1. Energy from Wind

2. Power Power: Rate at which energy is delivered Power = Energy Time Measured in Watts (W), kilowatts (kW), or horsepower Power is an instantaneous quantity Power does not accumulate Think gallons per minute

3. Energy Energy: Ability to do something Measured in kilowatt  Hours (kWhrs) Why? – Since Power = Energy/Time, then Power  Time = Energy Energy does accumulates over time Think gallons Gallons = (gallons/min)  minutes

4. Power kW (kilowatts) Energy kWh (kilowatt hours) Think gpm Think gallons

5. Wind Resource At any instant, the only question that makes sense is “What’s the power of the wind?” Answer depends on 2 quantities – Instantaneous wind speed, v – Air density, , which depends on Elevation Temperature Weather At sea level and 77  F (standard conditions), air density  = 1.225 kg/m 3 At 5,000 ft elevation,  is ~16% less than at sea level

6. Power Density of the Wind Power Density: P/A P/A = ½  v 3 (in W/m 2 ) Example: Suppose the wind speed is 8.0 m/s, and air density is 1.0 kg/m 3, then P/A = ½ (1.0 kg/m 3 )(8.0 m/s) 3 = 256 W/m 2 – For each square meter of area, there are 256 W of power – Use Metric Units! – If wind speed doubles, power density increases by 8

7. Swept Area The single most important parameter of a wind turbine is its rotor’s swept area A

8. Power of a Wind Turbine The power of a wind turbine is P = ½  v 3 A C P A: swept area of rotor C P : rotor efficiency Example: A 2.5 m diameter turbine with a 25% efficient rotor in our 8.0 m/s wind will have P = ½ (1.0 kg/m 3 )(8.0 m/s) 3 [  (2.5 m/2) 2 ](0.25) = 314 W

9. How NOT to estimate energy in the wind How much energy can this turbine produce? Need a constant wind speed and time Example: If the wind speed is a constant 8.0 m/s, then in 1 month our turbine will produce – (314 W)(30 days)(24 hrs/day) = 226 kWhrs/month – The average home in NC uses around 850 kWhrs/month The wind speed is not constant

10. 10 Minute Wind Data

11. Wind Speed Distributions

12. Using the Annual Average Wind Speed to Calculate Energy Production is Problematic Using the average Annual wind speed will under estimate energy production because of the cubic relationship between wind speed and power. Need to cube each 10 minute wind speed The average of the cubes is greater than the cube of the average

13. Cube of Average vs Average of Cubes for site with 6.5 m/s average annual wind speed Cube of the Average – Class 3 site @ 30 meters = 6.5 m/s – P/A =.6125 x 6.5 3 – P/A = 168 watts/m 2 Too Low Average of Cubes P/A of 10.0 m/s = 612 P/A of 5.0 m/s = 76.56 P/A of 4.6 m/s = 59.6 19.6/3748.16/3 6.5 m/s 249 w/m 2 Energy Pattern Factor (EPF) = Average of cubes / cube of average = 249 / 168 = 1.48

14. 10 minute data mph20stddirF 10/1/2006 0:001.000.6050 10/1/2006 0:101.000.520250 10/1/2006 0:203.10127050 10/1/2006 0:303.600.924850 10/1/2006 0:404.001.622551 10/1/2006 0:506.702.422553 10/1/2006 1:005.502.120254 10/1/2006 1:108.902.520254 10/1/2006 1:208.502.220255 10/1/2006 1:307.502.822555 10/1/2006 1:405.401.922555 10/1/2006 1:504.501.922555 10/1/2006 2:004.00227055

15. Average of Cubes is Greater than Cube of Average

16. Energy Pattern Factor Average of Cubes divided by Cube of Average 62.86 / 59.69 = 1.05 EPF = 1.05 Typical EPF = 1.9 Multiply power density calculated from average annual wind speed by 1.9 to get more accurate average annual power density

17. Estimating Average Annual Power Density from Annual Average Wind Speed What would be a reasonable estimate of an annual average power density when the average annual wind speed was 12 mph (5.35 m/s) and elevation was 4,000’ Annual Average P/A = ½ Density x V 3 (in meters/sec) x 1.9 AA P/A of 12 mph = ½ (1.225 x.88) x 5.35 3 x 1.9 AA P/A of a 12 mph wind at 4,000’ = 156 watts/m 2

18. Air Density Changes with Elevation

19. Swept Area Method of Estimating Energy Production (AEO) AEO = (Average annual power density x 1.9) x area of rotor (m 2 ) x efficiency x hours/year

20. Swept Area Power is directly related to the area intercepting the wind Doubling the swept area will double power available to it Nothing tells you more about a wind turbines potential than area swept by rotor Area = πr 2 or πd 2 /4 Relatively small increases in blade length produce large increase in swept area Doubling diameter will quadruple swept area

21. Credit: Paul Gipe Swept Area A = Pi D 2 / 4 1 m = 3.3 ft Area = πr 2

22. Swept Area of Bergey XL.1 Bergey XL.1 has three blades each 4’ long and a rotor diameter of 8.2’ 8.2’ / 3.28 (ft/m) = 2.5 meter diameter Radius = 1.25 meter Area = πr 2 Area = πr 2 = π 1.25 2 = 4.9 m 2

23. Power Intercepted by Bergey XL1 with 4.9 m 2 of Wind Power at 4,000’, 0 0, in 7 m/s wind Power = ½ density x area x velocity 3 Power = ½ (1.218 kg/m 2 ) x 4.9 m 2 x 7 3 Power =.609 x 4.9 m 2 x 7 3 Power =.609 x 4.9 x 343 Power = 1,023 watts

24. Estimating Annual Energy Output of XL.1 with Swept Area Method @ class 3 site; 6.5 m/s @ 5,000’ AEO in watts = Annual Average P/A x Swept Area x efficiency x hours per year AEO = (1/2 air density) x (v 3 ) x (1.9) x 4.9 x.20 x 8760 AEO = ½ (1.225 x.860) x (6.5 3 ) x 1.9 x 4.9 x.20 x 8760 AEO = 2,359 Kwh

25. Power Curve Method or Method of Bins 2 Things Needed  Need to know (or approximate) your wind distribution  Power Curve of turbine

26. Wind Distribution Wind is known to follow a Weibull distribution

27. Wind Distribution Wind is known to follow a Weibull distribution =WEIBULL(c, k, vavg) Rayleigh Distribution if k=2 Credit: Paul Gipe

28. Wind Speed Distributions Wind is empirically known to follow a Weibull probability distribution Weibull curve: has shape parameters: c & k Average k in US: k = 2 (Raleigh distribution)

29. Method of Bins Wind Distribution: From your logger!

30. Power Curve The turbine’s manufacturer will provide you with its power curve Bergey XL.1

31. Whisper Power Curves

32. Utility Scale Power Curve (GE)

33. Method of Bins Power Curve (kW) Wind Distribution (hrs) AEO (kWhrs) Hours Energy (kWhr)

34. Method of Bins Calculate Energy = Power  Time for each wind speed bin Sum ‘um up!

35. Charts from Manufacturer