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Power for Pumps

  1. Jan 20, 2013 #1
    Hi
    I am doing a mine engineering degree by correspondence and I have an assignment to do and the lecture notes do not provide any assistance on some of the cals.
    I am hoping some of you can help me
    I will start with these two questions and see if you can help before posting more:
    Q1)
    Determine the reqired power to pump 11.36 l/s against a total dynamic head of 112.78m if the pump operates at 70% efficiency
    Q2)
    What is the required power to pump 4000 litres per minute against a total dynamic head of 120m if the pump operates at 75% efficiency

    The formula the notes gives me is
    P = Q.γ.H / 10 power 6
    But I can't work out what this means
     
  2. jcsd
  3. Jan 21, 2013 #2
    This is a decent little paper on pump requirements.

    I've never seen a power formula with 10^6 in the denominator.

    Generally, the equation is some form of: [(Flow rate)*(Head)*(Specific Gravity)] / [(multiplier)*(% efficiency/100)]
    Of course care must be given to units to determine the value of the multiplier in order to get your power in the correct units as well. Though, again, I don't know how the multiplier could go as high as 10^6.

    Nor do I know what the y is in this equation (is it SG? If so, where is the efficiency considered?).
     
  4. Jan 21, 2013 #3

    rude man

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    Gold Member

    I assume we're talking about water.
    Q1:
    1st, what is the energy (in Joules) needed to lift 11.36 l over a height of 112.78m?
    2nd, what power is required to do this in a time of 1s?
    3rd, how do you adjust the required power in light of the 70% pump efficiency?

    Q2: same thing except convert l/min. to kg/s.
     
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