Power and wind speed averages

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Power and wind speed averages

Postby windgat on Wed May 13, 2009 8:41 pm

One day the wind blows constantly at 5 m/s for 24 hours. The next day it blows at 3 m/s for 12 hours, and 7 m/s for the other 12 hours. What is the difference in the average maximum available power for each day?
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Re: Power and wind speed averages

Postby Greystoke on Thu May 14, 2009 11:12 am

OK,
Judging by this (real) power curve, the average maximum power @ 7m/s over 24 hrs = 300Watt.
At 3m/s over 12 hrs the average maximum power = 50 Watt, and at 10 m/s over 12 hrs = 375 Watt.
The average over 24 hrs is therefore:(50x12 + 375x12)/24 = 212.5 Watt.
The difference is 87.5 Watt less.


Image
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Cor
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Re: Power and wind speed averages

Postby windgat on Thu May 14, 2009 1:26 pm

Why do you refer to 10 m/s :?:

The point of the speeds given is that the average wind speed is the same on both days...
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Re: Power and wind speed averages

Postby Greystoke on Thu May 14, 2009 3:22 pm

I forgot my glasses Image

Lets try this again:
Power @ 5m/s = 125 Watt 24 hrs

Power @ 7m/s = 300 Watt 12 hrs
Power @ 3m/s = 50 Watt 12 hrs
Average power = 175 Watt

Difference = 50 Watt more on day 2.
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Cor
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Re: Power and wind speed averages

Postby windgat on Sat May 16, 2009 11:54 pm

So you get 40% more power on day 2.

Using the approximate equation (given in my newsletter)
max P = r squared x v cubed

Using a radius r = 1, on day one, average max power is 125W.
On day 2, average for first 12 hours is 27W, and for the next 12 hours, 343 watts, so the average max power is 185W.
60W more is 48% more on day 2.

Any suggestions as to why the theoretical difference is greater than the difference according to that (real) graph?
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Re: Power and wind speed averages

Postby Greystoke on Sun May 17, 2009 10:04 am

The relationship is not linear. The unit generates relatively more power at higher speeds.
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Cor
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Re: Power and wind speed averages

Postby windgat on Sun May 17, 2009 11:49 am

Get those glasses out :D
The theoretical formula is a cubed relationship - nothing linear there!
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Re: Power and wind speed averages

Postby Greystoke on Sun May 17, 2009 2:54 pm

windgat wrote:Get those glasses out :D
Who's glasses? Image
I said the relationship is NOT linear.
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Cor
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Re: Power and wind speed averages

Postby windgat on Sun May 17, 2009 4:06 pm

:roll:
The (real) graph, is not linear, the formula is not linear, nothing is linear!

Let me rephrase:
Any suggestions as to why the theoretical difference between the two days from the non-linear formula is greater than the difference according to that (real) non-linear graph?
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Re: Power and wind speed averages

Postby Greystoke on Mon May 18, 2009 6:15 am

OK,
I misunderstood. :oops:

I think the main reason that the real graph gives a different answer is the fact that it does not follow the theory, mostly because the graph is offset from zero.
It follows a third order curve alright, but it appears to have been moved to the right of the graph (it tends to minimize @ v=±2m/s. In other words: the turbine stops below that windspeed)
This can be translated in the theory by a power loss at low speeds, but which loses its significance at the higher speeds. I think the loss is about 25Watt.
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Cor
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Re: Power and wind speed averages

Postby windgat on Mon May 18, 2009 10:07 am

I have been trying some curve fitting, with some puzzling results.

Graph says 50W for 3m/s. using the approx. formula for max power,
P = r^2 x v^3
r^2 = 50/27 = 1.8519
So blades must have a minimum of 1.36m radius.

Given that, the (real) graph shows half the theoretical max power at 6m/s.

