How Much Power is Needed to Pump Water Up 10 Meters and Out at High Speed?

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Discussion Overview

The discussion revolves around calculating the power required to pump water from a river 10 meters below the engine level and discharge it through a nozzle at a specified speed. Participants explore the problem in terms of theoretical calculations, efficiency considerations, and the geometry of the nozzle. The scope includes mathematical reasoning and conceptual clarification related to fluid dynamics.

Discussion Character

  • Exploratory, Technical explanation, Homework-related, Mathematical reasoning

Main Points Raised

  • One participant presents the problem of calculating power with and without efficiency losses, providing specific parameters such as water density and gravitational acceleration.
  • Another participant suggests breaking down the problem by considering the volume of water discharged in one second and its mass, prompting further calculations related to energy required to lift the water.
  • Several participants discuss the calculation of the area of the nozzle based on its diameter, with one participant providing a detailed calculation and another suggesting to keep the answer in terms of π to avoid rounding errors.
  • Some participants express uncertainty about how to proceed with the calculations related to the diameter and volumetric flow rate, seeking clarification on the relationships between area, velocity, and flow rate.

Areas of Agreement / Disagreement

Participants generally agree on the approach to solving the problem but express varying levels of understanding regarding specific calculations and concepts. There is no consensus on the final calculations or methods to be used.

Contextual Notes

Limitations include potential rounding errors in calculations, dependencies on definitions of volumetric flow rate, and unresolved mathematical steps related to the problem.

Who May Find This Useful

This discussion may be useful for students or individuals interested in fluid dynamics, engineering applications, or those seeking help with similar physics problems involving power calculations and fluid flow.

Jerome1
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An engine pumps water from a river 10m below its own level and discharges it through a nozzle of diameter 10cm with a speed of 50ms^-1. Find the power required assuming (a) no losses. (b) 70% efficiency. Water weighs 10^3kgm^-3. (g = 10ms^-2)

Please help me with this problem, I've bever come across such a problem before..
 
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Jerome said:
An engine pumps water from a river 10m below its own level and discharges it through a nozzle of diameter 10cm with a speed of 50ms^-1. Find the power required assuming (a) no losses. (b) 70% efficiency. Water weighs 10^3kgm^-3. (g = 10ms^-2)

Please help me with this problem, I've bever come across such a problem b4..

Hi Jerome! :)

I am not sure where you are stuck, so I'll just give some hints.

Suppose we discharge water for 1 second.
What is the volume that comes out of the nozzle during this 1 second?
What is its mass?
How much energy does it take to transport that mass 10 meters up?
 
how to go about the diameter thing!
 
Jerome said:
how to go about the diameter thing!

Well, what is the area of a circle disk with the given diameter?
 
Here...
d = 2 * r
10 = 2 * r
r = 10/2
r = 5cm
Area = PI*r*r
Area = (22/7) * 25
Area = 550/7
= 78.57 ~
am i correct?
 
Jerome said:
Here...
d = 2 * r
10 = 2 * r
r = 10/2
r = 5cm
Area = PI*r*r
Area = (22/7) * 25
Area = 550/7
= 78.57 ~
am i correct?

That's a close estimate (and the right way of working) but in order to avoid rounding errors it's better to leave your answer in terms of $$\pi$$ at this stage and say $$A = 25\pi \text{ cm}^2$$

Now that you have the area of the nozzle can you work out the volume of water that passes through it per second? If you don't know the equation remember that volumetric flow rate has dimensions of $$\text{[length]}^3 \cdot \text{[time]}^{-1}$$

[math]Q_V = Av[/math]
 
i don't know how to work it out
 
Jerome said:
i don't know how to work it out

As I said in my last post the volumetric flow rate has dimensions of $$\text{[length]}^3 \cdot \text{[time]}^{-1}$$

Area has dimensions of $$\text{[length]}^2$$ and velocity has dimensions $$\text{[length]} \cdot \text{[time]}^{-1}$$

Can you work out how to get the volumetric flow rate using these dimensions?

edit: When working out be wary of units
 
Last edited:

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