Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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.

share|improve this question
add comment

2 Answers

up vote 3 down vote accepted

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 $$E = 1000\frac{\mathrm{kg}}{\mathrm{m}^3}\times X\text{ m}^3\times 9.81\frac{\mathrm{m}}{\mathrm{s}^2}\times Y\text{ m}\times\biggl(3.60\times 10^6\frac{\mathrm{kWh}}{\mathrm{J}}\biggr) = 0.00273XY\text{ kWh}$$

In practice, of course, you wouldn't get that much energy out. In order to do so, you would have to extract all kinetic energy from the water as it falls - that is, your generator would have to stop the flow of water completely, and no real generator can actually do that. The closest you can get is probably a backshot water wheel, although water wheels suffer from other inefficiencies that reduce the amount of energy you actually get down to about 60% of the theoretical maximum. Although it doesn't actually stop the water, a turbine is mechanically better and actually can get you a higher efficiency overall.

share|improve this answer
add comment

Water turbines are a matter of pressure and flow if the pipe coming from the tank has 90s in it this will restrict the flow and the efficiency goes down. But the point to consider here is that your method of storing energy “water in a tank” is cheep and efficient in the long run. I helped some friends make a system for storing excess energy this way. When the solar panels have charged the batteries and the electricity is not needed the pump to reservoir is turned on. Late night when the wind turbine kicks in it does that same. This method made it possible to need 0 power from the grid. The point I’m trying to make is that the choice of turbine “efficiency” the height of the reservoir “pressure” and the design of the pipe system “flow” are THE determining factors way before you do the math. In your example a standard garden hose is ½” in diameter you must include the light and the layout of the hose. At lower pressures you will get more flow from a hose that is straight rather than in a coil. You need to monitor all this factors in your experiments.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.