# Bernoulli’s equation for a rotating fluid

• I

## Summary:

In a rotating fluid the pressure and kinetic energy density increase as you move outward, seemingly a violation of Bernoulli’s equation?
Consider a cylindrical container filled with an ideal fluid. Let it rotate at a constant angular speed (about the symmetry axis which is oriented vertically) and let the fluid be in the steady state.

Lets just talk about a horizontal slice so that the gravitational potential is constant. The pressure must increase as we move out from the center in order to supply the centripetal acceleration.

In the frame of reference rotating with the cylinder, everything makes sense. We have a fictitious centrifugal potential which decreases as you move from the center, and the pressure increases as you move out, so Bernoulli’s equation is satisfied.

I am confused how it works in the inertial frame. We no longer have any potential. The kinetic energy density increases as we move out from the center, as does the pressure. So how can Bernoulli’s equation be satisfied if they both increase?

What am I overlooking?

Related Classical Physics News on Phys.org
anuttarasammyak
Gold Member
Bernoulli's principle stands along the streamlines. I do not think it does not include radial direction that you are worrying about.

Bernoulli's principle stands for non viscous fluids. If water were non viscous, cylinder rotates but water would stay still.

etotheipi and Hiero
Bernoulli's principle stands along the streamlines. I do not think it does not include radial direction that you are worrying about.
Thank you! I never learned this but I actually was considering that it only works along streamlines.

The reason I thought this is from how we derive the Bernoulli equation from the equation ##\rho \frac{d\vec v}{dt} = \vec f - \nabla P##
We must integrate over space to get:

$$\int \rho \frac{d\vec v}{dt}\cdot d \vec x = \int \vec f \cdot d \vec x- \int \nabla P \cdot d \vec x$$

Assuming the force/volume is conservative, the right side gives the (negative) change in potential and pressure, but the left side can only be simplified if we take ##d\vec x = \vec v dt ## which is to say we integrate along a streamline. Then we can use ##\frac{d\vec v}{dt}\cdot \vec v= 0.5 \frac{d(v^2)}{dt}##

I suppose since the flow is assumed steady, we can say ##[\vec v =0] \implies [\frac{d\vec v}{dt}=0]## (otherwise ##\frac{\partial \vec v}{\partial t}\neq 0## which goes against the steady state assumption) and so that integral disappears which allows us to also use Bernoulli’s equation in regions with no flow.

Thank you. I’m sure there are more subtleties but I feel I understand it better now.

I must say I find it a bit odd that we can apply the Bernoulli equation in the vertical direction to find the height of the surface.

In the rotating frame it’s fine as there’s no flow so we can apply it everywhere.

But in the inertial frame the vertical direction is not along the streamlines so it’s really unjustified. (Right?)

Usually it is a choice if we want to analyze things in a non-inertial frame, but in this problem is seems more like a necessity!

anuttarasammyak
Gold Member
Though I am not so familiar with the topic, you may have to count centrifugal force as non conservative force, non zero vorticity and friction by viscous forces as I already mentioned in post #1 for the application of Bernoulli's principle in your case.