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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.

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    $\begingroup$ 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. $\endgroup$
    – Robert DiGiovanni
    Commented Jan 2, 2021 at 17:58

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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 $\frac{m}{s}$).

Propeller with velocities

(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:

Speed triangles

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)

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  • $\begingroup$ If someone could enlighten me, how to place the pictures centered, that would be great. $\endgroup$
    – U_flow
    Commented May 6, 2021 at 12:27
  • $\begingroup$ Sorry, no advice about centering pictures besides stuffing them with empty space on the left and then trial and error, but on a propeller the centrifugal forces increase with radius. This lets the slowed-down boundary layer accelerate outwards so it is much harder to achieve flow separation on the outer sections of a propeller than it is on a wing. $\endgroup$
    – Peter Kämpf
    Commented Feb 4, 2022 at 5:53
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$$ \arctan \left( 2 \pi \times n \times \frac{r}{v} \right) $$

  • 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 $$ \tan \left( \frac{\omega \times r}{v} \right) $$

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    $\begingroup$ 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. $\endgroup$
    – Ralph J ♦
    Commented Jan 4, 2021 at 15:56
  • $\begingroup$ actually, I think you mean $atan$ instead of $tan$ $\endgroup$
    – U_flow
    Commented May 6, 2021 at 12:24
  • $\begingroup$ Cotangent instead of tangent... $\endgroup$
    – xxavier
    Commented Jan 31, 2022 at 16:50
  • $\begingroup$ atan instead of tan, well, I tried to explain it in the text, but it's true that the redaction can be improved $\endgroup$
    – Fern
    Commented Jun 13, 2022 at 14:16
  • $\begingroup$ contagent instead of tagen, no, it's tangent $\endgroup$
    – Fern
    Commented Jun 13, 2022 at 14:19

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