14 FT. watter wheel.

[Jerry]

I have a freind who has a 14 ft overshot watter wheel. He's had it for many years and recently diceded to make power with rather than just watching it spin.

The wheel has a 36" V-belt pully. From this a belt goes to an 8" pully on a transmision. The transmision gears up the speed again to another 18" pully on the output of the gearbox.

This 18" pully is then belted to a 4" diameter pully on a converted ac motor.

The motor in our first test is a 3 phase, 1 hp, 1725 rpm, 4 pole. We wired it for 208 volts. Then using a 3 phase diode pkg. for dc output.

He wants to use the power to heat his house basement. Just to take the chill of.

Were thinking around 1,000 to 1,500 watts.

Here are some preliminary power #s from this home brew hydro plant.

For each wheel rpm we get 20 rpm on the 18" pully. He has a 4" pipe as well as a flume.

With both of these open the wheel rpm is 11 with 1250 rpm at the gen. At this rpm we're getting 1250 gen rpm and 1,000 watts.

Adding additioal load did not slow the rpm but voltage began to drop.

I think that we reach the max power of that alt.

We will be replacing the 3 phase pma with an 90 lb. Magnatek industiral dc motor.

I have 9 of these and they are awsome.

Motor specs. shunt wound, 2 hp, 1750 rpm at 180 dc, 9.9 amps. feild 1 amp at 100 volt or .5 amp at 200 volt. 7/8" shaft.

I used another motor (same) with my high amperage dc power supply test drive this motor to see if it would act as a generator. I simply flashed the feild with 24 volt dc, then reconected as normal and WALA. It makes power. Much easyer than the PMA.

With the PMA I had to rev it to 190 volts no load to get it to drage down to 120 volt, 1000 watt under load.

With the Magnatek rev it to 135 volts for 120 volts 1000 watts.

I'll try to get some pix at the hydro site once we've got it installed.

I think this genny will handel 1500 watts with ease.

                    JK TAS Jerry

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[ghurd]

I understood the 'flashed motors' are not that efficient as gennies. (?)

If he insulates the outside wall, between the first story floor and the basement wall, the heat can only go up into the home. I bet it will be a very noticable reduction in heating units he needs to buy.

Normally that is the worst insulated place in the whole house, with the fastest payback for insulation. Payback in maybe less than 1 year! Worth $40 either way.

With 1500W 24/7 there is a lot of heat.  It would be a shame to have it rise, turn sideways, and shoot out of the house.

Side note. It's not even standard practice to insulate there at all in N.E. Ohio.

They leave it a bare 2x8. And our gas is going up another 70% this fall.

G-

[finnsawyer]

Since power is equal to V^2/R, you should calculate the power for different values of load resistance and plot it versus load resistance to find the load value for which maximum power is obtained.  You may be surprised.  You may, for instance, be able to get 1500 watts by putting two 1000 watt loads in parallel.  As you reduce the load resistance monitor the temperature of the alternator to make sure you don't burn it out.

[Jerry]

Hi GeoM

Once this new generator is installed we'll try a bunch of loads. I used this very large motor to power a 5KW generator. It never got much above warm.

I think these things are very under rated and built very tuff.

We'll load it down very heavy while trying to find the max load for the 14ft wheel.

                    JK TAS Jerry

[Jerry]

Hi g.

This one seems to be more eficient then the PMA I built. It takes less rpm to get the same power out.

I'm not sure about his insulation? We rearly see temps below 30% F. A few days in the teens but not many.

                        JK TAS Jerry

[Nando]

Jerry:

The AC Induction Generator has not reached its maximum power, it has reached the maximum  supplied power, the RPM is too low, it should be around 1875+ RPM to obtain maximum power.

The DC Shunt motor needs higher RPM than 1750 to obtain maximum power output, (I am guessing around 2200 RPM) and this motor can be controlled to obtain peak power, using a simple MPPT circuit to maximize the harvested power.

Also, it would be simple to add a DC/DC converter to charge batteries in case of Grid power losses emergencies, utilizing inexpensive AC/DC converters running the AC input in DC mode -- simple to do.

If you can measure the water volume and the head I could give you the power available estimate.

Also, if that can not be defined I could build a small circuit that calculates and control the peak power available = MPPT heating controller.

Regards

Nando