The realm of plant-life on Earth, including the vast lawns of phytoplankton that lie in the first 50 - 100 metre depths of the oceans, in places, exists through photosynthesis. By definition, this is a process in which light from the Sun is harvested by chlorophyll (the green colouring matter in plants) which absorbs wavelengths in the range 400 - 700 nm (nanometres), a spectroscopic region known often as the PAR (photosynthetically active region). This accounts for around 47% of the total energy from the Sun, another 2% being from ultraviolet (U.V.) wavelengths, and the remainder mostly infra-red (heat). The initial light excitation events have quantum yields of close to 100%, but there are so called "dark reactions" too, and the energy losses in the various biochemical steps involved in using light energy to "fix" CO2 into carbohydrate and thence other components of life such as proteins and lipids accumulates to an efficiency often quoted as 6% of the total solar irradiance being converted usefully to biomass.
The detailed process of photosynthesis is very complicated, but it may be represented in summary by the following chemical equation:
CO2 + H2O ---> [CH2O] + O2
It takes an average of eight photons of light to "fix" each molecule of CO2, and assuming a mean energy of 217.4 kilojoules per mole (kJ/mol) for each photon, and a mean enthalpy (heat) of combustion for a CH2O unit of 467 kJ/mol (obtained as one sixth of the heat of combustion of glucose, C6H12O6, an hexamer of CH2O = 2801 kJ/mol), we may deduce that the process has an efficiency of:
(467 x 100)/(8 x 217.4) = 27%. So we might expect an overall photosynthetic efficiency of 47% x 27% = 12.7% as an absolute maximum. [Obviously the photon energies vary within the range 400 - 700 nm, they are not all absorbed with equal efficiency, but this is a rough upper limit, i.e. it won't get better than this!].
In reality, there is an energy required for translocation and respiration that amounts to about 33% of the total and in a forest, say, the presence of a tree "canopy" limits the available light to about 80%, so we might expect nearer to 12.7% x 67% x 80% = 6.8%, which is the commonly quoted mean. In reality this varies enormously, and sugar-cane turns over at around 8% whereas some plants may be as low as 0.1%, while for most crop plants, 1 -2 % is typical. Nonetheless, I am going to use the 6% value now.
Biomass Production Reckoned in Terms of Sugar.
To form one molecule of sugar (glucose) by photosynthesis requires locking-in 2801 kJ/mol worth of energy, according to the process:
6CO2 + 6H2O ---> C6H12O6 + 6O2. It is clear that the combustion of glucose involves the reverse of this process, and hence we may deduce the amount of energy for the forward step as being the negative of the heat of combustion, i.e. 2801 kJ/mol.
If we assume that our crop will be grown somewhere sunny, a mean solar irradiance of 5 kWh/m^2/day is reasonable. It is often quoted that the Sun's energy falling on earth amounts to 1,400 W/m2, but this is only true at the top of the atmosphere, and directly overhead. Below this, the absorption of radiation by the molecules of atmospheric gases attenuates the solar flux, and so the "solar constant" of 1.4 kW/m2 is geared-down to about 1000 W/m2.
The mean irradiance of a point on the Earth's surface depends on both the attenuation, the latitude and the cosine of the angle (theta) between the sun's ray and an imaginary "normal" drawn out from the surface point. Hence to get the average amount of sunlight falling on the surface, noting that at any instant half of it is in darkness, we need to integrate cosine theta over the surface of a hemisphere (i.e. between 0 and 180 degrees which comes out as 1) which multiplied by pi x r^2 gives just pi x r^2 or half the two-dimensional area of the hemispherical surface (i.e. 2 x pi x r^2. The surface area of a full sphere is 4 x pi x r^2). The integration allows for the fact that while cos theta = 1 at theta = 0, it is 0 at theta = 90 degrees. Put another way, while the irradiance is at a maximum when the sun is directly overhead (theta = 0), it is practically zero at the horizon (theta = 90 degrees). The integral of cosine theta = sine theta which amounts to 1 as taken between the limits of 1 and 0.
The upshot is that the average irradiance over the year is one fourth of the solar constant (roughly since this varies by a few percent over the duration of the annual orbital cycle of the Earth around the Sun) or about 350 W/m^2; i.e. 8.4 kWh/m^2/day. As noted, the value depends on latitude and is close to 5 kWh/day near to the equator or in the tropics, but about half that (it is reckoned at 100 W/m^2 or 2.4 kWh/m^2/day in northern Europe, e.g. the UK).
