Friends,
Responding to myself:
David House wrote:
> One can assume a standard sea level air density (0.0024 slugs per
> cubic foot), in which case the equation becomes 0.0001423 AV^3 , where
> A is expressed in square feet and V is in MPH. This results in a
> figure for instantaneous power in watts. Area is of course pi times
> the radius squared.
My math is wrong there. I was taking information from something I had
written 30 years ago ("Wind and Windspinners", p. 99), and 0.0001423
AV^3 assumes an efficiency of 20% in the turbine. One hundred percent
efficiency would be 0.0012 AV^3 .
My apologies for the error. The basic point remains, which is that there
is not much power in low winds to extract, and thus very modest reasons
for trying to do so.
d.
--
David William House
"Make no search for water. But find thirst,
And water from the very ground will burst."
(Rumi, a Persian mystic poet, quoted in /Delight of Hearts/, p. 77)
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