Author Topic: Matching prop to alternator  (Read 2573 times)

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Waterfront

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Matching prop to alternator
« on: December 15, 2005, 10:30:45 PM »
Hey everybody! Love the board by the way!


So, just checking if I have this figured out ok, when you reach cut-in speed, (for battery charging) you want your prop to have more power then your alternator, so it doesn't stall.


There's quite a nice calculating site here: http://www.alton-moore.net/wind_calculations.html and it seems to be able to give the prop power and RPM for a given wind speed. So, to make sure your alternator is not too powerful, you'd test it at this RPM and compare the power value with the one in the chart?


I know one way to do this without building a whole stator would be to test with just one coil, a «test coil». Since power is volts times amps, would you need to load the coil with a resistance so that amps would flow?? or would you leave it open circuit?? I saw a page on otherpower.com about an experiment where I think it's DanB was loading his coil with a 1 ohm resistance, would this be the ideal to go with??


Thanks,

Waterfront

« Last Edit: December 15, 2005, 10:30:45 PM by (unknown) »

willib

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« Last Edit: December 15, 2005, 09:57:42 PM by willib »
Carpe Ventum (Seize the Wind)

Flux

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Re: Matching prop to alternator
« Reply #2 on: December 16, 2005, 02:15:32 AM »
Don't worry about stall at cut in, you won't.


Use the test coil to find the number of turns for the cut in speed you have chosen.


If you have chosen a prop with nominal tsr of 7, you would probably take the cut in speed as if it was tsr8 at 7 mph wind speed. As you load the prop the tsr will fall as speed increases. You need to check the loading to make sure you don't fall below about tsr5 at full output.


If you are a good at calculations you may possibly be able to predict the output from load testing one coil, but I have never found it a useful thing to do.


It is much better to measure the resistance ( or calculate it) of the test coil. From that you can predict the winding resistance of the complete machine.


Knowing the open circuit voltage, battery voltage and winding resistance you can then predict the output and see how it compares with the prop power.


If you cut in at 200 rpm for 12v, at 600rpm the o/c voltage will be 36v. You have 24v in excess of the battery so the current into the battery would be 24 divided by the winding resistance. This current x 12v gives you power out.


Power lost in the alternator will be current^2 x winding R. From this you can find the input power ( output + losses) and compare this with your blade power.


Flux

« Last Edit: December 16, 2005, 02:15:32 AM by Flux »

Waterfront

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Re: Matching prop to alternator
« Reply #3 on: December 16, 2005, 05:18:36 AM »
Wow! Thanks Flux! That's exactly the explanation I was looking for!!
« Last Edit: December 16, 2005, 05:18:36 AM by Waterfront »