## Horizontal Axis Wind Turbine Rotor Design

**The Basics:**

Objects in the path of a wind stream experience a downwind force known as drag. Some of the earliest wind turbine designs utilized this force and are called vertical axis wind turbines (VAWT). While drag is easy to harness, VAWT's are limited to performing at much lower rpm's than horizontal axis wind turbines (HAWT) because as the blades rotate, one half of the rotor is always travelling upwind against the force of the wind stream in order to turn the blades into a position to harness the downwind drag again.

Wind also produces another force called lift, which always works at right angles to the wind direction and is the force utilized by HAWT designs.

Like turbines, there are also two main types of wind: clean wind and dirty (turbulent) wind. Clean wind originates from the upper atmosphere and is pulled down to open land and water by temperature and pressure differentials between the atmosphere and the earth's surface. These winds are considered clean because there are no obstructions in the path of the currents to collide with, and thus little to no turbulence is created. Because of the complicated measures involved with HAWT rotor design due to their lift based designs, HAWT's are strictly employed in clean wind environments. Surface currents that are travelling back to the upper atmosphere generally encounter resistance from things like trees and buildings, which slows the currents down and creates a lot of turbulence that can be difficult if not almost impossible for a lift based turbine design to harness. VAWT's are most suitable for such low, turbulent wind environments due to their simple drag based designs.

**Tip Speed Ratio:**

The first step in wind turbine rotor design is to select a tip speed ratio (TSR). The TSR refers to the ratio of the speed of the blade tips to the wind speed. The TSR is important because if the blades spin too slow at a given wind speed, then most of the wind will pass through the rotor with little being harnessed. But if the blades spin too fast, then they will 'blur' and act like a solid wall blocking the wind, commonly known as the 'wind wall effect'. Thus, it's necessary to match the angular momentum of the blades to the wind speed in order to maximize the rotor efficiency. The TSR required for a particular rotor will depend on the average speed and type of wind (clean or dirty) that's to be harnessed in a given location, and is determined by the aerofoil profile type used in the blade design as well as the number of blades. Generally, a high tip speed ratio is used for HAWT's as the high rotational frequency of the rotor shaft is needed for turning an electrical generator. In cases where more torque is desired over speed - such as for water pumping - or to harness low wind speeds, then a low TSR is more suitable. As advantageous as a high TSR is for performance in electrical generation, it can also come with drawbacks such as excessive noise, vibration/potential blade throw, or erosion from air/dust particles and precipitation contacting the leading edge of the blades, so care must be taken when establishing a TSR so that the maximum rotational frequency of the blades doesn't exceed the structural threshold.

**How Many Blades?**

If a small number of blades are used, then the blades must travel faster to maximize the energy harvest and thus rotors with high TSR's also have few blades and perform best in medium-high wind speeds. If a large number of blades are used, then rotational frequency can be slower. However, extra blades also increases the risk of the previously mentioned wind wall effect occurring at lower frequencies, so rotors with low TSR's and a large number of blades are limited to low wind speed applications. In general, a typical HAWT will have 3 blades and a TSR of somewhere between 5 and 7.

**The Wind Power Equation:**

The amount of energy that can be harnessed by a wind turbine rotor ultimately depends on the rotor size or 'swept area', the wind speed, the air density in the wind stream, and the turbine's efficiency. Like a sail, the larger the rotor's swept area is, the more wind energy it can harness. If you double the rotor's swept area, then the subsequent power potential will increase by a factor of 2. If the wind speed doubles, the power potential increases by a factor of 8; rotor size and wind velocity are the two main determining factors for theoretical power potential. An approximation of theoretical power potential from a wind stream at a given speed can be calculated with the following equation:

P = 1/2pAV^3

where:

P = power in watts

p = density of the air in the wind stream in kg per cubic meter

A = rotor swept area in square meters = πr^2

V = wind velocity in meters per second

Example using a 2.4m diameter rotor performing in a 5m/s wind stream with a typical air density of 1.23kg/m^3:

Swept area = πr^2

= 3.1459(1.2 x 1.2)

= 3.1459 x 1.44

= 4.5m^2

Power = 1/2(1.23 x 4.5 x 5^3)

= 1/2(1.23 x 4.5 x 125)

= 1/2 (691.9)

= 345.9 watts

The turbine's overall efficiency also has a major effect on potential power. Like VAWT's, performance from HAWT's is also limited but in a different way, by something called

