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View Full Version : So, how much power will my wind turbine make? An informational thread.


Ben Colla
16th October 2010, 22:51
I can't answer that for you. For three very good reasons.
1. I don't know where you are
2. I don't know what diameter turbine you are asking about.
3. I don't know how tall your tower is

But, with some thinking we can come up with a reasonably good answer.

I’ll refer to the following web sites and or pages
1. http://www.talentfactory.dk/en/tour/wres/index.htm
2. http://www.talentfactory.dk/en/tour/wres/calculat.htm
3. http://www.reuk.co.uk/Calculate-kWh-Generated-by-Wind-Turbine.htm
4. http://www.solacity.com/SiteSelection.htm
5. http://www.solacity.com/Economics.htm
6. http://www.sustainability.vic.gov.au/www/html/2123-wind-map.asp
7. http://www.otherpower.com/bottom_line.shtml
8. http://www.talentfactory.dk/en/tour/wres/enrspeed.htm


Everything here is aimed at personally owned turbines. So probably no bigger than 10 kW (or about a 7 meter diameter). Though it probably holds true for the commercial sized turbines as well.

Those three things I listed are the most important list of things, for wind turbines. They are listed in order of importance. The most important, sadly, is the one you can do the least about. *Sigh*.

1. Location Location Location. If you are not in a windy location, nothing else matters. You may as well stop reading now, and flick over to the solar panel section. There are numerous places to find wind maps online. The one relevant to me is link 6. But of course, that only helps if you are in Victoria, Australia. So, go find a wind map for your area. If the average wind is less than 4 m/s (8mph or 14 kph) you are not going to get much out of a turbine. The power in wind goes up with the cube of the speed of the wind. So you need wind. You need wide, flat open windy terrain, in general.

2. Diameter. Bigger is better, but big is more expensive. How big can you afford? How big can you legally go? How big can you handle (big = heavy). The power you can generate goes up with the square of the diameter. So, wind speed is better than turbine size. (besides, wind is free, big turbines are not)

3. The higher you go, the higher the wind speed. And as wind power increases with the cube of the speed, you need to go up to get power. But bigger towers cost a lot more and need a bigger area around them. Also better foundations and probably decent guy wires. Tall towers get expensive.

Let us pick my (proposed) site for a turbine for my example.
From the Victorian wind map (link 6), I know that my average wind speed is 7.3 – 7.4m/s. This is truly awesome. I am looking at a turbine rated at 5kw with a diameter of 5 meters. So if you use link 3 you get somewhere around 24,000 kWh/year (I’ll explain this page in a bit). Holy smokes batman!!! But, there is a problem. That sounds too good to be true. So let’s check the fine print on that wind map. Hmmm ... O here it is. Wind speed measured at 65meters above ground level.

65 meter tower. Ummm, I don’t think the local council would like that, the neighbours certainly won’t, and the bank manager is just going to laugh at me. You cannot directly translate wind speed at 65 meters altitude to the wind speed at ground level. It slows down the closer you get to the ground, until it’s zero at ground level.

You’ll have the same problem. Wind speed measured at a different height to your tower height.
But link 2 lets you estimate the ‘wind shear’ factor. That is, how much the wind speed varies depending on altitude.

So open it up in a different tab. I’ll use the ‘1.5’column (the middle one) as it’s a reasonable one to use. Read up on ‘roughness’ on the same site to find out why. I put 7.2 in the 1.5 column at 60 meters altitude and hit ‘calculate’. There it is, nice and easy, the height at 10, 20, 30 .... and so on meters. This will also spit out a pretty graph. So plot one.

Look at that graph. It shows you the wind speed depending on the altitude.
Notice how above 20 meters, it’s a lot more vertical, and below 10 meters it’s a lot more horizontal. You want to go as high as possible, but there are diminishing returns. Going from 100 to 150 meters gets you 1 m/s speed, and so it’s probably not worth the money to add the extra 50 meters height to a 100 meter tower. Going from a 10 meter tower, to a 20 meter tower gets you only ½ m/s. But the extra tower cost is marginal and may well be worth it. The cost/benefit ratio changes somewhere in this range, for small turbines. Higher is always better, but the cost adds up. (Unsubstantiated thought ... you want your tower height to be at the 45 degree part of this graph. Lower, means you are leaving easy picking’s wind speed on the table, and higher makes the tower to costly, while adding little to wind speed. Discussion?)

