## CHAPTER

1. See Figure 1.7. Solidity is a ratio of area of blades to area of rotor. Notice that the solidity of the Savonius is 1 or greater. Assume 1.

The giromill has a low solidity, around 0.2. Therefore, the difference in blade material is around 5/1.

Another way, use rotor swept area, 10 * 10 = 100 m2. Blade area of Savonius is 100 m2 or greater. Giromill has three blades; assume blades are 10 m long, 0.5 m wide. Blade area = 3 * 10 * 0.5 = 15 m2. Therefore, the ratio of Savonius to giromill is 100/15.

2. Assume capacity factor = 0.3. AKWH = 0.3 * 300 kW* 8,760 h/year = 788,000 kWh/year.

3. Assume capacity factor = 0.35. AMWH = 0.35* 1.5 MW* 8,760 h/year = 4,600 MWh/year.

4. HAWT, r = 50 m. Area = n r2 = 3.14 * 2,500 = 7,850 m2.

Choose locations: (a) one near Amarillo, Texas, P/A = 400 W/m2; (b) one near Comodoro Rivadavia, Argentina, P/A = 800 W/m2.

Use Equation 5.3; assume a capacity factor of 0.35 for Amarillo and a capacity factor of 0.40 for Comodoro Rivadavia.

a. AKWH = CF* Ar* WM* 8.76 = 0.35 * 7850 *400 * 8.76 = 9,627,000 kWh/year b. AKWH = CF * Ar * WM * 8.76 = 0.4 * 7850 * 800 * 8.76 = 22,000,000 kWh/year

5. Darrieus turbine, 34 x 42.5 m. Area = 0.6 H*D = .6 * 34 * 42.5 = 867 m2.

Use capacity factor of 0.3 and wind map values for Denmark, 500 W/m2, and mid-Germany, 300 W/m2.

AKWH = CF* Ar * WM * 8.76 = 0.3 * 867 * 500 * 8.76 = 1,139,000 kWh/year AKWH = CF* Ar * WM * 8.76 = 0.3 * 867 * 300 * 8.76 = 684,000 kWh/year

6. Giromill 10 x 12 m. Area = H*D = 10*12 = 120 m2.

Use capacity factor of 0.25 and wind map values of 250 and 300 W/m2. Small giromill will not be on very tall tower.

7. Estimate is around 3.3 GWh/year.

8. Use Figure 5.13, wind speed = 20 m/s. Use Equation 5.1, P = Trn.

a. At constant rpm (line A in figure), 160 rpm, torque estimated at 15,000 Nm.

P = Trn = 12,000 Nm* 17 rad/s = 204,000 Nm/s = 204,000 W = 204 kW

b. At maximum power coefficient (line B in figure).

T = 11,000 Nm, œ = 220 rpm = 23 rad/s P = Ta = 11,000 * 23 = 253,000 Nm/s = 253 kW

c. At constant torque (line C in figure).

T = 6,000 Nm, œ = 310 rpm = 32 rad/s P = Ta = 6,000 * 32 Nm/s = 192,000 Nm = 192 kW

This is not a viable operating point because of the high rpm. Also notice that the power is largest for answer b, line of maximum power coefficient. 9. Torque = 6,000 Nm, rpm = 160 = 17 rad/s. P = Trn = 6,000 * 17 Nm/s = 102 kW. 10. Frequency = number in the bin divided by total number of hours.

m/s |
Frequency |

1 |
0.014 |

2 |
0.043 |

3 |
0.068 |

4 |
0.087 |

5 |
0.099 |

6 |
0.104 |

7 |
0.103 |

8 |
0.097 |

9 |
0.086 |

10 |
0.074 |

11 |
0.061 |

12 |
0.048 |

13 |
0.036 |

14 |
0.027 |

15 |
0.019 |

16 |
0.013 |

17 |
0.009 |

18 |
0.005 |

19 |
0.003 |

>20 |
0.005 |

25 |

11. Mean wind speed is 8.2 m/s. Rayleigh distribution is calculated from that mean wind speed, bin width of 1 m/s. Estimated annual energy production is calculated by multiplying power curve times number of hours (wind speed histogram). It is always better to have actual data for the wind speed histogram, rather than a calculated distribution. Anemometer height for power curve measurements and anemometer height for wind speed are not known, so assume that both are at hub height. Power curve values should be at the midpoint of the wind speed bin.

From spreadsheet, energy = 3,400.000 kWh/year. Need to reduce value for availability, use 95 to 98%. At 95%, estimated energy is 3,200 MWh/year. May have to reduce value for air density, for example, Panhandle of Texas at 1,100 m, has air density of 1.1 kg/m3. This would give a 10% reduction to 2,900 MWh/year.

Class j |
Speed m/s |
Frequency |
Hours |
Power Curve kW |
Energy kWh |

1 |
1 |
0.023 |
202 |
0 |
0 |

2 |
2 |
0.045 |
391 |
0 |
0 |

3 |
3 |
0.063 |
553 |
0 |
0 |

4 |
4 |
0.078 |
679 |
0 |
0 |

5 |
5 |
0.087 |
764 |
34 |
25,983 |

6 |
6 |
0.092 |
807 |
103 |
83,073 |

7 |
7 |
0.092 |
808 |
193 |
156,031 |

8 |
8 |
0.089 |
776 |
308 |
238,860 |

9 |
9 |
0.082 |
715 |
446 |
319,071 |

10 |
10 |
0.073 |
637 |
595 |
378,872 |

11 |
11 |
0.063 |
548 |
748 |
410,011 |

12 |
12 |
0.052 |
457 |
874 |
399,557 |

13 |
13 |
0.042 |
370 |
976 |
361,014 |

14 |
14 |
0.033 |
291 |
1,000 |
290,645 |

15 |
15 |
0.025 |
222 |
1,000 |
221,968 |

16 |
16 |
0.019 |
165 |
1,000 |
164,870 |

17 |
17 |
0.014 |
119 |
1,000 |
119,167 |

18 |
18 |
0.010 |
84 |
1,000 |
83,853 |

19 |
19 |
0.007 |
57 |
1,000 |
57,465 |

20 |
20 |
0.004 |
38 |
1,000 |
38,366 |

21 |
21 |
0.003 |
25 |
1,000 |
24,960 |

22 |
22 |
0.002 |
16 |
1,000 |
15,828 |

23 |
23 |
0.001 |
10 |
1,000 |
9,785 |

24 |
24 |
0.001 |
6 |
1,000 |
5,899 |

25 |
25 |
0.000 |
3 |
0 |
0 |

0.998 |
8743 |
3,405,281 |

12. Cut-in wind speed = 5 m/s, rated wind speed = 14 m/s.

13. Cut-in wind speed = 4 m/s, rated wind speed = 10 m/s.

14. Primary method of control for power output is control of rpm.

15. Primary method of control for shutdown is feathering of blades, full-span pitch control.

16. Time for shutdown or to reduce power output is 3-6 s.

17. Any system. If students have read ahead, there are examples in Chapters 8 and 10.

18. Problems with tethered wind turbine: cable (what are size, weight, and length?), warning lights, small planes, winch to reel in or lower system, high wind, and speed control.

## Renewable Energy Eco Friendly

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.

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