🚀 go-pugleaf

Sheridan Mayo <scmayo@rsc.anu.edu.au> wrote in article
<604fsn$g6d$1@clarion.carno.net.au>...
> In article <342309F2.459A@mail.atl.bellsouth.net>,
> John Richardson  <john_rr@mail.atl.bellsouth.net> wrote:
> >Hyperion Systems wrote:
> >> If we had spent as much money devloping solar and wind power as we
have
> >> spent to get oil out of the ground and trying to decide what to do
with
> >> nuclear waste, we would all be probably paying next to nothing for
> >> non-polluting energy by now.  THOSE are clean energy.  It's amazing,
you
> [snip]>> 
> >
> >Not stupid enough to think we can replace all or most of our current
> >generating capacity by solar or wind power.  Have you given any 
> >thought to the amount of land required to replace just one power plant?
> 
> Have you?
> The energy density of sunshine at the earths surface is 400W/square
meter.
> That's 400 MW per kilometer square. OK this is on a sunny day, but in
Australia
> (in many parts anyway) sunny days are the norm. The most efficient solar
> cells are currently about 23% (according to UNSW solar cell research web
> page). Do the maths yourself, but I think you'll see the land areas
> involved in solar power needn't be outrageous compared to the size
> of a conventional plant. 
> 
<snip>

I though the insolation was about 1kW/m^2.

None-the-less, let's not forget a few things:
(1) That rate falls off as the cosine of the latitude.
(2) That rate falls off as the cosine of the sun's angle from zenith.
(3) The insolation that reaches the surface falls off with theta due to
greater atmospheric absorption and reflection.
(4) A generating plant must produce electricity 24 hours a day.
(5) Efficiencies of large area modules is significantly lower than single
cells.
(6) One square kilometer may not be a lot of land when it's mostly bare.
But try to cover it with something and it's a very big hunk of real estate.

The obvious solution to (1) and (2) is to tip the panels in order to be
perpendicular to the sun. That's fine for a rooftop size panel, but for an
array that's 1km on a side? 

You can tip the individual panels to get around (1) at the expense of
increased panel spacing (and hence lower fill factor). 

Getting around problem (2) is a lot more difficult since your panels have
to be spaced at ever greater distances to collect early morning and late
evening energy without some panels shadowing others. In practice you have
to space the panels such that you can capture all the light up to a certain
angle and then accept the loss due to shadowing. The point at which you
make this cutoff is going to be largely influenced by problem (3) and
anything over about 60 to 70 degrees is probably beyond the point at which
you're eating up real-estate to capture energy that just isn't there. I'll
call this angle the shadow angle.

Assuming that you can capture all of the incident light from sunrise to
sunset without regard for additional atmospheric attenuation with angle
your average daily collection is at most 32% of the rate at zenith. So even
using my 1kW/m^2 that's an effective collection rate of only 318W/m^2. I
wonder if your 400W/m^2 comes from a similar calculation.

Now, assuming a 25% end-to-end efficiency, our final output is down to
80W/m^2. And keep in mind that this is per unit area of solar panel, not
per unit area of ground. The land requirements go up as the product of
(1/cos(latitude)) and (1/cos(shadow angle)). So a plant at 30 deg latitude
and wanting a 60 deg shadow angle is going to require 2.3m^2 of land for
every 1m^2 of panel area. That puts the land use efficiency down to
35W/m^2. 

I went to the UNSW page and looked at their tables. With one exception, all
of the entries that had efficiencies over 20% were on the order of 4 cm^2.
There was a 22.7% crystalline Si entry that was 778 cm^2. The tables made a
special mention that the 10% milestone had finally been achieved for large
area modules (a Siemen's module at around 4000 cm^2 (0.4m^2). I suspect
that any large scale project is going to have to use large area devices.
But I'll continue working with a 25% total conversion efficiency.

I also went to Siemen's site and looked at their area factor map. The vast
majority of the world is shown as being able to produce between 4 and 5
kW-hours per day for every kW that a module is rated for at 1000W/m^s
normal to the panel. A few small areas were shown as hitting six. My ideal
calculation that I'm using came out at 7.6 which sounds reasonable. But it
does mean that my results are overly optimistic by something on the order
of 40% to 50%.

Fortunately the material costs are based on panel area and not land area.
So how much would it cost to get 400 MW daily average? Well, at 80W/m^2
we'd need 5 square kilometers of panel area (and probably about 10 km^2 of
land). Using 10% efficient large area modules and adjusting for the roughly
40% optimistic nature of my base calculations we could easily need six
times this area.

I couldn't find any sites that had costs, let alone costs for large scale
projects. So to be really optimistic (I hope) I'll use the lowest cost I
can find in DigiKey (which is $0.38/in^2 ($US) which works out to six cents
per square cm or about $600/m^2) and assume that between economies of scale
and future improvements we can lower that cost by a full factor of 100. So
now we are talking about what is probably a wildly optimistic price, at
least in the near future, of $6/m^2 which is comes out to $30 million
dollars in panels alone - perhaps as much as $180 million. Not to mention
the cost of all of the support structures and control systems for each
panel to track the sun and the several thousand kilometers of interconnect
wiring spread out over ten square kilometer of ground. And what about the
labor costs to construct all of this? If each panel is 10m on a side
(composed of several modules) then you are talking about 50,000 panels. The
support structure and other hardware is probably going to run about a grand
and it's going to take a crew of people a couple of hours to erect it so
there goes another $50 million or more. And the panels are subject to the
elements and have very finite lifetimes so you have to have systems that
can monitor the grid and identify specific panels that are failing. If each
module is 10,000 cm^2 you have 5 million modules (and maybe 30 million). If
the mean lifetime of each module is 50 years (a number pulled out of the
hat but one that I think is wildly over optimistic) then your modules will
have to be replaced at the average rate of between 300 and 1800 a day.

And what are you doing to the local ecosystem when you're extracting that
much energy out of the area?

Just my $0.02