Freeenergy Planet Solutions Chapter 12 Wind Energy

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.

Renewable Energy 101

Renewable Energy 101

Renewable energy is energy that is generated from sunlight, rain, tides, geothermal heat and wind. These sources are naturally and constantly replenished, which is why they are deemed as renewable. The usage of renewable energy sources is very important when considering the sustainability of the existing energy usage of the world. While there is currently an abundance of non-renewable energy sources, such as nuclear fuels, these energy sources are depleting. In addition to being a non-renewable supply, the non-renewable energy sources release emissions into the air, which has an adverse effect on the environment.

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