The Jacobs Wind Energy Systems 31-20 Wind Turbine that Wright-Hennepin Electric put up has a year to date output of 5981kWh.
This works out to 18.4kWh per day this year or 766.66Wh per hour.
If you calculate the power using a cP of 25% and the Rockford, MN average wind speed of 5.3m/s this 31-20 Wind Turbine should get get 38.4kWh per day.
HWP = 0.5 x 1.23 x SA x WV^3 x cP
(HWP) Harvestable Wind Power (w/hr) = 0.5 x AD (air density kg/cu m) x SA (sweep area m^2) x WV^3 (wind velocity m/s) x cP (coefficient of performance Betz limit 59.26%, R^1 or Copper loss, linear power curve to cubic curve matching — 0.25 for most systems)
HWP = 0.5 x 1.23 x 3.14 x 15.5 x 15.5 x 0.09260304 (69.853) x 5.3 x 5.3 x 5.3 (148.877) x 0.25 = 1599Wh
1.599Wh x 24 = 38.4kWh per day.
So lets work out the cP of this unit.
So if HWP = 0.5 x 1.23 x SA x WV^3 x cP
then cP = HWP / 0.5 x 1.23 x SA x WV^3
cP = 766.66 / 0.5 x 1.23 x 69.853 x 148.877
is a cP of 0.119 which is not very good.
December 9th, 2009 at 2:31 pm
This calculation is over simplified and is not an accurate way to estimate the CP of a wind turbine.
The amount of energy a turbine generates is the sum of all the energy preduced at each and every wind speed in the wind regime it operates in.
So to calculate the amount of energy you would expect a given machine to produce you need to know the shape of your wind speed distribution, AND the shape of the turbines power curve.
Saying that, it does sound like quite low generation for the size of turbine. Are you confident the mean annual windspeed is actually over 5m/s? Was that measured on the site by data logging for a year?
December 9th, 2009 at 4:45 pm
Hi Mark,
I agree that the calculation should take into consideration the different WV groups. But you can work out the cP of a wind turbine with out knowing the shape of the turbines power curve because you are working from the kWh produced by the wind turbine.
The idea of this calculation was to offer the cP for this wind turbine based on the WV annual average for that area. In this way people that are trying to get a real world idea of how much power a HAWT will product can use the average WV where their turbine will be place and a cP that is more realistic that what the manufactures offer.
Some wind turbines advertising uses the Betz limit (59.26%) for a cP. I have seen some that use a cP of 100%. So I am saying that according the kWh this wind turbine is producing if you use the average wind speed for that area the cP is 12%. This will give you more realistic power numbers.
Thanks!
Bob.
November 8th, 2010 at 3:17 am
Bob worst calculation ever.
Power extracted by a wind turbine is a cube of the wind speed. Therefore you must know what the Weibull distribution is of the site in question - otherwise, you are just trying to pull the wool over peoples eyes - like this whole blog.
Samson!