Aurora Grid Tie

[SparWeb]

Woops  can any one flip that for me  bet know one has ever seen one upside down ha ha 

Nah they all look like that in Australia!   

Menelaos - I thought you were in Germany?!

[Menelaos]

Thats right 

[Rob Beckers]

Menelaos, sorry for the delay. Business is really busy these days.

Using the following assumptions for your genny:

Blade diameter: 6m
Rotor eff: 42%
TSR: 6
Poles: 16
RPM-to-Voltage ratio: 1.174 (unloaded AC)
Phase resistance: 0.95 Ohm (each Wye-leg of alternator)

The RPM-to-Voltage ratio is based on your statement of 1.66 V DC per RPM. My calculations work from the alternator's perspective, and I need unloaded RMS AC Voltage. I'm assuming that unloaded DC is very near to peak value of the AC waveform. Assuming a sine this means RMS AC value is DC / sqrt(2), or 1.174 in case of your alternator.

This results in the following values:

Code: [Select]

Wind (m/s) Watt Rotor RPM Hz Unloaded AC I phase Loaded AC P loss Pout Eff. % DC Voltage Watt Out
0 0 0 0.0 0 0 0 0 0 100% 0 0
1 7 19 2.5 22 0.2 22 0 7 99% 29 0
2 56 38 5.1 45 0.7 44 1 55 97% 58 12
2.5 109 48 6.4 56 1.1 54 4 105 97% 72 60
3 186 57 7.6 67 1.6 65 7 179 96% 85 130
3.5 293 67 8.9 78 2.2 75 13 280 95% 99 226
4 434 76 10.2 90 2.8 85 22 412 95% 112 351
4.5 615 86 11.5 101 3.5 95 35 580 94% 125 511
5 839 95 12.7 112 4.3 105 53 786 94% 138 706
6 1436 115 15.3 135 6.2 124 108 1328 92% 162 1222
7 2264 134 17.8 157 8.3 143 198 2066 91% 187 1923
8 3356 153 20.4 179 10.8 162 333 3024 90% 210 2833
9 4751 172 22.9 202 13.6 179 527 4224 89% 232 3973
10 6483 191 25.5 224 16.7 197 794 5689 88% 254 5364
11 8587 210 28.0 247 20.1 214 1152 7436 87% 275 7024
12 11100 229 30.6 269 23.8 230 1617 9483 85% 295 8969

First column is the wind speed. Second one is the power extracted by the rotor, based on efficiency and energy in the wind. This is the amount of power the load (inverter, plus losses) has to match for the rotor to neither speed up or slow down. Next is RPM based on the target TSR. Next is the frequency of the alternator, based on number of magnet poles and RPM. The unloaded AC column is volts from the rotor, RMS AC voltage, assuming it's a sine waveform. The phase current is next, based on the rotor power and unloaded voltage; P = V * I * sqrt(3). The loaded AC is derived from the unloaded AC and current, taking the voltage loss across the internal resistance for each phase leg into account; if you work out the equation the voltage loss is V = R * I * sqrt(3), and yes, this takes care of the resistance of both phase legs. Next are power losses in the alternator, based (only) on the internal resistance; it's simply P = 3 * I^2 * R (three times the resistive loss in each phase leg). Actual losses will be larger, since things like eddy current and mechanical losses have not been taken into account. This is just to have good idea of heat produced in the stator, to draw a line you don't want to cross (hopefully before things melt). The DC voltage derives from the loaded AC voltage, for 3-phase it's close to 1.33 * Vac. Finally the last column is what's coming out of the inverter, taking inverter losses into account.

There are a few refinements at work in the calculations, as well as some approximations. For power I have a factor that corrects for the rotor not quite following the cube law. Similarly, I assume alternators get a bit less efficient at the higher RPMs, and add that to the mix. A (large) approximation in all this is that I'm not taking phase angle of the current vs. AC voltage into account. A regular 3-phase bridge has a piss-poor power factor, and the actual AC current is actually quite a bit larger than what's indicated here. Also, wareforms for alternators are usually not quite a sine wave, which is assumed here. Still, it's a good place to start for a power curve, and most blade profiles are just not sensitive enough when run off-TSR to make much of a difference as long as the power curve is in the ballpark.

