Here are some comments on the article in OGJ, July 10, 2000, p. 80, "The solar photovoltaics business: Has its time arrived?"
Fig. 4 "Global Energy Mix Outlook," shows renewable energy contributing 4 or 5% to global energy supplies during the 1990s. OGJ's 1989 Energy Statistics Sourcebook (pp. 394-96), however, shows that the "renewables" category contributed about 2%. Renewable energy, at this time, is the exclusive domain of hydroelectric generation. The combined output from all other renewable: solar, wind, geothermal, and biomass, has been negligible and has not been shown in energy statistics.
Using OGJ's statistics for 1987: converting hydroelectric production to btus and dividing it by worldwide primary energy consumption, hydro's contribution is 1.93%. Moreover, the hydro category has little growth potential, borne out in Fig. 3, which shows nearly constant hydroelectric output from 1965 to 1997.
Fig. 5 "Future Energy Path Options" shows how renewable energies might contribute anywhere from 30 to 40% of global energy by 2050, depending on technology development and policy changes. Fig. 5 is obviously in error by showing 20% of global energy coming from renewables during the 1990s, whereas OGJ's statistics show 2%. With little growth potential for hydro, future growth of renewables will have to come from solar, wind, geothermal, and biomass.
Let's examine the resource potential of photovoltaics, unconstrained by technical, economic, and policy questions, thus determined only by the amount of electricity that can be generated by sunlight reaching the earth's surface.
From a plot of "undepleted insolation," one can read the amount of solar energy striking the earth's surface for any month or latitude. "Undepleted insolation" means that the sky is free of clouds and haze. My reference plot, from a meteorology textbook, shows undepleted insolation as a family of curves ranging from zero to 1,100 cal/sq cm per day, over the entire year and for all latitudes.
As examples, I averaged undepleted insolation over 12 months for 40° latitude (a typical location in the US), and 50° latitude (typical for Europe). Converting units, the sun provides 325 w/sq m for an average day of the year at 40° latitude, and 281 w/sq m at 50° latitude.
Insolation (w/sq m/d) varies throughout the year: 484 in June, and 155 in December at latitude 40°; 484 in June, and 92 in December at latitude 50°. The "test conditions" of 1,000 w/sq m, mentioned in the article, would be available only during midday hours on a cloudless day in May, June, or July.
Applying this information to the rooftop system in Switzerland, shown in the photograph on page 80, we can estimate what wattage can be generated. From the information given regarding the size of the solar panels, total surface area of the panels is roughly 370 sq m. Assuming 15% conversion of insolation to electrical output, and a sunlight factor (determined by cloudiness) of 60%, the system will generate (281)(370)(0.15)(0.6)= 9,360 w each day over the course of 1 year. During a typical day in June, the system will generate (484)(370)(0.15)(0.75)= 20,150 w. In December, by contrast, with short days, low sun, and more clouds, generation will drop to (92)(370)(0.15)(0.35)= 1,800 w/day.
The drop in electrical generation by a factor of 10 from June to December means that either conventional power generation will be needed to fill the shortfall, or some electricity generated in the summer might be stored for use during the lean winter months. The photovoltaic system therefore, cannot operate independently while supplying energy on the basis of demand.
Looking ahead a couple of human generations, to a time when a large power plant might be developed using photovoltaics, we can estimate its requirements for solar panels and land. Let us specify that the hypothetical plant is capable of generating 1,000 Mw during the month of June. For a location in the desert of western Utah at 40° latitude, the sunlight factor might be 90% in June. The area for solar panels can now be estimated:
Solar area = 109 w/(484 w per sq m) (0.15)(0.9) = 1.53 x 107 sq m
Therefore, 15.3 sq km of photovltaic surface is required. Estimating an additional 50% of land coverage for non-producing surface, access roads, and auxiliary facilities, the land requirement would be 23 sq km, or a square parcel nearly 5 km on a side. What kind of environmental impact would such a facility have?
Wintertime generation would be a small fraction of that in the summer. In December, with somewhat more cloudiness, the plant would produce:
(155 w/sq m)(15.3 x 106 sq m)(0.15)(0.6) = 213 x 106 w
That is 21% of the output in June. In order to maintain consistent output, 787 Mw of supplementary generation would be needed to make up the wintertime shortfall, but it would be largely idle during May through August when output from the solar facility was near maximum.
By comparison, if the plant was located at 30° latitude (the best sites would be near Big Bend National Park, Texas), land requirements would be the same, but the drop in wintertime generation would be less. June insolation is 484 w/sq m/day; December is 213. If cloudiness in June and December are equal, then the required generation makeup for December would be roughly 560 Mw, while the solar facility generates 440 Mw.
Ther foregoing analysis shows that the role of solar power generation will be limited to supplementing conventional power. For example, solar generation would allow for greater peak consumption during midday hours, and would be insurance against rolling blackouts during heat waves. Nevertheless, the 100% variation of generated solar power, which is determined by season, cloud cover, and hour of the day, requires an equal amount of conventional power generation to offset that variation.
E:mail comments can be sent to: [email protected]
Tom Standing
San Francisco
