# Does Bernoulli's equation apply to a fountain?

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Can I use Bernoulli's equation to calculate certain measures of a fountain, or does it only apply to fluids in pipes?
Also if so, how could I calculate the pressure inside a tube used in a fountain?

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Update: I want to use the equation for the water flowing out of the fountain.

By "certain measures of a fountain", do you mean when the water leaves the pipe?

Edit: Okay just saw your update

Pressure within the pipes should be governed by the equation, and well you should be able to calculate it based on the velocity of water in the pipe (or leaving its mouth)

For the water leaving the fountain, there is no longer an "area" we are looking for, and since we're introducing another fluid (air), it's gonna be more complicated than just the equation.

How would I calculate the velocity in the pipe?

Off-hand what I can think of is that you measure the velocity of the water leaving the mouth of the pipe, and from there work backwards if the pipe is not of uniform diameter or has branched into others

Ok, using the velocity of the water which would be constant in a pipe of uniform diameter, how would I calculate the pressure inside this pipe (which would be uniform also throughout the pipe?).

This will not require Bernoulli's equation. Rather, you can consider pressure P = F/A

Since force is constant with constant velocity (given by constant momentum of the water flowing through per second of a section of the pipe),

momentum p = mv

Therefore, F = p/t

Since t = 1s in your assumptions, subbing into P = F/A,

P = mv/At, where t = 1s

m of the water can just be A x v x density of water if we look at mass of water flowing in that section of the pipe per second

F=p/t doesn't this equation apply when momentum is being changed, but velocity is constant p=mv.

It applies to find average force during a change in momentum, but since momentum is not changed at all during this flow, the area under the F/t graph will be a nice rectangle. Hence average force = force we use to compute :)

Oh, thank you!

• Alloymouse