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.

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?