Power for Pumps


by Poco Express
Tags: power, pumps
Poco Express
Poco Express is offline
#1
Jan20-13, 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
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Travis_King
Travis_King is offline
#2
Jan21-13, 10:12 AM
P: 763
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?).
rude man
rude man is offline
#3
Jan21-13, 11:39 AM
HW Helper
Thanks
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P: 4,390
Quote Quote by Poco Express View Post
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
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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