Oil From Algae - How Big Do the Tanks Need to Be?

The comment reproduced below to my article "Wednesday, April 09, 2008; Oil from Algae: Photosynthetic Efficiencies?" prompted some very salient points regarding the size of the tanks or tubular reactor vessels that would be required to grow algae on a large scale, and the quantity of the materials needed to make them from. I have therefore done a few sums, which I hope are illustrative:

I agree completely with Nic's figures which are very close to those I have deduced here previously, and I shall use them now. First I want to consider tube-reactors, and I am assuming that the tube is about a foot (30 cm) in diameter.

(.3 m x pi =) .94 m (circumference of tube) x 3.3 x 10^12 m (length of tube) = 3.1 x 10^12 m^2
as the area of plastic needed to fabricate 3.3 billion km of it! If we assume that the wall thickness is 2 mm (0.002 m), we have a volume of plastic = 3.1 x 10^12 m^2 x 0.002 m = 6.2 x 10^9 m^3.

Taking the density of the material as 0.85 (t/m^3), that amounts to 0.85 x 6.2 x 10^9 = 5.3 x 10^9 tonnes. If there are 7.3 barrels of oil to a tonne, that is the equivalent of 38.7 x 10^9 barrels of oil to make the plastic (this is about 3% of all the oil left in the ground). The world uses 30 x 10^9 barrels in total annually, and so this is more oil than the world gets through in 16 months all told.

I agree with Nic that it probably wouldn't be possible to make more than few million km of pipe per year and so it would take centuries to make it. Well before then we won't be recovering significant crude oil anyway, given the total of 1.2 x 10^10 barrels there is thought to be recoverable. On this basis, I think making biodiesel from algae on a large scale in tube-reactors is unlikely.

So, what about the open (raceway) ponds? If we need 1 million km^2 of them, that's 1 x 10^12 m^2. I am going to assume they are each 1 hectare in area. That's 100 million of them. I shall also assume that they are 1 metre deep. The area of the walls is 100 m x 1 m x 4 walls = 400 m^2. But the floor area is 10,000 m^2 (i.e. 1 ha). hence I shall forget the 4% of the total pond area made up by the walls and concentrate on the larger, floor area of 10,000 m^2 for each one.

The ponds could be lined with either concrete or plastic. Let's consider the concrete option first. If the floor is four inches, 10 cm thick = 0.1 m. So we need 10,000 m^2 x 0.1 m = 1,000 m^3 of concrete for each pond. So 100 million of them needs, 100 x 10^6 x 1,000 m^3 = 1 x 10^11 m^3 and at a density of sat 2.3 t/m^3, that's 2.3 x 10^11 tonnes of concrete, or 230 billion tonnes. In 2007, the world produced 2.6 billion tonnes of concrete, and again the job would take centuries assuming that an equivalent amount of additional concrete could be made to current consumption, or some of it diverted from existing purposes. The global warming effect would be pretty severe since making concrete releases huge amounts of CO2.

What about plastic lining? Well, the area is the same, at 10^12 m^2 and I shall assume that a 2mm thickness is O.K. again, so the volume is 10^12 m^2 x 0.002 = 2 x 10^9 m^3 of plastic.
This equals around 12.4 billion barrels of oil, which is about 1% of the world's total recoverable of 1.2 trillion barrels, or 1/3 that for the tubes, and it might be easier to make plastic sheets than tubes, but even if we could make say, a few thousand square kilometers (a few billion m^2) of plastic sheeting annually, the job would once again take centuries. We are dealing with a classic rate of flow (rate of conversion, rate of recovery) problem, and while not hampered necessarily by the resources per se, the engineering can't be done in time to save us from running out of cheap crude oil.

On Nic's final pint about instead using electric vehicles. The best figure I have is that you need a bare minimum of 0.08 g of lithium/Wh of power. But for real Li systems, it is more like 0.15 - 0.32 gLi/Wh. I shall take the mid-range of 0.23. An average PHEV needs around 9kWh. So we need 9,000 x 0.23 g = 2070 g = 2 kg = 0.002 tonnes per vehicle. Now there are supposed to be around 700 million vehicles on the roads. There are something like 30 thousand tonnes of Li recovered per year and so we would need to double production capacity to get something reasonable, i.e. that amount again to make the electric cars etc. [There is plenty of lithium in the ground, 28 million tonnes, I think, but it's the rate of recovery that is the limiting factor].

700 x 10^6 x 0.002 t = 1.4 x 10^6 tonnes to make that equivalent car fleet from PHEV's. Well at 30,000 t/year it's going to take 46 years. So, in principle this is the best bet. However it would mean making 15 million cars a year. We'd need a lot of new mining, processing and engineering
though. In all likelihood the world will have far fewer vehicles in the future, and we can supplant urban transport with trams and light railways so long as we can make enough electricity to run them and so need less anyway.

In conclusion I take your point about cars Nic, but still since aviation consumes more fuel per year than we can tool-up for with algal biofuel within decades (centuries) plane-flight does seem doomed in its present form and scale.

But whatever we do it's got to be done against that backdrop of cheap oil running out - yes, I know its cheap now because the world financial system is melting down - but it wont be for much longer; also the economic turndown means that there is little fiscal incentive and little money to be borrowed from the banks in their nervous condition, to build new PHEV factories and mining and processing plants for Li. PHEV's are expensive to make too and that is also bad is the current state of credit-crunch. There are other resource issues to PHEV's too, since they use various other metals, cobalt etc. depending on the exact battery design

Thanks very much,

Chris.


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 < id="54033"> 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)
Assume:
> 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.