2.1: Fundamental energy and volume requirements of CDR

How much do we need to sequester

The 2019 National Academies roadmap thinks we need to be sequestering ~10 GtCO2/year globally by around 2050 and ~20 GtCO2 per year by end of century, in realistic situations where we abide by the Paris Agreement targets:

“Approximately 10-20 Gt CO2e of gross anthropogenic emissions come from sources that would be very difficult or expensive to eliminate, including a large fraction of agricultural methane and nitrous oxide. Most scenarios that meet the Paris agreement… thus rely on CO2 removal and storage that ramps up rapidly before midcentury to reach approximately 20 GtCO2 by century’s end… If the goals for climate and economic growth are to be achieved, negative emissions technologies will likely need to play a large role in mitigating climate change by removing ~10 Gt/y CO2 globally by midcentury and ~20 Gt/y CO2 globally by the century’s end.”

The Lawrence paper, which reviews a range of carbon sequestration and geoengineering options in the context of the Paris agreement, gives a reference figure for roughly how much they think actually needs to be removed, which they call CDRref:

“…in the context of the Paris Agreement, useful reference values can be defined based on the difference between the 2 °C versus the 1.5 °C limits (see Methods): CDRref ≈ 650 Gt(CO2) for the cumulative CO2 budget

⊕650 Gt of CO2 is ~83 ppm of atmospheric CO2

¹ Which would be …. Ppm lower than where we currently areSo, according to the Lawrence paper, CO2 removal methods can be crudely evaluated on whether they can feasibly approach removing CDRref ~ 650 Gt of CO2. This is not as much as would be needed to return to preindustrial levels¹, but it still serves as a useful order-of-magnitude benchmark of what would make a truly significant dent in the problem.

So the key figures to keep in mind here when evaluating scale are that we will likely need:

• 10Gt CO₂ removal a year by 2050
• 20Gt CO₂ removal a year by 2100
• A portfolio of options that combined will be able to remove ≈ 650 Gt CO₂

Energy

CDR, it turns out, is amenable to an easy back of the envelope calculation in terms of its fundamental energy requirement. Below are four methods, getting increasingly accurate, to get a sense of the energy needed.

Method 1

Here’s a simple way to estimate it: What we are asking to do is to compress a dilute gas, and doing so in this scenario requires pressure-volume work at constant temperature. If you know/remember/want to learn some physics, the energy associated with P-V work at constant temperature is given by a logarithmic formula:⊕n= number of moles of CO2 removed
R = gas constant
T = temperature in Kelvin
$V_f$ is the volume of CO2 after capture
$V_i$ is the volume of CO2 before capture.$nRT*ln(\dfrac{V_f}{V_i})$⊕³ I will ignore the small differences between long, short and metric tons, i.e., tonnes
$(~1 atm) * (\dfrac{400}{1e6} \space ppm) = 0.0004 \space atm$
Wolfram Alpha

density of liquid CO₂ ≈ 1.84 kg/m³

$(4.2 bn \space km³) * (1 atm) * \dfrac{400}{1e6} * Ln \left( \dfrac{ \left( \dfrac{3000 Gt}{1.84 kg/m³} \right)} {4.2 bn \space km³} \right)$To remove all of the CO2 in the atmosphere, around 3000 GigaTons⁴, by compressing it from the (~4.2 billion km³) volume of Earth’s atmosphere, where it lives at a pressure of 0.0004 atm to the volume of its liquid form, we therefore need ~1.3e21 J.

Method 2

Klaus Lackner, one of the inventors of direct air capture of CO2, used a version of this formula in his first paper on the subject, which you can convince yourself from the ideal gas law (PV=nRT) is equivalent: the energy requirement per mole of CO2 removed is⊕P0 = initial CO₂ partial pressure ~ 400/1e6 * 1 atm
P = desired pressure of the output CO₂ capture stream, say 1 atm.

