Measuring power output

[tonig123]

Hi, I am a bit confused about the correct way of measuring the power output of a wind generator. I see some strange figures from manufacturers and others who measure open volts and then measure amps with a load and figure watts multiplying both numbers. It just doesnt seem correct to me.

Also, if charging a battery, it will keep the voltage down further affecting voltage readouts.

Is there a correct or standard way of measuring output of a wind generator?

[Countryboy]

Hi tonig,

  The correct way to measure output is to measure both amps and volts under a load.

Measurring open circuit volts is for figuring out the cut-in speed. (the speed at which the alternator reaches charging voltage)

[hiker]

i use 12v -50watt headlites and a voltmeter...

i keep on adding headlites tell the volts drop to 12volts..

then i know the the watts my alt is putting out..of course you have to conseder the rpms.....................

[Flux]

For a battery charging wind generator, the power out is the current x battery voltage measured at the battery.

This is how all RESPECTABLE manufacturers measure their power. Ignore the idiots that you refer to.

Flux

[jmk]

 If the battery volts go up as amps go up, do you use that voltage, or the voltage the batteries read before the high amp imput?

[Flux]

As far as the wind generator is concerned you use the battery voltage as it is. You will not be able to reclaim the power from the battery at the higher volts but the wind generator doesn't know that.

Flux

[jmk]

 So than use the battery voltage that it would show if everything was disconected for a while?

[SparWeb]

Tonig123,

"Is there a correct or standard way of measuring output of a wind generator?"

The answer is no, at least, not among us tinkerers, here.

Your test may be as thorough as you like, it depends on how much information you want, and how much you can handle.  For example, Flux has gone to the trouble of brake-testing various alternators to compare performance, and created power curves at comparable conditions to display the results.  At the other end of the spectrum, many people just measure open-circuit voltage in their drill-press (unknown RPM's), then bolt the prop on, and hoist it up the tower.  They don't worry too much about matching prop size to alternator size unless problems arise, and often they're lucky.  Either approach is what suits the particular user.

Where on the spectrum do you lie?

[finnsawyer]

Considering that these machines are three phase you could connect an oscilloscope to one phase and measure the time for one cycle and get the rpm from that.  Another way would be to put a small resistor in series with the battery and use the scope to measure the ripple across the resistor.  Keep in mind that that ripple is three times the frequency as it comes from the action of all three phases.  That is, three pulses of current are shot into the battery during one revolution of the alternator.  Once you can pick off the ripple you have the option of shunting it through a capacitor to an amplifier, if necessary, and using a frequency meter to measure the frequency.  

[tonig123]

I'm a beginner of the type that just put a prop on a generator and bring it up. This is because of my lack of good knowledge, but I am willing to learn through experimentation. I just put up my first ametek generator and was playing around with it and by watching the output and reading through the board I just realized that my blades are too small (3 feet) and are stalling.

[kitno455]

uhm, i think its a bit more involved than that, the fullwave bridge rectifier gives two 'bumps' per phase, and the more poles a gen has, the more bumps per rotation. you cannot just multiply by three.

allan

[SparWeb]

Since we're talking about measuring RPM, I thought I'd put in that the easiest way for a beginner to do it is to get a bicycle spedometer and program it accordingly.  I have written about this in my diary, and I can give more details if you want.

[Ungrounded Lightning Rod]

For rating the generator you use the amps times the battery voltage during charging.

For estimating the power you get back you use the amps times the battery voltage during discharging.

The ratio is the battery's efficiency.  It's the battery's fault, not the generator's.  So it doesn't affect the generator's rating.

You just have to realize it happens when you are figuring out how big a genny and how big a battery you need.

[Slingshot]

If you are trying to calculate the power output of the genertator, you must use the voltage and current that occur simultaneously.

[finnsawyer]

Full wave bridge rectifier?  Super full wave bridge rectifier maybe.  The three phase alternator uses six diodes two of which are connected to each lead from the alternator.  The diodes at each lead are arranged such that one allows current to flow from the lead into the positive terminal of the battery and the other allows current to flow from the negative terminal of the battery into the lead.  If we take the negative terminal of the battery as the common or ground for the circuit then the action of the diodes constrains the lead voltage to be between -Vd and Vb + Vd.  If the voltage between two leads such as Vab becomes sufficiently large in a positive sense, current will flow from lead a through a diode into the positive terminal of the battery.  Similarly if Vab becomes sufficiently negative current will flow from lead b through a diode into the positive battery terminal.  This will be true for both the delta or star configuration of the alternator. The three phase relationship between the alternator voltages means there will three positive maxima and three negative maxima per cycle.  So, the correct answer is six pulses.  I had neglected to take into account the effect of the negative pulses.

You also need to figure how many cycles per revolution.  The standard 3:4 design with three coils for every four magnets generates a number of cycles per revolution equal to the number of magnets divided by two.  The design I have proposed in my diary, the 3:2, which has three coils for every two magnets generates a number of cycles per revolution equal to the number of coils, or 1.5 times the number of magnets.  So it's (measured frequency)/(6x(cycles per revolution)) =  (revolutions per second).