Anyway, we have converted wind speed at X height, to wind speed at your tower height. Now we can calculate how much power you’ll actually get.

Open up link 3 again.
OK excellent.
It asks for 6 numbers. Some of these are simple.

Rotor diameter, is the blade diameter. Put in your blade diameter.
Mean wind speed. Or the average. We figured that out just a little bit ago. So put it in. Better to round down than up, so as to not inflate your generated power.
Cut in speed. This is the minimum wind speed that your turbine will make any power at. If you’re unsure use 3.5 (there is bugger all power in wind below 3.5m/s anyway).
Cut out speed. Is the highest wind speed that your turbine will make more power at. Surprisingly, it makes very little difference if you change this from 15 to 25.
Turbine effiency. This is tough to get an accurate answer to. It relates to how much power there is in the wind, to how much electricity your turbine can actually make. This CAN NEVER exceed 60%, no matter how magically awesome your turbine is, and probably won’t exceed 30%. Use 30% unless you know better, and if someone is telling you that their turbine they are trying to sell to you is better than 40%, they are probably lying to you.
Weibull Shape Parameter. Tough to explain. Go hit Wikipedia. OK. Good, your back. Leave it at 2. That was easy.
Hit calculate.
And you have your answer.
For me, it’s somewhere around 10,000 kWh’s / year, with a 5 meter diameter set of blades.

Fiddle with some of the numbers. I find it especially interesting how the ‘cut out’ speed makes very little difference. Even though the power in the wind increases with the cube, you’d think that a high cut out factor would be critical. But, very high wind speeds are very uncommon, so they add very little to total power generated. With these high wind speeds, you don’t want the turbine to blow up trying to capture the power, you just want it to survive.

Cut in is also interesting. There is so little power in very low winds, it’s not worth even trying to capture it. You can ignore wind below about 3m/s because there is no power there anyway. Hit up link 7, and read down until you see the chart. At 6 mph even 6meter blades will only be generating just over 100w. Link 8 also shows this. Low wind is low energy, and there is nothing you can do to fix this.

Changing the mean wind speed by even 0.2m/s makes a significant difference. There is that cube power increase working again. Varying the wind speed from 5.5 m/s to 5.7 m/s (5 meter diameter) adds 1,100 kWh/year

Now, you’re probably asking, how do I know those figures are correct?

Well, .... I don’t have a good answer. I have several tolerable answers though.

1. The Australian Government has a Renewable Energy Certificate scheme. Each REC is worth 1 mWh of generated electricity. If you visit https://www.rec-registry.gov.au/sguCalculatorInit.shtml and put in a 5 meter wind turbine at the default 2000 hours, is spits out 47 RECs, or 47 mWhs, over a 5 year period. Pretty close enough.
2. Rob Beckers has a spread sheet available, that does the same sort of calculations, and it spits out similar results. With luck he'll link it for us. [edit] He doesn't have to. It's right at http://www.greenpowertalk.org/showthread.php?t=7 in the first post.
3. http://www.solacity.com/Economics.htm has some charts, and if you follow them through, you also get similar results. However, both 2 & 3 here refer back to the same person eventually, so it’s slightly suspect. Not that I’m claiming that Rob is lying, just that using the same person twice for the same task inherently has a ‘they ain’t completely independent’ problem.

If I've made a mistake, let me know, and I'll correct it, but I think it's generally correct.

