Idea to Store Solar, or Wind Energy in Tanks in Central, South American, Africa

 I go to Togo, West Africa, and they already use, or need the black water tanks. they can have the water pumped using solar, windmills, or rope pumps, then allow it to fall, and light their houses with LED lights, and charge their cell phones.

So what do you think, anyone have experience using these cheap hydro generators?

Do they work?

Do they last?

Is there enough water pressure?

Thanks, Andy Lee Graham

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hobotraveler (author) 23 days ago

Hello, thanks, I think the answer is, this tank would empty way too quick to be of any real energy value. Here is a link to the super cheap, low as 5 dollar 12 volt generator. All Africans need to do is charge a cell phone....

Hey, I thank you for giving me a link to an example of one of these, um, what are we calling them? "12V hydroelectric generator" That is probably a pretty good description, for purposes of finding them on eBay, or reviews, pictures, etc, of them on various forums.

As I demonstrated earlier, big water towers are not going to be probably not going to be able to compete with a good battery, in terms of specific energy (kilojoules per kilogram) or in the cost of the energy storage like in dollars per kilojoule.

Of course the cost gets better, if you can use an existing source of pressurized water. Like municipal water. I mean, in that case, someone else has already paid the expense of building the water tower.

So I am kind of curious what kind of tech these new, mass produced, "12 volt hydroelectric generator", things have inside them. I tried doing an image search to see if I could find some pictures of one of one of these artifacts disassembled.

The best example I found so far, that includes someone taking the cover off one of these things, is this Youtube video,

This video also includes some video of the narrator hooking this thing up to his shower head, and trying to get some watts out of it. If you have not seen this video, it might be relevant to your original topic; i.e. "anyone have experience using these cheap hydro generators?"

By the way, if you want kind of an overview of every known form of energy storage, with a table showing numbers for joules per kg (specific energy) and joules per liter (energy density), the Wikipedia article for "Energy density" has this.

Although, many of the entries seen in these tables are a little too abstract to be useful, including entries like "anti-mater" and "ham sandwich"

However, I noticed one of these, that is not a kind of battery, and like the water-in-a-tower idea, it is something purely mechanical. That is the entry for, "compressed air".

The quote of 300 bar is kind of scary. That is a lot of pressure. A bar is about 15 PSI, and a typical garage, or construction type, air compressor has a tank strong enough to withstand about 5 to 10 bar (or 75 to 150 PS.

Just, intuitively, I am thinking the amount, and expense of plumbing required for a water tower, or compressed air tank, might be similar.

Also there are plenty of existing power tools designed to run on compressed air directly.

As a plus, even in places on Earth where water is scarce, atmospheric air is still plentiful.


I do not know these, "cheap hydro generators", to which you refer, but then I am not sure I have to.

The principle of storing energy, by simply lifting mass against gravity, that principle is well known, and can be summarized in a single equation:

U = m*g*h

Where U is stored energy (in joules), m is mass (in kilograms), g is the local acceleration due to gravity (typically 9.8 m*s^-2 ~=10m/s^-2, on Earth), and h is the height (in meters) through which the mass is lifted.

Anyway, a few kilograms of water and a few meters of height, will give you a few 10s of joules of stored potential energy.

In contrast, lithium-ion batteries can store kilojoules (i.e 1000s of joules) of energy, in cells that mass a few 10s of grams.

The formula for energy stored in a battery is, approximately,

U = V*Q = integral(V*I*dt)

where, again U is stored energy (in joules), and V is the battery's voltage (in volts), and Q is current capacity, which is roughly the time integral of current( Q=integral(I*dt)).

Usually battery manufacturers put their current capacity quotes in units like mA*h (milliampere*hours) or A*h (ampere*hours), however if I want to get an estimate of for a battery's stored energy in joules, I have to convert that number to ampere*seconds, to get the units to work out correctly,

i.e. (1 joule) = (1 volt)*(1 ampere)*(1 second) = (1 watt)*(1 second)

As an example, consider the 18650 size Li-ion cell. I've got one here on my desk. I think I pulled it out of a broken laptop battery. It is a common size. The same sized cell, can be found in batteries for other consumer toys, like cordless drills, and maybe other things.

This cell is cylindrical, like 19 mm in diameter, and 65 mm long. According to the Wikipedia page, "List of battery sizes", the current capacity for this cell is 1500-3600 mA*h.

The mass of this cell is about 45 grams, and that number is the number I got from putting one of these cells on a scale (balance), and weighing (massing) it.

Just as sort of a guestimate, I am going to take the typical cell voltage, of 3.7 volts, and multiply that by 1500 mA*h = 1.5 A*h, the lower number on that range given in the table.

(3.7V)*(1.5A*h) = 5.55 W*h = (5.55J/s)*(3600s) = 20000 J = 20 kJ

Next I am going to ask, how many kilograms of water, lifted to a height of 1 meter (or 10 meters) is needed to store the same amount of energy, that is 20 kilojoules.

Well, first I'll consider a 1 meter high tower. U = m*g*h. Solve for m.

m = U/(g*h) = ((20000J)/((10/m^s-2)*(1 m)) = 2000 kg.

or about 2 metric tons of water (which takes up 2.0 m^3 volume), for a 1 meter high tower.

Proportionally less water is needed for proportionally higher tower. In other words, if I build a tower 10 times as high, I only need to use 1/10 the mass; i.e.

m = U/(g*h) = ((20000J)/((10/m^s-2)*(10 m)) = 200 kg.

By the way, this simple calculation ignores losses due to inefficiency in the pump that does work to lift the water, or in the what you call "hydro generator" that receives work from falling water.

If both of those were only 70% efficient. The efficiency of both combined would be about 0.7*0.7 = 0.49 ~= 50%, meaning the actual amount of water I need to store, would be roughly twice that for the ideal case.

So what is the specific energy density for water stored in the 10 m tower?

20 kJ/ 200 kg = 0.1 kJ/kg

The specific energy of Li-ion batteries, from the sidebar on the Wikipedia page for "Lithium-ion battery", is 0.36-0.875 MJ/kg = 360-875 kJ/kg.

The ratio between those two is what? Like 3600 to 8750? So 1 kg of Li-ion batteries stores the same energy as 3600 to 8750 kg of mass on a 10 meter tower? Something like that.

But, you know, you should check this math yourself. I've been known to make mistakes on this back-of-an-envelope type stuff.

GregJnFla24 days ago

I have a 250 Gal (946L) poly tank but it drains through a 2" (5cm) pipe in a matter of minutes. Even at a slower flow I figure the most I can get out of it is a few hours of a low capacity system. Maybe a light or two and a cell phone charge but not much more.