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

  • Thread starter Thread starter jprockbelly
  • Start date Start date
  • Tags Tags
    Flow Hole Rate
AI Thread Summary
The discussion centers on calculating the rate of water flow from a bucket with a hole at its base. The key equation for flow rate is Q = A*(2*h*g)^0.5, where A is the hole area, h is the water height above the hole, and g is the acceleration due to gravity. It is clarified that while atmospheric pressure acts on both the water surface and the hole, it cancels out and does not affect the flow rate. Instead, the pressure from the weight of the water column above the hole, given by gh, is the crucial factor in determining flow. The conversation emphasizes understanding the role of pressure in fluid dynamics, particularly in this scenario.
jprockbelly
Messages
5
Reaction score
0

Homework Statement



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?


Homework 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.

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 realized this should have been postd in the physics, not maths forum}
 
Last edited:
Physics news on Phys.org
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.
 
jprockbelly said:

Homework Statement



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?


Homework 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.

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 realized this should have been postd in the physics, not maths forum}
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.
 
Thanks to both of you for your replies.

HallsofIvy, I am somewhat confused by your reply. Are you saying that the pressure IS important, or are you simply elucidating for my benefit?
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
Back
Top