Rob Beckers
17th October 2010, 08:57
Nice writeup Ben, thanks!
The wind turbine production spreadsheet (http://www.greenpowertalk.org/attachment.php?attachmentid=247&d=1189380768) I wrote is simply based on physics and math. There are many versions floating around the Internet of similar spreadsheets/programs. There is no bias inherently in there, at least none coming from the writer, it just calculates power in the wind, and from there (using a reasonable efficiency) how much energy it produces for a given rotor size. The wind assumes a Weibull distribution. All these are well-established standard ways to estimate energy production for wind turbines large or small. The bias that is present is that real life wind does not have to follow a Weibull distribution, most sites should be close enough, some are very different, and the conversion of wind speed vs. height does not account for effects created by the ground or obstacles. Turbulence will be the biggest unknown factor, throwing a wrench in the energy production works. People insist on putting turbines on short towers, which means that even if there are no obstacles you'll get ground-induced turbulence. Throw some buildings/trees and other things into the mix and it gets worse. That really is the main reason why almost all small wind turbines produce so far below the estimates given to people by installers.

My page on wind turbine site selection (http://www.solacity.com/SiteSelection.htm) tries to give people guidelines on how to avoid the worst of the turbulent air. Problem is that people don't believe what is in there (I didn't come up with those rules-of-thumb, they are accepted industry practice used by the more honest installers out there). They continue to believe that their site is 'special' and they have a howling wind just a few meters above ground (with no turbulence of course). They can't believe that a 60' tower really is the absolute minimum to get you out of ground-induced turbulence, and then only at a site that is very smooth and very flat (or a hilltop). For most sites 60' will still find the turbine in very turbulent air. Then they are disappointed after spending tens of thousands to put up a turbine that it produces next to no energy...

-RoB-

Steven Fahey
19th October 2010, 09:37
Good write-up, Ben. Now to make sure that everybody selling/promoting/buying wind turbines gets a copy!

Rob,
I took your spreadsheet and plugged the numbers from my turbine in a while ago, and found a realistic estimate of the energy I've gained from it. Due to my rather poor site (trees) and short tower (45feet), my 8-foot turbine probably doesn't collect more than 400 kWhr per year. This is more than enough for the purpose I've set it to (lights in the barn and the odd power tool), but someone expecting a return on the investment would be sick.

Now when anybody asks me how much money the turbine saves me, I say "50 bucks a year". They are usually confused or think I'm joking (I am joking, kind-of) - popular perception is that the purpose of wind power is to make money. If they are still listening to me at this point, it is the moment to inform people that wind power that does make money does not look like my turbine - it looks like giant wind farms instead.

Ben Colla
23rd March 2011, 04:25
I can't edit my original post, so I'll add this.

A couple of the links no longer work. So here are the things, newly edited, working. It's links 1, 2 & 8 that have changed.
Link 1 is very much worth the effort of looking at a lot.

1. http://guidedtour.windpower.org/en/tour/wres/index.htm
2. http://guidedtour.windpower.org/en/tour/wres/calculat.htm
3. http://www.reuk.co.uk/Calculate-kWh-...nd-Turbine.htm (http://www.reuk.co.uk/Calculate-kWh-Generated-by-Wind-Turbine.htm)
4. http://www.solacity.com/SiteSelection.htm
5. http://www.solacity.com/Economics.htm
6. http://www.sustainability.vic.gov.au...3-wind-map.asp (http://www.sustainability.vic.gov.au/www/html/2123-wind-map.asp)
7. http://www.otherpower.com/bottom_line.shtml
8. http://guidedtour.windpower.org//en/tour/wres/enrspeed.htm




I also want to emphasise that you need a windy spot, and that these aren't common.

Without wind you got nothing.

Have a look at the map I link, for Victoria, Australia, which is link 6. To get good wind you need one of the brown bits, and note how little there is available. Just for scale, Victoria is the smallest state in mainland Australia and is approximately half the size of france, or the same size as Great Britain. Windy spots are not common.

Rob Beckers
23rd March 2011, 07:24
Hi Ben,

There's one I think all those considering a small wind turbine should read before making any decisions (I send all my prospects over to read it, so they don't have unreasonable expectations): http://www.solacity.com/SmallWindTruth.htm

-RoB-

P.S. Posts can be edited for a month after first posting. There are actually good reasons for blocking unlimited editing, even if that is a hassle at times.

