Conference 7.286::space

From:	US1RMC::"[email protected]" "Mitchell James" 28-MAY-1993 10:38:03.00
CC:	distribution.%[email protected]
Subj:	Solar Power News via Internet

----- Begin Included Message -----

>From @cunyvm.cuny.edu:[email protected] Fri May 28 10:28:13 1993
Date:     Fri, 28 May 1993 10:27 EDT
>From: USRNAME <[email protected]>
Subject:  Solar Power News via Internet
To: [email protected], [email protected]
Original_To:  SPS,IN%"[email protected]"
Content-Length: 4474

********** Solar Power News via Internet *************
May 28,1993
your friendly editor, G.E. Canough

Any size space solar power station likely to be built in the near
future will most probably use photovoltaics (PV). This is just because,
we have the most experience with using these in space. I expect that
further down the road, other means of using sunlight will also be
developed, such as solar dynamic. But for now, PV is "it". This means
that we (the SPS community) should become informed as to what is going
on in terrestrial PV. Bob Forstrom (whom I met at the latest Princeton
space manufacturing conference) designs and builds solar powered
houses and he stressed to me the need for getting informed about
terrestrial solar. He also supplied me with a pile of information on
just what is going, so I figured I'd pass it on here. 

As it turns out, the USA may be in for the dawn of a new day in the
area of solar energy. Some of the programs the US Dept. of Energy
(DOE) used to do have been closed out (nuclear weapons and nuclear
reactor research) and so DOE has intelligently increased funding for
PV research and development by 25%. 

Here's a sample of what is going on:

UPVG and SMUD
Power companies in the USA have just formed (as of Dec. 1991) the
Utility Photovoltaic Group, UPVG) in order to foster development of PV
systems for use as power plants and for individual buildings. One
utility, the Sacramento Municipal Utility District (SMUD) has already
requested proposals and received proposals for a program which includes 
400 kW worth of residential systems, 100 kW worth of commercial (at 
individual buildings) systems and a 200 kW PV substation. 

PV producers:
United Solar, Inc. This company is building a new factory in Virginia.
It plans to manufacture 10 MW per year worth of PV. 

Photocomm in AZ has received a $1M order for PV systems Golden Photon,
Inc. is installing a factory to build CdTe PV modules, 2MW worth per year. 

Batteries:
A new type of battery has just been invented. It can be recharged in
just 15 minutes. 

Races:
Several solar powered vehicle races are taking place this year.

To get all the latest info on PV developments, subscribe to one or
both of these newsletters: 

PV News, Paul Maycock PV Energy Systems Inc. PO Box 290,
Casanova, VA 22017 phone = 703-788-9626

Photovoltaic Insider's Report, Richard Curry 1011 W.
Colorado Blvd, Dallas TX 75208 phone = 214-942-5248

You are also encouraged to subscribe to the SUNSAT Energy Council
Newsletter where all the latest on SPS and wireless power transmission
will be reported! [SUNSAT Energy Council Newsletter c/o ETM, Inc., PO
Box 67, Endicott, NY 13761, $25/yr.] 

The increase in PV production is a definite plus for solar power,
since the main reason PV is expensive is that it is not mass produced.
We will see the prices drop as the production increases. 

Although we often cast the power companies in the role of "the bad
guys" for burning coal, many of them are very concerned about the
affect of fossil fuel burning and are funding efforts to develop solar
energy. A local example of that is New York State Electric and Gas
(NYSEG), which is funding a large PV array to be installed at the
Kopernik Observatory.  This array (4kW) will be used to power the
lights at the newly installed science center at the observatory. And
there are 62 power companies who are members of the UPVG. So there is
a lot of concern and money is being spent. 

On May 10, there was an IEEE PV conference in Louisville, KY. I found
out about it too late to attend. If any of you went (Lewis PV people?), 
I would really appreciate getting a summary of the conference posted here. 

Ground based PV won't be able to meet all the energy needs of Earth,
but it can be used much more than it is now, and this is all to the
good. Producers of PV are likely to be very interested in the prospect
of space solar power stations, since these larger stations represent
an very large market for PV. 

I have also just learned some new info on climate change.  I have not
been a real alarmist about this in the past, but recent data is quite
disturbing.  So stay tuned for the next edition of The Solar Power
News, via Internet! 