$RT*ln \left( \dfrac{P}{P_0} \right)$

Wolfram Alpha

$(gas \space constant) * 300 Kelvin * Ln \left( \dfrac{(1 atm)}{\left( 1 atm * \dfrac{400}{1e6} \right)}\right)$

≈ 20 kJ/mole

Wolfram Alpha

Molar mass of CO₂ ≈ 44g = 4.4e-5t

Mass of CO₂ to remove = 3000 Gt or 3e12t

$20KJ/mol * \dfrac{3e12t}{4.4e{-5}t} ≈ 1.36e18 KJ$

$= 1.36e21J ≈ 1.3e21J$

⊕In reality, liquid CO₂ forms only at pressures over 5.1 atm, but this doesn’t change the answer much,since

$(gas \space constant) * 300 Kelvin * Ln \left( \dfrac{(5.1 atm)}{\left( 1 atm * \dfrac{400}{1e6} \right)}\right)$

≈ 23.5 kJ/mole versus 20KJ/mole.Assuming that we condense the CO₂ to about 1 atm pressure, from its initial 0.0004 atm pressure to remove all of it from the atmosphere would take 20KJ/moleand to remove our 3000 GigaTons of CO₂ would need the same ~1.3e21 Joules.

Method 3

A somewhat more rigorous way to arrive at this is to use the entropy of mixing, as is done here and here and here. The energy associated with the entropy of mixing is given by:

⊕n is now the total number of moles of all the molecules in the atmosphere
x1 is the fraction of gas 1 (# gas 1 molecules / total # of molecules)
x2 is the fraction of gas 2.$-nRT(x_1ln(x_1)+x_2ln(x_2))$

Let x1 be the fraction of CO2, and x2 be the fraction of the rest of the air. For x2 close to 1.0, the x2 term is negligible, which makes sense: when the rest of the atmosphere expands into the tiny space formerly occupied by the CO2, its entropy doesn’t increase significantly.⊕² i.e., $x_1=\dfrac{400}{1e6}$But for x1 far from 1.0,² we have a significant contribution from the first term. So let's look at that term.

Wolfram Alpha

Molar mass of CO₂ ≈ 44g

Moles of CO₂ in the atmosphere

$= \dfrac{3000 Gt}{44g} ≈ 6.8e16 \space Moles$

Moles per m³ of CO₂ ≈ $\dfrac{6.8e16 Moles}{4.2b \space km³}$

≈ 1.62e7 Mol/km³ = 1.62e-2 Mol/m³

Mulitply this by ratio of air to CO₂, i.e $\dfrac{1e6}{400}$, ≈ 40.5 Mol/m³ of airwolframalpha

$n = 40.6 \space mol/m³ * (4.2 bn \space km³)$
= 1.7e20  moles

$x_1=\dfrac{400}{1e6}$, $x_2 ≈ 0$

Energy = $-nRT x_1 ln(x_1)$

$= -1.7e20 * gas \space constant* (Earth’s \space temp) * \dfrac{400}{1e6} * Ln \left( \dfrac{400}{1e6} \right)$

= 1.27e21 JoulesFirstly, how many total moles of air are in the atmosphere? For consistency with the above, I'll take our ~3000 GigaTonnes of CO2 and multiply by 1e6/400 to get the total number of moles of all molecules in the atmosphere, and also use our same estimate of the atmospheric volume of 4.2 billion cubic kilometers.That's roughly 40 mol/m³ of total air in the atmosphere, So to un-mix all CO2 from the atmosphere it would take 1.27e21 Joules  (That’s our 1.3e21 Joules from above)

Method 4

This entropy of mixing framework can be extended to other scenarios like ones with interactions between the molecules, desalination of liquids, and so on. It also leads to a proper calculation for the case where we only remove some of the CO2 from the atmosphere, not all of it, which is what we’d want to do in practice. On a per mole of CO2 removed basis, we have an energy cost for partial removal of:
⊕P1 is the initial pressure of CO2 in the atmosphere
P0 is the overall atmospheric pressure
P2 is the pressure of CO2 remaining in the atmosphere after we do our removal
$∆G ≈ RT \left(ln\left(\dfrac{P_1}{P_0} \right) - \left( 1- \dfrac{P_1}{P_0} \right) * \dfrac{P_2}{P_1 - P_2}ln \left( \dfrac{P_2}{P_1} \right) - \left( 1- \dfrac{P_1+P_2}{2P_0} \right) \right)$


Wolfram Alpha
Normalising to ppm we have:
$P_0 = 1,000,000$
$P_1 = 410$
$P_2 = 278$Let’s say we want to bring the partial pressure from 410 ppm down to the 278 ppm pre-industrial level. Plugging into this formula I get 19.91 kJ/moleof CO₂ removed, very similar to the above.