Chris Olson
23rd March 2011, 12:48
Rob,

I had a little bit of trouble understanding your annual energy production numbers in that chart in that article.

I build 12 foot off-grid turbines so I'm most familiar with that size in being able to pull numbers off the top of my head.

So a 12 foot rotor running at Cp .35 in a 12 mph wind at 1,000 foot elevation makes 352 watts at the shaft. That's a little over 3,000 kWh in a year. If you run that rotor at Betz you'll have about 593 watts at the shaft, or about 5,200 kWh/year.

Now this is shaft power, not generator output.

Your chart shows a "good" 12 foot wind turbine is going to make 4316 kWh/year at 12 mph. With the efficiencies in most turbine generators running at moderate output that means the rotor is running at close to Betz.

What sort of rotor blades you got on these things to come up with those sort of numbers? The actual output of a 12 foot turbine with the generator running at 90% power efficiency at 12 mph is closer to 2,700 kWh/year.

I'm hoping you'll tell me that you took that chart off some manufacturer's website someplace. And that's one of the big problems in the wind power industry - I've seen power curves published from manufacturers like Southwest Windpower for the Air 403 that plot the turbine's output at Betz.
--
Chris

Ben Colla
23rd March 2011, 23:37
Hi Ben,

There's one I think all those considering a small wind turbine should read before making any decisions (I send all my prospects over to read it, so they don't have unreasonable expectations): http://www.solacity.com/SmallWindTruth.htm

-RoB-

P.S. Posts can be edited for a month after first posting. There are actually good reasons for blocking unlimited editing, even if that is a hassle at times.
I read that when you announced it. I thought it was a good read, and certainly not worth the controversy it generated over on otherpower, for 5 pages.

I agree, in general Wind Turbines are not suitable for most people, the power just isn't there, for most people. I'm not entirely convinced that one is suitable for myself, but all my research says I should get 10,000kWh/year at which point, with out current and expected power bills, it's financially viable. If I get more it's a bonus, if less, it's quite close. And I'm in one of the windiest spots in the state.

Rob Beckers
24th March 2011, 08:37
I read that when you announced it. I thought it was a good read, and certainly not worth the controversy it generated over on otherpower, for 5 pages.

Fully agree!
As it turned out there was one guy who was pushing his own Chinese import turbines relentlessly and wouldn't hear of any criticism that might detract customers. Then there were a few that kept insisting on comparing apples with oranges (as in DIY vs. having a turbine installed etc.).

Chris, where did you get that kWh number from?
It looks like you're multiplying 352 Watt by the number of hours in a year, and then converting it to kWh's. If so, that won't work to get the energy production of a turbine. The wind speed in my table (ie. 12 mph) is the annual average wind speed. Sometimes the wind blows harder, sometimes there is not quite as much wind. It assumes a Weibull distribution with a K of 2 (which works reasonably well for inland locations). Since power in the wind follows the cube (ie. ^3) of the wind speed, it is those stronger winds over 12 mph that disproportionately contribute to the overall annual energy production.

At an instantaneous wind speed of 12 mph, 30% efficiency, sea-level, I calculate an output of 300 Watt exactly out of a 12' turbine. That is exactly in line with your number.

The actual calculation of energy is a combination of power at a certain wind speed, multiplied by the time that the turbine spends at that wind speed during a year. So, for 12 feet (= 3.66 meter) and 12 mph (= 5.37 m/s) annual average wind speed


wind (m/s) Power (Watt) Weibull (%) kWh/year
0 0 0
1 2 5.34% 1
2 15 9.84% 13
3 52 12.87% 59
4 124 14.16% 154
5 242 13.82% 293
6 418 12.26% 449
7 663 10.01% 582
8 990 7.58% 657
9 1409 5.34% 660
10 1933 3.52% 597
11 2573 2.18% 491
12 3341 1.26% 370
13 4247 0.69% 256
14 5305 0.35% 164
15 6525 0.17% 98
16 7918 0.08% 54
17 9498 0.03% 28
18 11275 0.01% 13
19 13260 0.01% 6
20 15466 0% 3

Total = 4947 kWh/year


So, that last column on the right is the power (second column) multiplied by the Weibull percentage (third column) multiplied by 8766 hours per year. That gives you the number of kWh's per year for each of those wind speeds. Add it all up, and you have kWh per year for the turbine, or 4,947 for your turbine.

