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Power output of a pump

  1. Oct 10, 2015 #1
    1. The problem statement, all variables and given/known data
    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)


    2. Relevant equations



    3. 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 :)
     
  2. jcsd
  3. Oct 10, 2015 #2

    TSny

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    Looks correct. "change in work" should just be "work". 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?
     
  4. Oct 10, 2015 #3

    SteamKing

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    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.

     
  5. Oct 10, 2015 #4
    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 :/
     
  6. Oct 10, 2015 #5

    TSny

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    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?
     
  7. Oct 10, 2015 #6
    It represents the total energy Emech = K + U, should it be equal to zero (conservation of energy)?
     
  8. Oct 10, 2015 #7

    TSny

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    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?
     
  9. Oct 10, 2015 #8
    It represents the change in kinetic energy
     
  10. Oct 10, 2015 #9

    TSny

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    Yes. And the second part represents the change in potential energy. So, what does the entire expression represent?
     
  11. Oct 10, 2015 #10
    Is it the change in work?
     
  12. Oct 10, 2015 #11

    TSny

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    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?
     
  13. Oct 10, 2015 #12
    K(final) + U (final) = K(initial) + U (initial) ?
    Sorry I don't think I am understanding the concept :(
     
  14. Oct 10, 2015 #13

    TSny

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    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.
     
  15. Oct 10, 2015 #14
    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 ?
     
  16. Oct 10, 2015 #15

    TSny

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    Let's be clear on where you are taking the initial position and where you are taking the final position.
     
  17. Oct 10, 2015 #16
    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
     
  18. Oct 11, 2015 #17

    TSny

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    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?
     
  19. Oct 11, 2015 #18
    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
     
  20. Oct 11, 2015 #19

    TSny

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    Yes, that looks good.
     
  21. Oct 11, 2015 #20
    Thank you for your help!
     
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