Figuring RPM's from frequency

[alvin]

 Hi all

 I have a single phase 8 pole alt. At 10 Hz it is producing 1 amp. I don't have a tach. So would 10 Hz times 60 seconds divided by 8 poles
equal 75 RPM be right ? 

Thanks
Alvin

[frackers]

Do you mean 8 poles or 8 magnets. If 8 magnets then its 4 poles (pairs) so you're actually seeing 150rpm.

Just got my controller to read RPM accurately and my Piggott 10ft 6 pole (Otherpower stator) 24 volt machine is cutting in at 150rpm and just over 1kw at 280rpm. Information I've managed without for 3 years but interesting none the less

[ChrisOlson]

I have a single phase 8 pole alt. At 10 Hz it is producing 1 amp. I don't have a tach. So would 10 Hz times 60 seconds divided by 8 poles
equal 75 RPM be right ?

The formula is:
rpm = freq x 120 / # of poles

--
Chris

[alvin]

Thanks frackers and ChrisOlson for the response. It is 8 magnets and 8 coils. 150 rpm's sounds more like it. The most I've seen is 10 amps. I did not measure the frequency at that point, but it must not be linear.

This is my first attempt. Sticker shock from the NEO's led me to use microwave magnets on one disk and 2.5"X1/8" round neo on the other. 
Alvin

[rossw]

Thanks frackers and ChrisOlson for the response. It is 8 magnets and 8 coils. 150 rpm's sounds more like it. The most I've seen is 10 amps. I did not measure the frequency at that point, but it must not be linear.

Frequency is completely linear with RPM. Double the RPM and you must double the frequency.

Current won't be linear function of RPM when charging a battery though.

[ChrisOlson]

Current won't be linear function of RPM when charging a battery though.

Alvin, to add to what Ross said, you can calculate what  you should get for current vs rpm.

You need to know the internal resistance of your winding (either calculate it or measure it with a DLRO).  It's usually just about as accurate to calculate it if you know the length the windings and refer to a chart on resistance/Kft of the magnet wire.

The amount of open volts (with no load on the generator) at any particular rpm minus the loaded volts divided by the internal resistance of the winding will give you the amps the winding will deliver.  If you calculate this and graph it in your favorite spreadsheet application you will see that frequency is linear and increases in direct proportion to increases in speed.  The current, however, is not linear because the loss in the winding increases with the square of the current flowing in it.  The amount of loss in the winding is given by:
loss (in watts) = current (in amps) squared x resistance of the winding

If you calculate the loss in the winding with increase in current you can then calculate the power efficiency of the generator and graph it as well.  The power efficiency is the actual power (amps x volts) delivered by the winding, divided by the (actual power plus calculated I²R loss).

If you are a first time generator builder it is well worthwhile to actually do this in a spreadsheet and graph it, because it produces a better understanding of "how it works".
--
Chris

[kitestrings]

The formula is:
rpm = freq x 120 / # of poles

Alvin,

The formula I've used yields the same answer, but is:

rpm = freq (hz) x 60 / (# poles/2)

I suffer a bit from the CRS syndrom, so something that I find helpful is to start from known relationships.  A 60 hz, 2-pole motor is 3600 rpm, a 4-pole motor is 1800 rpm, a 6-pole motor would be 1200 rpm, etc. You can back into the formula from this starting point.  If you're on a 50 hz system, the common speeds will be: 2-pole, 3000 rpm, etc., but the formula is the same.

Chris, I like your explanation of calculating and plotting current based on stator resistance, and speed.  The modeling (what little) I've done is similar, but looking at the 'forcing voltage' over and above the nominal bank, and similarly calculating current (and I^2R losses).  If you have the generator to play with, I assume this is a more precise method, particularly for looking at efficiency.

Question: When you compare unloaded vs. loaded, do you put a set, fixed, larger than output load on it, or try to hold the voltage at a set level, or does it meatter?

~kitestrings

[alvin]

Thanks rossw and ChrisOlson

I meant the rpm's to amps as linear. I'll try to soak up this info. 
and take some more measurements. I am using a shunt that I had for an old motor controller to get my amp reading. It is a deltec that is 400amps = 100 millivolts. So I was calling 1 millivolt 4 amps. Do you think this may be an error?

 There is no wind right now, but it is 68 degrees out right now. I think I'll go till the garden.

Thanks for your formula too kitestrings.

Alvin

[ChrisOlson]

Question: When you compare unloaded vs. loaded, do you put a set, fixed, larger than output load on it, or try to hold the voltage at a set level, or does it meatter?

It doesn't really matter.  The open voltage of a permanent magnet gen where field flux never changes is known by rpm/volt.  For the loaded voltage line resistance, of course, must be taken into account.  With a 24 volt battery bank, for instance, even at 28.0 loaded volts you may well be running at 38-40 volts at the generator due to voltage drop in the line and rectifier.  And that can be figured in too, if you know the resistance of your wire wire run and add 1.4 volts forward drop on the rectifier to all your calculations.

If you put a larger than output load on it, assuming you have unlimited input power available to keep the rpm the same, you will just get a huge voltage drop, and hence more power right up to the point where the I²R loss in the winding consumes all the input power (generator in a dead short).
--
Chris

[rossw]

The battery voltage is a large chunk of the "nonlinearity" I was initially referring to, because the first (x) RPM will provide some voltage, but no current (until you are actually over the battery+rectifier voltage).

That's the first "step" in the curve.

From there on, there are nonlinearities in several places, including:
 - resistance in the stator
 - resistance in the external wiring
 - voltage drop in the diode (which changes slightly with current)
 - non-linearities in the batteries themselves. More charging current causes a greater voltage rise

As has been said already, with a proper understanding of these factors, they can be almost entirely "calculated out".

[kitestrings]

If you put a larger than output load on it, assuming you have unlimited input power available to keep the rpm the same, you will just get a huge voltage drop, and hence more power right up to the point where the I2R loss in the winding consumes all the input power (generator in a dead short).

I was assuming, maybe incorrently, that you were connected to a battery bank, for the loaded testing, and either trying to hold the voltage constant (so you added load as you increased rpm, and output).  Otherwise, doesn't the load on the alternator start to fall off as the battery voltage increases?  Just trying to understand how you are obtainiong the "loaded volts' readings.

~kitestrings

[ChrisOlson]

Loaded volts at the generator will always be measured voltage at the load plus voltage drop in the rectifier (if used) and voltage drop in the transmission line from the generator to the load.

If you are charging batteries, and voltage of the batteries comes up, then indeed you get reduced load at the same rpm.  But with a wind turbine, the speed of the turbine will increase, raising the voltage.  So the load still ends up being a function of how much power the blades can make minus the efficiency of the generator and transmission line.
--
Chris