Trying to calculate max energy that a wind turbine can get



#6

Hello,

They have installed some wind turbines next to my house. the monsters have a 80m tall mast and 70m diameter blades!
http://www.enercon.de/en-en/61.htm
however, what surprised me the most is that they claim that they deliver a max power of 2.3MW with a 45kph wind...

I was trying to calculate how much energy you could get form such a wind and could not come up with anything remotely close to that 2.3MW figure...

here is what I assumed:
- The energy is obtained from the wind kinetic energy
- the wind speed will drop by 1/2 (releasing 3/4 of it's kinetic energy to the mill).

- mass of air acting on mill = blade surface (estimated at 5% of the blade disk=3.14*(70/2)²*0.05=192m²) * 45000/3600 (speed in m/s) * 1.225 (air density) ~= 3Tonnes / second!

- kinetic energy of 3 tonnes at 45 kph = 1/2*3000*(45000/3600)² = 234_KJ

so, even if the darn thing was able to drop the wind to 0_Kph, it seems that it would be short a factor 10 on what the manufacturer pretends...

increasing the blade surface assumption would help, but since I do know that the blade cover less than 50% of the disk area, I do not think that this will cut it either..

any idea where I have things wrong?

cyrille


#7

Hi Cyrille,

The right formula is :

(Bernoulli and Albert Betz laws - see Wikipedia for example)

I get a theoric max power of 3,4MW

#8

The theoretical maximum energy you can extract appears to be 59% or so (query "Betz's law) not your 5%, and really good turbines apparantly can get 30-40% (see this or google/yahoo something like "how much energy can I extract from wind?")

Although the blade cross section may be only 5% or so, as it spins it is affected by a larger fraction of the air flow. The effect of a parcel of air extends beyond its physical cross section, due to air pressure effects (try to suck a vacuum, and nearby air rushes in to fill the void). While we like to think of air as (relatively) frictionless or incapable of exerting effects over a distance, it is "stiffer" than a vacuum!

#9

Betz' law allows no more than 59.3% of the kinetic energy of the wind to be captured. I've seen some claims that modern designs may reach 80% of that theoretical limit, but I've also seen claims that the overall system efficiency only reaches about 18%.

The blades generally cover *far* less than 50% of the disk area; just guessing based on the ones I've seen, I think it's under 10%. Apparently having two or three relatively small blades must do such a good job of approaching the Betz limit that adding more or larger blades is not cost-effective.

On a road trip on US I-40 and I-5 last year, I saw the largest items I've ever seen transported on public highways, which were 164-foot (50m) long blades for wind turbines, being transported on 160-foot trailers. I got to watch several of these exiting and later reentering the freeway at the I-5 interchange at Buttonwillow, and it was quite amazing seeing them negotiate a moderately sharp 90 degree turn. Obviously the route had been carefully planned beforehand to ensure that the trailers could negotiate the turns.

Unfortunately the only camera I had at hand was a cell phone, and the pictures didn't come out very well.

#10

Your problem is with the blade surface area assumption. Rather counter-intuitively, the blades do not need to be a large fraction of the total area in order to extract energy from across the whole area. The blades turn quickly enough so that each parcel of air passing through them will get close to at least one blade as it does so.

There's a theoretical maximum efficiency called the Betz limit, which says that the turbine can extract at most 16/27 of the incident power from the air.* Using your figures, the claimed power output represents (suspiciously precisely) an efficiency of 50%, which is within the Betz limit. So the claimed figure may well be correct!

Nigel (UK) (who once proof-read a chapter dealing with wind turbines from a book about the physics of alternative energy sources, and is not in any other way an expert on this subject!)

*The "incident power" here is (1/2) x density x pi x radius^2 x speed^3. This isn't the power that actually reaches the circle of the blades, because the airflow is spreading out as it passes the blades. The true incident power is significantly less than this.


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