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William James and Essays on Pragmatism | 13 comments (13 topical)

Re: Newtonian Mechanics Of Wind Energy (3.00 / 0) (#8)
by finnsawyer on Mon Jan 02, 2006 at 09:33:39 AM MST

So, you're coming around. No one is saying you can't use Newtonian Mechanics. But if you do, you need to make an honest effort, and that includes using Newton's Calculus. As you like to point out, the velocity of a mass is a vector quantity. The time rate of change of the velocity is the acceleration and is itself a vector quantity. This is the acceleration you need in your force equation. The acceleration vector can point in any direction in space depending on the forces involved. As such, it can be resolved in a component at right angles to the velocity vector and a component parallel to the velocity vector. The component at right angles to the velocity does not change the magnitude of the velocity. If that were the only component of the acceleration it would mean the energy of the mass would not change (0.5xMxV^2 doesn't change; V is the magnitude of the velocity vector). Only the direction of the velocity vector would change. The component of acceleration parallel to the velocity vector causes the mass to either speed up or slow down and is what allows energy to be added to or extracted from the mass. In the case of an air flow which is distributed in space you need to break the air flow into small units of mass and consider how each one contributes to the overall effect. You need to integrate all these little effects over the entire air mass. You have a great deal of work to do.

As far as your "bent air" equation is concerned, your coefficient C is conceptually the same as the lift coefficient Cl, as it must be determined for each air foil. The equation itself describes the "average" behavior of the airfoil, and you hope C will prove to be a constant over some useful range of angles just as Cl is. In fact, this is beginning to look more and more like the lift equation rehashed. I wouldn't be surprised if your C was related to Cl by a constant multiplier. Deductive reasoning? When it leads right back to where we already are?

In the real world an integral part of the behavior of air foils involves drag. You don't consider it. Not doing so means any results you obtain will overstate the efficiency of the systems.
GeoM
[ Parent ]

Re: Newtonian Mechanics Of Wind Energy (3.00 / 0) (#9)
by finnsawyer on Tue Jan 03, 2006 at 09:07:22 AM MST

I took another look at his equation. His force is not the lift force but is shown opposite to the drag force. So, C is related to the drag coefficient Cd? Well, not quite. Drag is pretty much constant over a small range of the angle he calls alpha. The problem is that 1 - cos(alpha) goes as (alpha)^2 for small values of alpha. In other words his force goes to zero dead on, which is is not supported by physical evidence. It's time to put this "theory" in the garbage bin where it belongs.
GeoM
[ Parent ]

William James and Essays on Pragmatism | 13 comments (13 topical)

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