In case that wasn't obvious: The last two columns are what should be programmed into the inverter as the MPPT curve. For a 6kW Aurora wind inverter (PVI-6000-OUTD-US-W) I would cut it off at 264V DC and 6,200 Watt, at which point the wind speed should be around 10.5 m/s and you're running just shy of 1kW in losses in the alternator.

For the Aurora inverter I would set the Vstart voltage (when the inverter switches on) at 60V for this turbine, and put a first datapoint in the table with (60V,0W). The next one would be (72V,60W) etc. For Pramp (how fast the inverter responds) I would put in 10,000 Watt/sec; we don't want the inverter to lag events, and inadvertently set up oscillations, this value is as fast as the inverter will do and seems to work well in practise. A Tprot time (how long the inverter stays online after the voltage drops below Vstart) of 180 or 240 seconds is good.

You may wonder why this would work, and why this would do "MPPT" for a wind turbine: If the numbers are correct (ie. if the assumptions and calculations are correct), the load imposed by the inverter (plus losses along the way) at any given voltage will be exactly equal to the power extracted from the wind by the rotor. That means the rotor will neither speed up nor slow down. If it would speed up the voltage would go up, causing the inverter to increase its load, and something along the same lines happens if the rotor slows down. The net effect is that if we get it right it will keep the rotor RPM exactly in the sweet spot (the target TSR) for optimal power efficiency, regardless of the wind speed.

Hope this helps!

-RoB-

[ruairihev]

Two quick questions for those in the know.
Can the Aurora inverter accept DC directly from the rectifier or does it need to be filtered with capacitors first?
The instruction manual says it can take 30 seconds to several minutes to connect to the grid, does the dumpload need to be connected at this stage or has the inverter some way of applying a load while waiting for the grid connection?
Thanks.

[Menelaos]

Tahnks a lot for your calculations and explainations. I have to take some time to go through all of that. Right now I am moving house...I hope it is ok for you if I come up with some more questions...

One Thing I didn understnad immediately...why do u calculate with a resistance of 0,9 Ohm? The resistance bestween 2 Phases is 1,9 Ohm! Or is that Number linked to one Phase then?

Max

[SparWeb]

... Business is really busy these days...
-RoB-

Of course, when you work for 17 cents/hour you're always busy.

(meaning, your willingness to help people for free is greatly appreciated)

[Rob Beckers]

Ruairihev, the Aurora inverters seem to work just fine directly off of a rectifier. For 3-phase AC there is actually not much ripple after rectifying, though I know multiple installations that have a single-phase alternator, bridge, and Aurora inverter. All working without issue. Now, this is not something you should try, but I even know of one installation where they pumped single-phase AC directly into an Aurora. It would fire up, produce power to the grid, just that they couldn't figure out why it was drawing such a large current from the turbine (30A if I remember right). What was happening is that the protection diode (anti-parallel to the DC input) was shorting out the negative halves of the sine wave, while the positive part would keep the inverter going. They were very clueless...

Menelaos, yeah, my spreadsheet needs single-leg resistance of the alternator. Once you work out the calculations that's actually easier too (loaded AC, power loss etc.), otherwise you're dividing by two all the time.

SparWeb, thanks for the kind words. Actually it's PV that's keeping me busy these days. Ontario has a 80.2 cents/kWh feed-in-tariff. We've been doing a number of PV installs this year, and then being a distributor for Power-One I've been selling PV inverters like there's no tomorrow. Business is getting in the way of the fun stuff...

-RoB-

[boB]

Ruairihev, the Aurora inverters seem to work just fine directly off of a rectifier. For 3-phase AC there is actually not much ripple after rectifying..
...

-RoB-

There is probably plenty of capacitance on the input of the inverter  already doing the filtering for you.

boB

[jarrod9155]

Maybe my installation was a freak problem but because of stray voltage i had to add 300kohm of resistance between the positive and negative to get the inverter to turn off in no wind maybe do to the 220 foot run from the mill to the inverter but was a real headach figuring out .

[nekit]

Menelaos, sorry for the delay. Business is really busy these days.

