The kinetic energy of each drop is
where m is the mass and v is the velocity when it hits the ground. Clearly we need some statistics about raindrop size; a quick search turned up this paper, from a collaboration between Brookhaven National Laboratory and a group of Chinese institutions. The important info we need is in Fig. 5:
The y-axis, volume-mean diameter, refers to the diameter of a perfect sphere with the same volume/surface area ratio as the drop. Coincidentally, this is also known as the Sauter mean diameter. Unfortunately, the paper doesn't provide a fitting function for these curves, but we can take a guess, estimating values from pixels:
x | y |
1.149825784 | 0.8509578544 |
3.066202091 | 0.9689655172 |
4.982578397 | 1.078544061 |
7.857142857 | 1.137547893 |
13.79790941 | 1.230268199 |
27.21254355 | 1.348275862 |
48.67595819 | 1.390421456 |
where ρ is the density of water. We still need to find v, but if we assume the drops hit terminal velocity for a sphere, we can use eqn 3 from this paper:
where g is the acceleration due to gravity, and I've substituted in the drag coefficient for a sphere. Putting these together, the energy becomes
This is the energy of each drop, so we need to multiply by drops per time to get power. The drops per time is related to the rain rate from above:
where A is the collection area. Putting everything together, we have
If we plug in values, this comes to
This one wound up being a bit dense, sorry. I'll try to pick a more accessible topic next time!
It would be interesting to think about how one could actually capture the kinetic energy... A big platforms on springs sounds like it would work, but I think it would just push the kinetic energy back into the water, unless you could engineer some kind of ratchet-y system which is out of my league. I think the trick is to get it to compress a heat engine.
ReplyDeleteI was going to talk about piezoelectrics as a possibility, but this ended up being a headache by itself.
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