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Airfoil-missing losses
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15
comments (15 topical)
Re: Airfoil-missing losses
(
3.00 / 0
) (
#15
)
by
finnsawyer
on Wed Nov 30, 2005 at 09:19:04 AM MST
Given your simplistic answer, I will take a stab at it. It is well accepted that off of the trailing edge of an airplane wing the air has a downward flowing component (your vector direction change?). As such, it can be resolved into a component perpendicular to the wing and one parallel. Now an airplane wing is not a good model for the windmill blade, as it is being dragged through the air. The primary force is drag, the secondary force is lift. In the case of a windmill blade the primary force is lift, the secondary force is drag. A better model for the windmill blade would be an airfoil stationary in a wind tunnel. If there is a change in air velocity parallel to the blade it would show up as a reduction, not an increase, due to drag. In the case of a blade assembly in a tube, then this backwash can be resolved into a component parallel to the incident wind and a component in the direction of the blade rotation, which would cause a vortex in the tube. There can be no component perpendicular to the tube walls (the blade assembly completely fills the tube). The vortex, if it exists, would quickly disappear in the tube as it encounters the tube walls. In free air the vortex would expand as it moves downstream until it disappears. The energy in the vortex would be energy lost. There may, in fact, be no vortex. As Dr. Feynman implies fluids do not always do as we expect. If that were the case, then there would be no change in the velocity vector's direction. Perhaps the "Newtonian Principle" can shed some light on this. But, as it happens, the drag coefficient accounts well for losses, while the lift coefficient accounts for the usable power from the blade assembly. This is why I recommend anyone involved in windmill blade design use the lift and drag coefficients.
GeoM
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Airfoil-missing losses
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15
comments (15 topical)
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