Using the following assumptions for your genny:

Blade diameter: 6m
Rotor eff: 42%
TSR: 6
Poles: 16
RPM-to-Voltage ratio: 1.174 (unloaded AC)
Phase resistance: 0.95 Ohm (each Wye-leg of alternator)

The RPM-to-Voltage ratio is based on your statement of 1.66 V DC per RPM. My calculations work from the alternator's perspective, and I need unloaded RMS AC Voltage. I'm assuming that unloaded DC is very near to peak value of the AC waveform. Assuming a sine this means RMS AC value is DC / sqrt(2), or 1.174 in case of your alternator.

This results in the following values:

Code: [Select]

Wind (m/s) Watt Rotor RPM Hz Unloaded AC I phase Loaded AC P loss Pout Eff. % DC Voltage Watt Out
0 0 0 0.0 0 0 0 0 0 100% 0 0
1 7 19 2.5 22 0.2 22 0 7 99% 29 0
2 56 38 5.1 45 0.7 44 1 55 97% 58 12
2.5 109 48 6.4 56 1.1 54 4 105 97% 72 60
3 186 57 7.6 67 1.6 65 7 179 96% 85 130
3.5 293 67 8.9 78 2.2 75 13 280 95% 99 226
4 434 76 10.2 90 2.8 85 22 412 95% 112 351
4.5 615 86 11.5 101 3.5 95 35 580 94% 125 511
5 839 95 12.7 112 4.3 105 53 786 94% 138 706
6 1436 115 15.3 135 6.2 124 108 1328 92% 162 1222
7 2264 134 17.8 157 8.3 143 198 2066 91% 187 1923
8 3356 153 20.4 179 10.8 162 333 3024 90% 210 2833
9 4751 172 22.9 202 13.6 179 527 4224 89% 232 3973
10 6483 191 25.5 224 16.7 197 794 5689 88% 254 5364
11 8587 210 28.0 247 20.1 214 1152 7436 87% 275 7024
12 11100 229 30.6 269 23.8 230 1617 9483 85% 295 8969

First column is the wind speed. Second one is the power extracted by the rotor, based on efficiency and energy in the wind. This is the amount of power the load (inverter, plus losses) has to match for the rotor to neither speed up or slow down. Next is RPM based on the target TSR. Next is the frequency of the alternator, based on number of magnet poles and RPM. The unloaded AC column is volts from the rotor, RMS AC voltage, assuming it's a sine waveform. The phase current is next, based on the rotor power and unloaded voltage; P = V * I * sqrt(3). The loaded AC is derived from the unloaded AC and current, taking the voltage loss across the internal resistance for each phase leg into account; if you work out the equation the voltage loss is V = R * I * sqrt(3), and yes, this takes care of the resistance of both phase legs. Next are power losses in the alternator, based (only) on the internal resistance; it's simply P = 3 * I^2 * R (three times the resistive loss in each phase leg). Actual losses will be larger, since things like eddy current and mechanical losses have not been taken into account. This is just to have good idea of heat produced in the stator, to draw a line you don't want to cross (hopefully before things melt). The DC voltage derives from the loaded AC voltage, for 3-phase it's close to 1.33 * Vac. Finally the last column is what's coming out of the inverter, taking inverter losses into account.

There are a few refinements at work in the calculations, as well as some approximations. For power I have a factor that corrects for the rotor not quite following the cube law. Similarly, I assume alternators get a bit less efficient at the higher RPMs, and add that to the mix. A (large) approximation in all this is that I'm not taking phase angle of the current vs. AC voltage into account. A regular 3-phase bridge has a piss-poor power factor, and the actual AC current is actually quite a bit larger than what's indicated here. Also, wareforms for alternators are usually not quite a sine wave, which is assumed here. Still, it's a good place to start for a power curve, and most blade profiles are just not sensitive enough when run off-TSR to make much of a difference as long as the power curve is in the ballpark.

In case that wasn't obvious: The last two columns are what should be programmed into the inverter as the MPPT curve. For a 6kW Aurora wind inverter (PVI-6000-OUTD-US-W) I would cut it off at 264V DC and 6,200 Watt, at which point the wind speed should be around 10.5 m/s and you're running just shy of 1kW in losses in the alternator.

