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re: wind from the sun
4 apr 2002
>...i did the calculation below a year ago, but looking again,
>it's badly flawed. anyone see how to fix it?
>here's a naive efficiency estimate for these "helio-aero-gravity" towers,
>huge chimneys with wind turbines at the top surrounded by acres and acres
>of greenhouses in deserts to make a "solar wind" up the chimney...
>suppose we start with our old friend the empirical chimney formula,
>cfm = 16.6 av sqrt(hdt), where av is the vent area at the top and bottom, 
>                         h is the height difference in feet, and
>			 dt is the (f) temperature difference. 
>the airstream's heat power in btu/h is roughly cfmdt = 16.6h^0.5dt^1.5
>for a 1 ft^2 chimney. with a constant power heat source (the acres and
>acres of greenhouses), this equation naively implies that the taller
>the chimney, the smaller the temperature difference and the larger the
>air velocity. smaller temperature differences are better, representing
>less waste heat. we might think of chimneys as transformers that increase
>the air velocity of a heat source and reduce the temperature difference.
>for a 1 ft^2 chimney, the power density is 16.6h^0.5dt^1.5 btu/h-ft^2,
>and the air velocity v = 0.1886h^0.5dt^0.5 mph.
>paul gipe's wind power book says wind power density is 0.05472v^3 w/m^2,
>where v is in mph. he says the best rotors achieve 40% efficiency (vs the
>60% betz limit, which may not apply for a chimney)... 90% efficiencies
>for the transmission, generator, and power conversion make the wind power
>density 0.01596v^3 w/m^2 or 0.00506v^3 btu/h-ft^2. 
>so the heat-to-electrical power conversion efficiency of the chimney
>is e = 100x0.00506(0.1886h^0.5dt^1.5)^3/(16.6h^0.5dt^0.5) = 0.0002h %,

there's the flaw: (dt^1.5)^3/dt^0.5 = dt^4, not 1.

for a 1 ft^2 chimney with a heatflow q btu/h on the order of
3412 btu/h-m^2, ie 1 kw/m^2...

q = cfmdt = 16.6sqrt(h)dt^1.5, so dt = (q/16.6h^0.5)^0.67, and 
cfm = 16.6h^0.5(q/((16.6h^0.5))^0.33 = (16.6h^0.5)^0.67q^0.33
= 6.51q^0.33h^0.33. v = cfm/1ft^2 fpm = cfm/88 mph makes 0.00506v^3
= 2.05x10^-6qh btu/h of electrical power, so the heat to power
conversion efficiency of the chimney is e = 100x2.05x10^-6qh/q
= 0.0002h %. hmmm. same numerical answer...

>where h is the chimney height in feet. e = 0.002% for a 10' chimney, 
>0.02% for 100', 0.2% for 1000', and 2% for 10,000'.

assuming i did it right this time...


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