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Wind force on rotor blades
By johnlm, Section Homebrewed Electricity Posted on Mon Apr 26th, 2004 at 10:14:10 AM MST
Calculating the wind force on the face of a prop.

After reading many of the postings on this site it seems apparent that folks would like to be able to get an idea of the force the wind puts on the face of the rotating blades. This is the force that will cause the blades to furl (or just plain try to push the whole assembly downwind). I copied this info during some research about 25 yrs ago and am not sure of its origin. I hope the formula formatting is understandable.
P=(1/2 Cd p A v^2 Fdl) / Gc
where:
P= Pounds of force on the rotor
Cd= Drag Coefficient = 1.57
p= air density = 0.076 Lbm/ft^3
A= Area swept by blade
v= wind velocity in ft/sec
Fdl= Dynamic load factor = 1.4
Gc= sec^2/Lbm-ft (coefficient of gravity?) = 32.2
The equation boils down to:
P= 0.0026 X (Area) X (wind velocity ^2)
I used this eq. to determine and set the furling point of a 5 ft dia prop (TSR=12) and it seems to be in the ballpark. One would think there should be some included factors for TSR as it seems likely a slow (low TSR) rotating blade would interfere less wind than a faster (high TSR) rotating blade. I need to do more experiments with this but as I said it seems to give reasonable values.
johnlm

Wind force on rotor blades | 3 comments (3 topical, 0 editorial)

Re: Wind force on rotor blades (none / 0) (#1)
by DanB (danb@*no spam*otherpower.com) on Mon Apr 26th, 2004 at 01:02:13 PM MST
(User Info) http://www.otherpower.com/

For me this is most interesting when considering loads on the tower.

Re: Wind force on rotor blades (none / 0) (#2)
by finnsawyer on Wed Apr 28th, 2004 at 10:14:21 AM MST
(User Info)

This is curious as it does not involve the lift coefficient Cl. One would expect that most of the axial force on the prop would be due to the component of lift in the direction of the incident wind. Conversely, only a small part of the drag is in that direction. Most of it is in the direction of rotation (tangential). What gives with this equation? Is it a catchall?
GeoM

Re: Wind force on rotor blades (none / 0) (#3)
by johnlm on Wed Apr 28th, 2004 at 11:23:48 AM MST
(User Info)

GeoM,
It might be a "catch all" as you mentioned. As I said, I just copied it out of a document I found many years ago that was giving info on wind loading on windmills and towers. It did not explain how the formula was derived. Im not exactly sure what makes up the "dynamic load factor" constant in the eq. I had ran across several comments on this board where people were stating they did not know how much force the wind loading was on the machine and I remembered I had this so I dug it out and ran some calculations on my 5 ft machine and built a tilt back (spring loaded) high wind furling capability into it and selected the spring force using this info. As I said it seems to give reasonable values as the head assembly would start tipping back at around the wind speed (estimated from the rotational speed measured on the prop) I had set it up to furl. The system I have up now uses a 1.25 in dia pipe for the tower so there is not much drag caused by the tower and most of the force is from the blades so I thought about connecting a fish scale type measuring scale (rated to 50 lbs or so) to the guy wire directly up wind and doing the trig calculations from the measured results to see what kind of real forces are present to compare to the values the eq gives. Note: I would do this on a day when high wind speed are not present.
As Dan B mentioned it is also useful in getting some kind of idea as to the loading forces pushing on the top of the tower, thus helping one design the tower and guying system to be capable of handleing the load.
johnlm

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Wind force on rotor blades | 3 comments (3 topical, 0 editorial)

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