There is much communicated on the prospect of growing oil-rich algae to convert into diesel as a replacement for the inevitably declining supply from crude oil. Now, PetroSun Inc. has announced that their Rio Hondo algae-farm in Texas will begin operations at its pilot commercial algae-to-biofuels facility . The farm comprises 1,100 acres of salt-water ponds from which it is thought will be produced 4.4 million US gallons of algal oil along with 110 million pounds of biomass, annually. Expansion of the farm is intended in order to provide fuel to run existing or putative biodiesel and bioethanol refineries, owned or part-owned by PetroSun. Such an open-pond design effectively consists of a nutrient-loaded/fed aqueous culture medium (see "Could Peak Phosphate be Algal Diesel's Achilles' Heel?", posted here on 6-4-08), in which algae will grow close to the surface, absorbing CO2 from the atmosphere in the process through photosynthesis - hence the other vital ingredient is sunlight.
I am interested in the likely photosynthetic efficiency (PSE) of such processes, and of crop-agriculture too, hence I shall now attempt some illustrative sums regarding their PSE viability:
4.4 x 10^6 gallons (US) of oil = 4.4 x 10^6 gal x 3.875 l/gal = 1.7 x 10^7 l.
1.7 x 10^7 l x 0.84 kg/l = 1.43 x 10^7 kg = 1.43 x 10^4 tonnes of oil.
If 110 million pounds of biomass is also produced, this amounts to 110 x 10^6 lb/2200 lb/tonne = 5 x 10^4 tonnes. Hence the oil is 22% of the total "mass" produced.
1,100 acres/ 2.5 hectares/acre = 440 ha.
Therefore, the oil-yield = 1.43 x 10^4 t/440 ha = 32.6 tonnes/ha. To "grow" this much oil would take: 7.19 x 10^8 tonnes/32.6 t/ha = 2.21 x 10^7 ha.
If the total area of the US is 9.8 x 10^8 ha (that's land plus water), and the US fuel consumption may be estimated as 0.25 (one quarter if the world's total) x 0.7 (proportion of oil used for transport) x 30 x 10^9 barrels/7.3 barrels/tonne = 7.19 x 10^8 tonnes, to grow enough algae to produce an equivalent amount of algal oil would take:
2.21 x 10^7 ha/9.8 x 10^8 ha = 2.3% of total US area.
For the entire world, the requisite land area is 4 x that = 8.84 x 10^7 ha, to be compared with a total surface area of around 500 million km^2 (about 147 million km^2 being land), or <0.2% of it.
If the UK switched all its transportation over to diesel engines, which are more efficient in terms of tank to wheel miles than spark-ignition engines that burn petrol (gasoline), we would need 40 million tonnes of oil (assuming oil = diesel, since the actual yields of diesel from oil are uncertain).
This amounts to: 40 x 10^6 tonnes /32.6 t/ha = 1.23 x 10^6 ha, which is about 5% of the total area of the UK mainland.
So, what about the photosynthetic efficiencies incurred?
(1) I am taking a good-mean of 5 kWh/m^2/day. Quite often, values of W/m^2 are quoted over the whole year, which is a bit misleading, since the solar radiation hitting the Earth (insolation) varies from region to region and according to time of day and the changing seasons. However, dividing 5 kwh/m^2/day by 24 hours gives around 200 W/m^2, to be compared with around 350 W/m^2 hitting the top of the atmosphere, as an annual mean. This would correspond to a sunny clime, e.g. Arizona, Australia or central Africa; it is probably less than half this for northern Europe.
Expressing the insolation as kWh/day is more realistic however. Hence:
5 x 10^3 Wh/day x 365 days x 3600 s/h= 6.57 x 10^9 W/m^2/year. Since W = J/s, we have:
6.57 x 10^9 W/m^2/year x 10,000 m^2/ha = 6.57 x 10^13 J/ha/year (i.e. the amount of solar energy falling on each hectare of land).
If we assume that "diesel" uniformly contains 42 GJ/tonne of energy, each ha yields:
32.6 x 42 x 10^9 = 1.37 x 10^12 J.
Thus, the PSE on oil is: 1.37 x 10^12/6.57 x 10^13 = 2.1%.
But only 22% of the total is oil and the rest is (other) "biomass". If we assume that 1 tonne of biomass contains 15 GJ of energy (about the same as for glucose), we have:
32.6 tonnes x 15 x 10^9 x (100-22/22) = 1.73 x 10^12 J/ha, which gives a PSE of:
1.73 x 10^12/6.57 x 10^13 = 2.6%, and so the overall PSE for the PetroSun process is 2.1% + 2.6% = 4.7%
(2) Now this seems quite reasonable, and is in the 6% "ballpark" usually given for the amount of the total solar radiation that is used by plants in photosynthesis, as I have described previously. However, at 32.6 tonnes of oil per hectare, the yield is rather less than the 5,000 - 20,000 gallons/acre quoted in Wikipedia  for algal oil production.
5,000 gals/acre = 12,350 gals/acre = 12,350 x 3.875 = 47,856 l/ha, and assuming a density of 0.84 tonnes/m^2, this amounts to 40.2 t/ha, as the lower estimate.
[Assuming, again, a sunny 5 kWh/day or 6.57 x 10^13 J/ha/year], if 50% of the algae is oil with an energy content of 42 GJ/tonne, it contains: 40.2 x 42 x 10^9 = 1.69 x 10^12 J, which equates to a PSE of 1.69 x 10^12/6.57 x 10^13 = 2.6%.
Then there must be another 50% of "other" biomass, which at 15 GJ/tonne, amounts to:
40.2 x 15 x 10^9 = 6.03 x 10^11 J (PSE = 0.9%), and so the total energy of this high-oil crop = 1.69 x 10^12 + 6.03 x 10^11 = 2.29 x 10^12 J, giving a total PSE of 3.5%, which is also reasonable.
I am worried, however, about the upper Wikipedia limit of "20,000 gallons/acre". At 4x the above this implies a PSE of 3.5% x 4 = 14%! which seems far too high, being above the theoretical PSE limit, at which around 12.7% of the entire insolation is usefully absorbed by a photosynthetic entity. It is conceivable that special methods are employed to increase the yields, e.g. CO2-injection or forced-UV conditions to increase the PAR of the solar spectrum. At any rate it could not be achieved in standard open-pond systems. If someone can prove me wrong here I should be most interested in knowing the details.