Dr. Gay E. Canough ETM,Inc. and BU-SUNY, dept.of physics
e-mail(Internet):  [email protected]
         (GEnie)   :  G.CANOUGH
phone/fax= 607 785 6499    voice mail = 800 673 8265
radio call sign:   KB2OXA

'Snail Mail:
ETM, Inc.
PO Box 67
Endicott, NY 13761

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% ====== Internet headers and postmarks (see DECWRL::GATEWAY.DOC) ======
% Date: Fri, 28 May 93 10:35:46 EDT
% From: [email protected] (Mitchell James)
% Subject: Solar Power News via Internet

  Why do we have to turn Sunlight directly into electricity??  Why not use
the Sun to heat water to produce steam to turn a turbine?  It would be just 
like a nuke sub power plant, but replace the reactor with a solar collector.  

  What would be the drawbacks to this?

 

What you describe is what .1 refers to as "Solar Dynamic" in the first
paragraph, where he justifies why PV will be what is used (namely, that is what
we have the most space experience with, and therefore is the least risky)

Burns

    And another issue is making the fittings/etc working in zero g.
    
    Also, maintenance of a dynamic system will be far higher than for a
    passive system like pv.
    
    Tony

    To start with photovoltaic systems are cheaper and more reliable
    for small systems anyway.   That is why they are used today, and
    will be for even the Space Station freedom.
    
    But for real  power satellites (i.e. for BIG systems) they are way 
    too expensive. To cut expenses designers can first concentrate the
    Sun light with mirrors, thus reducing the photovaltaic area needed.
    But in the end  unless  production  costs  come  way down and they 
    solve the degradation proplems that limit their life, such systems
    are not the final answer.
    
    Mirrors are cheap to build and launch, they can be just very thin
    "mylar" sheets or some such. So they will no doubt be part of the
    final solution. As it is essential to concentrate the Sun light.
    
    But what  will be the power satellite's  power generator, I can only
    guess.  Yes it could be a photovaltaic system, but much more likelly
    it will be some kind of dynamic vapor turbine or some kind of termal
    engine. As such should be cheaper (even in space) and they will last 
    a lot longer.
    
    Gil

    re .5
    
    I think your assessment of photovoltaic systems is unduly pessimistic.
    
    Your analysis is based upon present photovoltaic technology, crystal
    silicon.
    
    For other types of photovoltaic, like amorphous silicon, the answer is
    very different.
    
    It would be very little additional cost to build an amorphous silicon
    solar cell than to build a mylar sheet mirror.  
    
    It is also possible to tailor solar cells for the radiation of space,
    which has different characteristics than the radiation that reaches
    earth.
    
    It is possible that you could have an arrangement where the solar cells
    were either a topping or bottominc cycle to a dynamic system.
    
    Tony

    Bonjour,
    
    I think taht the main argument for PV cells is the reliability. The
    space conquest has always favoured reliability against efficiency. A
    power plant in space is almost impossible to imagine. It has to be
    frequently stopped to preventiv maintenance and if you want to be able
    to survive, you must have a backup system, either another plant either
    PV cells. So the cost would be, IMO, to big to be interresting. While
    the thermodynamic solution is, even a bit, week it has no place in
    space. It can be a solution in very large station in space or on ground
    (Mars, Moon...) for long term ("Forever") installations, and with
    backups. You can't risk N.Y. great electricity cut in space.
    The solution of hgh power cells in new technologies, with mirrors or
    with multilayered cells capturing different level of energy will have
    for a while the preference in solar electricity porduction.
    
    Ciao,
    
    jef
     

    re .7
    
    Note that PV needs a backup too since there are "eclipses" and "nights".
    This of course applies to other solar users too.
    
    Typically the backup is provided using batteries however, looking ahead
    a bit, batteries don't seem like the answer for a moon base during its
    two week night.
    
    One largely fantasy solution I came up with is to have two
    diametrically opposed power stations using sun-light (in whatever
    manner) so that one would always be in daytime (ignoring "lunar"
    eclipses). This would require long (thousands of km) transmission lines
    although there is a trade-off between "latitude" and amount of cabling.
    (Higher latitude means lower angle of incidence of the light hence
    lower efficiency but shorter cables.) I would think that the raw
    materials for the cables would have to be mined and processed on the
    moon.
    
    On orbit power stations that beam their power down to the surface might
    be an alternative and some of the objections to use on Earth would be
    less of a concern on the Moon.
    
    Has anyone seen any real proposals for power supply to a moon base?

