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Power for Pumps 
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#1
Jan2013, 03:27 AM

P: 6

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
Jan2113, 10:12 AM

P: 851

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?). 


#3
Jan2113, 11:39 AM

HW Helper
Thanks
PF Gold
P: 4,914

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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