Da, sigur e o soluţie.

Să facem calculele:

Energia solară la nivelul pământului la latitudinea de 45 de grade ajunge la 1000Waţi /oră / metru2. Vara.
Eficienţa conversiei energiei solare în energie electrică este de 13% cel mult. În ultimii 30 de ani această eficienţă a crescut de la 9%  la 13%.
Asta înseamnă că pe fiecare metru2 cultivat cu celule fotovoltaice obţinem 130 waţi pe oră. La 12 volţi. Sau 24. Curent continu.
Pe ansamblul unui an numărul mediu de ore însorite la latitudinea 45 este de 3,6 pe zi. Asta înseamnă 3,6*130W = 468W /zi /m2. Adică consumul normal pe o zi într-o casă foarte, foarte săracă, fără niciunul din aparatele de mai jos. Doar câteva becuri chioare.(de 25 de Waţi la 12 volţi)
Consumul câtorva aparate casnice: (Waţi/oră) la 220 V
Frigider 350-600
Televizor 120-180
Maşină de spălat 700-1.000
Calculator ( PC) 250-350
Aspirator 700-1500
Fier de călcat 750-1600
Ca să putem să ne uităm la televizor doar, ne mai trebuiesc vreo 2-3 metri2 de celule solare. Vorbim de un televizor mic, de voiaj.
Uitaţi de boiler. Discutăm de 2-3.000 de Waţi. Adică 20-30 de metri2 de fotovoltaice.
Şi să nu uităm, discutăm de 12 volţi sau 24 volţi, nu de 220. Curent continu.
Preţul ?
Panourile fotovoltaice propiu-zis nu costă enorm. Metrul2  instalat ajunge pe la 300-500 de euro. Dar sistemul de stocare al energiei electrice e altă poveste. Pentru că discutăm de perioade lungi în care soarele nu apare deloc. Chiar şi vara se întâmplă să fie înorat mai multe zile la rând.
Preţul în 2003 al unui sistem fotovoltaic de 1 kW variază între 4500 euro şi 6500 euro. Pe scurt 4-6 euro / watt
Repet e vorba de 12 volţi.
Una din firmele care vinde în Romania celule olandeze. http://www.kerychip.dk/EnergieSolara.htm
Preţul unei instalaţii mari fotovoltaice 1000 MW, instalare şi operare timp de 20 de ani,  este de aproximativ 47 de ori mai mare decât al unei centrale pe cărbune - incluzând şi cărbunele. Excluzând bineînţeles impactul asupra mediului, impact pe care nu îl putem calcula. Pentru 30 de ani de operare, preţul devine doar de 33 de ori mai mare. Ted Trainer
Calculul rentabilităţii unei instalaţii solare arată că în momentul de faţă o instalaţie medie (450W)  în SUA ( cea mai ieftină piaţă - sponsorizată de guvern) de aproximativ 8500$ se amortizează în aproximativ 150 de ani sau mai mult dacă comparăm preţul plătit acum pentru electricitatea din cărbune. Ted Trainer
Dar înainte de a vă face mai multe iluzii, iată ce crede preşedintele Exxon-Mobil LEE Raymond  despre subiect:
Solar energy is not a viable replacement. It would take 84 square miles of solar panels to replace the energy that a single gas station sells on a single day. It would take solar panels covering all of New Jersey to replace the energy dispensed by just 100 gas stations in a single day. Exxon has 3,000 gas stations in the US and only has 13% of the market, and in his words "Do the math" noting else is feasable to replace oil as an energy source!
Mai jos există un studiu foarte bun despre celule fotovoltaice (in engelză)

Direct conversion of sunlight to electricity by photovoltaic cells is one possible road to success.  Under ideal conditions the theoretical maximum efficiency of a single-crystal silicon cell is about 25%, while the practical maximum efficiency is about 22%.  However common sense suggests some practical problems:

1)         Without expensive tracking gear the sunlight will rarely fall directly on the receiving surface.

2)         Protective glass will absorb some radiation, and unless the receiving surface is kept clean the conversion efficiency will deteriorate.

3)       While it is conceivable that the cost of the actual photovoltaic cells could fall as low as US$100 per kW of generating capacity — that is to about two percent of 1995 costs — the cost of the housing and supporting structures, even if not sun-tracking, will be substantial.  Some panels could be installed on the roofs of houses; but in any realistic appraisal, cleaning and maintenance problems would need to be taken into account.

