Wednesday, May 16, 2007

Photosynthesis and Peak Oil.

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!

Related Reading.
(1) http://en.wikipedia.org/wiki/Algaculture
(2) Thomas L. Weyburn, "A Report on My Recent Investigations of Solar Energy Harvested by Photosynthesis in a Controlled Environment." (just google it!)
(3) http:www.theoildrum.com/node/2531
(4) http://www.life.uiuc.edu/govindjee/whatisit.htm
(5) http://www.upei.ca/~physics/p261/content/sources_conversion/photo-_synthesis/

12 comments:

Anonymous said...

Dear Chris,

This is a very interesting analysis. Could you precise one piont for me regardin the calculation on the algae biodiesel.

You calculated 12.5 kg biodiesel/m2 can be acheived at a photosynthetic efficiency of 4.7%. However, should you not take into consideration the yield of algal-oil harvesting and processing into biogas as well. In order words, should not you base the calculation on the energy content of the parent oil or, by taking into account the production yield. I would be very happy if you know these values!

Thanks

energybalance said...

Hi,

now the figure I have quoted is that given for "biodiesel"/m2, and so the quantity of algae/m2 must be somewhat greater.

So the fraction of one to the other would give the yield of the process.

If the yield is 50% say, then we could double the photosynthetic efficiency based on algae?

Do you mean also energy costs incurred in the processing of algae-biodiesel itself?

But in terms of making an estimate of the amount of biodiesel that could be grown, I presume any lose have already been factored-out in e.g. the wikipedia values of up to 20,000 gallons/acre?

So yes, the calculation is based on both the energy content of the fuel and the production yield of "biodiesel" directly, not of algae.

Regards,

Chris.

Anonymous said...

Thanks,

I have found data suggesting the yield of biodiesel production the algae parent oil transesterification) is about 80%. Thus, producing 12.5 kg of biodiesel/m2 requires 15.6 kg oil/m2. Assuming (optimist) a oil content of 50 % (w:w), this would require an algae productivity of 31.25 kg/m2. Assuming 1 g of algae requires 1.83 g of CO2, this translates into a CO2 fixation rate of 1.3 Kmole/m2, or an electron requirement of 10.4 Kmole/m2. This is also equivalent to 10.4 kmole/m2x217.4 kJ = 2261 MJ/m2 = 2261/(3600x8760) = 71.7 W/m2

This represents 20% of 350W/m2 PAR, which seems unrealistic.

Would you agree with my calculations? It is all a bit confusing to me.

Following this, I would like to know how much area is necessary, for instance, to fix 1 kg CO2/year?

I apologize if I must remain anonymous for professional reasons,

energybalance said...

Hi! Here are a few thoughts. O.K., 47% of 350 w/m2 total solar flux is PAR = 164.5 W/m2. So that is the amount in principle available to be absorbed.

I think I see what you mean, and the critical number seems to be the 8 (photons) x 217.4 kJ/mol = 1763 kJ/mol.

So, we can work out that 31.25 x 1.83 = 57.19 kg of CO2/m2 = 1,300 moles of CO2/m2.

Now that suggests (as you have it) that 1763 x 1,300 = 2.29 x 10^6 kJ/m2 = 2.29 x 10^9 J/m2 = 2.29 x 10^9 W/m2/year, and dividing by (3600 x 876) gives us an absorbed power of 72.6 W/m2, which is about what you get. (Now that's only 72.6/164.5 = 44% of the total available which seems quite reasonable to me. In a forest it could by 67% (losing one third for other metabolic processes) x 0.8 (for canopy effects) = 54%. So it's less than this.

However, the process only recovers 27% of that energy in terms of biomass even at 100% light harvesting efficiency. Hence the overall efficiency is 44% x 27% = 11.9%.

Now that is still a bit high (>6%), but from reading some of the literature, those biodiesel figures were arrived at from experiments done on 1000 m2 ponds into which CO2 was being actively pumped, 90% of the CO2 pumped in was absorbed apparently, and so the yield was probably artificially forced-up, rather than just leaving the ponds open to the CO2 concentration from the air.

Regarding how much area would be required to fix 1 kg of CO2 per year, just work it out according to your figures. So 57.19 kg of CO2/m2; hence 1 kg would take 1/57.19 m2 = 0.0175 m2 = 175 cm2 or a patch of about 5 inches square.

Hope this helps.

Chris.

Anonymous said...

Hello,

I was just wondering where your figure of 8 photons per CO2 comes from.

Thanks

energybalance said...

Hi,

sorry I missed this one - about where does my figure of 8 photons come from. I just knew it is the truth! But it's quited on various web-sites including the wikipedia one about photosynthesis.

Hope this helps.

Regards,

Vhris.

Anonymous said...

The calculations sewem to imply that the energy flux is assumed to be 350W/Sq M. then this is multiplied by 8760 hours per day.

I doubt that this is realistic, as I look out the window now and it is quite dark. You have to assume a factor for the percentage of time the sun shines.

Anonymous said...

Ok 8760 hours per year

energybalance said...

The 350 W/m^2 figure does specifically refer to the irradiance at the top of the atmosphere, but it IS averaged over the light and dark parts of the day/year, i.e. 350 W/m^2 is 1/4 of the solar constant of around 1.4 kW/m^2.

I have amended the calculation to make it a bit more obvious and to allow for the attenuation of the sunlight as it passes through the atmosphere.

The figures given refer (as I say) to somewhere in the tropics or closer to the equator for which I am assuming a generous 5kWh/m^2/day or a yearly average of 208 W/m^2.

Both units of kWh/day and mean annual W/m^2 are often quoted so it's helpful to have both to hand.

I hope this is clearer now.

By the way, there are some much more recent postings about photosynthetic efficiencies and algae production which might be of interest to you, based on some quoted yields by a company called PetroSun.

Regards,

Chris Rhodes.

billahrens said...

OK, now I understand energy flux numbers. 208 W/sq meter is reasonable average at equator.

Bill Ahrens (former anonymous)

Unknown said...

Chris,

I enjoy your blog and read it quite often. I have been sailing around the South Pacific now for 3 years and find myself in Bundaberg OZ where there is ample amounts of cane growing.

Having visited the sugar processing factory here, I believe you are incorrect in this statement:"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"

Cheers,

Ken on S/V Trim
I believe one would have to utilize the sugar itself to get the numbers you quote.

Chris Rhodes said...

Hi Ken,

thanks for your messages. You're on a boat... wow! Well, I'm glad you are still reading my blog as you go on your voyages.

I think I see what you mean. I'm not talking specifically about sugar-cane but for the purposes of illustration assuming that cellulose plant matter (chaff) can be converted to sugars and fermented into ethanol.

For sugar cane you are quite correct of course, that it is the actual sugar that is extracted and this is fermented and converted to ethanol.

There are considerable efforts to break down cellulose to simpler sugars that can be fermented to ethanol, which is what I was trying to illustrate, in terms of the likely amount of plant mass that can be grown given prevailing levels of sunlight, and the ethanol, yield that might be got if that cellulose component of the plant were turned into ethanol at typical yields for sugar fermentation.

Does this help?

Regards (from a rainy office in South East England!),

Chris