    Power on a monbase?
    
    EASY
    
    Solar power at the poles.
    
    The moons poles are just as good a place as any to have a base.  And
    they get sun all the time, save for tera eclipses :)
    
    Tony

    And at the while angle of incidence means something on the earth, I'm
    not sure it means near as much on the moon.  On the earth, the problems
    with angle of incidence are two fold, one the higher the angle of
    incidence, the greater the attenuation of the solar energy through the
    atmosphere and secondly cost constraints, land on earth costs a lot, or
    is too far away from where you want to use the power.  On a moonbase,
    both these concernes are nullified.
    
    Now, if someone knew the axial tilt of the moon, we could determine if
    the poles would remain in sunlight all the time (or if not the poles, a
    point only slightly higher than the poles 100 meters?)
    
    Tony

    re .9,.10
    
    Yes, you are right of course. Angle of incidence is not really a
    problem on the moon. (My brain is a "land-lubber".)
    
    Even so, if you put your power station at the poles and you don't want
    lots of cabling then you have to put your base(s) at the poles and this
    may not be desirable. Perhaps one of the poles would be a good starting
    point, particularly if speculation about water ice there proves to be
    correct.
    
    There are two lunar eclipses a year, each lasting several hours, to ride out
    with battery backup or whatever - which doesn't sound too onerous.

    Re .9, .10, .11
    
    Puting bases in the poles of the moon. Has one draw back "propultion cost",
    it takes more propultion energy (i.e. Delta-V) to put a spacecraft in the
    poles of the moon then at its equator.
    
    Its for a similar reason that Earth launch sites are also as close to the
    equator as possible.   It requires less powerfull systems to put the same
    amount of cargo in orbit.
    
    Putting a power station there is another story, it does have 2 advantages,
    over two stations on oposite sides of the moon's equator. 
    1/ More Sun-Light.  And
    2/ Half the cabling distance, from the pole to the equator is only one
    quarter of the moon's perimeter, compared against half a perimeter.
    
    * And who knows, water ice may exist at the poles, in the permanently
    shadowed places. If so, the poles would also become the site of the moon's
    most valueable mines. Water/Hydrogen mines.
    
    Mind you, for the very first small moon base, none of this matters. Its 
    not a solar power station they will use but a nuclear one. And going to 
    the poles get water (if it exists) will not be ecomomical or pratical
    right away, so they will either find it locally or import it from Earth.
    
    Gil

    RE .12
    
    Yes, it does take more delta-V to get to the moons poles, but that
    much?
    
    I thought the decreased delta-v on the earth for equatorial launches
    was due to the rotational speed of the earth.  If that is true, the
    rotational speed of the moon is rather slight....
    
    Tony

    Re .13
    
    Yes it makes a difference,  the moon's rotation is on the same plane
    as its orbit around the Earth, wich means that Earth spacecraft when 
    they arive go in Low Moon Orbit around the moon's equatorial plane 
    (v= 1.6Km/s). 
    
    Thus the difference (one way) is 3.2 Km/s in Delta-V, wich means if
    using H2-O2 rockets that the spacecraft will have to be twice as big, 
    when going from the Earth to the Moon. And twice as big coming back
    as well.
    
    Gil

RE: <<< Note 848.14 by MAYDAY::ANDRADE "The sentinel (.)(.)" >>>

>    the moon's rotation is on the same plane
>    as its orbit around the Earth, wich means that Earth spacecraft when 
>    they arive go in Low Moon Orbit around the moon's equatorial plane 

What is the energy advantage gained by entering orbit about a distant body
about that body's equator, as opposed to any other orbital inclination? I
can't see how the rotation of the body makes a difference.
						-- Tom

You don't have to slow down as much to be stationary over a point if the 
rotation about the equator is already negating some of your speed.

RE: <<< Note 848.16 by GAUSS::REITH "Jim 3D::Reith MLO1-2/c37 223-2021" >>>

>You don't have to slow down as much to be stationary over a point if the 
>rotation about the equator is already negating some of your speed.

Ok, yes, silly me. I was hung up on simply reaching orbit (and forgetting about
the landing aspect).
						-- Tom

    Rotation speed is less then 5 m/s,  what really makes a big difference
    is the Moon's orbital velocity around the Earth (1.0 Km/s) in the same
    plane as the Moon's equator.   Taken together with the velocity needed 
    to orbit the Moon itself (1.6 Km/s).
    