4)       The seasonal problem:—  As a basic geometrical principle, the intensity of sunlight will vary according to the cosine of the latitude.  Thus the ratio of light intensity in winter at 20 degrees from the equator,[ii] to that at 50 degrees from the equator, is 2.5 to 1.  But this is only a minor reason that those of us who live in temperate climes should not pay much heed to results which emanate from subtropical regions.  The geometrical effects are compounded by the fact that, away from the equator, the days are shorter in winter.  Also it tends to be cloudier in winter than in summer.  It follows that any estimation of the potential of photovoltaics needs to be related to latitude and to winter weather.  Since cloud cover is variable, outputs averaged over the whole winter will not suffice; a daily picture is needed. 
One solution to these wintry problems would be to have a huge excess of generating capacity during summer days, storing the excess energy as hydrogen (perhaps to be used for transport, thus avoiding conversion back to electricity).  However, common sense demands that the cost of such schemes should be fully evaluated before they are presented as serious possibilities. 
If the end result of such studies demonstrate that use of photovoltaics is not viable in Europe, for example, then an alternative plan might be to generate the electricity from a region lying within, say, 25 degrees of the equator, and then convert the electrical energy to hydrogen.  This could be viable, but common sense demands that full account be taken of the efficiency of conversion to hydrogen (0.70), the energy loss in compression (about 30% for liquidisation), the transport costs, and the efficiency of conversion back to alternating current electricity (about 0.40 using a fuel cell to produce DC electricity, which can then be converted to AC with an efficiency of 0.93).  Overall this gives an efficiency of 0.75 x 0.70 x 0.40 x 0.93 = 0.20.  These figures might suggest an increase in electricity cost of, 1 / 0.20, five times.  But transmission and distribution costs are about the same as production costs (with fossil fuel generation), so the foregoing calculation does not inevitably lead to a five-fold increase in cost to the consumer.  However by time the extra costs involved in the double-conversion process, and transport costs, are taken into account, a five-fold factor may well be a useful starting guesstimate for item 6 below.

5)       The diurnal problem:—  Photovoltaics will only generate electricity during daylight, so there are two options: (a) a standby system could replace the photovoltaics during night time; (b) part of the energy generated during daylight could be stored.  The trouble with (a) is that it would require a completely separate generating capacity based on some unspecified source of energy.  The trouble with (b) is either that there is a loss of at least 1 - (0.75 x 0.40 x 0.93) = 72% in the processes of conversion, storage and regeneration (as we have seen), or that excessive amounts of raw materials are required to achieve direct electrical storage; e. g. lead or vanadium for batteries. 

6)       Any estimates that are made today about capital equipment costs, while we have the benefit of cheap energy from fossil fuels, will need to be substantially revised when we have to rely on more expensive renewable energy sources.  In other words, calculation of renewable energy costs has to start from a likely guesstimate of how much energy will cost when we have to rely exclusively on renewable sources.  Only on that basis is it possible to make realistic assessments of the capital costs of such things as power stations, transmission lines, wind turbines, and electrical storage units.  In other words, the calculation is similar to the type which has to be made for an aircraft when planning a very long distance flight: only after guesstimating how much fuel will be consumed on the flight can the average weight of the aircraft be determined, and hence average fuel consumption assessed.  If the guesstimate proves wrong it has to be revised and the calculation done again.  For renewable energy the calculation is far more difficult because of the uncertainty surrounding the eventual cost of renewable energy; as we have seen above, the increase could be five-fold

The foregoing list may serve as a sort of check-list for editors of academic journals.  Any submission which fails to address the issues listed above should, of course, be sent back to the author for further development!

       Nearly all data related to renewable energy are taken from Dr. F. E. (Ted) Trainer’s 1995 paper: Can renewable energy sources sustain affluent society?  However data on theoretical efficiency of silicon cells is taken from the McGraw-Hill Encyclopedia of Science.     In mid-winter the angle of inclination of the sun to the equatorial plane is about 23°.  The calculation of relative insolation is therefore cosine (20 + 23) to cosine (50 + 23), or 0.731 to 0.292, or 2.5 to 1. 


