re: have any try to wind power generator system out of car altenator and would u please...
30 apr 2001
"steve spence" wrote:
>car alternators require a fairly high rpm in order to generate reasonable
>output. wind's lower rpm characteristics aren't a good match. gearing (with
>belts or chain) would have to be used, and contributes high friction losses.
that seems like an illogical answer, since this "low rpm" problem gets
_worse_ with larger wind turbines... and belts and chains seem at least
80% efficient (10 hp is 25k btu/h, but the only time we see high temps
or smoke from a riding mower belt is when it is stuck and the pulley is
moving.) what's wrong with 80% efficiency? a $35 datsun rear end with the
spider gears welded together (and some oil) might make a more efficient
4.5:1 right angle stepup drive (as a base for a vertical-axis design...)
>there have been many posts asking basically the same question about using
>an automotive alternator. why is it that no one gives any straight answers
>to the question, instead saying i.e.: "go read ?book/manual", "they are
it would be nice to see more logic around here, vs saying "go see the
website" or repeating some story. people have had trouble using auto
alternators with inadequate controls that waste power keeping the field
energized in low winds, but there are many ways to do something badly,
and clever people with pics and welders might do it better. alternators
aren't that efficient (60%?), but $/kwh seems more important, since the
fuel is free. junkyard alternators seem cheap and durable. we might pay
$10 for 3000 hours at 1000 rpm for 10k miles/year x 10 years at 30 mph.
>let's use a 50-60 amp, 12 volt alternator as an example as this is about
>the most common. are there any assumptions or "best bets" that can be
>made in regards to blade diameter, number of blades, using a straight
>blade with rough clark-y airfoil, fixed pitch mounting, etc?
ok. say the alternator makes 60 amps at 12 v (720 w) at 1000 rpm, and
a 10 mph wind has 0.0051x10^3 = 5 w/ft^2, or 3, after the betz limit.
then gathering 720 w requires 240 ft^2, or one 17.5' diameter blade in
a 55' circumference, and a low-solidity clarke y (with a flat surface
on one side) might have a 6:1 tip speed ratio, so the tips move at 60
mph (5280 feet per minute) in a 10 mph wind, and it rotates at 5280/55
= 96 rpm, and the alternator makes 96/1000x12 = 1.15 volts at 60 a, ie
69 watts. not much... and it's hard to use 1.15 volts. we might charge
one battery with another in series with the alternator.
we could match the available alternator power with the swept area... if
a d foot blade collects 3pid^2/4 = 2.36d^2 watts of wind and rotates at
5280/(pid) rpm, and the alternator provides 5280x12x60/(1000pid) watts,
d = (5280x12x60/(2360pi))^1/3 = 8', with a 150 watt output. not bad.
the blade might be a pressure treated 2x6 or 2x8 or some foamboard cut
with a hot wire and covered with fiberglass.
we might get more power by mounting a row of these along the ridge of
a house roof, and put them in series for a higher voltage. or use a wind
sock or a kite that funnels air into the prop, or make a multi-pinwheel
or ferris wheel or vertical-axis version with several alternators with 8'
blades and the datsun rear end at the bottom. slip ring assemblies seem
expensive. can we bore the shaft of an alternator to make a rotating