Water Based power Storage System

[tcmtech]

Love reading the stuff on this site. More interesting ideas to explore on my own time!

Anyway, anyone work with water storage systems used to store energy?

The reason I ask is that I have a stream less than 150 ft from my house and about 40 ft lower in elevation. Plus I have hill behind my house with about 60 ft higher elevation at the top.
The hose run is around 400 ft top to bottom.
I already have brass/stainless steel positive displacement gear pumps that also make good water motors and start at less than 5 psi head pressure.
And I have a lot of 1.5 inch ID irrigation line too so flow rates can be reasonable.

How much power can I produce from the elevation drop of 100 ft?

Say I want to produce 1 kw returnable energy at 70% system return efficiency. How many gallons of water have to flow down the hill to get that 1 kw of usable power?

Any thoughts?

This is just a feasablility question. I Dont plan to build it unless its going to give a resonable energy storage without having to use a million gallons of water to make it work!

 

[Pommie]

If you work in SI units then it is much simpler.

Energy stored = mgh.
m=Mass
g=gravity (9.81m/s²)
h=height (30m)

So, to store 1000 watts (Joules) would require 1000/9.81/30 Liters of water = ~3½ Liters.

However, I suspect that you actually mean 1kWh which would require 3,600 times as much or roughly 12,500 Liters. (~3000 gallon)

That amount seems excessive, maybe someone should check my maths.

Edit, Maybe someone that understands imperial units could confirm the answer.

Mike.

 

[mneary]

Certainly much more practical to do with a pond than with a tank. If you have some land at the 'top', a pond isn't that huge... 1/2 acre-foot (1/10 acre 5ft deep) is 0.5M liters +/-, or 1kW for 150+ hours. Losses in 400ft of 1.5 inch line might be substantial (you're moving 60 gpm), and I didn't account anything for motor or generator losses. And your soil might take some too.

 

[user_88]

1Kw flow rate ....

Approximate calculation of water flow that would be required in order to generate 1 kw, at 100 ft. vertical drop .....friction loss not included:

1kw=738 (ft-lbf)/sec = P lbf/in2 * 144in2/ft2 * Q ft3/sec

use P = 43 psi, corresponding to 100 ft, and solve for Q

giving Q = 0.12 ft3/sec = 54 gal/min

However, at 54 gpm flow rate, and a 400 ft. pipe run, using 1.5 in pipe, the head loss due to friction is prohibitive. See:
Pipe loss
By using 3" PVC the friction head loss at 50 gpm, and with a 400 ft pipe length is only about 3.6 ft, or equivalently, 1.6 psi loss.

Generator efficiency should be reasonably high, and not require any additional system modifications.

 

[tcmtech]

Ah... Interesting!

I should have stated the return was 1 kwh at any time frame of reason. Not 1kw per hour!
I do have enough space for a small pond say 60' * 60' * 10' plastic lined of course.
Actual volume estimates at 36000cu ft. or 269298 gallons.

Asuming the 3000 gallon per kwh is close I would have roughly 89 KWH storable energy.
Subtract line loss and motor/generator efficiency and I could possibly expect say 50 to 60 kwh usable?

Math related: We are americans, we avoid simple and practical at all cost! just look at our pollitics, and product designs! Ever wonder where Rube Goldburg got his inspiration?

 

[bryan1]

Eh tcmtech,
Go read the water forum on the Otherpower.com Discussion Board || Make your electricity from scratch! and I'm sure you'll find in the archives everything you need to know.

Cheers Bryan