Betz's law. According to Betz's law, no HAWT can capture more than 59.3% of the kinetic energy in a wind stream. The factor 16/27 (0.593) is known as Betz's coefficient, or the Betz limit. Practical utility-scale wind turbines achieve at peak 75% to 80% of the Betz limit, so a reasonable approximation for an actual coefficient would be around 45%, and is incorporated and expressed in the above power equation as:

P = 1/2ξpAV^3

where ξ = turbine coefficient = 45%

Example:

Power = 1/2(0.45 x 1.23 x 4.5 x 5^3)

= 1/2(0.45 x 1.23 x 4.5 x 125)

= 1/2(311.3)

= 155.7 watts

**Blade Design:**

The most effective approach to blade design would utilize what's called blade element momentum theory (BEMT), which combines blade element theory with momentum theory, and is used to divide the span of a turbine blade into multiple elements to calculate the local forces on each section of the blade at a specific design wind speed, and establish optimal chord width, thickness, and twist distributions accordingly. A simplified calculation method for custom DIY blades using BEMT is detailed in an example below, and a CAD model of the blade design used can be

__downloaded here__.

__8ft Horizontal Axis Wind Turbine Blade Design Using Blade Element Momentum Theory__

Rotor Spec's:Diameter: 8ft, 2.4m Swept area: 50.33ft2, 4.7m2 Tip speed ratio: 6 Blades: 3 Design wind speed: 18kph, 5mps (based on annual local average) Equations:Tip speed ratio per element λi = λ(r/R) Relative wind angle per element φi = 2/3Tan-1(1/ λi) Pitch per element σi = φi – α Chord width per element Ci = (VR2)/(βCLr λ λ) Chord thickness per element Cti = 1/8Ci |
Legend:CL = lift coefficient = ~85% (approximation; actual factor depends on chosen aerofoil profile and determined via wind tunnel testing) Ci = chord width per element Cti = chord thickness per element σi = pitch per element φi = relative wind angle per element λ = tip speed ratio = 6 λi = tip speed ratio per element R = rotor radius = 1.2m r = radius per element V = design wind speed = 18kph, 5mps A = rotor swept area = 4.7m2 α = angle of attack = 5° β = number of blades = 3 |

Element 1:λi = λ(r/R) = 6(0.31/1.2) = 1.6 φi = 2/3Tan-1(1/ λi) = (0.67)Tan-1(0.63) = 21.6° σi = φi - α = 21.6 - 5 = 16.6° Ci = (VR2)/(βCLr λ λ) = (5*1.44)/(3*0.85*0.31*6*6) = 7.2/27.5 = 0.26m, 10 1/4” Cti = 1/8Ci = 1/8*0.26 = 0.033m, 1 5/16” |
Element 2:λi = λ(r/R) = 6(0.38/1.2) = 1.9 φi = 2/3Tan-1(1/ λi) = (0.67)Tan-1(0.53) = 18.7° σi = φi - α = 18.7 - 5 = 13.7° Ci = (VR2)/(βCLr λ λ) = (5*1.44)/(3*0.85*0.38*6*6) = 7.2/38.9 = 0.19m, 7 1/2” Cti = 1/8Ci = 1/8*0.19 = 0.024m, 15/16” |
Element 3:λi = λ(r/R) = 6(0.46/1.2) = 2.3 φi = 2/3Tan-1(1/ λi) = (0.67)Tan-1(0.43) = 15.6° σi = φi - α = 15.6 - 5 = 10.6° Ci = (VR2)/(βCLr λ λ) = (5*1.44)/(3*0.85*0.46*6*6) = 7.2/42.2 = 0.17m, 6 11/16” Cti = 1/8Ci = 1/8*0.17 = 0.021m, 13/16” |
Element 4:λi = λ(r/R) = 6(0.53/1.2) = 2.7 φi = 2/3Tan-1(1/ λi) = (0.67)Tan-1(0.37) = 13.6° σi = φi - α = 13.6 - 5 = 8.6° Ci = (VR2)/(βCLr λ λ) = (5*1.44)/(3*0.85*0.53*6*6) = 7.2/48.7 = 0.15m, 5 7/8” Cti = 1/8Ci = 1/8*0.15 = 0.0188m, 3/4” |

