Work done to lift a mass of water

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SUMMARY

The discussion focuses on calculating the work done by a pump to lift 800 kg of water from a depth of 14.0 m and eject it at a speed of 18 m/s. The work required to lift the water is calculated using the formula W = mgh, while the kinetic energy imparted to the water is calculated using W = (1/2)mv². The total power output of the pump is determined by summing the work done in both lifting and giving kinetic energy, then dividing by 60 seconds. The net work done is equivalent to the change in total mechanical energy, which includes both potential and kinetic energy changes.

PREREQUISITES
  • Understanding of basic physics concepts such as work and energy
  • Familiarity with the formulas for gravitational potential energy (W = mgh) and kinetic energy (W = (1/2)mv²)
  • Knowledge of power calculations (P = Net Work / time)
  • Concept of mechanical energy conservation in nonconservative systems
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  • Study the principles of work and energy in physics
  • Learn about the conservation of energy in mechanical systems
  • Explore the relationship between power, work, and time in fluid dynamics
  • Investigate the effects of nonconservative forces on mechanical energy
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Students studying physics, particularly those focusing on mechanics, engineers working with fluid systems, and anyone interested in understanding the principles of work and energy in practical applications.

henry3369
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Homework Statement


A pump is required to lift 800 kg of water (about 210 gallons) per minute from a well 14.0 m deep and eject it with a speed of 18 m/s
(a) How much work is done per minute in lifting the water?
(b) How much work is done in giving the water the kinetic energy it has when ejected?

Homework Equations


Work = change in energy

The Attempt at a Solution


I don't understand the different between part a and part b.
Wouldn't the work done to lift it also be the work done to give it the final kinetic energy?

W = -ΔU or ΔK
 
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henry3369 said:
I don't understand the different between part a and part b.
Wouldn't the work done to lift it also be the work done to give it the final kinetic energy?
They are breaking the total work required into two pieces:
(a) How much work is required just to lift the water? (Without imparting any kinetic energy.)
(b) How much additional work is needed to give it some kinetic energy?
 
Doc Al said:
They are breaking the total work required into two pieces:
(a) How much work is required just to lift the water? (Without imparting any kinetic energy.)
(b) How much additional work is needed to give it some kinetic energy?
(a) W = mgh
(b) W = (1/2)mv^2

Ok so I got those two. The next question asks for the total power output of the pump.
It seems that the answer comes from adding (a) and (b) then dividing by 60 seconds.
I'm confused because:
P = Net Work / time
Net Work = -ΔU or ΔK, and the former gives (a) while the latter gives (b).

Additionally, I tried using conservation of energy with nonconservative forces:
Ki + Ui + Wnc = Kf + Uf
Wnc = Kf + Uf. Using this for the power equation gives the correct answer, but this is the work due to ONLY the nonconservative forces. If I want to find the total power output, wouldn't I need net work?
 
Doc Al said:
They are breaking the total work required into two pieces:
(a) How much work is required just to lift the water? (Without imparting any kinetic energy.)
(b) How much additional work is needed to give it some kinetic energy?
I found another post that explained that work is actually the change in total mechanical energy. So does that mean Wnc = Net Work?
 
henry3369 said:
If I want to find the total power output, wouldn't I need net work?

henry3369 said:
So does that mean Wnc = Net Work?
Yes and yes.

I would put it this way: The net work done must equal the change in mechanical energy (ΔU + ΔKE) of the water. (We are ignoring details like friction and so on.)
 
Think about it this way, it is not just lifting that water and then moving it, the pump performs a constant process in which it is both lifting some water while pushing the last water it lifted at the same time... if it was just talking about a specific 800kg of water it would be different, but the pump has water constantly moving, i had the same question in my study guide and that is how i thought about it after we went over the answers
 
Thank you for your contribution. Please note that this thread is 8 years old.
 
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