    Spacecraft coming from Earth, have to get to the Moon wich is travelling 
    at 1.0 Km/s and by using that velocity wisely they can go into Low Moon
    orbit with only an additional delta-V of (0.6 Km/s) less if you count 
    the Moon's own gravitational atraction, wich adds some velocity to the
    incoming spacecraft.
    
    However if they want to go into a Moon polar orbit, they have to cancel
    all their velocity in the equatorial plane (from whatever source) 
    (-1.6 Km/s) and add it into a polar plane (+1.6 Km/s).
    
    Gil

RE:          <<< Note 848.18 by MAYDAY::ANDRADE "The sentinel (.)(.)" >>>

*    Rotation speed is less then 5 m/s,  what really makes a big difference

I assume you mean 5 Km/s if you want to come up with the 1.6 Km/s number?


*    However if they want to go into a Moon polar orbit, they have to cancel
*    all their velocity in the equatorial plane (from whatever source) 
*    (-1.6 Km/s) and add it into a polar plane (+1.6 Km/s).

Excuse me?  A spacecraft following a ballistic orbit just needs to drop its
velocity below escape velocity to ensure capture by the body, doesn't it?  I 
don't think the apollo missions went into lunar-stationary orbit, did they?
And in any case do polar launches here on earth bother to kill their E-W
velocity vector?  Wouldn't it be a case of it depends on what orbital dynamics
you want as to whether you worry about bleeding off v in some orbital planes?
So if you are captured by the moon then just add v-vectors to put you over the 
'poles' - whether you pass over the same point on the moon's 'equator' each
orbit depends on if you want to.

    Re .-1
    
    You seem to be a bit mixed up ...
    
    The moon rotation speed at its equator "is less then 5 m/s" not 5 Km/s
    As per 	pi*3476000/(27.322*24*60*60) = 4.62598 m/s
    The Moon's diameter is  3476 Km and ONE rotation takes  27.322 days
    
    The Moon's orbital velocity around the Earth is 1 Km/s, just like a
    Space Shuttle in Low Earth Orbit has an orbital velocity of 7.7 Km/s.
    Both are orbiting the Earth, and to get to them not only will you 
    need to get there, but match their velocity and direction as well.
    
    And both the Moon's Rotational Velocity or its Orbital Velocity have
    nothing to do with the velocity needed to go into Low Moon Orbit
    1.6 Km/s <Its the Moon's mass (7.348e22 kg) and Diameter (3476 Km)
    that produce this number>. As per SQRT(4.903e12/1738000) = 1679.6 m/s 
    
    What I said in re.18,  was that since you have to match the moon going
    at 1 km/s around the Earth, you can also use this same velocity to put 
    you in Low Moon Orbit by just adding 0.6 Km/s, but it has to be in the 
    same direction and in the Earth-Moon's orbital plane, "the equator".

    Actualy,  its more complicated then this,  because the Moon's gravity 
    affects the velocity of the aproaching spacecraft. But the basic fact 
    remains that getting to the Moon's poles is more expensive then getting
    to the Moon's equator, and the reason is that the Moon equatorial plane
    is the same as the Earth-Moon's orbital plane. You will notice that
    all Moon landings were petty close to the Moon's equator.
    
    Gil

I am still not happy with the way you are trying to explain this. 

You need to achieve 1Km/sec orbital speed * RELATIVE TO THE EARTH * to be 
* STATIONARY * relative to the moon.
(assuming you are trying to orbit in the same direction!)

If you wish to directly enter orbit about the moon, you need to factor in 
2 components. 

1. 1Km/sec relative to the earth
2. 1.6Km/sec relative to the moon.

How you fit these together depends entirely on your chosen orbit about the moon.

I think I begin to see what you are getting at. 

(next unseen here if you already fully understand this problem!)

Imagine that you aim well in front of the moon, and arrange things so that when 
you reach the moons orbit you are dead still in space RELATIVE to the EARTH. 
The moon is still coming towards you at 1Km/sec. You only need to find 0.6Km/sec
towards the moon in your own velocity to achieve the required 1.6km/sec orbital 
speed about the moon. You just then have to aim carefully to miss by the correct
distance and the moon's gravity does the rest for you. 
In reality, you do not go out to the moon's orbit, turn 90 degrees and aim to 
just miss the bullseye. If this * were * the case there would be essentially no
difference what orbital inclination you chose. Just aim for the latitude of the 
orbit you wish to achieve. But since it is NOT the case, the real world needs to
be considered.

