Hey all just found this place and figured it's as good as any to ask my questions.While I have not acted on this at all yet , I want to make a waterwheel.The hope is a 24 sided Icositetragon overshot 12 ft in diameter with 8 foot head.The flow of the creek is always constant at 2000 (rough measurement) gpm.
I was wondering how wide the wheel needs to be?I assume the "buckets" need to hold as much water as possible on the given opportunity they are given during rotation.How big should the shaft be?Should the shaft and wheel axle be the same thing or just connected?Should I make the water wheel as light as I can or should it have some weight to it?Is there a better shape , like just making a circle/will it harm it if it's not completely circular.
My main goal is to power something like a 12kw generator , but I dont know of it's possible.Thanks to all who take time to help.
How can a 12 foot wheel fit under 8 feet of head, given an overshot wheel, unless the lower 4 feet of the wheel are under water? That wont spin.
2000 gpm is a LOT of water. Are you sure about that estimate? Here is a weir calculator ... WSU weir.
A weir opening 4 feet wide and 6 inches deep will produce 2052 gpm...
So as a first guess, I would say your wheel needs to be at LEAST 4 feet wide, maybe a lot more, or else each bucket will fill before the next one rotates into position, thus spilling water over the wheel without producing power.
The physics equation for hydropower:
P(kW) = Flow(gpm) x Head(ft) / 5.6 = 2000 x 8 / 5.6 = 2.857kW
Count on at least 50% losses to friction and overshot wheels being inefficient, losses in gearing up low RPM wheel to high RPM generator... ballpark 800watts is my guess. That is a good amount of power, but the investment to make a wheel, building, flume, etc capable of handing 2000 gpm... that's a big waterworks.