Solar energy. This is a favorite possible source of future energy for many people, comforted by the thought that it is unlimited. But, quite the contrary is true. The Sun will exist for a long time, but at any given place on the Earth's surface the amount of sunlight received is limited--only so much is received. And at night, or with overcast skies, or in high latitudes where winter days are short and for months there may be no daylight at all, or available in small and low intensity quantities. Direct conversion of sunlight to electricity by solar cells is a promising technology, and already locally useful, but the amount of electricity which can be generated by that method is not great compared with demand. Because it is a low grade energy, with a low conversion efficiency (about 15%) capturing solar energy in quantity requires huge installations--many square miles. About 8 percent of the cells must be replaced each year. But the big problem is how to store significant amounts of electricity when the Sun is not available to produce it (Trainer, 1995), for example, at night. The problem remains unsolved. Because of this, solar energy cannot be used as a dependable base load. And, the immediate end product is electricity, a very limited replacement for oil. Also, adding in all the energy costs of the production and maintenance of PV (photovoltaic) installations, the net energy recovery is low (Trainer, 1995).

Energy is essential to life, and energy from the sun is the ultimate basis for all life.  Well,  As one in the solar biz, here are the calculations used in the field. I live in New England and we use insolation tables that show average  sun hours (1 sun hour is equal to 1000w/m squared). At 13% efficiency  per module. That is 130 watts per sq meter of electric output for 1  sun hour. The average annual daily sun hours here is 3.6 per day.  So 130 watts * 3.6 * 365 gives the answer.  170,820 watt hours per  year per sq meter of module. 170 kWh / year per sq meter (rpughly 10  sq feet...100 sq feet is 1 kw of PV installed). Also note that the most beneficial use of PV is in the south for peak aircon load. Measured insolation in Arizona and much of Texas is in excess of 2200 peak hours per year. Module efficiencies of 18% are now realistic, giving conversion to a/c of 14%. With even a 20 sun concentrator these values go to about 22% and 17% respectively. So in much of the SW we can expect 370kWh/yr/sq. m. a/c out. MurrayÎn Australia  on average there are 5.5 hrs of usable sunlight per day, x 365 days = 2000 hrs (round figures) of sunlight.2000hrs x 1000W/m2 = 2MWhrs/m2/yrassume 12% efficiency = 240kWh/m2/




by Andrew Ferguson

 Photovoltaics (PV) is associated with several intractable problems.  The reader is likely to be aware of most of them: (a) high cost; (b) variability of output between summer and winter; (c) intermittency due to heavy cloud and the lamentable failure of the sun to shine throughout the night!  Pervading all these problems, and amplifying them, is a problem of comprehension which exists within the mind of almost everyone who pontificates on PV. 

In order to explain just how pervasive the problem is, I need to make a slight diversion.  A marvellous series of booklets is issued under the title Understanding Global Issues.1  The editor — who writes nearly all issues — is Richard Buckley.  To give an example of the extent of the coverage, here are the titles of the last five: (1) Democracy in America: taking power from the people; (2) Losing the Earth: land abuse and soil erosion; (3) United Europe: risks and rewards of enlargement; (4) The Battle for Oil: diplomacy, politics and big business; (5) The State of Israel: living with danger

So Richard Buckley has a mind to marvel at.  His grasp of every one of these issues is so comprehensive that he appears to have spent about a couple of years on research — yet there are ten issues a year.

When issue 117, Beyond Petroleum: the renewable energy revolution, came through my letter box, I wondered whether Buckley would have succeeded in penetrating the fog of misinformation that surrounds renewables, particularly PV.  When I found that he had not succeeded, I concluded that there must be a pervasive problem with PV, which can be summed up as total confusion on the subject by virtually everyone who writes about it.

Buckley covers other renewables in this issue of UGI beside PV, but in this article we will focus on PV. 

In the introduction, Buckley states, "A typical power station using nuclear energy or fossil fuels has a 1,000 MW [million watts] generating capacity.  To match this output, a wind farm would need 500-1,000 turbines, a solar farm would need 500 billion PV cells." 

These assertions take some unravelling.  The meaning of "output" is ambiguous: does it mean rated output or does it mean the actual output?  Anyone not familiar with the term 'rated output', or its synonym 'rated capacity', need only think of the 'rated power' of a car engine, which signifies its power at or near maximum output.  We can deduce that Buckley is referring to rated output, because he would more or less certainly be referring to wind turbines with a rated capacity of 2 MW to 1 MW, making the figures that he gives for the number of wind turbines (500-1,000) correct in terms of an assessment based on 'rated capacity'. 