You set off aiming for where the moon will be when you get there, with just 
enough velocity to crawl over the gravity balance point (where the gravitational
attraction of both bodies * ON YOU * are equal) then the moon starts sucking...
At this point you have NO (significant) * orbital * velocity relative to the 
EARTH, and to achieve equatorial orbit with only the minimum possible amount
of fuel useage, you just have to aim correctly and ensure you arrive with just
enough speed to add to the moon's approach speed to achieve orbit. 
I.E. the moon's gravity bends your path so that you are going in the SAME 
direction as the moon, and with the right speed to maintain orbit. 

BUT... If you aim over the top, the moon will bend your path, so that your orbit 
is at maybe 90 degrees to the path of the moon, but the moon will keep going and
leave you behind. 
This is just the same as what ULYSSES is going to do at JUPITER to achieve an 
orbit over the SOLAR poles. 

In order to achieve a POLAR orbit, you need to make a manoever very much like a 
scaled-up version of a geo-syncronous transfer orbit, but reaching out to the 
moon's orbit, then accelerate towards the moon to achieve the combined 1.6Km/sec
orbital velocity. This is clearly a significantly more expensive manoever, you
need to be going faster away from the Earth to be able to reach the moon's 
orbit without a gravity assist from the moon, then add most of 0.6Km/sec towards
the moon. This latter will have some gravity assist, as the moon comes racing 
towards you, but I have not tried to calculate how much.


	John Travell.

re .21
    
>This is just the same as what ULYSSES is going to do at JUPITER to achieve an 
>orbit over the SOLAR poles.
    
    Has already done.

Article: 2531
From: [email protected] (Sean Morgan)
Newsgroups: sci.nanotech
Subject: Solar Power (long)
Date: 13 Sep 93 22:29:24 GMT
Sender: [email protected]
 
(( In March Phil Goetz posted a long discussion on nanotechnology for Solar
   Power.  I tried to follow-up at that time, but was not aware at that time
   that posts had to be mailed in.  Much overdue, here is that discussion.
 
   The remainder of this message is in three parts:
 
   1) Phil Goetz original article (long).
   2) My comments (short).
   3) Phil Goetz's endorsement of my comments.
 
Sean Morgan
))
--------
From: [email protected] (Phil Goetz)
Newsgroups: sci.nanotech
Subject: Solar power
Date: Tue Mar 16 17:13:36 1993
 
When I read Eric Drexler's notion that we could supply our energy needs
by covering our highways with solar cells, I was skeptical.  So I calculated
how much land we would have to cover with solar nanocells to provide the
US's current electricity needs.
 
Electricity produced in the US in 1990: 2805 billion kilowatt-hours
Source:  _1992 World Almanac_, p. 195
 
Energy from the sun at Earth's distance (1 AU):  1358 W/m^2
Source:  _Space Mission Analysis and Design_, ed. Larson & Wertz, 1992, 3rd
front inside leaf page.
 
Energy which would fall on 1 m^2 on Earth's surface in one year
if there were no atmosphere:
 
  I will approximate this by assuming that the Earth has 0 inclination.
  (Actual inclination 23.439 degrees.)
  This will underestimate the light reaching the poles (as zero)
  and overestimate the light falling on the equator.  This assumption
  makes every day in the year have an equal amount of sunlight (11.967 hours),
  so we can just do the calculation for one day and multiply by 365.25.
 
  Let V be the vector from the center of the Earth to the square meter M
  we are examining.  Define a left-handed coordinate system centered on
  the Earth's center, with the X-axis increasing towards the sun and
  the Y-axis in the plane of the orbit, increasing towards sunrise.  Our
  unit of distance is earth radii.  At any one moment, the energy falling on
  M is 1358 W/m^2 times the cosine C of the angle between V and <1,0,0>.
  C = ( V dot <1,0,0> ) / ( |V| times |<1,0,0>| ), which evaluates to the
  x-coordinate of V.
 
  Define L = latitude (constant), r = rotation in radians from sunrise.
  Let r range from 0 (sunrise) to pi (sunset).  V is (if you're curious)
  <cos(L)sin(r), cos(L)cos(r), sin(L)>.
 
