wire size

[outback]

i have a size question here.1 about what size wire to use to run 700 feet from my tower to the house or battery bank at the house so as not to have to much loss. 24v system mill estimated at 500 watt keeping in mind that i would like to add more as i go along.2 also what size wire should i use to run down the tower?and  3 will ac or dc be easier to handle in that long of a run. what are my options? expense is an issue {wife}need say more.if i pull this one off though i think i can get her onboard.thanks for any help. outback

[veewee77]

wow. . . 700 feet. . .

The biggest you can afford. . . Others here can tell you better than I can but bigger is better in low-voltage, long-distance runs.

AC usually travels farther, but it depends on where you intend to put your batteries.

If you put them at the house-end, run AC and rectify it at the batteries. If they are out at the mill site, rectify it there and charge the batteries there.

If you are going to be using an inverter, (and several other factors will be at work here) and no DC loads in the house, put the inverter, batteries and all at the mill site and then you can run smaller wire with the 120VAC from the inverter.

As you can see, there are many factors that affect wire sizes, and without knowing everything about your system, it will be very hard to give you a cut-n-dried wire size.

Guys?. . . have I led him astray. . .?

Doug

[ghurd]

wow again. 700'.

With #4/0, 24v, 2% loss, you can put about 6.5 amps through.

But 500w is about 20 amps.

The 2% is usually for solar. Wind could probably stand a little more loss.

With #4/0 and 20 amps at 24v and 2% loss is shown as 232 feet.

http://www.solarseller.com/dc_wire_loss_chart__.htm

http://www.windsun.com/Hardware/Wire_Table.htm

Big bucks for the wire.

G-

[chux0r]

I've read (on this board, I believe) that you should notice no real difference between DC and AC, but I don't remember the reasoning (nor why we traditionally think of AC as the "long haul" variety).  I know AC is way safer at 13.5kV, since a DC arc at that voltage would melt down the whole tower before you could get it extinguished, so maybe that's one reason it's used.

Anywho, if your generator is 3 phase, then you might be served by moving DC just because you have two wires instead of three!  But of course those wires should be thicker because each will be carrying more current, so that might cancel out.

Big thick wire will sure be cheaper than buying 6 transformers to transmit a higher voltage.  But transformers are sure possible.

You could also build your stator to produce a higher voltage (say 48V for your 24V example), then use one of the DC-to-DC circuits on this board to bring the voltage down once it's in your house.  That would cut your current in half.

Or, if the weather's nice in your area, and you don't mind a little construction, build a "power shack" at your tower.  Put your batteries, charge controller, and inverter there, and run 120VAC back to your house.  This requires that you can keep everything cool (without freezing!) and dry.

I saw a post on this board about sending EMF pulses to transmit power, but I can't find it and it was quite over my head.  It sure sounded cool, though, and I wanted someone to try it.

[ghurd]

Reasoning is higher voltage, lower losses.

And AC is easier to reduce to lower voltages.

Some of the really long hauls are DC, like under the English Channel.

G-

[kitno455]

you did not mention just what speed your genny runs at, and how many poles and phases it has. the reason i bring that up is that your frequency is controlled by all those things, and frequency ends up being the deciding factor in your ability to use transformers. the core material in a transformer has a relatively limited window over which it will work well. if the freq is too low, the transformer will heat more, if it is too fast, hysteresis loss kicks in, and the thing gets 'sluggish'. since a wind genny is variable speed, you are likely to excede the transformers freq rating on both ends so i dont think a transformer will work right off the gen.

your other options involve boosting the voltage from the gen, moving the battery bank to the tower, moving the tower, etc.

allan

[pyrocasto]

the power drop should just cause the windmill to spin a little faster, until it gets the voltage up enough to get through the line loss, and charge the batteries. Though you will get more loss, it may be better than the extra $ for huge wire.

Am I wrong?

The other option is building maybe a higher voltage genny, and using a transformer at the batteries.

[Vernon]

That run is impractical .. too long for 24V. At 30 AMPS the drop would be 5V using big phat expensive 4/0 cable !!! At 30 AMPS you wind up with 19V at the user end.

I would evaluate using #4 down the tower and building a battery/inverter shed at the base of the tower. Step the voltage up to 480V and use 2C#6 wire for the 700 foot trip to the house. If you send low voltage DC you have to take big losses or own a copper mine.

[phil b]

Check http://www.electrician.com/ for voltage drop charts. They are really good for 'what if' stuff.

Also, I've asked several questions along the same lines because I also I have a long line to run. I'm hoping this summer.

http://www.fieldlines.com/story/2005/1/7/155314/4212

http://www.fieldlines.com/story/2005/2/14/5212/58716

http://www.fieldlines.com/story/2005/4/21/145915/034

I hope this helps.

phil b

[Cinosh07]

Try this site they calculator for wires.

http://www.powerstream.com/Wire_Size.htm

Cinosh07

[Victor]

"At 30 AMPS you wind up with 19V at the user end."

