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dazz Regular Member Joined: 15/04/2008 Location: Posts: 78
Posted: 06:20am 19 May 2008
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Hi All,
I have just finished converting an 80S into an 80SP.
The stator is in good condition, except for some obvious green oxidation in some of the crossovers of coil wires. So the laquer coating must have gone at these points. Wherever i noticed this, I siliconed a piece of cardboard between the wires to stop any possible shorting.
Once finished i was getting 0.4 ohm for each phase.
And on testing a single phase at 200RPM with a load of 27.3 ohm, the rectified DC output was 5.2V RMS(I checked that the meter was reading correctly by putting a cap across the output and it showed 7.4V indicating peak voltage, which does correspond to 5.2V RMS)
Does all this sound ok for the load I'm using?
It's only about 1W and I would have thought it should produce more per phase at that speed and load.
I'm just wondering whether the corrosion might have been more of a problem than i thought
Cheers, Daryl
dazz Regular Member Joined: 15/04/2008 Location: Posts: 78
Posted: 09:38am 19 May 2008
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I got brave and connected a 12V halogen globe of the type used in car headlamps. This is a much increased load somewhere around 0.2 ohm. It's a bit hard to tell as the meter only goes to 10'ths of ohms, but it fluctuated between 1 and 2, mostly on 2.
this time it produced 3.5V RMS at 160RPM on the single phase which amounts to about 63W if i assume 2 ohm for the load. This is not bad because all three phases would be handing out about 180W, probably a bit more.
Not bad for the speed
Daryl
GWatPE Senior Member Joined: 01/09/2006 Location: AustraliaPosts: 2127
Posted: 12:10pm 19 May 2008
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Hi dazz,
I do not believe you are measuring correctly. You should check the wattage, usually stamped into the base of a QI bulb base. Most bulbs would be 50/55W, or on a modern car, 60/100W. I would guess that the bulb was probably the 50/55W type. A bulb of this type would require 12Vrms to provide the 50/55W rated output, not 3.5Vrms. power = voltage^2 / resistance. power = 3.5 x 3.5 / 2 = 6.1W Do not despair, light bulbs are good indicators of output, but load resistors provide more useful data. A fully loaded QI bulb hurts the eyes to look at, at close range. When I was first testing my windmill, I blew many 300W, 28V projector bulbs. I replaced them with a bank of 100W audio dummy loads. An 8 ohm resistor across 24Vdc gives 3A, I had up to 5 in parallel on my windmill. I did have them switched in at slightly higher voltages. I replaced with a PWM cct. This is now replaced with a grid diversion regulator.
Out of curiosity, what is the stated wattage of the bulb you are using? If it is 50W 12V, then this would draw a little over 4A and the resistance would be around 3 ohms at the rated power. .. .. Gordon.become more energy aware
dazz Regular Member Joined: 15/04/2008 Location: Posts: 78
Posted: 02:04pm 19 May 2008
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Hi Gordon,
You are absolutely correct. I pulled the bulb out of it's holder and I'ts an H3 12V/75W which puts it at 1.92ohm
I should have twigged that 0.2 was way too low for a bulb
But you know how it is. you take a reading like that on your multimeter and don't think twice
The meter is a good Fluke but 15 years old and hasn't been calibrated since it left the factory. Time to do so I think.
This also means my measurements of the phase resistance were out, which also makes sense.
assuming the RMS voltage reading isn't out too much, then the actual power was 3.5X3.5/1.92 = 6.3W
That would be a total of 19W for the three phases at 160RPM
I'm no closer to knowing whether I'm getting reasonable power form the generator with this load, but considering the improvememnt from the 27 ohm load then I guess it could be about right.
I still have some issues with the bulb itself though. It says clearly stamped that it is an H3 12v/75W
The torch i got it from says it's a 75W torch but I thought H3 was supposed to be 12V/55W
Perhaps I should just go out and get those resisters you mentioned(and maybe a new multimeter). They are wire wound I presume?
Thanks, Daryl
Cheers
herbnz Senior Member Joined: 18/02/2007 Location: New ZealandPosts: 258
Posted: 07:29pm 19 May 2008
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Hi Daryl
Its not your meter
incandescent lamps have cold resistanc much much lower than operating resistance. Temp coeff of filiament when changing from 20 c to 1200 c
Reason there is strong advice not to use as dump loads.
initial current 8 times final.reason lamps always blow when first turned on.
