hydro power

[John Biggers]

I'm resreaching floating waterwheel power and need some help with the hydrology/kilowatts, i have a usgs station about a mile upstream at the low flow of the river is about 200 cubic feet per second. here are some of my equations. low flow of the river at 200 cfps
waterwheel size 8 feet
paddle size 18 or 24 inch pvc pipe cut in half install at 30 degrees to the approaching water to deture drag
paddle lenght 10 feet
the math at 200cfps divide by 6, 6 sides in a cube =33.33fps multipl the sum by .66 efficiency = 22 fps, divide the sum by the diameter of the wheel 25.12 = 0.88 feet, this is the distance that the wheel should turn in one second. now multiple 0.88 x 60 seconds = 52.80 rpms. if this is correct? this will asist in the gearing of a low rpm generator which is not a problem. in other reseach i came across a mill rights booklet on the web how to determined kilowats and horse power this is where I need help. is there a engineer out there to assist or a field person to help

[Dale Sheler]

The first thing I would look into is whether it would be legal to put such a unit in the river.

[John Biggers]

Thanks for the concern, ncgs 77-13 statues states that a stream or river may not be block obscured, except for the reasons of motive power. Taking in mind the size of this project i have contacted our company attoney with my intentions so far so good. I am still researching for some engineering help or the formulas to complete the design of the project. Wasserrad-mit-integriertem-generator this is reference to the construction

[Dave Turpin]

Floating waterwheels are a bit difficult to calculate. Most waterwheels, whether they be breastshot or overshot, use the height of the weir to determine the potential energy in the system. A floating wheel has zero head height, so most calculations will say that it will generate no power at all.

However, if you use Bernoulli's equation, you can find the extractable energy based on the kinetic energy of the water in the stream. This is of course wholly dependant on the flow velocity, not just the flow rate.

Here is the difficult part: When you punch the numbers, you realize that the only way to extract energy from this system is by SLOWING DOWN the water in the stream. How much this is possible depends immensely on the size of the wheel. It should not allow any place for water to divert around the wheel, otherwise it is only gaining energy by DRAG, and this is a miniscule amount of the energy in the water. (although, the Australians have been successfully using propeller systems in slow-moving billabongs for some time) Furthermore, the speed of the wheel, and thus the RPM of the generator, is difficult to calculate. At ZERO load, the wheel's tangential velocity should be only slightly slower than the stream velocity. But... Once you put a load on the system the wheel slows down, since that drop in speed is where the energy comes from.

So... If you are making a wheel 10 feet wide, you need to put it somewhere where the stream is 10 feet wide. Ideally the shore should be reinforced to prevent water from moving around the wheel. The height of the wheel is only important for setting the RPM. (the smaller the wheel, the higher the RPM), and you need to select a generator that will produce the power you want with a LARGE RANGE of RPMs. As for the gearing, when I was designing a wheel I was just going to use a motor gear reducer in reverse to step up the speed. Just make sure it isn't a worm gear. Worm gears don't work in reverse.

[Krista Hiles]

I agree with Dave, The stream has to be minimum 10 feet wide.

[Peter Mckinlay]

I'm resreaching floating waterwheel power and need some help with the hydrology/kilowatts, i have a usgs station about a mile upstream at the low flow of the river is about 200 cubic feet per second. here are some of my equations. low flow of the river at 200 cfps
waterwheel size 8 feet
paddle size 18 or 24 inch pvc pipe cut in half install at 30 degrees to the approaching water to deture drag
paddle lenght 10 feet
the math at 200cfps divide by 6, 6 sides in a cube =33.33fps multipl the sum by .66 efficiency = 22 fps, divide the sum by the diameter of the wheel 25.12 = 0.88 feet, this is the distance that the wheel should turn in one second. now multiple 0.88 x 60 seconds = 52.80 rpms. if this is correct? this will asist in the gearing of a low rpm generator which is not a problem. in other reseach i came across a mill rights booklet on the web how to determined kilowats and horse power this is where I need help. is there a engineer out there to assist or a field person to help

The most efficient water wheel was designed by German Army Captain in the 1800's it placed on the Rhine and powered the German woollen mill. The shape of the blade is unique!

Please be aware floating turbine power generators do not exist. They must be tethered and stabilised! Water strike shall continually act to lift the device from its site.

Cheers

Peter