How to calculate the deflection angle of the rotor disk wake...?

XXavier

Member
Joined
Nov 13, 2006
Messages
1,481
Location
Madrid, Spain
Aircraft
ELA R-100 and Magni M24 autogyros
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913 gyro (June 2023)
In a French forum, I've seen this illustration by Jean Claude:

Captura de pantalla 2020-10-05 a las 10.00.39.png

He compares the case of a circular wing with that of a rotor. For the same diameter and airspeed he shows that the lift is identical if the AoAs are conveniently chosen. He explains how he calculated the deflection angle of the wake for the circular wing –the key magnitude for calculating the lift– but doesn't explain how did he get the wake deflection angle for the rotor disk...

That's what I'm asking...

Or, put in a different way, how did he know that CL = 0,22 for the rotor disk @ 11,5º AoA...?
 
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Xavier,
If CL = 0.22, then induced angle e = 3.1 degree about like an FW.
But for produce the rotation losses, the rotor need an airflow through the disc about S *3.5 m/s
At 25 m/s, this is an extra angle of 3.5 /25= 0.14 rad or 8 degrees.

(S disk *3.5 m/s is depending of the profile losses, but seems a usual value on our rotors)
 
Thanks for your answer...

However, there's a point I would like to have clear. Suppose that I wish to calculate the lift L produced by a given rotor, of certain diameter D at an angle of attack and a relative wind V. In the case that the airflow through the disk was 3,5 m/s and V = 25 m/s, so that 'extra angle for the profile losses' will be 3,5/25 rad.

I would then add that extra angle to the AoA of a flat disk producing the same lift L under the same conditions... But how could I calculate the flat disk AoA for that lift L?

I would need a CL versus AoA curve for the flat disk...
 
I would need a CL versus AoA curve for the flat disk...
You just need added the induced angle = CL /π.A (rad)
"A" is the aspect ratio = b²/S = 1.27 for a disk

So, if CL is 0.22 , induced angle is 0.055 rad or 3.16 degrees .
Behind the rotor, the wake deflection is twice as much ie 6.32 degrees, as showed in my sketch
 
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You just need added the induced angle = CL /π.A (rad)
"A" is the aspect ratio = b²/S = 1.27 for a disk

So, if CL is 0.22 , induced angle is 0.055 rad or 3.16 degrees .
Behind the rotor, the wake deflection is twice as much ie 6.32 degrees, as showed in my sketch


Thanks again... I still don't know yet how to derive the CL of the flat disk for any AoA, but I'll try to find a chart for that...
 
Xavier,
(dCL/di)A = (dCL/di) /[(1+ (dCL/di) /πA)]


Since (dCL/di) ~ 5.7/ rad, then dCL/di)1.27 ~ 5.7/ (1+5.7/4) = 2.35/rad


So, for CL = 0.22, the disk need about 5.3 degrees attack angle
 
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Thanks for your help...

I was obsessed with the need of a chart of CL vs. alpha, and now I see that there's no need for that, and I've been able to replicate your example of the flat disk by assuming a stream tube of the same diameter (7m, r=3,5) as the rotor, v = 25 m/s, rho = 1,2 so that the
deflection angle d = arc sin 3175/(25^2 x 3,14 x 3.5^2 x 1,2) = 6,32º

Besides, and 'from the books', I know that sin d = 4 sin AoA/(2+A)

So, sin 6,32º = 4 sin AoA/3,27 => AoA = 5,15º

That, for the flat disk, according to your sketch. Now, to account for the rotation losses, the rotor needs an extra flow. In order to fit the figures of your sketch, (AoA = 11,5º) I would need an additional 6,35º of AoA. Those extra 6,35º would mean an extra flow of 2,76 m/s, since arc sin 2,76/25 = 6,35º

So, I have now all the figures for the disk and rotor as in your sketch: In order to have the deflection angle of 6,32º, the AoA is, for the rotor, 5,15 + 6,35 = 11,5º.

Now, I hope you are patient enough for an additional question: can this calculation method be used for other values of lift and airspeed?

Thanks in advance...
 
I do not agree with your calculation principle, Xavier.

Yes, the deflection is the same for the flat disk and for the rotor (3.16° at the middle of wing). Thus, to calculate the A.o.A of tip path plane, you just have to add to that the angle corresponding to the flow through the rotor (ASIN 3.5 /25) = 8.05°.
Hence tip path plane A.o.A is 8.05 + 3.16= 11.2°.
This calculation method be used for other values of lift and airspeed, as long as A.o.A no exceeds about 20°, because beyond that, the lift starts to come due to the blocking rather than only the deflection of the airflow


dCL/di of the flat disk being unrelated to the operation of the rotor, it does not seem correct to me to add the AoA of the flat disk to the angle of an false airflow ie ASIN (2.7/25)
 
I do not agree with your calculation principle, Xavier.

Yes, the deflection is the same for the flat disk and for the rotor (3.16° at the middle of wing). Thus, to calculate the A.o.A of tip path plane, you just have to add to that the angle corresponding to the flow through the rotor (ASIN 3.5 /25) = 8.05°.
Hence tip path plane A.o.A is 8.05 + 3.16= 11.2°.
This calculation method be used for other values of lift and airspeed, as long as A.o.A no exceeds about 20°, because beyond that, the lift starts to come due to the blocking rather than only the deflection of the airflow


dCL/di of the flat disk being unrelated to the operation of the rotor, it does not seem correct to me to add the AoA of the flat disk to the angle of an false airflow ie ASIN (2.7/25)


I see...

For the same airspeed, and for angles of attack of less than 20º, is there any general rule to estimate the increment of AoA of the rotor disk with respect to the AoA of a circular disk of the same dimensions generating the same amount of lift...?

Perhaps taking half of the total deflection in the case of the disk (in the example, 6,32º/2 = 3,16º), as you point out, and then adding asin (3,5/airspeed)...
And that 3,5 m/s, is it valid for all cases, for any airspeed & rotor diameter, (always below 20º AoA)...?



Thanks for your patience...
 
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Xavier, the airflow S* 3.5 m/s through the rotor seems quite valid for our gyros usually loaded around 7 kg/m2.
The forward speed has practically no effect on this airflow, since it has almost no influence on the rpm.

It is the friction of air on the blades due to the rpm that requires this airflow: With no friction, no flow needs to pass through.
 
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Xavier, the airflow S* 3.5 m/s through the rotor seems quite valid for our gyros usually loaded around 7 kg/m2.
The forward speed has practically no effect on this airflow, since it has almost no influence on the rpm.

It is the friction of air on the blades due to the rpm that requires this airflow: With no friction, no flow needs to pass through.

Many thanks for your attention to my queries. I really appreciate your explanations...
 
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