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## How many blades should a HAWT have?

We know from physics that the power is the work done in the unit time. Unit time being a second, the more work a turbine does in a given second the more power we can extract from it. Lets us ask the following question to ourselves. What is the lowest RPM (Revolution Per Minute) a single bladed turbine can make without losing its efficiency drastically? Please note the emphasize on word lowest. It is 60 RPM or one revolution per second. Why? Because it should sweep whole rotor area in one second. If turbine rotates slower than this, it will miss some air particles and its output power will be low. However, if it rotates faster than one revolution per second it will produce more power which is better. But faster-rotating blades create more stresses on the overall structure. Therefore one bladed turbine should make at least one revolution per second to cover all the rotor surface to produce optimum power. However, if rotates faster, it will produce more power.

What will be RPM of two-bladed turbines to produce optimum power? With same thinking, it would be 30 RPM, that is half revolution per second for the two-bladed turbine. This is because while one blade sweeping one-half of the circle area, the other blade is sweeping the other half. By using the same analogy a three bladed turbine will make 20 RPM (one-third of revolution per second) and a four bladed turbine will make 15 RPM (one-fourth revolution per second). I will STRESS one more time that, if turbines rotate faster then these, up to a certain point, they will be more efficient. For example, if a single bladed turbine making 2 revolutions per second will be more efficient than same turbine making one revolution per second. This is because wherever the blade is during its revolution, the wind is escaping in the opposite end, by turning faster it will catch some of them. But there is an upper limit how fast a blade should turn, if it turns much faster, it would act like a solid surface and stresses will be enormous in the system. Following video shows what happens when a three bladed wind turbine rotation speed exceeds its design limit for a given wind speed.

Now we know that the more blade the turbine has the slower it can rotate to cover the rotor sweep area. Since the rotor rotation also depends on wind speed and TSR, now we know that there is some relationship between wind speed, TSR, and the blade numbers. But we still don't know how many blades we should have for a given wind speed and TSR. The wind speed and TSR are the two constraints to answer this question. The wind speed is the biggest and the most important constraint nature put for the wind turbine and we do not have any control on it. You can build worlds most efficient and powerful turbine and put it in a location where there is no wind, you will get nothing from it. The second constraint is the TSR and PLEASE DO NOT FORGET THAT TSR IS NOT APPLICABLE TO DRAG TYPE VAWT. Unlike to the wind speed which we have no control, TSR constraint is set by the designer of HAWT. The TSR number for a given wind speed is always taken as high as possible so that we extract more power due to high rotation. Please note that for 10m/s wind speed TSR could be 7 or 8 ( I do not know actual number) but say for 30m/s wind speed, TSR could be 1 or less. This is because high wind put a lot of pressure on the system. If we kept TSR at 8, blade tip speed would be 240m/s. I hope you watched the wind turbine destruction video given above. That destruction happened because TSR was high for that wind speed. I assume that this happened because pitch control mechanism failed to adjust TSR to lower values. By changing pitch angle in one direction the blade TSR can be reduced and turning it in the opposite direction TSR can be increased. Now by using the wind speed constraint put by nature and TSR constraint set by the designer we now can find maximum rotor diameter for the given number of blades.

For the argument sake lets build a HAWT which will work with optimum performance for the wind speed in between 10m/s to 20m/s with TSR being 6 ( this is our choice, we as well chose another number such as 5 or any reasonable number ). Since this turbine will work between 10m/s and 20m/s and our turbine does not have pitch control mechanism, we should design the turbine for average wind speed of 15m/s.

```The tip speed of a given  blade can be found seperatly
from two formula given below.

TSR = V_tip_speed/V_wind_speed = 6

V_tip_speed =  6 * V_wind_speed = 6* 15 = 90  (1)

V_tip speed = Angular_velocity * R (2)

and

Angular_velocity = 2*PI*n/60  (3)

where n is being revolution per minute (RPM) and R is the radius of the blade.

By using formula (1), (2) and (3) we can find that

2*R*PI*n/60 = 90 ==> n*R = 860 (approximetly) (4)

```
From our previous discussion we know that for a single bladed turbine lowest RPM is 60.

```Lets find optimum diameter for 1,2,3 and 4 bladed HAWT.

By using the formula (4) and minimum RPM for a given blade number we can conclude that

60*R = 860 ==> R = 14.33m.

30*R = 860 ==> R = 28.66m.

20*R = 860 ==> R = 43m.

And for four bladed turbine it is,

15*R = 860 ==> R = 57.33m.
```

To demonstrate this concept following flash animation for 1,2,3 and 4 bladed HAWT has been made. Please note that from the time you press "Rotate" button, to the time the rotation stops, represent 1 second time interval. Note that during this time interval 1 bladed turbine makes 1 revolution, 2 blade turbine makes half revolution, 3 bladed turbine makes one third and 4 bladed turbine makes quarter rotation. However in each case whole rotor area swept by the blades. It is also important to note that tips of every blade travel a distance of 90m. This is because V_tip_speed is 90m/s and time interval is 1 second. The circumference of each swept area is 90,180,270 and 360 respectively for 1,2,3 and 4 bladed turbines for this design.

Please note that for a 4 bladed turbine it has been said that, optimum diameter should be 57.33m, but we may see 3 bladed turbine rotates in this range. This is because, while due to less area the 3 bladed turbine sweep it might louse some power, however by increased annulus around optimum radius it gains additional power. Since annulus increase by square of the radius gain will be much higher than lost. If the wind speed, TSR, density of the air and efficiency kept constant the only one variable determines the output of the turbine is the area that extract power from the wind. Note that I did not say the rotor sweep area. Assume that we have a 3 bladed turbine operating its optimum radius of 43m. The power output of this turbine will be C*43**2 where C is a constant including in it air density, PI and cube of the wind velocity which we assumed all fixed values.

```Therefore power output of three bladed turbine in its optimum radius will be,

P = 43*43*C = 1849*C Hp.

Now assume that we used same 3 bladed turbine in the optimum diameter
of 4 bladed turbine which is 57.33.

What will be the power output? Is it

P = 57.33*57.33*C = 3286.73*C Hp ?

We think that it will be,

P = 3*3286.73*C/4 = 2465*C Hp.
```

Why? This is because blades are not sweeping whole rotor area. Please look at following animation and figure out what will be the output of one and two bladed turbines if their rotor diameter is taken to be 57.33m.

Hint: For 1 blade power output will be 821.7*C Hp and for two blade turbine it will be 1643.4*C Hp.

What will happen if we keep TSR at 6 and change wind speed to 30m/s? Since wind speed doubled and TSR kept constant, the tip speed of blades will be 180m/s rather than 90m/s. This will cause rotation speed (angular velocity) to double too. Following flash animation demonstrates what happens when the rotation speed is increased.