Model the pressure of a zero-gravity simple fluid system

[member 428835]

Hi PF!

A circular cylinder contains an inviscid liquid, which is softly disturbed (i.e. velocities are small). There are no body forces, which implies the equilibrium free surface is that of a spherical cap with radius $R$ (see figure A). Inviscid implies we can assume potential flow, such that the liquid velocity $V$ can be expressed in terms of it's potential such that $V = \nabla \psi$. The pressure throughout the liquid domain is $P = -\nabla \psi$ (recall velocities are small, so the non-linear term drops). The pressure at the free surface is governed by the Young-Laplace equation, which implies $P \sim \sigma/R^2$ there (actually this pressure term turns out to be VERY mathematically complicated, which is why I simply use the proportional notation). So the pressure balance at the free surface can be expressed by equating the two pressures at the interface.

Now let's assume the base of the cylinder is open such that flow can enter, shown in B). What is the pressure now in the liquid domain? I believe it is no longer $P = -\nabla \psi$, but may also include a reservoir pressure from the inflow. How would you model this, given the small disturbance of the transient interface from equilibrium is some function $\xi(x,y,z,t)$? There's a way I've been shown, but it doesn't intuitively make sense to me. Any help? Seems like @Chestermiller might know this one?

 

[Chestermiller]

What makes you think that, without gravity, the equilibrium shape will be as you have drawn it?

 

[member 428835]

Without body forces surface tension dictates the displacement. According to the Young-Laplace equation the radii of curvature will be equal to minimize energy. I drew it this way for some contact angle, which looks hydrophilic, but for sure it could be anything which changes the shape of the interface but maintains the spherical cap shape.

 

[Chestermiller]

It seems to me the cap should be convex. After all, the liquid will form a sphere if it is small.

 

[member 428835]

It seems to me the cap should be convex. After all, the liquid will form a sphere if it is small.

The liquid forms a sphere if it's in a pool of gas, but gas would also form a sphere if in a pool of liquid.

Ultimately this comes down to contact angle. If the liquid is hydrophilic (silicone oil on glass) we'd expect concave; if it's hydrophobic (mercury on glass) we expect convex. I assume the open flow scenario of B) is occurring slow enough that the shape remains spherical.

 

[Chestermiller]

The liquid forms a sphere if it's in a pool of gas, but gas would also form a sphere if in a pool of liquid.

Ultimately this comes down to contact angle. If the liquid is hydrophilic (silicone oil on glass) we'd expect concave; if it's hydrophobic (mercury on glass) we expect convex. I assume the open flow scenario of B) is occurring slow enough that the shape remains spherical.

I think the contact angle is an effect localized to the region near the wall. If there were no wall, the glob of fluid would be a sphere.

 

[member 428835]

I think the contact angle is an effect localized to the region near the wall. If there were no wall, the glob of fluid would be a sphere.

Without body forces, equilibrium is only achieved when all free surfaces have the same radius of curvature. Otherwise there will be a force imbalance according to the Young-Laplace equation. But for large scales (say well beyond the capillary length scale) in gravity, I agree with your statement, and in fact the interface will be approximately flat everywhere.

But here we consider no body forces, so the equilibrium interface will always be curved according to the contact-angle.