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

Tue Oct 25 10:48:48 UTC 2005

On Mon, Oct 24, 2005 at 02:11:56PM -0700, Adrian Tymes wrote:
> Math errors detected.

Very possible. I hacked that out in a few minutes at breakfast
before dashing off to work.

> --- 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)

These were U.S. figures. If you look at the Solarhaus Freiburg
http://www.oekosiedlungen.de/energieautarkes-solarhaus/steckbrief.htm
(look at the picture http://www.oekosiedlungen.de/energieautarkes-solarhaus/79_4_Autark_H.jpg
http://www.informatik.uni-bonn.de/~rhino/tourguide/bilder/solar.jpg
http://www.informatik.uni-bonn.de/~rhino/tourguide/bilder/solar2.jpg
no extra collectors but roof and facade) that's 100% self-sufficient.
Summer surplus is stored as hydrogen in the tank, and is used in a fuel
cell. 

The reason this is possible is because most energy is used for
heating, and thermal solar collector efficiency is quantitative
(>>85%, some do over 90%).

The reason this is not cost effective is that the house is 3x as expensive
as non self-sufficient analog.

> would lead to a much lower yield.  But let's overlook that for now.

Actually 6 kWh is reasonably conservative for 
http://www.eere.energy.gov/solar/images/map_solar.gif but this is order
of magnitude, anyway. 

> >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

There is a range of U.S. household consumption cited, ~1000 kWh being
at the top. The trend is up.

> > (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

Math error allright, however insignificant.

> 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.

I'm not sure this is something which can be blamed on an a single
administration, or a couple. It must be a national trait.

> > 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.

Okay, 357031.0 km^2 * 0.12 = 42843.72 km^2 paved.
42843.0 km^2 / 82500000.0 people = 0.0005193 km^2 = ~520 m^2 paved /person.
Using http://www.solarserver.de/pvrechner/index.php at 10% efficiency I get 
49936 kWh/year where I live. According to http://de.wikipedia.org/wiki/Energieverbrauch
for Germany total energy consumption (i.e., not just the household number) is 
48594 kWh/person.

This means that those 12% can completely cover the total energy
consumption of Germany at current 10% efficiency.

According to http://www.boxer99.de/subs/Tabellen/stromverbrauch_europa.htm
6000 kWh/year electricity person. This means 1.45%, not 12% of land area is 
enough at 10% efficiency. 2.9%, not 12% enough at 5% efficiency. Heating
costs by solar thermal collectors are not included, but are negligible,
given quantitative efficieny.

> 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

Only if you want to completely cover the energy input, industry included.

> 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.

Again, the efficiency is irrelevant. Total integrated cost over lifetime
is, and efficiency is only a small factor there. If you can have
high efficiency effectively for free (e.g. by scaling the rectenna
approach into solar (NIR/VIS/UV) spectrum), the more power to you.

-- 
Eugen* Leitl <a href="http://leitl.org">leitl</a>
______________________________________________________________
ICBM: 48.07100, 11.36820            http://www.leitl.org
8B29F6BE: 099D 78BA 2FD3 B014 B08A  7779 75B0 2443 8B29 F6BE
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