For the Aurora inverter I would set the Vstart voltage (when the inverter switches on) at 60V for this turbine, and put a first datapoint in the table with (60V,0W). The next one would be (72V,60W) etc. For Pramp (how fast the inverter responds) I would put in 10,000 Watt/sec; we don't want the inverter to lag events, and inadvertently set up oscillations, this value is as fast as the inverter will do and seems to work well in practise. A Tprot time (how long the inverter stays online after the voltage drops below Vstart) of 180 or 240 seconds is good.

You may wonder why this would work, and why this would do "MPPT" for a wind turbine: If the numbers are correct (ie. if the assumptions and calculations are correct), the load imposed by the inverter (plus losses along the way) at any given voltage will be exactly equal to the power extracted from the wind by the rotor. That means the rotor will neither speed up nor slow down. If it would speed up the voltage would go up, causing the inverter to increase its load, and something along the same lines happens if the rotor slows down. The net effect is that if we get it right it will keep the rotor RPM exactly in the sweet spot (the target TSR) for optimal power efficiency, regardless of the wind speed.

Hope this helps!

-RoB-

Rob
What formula did you use to get the last column, Watts Out?
Thanks

[Rob Beckers]

Rob
What formula did you use to get the last column, Watts Out?
Thanks

Output power (to the grid, which is what the inverter expects in its MPPT table) is power extracted by the rotor (second column) minus stator losses (8th column), times inverter efficiency (95% is reasonable, the Aurora inverters get up in the mid-90's really fast, their 'sweet spot' is very wide), minus the base-line losses of the inverter, to run the H-bridge etc. (around 40W, which should actually overestimate it).

As mentioned, all this ignores other forms of loss, such as eddy currents, mechanical losses (though the rotor efficiency accounts for some of that). If you have any idea of how much they are this could be accounted for by setting the inverter efficiency lower. If you have an anemometer (doesn't have to be calibrated, as long as it gives repeatable measurements), preferably close to the turbine hub, it's is straight forward to optimize the MPPT curve with real-world data. Simply upload a trial curve, run it for a while and log datapoints (wind vs. output power), derive the turbine power curve from that. Then repeat the same for an MPPT curve that's 15% higher and lower, and see where the turbine power curve moves. That will tell you what's the best; you want the 'highest' power curve for the turbine (most power output vs. wind speed). It may well be that the initial MPPT table is close, since blade profiles are just not that sensitive to being run off-TSR. In other words, you can get it wrong by quite a bit without making much difference in output.

As to Jarrod's note regarding the input resistor: We had seen this issue with others but never quite figured out why. Jarrod was the one to solve the riddle (thank you!), and I now recommend putting a 100kOhm 5Watt resistor over the DC input of the inverter to bleed off stray voltage. This is either leakage seeping through from the output side when the inverter is doing nothing, or voltage coupling into the wiring and somehow being rectified. It is not a problem when the Power-One PVI-Wind-Box is used as a rectifier, because it has bleed resistors build in. When you use your own rectifier, without any resistance over the DC input, the voltage never drops below 50 - 60V DC, keeping the inverter switched 'on' if Vstart is set low, even when there's no wind. The extra resistor solves that.

-RoB-

[jarrod9155]

Menelaos,
                     Im not sure of your  rotor or stator diameter but I do see your blade diameter is 6m or 19.68 feet . I would be worried about heat in the stator around any thing above 3,000 watts . My first stator I pulled off had exstreme heat evidence I had painted it white and some of the coils had turned the paint a burnt yellow color and bubbled the electric tape that holds the coils after winding . Now the Aurora locks in the highest pout it has put out watts  it was 3510 , So what I am getting at is anything above that is pushing the heat in the stator . I also have my magnets pinned and glued not incased to help with coilling . Durring a thunder storm with powermax blades it didnt furl right and that caused my 3510 pout . Now I furl around 25 mph staying at 2,000 watts with 18 foot rolay wind got 222.


New personal record this weekend 10,000 watts in one day  , this month it  is looking like a 100 kilo watts  pruduced .

[DamonHD]

Do you mean 10,000Wh (Watt.hours) and 100kWh (kiloWatt.hours)?

Rgds

Damon

[jarrod9155]

Do you mean 10,000Wh (Watt.hours) and 100kWh (kiloWatt.hours)?

