Stuck on a 1500hp Pump Problem: Solve for Velocity & Efficiency

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Homework Help Overview

The problem involves a 1500hp pump that propels water to a height of 125 meters. Participants are tasked with determining the speed of the water as it exits the pump and calculating the flow rate based on the pump's efficiency.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the need to find the velocity of the water without knowing the pipe size and question if they should use motion equations. There is a suggestion to consider kinetic and potential energy in the analysis.

Discussion Status

Some participants are exploring the relationship between kinetic energy and potential energy, while others express confusion about the formulas and their application. Guidance has been offered regarding the relevant equations, but there is still uncertainty about how to proceed.

Contextual Notes

Participants note a lack of clarity regarding the necessary formulas and express concern over their understanding of the physics concepts involved, particularly after a long time away from the subject.

crslax9
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I have been working on this problem for 2 days and can't seem to figure it out..

A 1500hp pump throws a jet of water 125 m into the air. (a) With what speed does the water leave the mouth of the fountain. (b) If the overall efficiency is 60%, how many liters of water per minute are thrown into the air?

Thank you for the help as I am pulling my hair out.

I think where I am getting lost is that I need to find the velocity without knowing the size of the pipe. Am I on the right track? Should I be using a motion equation to first get the velocity?

I am just not sure why I am getting so stuck on this problem.
 
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crslax9 said:
I think where I am getting lost is that I need to find the velocity without knowing the size of the pipe. Am I on the right track?

No, you're barking up the wrong tree. Think energy, as in:

How much kinetic energy must, say, 1kg of water have had in order to make it 125 feet in the air?
 
Tom Mattson said:
No, you're barking up the wrong tree. Think energy, as in:

How much kinetic energy must, say, 1kg of water have had in order to make it 125 feet in the air?

so if kinetic energy is (1/2)mv^2 all over v what numbers go where? Maybe this is where I am getting lost? If the v on the top of the equation is velocity and the v on the bottom is volume correct?
 
crslax9 said:
If the v on the top of the equation is velocity and the v on the bottom is volume correct?

I have no idea of what you are saying here.

You're interested in the speed with which the water leaves the pump, right? Well then wouldn't it make sense to consider the KE at the location of the pump, and the PE at the top of the arc?
 
Tom Mattson said:
I have no idea of what you are saying here.

You're interested in the speed with which the water leaves the pump, right? Well then wouldn't it make sense to consider the KE at the location of the pump, and the PE at the top of the arc?

Yes it does make sense but like I said I am lost on what formulas to use and how to use them. it has been a long time since I last had physics.

If you can give me an idea of what to use to begin this equation I think I may be able to go from there.

Thank you for your help so far
 
crslax9 said:
If you can give me an idea of what to use to begin this equation I think I may be able to go from there.

You need the equations for kinetic energy and for gravitational potential energy.
 
okay I know both of those equations, but why do I need to know gpe if I only need the speed of the water at the pumps mouth?
 

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