[extropy-chat] Solar math (was: Nuke 'em)

[Adrian Tymes]

Math errors detected.

--- Eugen Leitl <eugen at leitl.org> wrote:
> http://www.eere.energy.gov/solar/pv_cell_light.html says it's about 6
> kWh/m^2/month 
> insolation, for U.S. June. At 5% efficiency, a 60x60 m would be about
> enough for
> the average 2005 household
> http://www.eia.doe.gov/emeu/reps/enduse/er01_new-eng.html

>From your figures:
6 kWh/(m^2*month) * 60m * 60m * 0.05 = 1080 kWh/month

Of course, that's for US June; yearly average and northern latitudes
(much of the US is south of Germany - far west, but also a bit south)
would lead to a much lower yield.  But let's overlook that for now.

>From the EIA page:
> In 2001, the 5.4 million New England households consumed 39 billion
> kilowatthours (kWh) of electricity

(3.9 * 10^10)/(5.4 * 10^6) =~ 7222 kWh/year =~ 602 kWh/month

> (notice that's *one third* of that in Germany at 333 kWh/month, so
> 30x30 m should 
> have been enough -- something is running quite wrong on your side of
> the pond).

(60*60)/(30*30) = 1/4, not 1/3

That which is running quite wrong over here can be traced in part to
the two main occupants of the White House, and their extreme reluctance
(until recently) to pursue energy efficiency and conservation as part
of national energy policy.  That's only one factor, though, and
digressing in any case.

> 12% of Germany land area is covered with artificial structures,
> that's about
> 42843 km^2. At 5% efficiency that's 257058000000 kWh, or enough
> energy for
> 771251125 people. As there are only 82.5 Mpeople, it would be enough
> to use
> 1.2%, one tenth of the entire covered area.

6 kWh/(m^2*month) * 42843km^2 * (10^6 m^2/km^2) * 0.05 =
12852900000 kWh/month.

Using the (unsourced, but plausible) German number:
(12852900000 kWh/month)/(333 kWh/(month*household)) =~
38597297 households

(8.25 * 10^7)people/38597297 households =~ 2 people/household, which is
about right.

So your figures appear to support covering all artificial structures -
the full 12% of the land that's already covered - with solar panels,
but not merely 1/10th of the structures.  And, again, that's not
correcting for the difference in insolation between US June and Germany
year-round...although perhaps improvements from the current 5%
efficiency could eventually make up for that difference:
http://www.solarserver.de/lexikon/lexikons-e.html
says the average annual insolation in Germany varies between 950 and
1100 kWh/m^2 depending on the region, so 30% efficient panels
(advanced, but not apparently physically impossible), plus power
storage to smooth out the differences in insolation over the year,
might do the trick.