Rgds

Damon

My Bad thats what I mint  to say . Thanks

[Menelaos]

OK, I did have a closer Look at the numbers in your Table, Thanks again for all that effort!

Now lets talk about safety.
As I already said I am Planning to maybe enlarge Rotor diamter to 7 meters. I think there are limits to Fulring.
Ich feel that if the Diamter of the turbine gets bigger and bigger, furling would not anymore work safely as it is slow in reaction on large turbines.
What do you think is the limit in Diamter to make a safe furling that still works fine?
Not regarding that I want to build my Pitch controll...well actually its stalling the Blades by reducing angle of attack.

I made calculations to roughly get all the angles of the mechanism an d the weights and springs needet.
I am planning to have the blades beginning to stall at about 200 RPM.
Yet I do not know how acact this will be. It might be 190 but also 220 RPM.

Lets assume I reach the maximum power output or the voltage /power in the curve that was suggested here. What will the inverter do then?

What I actually want it to do is, do then stall the blades by quickly taking a high load from them and put it on a resistance or whatever so that the blades stall, schlow down and losse Cp.

Ist this possible with the aurora and how is it done?
I read something about the possibility to apply a dump load to the inverter...but I am not sure, there is not much german literature on that item.

So at a set power maximum of lets say 6,5 KW I want the aurora to load the blades with lets say 10 oder 15 KW to slow them down so that the slow furling or in my case the pitch controll would only be backup safety in case of really high winds or gusts that the inverter couldn't deal with anymore.

This would be a very nice feature. In my case, the generator would hardly ever make more that 350 Volts as the Pitch controll wouldn't allow it to do so, even if not exactly adjustet on the spot.

What do you say Rob?

[jarrod9155]

   Today was scary for the wind mill 20 40 mph hour gusting . My wife called me and said get home the wind mill shaking come to find out the wind was coming from the west and had stretched the top guide cable  because the east side had slack on it . I used a ratchet strap and tighten down the west side and stablized the swaying all the connectors and turn buckles look good I think a mount in the ground bent over . Ok Here were the aurora part of this comes into play , Durring the time I was out there fixing the guide problem I noticed the windmill furl correctly and then come out of furl working perfectly then I saw what caused my problem when ever a quick gust of wind would come the mill wouldnt furl  the tsr would just go throw the roof and the blades would get real loud and last 10 feet of scedual 40 pipe would just start flexing right over these gust were well over 40 mph . It seems the aurora wouldnt applie the load quick enough and would alow it to over spin and get away and then my omron voltage relay would kick on  my base board heater dump load and the mill would furl . Now this would only occur durring quick burst of high winds not only was it bad for over voltage but the pole 60 feet up didnt like this . The theory is a mill unloaded wont furl its true and a 18 foot circle of wood in 40 plus winds is scary . Now the aurora has a setting for how quick it ramps up mine is maxed out fast as it can go the aurora wants to track buy appling load to voltage and as the voltage climbs over the the set load per voltage it tracks the mppt and so on but quick gust alow it to take away and the aurora in a since cant applie load quick enough so furling starts to fail . So any one getting started with the aurora and a mill that uses furling as a govenor be sure to have a dumpload set up at what you want the mill to furl at . I lowered my dump load setting and the mill started furling alot better , plus I added some more load to the top end of the mppt curve  to help slow the tsr down .  

[Menelaos]

Thats another reason for why I will build with Pitch. If I have Problems with the Generator which might (caused by whatever) not load the prop, I am in big trouble as the machine would not furl properly. If I imagine a 6 or 7 m prop spinning up like hell with nothing to brake it down...no, I better do not imagine 

[Rob Beckers]

Menelaos, I have good experiences with variable-pitch turbines, that move the blades to stall using an RPM governor. If made well, this is a very reliable, and very fast mechanism to shed power, and keep the turbine RPM under control. We use Aurora inverters with 6kW Scirocco wind turbines, without any overvoltage protection, dumpload, nor furling. Works like a charm! If the turbine does not have a blade-pitch governor, and relies on furling, I would absolutely use an Omron relay plus contactor to disconnect the inverter when the voltage exceeds a set limit. Furling alone is not fast enough to protect inverters. Of course, when in doubt it is cheap insurance to add a voltage sensing relay and contactor to disconnect on overvoltage.

