Wednesday, April 09, 2008

Oil from Algae: Photosynthetic Efficiencies?

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 [1]. 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 [2] 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.

Related Reading.


Anonymous said...

I have seen recently na article on CNN about a company that grows algae in "vertical bioreactors". The link is
Unlike solar panels that require to be spread out in the sun, I wonder if this technique can bring a better oil per square meter yield. Is that one of the innovations you were already aware of?

Professor Chris Rhodes said...


no, I didn't know about this technology. I was thinking about open-pond systems. I shall read-up on these "vertical" varients.



Anonymous said...

The 20000 number is definitely referring to photobioreactors, which have about a 28 fold increase in algal density and allow for the use of monocultures of algae with much higher lipid contents.

Sorry about posting anonymously, I'm writing a review of all of this information as we speak, and happened upon this page while trying to find information about the energy balance for algae-based biodiesel (hint: no one has done it because no one knows how it's going to be processed yet).

Professor Chris Rhodes said...


thanks very much! That clears that one up! You say no-one has done it yet - does that include the various companies that are claiming to be producing algal biofuels (or are about to be!)?

(hint: we seem to be running out of time to replace oil).

My understanding is that extracting the algal-oil is not clear-cut and may involve using organic solvents (made from oil) or enzymatic degradation of the cells. If that can be done then presumably transesterification would turn it into biodiesel. But the scale of the technology would have to be very big overall!

I should very much like some details of your review once you have finished it - if that's possible?



Anonymous said...

Hi energybalance:

Not sure if you're still reading, but regarding the extraction of algal oil and transesterification, this is a good paper on it:

- Different Anonymous Guy

Anonymous said...

Also, for a more scientific look at algal yields and species, see the excellent review/compilation by Chisti:

Professor Chris Rhodes said...

Thanks Anonymous,

just downloaded it, and it looks fascinating.


NikC said...

Hi Chris

I was checking some background info on Algae after reading an article “Algae on the Move: The 2008 Algae Biomass Summit Wrap-up” in Renewable Energy World <> when I came across your site.

You’re right 14.7%+ is too high for PSE, its limited by the nature of PS process. I can’t remember the exact figures but the maximum practical yield, under ideal conditions, is between 5 and 10%. High concentrations do not improve the situation as only the algae near the surface will photosynthesise and if the light’s too strong they can’t use it all so high agitation is also required.

I attended the International Society for Applied Phycology (algae) conference this summer in Galway and picked up some interesting data on algae, the best algae brains in the world were there so I am fairly confident that their figure of just over 10% absolute maximum for photosynthetic efficiency is correct. If I recall correctly, figures for open (raceway) pond reactors of a few percent would be about right.

I was mulling over the figures for replacing all our oil with algae biomass rather than just diesel, thought you may find my workings of interest:

World oil consumption is ~= 85Mbboe/day @ 6.1GJ/bboe (bboe = barrels of oil?)
= 5.185e17J/day = 12.345Mtoe/day (@42GJ/t)
> Most of world’s oil energy is used for fuel.
> Use total algal biomass is for fuel manufacture e.g. via hydrothermal liquefaction (a form of pyrolysis).
> Crude oil to fuel (well to tank) losses ~10%
> Algal biomass to fuel (well to tank) losses ~25%

Realistic practical PSE for algae is about 5%, possibly slightly higher for enclosed reactors and probably about half for raceway ponds, assuming they run 24/7:
,’, @ 5KWh/day energy yield/capture by algae daily yield is
5,000 * 0.05 * 3600 = 900,000J/sq-m

Production space:
Therefore algal biomass energy content needs to be ~15% more than crude production then
Required algal biomass production per day = 1.15 * 5.185e17 / 900,000
= 662,528 sq-Km of bioreactor area
Add to this access roads, space between reactors and space for processing and storage then the total area required is of the order of 1 million square kilometres, or about 10% of the Sahara desert.

Production mass:
Micro algae energy density (HHV) depends very much on relative oil, protein and carbohydrate content but would typically be around 20MJ/Kg (~half of oil)

,’, total algal mass required = 1.15 * 5.185e17 / 20e9 ~= 30 Million tonnes/day

The burning questions are:
 How fast could we build this quantity of algae production and processing facilities?
 How much will it cost?, is it economically viable etc.
 Can we build this amount of infrastructure? Algae ponds will probably need to be concrete or plastic lined over an area of 1million sq-Km plus, closed reactors would probably require (using AlgaeLink system) about 3.3Billion kilometres!!! of bioreactor tubing (my guestimate is that at best we could produce a few million Km of tubing per year so we a talking of a multi century project).
 Is there a better solution?

