# How would flow angle change along the length of a propeller airfoil?

Let's assume I have:

• windspeed of 15 m/s
• blade length of 10 m
• rotational speed of one revolution per second

Can someone please help me calculating how the flow angle would change along the length of an airfoil? I have been trying to do the process for the last couple of days but I don't know how to do it.

• A turbine with 10 m blades scribes roughly a 60 meter circle (pi x 2 x 10). The tips would be going 60 meters per second, the hub, much less. Holding the 15 m/s wind as constant, you should be able to determine relative wind with vector triangles. Remember to include blade AOA, which is typically less near the tips. – Robert DiGiovanni Jan 2 at 17:58

In order to explain the situation, I have drawn a couple of pictures.

Consider a propeller rotating with the rotational velocity $$\omega$$ (in $$\frac{rad}{s}$$), which is flying at a velocity of $$u$$ (in $$\frac{m}{s}$$). (Please excuse my crude drawing skills...). If we cut the propeller at four places along the radius $$r$$, we get the following velocity triangles (at the respective positions 1, 2, 3 and 4). Note that in the following we look at the propeller from the tip in direction of the hub: Note that the tangential velocity at each crosssection rises, however the forward velocity stays the same! The tangential velocity can be simply calculated by the factor $$\omega \cdot r$$.

Now everything which is left to do is to calculate the angle $$\alpha$$, which can be simply done by taking the $$atan()$$ function (btw, I believe that Fern got this point wrong)

$$\alpha = atan(\frac{u}{\omega \cdot r})$$

Therefore in your specific case with $$n=1 \frac{rev}{s}$$ we get:

$$\alpha = atan(\frac{15}{2\cdot \pi \cdot r})$$

with $$r$$ going from 0 to 10 meters (in order to remain practical from 0.1 to 10 meters)

• If someone could enlighten me, how to place the pictures centered, that would be great. – U_flow May 6 at 12:27

$$\arctan \left( 2 \pi \times n \times \frac{r}{v} \right)$$

• n is the rotational speed in rps
• v is the wind speed in m/s
• r is the radius in m

The relative wind seen by the blade airfoil has 2 components:

• on in the direction of the wind, which value is the actual wind speed $$v$$
• perpendicular to the wind due to the motion of the blade, which value is the lineal speed of the blade section: $$\omega \times r$$

So, the angle along the blade will be the one with tangent $$\tan \left( \frac{\omega \times r}{v} \right)$$

• Can you explain why this is the correct answer? A "magic equation" in isolation doesn't help the OP understand how his problem can be solved, or what the limitations are on this solution. – Ralph J Jan 4 at 15:56
• actually, I think you mean $atan$ instead of $tan$ – U_flow May 6 at 12:24