Switching on a dumpload automatically is part of the Power-One rectifier box, it goes under the name PVI-WIND-BOX (or the older designation PVI-7200-WIND-INTERFACE). While it works, it's not great: The voltage at which the dumpload comes on is essentially fixed at 530V DC. Much too close to the 600V limit IMO, and most turbines are by that time already run away so far that stopping them with just a dump load is very difficult. The other shortcoming of the wind box is that it only allows 17A per phase maximum. The 6kW inverter allows 36A in, and with the poor power factor of a bridge rectifier the AC phase current is really not much below that either, so the inverter and wind box are a poor match from that perspective.

For the 6kW Aurora I wouldn't run it above 6.2kW output maximum. The firmware doesn't seem to have a build-in limit (until reaching the maximum current where it throttles the IGBTs), and some run them up to 6.9kW, but in discussions with Power-One the consensus is that 6.2kW is as high as you want to go for reliable long-term operation.

When the inverter reaches maximum power, and the voltage keeps going up, it will simply throttle back the input current so the output power continues to stay at the maximum power value programmed with the MPPT curve. The inverter can change output power quite fast, close to 10 kW/sec, so that should keep up reasonably well with the changes in wind. But if the maximum programmed inverter output is reached it will not increase the load, and for non-governed rotors that could cause it to spin up quickly (before it has a chance to furl).

-RoB-

[Menelaos]

Hi Rob,

Thanx for that explaination.
If the inverter does not extract more that the power set in the curve that would be good. I still have plenty space to those 600 Volts. So thats not a problem at all then. I am no more concerned about blowing the inverter then. So now all I need to do is make the governor and chosse the right springs and stuff to make it react at between 10.5 and 12 m/s. Thats quite a range and will be relatively easy to make.

I then do not need that funny box as well :-)

I have already startet building the governor. I will post pictures when it is finished.

Is it possible to order the inverter from you?
Do you have the 220 V version as well and could you arrange delivery to Germany when it is time?

Max

[Menelaos]

Anyway it is probably not a bad idea to set up the pitch controll that way that it does not react before 12 or 13 m/s.
By doing that the voltage of the Genni would be higer in higher winds and thus I will have less losses in the stator at continious power output of 6.2 KW.
OK, the blades will get a slighly higer TSR then but that shouldn't bother me as they cann not really run away due to pitch control. So above 10.5 m/s the efficiancy of the alternator would rise again. I just have to see if the blade will the become noisy but as long as TSR doesen't get too high this shouldn't be a problem :-)

Max

[Rob Beckers]

Is it possible to order the inverter from you?
Do you have the 220 V version as well and could you arrange delivery to Germany when it is time?

Hi Max,

Alas, I can only sell the North American version. The German one is not only for a different voltage and frequency, but has different listing requirements as well (we have CSA and UL over here, you have CE and VDE). You'll have to get one locally for it to work for you.

I'd love to see pictures of the pitch control mechanism, so please post!

-RoB-

[Menelaos]

I have built a control before...That one was for a 3 m 2 bladed prop which looked like that:



The principle of that stall controll is shown in the picture below that I once found here on the forum as this is very close to the way I build my controlls.
I inserted the formula I made up to calculate the forces acting on the spring since that is a little tricky to handle if done propperly



I am now building a 3-bladed prop with 6 m diameter so the construction needs to be more stable.

I am using trailer Hubs to hold the blades. Each one is a 3000 pound hub :-) I like to be on the safe side :-)



I have a Hub for the Blades lasercut, cnc bend and weldet to fit the trailer hubs.



On the inner side (this side of the hub where the lid is) I have made an M12 thread. There I can adjust the preset angles of the flyball controll mechanism and also allow to chanche TSR of the prop later if needed.
Now imagine to have the trailer hubs fitted to the side which are connected to the pipe that the blades will fit to...and in the inside the construction is close the one shown in the first picture




The hardest thing is to find the right spring.
There are some thing to take into account. The spring has to be pretensioned and must release at a certain pressure. BUT at the same time ist must not be to easy or hard to push further. If it goes on too easy, the blades would turn very quicky and fully which we do not since in strong wirds the prop would alway spin up and slow down, spin up again and slow down...But I want it do adjust and run on fixed speed. So the hard part is to make an exact fit of weights and Spring. It takes quite an amount of calculation but then usually works out on the spot :-)

To make that work out, I have catalouges with thousends of springs to choose the right one from :-)
Whe the whole thing is finished, I can post some more pictures if you like.

