Keith Henson wrote:
On Tue, Oct 12, 2010 at 11:00 AM, "Dan Minette" <danmine...@att.net> wrote:
To: "'Killer Bs \(David Brin et al\) Discussion'"
We probably will never know if this StratoSolar method works.
...
David Hobby <hob...@newpaltz.edu> wrote:
I see bigger problems with losses in the light pipe.
The plan seems to be to have a flexible tube lined
with reflective material to guide the solar radiation
down to steam turbines or whatever on the ground.
Most of the light would have to reflect off the sides
many times, losing at least a few percent of its
intensity at each reflection. So nothing makes it
to the ground, and the light pipe melts. There may
be solutions to this too, but they're going to be
tricky.
How many reflections are you assuming light will make
as it goes down the pipe, and how glancing are they?
Keith--
Hi. Thanks for the details. I started thinking about
the problem.
It depends on the acceptance angle and the diameter of the light pipe.
I'll give you that the spread for light come out of the
whole array and into the pipe is 30 minutes, the same as
the sun subtends in the sky. So that would be an average
deviation of something like 10 minutes, or .003 radians.
(Actually achieving that may be a headache, but I bet it
could be done if it mattered. Although I believe that
it doesn't matter that much, since even if light went into
the pipe with only small angular errors, the average incidence
angle would rapidly increase due to somewhat random reflections
off the walls. See below.)
This stuff:
http://www.revelationlighting.co.uk/OLF%20Spec.pdf
has a .99 reflectivity for angles less than 27 deg,
That's pretty good reflectivity. Plastic tends to
crinkle, though, so you'll need some sort of backing
to help keep it flat.
> and almost all the
loss comes from the points not being sharp.
Lost me there. What points?
At .999, which the
optical guys say is not hard, and a 30 meter diameter light pipe, the
loss is about 7%. One option is to fill the pipe with argon which
reduces the Rayleigh scattering.
Working backwards, you're assuming around ln(.93)/ln(.999) = 73
reflections? For a 30 km light pipe, that's around one reflection
every 400 meters, for an average angle of 30/400 = .075 radians,
or 4 degrees. It would take a thorough analysis, but I'm betting
that successive reflections from the slightly crinkly walls of the
light pipe would gradually increase the average incidence angle,
pretty much like a random walk.
O.K., I'll buy that, if you can get .999 reflectance at angles
of a few degrees.
There is 4 GW coming down the pipe. At 7% loss, 280 MW. The area of
a 30 meter x 20 km pipe is 2 million square meters so the loss would
be 140 W per square meter. In open air it is only going to get
slightly warm.
O.K., but what about localized losses? Suppose there's a sharp
bend when the light pipe hits the jet stream, or something?
If the pipe bends something like 45 degrees over 300 meters, then
you'd have basically all the light hitting one side of the pipe
over around 100 meters. And it would hit at a 10 or 15 degree
angle, which probably decreases reflectivity to .995 or so?
Then you've got .005 of 4GW hitting an area of around 100*30
square meters, giving .005*4GW/3000 = 7000 watts per meter.
So that's as hot as grabbing a 60 watt incandescent bulb?
It might still work, but things are getting tricky.
For instance, after that one bend the average light ray
is going to be hitting the sides of the pipe at 10 or 15 degree
angles all the way down. (Unless you've got a mechanism to
"straighten out" rays that are bouncing off the sides too much?
I can't think of an easy one.)
If you have a ray permanently at an angle of .2 radians, it
hits every 150 meters, which would be around 100 times over 20 km.
And if reflectivity is down to .995 at that angle, you're left
with .995^100 = 60% of the light at the bottom.
Another problem could be "fluttering". If you have enough transient
surface waves running over the light pipe, each one giving large random
reflections to rays unlucky enough to hit it, you could rapidly
have almost all of the rays bouncing off the walls at 20 or 30
degrees. That gives you more reflections per ray, each at larger
angles with lower reflectance. Something like that could really
cause big losses.
It's an interesting problem. Thanks.
---David
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