Re: Mobile S-Rotor!


"TomGee" <lvlus@xxxxxxxxxxx> wrote in message
news:1133876928.687033.286780@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
> Why, Chris, in the old days, we used downshifting to slow down on hills
> in order to save on using up the brakes. But your comment about being
> useful in crosswinds shows that you don't have a good picture of my
> device yet. It is totally enclosed except for the wind entrances and
> exits. A 4-foot long device will have 4 rotors each 6" in diameter.
> The front of the device is an open 8" that funnels the wind into a 3"
> opening at the rotor the length of the device, allowing it to exit at
> the rear 3" opening. Thus, the already extreme power of a 60 mph wind
> is compressed into an even more powerful packet to produce the torque
> to turn the load.
>

If the intake is 8" diameter, at 88 fps flow, you have an air mass flow rate
of about 2.27 lbm/s (x-section of 0.349 ft^2 * 88 fps 1lbm-air/13.5 ft^3).
If the air initially enters at 88 fps, each lbm of air has a kinetic energy
of 1/2(88^2)/32.2 = 120 ft-lbf. So the rate of energy flowing into the
system is 2.27*120 = 274 ft-lbf/s = 0.5 hp.

But not all of that energy is recoverable. The air must leave the tube
sooner or later, and when it does it will carry away some kinetic energy. A
famous limit known as the 'Betz limit' says you can't extract more than ~59%
of the power from the 'wind', so that means that ideally you could extract
0.29 hp.

The outlet pressure of your tube device would be lower than the inlet
pressure, so it takes a certain force to push this tube through the air at
88 fps. If it takes more than 1.8 lbf to push it through the air (or it
adds more than 1.8 lbf drag to the car), then you have a net loss of energy.
(.29hp = 162 ft-lbf/s; 162 ft-lbf / (88 ft/s) = ~1.8 lbf)

Can you push an 8" wide funnel, with a 4-foot long tube through the air at
60 mph with less than 1.8 lbf?? Or have it add less than 1.8 lbf of drag to
the vehicle??

daestrom


.

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