Storing potential energy in raised water tank, use it later

[ert345]

Hi,

I am in a situation where I have to pump water from a deep well(150m depth, 30m static level) using solar power - no batteries. The water is pumped slowly during the day to fill a tank, say 4 metres high (or more because there is a hill 15 m high), 8m3 capacity. During the night an automatic system will water the plants.

Since I have to fill in the tank and since the tank must be raised can I recover some of the energy spent in pumping the water from the well to the tank by installing a small hydro-generator at the outlet of the tank, producing electric energy as I water the plants?

We can assume that 8m3 of water will come out of the water tank every night.

Any comments on this (maybe stupid, sorry) idea?

[wdyasq]

I think you should go ahead with the project. All of the free research for and by others on this site indicates it would be a stupid and/or ignorant idea. But, what do they know.

A good search would have brought up the previous remarks and probably a pile of calculations. A great failure proving the 'conservation of energy' parts of physics will be interesting to read. And building a project is a lot of fun until one realizes  they should have done a bit of research, rather than relying on a forum to do the research. (research is less expensive too)

Ron

[TomW]

Ron;

Uh, I think you missed the part where he says he pumps and releases the water for plants daily, anyway. Seems he is just hoping to recoup some of that energy as he releases the water.

Here's a fiver, find a cup off coffee you old grump.

Granted, the low head to the plant beds will limit the available power. 93  cubic meters [24,568 gallons] seems like a fair bit of water but likely not much energy to be had with low head. Could also float a weight on the water and capture the energy released as it falls in one quick go after the water has been released to the plants. I could go on but its early and the ideas won't make sense.

I submit that it would work at some level but likely very minimal power recovery. Like maybe enough to charge a big cap to play a radio for the plants awhile each day.

Just another angle.

Cheers.

TomW

[TomW]

No idea where I got 93 cubic meters? 8 cubic meters will be pretty useless for power. So I retract any claims useful power will be recovered, altho some power may be recovered.

Duh.

Cheers.

TomW

[GaryGary]

Hi,

8. m^3 is 2112 gallons, or 17500 lbs.
   
4. meters is 13 ft
   
   

So, total energy is (13ft)(17500 lbs) = 228000 ft-lb or about 0.1 KWH

Not much.

Gary

[tecker]

The best Idea I 've heard is to  start with the a conical concentrator on the bottom of your tank to bring all the weight of the tank to a head and set a horizontal turbine at the head with a stir to develope a vortex in the tank as the water moves through the head.a little engineering but sound like a fun project.  the stir adds a lot of torque the shaft but slows the shaft down . I'm sure a conical extention to your tank would start to spin with out the shaft and give higher shaft speed .I would like to do the same to my rain barrel but would have to set another barrel to drip with .

[richhagen]

If there is excess pressure (height) in the tank, one would be better off to lower the tank and save the power in pumping.  

The potential energy in stored water is M*g*h where M is the mass, g is the acceleration of gravity, approximately 9.8 m/s^2, and h is the height.  If one used Kg for the mass, and meters for the height, then the result of multiplying out the equation will be in Kg*m^2/s^2 which is the definition of Joules, a unit of energy.  A Watt is equivelant to one Joule per second.  One KWH is equal to 3,600,000 Joules.  

You can do the math for your situation, and see the energy available in your tank.

As Ron has correctly pointed out in his own special way, considering the effort involved to generate power this way, one would generally be looking for a greater height differential, and a greater mass of water to make such a project more practical.  

Double check my math, but if one had even 100 meters of height differential, one would have about a KWH of stored energy per every 970.4 gallons of water.  If you recover 75% of that you would need about 1294 gallons of water per KWH.

Make your project work for you - Rich Hagen

[tecker]

I missed the 23000 gal part .Maybe design around a 55 gal drum cutting a second drum to a cone ,concentrting to 4 inch with the turbine set on a slightly leaking shaft no seal but a close tolerence on the shaft hole  and a large pulley . the tank and concentrator would around  600 lbs of water to a 4" head .

