## Maximum Cp For Drag Device

1. Blade, r to center of mass = 5 m, mass = 500 kg, F = 320 Nm.

2. Use Equation 6.9, P = F* v, and substitute force from Equation 6.8.

Notice that power loss is proportional to velocity cubed. CD = 1, and assume p =1 kg/m3. Strut is 4 m x 0.025 m, rpm = 180.

Area for 1 m section = length * width = 1 m * 0.025 m = 0.025 m2.

Power loss for each section, P = 0.5 v3*A = 0.0125 v3, watt.

Take r at the midpoint of the section. Have to find velocity at center of each section, v = ffl * r.

Sum 5,150

For three struts, the power loss would be 15 kW, which means do not use struts, have blades.

3. rpm = 80, ffl = rpm * 2^/60 = 80 * rc/30 = 8.4 rad/s. Show diagram. Strut has radius from 0.25 to 2.25 m. CD = 1, and assume p =1 kg/m3.

Take differential area of strut, dA = D*dr = 0.05 dr, where D is the diameter. For one strut, P = 0.5 v3 * A, dP = 0.5 (®r)3 0.05 dr.

2 r4

P = 0.025 * 8.43 J r3dr = 14^^ = 3.7(25- 0.004) = 92 W

Power loss for four struts would be 370 W.

Can do calculation numerically, similar to problem 2.

For one strut:

Section Radius m v = ffl*r m/s Power W

Sum 93

Notice how close the numerical result is to the calculus result. 4. Power out is from Equation 6.12; substitute into Equation 6.7 for power coefficient. From calculus take derivative, dCP/du, and set equal to 0. Solve for u.

Substitute u = v0/3 in Equation 6.12 and you have maximum CP = 4/27 CD. For CD = 1 the efficiency is 15% for a drag device.

Numerical solution: calculate CP = u(1 - u)2, where u goes from 0 to v0 by tenths.

0.1 |
0.08 |

0.2 |
0.13 |

0.3 |
0.15 |

0.4 |
0.14 |

0.5 |
0.13 |

0.6 |
0.10 |

0.7 |
0.06 |

0.8 |
0.03 |

0.9 |
0.01 |

1 |
0.00 |

Maximum occurs between 0.3 and 0.4, so do that by hundredths.

u |
Cp |

0.31 |
0.14759 |

0.32 |
0.14797 |

0.33 |
0.14814 |

0.34 |
0.14810 |

0.35 |
0.14788 |

0.36 |
0.14746 |

0.37 |
0.14685 |

0.38 |
0.14607 |

0.39 |
0.14512 |

0.4 |
0.14400 |

This shows that CP maximum occurs for u = 0.33. Or from a plot, estimate value of u for peak CP.

5. If the solidity increases, the peak of the aerodynamic efficiency occurs at lower tip speed ratios.

6. a. If it operates at constant tip speed ratio, then it operates at maximum efficiency at any wind speed.

b. If it operates at constant rpm, it reaches maximum efficiency at only one wind speed. See Figure 6.13.

7. Maximum efficiency is 59%. This was calculated by using conservation of energy and conservation of momentum.

8. Low-solidity rotors reach their maximum CP at higher tip speed ratios.

9. Take the derivative of Equation 6.18 with respect to a and set equal to 0. Same as problem 4, value is a = 1/3.

Substitute that value in Equation 6.18. Maximum CP = 16/27 = 59% for a lift device.

Numerical solution: calculate 4a(1 - a)2, where a goes from 0 to v0 by tenths. Same as problem 4, except CP is four times larger.

For lift and drag devices the peak of CP occurs at the same value, (wind speed)/3; however, the efficiencies are different. 10. Tip speed ratio = 7, TSR = speed of tip of blade divided by wind speed.

Have to change from rad/s to rpm.

Table of rpm for various radii and rotor tip speeds.

Radius Tip Speed, m/s m 70 140 210

Radius Tip Speed, m/s m 70 140 210

2.5 |
268 |
535 |
803 |

5 |
134 |
268 |
401 |

25 |
27 |
54 |
80 |

50 |
13 |
27 |
40 |

Notice that as radius increases, rpm gets smaller, even though the TSR is the same. As you can see, at high wind speeds, you cannot let the rotor operate at maximum CP, since the rpm would be too large. It would fly apart, unless you had a very strong structure (much more cost). If wind turbine operates at constant rpm (induction generator), the design point of maximum CP determines the rpm.

