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Forum Index : Windmills : Wire Size Calculation & Seller
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norcold Guru  Joined: 06/02/2011 Location: AustraliaPosts: 670 |
[line loss]http://www.nooutage.com/vdrop.htm[/URL] An online line loss calculator for both Ac, single or 3 phase & DC We come from the land downunder. Vic |
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Downwind Guru Joined: 09/09/2009 Location: AustraliaPosts: 2333 |
Simple it come back to square area of copper required to give equal loss. For example using Norcold's calculator HERE shows that for 3 phase using 3 x 10mm2 cables gives around the same loss as DC using 2 x 25mm2 cables, calculated over a distance of 100 foot, at 24v and 70 amp. 3 x 10mm2 = 30mm2 2 x 25mm2 = 50mm2 So the cost of cable for DC is almost double than 3 Ph, for the same loss. If you add to this the gain from using a rectifier (not resistive loading) than 3 phase wins in lesser mm2 total cable area required and lower cost and greater flexibility. Pete. Sometimes it just works |
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| norcold Guru Joined: 06/02/2011 Location: AustraliaPosts: 670 |
Precisely, that is why the 4mm sq x 3 run on my 48v mills works well. Idealy I should be probably 6mm sq or better but I had 4mm sq at hand. Also as I understand it the voltage on the AC rises with revs (as does amperage) whereas with DC it`s pegged at the battery voltage. Batteries down a bit = more line loss, just what you don`t want. Also I believe extra line resistance helps stop mill over revving, as seems to be the case in my mills. They survive some high gusts during tropical storms. Of course too much resistance and you`ve lit up a Viscount.  We come from the land downunder. Vic |
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wallablack Senior Member  Joined: 10/08/2011 Location: AustraliaPosts: 164 |
Thanks guys, This has been a very good learning curve. I think it was always my assumption that AC travels further than DC with lees line loss. I too have always run the AC from the turbine to the battery/grid connect. These have always only been short distances in the past but this one is a bit longer than I have ever needed to deal with. Norcold's calculator is really good.... What is an "MCM" cable measurement though? Foolproof systems do not take into account the ingenuity of fools. |
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| Warpspeed Guru Joined: 09/08/2007 Location: AustraliaPosts: 4406 |
No it isn't that simple. That would certainly be true if the ac currents flowing to the rectifier were pure sine waves. If you have 20 amps of dc, it is exactly the same as having 20 amps rms ac. That is by strict definition, that is what the rms value is, the exact dc equivalent. The extra cable loss come from the highly distorted current waveforms in each phase. The combined rms phase currents are considerably higher than the net total dc output current from the rectifier. Don't just take my word for it, measure it for yourself. It must be with a good quality multimeter that reads TRUE rms (most cheapy multimeters don't) and read the individual phase currents going into the rectifier. Then read the dc current coming out of the rectifier. You will discover that each of the true rms phase currents will be about 82% of the dc current. It won't be exact because the mill voltages are probably not pure undistorted sine waves, especially when loaded by a rectifier, but it will be close. Try using your cable loss calculator assuming 82 amps rms per phase, and compare that to 100 amps dc. You will find 3 cables that each have to carry 82 amps will have a higher loss than two cables that each carry 100 amps, assuming the same total copper volume in each case. It's 82+82+82 versus 100+100 Or 246 versus 200, about 23% higher current density given the same total cross section of copper. But, losses are proportional to current density squared 1.23 x 1.23 = 50% higher loss ! And THAT is the problem. Don't assume that just because everyone fits the rectifier at the load end it is the correct and most efficient way to do it. Cheers, Tony. |
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