[tecker]

2110 gals I'm a dork yes

[spinningmagnets]

This is one of the wacky ideas I am personally interested in. No matter how small a source of energy it provides, I think this can be done fairly cheap. The big problem is wind is highly variable, sometimes there is even a week with almost no significant winds. Storing variable wind energy in batteries is something I'll probably end up doing, but after a while the battery wears out and has to be (expensively) replaced. I want to get as much energy as possible directly from micro-hydro + something like this, then add other sources over time.

I can get 12 volts from an elevated water tank spinning a water wheel that runs some type of alternator. I know gearing (to increase RPM's) adds inefficiency, but the fuel is free. I might only use it for music, lights, and TV in the shop.

You may be interested in considering a "sucker pump" (piston with a check valve flapper on a cable at the bottom of a pipe well) to lift water from deep levels to an elevated tank (like the rocking crude oil pumps). By using a rocking beam with a counter weight equal to the weight of the column of water you're lifting, the actual windmill force needed will be small. A common old-style windmill (purchased, or made, with many thin blades) can lift lots of water 24 hrs a day in very light winds. (This might free up your solar panels for battery charging or another use)

If you're going to drain some water out of your elevated tank each night whether you get any energy from it or not, I vote for getting some energy from it, even if it just powers an LED. Please keep everyone posted on what you end up doing!

[craig110]

Let me "second" this comment.  Since there is no free energy, any energy that you can extract from the water flow results in the water having less energy - which is the equivalent of having the storage tank at a slightly lower height.  If that lower effective height still provides enough pressure to water your plants, then you are wasting the solar power pumping the water up to an excessive height and your best power savings would be from lowering the tank.  (Trying to extract the excess energy is limited by inefficiencies.  Eliminating the excess energy needs in the first place doesn't have inefficiencies and so gives you a better overall return.)  If the tank is already at the minimum height needed to drive the plant-watering mechanism, then trying to extract electric power from the flow will cause the watering to fail.

On the other hand, if you are just interested in playing around with low-head water turbines to see what they can do, then ignore whether this is worth it and just have fun!

[DamonHD]

As Franklin might have said: a 1 Watt not spent in pumping is 1 Watt generated...

[Chagrin]

You're violating a fundamental law of physics here. GaryGary has already calculated the ultimate potential energy of the water; how you release that energy (your whirlpool) is relevant only in reference to how much efficiency you can attain. You won't beat the ~.1 KwH.

[thefinis]

Alas not much energy to be had. One problem I see is that if you try and take the energy from the water flow it will reduce flow/pressure and may change the watering system you already have set up.

The best bet might be to try and take energy from the flow and pump more water. It will not be much 30 m to water 15m to top of hill and 4m to top of tank so best guess for pumping head is say 50 m minimum and best guess flow head is maybe 15 m. You are already looking at less than 1/3 recovery even if you could recover 100% and it is doubtful you will get 1/10 of what it took to pump it up. However if you could setup a system that pumps up water as it releases it you might free up a little solar in the daytime if electricity is what you are after. If it is more water you are want then a water pumped for water released system might have a better return especially if you do not pump it into the tank but release it at plant level a 20 m head savings.

Finis

[DanG]

Using a DIY hydro site calculator I found after 50% system losses and 20% pressure losses from pipe friction you probably have 8 watt output available steadily for 16 hours or 100 watts for 100 minutes.

He did mention a 15 meter hill, so the conversions were for that 15m or 49 feet of head and with an ~8 liters per minute or ~80 liters per minute water release...

Sounds like a good way to have lights to do late-night bug killing or some irrigation gate controls etc., without borrowing energy off the pumping arrays...

see http://www.nooutage.com/hydroele.htm to check my calculations...

[tecker]

Right Ie how much of the mass you can spin and what speed it will spin .

[ert345]

Thanks for the comments.

Obviously this is not a good way to recover energy because of the losses. And of course the law of energy conservation still holds for those who panic.

It could be interesting to try out only when the water must be lifted that high. Problem: not a lot of energy return.

Another problem is that the output water pressure will be less as a result of the hydrogenerator and this will mess up the plants watering system.

Maybe:

If there are any small devices attached to the taps of a house (for data logging or just plain LED flashing when the water flows i've seen those around, or temperature controllers) then I guess they could be powered by the water flow.

andreas

[richhagen]

How about this one:  A Joule saved is a Joule made.  

Rich

[DamonHD]

A jewel of brevity!