fluid and of incompressibility ? Let us assume that inside the confined space, you have an infinitesimal pressure meter made of a cylinder and a piston pushed by a spring (inside the cylinder there is only vacuum, and the piston prevents the fluid from entering inside the cylinder). Now, to augment slightly the pressure inside the confined space, you have to make something penetrating inside the confined space (e.g. another piston), otherwise, since the piston of the pressure meter is free to press the spring, an empty space would be created, zeroing the pressure. Since the fluid is incompressible, the volume of the "entering thing" MUST be equal to the volume left by the pressure meter piston after it has pressed the spring, and this is independent of WHERE the pressure meter is located. So, the difference in pressure must be felt the same at every point.
If, in place of making something enter inside the confined space, you make the confined space smaller (e.g. a balloon can be pressed), the same argument holds.
Now, to bring the confined space to some pressure, you have to augment the pressure more and more, starting from 0. Since the dP is the same at every point during this augmentation, it follows that the P is the same at every point.

 

[coquelicot]

Another answer. See this article, where the law $dP = \rho g dh$ is derived from Newton laws (or the Navier-Stokes equations):
https://en.wikipedia.org/wiki/Hydrostatic_equilibrium
So, without gravity, you have $g = 0$, hence $dP = 0$ hence $P ={\rm Const}$.

This seems to confirm what I always believed: all or almost everything in fluid mechanics can be deduced from the Navier-Stokes equations (despite the one I wrote above is indeed not relevant).

 

[boneh3ad]

Draw a tetrahedral differential fluid element with three triangular sides normal to $x$, $y$, and $z$ and the fourth side completing the shape. Assume they all have different pressures and sum the forces. What condition must apply for this element to be in equilibrium?