Max

[Menelaos]

Ok, got it right with the pictures now :-)

[SparWeb]

Whe the whole thing is finished, I can post some more pictures if you like.

Max

Yes!  Please do.  (It would be better to create a new thread at that time - the topic has drifted away from the Aurora now.)

[Menelaos]

Hi Together, Hib Rob

Rob, I made a small excel sheet with the formulas you stated for your calculation table to make the power curve for the inverter. I try to attach it to this post and hope it works.

To check, I typed in the same values you used for the calculation and I basically get the same numbers. Your xalculation of rotor power seems to be a bit different, so I reduced efficiency in my shhet to get the same power outout to male the other numbers comparable. So far so good...until I discoverd something that throws up some questions.


Pleas check te tool an dthen type in lets say 15 Ohms for the resistance between 2 phases. Now the numbers get strange...
From a certain point on, the values for loaded AC voltage start decreasing again. So the power output also decreases and voltage numbers even start to become negative.

So there must be something really wrong here.

Things start getting odd in column "G" with loaded AC-voltage.
You stated the formula for the voltage loss to be " V = R * I * sqrt(3)".

I apllied that. Your numbers show that you really use this formula. Since I git these strange figures, I tried to change the formula to  V = R * I / sqrt(3), so I devide by root of 3 instead of multiplying....

If I do so, I do not get negative numbers anymore...is maybe this the correct formula? Anyway I do not quite understand what this sqrt(3) is applied for anyway...

So what has happened? Are you wrong with your formula or is it correct  and the mistake is elewhere and if so, what could it be???

Would be nice if you find the time to check this out and answear. :-)

I know 15 Ohms is a lot...but I have made an alternator for a big Lenz turbine that has a resistance of 14 Ohm, so I discovered this strange behavior of the tool...

Max

 Netzkennline Tool.xls (14.5 kB - downloaded 171 times.)

[jarrod9155]

Is there a way you could put a link up to down load this excel program once you get the bugs worked out  .I also have a aurora grid tied and every time I change something it would be nice to just put in the numbers and go and not have to bug Rob every time .

[Menelaos]

You can download it in this threat...after everything is sorted out and the explaination for thise strange numbers found, I will post the latest version here for everyone to download....but first lets see what  Rob will reply...

Max

[SparWeb]

Assumptions...
Your spreadsheet shows current based on the unloaded voltage, however once the current applies a load, the voltage across the output of the alternator will drop.  The Aurora will actively, or with a look-up table, find the optimum load to apply so that the voltage doesn't fall too much, seeking the maximum power point.  To model an active component like the inverter, you need a lot more steps to account for the effects of current load.  The rest of the analysis will not add up until you do that.

[Menelaos]

Of course the spread sheet shows current based on unloaded voltage...I can not make it relate on loaded voltage as loaded volate depends includes the voltage drop which again is is calculated on current.

The Alternator I made was tested with loads. So I have a complete table of performance on different RPM and diferent loads at those RPM.

Today I tried to change the formulas so that the results will match the spreadsheet. It took me ours...and Then I finally got it right :-)
It seemed to work until again on some points the same problem occured. The problem is always that the voltage drop will at some point be enormous and then figures start to get negative...As I have the table of my alternator, I can compare to see that the tool makes bull$#|+ then...

The aurora will not find the correct load automatic. It needs the table to find the right voltage...this is what I have to type in....it will the apply the correct load because I tell it what power it is to extract at which voltage...and that voltage will not be open voltage. So in order to get it right, I must calculate the loaded voltage or things will not be right, especially in high winds.

I do not want to model the inverter, I just want to calculate the voltage drop so that the inverter can be set up right and there is a bug somewhere which has to be found now.

The second version I made today is far closer to the real results...but there still are mistakes in only some calculations. This only happens if the alternator load is verry high on verry small RPM. This usually is a section that does not apply to our machines so it can basically be used and for most situations it will be fine!!!