11. In the final analysis, you want to produce as much energy, kWh, as possible at the lowest cost. The cube of the wind speed histogram is proportional to the energy. So you would probably choose the maximum efficiency at below or near the peak of the energy curve for a site. Use Figure 3.12 as an example. This would be in the range of 8-12 m/s. However, with larger megawatt wind turbines, at higher hub heights, it would be in the range of 10-15 m/s.

Data for problems 12-18: rpm = 65, rated power = 300 kW at 18 m/s, hub height = 50 m, tower head weight = 3,091 kg, mass of one blade = 500 kg; rotor: hub radius = 1.5 m, radius to tip of blade = 12 m.

12. Velocity of the tip, v = (0*r = 6.8 * 12 = 82 m/s.

13. Velocity of the blade root. Since r is small, v will be smaller.

14. For center of blade, v = (0* r = 6.8*6 = 41 m/s.

KE = 0.5 mv2 = 0.5 * 500 * 412 = 4.2 * 105 J If you did it by calculus, would the answer be larger or smaller?

Blade length = 12 m - 1.5 m = 10.5 m Divide into ten sections, each section is 1.05 m long, mass of each section = 50 kg.

50 | |||

Blade |
r |
v |
KE |

Section |
m |
m/s |
J |

1 |
2 |
13.6 |
4,624 |

2 |
3.05 |
20.7 |
10,754 |

3 |
4.1 |
27.9 |
19,432 |

4 |
5.15 |
35.0 |
30,660 |

5 |
6.2 |
42.2 |
44,437 |

6 |
7.25 |
49.3 |
60,762 |

7 |
8.3 |
56.4 |
79,637 |

8 |
9.35 |
63.6 |
101,060 |

9 |
10.4 |
70.7 |
125,033 |

10 |
11.45 |
77.9 |
151,554 |

Sum |
627,954 | ||

KE = |
6.3 *105 J |

16. F/A = 0.5 p v2, let density = 1 kg/m3 F/A = 0.5 (100) = 50 N/m2

Total force = 50 * area = 50 * 3.14 * 122 = 23,000 N.

17. Energy = 800,000 kWh, area = n r2 = 3.14 (12)2 = 452 m2

Specific output = annual energy/area = 800,000/452 = 1,770 kWh/m2.

This is large compared to actual industry values (see Chapter 8). Energy/swept area depends on efficiency of system and also on the wind regime.

18. Output/weight = 800,000 kWh/3,091 kg = 260 kWh/kg

Data for problems 19-25: rpm = 21 = 2.2 rad/s, rated power = 1,000 kW, rated wind speed = 13 m/s, hub height = 60 m, tower head weight = 20,000 kg; rotor: radius to tip of blade (rotor) = 28 m, hub radius = 1.5 m, mass of one blade = 3,000 kg.

19. Velocity of the tip, v = a>* r = 2.2 * 28 = 62 m/s.

20. Velocity of root of blade, v = ®*r = 2.2*1.5 = 3 m/s.

21. For center of blade, v = (0* r = 2.2 * 14 = 31 m/s.

Hub is at 1.5 m, so blade is 28 - 1.5 = 26.5 m long. Divide that into ten sections and calculate for midpoint. Mass of each section = 300 kg. However, for actual blades, mass will change along the blade, which means inner sections would have more mass.

50 | |||

Blade |
r |
v |
KE |

Section |
m |
m/s |
J |

1 |
2.8 |
6.2 |
5,692 |

2 |
5.5 |
12.0 |
21,564 |

3 |
8.1 |
17.8 |
47,633 |

4 |
10.8 |
23.7 |
83,898 |

5 |
13.4 |
29.5 |
130,361 |

6 |
16.1 |
35.3 |
187,019 |

7 |
18.7 |
41.1 |
253,875 |

8 |
21.4 |
47.0 |
330,927 |

9 |
24.0 |
52.8 |
418,176 |

10 |
26.7 |
58.6 |
515,622 |

Sum |
1,994,767 |

KE = 2 * 106 J. With calculus would get better answer.

23. F/A = 0.5 pv2. Let density = 1 kg/m3. F/A = 0.5 (225) = 112.5 Nim2

Total force = 112.5 * area = 112.5 * 3.14 * 282 = 2.8 * 105 N.

Specific output = annual energy/area = 2,800,000/2,460 = 1,100 kWh/m2.

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