Calculating Pump Power and Fountain Height from Water Ejection Speed

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Homework Help Overview

The discussion revolves around a problem involving a pump drawing water from a well, specifically focusing on calculating the power output of the pump and the height of a fountain based on the speed of water ejected. The subject area includes concepts of energy, specifically potential and kinetic energy, as well as power calculations in the context of fluid dynamics.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculations of potential energy and kinetic energy, questioning the correctness of the power output derived from these energies. There is also an exploration of the relationship between kinetic energy and potential energy in the context of the fountain height. Some participants express uncertainty about the implications of negative values in their calculations and the proper setup of equations.

Discussion Status

The discussion is ongoing, with participants providing feedback on each other's calculations and questioning assumptions made in the problem setup. Some guidance has been offered regarding the interpretation of energy equations and the conservation of energy principle, but no consensus has been reached on the correct approach to finding the fountain height.

Contextual Notes

Participants note potential issues with unit consistency and the implications of drawing water at a specific rate. There is also mention of homework constraints that may affect how the problem is approached.

amanda.ka
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Homework Statement


A pump draws 500 kg of water/min from a 12 m deep well. The water is ejected with a speed of 20 m/s.
a) What is the power output of the pump?
b) How high is the fountain (ignoring air resistance)

Homework Equations


The Attempt at a Solution


For part a I found the potential energy and kinetic energy:
PE: m*g*delta h = (500)(9.81)(12) = 58860 J
KE: 1/2mv^2(final) - 1/2mv^2(initial) = 1/2(500)(20)^2 - 0 = 100, 000 J
Then I used Power = change in Work/change in time
= KE+PE/60 seconds
= 2647.7 W
Is this correct?

And I'm stuck on part b, here is what I did but I am not sure if it is correct:
[1/2mv^2(final)-1/2mv^2(initial)] + [mgh(final) -mgh(initial)]
-1/2mv^2(initial) = mgh(final)
-1/2*20^2/9.81 = h (final)
so height final = 20.4 m

Any help is appreciated :)
 
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amanda.ka said:

The Attempt at a Solution


For part a I found the potential energy and kinetic energy:
PE: m*g*delta h = (500)(9.81)(12) = 58860 J
KE: 1/2mv^2(final) - 1/2mv^2(initial) = 1/2(500)(20)^2 - 0 = 100, 000 J
Then I used Power = change in Work/change in time
= KE+PE/60 seconds
= 2647.7 W
Is this correct?
Looks correct. "change in work" should just be "work". Does your instructor care about significant figures?

And I'm stuck on part b, here is what I did but I am not sure if it is correct:
[1/2mv^2(final)-1/2mv^2(initial)] + [mgh(final) -mgh(initial)]
Is this supposed to be an equation? I don't see an equal sign.

-1/2mv^2(initial) = mgh(final)
-1/2*20^2/9.81 = h (final)
so height final = 20.4 m
You can see that something went wrong here. Your next to last equation implies that h is negative. Did you just arbitrarily drop the negative sign to get your final answer?
 
amanda.ka said:

Homework Statement


A pump draws 500 kg of water/min from a 12 m deep well. The water is ejected with a speed of 20 m/s.
a) What is the power output of the pump?
b) How high is the fountain (ignoring air resistance)

Homework Equations


The Attempt at a Solution


For part a I found the potential energy and kinetic energy:
PE: m*g*delta h = (500)(9.81)(12) = 58860 J
KE: 1/2mv^2(final) - 1/2mv^2(initial) = 1/2(500)(20)^2 - 0 = 100, 000 J
Then I used Power = change in Work/change in time
= KE+PE/60 seconds
= 2647.7 W
Is this correct?

You need to examine the units of your calculations.

The pump draws water at the rate of 500 kg/min, which is not the same as 500 kg.
You are then multiplying 500 kg/min by v2 in (m/s)2. I don't think the end result of this calculation is Joules. Ditto for the calculation of PE.

And I'm stuck on part b, here is what I did but I am not sure if it is correct:
[1/2mv^2(final)-1/2mv^2(initial)] + [mgh(final) -mgh(initial)]
-1/2mv^2(initial) = mgh(final)
-1/2*20^2/9.81 = h (final)
so height final = 20.4 m

Any help is appreciated :)
 
TSny said:
Looks correct. Does your instructor care about significant figures?Is this supposed to be an equation? I don't see an equal sign.You can see that something went wrong here. Your next to last equation implies that h is negative. Did you just arbitrarily drop the negative sign to get your final answer?
TSny said:
Looks correct. Does your instructor care about significant figures?Is this supposed to be an equation? I don't see an equal sign.You can see that something went wrong here. Your next to last equation implies that h is negative. Did you just arbitrarily drop the negative sign to get your final answer?
For my equation i tried to do -1/2v^2 = mgh then solve for h but I think that is incorrect because I did get a negative :/
 
Go back to your starting expression [1/2mv^2(final)-1/2mv^2(initial)] + [mgh(final) -mgh(initial)]. What does this represent? What should it be equal to?
 
