## Freeenergy Planet Solutions Chapter 12 Wind Energy

Energy2green Wind And Solar Power System

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1. kWh/kW is the energy produced divided by the rated power (size of the generator). So, numbers range from 412 for Flowind to 2,833 for the Nordtank. In reality, this calculation is in hours, or the equivalent number of hours that the unit was at the rated power. There are two main factors: (1) what wind regime (energy produced) and (2) size of the generator. If it is a poor wind regime, lower kWh/kW, and if the generator is too big for the rotor size, lower kWh/kW.

2. Average power = annual kWh/9=8,760 hour, capacity factor = average power/rated power.

Fayette, CF = 41,000/(8,760 * 90) = 0.052. Vestas 23, CF = 434,000/(8,760*200) = 0.25. Bonus 120, CF = 276,000/(8,760* 120) = 0.26.

3. Average capacity factor = 17%

4. Enertech 44/40 means it is 44 ft diameter and rated at 40 kW. Area = ^r2 = 3.14 6.72 = 141 m2.

a. Specific output = annual kWh/area = 49,467/141 = 350 kWh/m2.

If change ratio for full year, specific output = 12/7 * 350 = 600 kWh/m2.

b. Specific output = annual kWh/area = 86,592/141 = 614 kWh/kW.

5. Enertech 44/25:

a. Specific output = annual kWh/area = 91,372/141 = 648 kWh/m2

b. Hours = 3,254/0.64 = 5,084, or number of hours for 7 months (April-October, hours = no. days * 24).

Average power = 49,467 kWh/5084 h = 9.7 kW. Capacity factor = avg. power/rated power = 9.7/25 = 0.39.

6. Enertech 44/60; 44 ft diameter, rated power = 60 kW.

a. Specific output 91,732/141 = 651 kWh/m2.

Capacity factor = avg. power/rated power = 10.5/60 = 17.5%.

7. May, specific output = 9,078/141 = 94 kWh/m2; August, 2,443/141= 17 kWh/m2.

Of course, specific output depends on the wind, so yearly values are a better way to compare wind turbines.

a. Specific output = 600,000/452 = 1,327 kWh/m2

c. Output per mass = 600,000/14,250 = 42 kWh/kg d. Output per installed cost = 600,000/180,000 = 3.3 kWh/\$

When you compare problems 8 and 9, you have to realize that Vestas is still producing wind turbines and Carter is not. Problems with maintenance and capitalization led to the company going out of business. Notice that the V27 has a larger diameter and a smaller rated power. Also, this is a comparison between lightweight two-blade and heavier three-blade wind turbines.

d. Output per installed cost = 500,000/225,000 = 2.2 kWh/\$.

10. Carter 300, avg. power = 600,000/(365 * 24) = 68 kW, CF = 68/300 = 23%. Vestas V27, avg. power = 500,000/(365 * 24) = 57 kW, CF = 57/225 = 25%.

11. Estimate for V90, wind map value of 500 W/m2 and CF = 35%, V90 rotor area = 6,360 m2. AMWH = 0.35 * 500 * 5,027 * 0.00876 = 9,700 MWh/year.

Estimate for V52, generator size method, CF = 35%. AMWH = 0.35 * 1.65 * 8,760 = 5,000 MWh/year.

12. From Table 8.5, need pump diameter of 1.8 cm. Could pump around 2 cubic m/h.

13. From Table 8.5, need pump diameter of 2.2 cm. Could pump around 3 cubic m/h.

14. From Table 8.5, need pump diameter of 1.0 cm. Could pump around 1 cubic m/h.

15. Multiply number of minutes times flow (L/min) for each bin and sum to get total volume of water pumped. 10,000 L = 1 m3.

Same spreadsheet is used for problems 15 and 17.

 V average 5 Farm Windmill Electric- Electric lass j Speed m/s Frequency No. Minutes Flow L/min Volume L Flow L/min Volume L 1 1 0.06 2,629 2 2 0.11 4,785 0.0 0 3 3 0.14 6,135 0.4 2,303 4 4 0.15 6,566 5.3 34,956 0.0 0 5 5 0.14 6,187 11.6 71,667 0.0 0 6 6 0.12 5,256 15.8 83,133 3.2 16,909 7 7 0.09 4,077 19.3 78,563 18.0 73,417 8 8 0.07 2,909 21.7 63,059 36.2 105,390 9 9 0.04 1,919 22.7 43,493 53.0 101,764 10 10 0.03 1,174 20.8 24,369 66.2 77,683 11 11 0.02 668 19.0 12,708 75.7 50,597 12 12 0.01 354 13.9 4,921 85.1 30,104 13 13 0.00 175 10.3 1,801 88.8 15,523 14 14 0.00 81 9.2 740 83.3 6,718 15 15 0.00 35 70.7 2,458 16 16 0.00 14 59.1 828 Sum 0.99 42,965 421,713 481,391

For farm windmill, water pumped = 42 m3.

Notice that if frequency distribution does not add to 1 (close), you have made a mistake. Student answers will vary due to estimation of flow from graph. With spreadsheet, easy to change average wind speed. At 6 m/s, farm windmill = 50 m3; electric-electric = 83 m3.

16. Bergey Windpower, Southwest Windpower, ?

17. Electric-electric system, water pumped = 48 cubic m. See problem 15 for spreadsheet.

18. Enertech 44/25, 68,000 kWh/year; 44/60, 104,000 kWh/year.

Student answers will vary due to estimation of power curves from graph.

 V average 6 Enertech 44/25 Enertech 44/60 lass j Speed m/s Frequency No. Hours kW Power kWh Energy kW Power kWh Energy 1 1 0.04 374 2 2 0.08 700 0.0 0 3 3 0.11 942 0 0 0 4 4 0.12 1,078 0 0 0 0 5 5 0.13 1,107 4 4,430 3 3,322 6 6 0.12 1,046 5 5,228 6 6,273 7 7 0.10 919 10 9,187 11 10,106 8 8 0.09 757 14 10,598 19 14,383 9 9 0.07 588 18 10,581 26 15,284 10 10 0.05 432 20 8,632 34 14,675 11 11 0.03 300 23 6,908 46 13,816 12 12 0.02 198 26 5,159 53 10,517 13 13 0.01 125 28 3,490 55 6,854 14 14 0.01 74 30 2,235 60 4,469 15 15 0.00 42 31 1,315 65 2,756 16 16 0.00 23 32 736 68 1,565 17 17 0.00 12 33 393 70 833 18 18 0.00 6 33 194 72 423 19 19 0.00 3 33 91 74 205 20 20 0.00 1 34 42 76 95 Sum 0.99 8,705 68,498 104,021

19. For 2005, 3,974,759,861 kWh = 3.97 TWh. Vestas, installed capacity = 563,985 kW. Kenetech, number installed = 3,598.

20. Power will be reduced from 30% to 50%.

21. Llano Estacado Wind, White Deer, Texas: Installed capacity = 80 MW.

From Figure 8.5, CF = 0.34. AMWH = 0.34 * 80 * 8,760 = 239,000 MWh/year.

24. Vortex generators, small vanes to mix laminar flow with boundary layer. Suction or blowing air through holes in the blade.

25. Active stall control, since blade pitch can be changed to still obtain power output. The other solution is to use airfoils, which are less sensitive to surface roughness. 