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Forum Index : Windmills : Wire Size Calculation & Seller
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wallablack Senior Member  Joined: 10/08/2011 Location: AustraliaPosts: 164 |
Good Morning All, Is there any good online wire size calculators? I am trying to calculate a cable run at the moment. I need 3 x AC cable runs at 45 metres including up the tower. 2Kw 24V Turbine (2800w Max) Each AC phase voltage is 20.74V. AC current is about 63A at each phase. I have calculated 10mm2 cable would only give me 2% loss. I am not sure if this is 100% correct, any thoughts? Does anyone know of a good cable supplier? Foolproof systems do not take into account the ingenuity of fools. |
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norcold Guru  Joined: 06/02/2011 Location: AustraliaPosts: 670 |
www.reuk.com.uk have a top online wire size calculator . Have found a compromise has to be reached between cost & loss. Remenber you will not be achieving 100% output very often. Try futurlec for cable, although have found Ebay can sometimes come up with cable. We come from the land downunder. Vic |
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Madness Guru Joined: 08/10/2011 Location: AustraliaPosts: 2498 |
If you see your local scrap metal yard you can get heavy cable at a good price. There are only 10 types of people in the world: those who understand binary, and those who don't. |
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Downwind Guru Joined: 09/09/2009 Location: AustraliaPosts: 2333 |
norcold, Your link is broken when i tried it, can you please try to use the Insert link button in the future. 
You might find it will bring up a yellow bar on the top of your page asking for permittion to trust the web site. 
click the yellow bar, then click the link button again, this time it will ask for the link, just follow the prompts. The forum inserts false charactors in a posted link to stop spamming so it pays to use the link button to prevent corrupt links. Pete. Sometimes it just works |
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| Privatteer Newbie  Joined: 09/06/2012 Location: AustraliaPosts: 39 |
I think you might have made a mistake or I am way to drunk. Perhaps used a 240v table? At 20.7v and 10% loss I came up with 50mm! At 20% 25mm2. I have assumed multicore cable mostly in a underground conduit. Edit: Just to expand on that with 16mm2 I arrived at 7.32volt drop or about 13.38v AC or roughly 18Vdc once rectified. At 50% ouput ie 30Amps that drops to just 3.3v lost or roughly 24Vdc. 10mm2 the values approach 12v lost which is too high. So unless I have made a mistake myself 16mm2 is ideal if you don't mind losing some potential power. |
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| norcold Guru Joined: 06/02/2011 Location: AustraliaPosts: 670 |
Sorry about that here is the address again, reuk.co.uk(not com) my intent was not to give a link. Go to that site than on through their links and you`ll find a lot of info regarding our interests, as well as an online wire size calculator. We come from the land downunder. Vic |
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| wallablack Senior Member Joined: 10/08/2011 Location: AustraliaPosts: 164 |
Privatteer - You may have been very drunk but your calculations were better than mine, Maybe I might have a few beers and try my math again. The reuk wire calculator is pretty good and can be found HERE A cable "10mm in DIAMETER" will give a loss of 3% which is a 78mm2 cable. A cable "10mm2" will give a loss of about 25% which is no good. Foolproof systems do not take into account the ingenuity of fools. |
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| Warpspeed Guru Joined: 09/08/2007 Location: AustraliaPosts: 4406 |
45m of 10mm^2 cable will have a resistance of : .0172 ohms divided by 10 (mm^2) multiplied by 45 (metres) = 0.0774 ohms At 63 amps rms per phase, power loss will be I^2 x R 63 x 63 x .0774 = 307.2 watts per phase. There are three cables and three phases, so total losses will be 921.6 watts Input power 63+63+63 amps @ 20.74 volts = 3,919 watts Output power 3,919 - 922 = 2,997 watts Efficiency 2,997/3,919 x 100 = 76.5% Voltage drop 63 x .0774 = 4.876 volts per phase Voltage at load 20.74 - 4.876 = 15.864 on each phase This all assumes a purely resistive load, which a rectifier is not. For 2% loss instead of 23.5 % loss you will need to reduce the resistance about roughly twelve times, to reduce the voltage drop twelve times. Or start thinking in terms of three 120mm^2 cables. Cheers, Tony. |
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| Warpspeed Guru Joined: 09/08/2007 Location: AustraliaPosts: 4406 |
Thinking about it a bit more, it is probable that you read somewhere that 45 metres of 10mm^2 three phase cable has a loss of 2%. That would be correct at around twelve times the voltage, maybe at an assumed 240v instead of 20v. You will still lose around 4.9 volts at 63 amps regardless. At 240v, a 4.9v loss is not all that much. At 20v, the same 4.9 volt loss becomes massive. Privateers figures look sound to me. Cheers, Tony. |
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| Downwind Guru Joined: 09/09/2009 Location: AustraliaPosts: 2333 |
[quote] At 63 amps rms per phase, power loss will be I^2 x R 63 x 63 x .0774 = 307.2 watts per phase. There are three cables and three phases, so total losses will be 921.6 watts [/quote] Is it actually 3 times the loss or only 2 times, what my thoughts are, with 3 phases only 2 of the phases are active at any peak period, meaning one phase will be at neutral (0v - supplying no power) and the other phases will be either high or low in respect to neutral, then the phases alternate, but still only 2 phases are at peak to peak of the wave at any time. So why is it calculated as 3 time the loss when the third phase is doing nothing. Pete. Sometimes it just works |