This needs a little correction, for air density (I used 300 meters I believe, making for a factor of 0.97 for air density), and turbulence (I used 10% derating, or a factor of 0.9). Multiply these factors with the annual production, and you get a corrected annual production of 4,318 kWh per year.

The table in the article shows 4,316 kWh/year; it comes from taking a number of 'curves' based on the above calculations, and then parametrizing those into that equation you see in the article:
Energy [kWh] = 2.09 • Diameter^2 [m] • Wind^3 [m/s]

Hopefully that explains better where the annual energy production numbers come from, and why they are most definitely not breaking Betz. The overall efficiency I used is 30%, which is pretty well the top that only the best of turbines will reach. It also doesn't take the effect of furling into account (above 12 m/s the turbine should be protecting itself and limiting output power to something like rated power at best).

-RoB-

Chris Olson
24th March 2011, 09:17
Fully agree!
Chris, where did you get that kWh number from?
It looks like you're multiplying 352 Watt by the number of hours in a year, and then converting it to kWh's. If so, that won't work to get the energy production of a turbine.

Ok, I see what you're doing there to come up with those numbers. I'm pointing out that is a common method used by the wind power industry. In practice, in the real world, with real turbines, on real towers, running in real wind, I've never seen it.

The actual output of a 12 foot turbine on a site with 12 mph wind speed, the wind speed logged by a MET pole @ 33 feet, is closer to 2,700 kWh/year with the turbine on a 80 foot tower. If you put it on a 120 foot tower you'll get about 3,500 kWh/year out of that same turbine. Never seen one yet make 4,300 on a site with a 12 mph average. That doesn't mean it can't be done if the aloft wind speeds are considerably higher than at 33 feet, but I've never seen a turbine actually do it.

I got reams and reams of logging data, gathered on real turbines in the real world.

I decided to modify my post and add that not all small wind companies do this. A Jacobs 31-20 is considered a 20,000 kWh/year machine on a site with a 12 mph average wind speed. Those turbines have been successful in the market because they actually make the advertised 20,000 kWh/year @ 12 mph. If Jacobs would've used the method used in your chart, it would be about a 36,000 kWh/year turbine and there would be lots of unhappy customers because there is few installations where it would actually make that.
--
Chris

Rob Beckers
25th March 2011, 08:20
Hi Chris,

Taking the numbers directly from the Jacobs' brochure (http://www.windturbine.net/documents/WTIC%20Brochure%202010.pdf): On a 'good' site they predict 20,000 kWh/year for 11 mph annual average wind speed.

When I plug the wind speed and rotor diameter into my equation it results in 22,230 kWh/year. IMO very much in line with Jacobs' prediction; it's within 10%, not so bad for a generic equation that knows nothing about the airfoil or turbine efficiency.

The equation, and tables, in the article were actually made to tell people not to expect too much from their turbines. They are far lower production numbers than what many turbine manufacturers claim. I also go on to show from studies that most real-life small turbines generally do no better than half - just 50% - of the predicted production numbers. And those are the good installations! Many do much worse! I'm not at all surprised by your production numbers, they are very much in line with what other small turbine installations are seeing.

Trouble is that much of it is very site specific: My impression is that the measly production (compared to what the turbine could do) is largely due to more turbulent air than expected, and in many cases an overestimation of the annual average wind speed for the site. Possibly the assumption of a Weibull wind distribution breaks down for some sites as well, though keep in mind that it was developed from actual measured wind data to be a good fit.

The methodology I used for predicting energy production is the exact (and I do mean exact!) same one used for all the large wind turbines as well. All they do is refine it by using an actual power curve (where I predict one based on rotor diameter), and they use the actually measured wind distribution (where I make do with a Weibull distribution, and for most small turbine installs the actual distribution is never measured). It is just basic physics: If you know the turbine's power curve that will tell you how many kW it'll produce if the wind blows at a YY m/s. Multiply that with the number of hours the wind blows at that wind speed each year, and you have kWh's produced by that wind speed. Do the same for the other wind speeds, and you get total energy production. Correct for air density, and turbulence, and you have a predicted energy production.

