Small scale water power, how does water volume and hight convert into electric energy?

I was playing a little bit with the basic physics behind water power production but I can't get the numbers right.

Let's say that I put a windmill that pumps water into a watertank on the top of my house, then I connect some kind of pipe with generator and starts to drain the watertank.

How much electric power (kWh) can I get out from a watertank with size $X\text{ m}^3$ placed $Y\text{ m}$ above the ground?

How does the formulas look like?

Let's put some numbers on this problem, and see where we end up:

Let's say the tank is $1\text{ m}^3$, and it is $10\text{ m}$ off the ground so the water will fall $10\text{ m}$ to the generator.

Let's connect the generator with a standard garden hose that has a 1 inch diameter, with an area of $2.54\text{ cm}/(2\pi) \sim 5.1\text{ cm}^2$.

And then I guess we would get a $10\text{ m}$ column of water pressure, that could be transformed with the area into the force the hight is putting on the system. Something like the earths gravity (9.82)*density*height = 9.82*1*10 ~ 98 Newton (???).

And then maybe use that we can find that pressure=Force/Area, but how to move from pressure to energy?

Thanks David Zaslavsky for the example, and in theory that would mean that to store 1kWh I need like 40m3 at 10m height. That more or less mean that if one would try to build something like this in real life things need to be quite big.

Also thanks Fortunato for illustrate the practical problem in extracting the energy, and that even thou it is hard to get hight numbers it can be worth the effort anyway.

• I would think that it would be more efficient to directly collect the energy from the wind, unless you're trying to create a reservoir for power generation when it isn't windy. – CoilKid Dec 19 '16 at 17:45

What you're looking for is actually energy, not power, and you can put an upper limit on the amount you can get by computing the gravitational potential energy lost by the water as it drops. If the volume of the tank is $V = X\text{ m}^3$ and its height above the ground (or more precisely, above the point where you extract the energy) is $h = Y\text{ m}$, the amount of energy you get is no greater than $$E = \rho V g h$$ where $\rho$ is the density of water and $g$ is the gravitational acceleration. If you put in all the numbers and unit conversions to get it in kilowatt-hours, that works out to $$E = 1000\frac{\mathrm{kg}}{\mathrm{m}^3}\times X\text{ m}^3\times 9.81\frac{\mathrm{m}}{\mathrm{s}^2}\times Y\text{ m}\times\biggl(3.60\times 10^6\frac{\mathrm{kWh}}{\mathrm{J}}\biggr) = 0.00273XY\text{ kWh}$$