Lessons · Engineering · Head, flow and the power a pump needs
Head, flow and the power it takes
Hydraulic power is density times gravity times flow rate times head. Divide by efficiency to get the shaft power the pump actually needs. Head is the height the pump must lift against, plus everything the pipework wastes.
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In beta. This lesson was written for Hone and has not yet been checked by a licensed engineer. Practice material, not professional advice. What that means.
What it is for
Pumps are chosen from a duty point -- a flow and a head together -- and quoting one without the other selects the wrong machine. The same pump at a different head moves a different amount.
How to think about it
Get flow in cubic metres per second and head in metres, multiply by rho and g, then divide by the efficiency the problem gives. Remember the head is not just the lift: friction in the pipe is part of what the pump is working against.
Worked example
Hydraulic power = rho x g x Q x H.Watts, with SI units throughout.
Shaft power = hydraulic power / efficiency.Efficiency is always below one.
Head = static lift + friction losses.The pipework is part of the duty.
A duty point is a flow AND a head.Either alone selects the wrong pump.
Your turn
rho=1000, g=10, Q=0.02 m³/s, H=20 m. Write the hydraulic power in watts.
1000 x 10 x 0.02 x 20 = W
Solve one, graded on the server
The trap
Sizing on the static lift alone. Friction can be the larger half of the head on a long line, and a pump chosen without it will not reach the far end.