It can thus be inferred that Buckley is referring to 'rated capacity' with respect to PV too.  Later on (p. 16) he tells us that, "A typical solar cell generates less than 2 watts."  There is some ambiguity even there, because it is not clearly stated that he is referring to 'rated capacity' — also known as peak power, with the watts sometimes being symbolized as Wp.  However, it is fairly obvious that he must be referring to peak watts, and it is thus clear that 1000 MW of rated capacity would require about 500 million cells to provide the same rated capacity.  Note that he makes a typographical error both there and on page 16, by referring to billion cells when he means million cells, but the 'billions for millions' mistake is one to which I too am prone, and we should think nothing of it!  The point merely needs to be cleared up before we proceed.

The matter that is really worth dwelling on is that it is vastly misleading to compare the rated capacity of nuclear and fossil fuel plants with the rated capacity of PV.  A nuclear or fossil fuel plant would operate in the range of 60% to 80% capacity factor (of course one reason for that is that demand is not always there, requiring production to be cut back). 

Let us take 70% as an average figure.  Thus a 1000 MW plant would produce power equivalent to 700 MW (i.e. 700 x 24 x 365 megawatt hours per year).  A PV array, using 17% efficient cells, packed into modules sufficiently tightly to provide an efficiency of 14.3% (what BP Solar call 'high efficiency' modules), operating in a good situation like Toledo, in sunny Spain, with an annual insolation of 200 watts per square metre (2000 kilowatts/hectare), would achieve an output, delivered to the grid, of about 14% of rated capacity.  Thus to produce the 700 MW of actual output referred to earlier, the PV capacity would need to be 700 / 0.14 = 5000 MW.  

At a cost of $5 per watt (see later) the cost of 5000 MW capacity would amount to $25 billion, twenty-five times the cost of the fossil fuel plant (which normally cost about $1 per watt of rated capacity, and hence 1000 MW would cost $1 billion). 

At a peak watt output of 143 watts/m2 (see later), the module area required to provide 5000 MW of capacity would be 35 million m2.  In 2000, BP Solar produced a booklet which showed, by way of comparision with photovoltaics, ranges of costs for cladding materials.  The booklet gave a cost range for "Glass Wall Systems" of $560-$800 per m2.  Taking an intermediate cost of $680/m2, 35 million m2 of glass wall cladding would cost $24 billion, i.e. twenty-four times as much as the fossil fuel plant.  It is because of this cost of basic materials, that claims for dramatic reductions in the future cost of PV cells are largely irrelevant, even if true.

Buckley should not to be blamed for his failure to unearth these facts, because the PV industrial lobby is singularly adept at hiding them.  Moreover, the lobby is aided and abetted by the likes of the World Watch Institute, which, although it produces otherwise impeccable books of statistics, keeps to the etiquette of the PV lobby, adhering to the rule that the only figure that should ever be mentioned pertaining to PV, at least in polite society, is 'rated capacity'!  Never, perish the thought, should it be breathed abroad that, even in sunny Spain, the actual output is only 14% of the rated capacity!

Buckley did his homework well, and searching through the PV lobby literature, he picked up the hoary old favourite (p. 6): "According to the National Renewable Energy Laboratory, the whole US electricity demand could be met from the 'solar energy resource in a 100-by-100 mile area of Nevada'."  As we shall see, that might be strictly true, but it is more impressive for what it omits than for what it states.  Let us note, for present, that the area considered is 26 billion square metres, but as an allowance has to be made to prevent shading, it represents about 13 billion square metres of module.

The trouble with solar energy is that it is intermittent, thus some of the electricity has to be stored.  Hydrogen is the method generally touted for storage, using fuel cells to recover the energy as electricity (expense being apparently no object!).  Let us suppose that half the 3.5 trillion (3500 billion) kWh of electricity used in the US each year can be used directly, while the other half has to be stored as hydrogen and regenerated.  Based on average US insolation, this would reduce the effective energy-capture of the PV cells to 116 kWhe/m2/yr.  For Nevada, it would be in the ball park to uprate that by 20%, to 140 kWhe/m2/yr.  Thus the area of module needed would be 3.5 trillion kWh divided by 140 kWh/m2 = 25 billion m2. The National Renewable Energy Laboratory, we may recall, suggested 13 billion m2 would do the job, but they did that by ignoring the fact that it would be completely impossible to replace US electricity by PV electricity, without being able to store some of it so as to satisfy varying demand, and so that everyone did not have to light their candles at sunset!

