calculating power drop from turbine location

[rijopiatt]

At what distance can I expect a significant power drop. I'm deciding where to locate my tower. I assume that next to the entry point to the building would be ideal barring turbulence issues. However, no structure wind interference would be better, but I understand that voltage/amperage drop will be experienced with increased distance away. Let's say I want to use a 24volt generator/alternator rated at 100 amps. and I would hope to get 1500 watts output how much power would I lose and at what distance. I also have heard that the cable can be quite expensive per foot.

[Flux]

At low voltage there is always the problem with volt drop. !00A is rather a large current for a 24v system and the cable will be costly if you want to go a significant distance.

With wind much depends on the type and design of turbine, some volt drop can in some cases be an advantage in that it could prevent stall with a powerful alternator. Commercial machines are usually designed such that any volt drop reduces your output but with some home built machines a 10 to 20v line drop at full load may not be a source of loss.

With such a high power set up at 24v the cable cost is likely to be a big issue if you need to get more than about 50ft from the battery. Higher power installations are better suited to 48v or higher.

There are calculators that you can use to find the volt drop but they may not give you the true answer unless you know how the turbine will be affected by loading. Any cable resistance will cause loss but if the gain in prop efficiency from better matching means more alternator output the overall loss may not be an issue. At 24v then any cable run that introduces more than about 2 ohms may mean extra losses and you may not be able to use short circuit braking so there are other issues to consider.

Flux

[BrianSmith]

100A thru 2 ohms is 20,000W of power (loss)....think much much much less resistance for delivering 100A of output.

[XXLRay]

Wire Resistance[Ω] = (Length[m] * ρ[Ω * mm²/ m]) / Area[mm²]
or
Wire Resistance[Ω] = Length[m] / (Area[mm²] * κ[m / Ω * mm²])

Copper:
ρ(rho) = 0,01786(Ω * mm²/m)
κ(kappa) = 56(m/Ω * mm²)
 
Aluminium:
ρ = 0,02778(Ω * mm²/m)
κ = 36(m/Ω * mm²)

Lossrate = (Wire Resistance[Ω] * Rated Power[W]) / (System Voltage[V])²

Don't forget to double the wire length as you have to consider the way to the turbine and back again.

[tanner0441]

Hi

The formula I used to use for fitting heavy current devices in boats was. 

Volts drop = Cable resistance for 1 meter  X  Max current  X  Cable length in meters.   All over 1000

It works for high voltage runs as well..


Brian

[imsmooth]

I would use a high voltage, low current system for power delivery.  This is what power utility companies do.

[wpowokal]

Using a high voltage transmission and I assume turbine has inherent problems for home built, insulation & stepping the wild AC down to mention only two, Flux has actually given the answer. It can be done but usually beyond the financial input of a "hobby".

Unless one is in an exceptional wind area any wind turbine rarely achieves full rated output and if it does all what Flux said applies and more.  I for one don't worry about sending power 30 M (100') it realy is not that far (read what Flux said). I use 16mm sq 2 core to transmit DC from my 10' duel rotor, I don't know what the losses are, what I do know is that mostly when the unit is at full power the system volts are high anyway.

I understand it is not easy for "newbies" and that is why we ask questions but I would suggest placing the turbine in good wind at a reasonable distance would yield a greater output on average, rather than getting bogged down on peak amps that will rarely be achieved or usable, clearly you don't want to melt the cable but!!!!!!!!

Allan