# How to calculate fountain GPM for a fountain height of 120 feet

## Homework Statement

How can I calculate the GPM (gallons per minute) of water needed to make a fountain shoot 120 feet in the air straight up? The diameter of the nozzle is 2 inches and friction is neglected. Also, what is the minimum liquid horsepower of the pump neglecting friction and assuming 100% efficiency. Please explain with formulas. Thanks.

## Homework Equations

Need equations. I'm sure there is a very simple equation to calculate this since there is only 2 given values, I just can't find it.

## The Attempt at a Solution

Need an equation to solve.

rock.freak667
Homework Helper
If you use conservation of energy, what velocity should the water exit the nozzle at?

Not sure how to get that. I know Q = VA, so if i knew the velocity I could get Q but not sure how.

rock.freak667
Homework Helper
Not sure how to get that. I know Q = VA, so if i knew the velocity I could get Q but not sure how.

Right so if you consider the kinetic energy of the water leaving the nozzle is being converted into the gravitational PE to each 120 ft. you will be able to get the velocity.

I don't know what that equation is though.

rock.freak667
Homework Helper
I don't know what that equation is though.

Do you know what the equations are for kinetic energy and potential energy?

KE = 1/2 mv^2
PE = mgh

rock.freak667
Homework Helper
KE = 1/2 mv^2
PE = mgh

and if you put them equal to each other, can you get the velocity?

Sure, but what do I use for m?

rock.freak667
Homework Helper
Sure, but what do I use for m?

You don't need 'm'

I see. So I got 88 ft/s for my velocity. Now Q = (88 ft/s)([PI x {2in/12}^2] / 4) = 1.9198 x 60 sec = 115.1917 ft/min? How do I get to GPM?

Nevermind, I figured it out. So I got 861 GPM. Now I need to figure out the liquid horsepower. I have the equation LHP = (Q x HP) / 3960. Not sure what to use for HP.