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

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SUMMARY

The discussion focuses on calculating the power required to pump water from a river 10 meters below the engine's level, discharging it through a 10 cm diameter nozzle at a speed of 50 m/s. The power is calculated under two scenarios: (a) with no losses and (b) with 70% efficiency. Key calculations involve determining the volumetric flow rate using the nozzle area and the mass of water being pumped, considering the density of water at 1000 kg/m³ and gravitational acceleration at 10 m/s².

PREREQUISITES
  • Understanding of fluid dynamics concepts, specifically volumetric flow rate.
  • Knowledge of basic physics principles, including gravitational potential energy.
  • Familiarity with the formula for the area of a circle.
  • Basic proficiency in unit conversions and dimensional analysis.
NEXT STEPS
  • Calculate the power required to pump water using the formula: Power = (mass flow rate * g * height) + (0.5 * density * velocity^2).
  • Explore the impact of efficiency on power calculations in fluid systems.
  • Learn about the principles of hydraulic systems and their applications in engineering.
  • Investigate the effects of nozzle diameter on flow rate and pressure in fluid dynamics.
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Engineers, physics students, and anyone involved in fluid mechanics or hydraulic system design will benefit from this discussion.

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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