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| norcold Guru Joined: 06/02/2011 Location: AustraliaPosts: 670 |
. . MODEL . . . . . < 150' . . . . 150' - 300' HY- 400/12V . . . . 08 AWG . . . . . 06 AWG HY- 400/24V . . . . 10 AWG . . . . . 08 AWG HY- 600/26V . . . . 10 AWG . . . . . 08 AWG HY-1000/24V . . . . 08 AWG . . . . . 06 AWG HY-1000/48V . . . . 10 AWG . . . . . 10 AWG HY-2000/48V . . . . 04 AWG . . . . . 02 AWG HY-3000/48V . . . . 04 AWG . . . . . 02 AWG Above is recomendations from Hyenergy Systems. 8AWG = 8.36mm sq 6AWG = 13.29mm sq. I have 2 x 500watt turbines at 40m and 30m and use 4mm sq on 48v. The father of small turbines Hugh has a post on the subject(Windempowerment forum) I read back a year or so ago, might be worth having a read of. We come from the land downunder. Vic |
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| Warpspeed Guru Joined: 09/08/2007 Location: AustraliaPosts: 4406 |
Pete, if you have three 100 watt light bulbs connected to three phases, are you suggesting there can only be 200 watts because one bulb is always off ? The rms figure accounts for the voltage and current cycling from zero to sine wave peak value. The rms value of alternating current is assumed to be the exact same as an equivalent amount of constant flowing direct current. RMS currents in each of three phases can be simply added to get the total power, and the losses can be added in the exact same way. Cheers, Tony. |
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| Downwind Guru Joined: 09/09/2009 Location: AustraliaPosts: 2333 |
Tony, its not a claim only a point of view that 3 phases and windmills dont workout the same as 3 phase mains. For starters we full wave rectify the 3 phases, the only real part of each wave cycle we are interested in is the peak to peak voltage, so any voltage less than the peak voltage contributes nothing to our DC supply. If we used a motor on the 3 phase i would see it different than just rectifing it, if we used light globes on each phase then we would need a separte neutral point, which we dont have or use with a wind generator. With rectified 3 phase would it not be correct that only 2 of the 3 phases are supplying full power through the diodes at any one time. Pete. Edit....... Looking at this wave image. 
My theory might be wrong as only 1 phase would be at peak voltage (either side of common) at any one time, remembering we are rectifing the power and using peak voltage. Sometimes it just works |
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| Warpspeed Guru Joined: 09/08/2007 Location: AustraliaPosts: 4406 |
Yes you are quite right, rectifiers have very different form factors to resistive loads. As I stated in my first post, my calculations were based on a resistive load and unity power factor, which is how cable losses and wiring sizes and temperature rises are all generally calculated. Now a six pulse three phase rectifier, as you have shown it, behaves very differently. The dc output voltage is not the RMS value, but the peak value, less about five percent (minus two diode drops). The current in each phase becomes a square wave with a peak value equal to the full dc output current, but it is only there for one 1/6 of a cycle in one direction, and 1/6 of a cycle in the other direction. For 2/3 of each cycle the current in each phase sits right at zero. This alters things considerably in many profound ways. The rms current in each phase will be 82% of the dc output current, not 33% as some people might expect. So if we take an rms (per) phase current of 63 amps, the dc output current from the rectifier is going to be 76.8 amps. What all this means is that we have three wires each carrying 63 amps rms to generate only 76.8 amps dc which is not a very efficient way to do it. A much better way would be to put the rectifier at the mill and feed the power down two cables each 1.5 times larger. That would require exactly the same amount of copper. In one case we have 63 + 63 + 63 = 189 amps circulating around three wires In the other case we have 76.8 amps flowing around only two wires which are 1.5 times the cross sectional area. Let's use some crazy numbers here. Say the dc wires would need to be 0.1 ohm each, and the three phase wires each 0.15 ohms (same length, same amount of copper both examples). I^R losses for each of three phases 63 x 63 x 0.15ohm = 595.35 watts. Total power loss all three wires 1786.05 watts. I^R losses for each of two dc wires 76.8 x 76.8 x 0.1 = 589.8 watts Total power loss for two wires 1179.6 watts. You can see, using the same amount of copper, placing the rectifier at the mill reduces copper losses by ONE THIRD. Three phases are a very efficient way to send power over long distances, but only where the load is resistive. If the load is a rectifier, or has a very low power factor, dc can be a much more efficient way to transmit power. Cheers, Tony. |
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| Downwind Guru Joined: 09/09/2009 Location: AustraliaPosts: 2333 |
Thanks Tony, I have always thought there was a bug in the general view of losses with 3 phase and windmills due to the fact of rectified power. Your example confirms this. If we look at your above example of 3 phase losses and DC losses there is about 1/3 the difference which then comes back to what i asked in the start, about 3 times the loss due to 3 cables or is it only 2 times the loss. I still think regardless of the rectifier location there will only be 2 times the loss over the 3 AC cables when 3 phase is used with a rectifier. Pete. Sometimes it just works |
