How to measure the power output of a PMA?

[zap1]

Hi,

I have a self-made PMA and I would like to create a power curve using RPM VS Watts. The PMA will be used on a wind turbine. Is there a standard way to measure the power output?

I use a tachometer to measure the rpm of the rotor. As for watts, I connect the 3-phase output to bridge rectifers (AC->DC), then to a 12V lead-acid battery with a diversion load controller to divert excess power with the battery is full. I use 2 DVMs to measure the battery voltage and the current going into it. Then apply P=VI to calculte the power output. However, if I use a battery of different brand or capacity, will the result stay the same?

Another way is to use open voltage, stator resistance and battery voltage to find the current going to the battery. Say, open voltage is 16V at 130rpm, stator resistance is 2ohms and battery voltage is 12.7V.

So I = (16-12.7)/2 = 1.65A

Power output at 130rpm = 1.65A X 12.7V = 20.955W

I did some testing using the above 2 methods and the results were quite different. Using method 1 at 130rpm, I only get 2W comparing to the second method which is ~21W.

I'm a little bit confused.

Thx

Zap

[Flux]

Hi Zap

Your first method is correct. Using different batteries will only change the voltage if it is below the set point of the charge controller and that will have a minor effect on the measured power. Above the point where the charge controller is operating any battery will give the same result.

Method 2 will never give you an accurate result, there are factors going on with the rectifier that you can't allow for, but usually the answer is a lot nearer than you have got.

What do you call stator resistance? Assuming it is star connected you should be using line resistance ( between a pair of leads), this will be twice the phase resistance.

You don't say what type of alternator it is, if it is an air gap type you should get answers a lot closer than you have. If it has a slotted iron core you will have leakage reactance and you may not be able to make much sense of the results, but if it is showing reactance limiting at such low output you are in trouble.

Another factor may be the accuracy of the resistance measurement if it is done with a multimeter, even at 2 ohms you may be well in error, check it by the volt drop method unless you have a Kelvin bridge.

Try some results at higher speed and power, you are close to cut in and rectifier drop is significant so that may be affecting the results.

If it is an air gap alternator, check that you are using line resistance, check the resistance value and check at a higher speed and come back with your figures and perhaps I can help some more. If it is slotted iron cored I may not be able to make much sense of the results.

My air gap alternators behave as though the resistance is higher than the measured value by a factor of typically 30% but when you take this into account the second method gives results good enough to predict output for design.

For true power measurements method one is the right one.

Interesting stuff.

Flux

[Devo]

I am probably wrong as the only formula's I know are from Flux,Ed & commanda but if memory serves me right I think Commanda posted once in my thread on the second way

16-12.7/2=1.65

Now I think the watts is the change in voltage X the ohms not the total voltage

3.3 volts X 1.65 ohms , still not 2 watts though so maybe I misread the post by Commanda

Flux am I wrong here?

Devo

[Flux]

Consider a dc generator for now, it's easier to follow.

The voltage you measure on open circuit is the emf. If you connect a load resistor the terminal voltage will fall.

The difference between the emf and the terminal voltage is caused by the internal resistance. The current flowing is the terminal volts divided by the load resistance.

This current also flows internally so the internal resistance is the difference between emf and terminal voltage divided by the current.

The load power is the current x load voltage. The internal loss in the generator is the current x the internal resistance.

With a battery the load voltage is fixed but the power in the load is still current x battery voltage.

The power loss in the machine is difference between emf and battery volts x internal resistance.

The alternator case is not as simple as it may have reactance and you will measure the impedance rather than resistance and you have the current switching between phases as the rectifier commutates, but with low reactance the current is carried by 2 wires at any instant.

With 3 phase ac loads you need to consider each phase separately.

Flux

[Flux]

Correction.

Paragraph 5  should have said  " internal loss in the generator is the current x the internal volt drop".  It is the current squared x the internal resistance if you want to work it that way.

Flux

[finnsawyer]

You have to ask yourself what you mean by power output.  If you mean the power you can get into a 12 volt battery than the first method is the correct one for that battery.  If you mean the maximum power you can get at that rpm then you should connect a series of resistors of decreasing value across the PMA.  You measure the voltage and use VxV/R to get the power.  You find the maximum value of power for that rpm.  Do this for a number of rpms and plot the power versus rpm.  Technically, that is the power curve of the PMA.  You need to be careful you don't overheat the PMA or blow the rectifiers.  In other words you need to limit the current.  What this means is that the power curve can be found for a range of rpms starting at zero.  I hope this helps.

In the second method you're trying to find the Thevenin Equivalent circuit of the PMA (a voltage source in series with a resistor), which is a valid approach, but ony if your voltage is pure dc or pure ac.  I'd say it is not pure dc.  In the first case you are measuring dc.  The current may actually have an ac component, but your meter ignores it.  By rights you should use an oscilloscope to avoid confusion.  I guess the moral here is to be careful when evaluating circuits involving rectification.

A little clarification is in order.  The rectifiers put out ac on top of dc.  When you connect the battery it acts like a huge capacitor eliminating most of the ac, so your volt meter measures only dc.  The ammeter measures only dc, so the results are correct.  In the second case when you measure the open circuit voltage the meter reacts to both the ac and dc.  It reads too high.  Hence the wrong result.  

[zap1]

Hi Flux,

It's an air gap type alternator. The stator resistance I measured is the resistance between the 3 ac wires, and they are all 2 ohms from each other. The generator is in star configuration. I'll try it at a higher speed.

Hi GeoM,

As for measuring the PMA power output, why do I need different values of resistors to measure the power? Can't I use a fixed resistance at different rpm to get the power curve? If the stator resistance is 2 ohms, what resistor should I use to measure the power? A 2 ohm resistor?

Thanks,

zap