  So the total energy falling on M in 1 year is
                                                              /r=pi
  365.25 days/year * (1358 * 11.967) W-hours/(m^2 * days)  * |  
cos(L)sin(r) dr 
                                                            /  r=0
  = 5935746 * cos(L) * pi/2 = 9324 * cos(L) kW-hours/year
 
  As a fraction of 1358 W/m^2 * 12 hours for important US cities:
 
        City            Latitude        Fraction
 
        Buffalo         42d52'           .42
        NYC             40d45'           .43
        Wash DC         38d53'           .45
        Dallas          32d47'           .48
        Miami           25d46'           .52
        equator         0                .57
 
Fraction of solar energy which passes through atmosphere: .72
Source:  A problem in _The World of the Cell_, p. 349.
This is not a good source, and might be a guess, but is the best I could
find.  It refers only to the growing season, so is an overestimate.
I'm sure that in Buffalo we don't get 72% of the sun's energy;
it's cloudy most of the time.
 
Solar cell efficiency:  From _Space Mission Analysis and Design_ (1992), p.
397: 
 
Cell type:      Silicon         Gallium arsenide        Indium phosphide
 
Theoretical
efficiency [?]     18%                  23%                     22%
Achieved in lab
(single cell only) 14                   18                      19
Assembled array    12                   15                      16
 
I don't think we'll use gallium arsenide or indium phosphate, but silicon.
Si is much more common, and I hate to think how tons of arsenic would be
manufactured and distributed for nanofabrication.
    
I've taken 15% efficiency as my optimistic estimate, assuming nanofabrication
is extremely accurate and inexpensive.  (The cells that SMAD refers to are
for use in space; since it costs ~$60,000/lb. to send them up, we can assume
they are the most expensive money can buy.  Your Radio Shark cells won't
reach these efficiencies.)
 
For those of us who favor the cellular approach to nanotech, here is the
efficiency of chloroplasts.
 
Photosynthetic efficiency (from _The World of the Cell_):
 
  10 photons of light per molecule of CO_2 fixed into glucose
  6 molecules of CO_2 per glucose
  686 kcal/mole difference between glucose and CO_2 + H_2O
    (1 mole = 6.023x10^23 molecules)
  About one out of 2 photons absorbed by a chloroplast (World of the Cell
    p. 320, with the assumption that an unlabelled axis ranging from 0 to 100
    represents percent)
  10*6*2 photons needed
  Energy of photon at 670 nm (most efficiently absorbed): 43 kcal/einstein
    (an einstein is a mole of photons)
  686/(10*6*2*43) = 13.3% efficiency
 
  The end product here is glucose, not electricity.  But if cellular nanotech
  is widespread, glucose may be a more useful power source than electricity.
 
Toting it up:  For Washington, DC, the total energy available per m^2 per
year is 9324 * cos(38.88) * .72 * .15 = 784 kW/hrs.
2805 billion kilowatt-hours / 784 kW/(hrs*m^2) = 3.6 billion m^2
1 m^2 = 1.196 yard^2, 1760 yards = 1 mile, so
3.6 billion m^2 = 977 sq. miles, or a circle 35 miles in diameter
to supply the energy needs of the US.  At last, here's a way Rhode Island
(1055 sq. miles) can earn all those Senate votes they get...
 
A 2-lane (each way) interstate is about 40 ft. wide.  So we need 129,000
miles of interstate.  Do we have that much highway?  I suspect we do.
 
This estimate should be worsened by factors for power loss in transmission,
sunlight loss due to snow and traffic, increased atmosphere effective depth
and edge effects (light is refracted away from but not onto our land) at
sunrise & sunset, etc.
 
If we started using solar power to replace coal, gas, and oil, we would
need much more land.  I have estimated the amount by observing for coal,
gas, and oil the percentage which was used to generate electricity, how
much electricity it generated, and how much electricity the entire US
consumption of that fuel could generate.  One one hand, this ignores the
inefficiency of converting fuel to electricity, and this causes the estimate
of needs to be too high.  But on the other hand, it ignores the fact that
electricity is _not_ used for the things fuel is used for because fuel is
more efficient for those purposes.
 
Fuel    billion kWhrs           Fraction of total       Billion kWhrs
        generated (1990)        usage used to make      total fuel consumed
                                electricity (1989)      could have made
 
Oil            117                      .043                 2721
Coal          1557                      .861                 1809
Natural gas    263                      .142*                1851
 
* - I found no figures for the fraction of natural gas used to make
electricity.  I instead calculated the number of kWhrs the natural gas
we used could have generated from the fact that we used 1.023 times as
much natural gas (measured in BTUs) in 1989 as coal, so 1809 x 1.023 = 1851.
 
