Author Topic: Calculate power from windmill on DMM readings?  (Read 852 times)

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robfred

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Calculate power from windmill on DMM readings?
« on: December 16, 2006, 08:06:22 PM »
Was wondering if anyone could help me with the equations to calculate power out from a windmill with a resistive load from DMM readings of voltage and amperage.  Also how does the load relate to the power output.  


For example, if I have a 12VDC 10amp rated motor/generator with a 2VDC 300mA load and I get voltage readings of 3 volts and amperage of 3.5 amps at 6.5 m/s what is the power being produced by the generator?


What is the effect on output is I increase the load to say 7.5V and 900mA?


please help

« Last Edit: December 16, 2006, 08:06:22 PM by (unknown) »

alancorey

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Re: Calculate power from windmill on DMM readings?
« Reply #1 on: December 18, 2006, 11:32:30 AM »
I'm not sure what you mean by output, most people mean wattage.  Wattage is amps * volts.


  1. VDC 300mA = 2 * 0.3 = 0.6 watts
  2. volts and amperage of 3.5 amps = 10.5 watts
  3. 5V and 900mA = 7.5 * 0.9 = 6.75 watts


You shouldn't be using a light bulb as a test load, even though it's nice visually.  Light bulbs have much lower resistance when cold than when they're glowing or at full brightness.  This can screw up comparisons.  Use a real resistor, or if you don't have one make one up from something like steel wire.  You can find wire tables on the web for different sizes of wire, but double check with an ohmmeter once you've got it cut.    Something like electric fence wire should work well although the galvanizing on the outside will make some difference from the pure steel wire in the tables.  Coat hanger wire would work if you've got patience enough to connect a bunch of them together (probably best done with machine screws and nuts).  Make up something that doesn't heat up with use and then you've got a good standard that shouldn't change much with time.  Don't let it get rusty.


  Alan

« Last Edit: December 18, 2006, 11:32:30 AM by alancorey »