Rainwater generator

[netshaman]

Looking at the operation of my rain gauge, I imagined a generator operating on the same principle.

A sort of cup connected by a strap to a magnetic counterweight sliding in a solenoid, this counterweight weighing just enough to counterbalance the weight of the empty cup and allowing it to rise as quickly as it descends when it is full of water.
When it rains the cup which is kept locked by a spring at the preset calibration fills with water, from a certain body of water on board, the locking is deactivated: the body of water drives the assembly towards the low and therefore the magnetic counterweight rises in the solenoid, and this generates a current.
Arriving at the bottom, the cup is emptied at once thanks to a system which makes it tilt: the counterweight descends again, the locking is engaged and the cycle begins again.
Of course this system can only work well when the rain is dense enough.
By putting several devices in parallel and synchronized, it will be possible to generate a continuous current which will charge a battery pack.

[Gaston]

The mechanism seems clever (although I do not think I understood it in full).

On the other hand, a small calculation of the amount of recoverable energy:

For a roof with a surface of 100m².
It receives approximately 1000 liters per m² per year (in France).
Assuming that the displacement of the magnet is 1 meter (that already makes a good solenoid), the recoverable energy is:
100 * 1000 * 9,81 * 1, about 1 MJ or 278 Wh annually

Roughly enough to charge a cell phone.

[netshaman]

It was just an idea like that.
Indeed it does not do much.

[dedeleco]

As much as possible, mounting the large rainwater tank on the roof is emptying it so full, this m3 quickly in a turbine with electric generator at good efficiency, the lowest possible low at the level of the sewer 5 to 10m in elevation), like EDF on large dams, which collect rainwater over tens of km2 and hundreds of m of falls, which changes the order of magnitude entirely ...
But in Paris it is 60cm of rain per year, 600l / m2 and 100m2 of roof for 5m of elevation gain giving 60m3 per year, we get 600x100x9,81x5 = 2943KJ divided by 3600s gives = 0,817KWh or a saving of 0,1 € !!

Even with 10m and more it is difficult to exceed the € savings.

So the hydraulic energy is really low as long as we don't take Km2 and 100m !!

We can also see it as a proof that the current energy is not expensive at all, measured in manual work !!