Calculating the mechanical power of a water pump 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.


*

*How is this force calculated mathematically?

*Assuming an ideal electric centrifugal pump, can we establish the electrical power required by the pump, given the difference in heights between the containers?

*Does this apply to all pumps, or just centrifugal pumps?

 A: 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.
