Work, energy and power question

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Homework Help Overview

The problem involves calculating the power developed by a pump that raises water through a vertical distance while also delivering it at a specific speed. The subject area includes concepts of work, energy, and power in the context of fluid mechanics.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculation of kinetic energy and work done against gravity, questioning how to incorporate the speed of the water and the significance of the vertical distance provided.

Discussion Status

The discussion is active, with participants exploring different aspects of the problem, including the relationship between kinetic energy and work done. Some guidance has been offered regarding the calculations, but there remains uncertainty about the interpretation of the problem's parameters.

Contextual Notes

Participants express confusion about the relevance of the vertical distance in relation to the speed at which the water is delivered, indicating a need for clarification on the problem's setup.

blackcat
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The question is:
A pump raises 1800 litres of water per minute through a vertical distance of 9m. Given that 1 litre of water has a mass of 1kg, find the power developed by the pump if the water is delivered at 12m/s.

My thoughts are as follows. If the pump raises 1800kg of water per minute, then it must raise (1800/60)kg of water per second, which is 30kg per second. It's raising this against the force of gravity, so Force = ma = 30*g. Work done = force*distance = 30g*9 = 291g. Power is work done / time, but I've modeled it per second so power = 291g.

Obviously that can't be right as the thing states it's delivered at 12m/s and I haven't used that. Please help.
 
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You need to work out the kinetic energy of the delivered water, and how much work is done accelerating it.
 
thanks for the reply and sorry i am late.

so the kinetic energy is (0.5*30*12*12) = 2160J. And the work done is against gravity, so E = Work done = Fd = (30g*9) = (291g)J.

So is the power equal to (2160 + 291g)J/s? Because I have modeled everything in 1 second.

Is this right?
 
wait. it says the water is delivered at 12m/s so work done = 30g * 12 instead of 9. but i don't understand why the vertical distance is given to me? and what does "delivered at 12m/s" mean?
 
blackcat said:
thanks for the reply and sorry i am late.

so the kinetic energy is (0.5*30*12*12) = 2160J. And the work done is against gravity, so E = Work done = Fd = (30g*9) = (291g)J.

So is the power equal to (2160 + 291g)J/s? Because I have modeled everything in 1 second.

Is this right?

Yes, it is. The pump does work, accelerating the water to 12m/s and also moving it up, doing work against the gravitational force G = mg.
 

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