The fact that some figures are from 1990 and some from 1989 does not
damage the data because only ratios are taken from 1989, and ratios stay
contant more than total consumption does.  All numbers are from the
1992 or 1991 _World Almanac_.  Electricity production is net production,
not counting station usage.
 
So the total billion kWhrs that our fuel consumption could have produced,
added to the electricity produced, is
 
2721 + 1809 + 1851 + 2805 - 117 - 1557 - 263 = 7249
 
So to use solar power to provide all our energy needs, we would need about
2525 sq. miles of 15%-efficient solar cells, which we could place on
333,000 miles of 2-lane interstate.  The greatest error in these figures
is the error in how much sunlight passes through the atmosphere.
I would guess that error is at most a factor of 2.
 
Phil [email protected]
 
[There are about 4 million miles of highway in the U.S.  It's proposed
 that higher conversion efficiencies could be obtained by using tuned
 full-wave dipoles in the optical frequencies.  However this has not
 been demonstrated...
 --JoSH]
-------------------
From: [email protected]
To: [email protected] (Phil Goetz)
Subject: Re: Solar Power
 
>Fraction of solar energy which passes through atmosphere: .72
>Source:  A problem in _The World of the Cell_, p. 349.
>This is not a good source, and might be a guess, but is the best I could
>find.  It refers only to the growing season, so is an overestimate.
>I'm sure that in Buffalo we don't get 72% of the sun's energy;
>it's cloudy most of the time.
 
Of course different frequencies have different attenuations, so a detailed
analysis would need to know what frequencies the cells used.
 
The rule of thumb used in the heyday of the solar power satellites (1970's)
was that collectors on the earth's surface were only 10% as effective as
space-based ones (night, seasons, weather, atmospheric albedo and
attenuation).  Unless they were biasing the figures to make their case look
good (or my memory is faulty?), your numbers seem to be high.
 
The only table I have close to hand (CRC, 56th edt., pp.F197-8) was
entitled "Total Monthly Solar Radiation in a Cloudless Sky".  OK for
Calgary or southern California I guess, but not much use elsewhere (ah,
maybe that's where your 72% comes in?).  The values for 40 deg N were 8.8,
12.2, 16.4, 20.3, 23, 24, 23.4, 20.9, 17, 13.2, 9.7, 7.7 for each month,
average 16.3833 kcal/cm^2 (per month, like the title said).  I make that
260 W/m^2 or 2284 kW hr/m^2/yr.  Your formula gives 9324cos40=7146, as I
said, a little high.  I will fax (or scan/e-mail) the table to you if you
like.
 
A big disadvantage of land-based collectors is that they have to be cleaned
(though it is probably safe to assume that nanotechnology can make the
roads self-cleaning and self-repairing -- no more pot holes or frost
heaves!)  Even so, it seems to me that these collectors are superior to
space-based ones because they don't require the construction of a
supporting structure, are easily accessible for installation and
maintenance (if required), and provide power close to the usage point
(assuming inter-continental relays for night-time usage).
 
Sean Morgan
---------------
Date: Fri, 19 Mar 93 14:51:53 EST
From: [email protected] (Phil Goetz)
To: [email protected]
Subject: Re: Solar Power
 
Your more accurate figures are important, since they indicate at least
3 times as much area is needed.  This puts the estimate at around 8,000
square miles for total energy needs.  (If half the energy were lost to
transmission, storage, or disuse, it would take us to about 16,000 sq. miles,
which is somewhat disheartening.)  Why not send it to the list?
 
Phil [email protected]
---------------
(( because I didn't know how!  Now done.))
 
-----------------+---------------+----------------------------------
Sean Morgan      | Integrated    |  ALBERTA  3rd Flr, 6815 - 8 St NE
403/297-2628     | Manufacturing | RESEARCH  Calgary, AB, Canada
[email protected] | Program       |  COUNCIL  T2E 7H7

Ahh, 15% efficiency?

Sorry, some people are confused here.

There is AM0 efficiency, which is reflective of the efficiency of a solar cell
in space (i.e. air mass zero) and there is AM1 efficiency, which is reflective
of earth based solar cell efficiency.

Silicon solar cells have achieved AM1 efficiencies of over 22%.

