Thread: Alternator Mods
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Old 07-16-2009, 02:14 PM   #48 (permalink)
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Quote:
Originally Posted by metalshark View Post
Im not saying Ive done it but kr4ua said he did. like I said before I would love to see pictures of it working.

It sounds like it would be plausible at speed over 50 with a big enough turbine.

I once had an idea of an alternator driven by exhaust force.. using a turbo exhaust housing and make custom fittings for an alternator to be place at a cool distance.
What would actually happen is that (possibly unless it's in a place of high turbulence already) the blades of the turbine or cage fan would just increase the overall drag coefficient, which would make the speed at which drag becomes an issue even lower than ~45mph.

The exhaust driven turbine-alternator has been done by a company... we're awaiting mass production or ultimate failure.

Regarding the wind-driven alternator -

Here's the formula you need to determine that it will or won't work:

P = 0.5 x rho x A x Cp x V3 x Ng x Nb

where:
P = power in watts (746 watts = 1 hp) (1,000 watts = 1 kilowatt)
rho = air density (about 1.225 kg/m3 at sea level, less higher up)
A = rotor swept area, exposed to the wind (m2)
Cp = Coefficient of performance (.59 {Betz limit} is the maximum thoretically possible, .35 for a good design)
V = wind speed in meters/sec (20 mph = 9 m/s)
Ng = generator efficiency (50% for car alternator, 80% or possibly more for a permanent magnet generator or grid-connected induction generator)
Nb = gearbox/bearings efficiency (depends, could be as high as 95% if good)


Assuming the best possible circumstances, and 100% of the Betz Limit -

(Work the formula from left to right)

0.5*RHO(1.225) = .6125
(That foglight hole isn't even .5M^2, but we'll use it anyway, since there are turbine designs that have more than "face value" for their area)
.6125*A(.5M^2) = .6125*0.25 = 0.153125

0.153125 * Cp(.59) (Using Betz Limit) = 0.09034375

0.09034375*V3(WindSpeed, or Vehicle speed, in this case, in M/S) (45mph = 20.1168 meters / second) (20.1168 meters / second * 3 = 60.4m/s

0.09034375*60.4 = 5.4567625

5.4567625*Ng (we'll assume the alternator is replaced with a permanent magnet generator, at 80% efficiency)

5.4567625*0.8 = 4.36541

4.36541*Nb (again, assume space age efficiency here)

4.36541*0.95 = 4.1471395

So P = 4.1471395, or about 4.15 watts.

4.15 WATTS!, and that's assuming the best case scenario.

I'm reasonably satisfied to say that this claim is COMPLETELY false, unless I've overlooked something.
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