BUT there must be a reason why it does give strange results on some occaisions and this is what I want to find out about. So I will not yet upload the new version...I will do that when I feel everything is fixed..I hope rob will join in here soon...

Max

[joestue]

you must start with the input power and work forwards, not backwards, as well as incorporate this normalized chart
http://books.google.com/books?id=VKGF-E738yUC&pg=PA83&lpg=PA83&dq=%22theoretical+curves+of+torque+versus+rpm%22&source=bl&ots=rkLeGvAjXX&sig=MCCavdh7sVdWR7nBT5lg53vNeOs&hl=en&ei=VPKwTbqiAc2ctwfOiunrCw&sa=X&oi=book_result&ct=result&resnum=1&ved=0CBgQ6AEwAA#v=onepage&q=%22theoretical%20curves%20of%20torque%20versus%20rpm%22&f=false

You cannot calculate the maximum power point without knowing what the slope of the torque versus rpm graph looks like, how much it differs between various blade profiles i have no idea.
if the alternator is rather efficient you can assume that peak power is close to the optimal tsr, but it may be significantly higher even with an 80% efficient system.
torque is approximately equal to amps assuming there's no jerry rigged rectifiers.

*LOL. that graph is all kinds of fail. anyone got a better one? google is failing me

[SparWeb]

The Aurora probably interpolates between the points in the table you give it (Yes Rob's input is necessary here).

For you to fix your spreadsheet, you could try a "lookup" table of your own.

If you are determined to produce a rational analysis, then you need to know a lot more about the Aurora and what happens as it regulates the power.

And here are some examples of torque and power curves, as Joe mentioned, generated by my own efforts to learn about WT rotor blades:





--edit--
Forgot to mention that the blades in this model have an abrupt stall.  Not all blades do that.  If the stall is relatively gentle, then the torque curve peak is more rounded, and the power curve isn't so steep at low RPM's.  They also have fixed pitch.  The curves for a rotor with adjustable blades, like yours Menelaos, will have very different torque curves from these.  Curves that are adjustable by the balance weight, too.

[Menelaos]

Thanx a lot for your replies and please don't get me wrong here...BUT what you stated does not help me at all.
I do not look at the problem from the blades side of view.

I do know the blades specifications! I have the exact curves of cp versus RPM and so on....
Of course it will not be a perfect match from the beginning but I need something to start with! For doing that I need to start from he altenators side of view Forget about the blades, Imagine there is a some kind of magic that gives me exactly the power that is calculated in the first collumn of my excel sheet...and now lets go on...

By the way...my blades are not really variable pitch. They work as fixed pitch blades. Only starting at a set RPM they will pitch which is ewual to furling point. Before that they do not move and behave like "normal" blades...and we can cnsider that in the power curve once we get the tool working

next...I know that TSR of the blades and thus their efficiency nad power output will change if not the optimum power is extracted from the blades...but again...this has nothing to do with those alternator calculations that I want to get right...

I have counterchecked the spreadshet in the lastest version, as I already said, it works pritty good and matches my practical measurements of the alternator except on extraction of verry high power on small RPM or with high resistance of the alternator. In those cases, the voltage drop will be calculatet to be verry high. The more power In want to extract, the higher it gets until loades voltage gets lower and lower althoug input power gets higher and higher...This is not correct anymore then.

I do know that this is not calculated right because ai have made those real life power tables....so basicaly the tool works on quite efficient alternatots but in he not so efficient ones, it puts out strange data. Please try the tool and see what I mean.

Max

[ruairihev]

Hi,
"I do not look at the problem from the blades side of view." I'm not sure that this is true. Your assumptions for power seem to be using blade diameter and TSR.
Anyway, I think the problem maybe that you are using a constant value of TSR = 6 and a constant power coefficient for your calculations. To get the tool to work you need to increase the TSR value as the power extraction gets higher, this stops the voltage going negative. If you keep the TSR constant the efficiency of the generator keeps dropping as the power extraction increases, not sure how this relates to real life but it explains the spreadsheet problem.
If you have the real life power tables then can't you just use them? Using them will give you exact real figures and not figures based on assumptions that the spreadsheet will give you.
Not sure if this helps you any.
ruairihev