TSny said:
Go back to your starting expression [1/2mv^2(final)-1/2mv^2(initial)] + [mgh(final) -mgh(initial)]. What does this represent? What should it be equal to?
It represents the total energy Emech = K + U, should it be equal to zero (conservation of energy)?
 
amanda.ka said:
It represents the total energy Emech = K + U, should it be equal to zero (conservation of energy)?
It does not represent the total energy. Let's take part of your expression: [1/2mv^2(final)-1/2mv^2(initial)] . What does this represent?
 
TSny said:
It does not represent the total energy. Let's take part of your expression: [1/2mv^2(final)-1/2mv^2(initial)] . What does this represent?
It represents the change in kinetic energy
 
Yes. And the second part represents the change in potential energy. So, what does the entire expression represent?
 
  • #10
TSny said:
Yes. And the second part represents the change in potential energy. So, what does the entire expression represent?
Is it the change in work?
 
  • #11
No. As you noted, total mechanical energy is the sum of kinetic energy and potential energy: Emech = KE + PE. So, if you add together the change in KE and the change in PE, what do you get?
 
  • #12
TSny said:
No. As you noted, total mechanical energy is the sum of kinetic energy and potential energy: Emech = KE + PE. So, if you add together the change in KE and the change in PE, what do you get?
K(final) + U (final) = K(initial) + U (initial) ?
Sorry I don't think I am understanding the concept :(
 
  • #13
amanda.ka said:
K(final) + U (final) = K(initial) + U (initial) ?
OK. This equation is correct. Note that this is equivalent to saying [K(final) - K(initial)] + [U (final) - U (initial)] = 0.

When you wrote the expression [1/2mv^2(final)-1/2mv^2(initial)] + [mgh(final) -mgh(initial)] in your first post, you were writing [K(final) - K(initial)] + [U (final) - U (initial)] . So, I was trying to prod you into seeing that the expression you wrote down represents the total change in mechanical energy, ΔEmech.

[1/2mv^2(final)-1/2mv^2(initial)] + [mgh(final) -mgh(initial)] = ΔEmech.

The law of conservation of energy tells you that the change in total energy is zero. ΔEmech = 0. Therefore

[K(final) - K(initial)] + [U (final) - U (initial)] = 0

You can then rearrange this as K(final) + U (final) = K(initial) + U (initial). This just says that the final total energy equals the initial total energy. You can solve the problem by starting with either this equation

[K(final) - K(initial)] + [U (final) - U (initial)] = 0

or, this equation

K(final) + U (final) = K(initial) + U (initial).

They represent the same idea of conservation of energy.

So, using either equation, try applying it to your situation and see if you can get a positive value for h.
 
  • #14
TSny said:
OK. This equation is correct. Note that this is equivalent to saying [K(final) - K(initial)] + [U (final) - U (initial)] = 0.

When you wrote the expression [1/2mv^2(final)-1/2mv^2(initial)] + [mgh(final) -mgh(initial)] in your first post, you were writing [K(final) - K(initial)] + [U (final) - U (initial)] . So, I was trying to prod you into seeing that the expression you wrote down represents the total change in mechanical energy, ΔEmech.

[1/2mv^2(final)-1/2mv^2(initial)] + [mgh(final) -mgh(initial)] = ΔEmech.

The law of conservation of energy tells you that the change in total energy is zero. ΔEmech = 0. Therefore

[K(final) - K(initial)] + [U (final) - U (initial)] = 0

You can then rearrange this as K(final) + U (final) = K(initial) + U (initial). This just says that the final total energy equals the initial total energy. You can solve the problem by starting with either this equation

[K(final) - K(initial)] + [U (final) - U (initial)] = 0

or, this equation

K(final) + U (final) = K(initial) + U (initial).

They represent the same idea of conservation of energy.

So, using either equation, try applying it to your situation and see if you can get a positive value for h.
Kf + Uf = Ki + Ui
= 1/2mv^2 + mgh = 1/2mv^2 + mgh
= 100,000 + (500)(9.81)h = 0 + (500)(9.81)(12)
= 100, 000 + 4905h = 58,860
41140 = - 4905 h
h = -8.387 m
My answer is still negative, is something wrong with the initial/final values ?
 
  • #15
Let's be clear on where you are taking the initial position and where you are taking the final position.
 
  • #16
TSny said:
Let's be clear on where you are taking the initial position and where you are taking the final position.
Would the initial U be negative 12 since it is below the well? If I take the top of the well as a reference point (zero) then the water would initially start -12 m below that
 
  • #17
My interpretation of the problem is that the water is projected upward with a speed of 20 m/s at ground level. So, if you are going to let your initial point be the point where the water has a speed of 20 m/s, then what would be your initial height?
 
  • #18
TSny said:
My interpretation of the problem is that the water is projected upward with a speed of 20 m/s at ground level. So, if you are going to let your initial point be the point where the water has a speed of 20 m/s, then what would be your initial height?
Kf + Uf = Ki + Ui
0 + (500)(9.81)h = 1/2(500)(20^2) + 0
Now I get the same answer as I originally got but with a positive sign
 
  • #19
Yes, that looks good.
 
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  • #20
TSny said:
Yes, that looks good.
Thank you for your help!
 
  • #21
OK. Good work!
 

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