The assumption that space based solar cells are the most efficient is false. 
They have the best combination of power, weight, size and reliability determined
for the mission of the space craft that money can buy.  They may not be the most
efficient cells that can be bought.

Now, if we are talking of nano-technology the whole efficiency equation is
turned upon its head.

It is trivial to assume the capability of multijunction solar cells, if you are
able to use assemblers.  Multijunction solar cells (without using vast
quantities of materials like arsenic) can obtain efficiencies of 40+%.  I think
the theoretical max efficiency (assuming infinite junction possibility, but
allowing for real electrical losses), is something on the order of 60%.  

Also, although I'm not entirely sure where the info is available now (perhaps
the NREL, National Renewable Energy Laboratory, in Golden, CO), fairly accurate
figures for monthly solar radiation throughout most of the U.S. exist.

The claim about highways is probably true, but begs the question, who would
cover highways with solar cells?  Obviouslyu no one who ever has been on the
Washington Beltway, or Rte. 128 in Boston or most highways in LA.

Highways are used as an example to show that the amount of land required to
provide solar energy for the nation.  It is also a number that can be obtained,
as opposed to the area of house roofs (which is a much more viable location for
solar cells).

The issue of space power satelites is a total CANARD.  No one has any viable
concept of how energy would be beamed down.  Beaming that much power on
microwaves is totally infeasible.  Imagine the disaster if the station was taken
over by terrorists.  They could kill hundreds of million within hours.  NO ONE
WOULD EVER BUILD A MICROWAVE BEAMING UNIT WITH THIS POWER.  THIS IS A MAJOR HOLE
IN ANY SPACE POWER SATELITE CONCEPT. 

Tony

From:	US4RMC::"[email protected]" "Andrew Yee, Science North"  
        1-AUG-1994 04:13:15.19
To:	[email protected]
CC:	
Subj:	New solar cells with record efficiency

European Space Agency
Press Information Note No. 07-94
Paris, France					29 April 1994

NEW SOLAR CELLS WITH RECORD EFFICIENCY

Under contract with ESA, European industry has recently developed high
efficiency solar cells for use in future demanding deep-space missions
such as the recently approved ROSETTA cometary mission.  The new solar
cells reach a 25% efficiency under deep space conditions.  The
eficiency is the ratio between the electrical energy produced by the
cell and the incoming solar energy.  The higher the efficiency, the
"better" the solar cell. 

Unlike telecommunications and Earth satellites which orbit near the
Earth and are normally powered by solar cell arrays, spacecraft
operating at a very large distance from the Sun (typically deep-space
probe) experience a solar intensity which is only about 5% or less of
that near the Earth.  This was the case for ESA's ULYSSES for instance
which, before reaching the Sun's poles had first to travel to Jupiter
at 780 million km from the Sun (Jupiter is five times further away
from the Sun than we are!).  Moreover, the equilibrum temperature of
solar arrays at those distances goes down to about -100 degrees
Celsius.  Current solar cells used all over the space world are not
generally made to operate at these low temperatures and solar
intensities.  They allow for 10 to 20% efficiencies in near-Earth
orbits but shows anomalous behaviour at deep space conditions. 

For this demanding environment deep-space probes have to use power
sources other than solar panels, because their electrical performance
degrades too much at these low light intensities and low temperatures.
Until now, deep space probes had to use thermonuclear power
generators, like the so called RTGs (Radioisotope Thermoelectric
Generators).  As RTG's technology is not available in Europe, ESA
therefore attempted to develop a power source based on very high-
efficiency solar cells. 

Under low-light low-intensity conditions, 25% efficiency has been
achieved on 6 x 4 cm Silicon cells.  The 25% mark represents the
highest efficiency ever reached worldwide with Silicon cells without
special optical concentration devices to increase the amount of
sunlight collected to be converted into electricity.  Another
breakthrough had already been reached by ESA a little over one year
ago with solar cells of a different technology, the Gallium Arsenide
(GaAs) type, where 23% efficiency was reached on 2 x 4 cm cells. 

This technology milestone in Silicon solar cells was reached by an
industrial team led by DASA (Heilbronn, Germany) with CISE (Milano,
Italy) as sub-contractor (CISE being also responsible for the
development of high efficiency GaAs solar cells). 

ESA expects that the new high performance Silicon cells could
profitably be used in deep space missions for Europe and that this
technology could also be of interest for near-Earth orbit space
applications as well as for Earth based ones.