Here is a piece I did about AC and DC in 1999. It explains why DC wiring is slightly more efficient than 3-phase AC wiring where the AC is feeding a rectifier (rather than grid AC which is at higher voltage). One way to understand why AC is less efficient is to notice that Ac current is intermittent, only using the wire part of the time, whereas DC uses it all the time. So **for the same peak voltage**, DC is going to work better.

There is some confusion about how to calculate losses in the cables of a 3 phase wind system. If the 3 wires are feeding a rectifier, charging a battery, then the current in the cables is dominated by the need to supply DC at battery voltage to the load.

The cables may change, but the current tends to remain constant.

*Paul Gipe’s question *

*What is the power lost in conductors from a 850 Watt permanent-magnet, three-phase alternator feeding a diode bridge rectifier delivering a nominal 24 VDC to a battery bank. The rpm, voltage, and current of the three-phase alternator varies with wind speed. There are three cables (conductors) between the wind turbine’s alternator and the diode bridge. There are two conductors from the diode bridge to the batteries. The conductors are #8 AWG with an AC resistance of 0.78 Ohms/1000 feet. There is 150 feet from the wind turbine to the diode bridge, and an insignificant distance from the diode bridge to the batteries.*

The DC current is found by dividing watts by volts. this gives 850/24 = 35 amps.

**If** you want to analyse situation **mathematically**, current in each conductor is 35 amps for 2/3 of the time. The rms current in each conductor is therefore (2/3)^.5=0.82 times 35 amps = 29 amps (rms). Resistance of each conductor is 0.78 times the cable run of 150/1000 feet, giving 0.117 ohms.

Power loss is I^2*0.117

which is 2/3*35^2*0.117

which is 1/3*35^2*.234 for each conductor

which is 35^2*.234=287W in total.

This is a 34% loss!

**An easier way** to analyse the situation (which also give the same answer) is to say that at any given instant the DC current 35 amps is flowing around a circuit path with resistance equal to 0.78*(cable run in feet/1000). Cable run is 300 feet for the full circuit.

Finally I should point out that the above is strictly only true if the internal loss is small. As loss increases, the situation becomes much more complex, since more than 2 wires will start to conduct at once during the changeover. But the above answer will be accurate enough for practical purposes, given that we are arguing about such large differences in our answers.

Hugh

October 1999

Well, then for this novice, it begs the question:

Is it more efficient to rectify the ac at the alternator and send dc down the pole to the charge regulator? Of course this would result in more expense in thicker cable, but it looks like you would need to do that with ac cable anyways. 34% energy loss one heck of a loss to pay for, no matter jow you look at it; heavier gauge ac or dc cables or inceasing alternator size thus possibly needing to increase overal turbine size.

Hughe maybe you can give us some positive advice?

Thanks,

Scott

hi

To be honest I mostly use 3-phase transmission so that I can have the brake switch at the house and brake on the AC side but for long cables I often put the rectifier at the turbine or at the base of the tower and use DC for most or all of the wire run. This is because it uses less copper and/or wastes less power that way.

Another piece of advice on this is that if your alternator is super-efficient axial and your wires are very thick you may stall the blades so it’s better to have some loss in the transmission cable anyway or you may stall the blades, and lose out that way. But it’s cheaper to use DC than 3-phase AC as you only need 2 wires and it has roughly the same loss for the same size per wire.

Have fun

Hugh