[Discuss] Eclipses Re: Great talks last night, however...

[grg-webvisible+blu at ai.mit.edu (grg)]

On Sun, Jul 23, 2017 at 12:46:22AM -0400, Richard Pieri wrote:
> > OK, so here you're saying that instead of a <10% charge/discharge
> > efficiency, batteries actually have a 75%-80% charge/discharge efficiency?
> 
> No. I'm saying that chemical batteries have *at best* a charge
> efficiency of around 75-80% in the real world.

OK, so you're saying that instead of single-digit percentages, there are
real-world battery installations which get 75%-80% charge/discharge
efficiency; meaning that if using them we'd only need to make 20%-25% more
solar power, not 1000% more, to compensate for the loss in batteries.

FWIW, here are some references which show measurements at 80%-90%, so I
still think we'd only really need to generate 10%-20% more solar power
instead of 20%-25%... but at these small numbers, these smaller differences
don't at all affect whether solar+batteries is technically feasible.

here's a report on an actual power grid battery system they built and
deployed for a wind farm, reporting a measured 84%-86% charge/discharge
efficiency:
http://www.sandia.gov/ess/docs/pr_conferences/2014/Friday/Session10/04_Vishwanathan_V_Powin_Dispatchable_Battery.pdf

for a wider range of design parameters (spanning 80%-90% efficiency) see fig.5:
http://umanitoba.ca/outreach/conferences/phev2007/PHEV2007/proceedings/PluginHwy_PHEV2007_PaperReviewed_Valoen.pdf


> > Agreed!  And Utah, and Arizona, and New Mexico, and large parts of
> > Colorado, Wyoming, Idaho, Oregon, and Washington by your map.  And don't
> > forget Great Plains states like Texas, Montana, North Dakota, South
> > Dakota... hey, I think we're over 0.15%!
> 
> There are three problems that I would consider breakers for these regions:
> 
> First, you just described the heart of Tornado Alley.

But somehow, 99.99% of people and corn and cows (not counting that
unfortunate animal in the movie Twister) have managed to survive there.
I'm betting solar panels will have a similar tornado survival rate, unless
we decide to install them only at trailer parks.


> Second, you can't charge Li-ion batteries when they are below freezing
> (0C) which makes much of these areas useless for Musk's storage systems
> for significant portions of the year.
> 
> And third, high temperatures (above about 25C) reduces efficiency, and
> it causes batteries to wear out faster than their published ratings
> which means you'll be replacing them that much more frequently if you
> set up your stations in the non-freezing areas.

One standard solution to weather exposure would be to house them below the
frost line, which is only 2'-3' deep in Kansas:
    https://www.ngs.noaa.gov/PUBS_LIB/GeodeticBMs/#figure13
You'll get a moderate temperature all year round.

Luckily, the 10,000 km^2 solar+battery farm will still meet the entire US's
energy needs even if you replace the batteries more frequently.


> > Right - as in my prior email, when you do the math it comes out to a factor
> > of pi (and 24/pi is 7.64 hours, within the range you give).
> 
> No. It's significantly more than that because a geostationary station is
> always at "noon" when it's exposed to the sun while a ground station's
> noon is only a fraction of it's exposure period.

Again, if you do the math, it's exactly pi.  The equator is a circle; the
sunlight incident on it is its shadow at this point in space, which is a
line that is the diameter of Earth - on that line every point is always at
"noon", and it would collect all the light the equator sees.  Will ascii
art help?

             -->         ___
             -->        / | \
    sunlight -->        | | |
             -->        \ | /
             -->         ---

The line has length Earth diameter D, the equator has length pi*D, and
they'd both collect exactly the same sunlight.  That D's worth of "always
noon" sunlight is spread across the equator's circle, and averaging over a
day every point on the equator sees the same amount, namely (D) / (pi*D)
= 1/pi of the "always noon" solar influx.  The factor you're looking for
is exactly pi.


> > FWIW, on that last non-technical bit, I and I wager many others on this
> > mailing list see very many places in all the named locales which have good
> > potential for solar.  And that's one of the great things about solar power:
> 
> Maybe good on small scales like homes and offices. Not so good for large
> scale like replacing global dependence on fossil fuels.

Actually, awesome for that.  This thread has shown that a total of 10,000 km^2
of solar panels + batteries will provide all of the US's electrical needs
using current technology, and that's true whether those 10,000 km^2 are in
one place or spread over 10,000 places or in 10 million places.  Arguably,
the closer you put a panel to where the electricity is used, the more
efficient your power distribution (a real advantage over nuclear plants,
dams, coal-fired plants, etc.).  In practice I expect a combination of some
large sites and lots of tiny sites with a whole spectrum (no pun intended ;)
in between, but as long as all the pieces add up to 10,000 km^2, for
electricity generation you've completely replaced fossil fuels with
solar+battery.

--grg