Herb
dazz Regular Member Joined: 15/04/2008 Location: Posts: 78
Posted: 03:43am 20 May 2008
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ahhh, thanks Herb! I think you just saved me $300
Sounds like another good reason to use wirewound resistors as a test load.
I went looking for them last night. I've never had need for anything more than the 5W ceramic type so I wasn't sure what they would look like.
I found a range of 100W devices at RS-Components. Very StarTrek looking things with the gold heatsink. And at around $16, can't really go to far wrong.
When I start taking serious measurements I'm going to uses voltage and current to calculate power and leave the resistance out of it. There is always going to be some heating and i can just see it being problematic.
I'm quite looking forward to setting up the current measurement. I'm using a current transducer with proportional DC output. Never used one of these before, so it should be interesting. It's fast response, so once hooked up to the MCU i should see the waveform nicely.
Cheers Daryl
herbnz Senior Member Joined: 18/02/2007 Location: New ZealandPosts: 258
Posted: 05:00am 20 May 2008
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Hi Daryl
I always just use my batterirs. The generators do have different characteristics on these rather than resistors and as charging is usuallythe finaluse you dont get mislead.
also works well cos the testing motor is being feed from the same batteries so only losing by the loses.
Good luck
Herb
GWatPE Senior Member Joined: 01/09/2006 Location: AustraliaPosts: 2127
Posted: 08:24am 20 May 2008
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Hi dazz,
the resistors you mention will probably need to be mounted on a heat sink to dissipate the rated 100W. I use wire wound air cored ceramic resistors, that do not need to be derated.
As herbnz says, you might be better starting with a battery. The battery presents no load up to the cut in voltage. It is very easy to overload the blades with a resistive load, unless there is a voltage sensing load controller. .. .. Gordon. become more energy aware
dazz Regular Member Joined: 15/04/2008 Location: Posts: 78
Posted: 09:12am 21 May 2008
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Thanks Guys
Gordon, yeh, I sort of suspected they would need extra heat sinking once i start playing with the higher power levels.
You both may well be right that I'd be better off using batteries as my load. I haven't tried that so far because I'm playing with voltages below cut in and also I worry how the state of charge of the batteries will mess with my results. The tests I'm doing mean that it's favourable to have a near constant load so i can get a clear picture of output Watts against RPM.
Cheers, Daryl
GWatPE Senior Member Joined: 01/09/2006 Location: AustraliaPosts: 2127
Posted: 02:11am 22 May 2008
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Hi dazz,
The resistance does not present a constant load, unless you modulate the power somehow. As the voltage increases with the rpm, the resistor presents a squared power relationship. For a constant resistance load, If the voltage output doubles, then the current would also double, therefore the power would quadruple.
Ideally, the rotor rpm should follow the windspeed. The power in the wind however has a cubic relationship to windspeed. This is why a load resistor calculated for maximum power overloads the windmill at lower power levels. .. .. Gordon.become more energy aware
dazz Regular Member Joined: 15/04/2008 Location: Posts: 78
Posted: 05:20am 22 May 2008
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Hi Gordon,
Sorry, constant load was a bad choice of words on my part. I should have said known load.
But I also see your point and without knowing it, I was actually trying to simulate the problems of maximum power and low power with my SMFIS project. One of the things I have been trying to do was stress the drive motor in the test rig with a low resistance load to gain better performance from the generator by reducing the magnetic fields over the stator.
The obvious extreme is where the generator won't actually turn until the magnetic field is reduced. At this point, the improvement in efficiency is infinite. But this is also near impossible to simulate with a motor drive because you can't have a motor that is so stressed it won't turn(well not for long anyway)
Anyway, I'm rambling. If you have the time to take a look at the SMFIS thread, I would appreciate your thoughts, particualrly with regard to maximum power theory and how SMFIS might be used to alleviate some of the low wind problems in a max power setup.
Notice. New forum software under development. It's going to miss a few functions and look a bit ugly for a while, but I'm working on it full time now as the old forum was too unstable. Couple days, all good. If you notice any issues, please contact me.
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