If you tell me the (real) blade radius, then I will post the graphs.
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Re: Power and wind speed averages

Postby Greystoke on Mon May 18, 2009 2:20 pm

Can't tell. That info is not given, but I understand your reservations.
My problem with the theoretical formula is the fact that it is based on the kinetic energy of a moving body, ie: ½mv², in which m = the mass of the moving air, and v = the speed of the air.
But the only way in which this energy is tranferred to electrical energy is if the air stops moving AFTER the turbine, having expended ALL its energy. Of course that is not the case, but we'll never know the true story as to what really happened.

50 Watt @ 3m/s (and nothing before that) is indeed strange, particularly if the unit generates only 300 Watt @ 7m/s. It doesn't gell. But then, . . . I've seen a number of these curves, and most leave some questions. I also think a lot depends on the shape of the vanes.
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Cor
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Re: Power and wind speed averages

Postby windgat on Mon May 18, 2009 9:27 pm

The full formula is:

P = 0.59 x 0.6 x pi x r^2 x v^3
so
P = 1.11 * r^2 x v^3
or approx:
P = r^2 x v^3.

So the Betz limit is taken into account already, i.e. I am not assuming the air stops behind the turbine!
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Re: Power and wind speed averages

Postby Greystoke on Tue May 19, 2009 6:36 am

Yes,
I understand that 0.59 is the Betz correction factor. Without it the formula would read: 0.6?r²v³, which represents the kinetic energy of a moving body (the flowing air), ie: ½mv².
Proof:
½mv² = ½1.22?r²v(=mass of the air through the turbine per second)v² = 0.6?r²v³

The Bletz derivation is quite complicated and these days done with computors. 0.59 does not take into account losses due to friction or Eddy currents, complicated blade shapes, etc. It is just a good average :D
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Cor
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Re: Power and wind speed averages

Postby aero energy on Thu Jun 04, 2009 7:55 pm

Station Name Mean (m/s)
BLOEMFONTEIN - STAD 3.39
BLOEMFONTEIN WO 3.77
FICKSBURG 3.23
VAN REENEN 4.27
VREDE 3.60
WELKOM 4.32
WEPENER 2.67
IRENE WO 3.50
JHB BOT TUINE 2.68
JOHANNESBURG INT WO 4.08
PRETORIA UNISA 2.98
PRETORIA WO 2.40
SPRINGS 2.50
VEREENIGING 3.65
CHARTERS CREEK 4.12
DURBAN WO 5.06
GAINTS CASTLE 13.94
GREYTOWN 5.00
IXOPO 5.82
LADYSMITH 2.80
MARGATE 4.15
MBAZWANA AIRFIELD 3.13
MTUNZINI 3.71
NEWCASTLE 3.28
ORIBI AIRPORT 2.92
PADDOCK 4.02
PIETERMARITZBURG 1.81
PONGOLA 2.14
PORT EDWARD 2.86
SEZELA 3.82
ULUNDI 3.50
VIRGINIA 3.80
VRYHEID 2.78
ERMELO WO 5.24
KOMATIDRAAI 2.33
NELSPRUIT 2.17
PIET RETIEF 3.42
RIETVALLEI 3.64
STANDERTON 3.44
WITBANK 4.27
BLOEMHOF 3.26
KLERKSDORP 3.72
LICHTENBURG 3.55
MAFIKENG WO 4.55
POMFRET 4.79
POTCHEFSTROOM 3.57
RUSTENBURG 2.36
PILANESBERG 4.08
TAUNG 2.91
ALEXANDERBAAI 4.80
BRANDVLEI 5.27
CALVINIA WO 4.76
DE AAR WO 6.88
KATHU 3.84
KIMBERLEY WO 4.58
KOINGNAAS 6.20
NOUPOORT 5.60
POSTMASBURG 4.91
PRIESKA 3.75
SPRINGBOK WO 8.27
UPINGTON WO 4.36
VAN ZYLSRUS 2.62
VIOOLSDRIF 2.95
ALLDAYS 3.08
CUMBERLAND 2.41
DOORNLAAGTE 2.25
ELLISRAS 2.33
GRENSHOEK TZANEEN 2.22
HOEDSPRUIT 4.42
MARKEN 3.26
PIETERSBURG WO 4.42
POTGIETERSRUS 3.11
THABAZIMBI 2.59
THOHOYANDOU WO 2.84
TSHIPISE 2.67
BEAUFORT-WES 5.04
CAPE TOWN TABLE BAY 4.10
CAPE TOWN WO 6.63
CERES EXCELSIOR 2.90
GEELBEK 5.62
GEORGE WO 3.00
HERMANUS 3.79
KNYSNA 1.74
LAINGSBURG 3.84
LAMBERTSBAAI NORTIER 4.22
LANGEBAANWEG WO 6.88
MALMESBURY 4.14
PAARL 3.27
PLETTENBERGBAAI 2.97
PORTERVILLE 4.03
RIVIERSONDEREND TYGERHOEK 2.86
ROBBENEILAND 3.07
SIMONSTAD KAAPPUNT 6.65
STILBAAI 4.13
STRAND 3.55
STRUISBAAI 3.73
VILLIERSDORP VYEBOOM 2.67
WORCESTER 3.09
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Re: Power and wind speed averages

