# Do 3 blade wind turbines generate more power than a 9 blade, all else being equal?

After doing some research, I have come to the realization that 3 blade design horizontal turbines have the greatest maximum coefficient of performance. Most sources seem to agree a Cp of ~0.45 at TSR=7 is optimal for 3 blade designs.

The power output of a turbine is P=0.5*CprhoA*V^3. So what I cannot wrap my head around is why, all other things being equal, more blades (of course up to a limit) does not equal a greater coefficient of performance and thus more power. How is only 3 blades this limit on something 200m in diameter?

It seems there would be a significant portion of the air passing freely undisturbed through the rotor.

I welcome any answers and explanations, but preferably backed by formulas.

• This video youtube.com/watch?v=1DOOMpBdj8c shows the vortex patterns from a full scale wind turbine. It gives a good idea how close together the wake patterns and vortices from each blade really are. In simple terms, there isn't enough "air space" to fit in any more blades, without the flow around the each blade interfering with each others. Commented May 30, 2019 at 16:24
• You can have wind machines with many more than 9 blades, but they are high torque low speed machines compared to high speed low torque machine with 3 blades - 2 bladed were popular for a while but the noise and vibration generated as the blades passed the tower was an issue. Commented May 30, 2019 at 18:11
• @SolarMike Would you describe a 5MW 3-blade turbine running at 5 or 10 RPM "high speed low torque?" And the basic problem with any 2-blade design is that the dynamics are a nightmare when you try to yaw the rotation axis while it is running. The moment of inertia resisting the yaw force goes from huge when the blades are horizontal to zilch when they are vertical. For any number of blades > 2 the moment of inertia is independent of the blade positions. Commented May 30, 2019 at 19:54
• @alephzero and the single blade turbines... Commented May 30, 2019 at 19:57
• @alephzero Your description above is based on a stationary rotor. Gyroscopic forces rule here. Page 20, formula 3.2, of Yaw Dynamics of Horizontal Axis Wind Turbines: Final Report gives the formula for yaw motion of a two-bladed HAWT. It's a bit more involved than you might expect. And that equation only considers the flap torque and not any of the net force imbalances. Gyroscopically, you have it backwards. The required torque would be less at the 90 - 270 than at the 0 - 180 position. Gyroscopes are weird that way. Commented May 31, 2019 at 1:58

Because in the real world, all else is never equal, but let's start there analytically.

## What does "all else being equal" actually mean?

When we add in an actual vortex model, the potential jumps in the flow change from a single jump at the disk to spiral jumps in the wake trailing behind each blade. These two models are highly compatible. Using a complex Fourier expansion of the vortex flow, you can separate the induced flow into a constant induced flow and a cyclic induced flow. Interestingly, the proportion of these two are constant radially, and they always have the same vector direction. This is why the actuator disk model appears to overperform it's rather pathetic assumptions, as does lifting line theory applied to modest AR wings, and Michell's integral applied to the wakes of not-so-thin ships. DTIC vortex rotor model evaluated with complex Fourier Analysis, 1964 The takeaway is that TSR needs to be matched to the blade number, and it has to be reduced for greater blade numbers. This combined with the result from the above paragraph puts a downward trend on best possible performance as the blade number increases. This performance limit vs TSR curve is called the Schmitz whirlpool power coefficient. See page 7 here - https://mragheb.com/NPRE%20475%20Wind%20Power%20Systems/Optimal%20Rotor%20Tip%20Speed%20Ratio.pdf (This result probably isn't the same as the power coefficient formula in the first link, but I'm not certain.)

## So what is missing from Momentum theory?

1. Aerodynamic drag

2. Tip losses due to the real vorticies not behaving like the highly concentrated ideal tip vortices.

3. Structural constraints related to performance.

4. Economic constraints and "rest of system" interfaces.

The aerodynamic drag limits HAWTs to TSRs best suited to at least three blades. The best two-blade designs don't quite match the best 3 blade designs because they can't quite run at their optimum speed due to excessive drag. Here I'm taking about skin friction and profile drag. These are the components of basic 2D airfoil L/D ratios. Vortex drag is handled next.

Very early on researchers realized that wing tips didn't work as well as the simple models predicted they should. Glauert and Prandl each had tip correction factor formulae. These favor more blades, not fewer, in an all else being equal scenario. But for high aspect ratio designs, the differences are quite small.

The biggest problem is structural. If you change from a three bladed to a four bladed rotor, you have to run a bit lower TSR to maximize performance. This means a higher torque at the same power. So the rotor ends up using more material with more blades. Also, the profile drag total is higher because you have to support that extra torque with thickness. And the blade area total has to be larger because with lower TSR, the apparent wind speed is less. So you don't save much on aero drag even though it is running slower (but this part can get tricky due to the complicated details of the geometry differences).

Rest of system issues - for rotors of any respectable size, you don't want to have to deal with slower. You already have gearboxes with 30 ton gears in two stage gearboxes. There are huge losses in gearing the rotor rpm up to a suitable generator rpm. This is one of the reasons why three-blade rotors are designed to run a bit quicker than the calculated aerodynamic optimum. It's a bigger problem for the large machines, but still an issue for kilowatt sized units. To save on gearbox cost and losses, you want to run at the fast end of the flattish Cp vs TSR area. So you can trade a bit of aero Cp off for a more manageable and efficient gearbox. Gearbox losses are typically proportional to the maximum torque rating of the gearbox, so looking at this a different way, you'd need a lower cutout speed with the higher blade count and the same gearbox torque rating.