Now let us look at the cost.  It is possible to envisage the cost of fully installed PV modules coming down to $5 per peak watt.  Thus 1 square metre of module, with a rated capacity of 143 watts, might come down to a fully installed cost off $715.  Thus the total cost of the aforementioned 25 billion m2 would be 18 trillion dollars. 

Of course no one is going to consider spending 18 trillion dollars in one year, so let's take it more easily, and consider investing just 200 billion dollars a year.  It would be in the ball park to assume that there are 200 million American families, so that would be an annual investment of $1000 per family.  On that 200 billion dollars a year basis, the number of PV cells would continue to increase year by year until 30 years had passed.  At that time, the cells built 30 years ago would become time-expired and it would be necessary to go on spending $200 billion a year to replace the modules built 30 years earlier, so as to maintain the level of PV electrical output.

And what would the level of output be?  Well after 30 years, at $200 billion a year, we would have spent 6 trillion dollars.  That is only 33% of the 18 trillion dollars needed to cover the entire electricity demand.  In other words, this rate of investment, if continued indefinitely, would replace 33% of the US present electricity demand.  However, at the current rate of US population growth, the increase in population after 30 years would be 37%; thus the investment would not succeed in matching the increase in demand occasioned by increasing population.  Moreover, we have much underestimated the cost by not including the electrolysis equipment and the fuel cells which would be needed to produce that half of the electricity, which we have assumed must come from stored sources; neither have we allowed ourselves any excursions into the painful subject of the energy inputs needed to make all this equipment.  Suffice it to say that, although I would put the figure slightly higher, 20% of the total output, over their 30 year life, of the modules is likely to be required as input when building not only them, but the associated installation equipment.  Moreover the output during the first year of module building (with, on average, half the modules producing) would be one sixtieth, 1.7% of their lifetime output (lifetime of 30 years assumed).  Thus it would be many years before the total output of the PV modules exceeded the total input needed to make them and the related installation equipment. 

It is apparent that there is a large gap between reality and the implications of the National Renewable Energy Laboratory.  That particular industrial lobby, for such it can only be assumed to be, is but one of several.  It was in order to respond to similar extravagant claims by the New York Solar Energy Industry Association, that I worked out the above mean energy-capture figure of 116 kWhe/m2/yr (for average US insolation), which I uprated to 140 kWhe/m2/yr for Nevada.  Note that both these figures are applicable to an assumption that half the electricity needs to be stored.   I admit that there could have been some fine-tuning of that figure for this particular analysis (perhaps less than half of the energy needs to be stored), but the unknown costs, as mentioned in the previous paragraph, are so great that it seemed unnecessary to strive for accuracy.

What to do about the pervasive problem

So we have established the existence of the pervasive problem, but what can the ordinary person do about it?  Understanding PV is like riding a bicycle.  It seems enormously difficult at the start, but dead easy once one has the hang of it.  The only thing that the ordinary person can do is to get the hang of it, and then never believe a word that comes out of the industry, and its protagonists, until one has checked their assumptions.  So let me finish by helping you get the hang of it.

Cells of 17% efficiency, packed sufficiently tightly to achieve modules of 14.3% efficiency, will yield 143 watts per square metre of module, when subjected to 1000 W/m2 of insolation.  In sunny Spain, with an annual insolation of 200 W/m2, the modules will deliver, to the grid, 14% of their rated capacity; that is to say, 14% of 143 watts, i.e. 20 watts, or (20 x 24 x 365) / 1000 = 175 kWh/m2/year.

So what happens in say Reading, England, where annual insolation is only 110 W/m2 (and incidentally 70 W/m2 in winter)?  The capacity factor drops proportionally; that is it drops to 14 x (110 / 200) = 7.7%, so the output would be 7.7% of 143 watts, i.e. 11 watts per square metre, or 96 kWh/m2/year.

You will have noticed that there is another way of looking at things.  20 watts is 10% of 200 watts and 11 watts is 10% of 110 watts, so it can also be said that, with a panel correctly angled so as to catch as much sun as possible, the PV module will deliver, to the grid, 10% of the insolation which would fall on a horizontal panel of the same area (i.e. the amount of insolation according to how insolation is measured). 

