Grid-Tied Inverter Opinion

[nekit]

I'm in the process of building a 17' wind generator and wanted to see if anyone else has an inverter that they would recommend.  It will just be grid-tied with no battery backup.  Also if you do, what output voltage would you wind your coils to.

Thanks for the help.

Rob

[imsmooth]

Power-one sells the UL approved Aurora grid-tie starting with 3.6kw units.  Input voltage range is 50v-600v.

[Rob Beckers]

Hi Rob,

Your choices are either the SMA WindyBoy series, or Power-One Aurora series. As far as I know those are the only UL/CSA listed wind grid-tie inverters available in North America (assuming that's where you're located). Both do MPPT through the use of a power curve; the WindyBoy uses either a 2 or 3 point 'curve' (ie. a straight line, or two straight lines) for MPPT, the Aurora has a 16-point table. Both can use voltage vs. output power, and the Aurora adds frequency vs. power to that as well as an option. The latter is preferred because it directly tracks turbine TSR, but either voltage or frequency will work well.

As to voltage, take a look at the inverter you intend to use first. The SMA inverters only cut in at 200 or 250V DC (depending on model), so you have to plan accordingly. The Aurora have a wider range, starting at 50V, though full output power requires a higher input voltage.

Selecting voltage is a trade-off between keeping it as high as you can to minimize wire losses (and cost, since thinner wiring can be used for a given distance to the inverter), while also having to absolutely guarantee that the DC input voltage to the inverter never, ever exceeds the inverter's limits (600V for the Auroras). This last bit is absolute, exceed it and you will turn your inverter into an expensive paperweight, guaranteed. Warranty does not cover the repairs either, the inverters log the event internally before blowing up. Keep in mind that the input voltage has to be under control also when the grid fails, so when there's no load on the turbine. Somewhere around 200V - 240V AC phase-to-phase seems to be the happy medium most turbine manufacturers choose for full power. This translates to around 270 - 325V DC after rectifying (Power-One offers a rectifier box as well, under the name PVI-WIND-BOX, it also generates the frequency signal that the inverter can use for MPPT).

Hope this helps!

-RoB-

[nekit]

Thanks for the help.  The Power-One seems like it might be my best bet.  I am building a 17' turbine to the Otherpower's new 17' specs.  I live in the midwest with a AWS of 13mph @ 100' height that the turbine will be.  Any idea on which size inverter would work well with this system.  Also does anyone have any experience with winding the coils on this size unit for the Power-One inverter, non-islanding.  I'm wondering wire size and # of turns.

Thanks for the help anyone can give.  I'm a professional fabricator, but this alternator stuff is new to me.

Thanks,

Rob

[Rob Beckers]

Hi Rob,

Others that have actually built turbines of this size and grid-tied them will be able to give you a better answer, but I can get you started from the engineering perspective: A 17' turbine, that is 30% efficient (a reasonable number), will produce around 5kW at 11 m/s (24.6 mph), and 6.7 kW at 12 m/s (26.8 mph). This is the wind speed where you definitely want to start furling or use some other means to keep RPM under control.

Of course, your actual output will be different, depending on the overall efficiency. The beauty of MPPT for a wind turbine is that you are not limited by the battery voltage dragging down the turbine; the power out of the wind turbine really is the wind power times the overall efficiency. What the above numbers say is that you'll be looking at a 6kW inverter or thereabout. There is a 6kW Aurora, type PVI-6000-OUTD-US-W, that works together with the rectifier box type PVI-WIND-BOX.

Don't worry too much about capturing that last bit of power at the high end. When you calculate the contribution of the top of the power curve to energy production it shows to be just minor. For most places on land the wind just doesn't blow hard enough and often enough for it to matter much.

-RoB-

[SDSUMetalHead]

I am considering similar options for a project. I believe the thing that will determine the # of turns and wire size will be what the input voltage for the inverter will need to be (or it might be output). So you will need high voltage for that and will need to wire for high voltage and lower amps (according to my unexperienced thoughts) 120V at 41.7 amps for 5 kW. I know there is a way to figure out the rest, but I don't know it off the top of my head.

[SnickersFS]

Well, Faraday's law is your answer.

V=N*B*A/time

These figures will be based on round coils. Using round coils for math purposes makes figures easier. In the Homebrew Wind Power Book they use an odd rectangular shape.

Teslas are based on KJMagnetics list of Surface Gauss. 10,000 gauss= 1 Tesla

BrMax is the strength of the field in the center of the magnet. Surface Gauss will help us get in the ballpark of truer to real world numbers.

All of the designs in Homebrew Wind Power use 500rpm as a max continuous rpm. So that will be the base time for our equation.

140 turns *.5 Teslas(5000 surface gauss) * .0182 sq meters / .12 seconds (about 500rpm)

Equals

10.61 Volts

Size of the wire doesn't matter for volts, but does matter for amps.

Higher resistance the lower the amperage.

Total volts divided by total resistance of that phase gives you the amps.

10.61 * 4 coils = 42.46 Volts / total resistance of the 4 coils = amps

To determine resistance of the coils you need to know how many feet of xwire are in each coil.

Lets use #17 AWG and the Wire Guage Chart from Appendix B of Homebrew Wind Power.

You don't want the coil to be thicker than 1/2 inch or 13 millimeters.

#17 AWG is 1.1495mm thick

1.1495 * 11 = 12.6mm(just shy of 1/2 inch)

140 turns / 11 = 12.76 or 13 thicknesses from I.D. to O.D.

13 * 1.1495mm = 14.94mm

14.94mm / 25.4 = .588 inches or 9/16ths inches+

O.D. is 6 inches or total area of the coil (.0182 square meters)

6 - 1.125 = 4.875 inches I.D. of the coil.

Pi * diameter = circumference or length for 1 turn.

3.1416 * 4.875 = 15.31 inches (close enough for horseshoes)

15.31 * 140 turns = 2143.4 / 12 inches = 178.6 feet of #17 AWG wire.

#17 AWG is 197.49 feet per ohm

1 / 197.49 = .00506 * 178.6 = .9 ohms

42.16 Volts / 3.6 Ohms = 11.71 Amps

After rectifying,

81.92 Volts and 35.13 Amps

To get more volts you would have to add more coils per phase, amps would stay the same. You would need to have 6 coils per phase to get 120 volts at the same amps.

If you use #14 AWG wire you change the total length of the coil and the Ohm's per foot value is different, so a recalculation is necessary. Also means a new Pi*D calculation.

If you change the # of turns a recalculation is necessary. You are altering both volts and amps.

There is a lot of extra math if you are interested in designing your own Generator.

The math isn't difficult, you just need to know what to plug in where.

As you can see it can be a little tedious to attemp a new design for output, but I hope this helps it to be a little more understandable.

Paper numbers will be different from the final product but this will put you in the right direction for design it yourself.

SnickersFS