 No, at 30 amps you would see ~30 volts at the TURBINE end. Battery voltage + voltage drop.

 Without going into long discusions about energy loss and wind speed distributions as it relates to the total budjet. I will just say that with a 500 watt turbine I would not hesitate to do that run with 1/0 or even #2 if that was the limit of the wifes budget.

 If the system was 48 volts the wire run becomes well within reason at #4 or #6

Make the wind fun!

Victor

[Vernon]

True, I was simply using the voltage he gave me as the output of his turbine. The bottom line is the fact that the monster 4/0 cable dissipates 150 Watts at 30 amps. Wind energy is not plentiful enough to blithely allow a 20% loss in a wire run. I used 30 AMPS because I got 700 feet mixed up with 700 watts. Using your idea of #2 wire and correcting to 20 AMPS gives a 10+ volt drop ... you almost need another turbine to supply the approximate 200 watt copper losses and I wonder what his wife would think of the expense of that item !

At the very least he should use that #2 wire with a step up regulator and higher voltage - perhaps 125V battery. I will note that other posts have expressed concern about the losses in the belt drive and field supply that would be required by an automotive alternator. I would not go through all the trouble of building a custom low loss direct driven PMA only to throw away all the gains in a too small wire run.

It would pay to transmit the 500W at 480VAC 60HZ.

 

[Victor]

Most of the energy delivered by a 500 watt wind turbine will probably be at a mean rate  of about 100 watts  say at  4 amps or around a 5% loss with #2 copper 1500 ft round trip. If the wire cost is more than 5% of the total project cost, using a larger size will make the total project payback worse.

 The above example is a guesstamate but serves to make my point. If you have an accurate power curve for the machine in question and wind speed distribution data for the site, you can nail down the most cost effective wire size.

 Doing this detailed of a study to determine wire size is usually an exersize of splitting hairs but when the wire run is a significant portion of the total cost of the project it may be warrented.

 If you look at  reccomended wire sizes for commercial wind turbines, it is not uncommon to see losses of 15% or more at full load. The key point is that the turbine is at full load for a very small percentage of it's run time. The manufacturers want there product to preform well and would have you spend more money on wire if it would have any effect but after a point you may as well just throw money in the trench.

 As a cautionary note. If the turbine in question requires an electrical load to keep from overspeeding (not all designs do), excesive wire resistance will lessen the load.

Victor Creazzi

Aerofire Windpower

www.aerofirewind.com

[wooferhound]

Here is a bunch of "Voltage Drop Calculators . . .

http://www.electrician.com/vd_calculator.html

http://www.stanselectric.com/vdrop.html

http://www.elec-saver.com/freetool06.htm

[outback]

thanks everybody,

            guess i have a lot of thinking to do.sounds like i better start putting some bread away for wire.

[Chiron]

Iron powder and Ferrite transformer cores can span several decades of usable frequency in power transformer applications. I've used small ferrite toroids for audio transformers that performed very well over the 20Hz to 20kHz audio range at as much as 100 Watts.

If your genny is 3 phase you could wind a 10:1 step up transformer for each phase and use 3 wires of smaller diameter to send the 240VAC to a matched set of 3 step down transformers on the battery end into rectifiers. You may have to expriment a bit to find the # of turns that works best for your unit.

[Tom in NH]

Chiron has a good idea. Would you consider putting your batteries and inverter at the base of your tower, then step up the voltage to 220vac or higher for the run to the house? This would allow you to cut line losses and use smaller wire because of lower current. --Tom

[richhagen]

Wow, 700 feet, that's over an eighth of a mile!  Because of the cost to double the diameter of the wire becomes significant at this length, consider a higher voltage mill if possible.  The power consumed in getting the power to your house is equal to the square of the current multiplied by the resistance in the wire(I^2*R).  Since the resistance is generally a function of the square of the diameter, doubling the diameter of the conductor increases the area by 4 times and cuts the resistance to 1/4th of its original value.  Doubling the voltage for the same amount of power cuts the current in half, this results in the power loss (I^2*R)being cut to a fourth of its original value as well.  If you havn't already gathered the components for your system, starting with a 48V system is probably the cheaper route.   The longer the wire runs, the more advantage to a higher voltage system.

