Homebrewed Electricity > Wind
New build- 20 foot diameter variable pitch windmill
SparWeb:
Hi again windy,
Here's the Oztules story I remembered: https://www.fieldlines.com/index.php/topic,129800.msg842966.html#msg842966
That was an ad-hoc test with 4 ohms in DC (after the 3-phase rectifier) hooked up to a 13-foot turbine. He estimated 5kW based on 145v and 35 Amps.
He also says they were turning 450RPM which puts the TSR at about 7 if his wind speed guess 30 mph is right.
This sounds like an interesting point of comparison for your project. Another important detail is that Oztules' 13-ft stator had about 3.9 ohms per phase in Star (not ".39" which I think it a typo) and he hooked it up to a 4 ohm load - the fact that they match was on purpose.
Let's scale up to your 20-ft blades:
20^2 / 13^2 = 2.37 more swept area, meaning you can expect 2.4x more power under the same conditions... 12 kW.
I hope you're prepared for that much power. In fact I think you should design your load closer to 15kW or more, so that it can withstand some "overshooting" as you tune the furling system.
When getting away from batteries altogether, the electrical system gets simpler. All the AC-DC conversions go away. The calculation you did, however is based on batteries and it won't be of much use. Instead, you want to focus on combinations of wire windings that deliver the power, voltage, and resistance range you need. These will trade-off with each other until you get to the combination that suits you. The load you attach to the 3-phase leads should match, or have less resistance than, the alternator's resistance. I expect the Star-phase resistance you should be aiming for is between 10 to 15 ohms. Depending on the corresponding voltage range you end up with. The line voltage will change A LOT with speed of the turbine, so abandon ALL notions of constant voltage that infects the minds of us battery-bound wind millers.
kitestrings:
--- Quote ---Another important detail is that Oztules' 13-ft stator had about 3.9 ohms per phase in Star (not ".39" which I think it a typo) and he hooked it up to a 4 ohm load - the fact that they match was on purpose.
--- End quote ---
I'm having, have had, trouble getting to these numbers to work - I'm missing something key. The numbers for Oz's as you've corrected look about right. I'll take one I know the properties for:
Ours is a 15' turbine. In a strong wind we'll see typical peaks in the 4-5 kW range. On a recent reading we saw 119VDC peak and 3,865w. So, if my math is correct, This should have been at an RMS Vph of 49 VAC; 84 Vline. If we have a balanced 3-phase load and 1.9 ohms we'd see 3,800w at about 26 amps; this seems about right (84 * 1.732 * 26 =). And, we dump into a 3.4 ohm load bank.
If I calculate the stator resistance, however, I come no where near this value. We have 2 in-hand, 14# CU coils. They each weigh 585 g (1.28 lbs), and there are 4/phase. This puts each coil at .065 ohms, or .26 ohms/ph; .52 ohms/line.
I get to similar numbers with Dan's 15'ter. He had about 21-22 lbs. of copper, 3 in-hand #15, 5 coils/ph. I'm coming up with only about 1/3 ohm per ph. Is the the difference purely found in the (transmission) line resistance, or the effect of heat on the windings as the power increases? What am I missing?
windy:
Kitestrings,
I see you are a bit confused too. I have been trying to figure some winding combinations for my idea and the numbers that SparWeb posted just don't seem to work. There is no way I can get that much wire in a coil to get to 1 ohm per coil and still fit in the stator. I posted a picture of what I believe SparWeb was trying to say. I found this formula on the internet.
My starting point is 10kw at 240 volts. 15 coil 20 pole generator using 240 volt water heater elements for the dump load. Could even go to 18 coils and 24 pole.
Wye connected:
10kw is 3.3kw between each phase to neutral
I through each 3.3kw resistor = 3.3kw/(240/1.732) = 3.3kw/138.57 = 24 amps
R = (138.57^2)/3.3kw = 5 ohms
Not sure what (between each phase to neutral) means. Does that mean to the star point or across the terminals.
Below is my sketch.
5- 1 ohm coils in series for a total of 10 ohms across the phases. What I can't figure out is if I use 12 gauge wire I would need 630 feet of wire to get a 1 ohm coil. 12 gauge wire is listed at 630 feet/ohm.
Any comments?
windy
SparWeb:
A couple of subjects to consider:
The per-phase resistance of a wind turbine alternator is set by many factors,
The sizing of a wind turbine alt for battery-charging is driven by different factors than when loaded by resistance heating,
Adding MPPT to the system means the alternator operates in-between-ish the two types of operation,
Battery-charge systems require rectifiers but direct resistance heating can work off pure AC,
I may have made a mistake myself...
(You know it's going to be a long post, when I start with an index ;) )
I gave the Oztules test as an example, but I fear I didn't provide enough context along with it. Oz is brilliant and in his way he gives lots of information but when he's not in "teaching" mode he leaves a lot of things out (such as his inverter construction stuff). I thought I'd filled the gap but not well enough. Oz was hooking up his resistors after a DC rectifier, not directly to AC. That throws a complicating factor into the mix. I tried to stay away from it before, but maybe I should have taken it on. Since you, windy, aren't intending to rectify to DC that effort's probably wasted.
