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Waterwheel. Power expectations?

 
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I have a 10 acre pond that produces a flow over the weir gate about 24" wide by 3" deep year-round. The water drops about 8' to the stream below. I built a waterwheel 5' high by 2' wide with 12 buckets. I plan to attach a generator that can produce up to 2000 watts at 1000 rpm. So far this has all been "gut-feel." Can anyone give me an idea if I have sufficient flow and head to spin the generator with a reasonable load? I'd love to get 1,000 watts or more. I plan to build a sluice that will direct outflow from the pond down about 3' where it will dump into the top of the waterwheel (overshot).
Waterwheel-Small.png
My waterwheel
My waterwheel
 
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The formula for hydropower is Power = Flow x Head / 5.6

Units:
Power = kW
Flow = gallons per minute
Head = feet

To calculate the flow, from this website ... WSU contracted rectangular weir calculator ... I get 366 gal/min

So Power = 366 x 5 /5.6 = 327 Watts

That is a non-negligible amount of power.

Now lets consider the efficiency of each part.
Overshot wheel is 70% efficient at best.
Friction losses in the flume cost another 10-20%
Gearing up the very low speed waterwheel to the high RPM required by the generator = 50% lost, as a guess.

Realistic power achievable = 327W x 70% x 80% x 50% = 92 Watts

A 2000 Watt generator is oversized for this, and may not work at all. It may just stall the wheel.

--------------------------------

Assuming a smaller generator, I think this is worthwhile. If that flow rate is available year round, that is 806kWh per year, 67 kWh per month. My small, very efficient grid-connected house averages about 400kWh per month, but that includes AC for hot Texas summers.

A small off-grid cabin with minimal power demands can easily get by on 67kWh per month, especially if you add in a few solar panels. Water and solar are often very complimentary power sources - when it it rains, solar is low.

Next step should be to verify the actual flow rate. Divert it into a 55 gal drum or a large trash can, time how many seconds it takes to fill it.


 
Stephen Woolstenhulme
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Thanks for the information. I'll report back when I get it installed. I do need to measure the actual flow.
 
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Hi B,

I have not crunched the numbers.  In your post you indicate power is KW, but in your calculations it looks as if you need up with watts.  Like I said, I am only glancing at this and may have missed a step,
 
B Beeson
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You're right, John. I was sloppy with the units. The equation is for kilowatts, and I converted to watts without showing the intermediate step.

P(kW) = F(gpm) x H(ft) / 5.6

P = 366 x 5 / 5.6 = 0.327kW = 327W
 
Stephen Woolstenhulme
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Do I pick up anything from the fact the water will gain a bit of head in the sluice as it drops 3' during the 20' run from the pond to the waterwheel?
 
B Beeson
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Pictures would help a lot!

Terminology can be confusing here... as the water drops 3 ft, it is loosing "head", and gaining velocity. English is not precise enough for this job.

Head represents potential energy of the water.  Going down the flume, it converts potential energy into kinetic energy. The fast-moving water will impact the water wheel, converting some of the water's kinetic energy into rotational energy of the wheel. But this is an inefficient process. Rapid flow makes more friction and turbulence, which robs some of the kinetic energy, converting it into wasted heat. This is one of the reasons water wheels are not very efficient.

Ideally, you want the water to flow down the flume with just enough slope to overcome the slight friction from moving slowly to the drop off point above the wheel. The wheel should be as tall as possible to capture the most head possible. If you have 8 ft available, use all you can. But since you already have the 5 foot wheel made, you can allow 3 feet of drop, gaining kinetic energy, but capturing that energy is tricky. You'll need to carefully shape the flume exit and the buckets to prevent water splashing out and wasting energy. To reduce friction, make the flume as smooth as possible. A round or square profile is better than a wide rectangle, then widen it out to match the wheel width right before the exit.


 
Stephen Woolstenhulme
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Thanks. I'm thinking pvc for the penstock.
 
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I have a feeling you would get so much more power from increasing the wheel size from 5' to 7' or so that the work and expense would pay you back, if it is possible to extend the sides and add new buckets.

5' to 7 1/2' would give half again as much power, pretty significant.
 
Stephen Woolstenhulme
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Great idea. I can make that work. Thanks.
 
Glenn Herbert
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Thinking about water volume and wheel speed... You want to use all of the water hitting the wheel and not have any spill over because the buckets can't hold it all at the speed the wheel is turning. As a rough first approximation, it looks like each bucket will hold several gallons of water, maybe 3 gallons. At 12 buckets filled per revolution, and 366 gpm flow, you would need to have the wheel turning about 10 rpm, or one revolution every six seconds. This seems practical; much faster and water would start to fly out of the buckets by centrifugal force. Maybe you could go twice as fast without problems.
 
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The US DOE website has some information on planning renewable energy systems, too:

https://www.energy.gov/energysaver/planning-microhydropower-system

US Department of Energy website wrote:To see if a microhydropower system would work for you, you will want to determine the amount of power that you can obtain from the flowing water on your site. This involves determining these two things:

Head -- the vertical distance the water falls
Flow -- the quantity of water falling.
Once you've calculated the head and flow, then you can use a simple equation to estimate the power output for a system with 53% efficiency, which is representative of most microhydropower systems.

Simply multiply net head (the vertical distance available after subtracting losses from pipe friction) by flow (use U.S. gallons per minute) divided by 10. That will give you the system's output in watts (W). The equation looks this:

[net head (feet) × flow (gpm)] ÷ 10 = W



As you see, they based it off 53% efficiency.  A homemade waterwheel might be much less.  The website also describes methods for estimating and calculating flow rate.
 
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