The potential energy of anything falling from a height is given by U = mgh, where:
- U is the potential energy in joules
- m is the mass in kilograms
- h is the height in meters
Now, 1000 liters of water has a mass 1000 kg and the value of g is 9.8 m/s^2.
The amount of energy 1000 L of water would have after falling 20 m is,
U = 1000(20)9.8 = 196,000 J
Given 1000 L of water falls 20 m every hour the amount of power available is,
196,000/3600 = 54.444 W
There are 3600 seconds in an hour & watts is the joules per second.
Allowing for friction and efficiency of a turbine, say 20 percent of the available power is converted to electricity - turbine manufacturers will be able to give a more accurate number. The power generated would be,
Electrical power generated = 0.2(54.444) = 10.888 watts per hour.
Note watts, not kilowatts. If you want kilowatts then increase the height, the amount of water flowing and the flow rate.
From Móż's answer linked in comments, 90% efficiency from the turbine is possible, and 90% from the generator is also possible with large hydro stations(pdf) coming close to 90% overall efficiency(pdf) but in specific conditions that a small system won't be able to replicate, because for example they use large synchronous AC generators.
(via wikidot - relative efficiency of various turbines as flow rate varies as a proportion of rated)
With a very small system like the one asked about efficiency will be lower because water viscosity and edge effects play a larger role. However, since costs are irrelevant some normally ignored solutions become possible - a bucket chain, for example, scales better and could be more efficient for a system this small. But those can't scale up (100 cubic metres per second falling 100m... in buckets?!?)
Sorry, the only easy efficiency references were pdfs in glossy brochures