An example comparing doubling the system voltage or the conductor size if you had 24 watts of power from a fixed power source. (a little different than a wind turbine, but suitable for an example)

Current = I

Source Voltage = V

Conductor Resistance = R

Power Consumed in conductor = P (this is what we want to minimize)

base example:  I = 1 Amp, V = 24 Volts,  R = 1 Ohm, then P=I^2*R = 1 watt

effect of doubling conductor diameter with everything else the same as in base example: I = 1 Amp, V = 24 Volts, R = 1/4 Ohm,  P = I^2*R = 1/4 watt

effect of doubling source voltage with same power from source and other factors same as in base example: I = .5 Amps, V = 48 Volts, R = 1 Ohm, P=I^2*R = 1/4 watt

You could locate the inverter and batteries near the tower and transmit 120 or 240 volt AC to the house.  This may make system maintenance/monitoring more difficult, and a minor consideration is that the byproduct of rectifying/transforming the electrical energy is heat, although probably only a few bucks worth of equivelant electric heat over a heating season and not much of an issue at all if you heat with wood.  Rich Hagen

[Flux]

If you are using a commercial turbine you will have to use cables within the makers specification but if you are building a dual rotor axial machine as most people here  seem to do, the situation is rather different and Victor has hinted at this.

Unless you have some means of tracking the generator characteristic by using some form of electronics you have to under power the alternator in high winds so that the blades work at a reasonable aerodynamic efficiency. You have to trade electrical efficiency for propeller efficiency and reach a balance.

If you use bigger magnets and thicker wire you improve the electrical efficiency of the alternator but you stall the blades and the high wind performance is hopeless.

In this case you could use that bigger alternator and throw the necessary losses into the cables rather than have the heat in the stator of the alternator.

Normally if you choose a low cut in speed for light winds you will only have an overall electrical efficiency of not better than 50% at full power. If you keep the alternator up at about 70% this will not increase the cost much and you can loose 20% in the cable.

As Victor pointed out the losses at low power where most of the wind is  will be relatively small.

This may seem a strange idea but just ask anyone who has built a machine that stalled badly how much better it works with a bit of extra resistance. You would aim to make it stall really badly and use a bit more resistance. I bet the bigger magnets will be cheaper than the thicker cable.

Regarding transformers they are not as big a problem as hinted at but at the low frequencies they need to have big cores to handle the flux. You will not have any problems at the higher frequencies as the flux density falls with frequency.

Remember that the large transformers will be very costly or not very efficient if you limit their size. They also may delay start up in low winds. I think that solution would cost more than the bigger magnets.

Flux

[ghurd]

Taking the low wind part a step more, or from another perspective.

As the losses increase, the amps will go down.

As the amps go down, losses decrease.

If the mill can or is making 500w, and a large percentage is lost in the wire,

the V will rise and the amps will drop.

When the mill is only making 25w, 1A, #8 wire will carry it with only a 5% loss.

Like Victor said, Vmill+Vloss=Vbattery.

And watts made minus losses = watts to the battery.

So The mill will be higher V and lower A.

Meaning it will balance out.  Somewhere.

Charging will still happen.

(Losses are still losses)

G-

[Ungrounded Lightning Rod]

big phat expensive 4/0 cable

No fooling on that expen$ive.  I'm wiring a new utility drop (upgrade to 200A) in one of my houses this week and just bought about thirty one feet of 2/0 - at $1.74 per FOOT per CONDUCTOR.  Copper prices have climbed in recent months, too.

Think of each wire as a stack of pennies the length of the run... for which you paid extra to draw into wire, wrap in insulation, and get approved by the Underwriters Laboratories.

[outback]

as i read these comments, i keep thinking about all the talk of losses.in my eyes it isn't really a loss,if that mill wasn't there, there wouldnt be anything coming down that wire. i havent made any comittments yet but am trying to get started off right as far as my wiring is concered be for i build another stator for my dual rotor machine.i'm using the 12mag 2x1/2" 9 coil 3phase design the stator i have wound up now 100 turns 17ga puts out apprx.250rpm  33v 8a into a 24v battery.and that is with just one rotor on.it doesnt double it with both rotors on so i was thinking maybe build two identical machines,single rotor.i have alot of options here and am looking for the best and simple way of doing it.

[Nando]

OUTBACK:

Stop and think. 700 feet is 1400 feet wire length.

First make the generator to produce a peak voltage (at maximum RPM) of 240 AC ( 3 phases Delta/Star).

Sent the AC to the load and at the load convert down with 3 transformers to the battery voltage charge controller operating voltage -- this would be the least loss.

And the Cheaper one.

Also, with room to grow with power -IT IS A GREAT MISTAKE to send such low voltage for such long distance, it will require a wire diameter that is quite too expensive.

There are other ideas that I am working with Amanda for these such cases ( high voltage wind mills) for wide operating voltage,power and distance.

If your coils were calculated for 24 volts, then place 10 times the number of turns (of course the wire size to have 10 times less current)

Regards

Nando