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

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SUMMARY

The calculation of gallons per minute (GPM) for a fountain height of 120 feet involves using the equations for kinetic energy (KE) and potential energy (PE). The velocity of the water exiting the nozzle is determined to be 88 ft/s. Using the formula Q = VA, where A is the cross-sectional area of the nozzle, the flow rate is calculated to be 861 GPM. The liquid horsepower (LHP) can be computed using the formula LHP = (Q x HP) / 3960, where HP is the pump horsepower.

PREREQUISITES
  • Understanding of basic physics concepts such as kinetic energy and potential energy
  • Familiarity with fluid dynamics principles, specifically flow rate calculations
  • Knowledge of the formula for cross-sectional area of a circle
  • Basic understanding of horsepower calculations in fluid mechanics
NEXT STEPS
  • Research the derivation of the equations for kinetic energy and potential energy
  • Learn about fluid dynamics principles, specifically the Bernoulli equation
  • Explore the calculation of liquid horsepower in pump systems
  • Study the effects of nozzle diameter on flow rate and fountain height
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Engineers, physicists, and anyone involved in designing water features or fountains, particularly those focused on hydraulic calculations and pump efficiency.

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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.
 
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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.
 
camino said:
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.
 
camino said:
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
 
camino said:
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?
 
  • #10
camino said:
Sure, but what do I use for m?

You don't need 'm' :wink:
 
  • #11
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?
 
  • #12
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.
 

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