Postby Greystoke on Fri Jun 19, 2009 10:00 am

Hi guys,
I needed to know how much power I can expect from the turbine I'm planning to make. For that purpose I needed a wind speed distribution graph (Weibull graph) for the PE area. It's a bit difficult to get, but . . with some "poetic license" I managed to get something that can be regarded as reasonably accurate.

Image

The figure of 311 Watt is an average level which I can expect to be exceeded 50% of the time.
Regards
Cor
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Re: Power and wind speed averages

Postby windgat on Fri Jun 19, 2009 11:38 am

There is a standard approximate factor (like maybe 1.3? wild guess...) I saw somewhere once (sorry, can't remember...) which gives an expected power based on average wind speed. Some manufacturers use this to beef up their marketing figures, by multiplying the steady wind speed performance by this factor when quoting performance figures.

It would be interesting to see what you get if you multiply the occurence % by the wind speed cubed, and sum the results, then take a cubed root...
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Re: Power and wind speed averages

Postby Greystoke on Fri Jun 19, 2009 12:03 pm

windgat wrote: . . It would be interesting to see what you get if you multiply the occurence % by the wind speed cubed, and sum the results, then take a cubed root...
Not a problem. It will end-up the same answer, I'm sure. I'll check it out now-now. :D


PS: Yep, . . . . aVerage windspeed becomes 4.3 m/s, which - with a 2m blade - will generate 311 Watt
Regards
Cor
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Re: Power and wind speed averages

Postby windgat on Fri Jun 19, 2009 1:45 pm

So did you calculate via the cube/cube root weighted average? I am interested int he difference between that and a linear average.
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Re: Power and wind speed averages

Postby Greystoke on Fri Jun 19, 2009 1:48 pm

Yes I did,
The linear (weighted) average of the windspeed is 3.5 m/s, and the "cube/cube root weighted average" is 4.3 m/s.
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Cor
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Re: Power and wind speed averages

Postby windgat on Fri Jun 19, 2009 2:00 pm

Wow. So the 'wind speed factor' 1.2, and the 'power factor' is 1.85.

That may explain some of the claims by manufacturers about the power at the 'rated' wind speed!
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Re: Power and wind speed averages

Postby CaptainJakes on Wed Jan 12, 2011 10:03 am

Hi
I live in Willow Park Mannor Pretoria and have set up a 2kw wind turbine and +- 300 watt solar pannels (BUT THEY SEEM TO CHARGE AT 1.5 AMPS ON A HOT DAY ) and it is runing very well .I am running my lights and 2 TVs fridges off of it and i have not yet run out of power yet .I am currently converting our stove to gas asswell . I have 10 x deep cycle bateries = 120 volts DC converter 120 volts to 220 volts .if you want you are welcome to give me a call and come have a look at my setup for feedback .I am still in the experimental stages but it seems to ok. My cell is 0761256057 Jacques.
I hope this helps. :D
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