# Rate of flow from a bucket with a hole in it.

by jprockbelly
Tags: bucket, flow, hole, rate
 P: 5 1. The problem statement, all variables and given/known data Consider a bucket with a hole, area A, near the base. If the bucket is filled with water to a height h above the hole at what rate will water flow out of the hole? 2. Relevant equations I would guess that the relative equations are the velocity head, v2 = 2*h*g, and the rate of flow, Q = A*v. However I am not sure that this is the correct way to go, as I would have thought that the pressure difference between the water at the base of the bucket and the atmosphere would be important. 3. The attempt at a solution If I am correct this can then be solved as: Q=A*(2*h*g)^0.5, as a volume per second Can anyone tell me if there is another way to do this? {sorry, just realised this should have been postd in the physics, not maths forum}
 P: 37 This is correct. Although the air pressure is pushing down on the water at the top of the bucket, it is also pushing on the water coming out the hole in bottom. Sort of like it is trying to push the water back up into the bucket. The effect is that these two pressures from the atmosphere cancel out.
Math
Emeritus
No, the "pressure difference between the water at the base of the bucket and the atmosphere" is NOT important. What is important is the pressure at the hole. The water in the bucket under the hole is irrelevant. Also the atmospheric pressure is not important because it is essentially the same at the top of the water and at the hole. On the pressure caused by the weight of the water above the hole is important. If the top of the water is h above the hole, then you have a column of water of volume hA above the hole. Its weight is $g\mu hA$ where $\mu$ is the density of water which, in g/cm3, you can take to be 1. The force at the hole is ghA so the pressure is gh.