Element 5:λi = λ(r/R) = 6(0.61/1.2) = 3.1 φi = 2/3Tan-1(1/ λi) = (0.67)Tan-1(0.32) = 11.9° σi = φi - α = 11.9 - 5 = 6.9° Ci = (VR2)/(βCLr λ λ) = (5*1.44)/(3*0.85*0.61*6*6) = 7.2/56 = 0.13m, 5 1/8” Cti = 1/8Ci = 1/8*0.13 = 0.0163m, 5/8” |
Element 6:λi = λ(r/R) = 6(0.69/1.2) = 3.5 φi = 2/3Tan-1(1/ λi) = (0.67)Tan-1(0.29) = 10.8° σi = φi - α = 10.8 - 5 = 5.8° Ci = (VR2)/(βCLr λ λ) = (5*1.44)/(3*0.85*0.69*6*6) = 7.2/63.3 = 0.11m, 4 5/16” Cti = 1/8Ci = 1/8*0.11 = 0.0138m, 9/16” |
Element 7:λi = λ(r/R) = 6(0.76/1.2) = 3.8 φi = 2/3Tan-1(1/ λi) = (0.67)Tan-1(0.26) = 9.8° σi = φi - α = 9.8 - 5 = 4.8° Ci = (VR2)/(βCLr λ λ) = (5*1.44)/(3*0.85*0.76*6*6) = 7.2/69.8 = 0.1m, 3 15/16” Cti = 1/8Ci = 1/8*0.1 = 0.0125m, 1/2” |
Element 8:λi = λ(r/R) = 6(0.84/1.2) = 4.2 φi = 2/3Tan-1(1/ λi) = (0.67)Tan-1(0.24) = 9° σi = φi - α = 9 - 5 = 4° Ci = (VR2)/(βCLr λ λ) = (5*1.44)/(3*0.85*0.84*6*6) = 7.2/77.1 = 0.09m, 3 9/16” Cti = 1/8Ci = 1/8*0.09 = 0.0113m, 7/16” |

Element 9:λi = λ(r/R) = 6(0.91/1.2) = 4.6 φi = 2/3Tan-1(1/ λi) = (0.67)Tan-1(0.22) = 8.3° σi = φi - α = 8.3 - 5 = 3.3° Ci = (VR2)/(βCLr λ λ) = (5*1.44)/(3*0.85*0.91*6*6) = 7.2/83.5 = 0.086m, 3 3/8” Cti = 1/8Ci = 1/8*0.086 = 0.0108m, 3/8” |
Element 10:λi = λ(r/R) = 6(0.99/1.2) = 5 φi = 2/3Tan-1(1/ λi) = (0.67)Tan-1(0.2) = 7.6° σi = φi - α = 7.6 - 5 = 2.6° Ci = (VR2)/(βCLr λ λ) = (5*1.44)/(3*0.85*0.99*6*6) = 7.2/90.9 = 0.079m, 3 1/8” Cti = 1/8Ci = 1/8*0.079 = 0.009m, 3/8” |
Element 11:λi = λ(r/R) = 6(1.07/1.2) = 5.4 φi = 2/3Tan-1(1/ λi) = (0.67)Tan-1(0.19) = 7.2° σi = φi - α = 7.2 - 5 = 2.2° Ci = (VR2)/(βCLr λ λ) = (5*1.44)/(3*0.85*1.07*6*6) = 7.2/98.2 = 0.073m, 2 7/8” Cti = 1/8Ci = 1/8*0.073 = 0.009m, 3/8” |
Element 12:λi = λ(r/R) = 6(1.1/1.2) = 5.5 φi = 2/3Tan-1(1/ λi) = (0.67)Tan-1(0.18) = 6.8° σi = φi - α = 6.8 - 5 = 1.8° Ci = (VR2)/(βCLr λ λ) = (5*1.44)/(3*0.85*1.1*6*6) = 7.2/101 = 0.071m, 2 13/16” Cti = 1/8Ci = 1/8*0.071 = 0.0088m, 5/16” |
Element 13:λi = λ(r/R) = 6(1.2/1.2) = 6 φi = 2/3Tan-1(1/ λi) = (0.67)Tan-1(0.17) = 6.5° σi = φi - α = 6.5 - 5 = 1.5° Ci = (VR2)/(βCLr λ λ) = (5*1.44)/(3*0.85*1.2*6*6) = 7.2/110.2 = 0.065m, 2 9/16” Cti = 1/8Ci = 1/8*0.065 = 0.008m, 5/16” |