The answer to last question is yes, change to electric technology for land transport, save biofuels for air transport. Electric vehicles far more efficient (~80% tank to wheel) than other technologies, best hybrid might ultimately achieve about 50% and renewable electricity easier to generate than biofuels, best PV nearly 40% conversion efficiency.

Anonymous said...

Hi energy balance, I just found your site whilst trying to find out what a fair figure for the PS efficiency might be. I had heard that 8% was reasonable, so perhaps they were right. The important point for me, who is involved in farming is that it is four times better than the best staple crops.To maximise the yields on a per unit area basis, the water plus plant nutrient medium might need to flow between layers of transparent plastic, with harvesting matching growth to optimise production. Being enclosed in plastic should provide quaranteen from potential pests and disease.

Professor Chris Rhodes said...

Hi Nic,

I agree with your figures and the questions you raise are very salient. I am about to post a new article on this with some sums about what to line the tanks with or to fabricate the tubes from.

It will be no mean feat to make them but then neither will installing so much battery capacity, I fear.

I'll put it there today or tomorrow (17-11-08).



Professor Chris Rhodes said...

Hi (the last) Anonymous),

that's very interesting. However, I think to install the technology on the full scale would probably need an awful lot of plastic sheeting.

I think it might prove useful to supply small areas maybe, but not that 30 billion barrels worth that we use globally.

See my note to Nic, I shall post some figures today or tomorrow.



Unknown said...

I just read your interesting report on algae sourced biomass fuel. This was the first paper I found on the topic (which I just heard about) and as with most papers on alternative source fuels it seems to be based on an "all or nothing" approach to the problem. It is true that with current technology/consumption rates algae sourcing is impractical but I think we should invest in the technology and begin developing it now so that it can begin to contribute to our total energy basket and as advancements occur it will perhaps one day become a practical sole fuel source.
In the meantime there are many unusable water sources that could become ideal breeding grounds for test projects such as the malodorous settling ponds at sewage treatment stations or contaminated industrial sites that would have additional benefits of algal remediation. Furthermore algae blooms are a serious problem in many of the worlds deep oceans and tidal flats as a result of nitrate runoff from farming and industrial pollution which is leading to problems with coral reefs and fisheries it is certainly technologically feasible to develop sail powered harvesters that could be deployed to capture much of this resource creating a win-win-win condition, to wit we get an new "clean" "sustainable" bio-fuel source, existing damage to the oceans aqua-culture is mitigated [perhaps even reversed], and new technologies are developed that will one day enable terra-forming of other planets.

Professor Chris Rhodes said...

Hi P,

don't get me wrong, I am a fan of algae and I agree that it doesn't have to be introduced on the full 30 billion barrels worth of oil a year scale to be useful, probably as part of an energy mix.

Liquid fuels do pose a particular problem in how to replace them, of all the energy considerations.

I am coming around to the idea od small-scale community production in a kind of "village pond" arrangement, i.e. to provide essential transport fuel for relatively small communities. I feel that biochar production too might be best done as an ensemble of small scale operations.

"Farming" the oceans, too, including growing algae on seawater does offer the potential for creating significant amounts of biomass.

We are running out of time though, and the world's governments need to make some clear decisions about what to go for on the grand scale. Or perhaps small communities will begin to break-off on their own accord to become sustainable from what can be produced locally, not the least of which is food.


Unknown said...

That is why the technology needs r&d. from the biology to the chemistry and to the engineering...a lot of work and pain in the "arse". total yield per ha would increase if done. in my opinion, an increase to 70% maybe. (Brotyccocus braunii sp. is 65% oil--from

Professor Chris Rhodes said...

Interesting idea, Kai. I think most of the R and D is being done in private Industry so it's not in the scientific literature necessarily.



Unknown said...

hi gentleman,

we r doing the project on same topic.
bt stuck up with the mass and energy balance. so could u do any help for us it will be lyk touching sky.......

thanking you

Anonymous said...

With reference from “Oil from algae; Photosynthetic efficiency “ Are the calculations under correct or should the figure be 24 times higher?
From the article:
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.

Shouldn’t it be:
5 x 10^3 Wh/day x 365 days x 3600 s/h (x 24h) = 1.58 x 10^11 W/m^2/year.
Or have I misunderstood?

Professor Chris Rhodes said...

Dear Arve,

the flux is given at 5 kWh/day, so the number of hours (i.e. 24) is not at issue. So, you multiply 5 x 10^3 Wh x 365 days/year x 3600 J/s/h = 6.57 x 10^9 J/year.