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| Warpspeed Guru Joined: 09/08/2007 Location: AustraliaPosts: 4406 |
Suppose you have three houses in the same street, each using a different single phase from the three phases available on the same power pole. House A draws ten amps and has a 50 watt loss due to cable resistance. House B also draws ten amps and has a 50 watt cable loss. And House C draws ten amps and has a 50 watt cable loss. I am saying the total power loss must add up to 150 watts. Are you saying the total power loss must only be 100 watts just because there are spread cross three different phases ??? Here is another thought experiment that might clarify. Suppose you connect a heating load to the end of a long cable that has one ohm resistance. There are two electric heating elements available in a water tank, one draws one amp, the other draws ten amps. Which would be the most efficient, connecting the one amp load for ten hours, or connecting the ten amp load for one hour ? In theory the amp hours, and watt hours into the tank would be the same. But what about the cable losses ? At one amp the I^2 R losses in the supply cable would be 1amp x 1amp x 1 ohm = 1 watt for ten hours At 10 amps the I^2 R losses in the supply cable would be 10amps x 10 amps x 1 ohm = 100 watts for one hour. In the first case we lose 10 watt hours in our cable, in the second we lose 100 watt hours. Cable losses will always lower for low continuous power, than very high intermittent pulsing power peaks, it the losses proportional to "current squared" part of the formula that kills efficiency. The problem with a rectifier is that it draws very high amplitude square wave peak currents of rather short duration. Even if time averaged out, the cable losses will always be far higher for a rectifier or any violently pulsing load, than with a steady continuously conducting resistive load. In fact if you order a transformer that is going to be used to directly drive a rectifier the transformer MUST be de-rated to prevent overheating. And the same limitations apply to wiring that supplies current to a rectifier. Cheers, Tony. |
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| Downwind Guru Joined: 09/09/2009 Location: AustraliaPosts: 2333 |
Tony, Im not trying to create an argument here, just asking about a theory with 3 phase and a windmill situation, due to my thoughts on using 3 phase rectified power, and not modeled on a standard 3 phase mains house supply which i feel most calculations are done by, which im questioning if that method is flawed when it comes to rectified 3 phase for the use of battery charging. With your last example of house A (A) house B (B) and house C (C), then yes you are correct, but if we adjust the example to better suit a 3 phase rectified situation, then "A" would switch his power off before "B or C" switched his power on, and again "B" would switch off before "A or C" switched on, so in reality only one source is on at any one time. To muddy the example a little we need a return path for all this to work (we have no neutral or common as you would in the house power example) so which ever phase is not active will be the return path, meaning if "A" was active then "B & C" would be the return. Then it gets a little deeper if this is the case, because if we have 2 cables as a return then the cable resistance on the return is half that of the active supply. Like i said its just a theory, and im asking for others points of view, as doing the calcs on the theory shows big differences in resistance losses to cable size, and experience with wind turbines suggest the standard 3 phase calcs are wrong, and im just trying to work out why to offer better advice. Pete. Sometimes it just works |
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| Warpspeed Guru Joined: 09/08/2007 Location: AustraliaPosts: 4406 |
Standard three phase calcs work fine with sine waves and a resistive load. With a rectifier load, the phase currents are nothing like sine waves and the whole thing becomes far less efficient. What I am saying here, is that given a certain amount of copper, I could get significantly lower power losses between the mill and the battery, if I run two thicker wires instead of three thinner wires, and place the bridge rectifier at the mill instead of at the battery end of the system. This is not some wild theory of my own, but standard electrical power engineering you can look up in any reference text book on rectifiers. It is not an opinion, it is fact. Cheers, Tony. |
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| Downwind Guru Joined: 09/09/2009 Location: AustraliaPosts: 2333 |
Rectifiying at the mill is not always the most suitable option when it comes to mill control and data logging, so to see DC cables as the best solution is not always an option. Personally i would not rectifiy at the mill, and would run 3 phase to my battery shed, where all the controls are located. Experience tells me the windmill 3 phase loss is not as great as the calculations that are used would suggest. This would mean the method of calculation is flawed, and over distance DC may well be higher loss than AC. Pete. Sometimes it just works |
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| Warpspeed Guru Joined: 09/08/2007 Location: AustraliaPosts: 4406 |
How can dc ever have a higher loss than ac ? Can you explain the mechanism. Cheers, Tony. |
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