freefall51

Events in Russia Could Push U.S. Toward a Clearer Energy Policy

Quick estimate what kind of sunroof it takes to power my 180 hp Explorer:

Assume 0.76 kW/hp
Sun is shining 60 deg on suncollector, efficiency following cosinus function cos (90-60) deg = cos (30 deg) = 0.866
1200 KW /m2
Sun/Power Conversion Efficiency 27 %
Inverter efficiency 75 %
Electro Motor efficiency 90 %.


Here we go

180 hp * 0.76 kW/ hp
----------------------... = 722 m2
(1.2 kW/ m2 * 0.27 * 0.75 * 0.90 * 0.866)

The sunroof is then 722 m2 * 11 ft2/ m2 ~ 7.945 ft2, just about the area of my lot at home. So, I am sitting in the dark, while my car is being fueled. This assumes the sun is shining of course.

Can we stop this please.