# How to calculate (pump) power to maintain a static head?

ag123
i've this question that i can't seem to work out an answer.

supposing that a (centrifuge) pump transfer an amount of energy (power) p to a column of water height h so that it maintains that as a static head. i.e. the water level in that tube is at a fixed height h due to the power transferred by the pump.

how do i calculate this height (head) h based on the power transferred into the water. note that the column of water is non moving.
if i assume the column of water has potential energy = mass * g * h/2 = density * area * h^2 * g / 2
where area is the cross-section area of the tube and the input power being constant. it would seem that the pump can support a much higher column (head) of liquid by simply reducing the diameter since:
h = sqrt(2 x power / ( density x area x g ))
h = sqrt(2 x power / ( density x pi x (d/2)^2 x g))
this calc would look absurd as if d diameter -> 0 , h -> infinity
so if i use a very small diameter, i can support a huge column of water with very little power
assuming no other forces act other than gravity on the water itself

i.e. how do i work the formula for how much power from the pump is required to maintain a static head a column of water height h?
(for that matter, you can assume all that power (energy) is transferred into the water, but it is a fixed amount of energy per sec watts)

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Homework Helper
the column of water is non moving
If the column of water is not moving, no power is transferred to the water column (a plug doesn't do any work either, and has the same effect).

ag123
well, the problem is that it isn't a plug, but a (centrifuge) pump spinning up to maintain the column of water at a particular head. so power is consumed to generate the pressure that maintains the column of water at that height. in an attempt to simplify the problem, i used the potential energy of that water column and assume that it does a free fall. so the total potential energy lost would be mass x g x height. I'm using 1/2 height to get to the center of mass of the water column. so the pump would need to maintain that amount of energy to keep the water column at that height. but there must be some errors somewhere, as it implies based on the first post that for the same water column, i can make the pump pump to ever greater heights simply reducing the diameter of the tube. and and the same amount of power/energy is supplied, but i get much greater heights without putting more energy but simply reducing the diameter of the tube. it looks strange and incorrect.

in a sense, if you consider the motor and pump efficiency as a percentage, then you can work out for that head, how much power you need. and if you exceed that input power, on top of maintaining that head the water would be pumped upwards overcoming the dynamic heads and losses.

in a sense that (static head of a particular height) would be a minimum to pump that water that high.
i'm not sure where else is wrong in this model, approach.

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Homework Helper
Gold Member
That pump would be working at zero flow regime.
There is a point on its operational curve provided by the manufacturer, as well as consumed power and efficiency at that point.
The diameter of the pipe would be irrelevant, static head or pressure is the only thing the pump must maintain.

In practical terms, that regime is to be avoided because increased heat into the fluid due to the action of the impeler on the fluid must be considered, as temperature tends to increase much with time of zero flow regime.

https://www.engineeringtoolbox.com/pumps-temperature-increase-d_313.html

https://www.engineeringtoolbox.com/centrifugal-pumps-minimum-continuous-flow-d_1445.html

https://www.engineeringtoolbox.com/centrifugal-pumps-d_54.html • ag123
Mentor
This is a case where theoretical physics is not sufficient to describe what happen, and we need to consider practical engineering realities. In the real world, centrifugal pumps are never 100% efficient. They are most efficient when operating near their best efficiency point, and progressively less efficient as the operating point moves away from the best efficiency point. This is illustrated in a pump curve such as the example below. The pump curve is adjusted to fit the requirements of a particular installation by trimming the impeller. The curve above shows the performance for impellers ranging from 7" to 9.5" diameter in a particular pump. For any impeller size within that range, you can read the operating curve, the power required, and the efficiency. The efficiency curves stop at 50% efficient because it is bad practice to operate this pump at lower flow rates. The radial load on the shaft is high, and bad things happen. Bad things like broken shafts, failed seals, and vibration.

The hydraulic power, which is the theoretical power at 100% efficiency, is calculated using:

HP = TDH X GPM / 3915, where
HP = horsepower
TDH = total head in feet
GPM = gallons per minute

The difference between hydraulic power and actual power (from the curve) is power lost due to inefficiency. It heats the water. At zero flow, there is no water flowing through the pump to carry that heat away. That causes further bad things to happen because the heat can be enough to make the water boil.

• BvU, ag123, russ_watters and 1 other person
Mentor
well, the problem is that it isn't a plug, but a (centrifuge) pump spinning up to maintain the column of water at a particular head. so power is consumed to generate the pressure that maintains the column of water at that height. in an attempt to simplify the problem, i used the potential energy of that water column and assume that it does a free fall. so the total potential energy lost would be mass x g x height. I'm using 1/2 height to get to the center of mass of the water column. so the pump would need to maintain that amount of energy to keep the water column at that height. but there must be some errors somewhere, as it implies based on the first post that for the same water column, i can make the pump pump to ever greater heights simply reducing the diameter of the tube. [emphasis added]
The error is that you made an arbitrary calculation to try an equate power and energy. Your assumption doesn't have anything to do with reality. Real-world pumps and water columns don't work that way.

• Lnewqban, BvU and ag123
Mentor
As others have alluded to, you need to know the so-called pump characteristic (provided by the manufacturer). This is the unique relationship between the pump head, the flow rate, and the rotational speed. Usually it is expressed as the flow rate as a function of the head and the rotational speed. The flow rate increases with increasing rotational speed, and decreases with increasing head.

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• Lnewqban