So it is all perfectly simple.  Now its over to you to challenge the hype of the industrial lobbies!  You will probably spend most of your time countering claims about fantastic breakthroughs, just around the corner, which will reduce everything to a fraction of the present cost.  The reality is that, as fossil fuel becomes more scarce, it is equally likely that prices will increase (in the USA, the first signs of natural gas shortage are plainly evident, with gas prices in June 2003 almost twice what they were a year ago, and deals in the UK to import gas from Russia are made against the background knowledge that by 2020 the UK will need to import 80% of its gas). 

1.             UGI on the web <www.global-issues.co.uk>  Subscriptions (10 issues), UK Ł29.50, rest of World Ł34.50.  Understanding Global Issues Limited.  The Runnings, Cheltenham GL51 9PQ, England. 

In a Kilowatt-Hour, there are 3413 BTU.   In a gallon of gasoline, there are 115,000 BTU.  
A good solar system 'might' generate a KWH of electricity per square meter IF you had 100% convernsion efficiency, and do it for an average of 8 hours a day.  However, you now have on the order of 15% efficiency?   So you might get 5000 watt-hours per day out of your sq. meter solar panel?   Or about 17000 BTU equivalent.
A good gasoline station can sell 30,000 gallons a day.   That is 30,000 times 115,000 BTU, or 3.45 billion BTU.
If your solar array generates 17000 BTU a day equivalent two hundred thousand  plus square meters of solar panel to replace the energy in one gas station, per day.
If the skies are cloudy for one day in five, you will, of course,  need more. So we up it by another 20% to allow for cloudy days.  IF you live in Seattle, well, better up it by 200%.  
Now, let's see....200,000 sq meters is a good chunk of solar  cells....what's the price per sq meter of solar cells?   and the mounting?  and the wiring?  and controllers?   And what do you do  with it to 'store it'?  
A sq KM has 1000 x 1000 sq meters, or 1 million sq meters.  So this  would take not more than a sq KM.      At 15% efficiency.   Assuming 8 hours of full sun a day, all the time, year round.  In a good location.    Likely, to compensate for shady days, rain, snow, fog, etc, you would need 3 times the solar collecting area.  

NOw if your gas station sells more, then you need a bigger solar array..some pump 100,000 gal a day.  

Calculaţi-vă singuri preţurile pentru orice variantă. Dar e vorba de California. După ce că au soare mult mai au şi subvenţii de la guvern.
The panels are about $10k, the inverter is a bit over $2k and the
mounting rails, connectors, breakers, wiring, etc. are a few hundred
each. They estimate $3200 installation by a certified dealer. Add a
little for shipping and you're at $17.5k, with a 57% panel/43% BOS
split. A local dealer might price the panels and hardware higher but
do the installation for "free", which would tend to distort BOS.
Note that total cost is just under $6k/Wp, which is about as good as
I've seen. This only makes economic sense in California or similar
locale where rebates abound and incremental electricity costs 18
cents/KWh. To compete straight-up with fossil fuels the cost needs
to drop below $2/Wp. Current crystalline PV technology won't get
there anytime soon, but startups pursuing thin-film and dye-polymer
cells have price targets at or below $1/Wp. SES claims Dish-Stirling
could hit $1/Wp in mass production vs. $10/Wp for the current hand-
built systems.

De fapt iluzia solară este de scurtă durată. Odată cu petrolul, industria care manufacturează, transportă şi vinde panouri solare dispare şi ea.

How to consider, then, the EROEI for the most complex sources of
energy? Just by inference? By indirect methods? By energetic
equivalents, linked to money, a substance almost as volatile as the
soul? If a solar photovoltaic panel is in place only after four
hundred fossil fuel powered factories provide the piece parts, which
in turn, need thousand of parts, materials and services coming from
more and more fossil fuel powered factories and companies, how to
calculate? Only by the percentage that the four hundred factories
devote in the materials and services to these solar activities? What
if one, only one of these 100% fossil fueled factories is stopped?
Could the solar panels be produced? If not, we need the 100% of this
fossil fuel factory, not only the percentage actually devoted to
produce parts for solar panels. And what about the thousands of
second level factories supplying to the first level factories and
service companies?

In summary, the present fossil fueled industry does not need the
solar photovoltaic or wind industry to exist. However, the solar
photovoltaic business needs the whole of the fossil fuel industry to
start with and probably to stand still. This forum touched some time
ago the subject of the solar breeder factories, but did not
elaborate. This will be the clear proof that wind, nuclear or solar
are sustainable. All other arguments remain doubtful.