# Speed of ejection of liquid from revolving tank

Consider a cylindrical tank filled with a liquid. Suppose it has an orifice somewhere at the bottom and the liquid is spurting out of this hole at a certain velocity. Now, if you rotated this cylinder about an axis passing through its centre, will the speed of ejection of fluid change?

I feel it should, because of a centrifugal force, but how do I prove this mathematically?

• What's causing the initial exit velocity (speed)? are you assuming a constant refilling of the tank to maintain the fluid level & thus the pressure head? Next, are you assuming the fluid has enough viscosity to be accelerated to the angular velocity of the tank? – Carl Witthoft Dec 12 '13 at 14:27
• Ignore viscosity. Also there is no constant refilling – user34304 Dec 12 '13 at 14:29
• Without viscosity there's no momentum imparted to the fluid, so no change in "ejection speed." But as the tank empties, the force decreases linearly with fluid height. – Carl Witthoft Dec 12 '13 at 15:19
• you are struck at centrifugal force. suppose a giant monster is rotating the whole tank in a big circle. from the ground frame of reference the speed of ejection of fluid Will not change. – user31782 Dec 14 '13 at 5:26

## 1 Answer

Suppose the cylinder is very wide. Then certainly centrifugal force would cause the fluid pressure to be higher at the perimeter than it is at the center. So if the hole is near the periphery, there is a greater "head" there, so fluid should be ejected at higher velocity.

Ignoring viscosity, the velocity should be proportional to square root of pressure.

EDIT: Got some drive-by downvotes, but here's the mental picture of the rotating tank. Anyway, in the $\omega = 0$ case, the pressure at the orifice is just proportional to height of the liquid above it. In the $\omega = high$ case, assuming the liquid is rotating at the same speed as the tank, then the radial acceleration of a unit of liquid is $r\omega^2$, which has to be integrated over the radius of the liquid from rim to surface, and multiplied by density $\rho$ to get pressure. Clearly that can be very high.

Then, if you want to know the ejection velocity, Bernoulli (much simplified) says $V^2 = 2p/\rho$.

I'm not sure but I suspect in the limit the ejection velocity would equal the rim velocity (just like throwing a baseball).

• Sir, how exactly did u get that expression? – user34304 Dec 13 '13 at 16:07
• @user34304: It's basic Bernoulli, and no need for "sir" :) – Mike Dunlavey Dec 13 '13 at 16:48
• @MikeDunlavey if firstly the wide cylinder was not rotating water was still in it let's say ejection speed was $v$. Now after rotation $v$ will increase but this $v$ will be in ground frame of reference not w.r.t the surface of rotating cylinder because both cylinder and water will have same tangential velocity.can you elaborate your answer by including speed w.r.t certain frame of reference. – user31782 Dec 14 '13 at 5:40