I agree with you that the predictions in the article may still be at the high side, despite the modest assumptions for efficiency etc. If I find the time I may change it to use 25% overall efficiency - that's a 16% lowering overall. There may be the odd turbine that does better than that, but most will not even reach those numbers, especially with actual installations generally at poor sites. Its just a bear of a job to redo those tables; I have to type in every number by hand (haven't found a way to copy and paste entire tables into MS-Frontpage or DreamWeaver).

-RoB-

Rob Beckers
25th March 2011, 08:28
Chris, one question: Is your turbine directly charging batteries?

The equation and tables are meant for turbines that are optimally loaded, with an MPPT type controller or inverter. Hooking a turbine directly to batteries has large consequences since it directly limits voltage, and through that rotor RPM, so the airfoil will run off-optimal for a large part of the wind speed range. By some measures the losses because of this can add up to 30% vs. an optimally loaded turbine. Funneling power through batteries takes another 5% or more (depending on battery state and age) that you'll never get out of those batteries.

-RoB-

Chris Olson
25th March 2011, 09:45
Chris, one question: Is your turbine directly charging batteries?
-RoB-

Yes, my machines are battery charging, off-grid installations only. I do not mess with grid-tie stuff because I don't believe renewable energy systems are feasible when grid-tied.

This is very interesting because there's a Vestas V82 that was installed at St. Olaf College in Northfield, MN several years ago. They used the same type of calculations to predict output of the turbine. That machine has made roughly 58% of the power predicted it would make on that site.

My wife and I toured that turbine installation last year and even though the machine has fallen short of expectations it's still a successful machine. St. Olaf College is proud of it because it's privately owned by the college and it has still provided about 1/3 of the total power required by the campus. The day we toured it, it was developing about 600 kW.

I know of two Jacobs 31-20 installations that only make 12,000 kWh per year and they're considered successful installations because they do the job for which they were installed for. The owner of one of those Jakes told me that the service intervals on the machine are longer and the lifespan of it will be longer before overhaul because it runs at lower output the majority of the time, and he's happy with it because it powers his shop and does the job.

I think the problem with wind power, if you want to call it a "problem" is the cost and the availability of cheap energy from fossil fuels. But I also think in the next 10 years you're going to see fossil fuel based energy become anything but cheap. Marcellus Jacobs closed the Jacobs factory in Minneapolis in 1956 because of the REA and he saw the market for his machines evaporating. When the energy crunch of the early 70's happened, he and his son Paul started building turbines again in Ft. Meyers, FL. Paul now lives in Corcoran, MN, and while Wind Turbine Industries Corp in Prior Lake bought the inventory, Paul is still developing windplants because he believes they're going to be important again.

Jacobs is the oldest wind electric power company on earth still building turbines. Aermotor is the other successful wind power company with water pumping mills and there was once 2.5 million of those Aermotor mills in North America. And they all worked on 35 foot towers. The wind still blows now just like it did then. What's changed is the perception of what works and what doesn't based on cost. It's "cheaper" to put in a deep well submersible pump and run it on grid power than it is to pump water when the wind blows and store it in a cistern.

But artificially cheap electricity is going to become a thing of the past. In fact all cheap energy sources are going to become a thing of the past. Civilization as we know it was built energy intensive on the power of oil and coal, both of which are finite resources. 70 years ago when Jacobs and Aermotor were throwing up mills all over the country civilization was sustainable. Switching to fossil fuels, and depending on them, has made man's civilization a house of cards. The people who don't want to depend on a house of cards are my customers. And I have no shortage of them.
--
Chris

Zed universe
13th April 2011, 05:08
Wind power is the need now. we need to tap this form of energy. it is one of the safest and cheapest form of energy. With the depletion of non renewable resources we need to harness this form of energy.
(Admin: link to product removed)