Homebrewed Electricity > Hydro
Home made hydro help
mab:
25 - 35 lps at 3m head is a nice resource to have.
In my experience a bigger generator tends to be more efficient so i suspect you're better off with the one you have. (unless the smaller generator needs fewer rpm for 24v, allowing reduced gearing - but i doubt that would be the case). Granted at 20w the efficiency benefits of a bigger genny won't be apparent, but it ought be possible to get more power with that much water.
--- Quote from: SteveB on October 24, 2023, 03:28:23 PM ---...Also the water flows down a sloping wall, should I concentrate the water into a pipe to reduce the resistance?
Any thoughts?? Please 🙏
--- End quote ---
Difficult to advise without seeing it - but I'm not sure if you can post pics if you're new here.
Do you mean that part of the 3m is with the water flowing down the sloping wall before falling on the wheel? That would waste a lot of energy if it's how I'm imagining it, (but I'm not sure a pipe would be better), letting it freefall might be better, but might make it splash around when it hits the wheel.
Is it an overshot waterwheel you have? With buckets to hold the water?
mab:
--- Quote from: SteveB on October 20, 2023, 11:43:24 AM ---...
So I live in a mill , which has a 3 M water shoot from a canal . The water is flowing enough to turn the wheel 15rpm . I have another wheel that has cogs ,that then turns a small cog wheel 11-1 ratio.
Attached is a 42 toothed bicycle sprocket.
I’ve taken off the blades from a 24v 800w wind turbine generator. On this I’ve put a small bicycle sprocket, with a ratio of 4-1.Giving me a total Rpm of 660.
It is 3 phase Ac , which passes through a charge controller converting to Dc .
With no load it rotates faster giving me a 3 phase V of 36v
So as soon as I put it on the load it slows to the speed I have mentioned, and only produces 20v .
--- End quote ---
I think i misread your original post before:
If I'm reading it right now you're getting 20v 1A at 660rpm, and open circuit at 36v it's spinning faster.
This suggests to me that really wants to be spinning faster for power at 24v, and higher gearing may be necessary (or better: a lower rpm generator).
Is there a power/rpm/windspeed chart for the original wind generator?
Did it ever produce good power as a wind generator?
joestue:
Sounds like this guy has a kilowatt of hydropower theoretically available.
I built a cross flow turbine using 24 strips of steel 2 inches wide cut from the side of a 16" diameter hot water tank.
The strips of steel were welded to the sides of 2, 12" diameter 24 tooth skillsaw blades, with a shaft bolted to thr blade through thr center hole.
One of the shafts was 1.5" diameter aluminum running on a wood v block lol, in one night it eroded like 1/16th of an inch off the aluminum.
10 feet isnt typically considered enough for a cross flow turbine, but 25 liters a second falling 10 feet should be enough to get 100 watts out of a 1 foot diameter cross flow turbine at several hundred rpm.
For the other side of the turbine, it hung from the generator by a v belt. No bearings needed other than in the motor
We were getting 50 watts out of a lot of water (1 to 2inch thick stream, 2 feet wide) , falling 1 foot plus 1 foot for the diameter of the turbine)
Adriaan Kragten:
There is a simple formula to calculate the potential hydraulic power Phyd available in the water. This formula is: Phyd = ro * g * H * q (in W). ro is the density of water which is 1000 kg/m^3, g is the acceleration of gravity which is 9.81 m/s^2. H is the height difference in between the high water and low water level in m. q is the flow in m^3/s.
So first you have to measure the height H and the flow q and then you can calculate Phyd. Your water wheel or water turbine has a certain efficiency and this reduces the mechanical power coming out of the water turbine. If you use a transmission, this transmission also has an efficiency and this reduses the mechanical power coming out of the gear box. The generator also has a certain efficiency and this reduces the electrical power coming out of the generator.
To get maximum electrical power, the generator has to match well with the water turbine. This problem is explained for a wind turbine in chapter 8 of my public report KD 35. For a water turbine, you must follow the same procedure. This means that first you have to determine the optimum cubic line of the water turbine. Formula 8.1 of KD 35 is the formula for the optimum cubic line. Next the generator has to be loaded such that the Pmech-n curve of the generator lies as close as possible to the optimum cubic line of the water turbine. I have followed this procedure for a small water turbine in public report KD 598. However, I used a water turbine for which the characteristics can be determined easily and a generator which has been measured for different load conditions. If you don't know the characteristics of your water wheel and your generator, it will be a process of try and error.
All my public KD-reports can be copied for free from my website: www.kdwindturbines.nl at the menu KD-reports.
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