So let's step back, and compare apples to apples: Oztules was experimenting with a 13-foot diameter turbine, and pumped 5kW into a resistance load, so I scaled up by swept are and left it at that. Kitestrings has chucked in more valuable numbers, also suitable, because he can also get 5kW from his 15-foot turbine. His system is quite different, though, and we can't forget that his MPPT controlled system is set up to approximate the ideal wind power curve for his turbine - not to replicate ideal resistance load matching. The power curve is similar, but it's not an apples-to-apples comparison any more. But you are in the same ballpark.
When sizing up the swept are of blades, the exact electrical system doesn't matter yet, because you can work out what kind of power is coming in regardless of what purpose it's put to. All you need is the wind power equation:
P = 1/2 * rho * V^3 * Area * Cp.
rho at sea level is 1.225 kg/m^3
Area of a 20-ft blades is 29 m^2
Cp varies a lot, but safe to assume 0.35
For that size then, we can work out the power curve for various wind speeds:
10 kph - 0.1 W
20 kph - 1.1 kW
30 kph - 3.6 kW
40 kph - 8.6 kW
50 kph - 16.8 kW
That is mechanical power at the shaft of the rotor blades, available to drive the alternator if it will take it.
The alternator presents a load which is the combination of useful work and losses.
Power is lost to mechanical friction (ignore that for now) and to heating of the alternator coils (I^2*R). The rest is useful power (I*V).
Next thing: The ideal power transfer from a source whose losses are mostly resistive, to a load which is also resistive, occurs when their resistances are equal.
That's why Oztules picked a coil with a 4 ohm resistance: it was a match to his alternator's resistance which was also 4 ohms.
(Again, ignore for now that Oztules had rectified to DC. It does matter, but I think he didn't need to care because it was, after all, just a funny gag he was doing)
This is convenient because it makes it simple to figure out the resistance of the load - it just has to equal the resistance of the alternator. If the alt is in Star, and the load is in star, then yes indeed, your diagram is spot-on, Windy. What you've drawn is realistic, and a good starting point. The terminology you need is that your alt is in "Star" or "Wye", and each "leg" has a resistance of 5 ohms. If you measure across any of the "lines" you would get 10 ohms because you'd always get two legs in series.
Here's a diagram of several 3-phase hook-up schemes. Star is the top left, and the NEUTRAL point is also called the "Star" point, at the middle where the 3 phases come together.
Yes it can be confusing if you don't stick to the terminology religiously. Use a motor for an example: Usually the connection scheme inside a motor is is fixed, and you often don't have access to the insides of motors: it was built in Star, and it's always going to be in Star, so in that case you wouldn't take measurements from the neutral point. You couldn't. You can easily measure line-to-line, which is where the phases come out of the alt. If you were troubleshooting a motor, you would measure voltage across the lines because you have those wires coming right out of it. To measure anywhere else would mean to crack open the case. If someone came along with a Delta-wired alternator, they don't even have a neutral point, so it would be pretty inconvenient for them to need measurements from a neutral point that they don't have. So ignore the neutral point for measurements and calculations.
The phase resistance as you have drawn it is 10 ohms. That's because 2 phases are in series anywhere you measure the lines.
OK
This is getting pretty long and I haven't tackled half of it yet. I will pick it up tomorrow.
I'll leave you with this: I know why you're talking about 240V, but that's not going to seem as important when we get to the end. That's the tail wagging the dog. Voltage will rise and fall with speed. From 0 to whatever maximum you get at furling. If you want the max to be 240 V then you will wind coils accordingly, but when the WT is in service you won't see that peak very often if you've done things right.
kitestrings:
I think you meant "swept area" early, more importantly
--- Quote ---The phase resistance as you have drawn it is 10 ohms.
--- End quote ---
I think you meant "line resistance" here, 5 ohms is the indicated phase resistance; two in series 10 ohms L-L, but otherwise I follow and can see we kept you up late here Spar ;)
I'm sorry if I've complicated it with my system, it just saved me from looking back at other alternators because I know our specs/details. Our is MPPT and it is battery charging. I've just always had trouble understanding why the calculated resistance for a given alternator was seemingly so low compared to what would be the matching balanced 3-phase load that it could support. I suspect that one of the variables, and you may get to this, is whenever we look at AC we have a reactive component to consider, and what we see/measure, or "apparent power" (VA or kVA) is related, but different from the "real power" (watts or kW), and I assume there are losses in both the MPPT conversion (there's certainly heat to dissipate from the FETs) as well as in the rectification with battery charging.
Back to Windy's project - I recall someone posting a few years back on direct, wind to dhw project for a fairly sizable turbine. Maybe you recall. When I have a minute I'll search it. Lastly, I've found this to be helpful among some of the tools and calculators I draw from:
https://www.watlow.com/en/resources-and-support/engineering-tools/3phase-delta-wye-calculator
Best, ~ks
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