[TowerTalk] mid-point loaded beam elements
Jim Lux
jimlux at earthlink.net
Mon Apr 7 12:32:18 EDT 2003
>
>
>I'm wondering if it would be feasible to enclose such coils in a plastic
>enclosure. The material would have to be reasonably UV-resistant, strong,
>somewhat flexible, and fairly easy to fabricate. The insulated gap in the
>elements would be fiberglass rod, and the element would be supported just
>outside of the loading coils by a Phillystran truss, but experience with
>this antenna in a linear-loaded form suggests that a good deal of element
>movement and flexing is inevitable.
>
>I'm aware, of course, that the wind area of the coil covers will introduce
>some additional load on the element. Have I missed other mechanical
>issues? Any suggestions for a good coil cover material?
The wind loading might be less with a suitable cover over the coil than
just the coil hanging in the wind. Small wires are quite draggy, compared
to a streamlined strut. A working Cd for a small wire is around 1.2, while
for a streamlined strut it could be as low as 0.3. Even a round tube,
because it's bigger, might have a lower Cd than the wires that it
encloses. A coil of wire/tubing with the space between wires comparable to
the wire diameter is going to be quite high drag, in proportion to the
actual cross sectional area, largely because of the turbulence and
interaction of the flows around all the wires.
I have a handbook (Blevins, "Applied Fluid Dynamics Handbook") that gives a
general equation for drag of a grille of wires with porosity alpha (i.e.
open area/total area) as
Drag force = Area * 1/2*density*V^2 * K
where K = beta*(1-alpha^2)/alpha^2
for 0.05< alpha < .8 (which seems cover the range for lots of coils)
beta varies from 1.3 down to 0.52 depending on Reynolds number (from 20 to
400). For AWG10 wire in a 10 m/sec wind (22 mi/hr), Re is about 1700, so
you'd use the 0.52 value for beta..
Running some sample numbers.. assume you want to compare two situations.. a
12" long coil that is 4" in diameter, made from AWG10 wire, spaced 0.2
inches apart (i.e. a 0.1 inch gap between windings). or the same coil
covered by a smooth cylindrical surface.
For the coil, we'll assume it's the same as two grilles that are 4"x12",
with 50% porosity.. K = 0.52 * (1-0.5^2)/(0.5^2) >> K=1.56
So, the drag force will be: Area*1/2*density*V^2 * (2 * 1.56) (because
there are two grilles..front and back side of the coil)
Now, the round tube... Reynolds number is 40 times higher (because you're
working with 4" tube instead of 0.1" wire) at around 70,000 (again, for
10m/sec wind). It turns out that this is about the worst possible
situation, because the Cd of the cylinder is 1.2 at Re=100,000 (and drops
to 0.4 at Re=1E6).. A slightly larger tube or a faster wind might actually
greatly reduce the wind drag.
Drag = area * 1/2*density*v^2 * 1.2
In both cases, the first part of the expression (area*1/2*density*V^2) is
the same..
The proportionality constant is 1.2 for the big round tube compared to 3.12
for the grille... and, if you make the tube a bit bigger in diameter (or
the wind is a bit faster), then even though the area is greater, the Cd is
much less..
Take home message: Tiny wires have huge drag, compared to their cross
sectional area! Airplanes use struts instead of wires for bracing for a
good reason.
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