However, let's assume sunny climes in the following but note we need to probably double the figures for northern Europe.
If we assume 5 kWh/m^2/day, or a mean 208 W/m^2 then 6% of that gives an irradiance of 208 x 0.06 = 12.5 W/m2 = 12.5 (J/s.m2). Over a year, this amounts to 3,600 s/hr x 8760hr/yr x 12.5 (J/s.m2)= 3.94 x 10^8 kJ/m2.
Over an area of one hectare = 10,000 m^2, we have 3.94 x 10^5 x 10,000 = 3.94 x 10^9 kJ/ha, which is enough to form:
3.94 x 10^9/2801 (kJ/mol) = 1.4 x 10^6 moles of C6H12O6. Since the molecular mass of glucose is 180, this equals: 1.41 x 10^6 x 180 = 2.53 x 10^8 g = 2.53 x 10^5 kg = 2.53 x 10^2 tonnes, or 253 tonnes per hectare. So, in principle, if cellulose-digesting/fermenting technology can be implemented in time, we could make an awful lot of fuel; let's assume half of that matter is cellulose (at least, cellulosic) and the fermentation efficiency happens with a typical 60%, giving around half the fermentable weight of ethanol , and so that would give us 253 x 0.5 x 0.6 x 0.5 = 75.9 tonnes of ethanol per hectare which is significant. It is particularly significant in that it is the chaff, the crop refuse that would be fermented meaning that no special crop need be planted in competition with food production, and in the oil-dearth era at hand, we will need to grow as much of our own food as possible, rather than flying it in from far flung regions of the world!
Biodiesel produced from Algae.
As I wrote about in the posting, "Biofuel from Algae - Salvation from Peak-oil?", there is the possibility to produce biodiesel from algae farmed in ponds (or special flowing-systems with extensive pipe-networks) on a scale that is very large indeed compared with the yield of fuels that might be extracted from fuel-crops, e.g. soya. The technology is yet to be implemented on the large scale, and the quoted yields, though optimistically impressive, are arrived at by scaling-up the results from much smaller situations. For example, the wikipedia entry (en.wikipedia.org/wiki/Algaculture) quotes a yield of biodiesel of between 5,000 and 20,000 gallons (US) per acre, which is the equivalent of 42 and 170 tonnes of it per hectare. In my previous posting on the subject, I quoted another study which indicated a yield of 125 tonnes of biodiesel per hectare, and I shall use this as a reasonable average among estimates.
The energy value of biodiesel is given as 35.7 MJ/litre, and at an average density of 0.84 kg/litre, this is 35.7/0.84 = 42.5 MJ/kg. On the basis of these figures, we can estimate the amount of useful radiation that is hitting the Earth per square metre. So, a yield of 125 tonnes/ha = 125 tonnes/10,000 m2 = 12.5 kg/m2, which is a quantity of biodiesel that contains:
12.5 kg x 42.5 MJ/kg = 531.25 MJ of energy = 531.25 x 10^6 J/(3600 x 8760 s) = 16.85 J/(s.m2) = 17 W/m^2.
If the irradiance is 208 W/m2, the photosynthetic efficiency amounts to: (16.85/208) x 100 = 8.1%, which is possible but on the high side of what is normally found for the plant kingdom. The upper estimate of 20,000 gallons/acre amounts to 9.2% (19.2 W/m^2), while the lower 5,000 gallons/acre translates to one quarter of this, i.e. 2.6% (5.4 W/m^2).
There are many problems associated with growing algae and making biodiesel from them as may be seen from the "oilgae" link (top left hand side of blog), but I think there are reasons to remain optimistic about cellulosic and algae based methods for producing biofuels in the oil-dearth era. I still don't believe that we can match current supply/demand for petroleum based fuel, but it won't be a lack of solar-energy that will let us down.
On a final note, it is interesting to compare the energy yields of different fuel materials. So, the heat of combustion of glucose (taken as representative of wood etc. type biomass) is 2801 kJ/mol x 1000/180 (g/mol) = 15,561 kJ/kg = 15.56 MJ/kg.
c.f. 42.5 MJ/kg for biodiesel as we have deduced above. This factor of 2.7 is very close to the ratio of calories obtained per unit mass of fats vs. sugars in the diet, quoted in attempts to help us to lose weight, emphasising the point that fuel is fuel and energy is energy, whether it is burned in mechanical or animal machines such as our own bodies!
(2) Thomas L. Weyburn, "A Report on My Recent Investigations of Solar Energy Harvested by Photosynthesis in a Controlled Environment." (just google it!)