Calculating Pump Power and Fountain Height from Water Ejection Speed

  • #1
amanda.ka
46
0

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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  • #2
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?
 
  • #3
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 :)
 
  • #4
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 :/
 
  • #5
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?
 
  • #6
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)?
 
  • #7
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?
 
  • #8
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
 
  • #9
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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Likes amanda.ka
  • #20
TSny said:
Yes, that looks good.
Thank you for your help!
 
  • #21
OK. Good work!
 
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