Say I want to pump water from one container to another. The water levels are 3 meters apart, and I want to pump 10 litres per hour. I figure the mechanical power necessary, assuming no losses, is:

$$ \require{cancel} \dfrac{10\cancel{l}}{\cancel{h}} \dfrac{kg}{\cancel{l}} \dfrac{\cancel{h}}{3600s} = \dfrac{0.0028kg}{s} $$

$$ \dfrac{0.0028kg}{s} \dfrac{3m}{1} \dfrac{9.8m}{s^2} = \dfrac{0.082kg\cdot m^2}{s\cdot s^2} $$

$$ \dfrac{0.082\cancel{kg}\cdot \cancel{m^2}}{\cancel{s}\cdot \cancel{s^2}} \dfrac{\cancel{J} \cdot \cancel{s^2}}{\cancel{kg} \cdot \cancel{m^2}} \dfrac{W\cdot \cancel{s}}{\cancel{J}} = 0.082W $$

But, I know from practical experience that real centrifugal pumps that can work at a 3m head are big and certainly require orders of magnitude more electrical power. What explains the difference?

Intuitively, I figure this must be because the pump must exert some force to balance the force of gravity from pushing water backwards through the pump, siphoning the water back to the lower container, then exert yet more force to accomplish what was desired, pumping to the higher container.

  1. How is this force calculated mathematically?
  2. Assuming an ideal electric centrifugal pump, can we establish the electrical power required by the pump, given the difference in heights between the containers?
  3. Does this apply to all pumps, or just centrifugal pumps?
  • $\begingroup$ Thinking about this more, I must be neglecting that the pump must work against the pressure of the column of water above it. I can figure the pressure of a 3m column of water, but if pressure is force per area, I'm not sure how I get rid of the "area" term to be left with just force, to calculate the power required. $\endgroup$ – Phil Frost May 7 '13 at 13:24
  • $\begingroup$ Does "fountain pump" mean "pump used in a fountain", or does it have some other technical meaning? I ask because your average fountain probably gets through quite a bit more than 10 litres per hour. $\endgroup$ – Nathaniel May 7 '13 at 13:34
  • $\begingroup$ @Nathaniel I just mean a pump sold for fountains. They are just submersible centrifugal pumps. Yes, they usually do get more than 10 litres per hour, but they usually aren't working against a 3m head: they are just squirting straight up with no head. $\endgroup$ – Phil Frost May 7 '13 at 13:39

The number you calculate is proportional to the flow rate you put in, so a ten times faster flow rate will require ten times more electrical power, so if you do the same calculation with a flow rate closer to what you'd get from a fountain, you'll probably get a lot closer to the right order of magnitude.

This calculation is actually the right way to calculate the power used by the theoretically most efficient fountain pump. Of course a fountain pump doesn't quasi-statically move water from one reservoir to another, but instead gives the water kinetic energy. But pretty most of that kinetic energy is then converted into potential energy by the water's inertia. (Some will be lost to heat instead, but this can be minimised by making the flow laminar, e.g. by using a flow straightener, especially if the flow rate is quite high.) So this calculation tells you the power you have to put into the water in the form of kinetic energy, in order for it to reach 3 metres in height.

However, there will be losses in the pump itself. Mechanical pumps can be made very efficient if they can operate relatively slowly, pushing water from place to place without much inertia involved. However, for a fountain you have to start with non-moving water and accelerate it to a high velocity, and I suspect it's a lot harder to do that efficiently.

  • $\begingroup$ I've removed the words "fountain pump" because my question really has nothing to do with fountains -- it applies to any centrifugal pump. I'm not making a fountain -- I'm simply lifting water from one level to another, say from a bucket on the floor to a bucket held over my head. If a centrifugal pump can't do this efficiently, what's an example of a pump that can? Why? $\endgroup$ – Phil Frost May 8 '13 at 11:20
  • $\begingroup$ I don't know an awful lot about pump design, but if your pump is sold as a fountain pump (as you implied in the comments on your question) then it's really not designed for lifting water from one level to another. It's like the difference between a bicycle pump and a desk fan. One is designed to pump air relatively slowly from low to high pressure, and the other is just designed to accelerate it, without doing work against a pressure gradient. If you tried to inflate a tyre with a desk fan you wouldn't get very far. ... $\endgroup$ – Nathaniel May 8 '13 at 12:23
  • $\begingroup$ ... The reasons why different designs are more suited to different tasks than others are interesting, but as I said my knowledge is a little limited in this area. Still, if I get a chance I'll update my answer with some information about it. $\endgroup$ – Nathaniel May 8 '13 at 12:23
  • $\begingroup$ The relevant technical terms are "positive displacement pump" for the kind that's good at slowly pumping up hill, versus "velocity pump" for the kind that's good at accelerating fluid. A centrifugal pump is a velocity pump, but for this task you want a displacement pump. Wikipedia has some information, but it's not very well organised, because it's interspersed between the entries of this long list of pump designs. $\endgroup$ – Nathaniel May 8 '13 at 12:34

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