Wolfram Alpha
Moles of CO₂ in a tonne = 1t/44g ≈ 22,700
20KJ/mol * 22,700 ≈ 450,000KJ = 450MJ

Wolfram Alpha
450MJ per tonne
Gt per 1ppm ≈ 7.81
ppm removed = 410-278 = 132
7.81e9 * 132 * 450MJ ≈ 4.6e5 MJ = 4.6e14MJ ≈ 5e20J

Wolfram Alpha
450MJ per tonne
500Gt
500e9*450e6 = 2.25e20J ≈ 2.5e20J

0.2 kWh/kg = 0.2 * 3600 kWs/kg = 720 kWs/kg = 720KJ/kg

720KJ/kg = 720*44g KJ/mol = 31.7KJ/mol ≈ 30KJ/mol

⊕By the way, MacKay has it in his book at about 30KJ/mole for the minimum energy cost. His 0.13 in this table kWh / kg in this table is our 20 kJ/mole, and he adds an additional compression cost that adds about a third (I’m not exactly sure where this comes from, maybe he’s pushing towards dry ice densities, but in any case it doesn’t add that much in the grand scheme):

“…energy requirements of three main technology components: (1) sustaining sufficient airflow through the systems to continually expose fresh air for CO2 separation; (2) overcoming the thermodynamic barrier required to capture CO2 at a dilute ambient mixing ratio of 0.04%; and (3) supplying additional energy for the compression of CO2 for underground storage. While components (1) and (3) can be quantified using basic principles, and several studies 61,62 indicate that combined they would probably require 300–500 MJ/t(CO2) (or ~80–140 kWh/t(CO2)), the energy and material requirements of the separation technology (2) are much more difficult to estimate. The theoretical thermodynamic minimum for separation of CO2 at current ambient mixing ratios is just under 500 MJ/t(CO2)62.”

wolframalpha
1 Mole of CO₂ = 44g = 4.4e-5t
500 MJ/t = 500 * 4.44e-5 MJ/mole = 0.022 MJ/mole
0.022 MJ/mole = 22 KJ/mol500 MegaJoules / tonne gives 22 kJ/mole of CO2. If you suck out Lawrence’s 650 GtCO2 you get 3.25e20 Joules, or 9e20 Joules at 35% thermodynamic efficiency.

In any case, compare our rough numbers, which are on the order of 1e21 Joules for a “full scale” sequestration of the atmosphere’s anthropogenic carbon, with the ~6e20 Joules per year total energy consumption of human civilization. We’re talking perhaps a year or two of civilization’s total current energy consumption, if we wanted to remove decades worth of CO2 emissions.

1 Joule/second = 1 Watt
31,536,000 seconds in a year ≈ 30,000,000 = 3e7
3e8 seconds in 10 years.
1 J/second for a decade ≈ 3e8 J = 3e8 Watts
5e20 / 3e8 ≈ 1.7e12 Watts = 1.7 TeraWattsThis is arguably not that much energy in the grand scheme. Even if we assume we need to do it basically twice to compensate for outgassing of CO2 by the oceans (see below), it is still on the order of 5e20 Joules to remove 500 GigaTons,which works out to 1.5 TeraWatts running continuously over 10 years, which is about 1/10th of civilization’s energy consumption if we were to do it over 10 years. Significant, yes, indeed huge, but not impossibly so. We’re talking on the order of solar panels covering a US state.

⊕Note that not all of the schemes we’ll consider require paying this energy “out of pocket”, as it were: for instance, biology-based schemes rely on the energy of the sun for photosynthesis to do the work, and mineral weathering schemes like Project Vesta rely on chemical free energy in the minerals they harvest.The Lawrence paper points out that reaching this theoretical minimum energy is far from easy, though (we’ll discuss why later):

“…thermodynamic minimum values are rarely achievable. Current estimates for the efficiency of DACCS are technology-dependent, ranging from at best 3 to likely 20 or more times the theoretical minimum 61, or ~1500–10,000 MJ/t(CO2), implying that removing an amount equivalent to CDRref by 2100 would require a continuous power supply of approximately 400–2600 GW…”

Volume

If you’re worried about storage space for all that CO2 underground, there is more than enough:

“The US Department of Energy publishes a national atlas of storage capacity by state. The calculations assume that even in areas that look promising for CO2 storage, only 1-4% of available geologic capacity will actually be used for CCS. Even with this limitation, the DOE still estimates overall potential for storage in the US to be at 3,600 to 12,900 billion metric tons of CO2. To put that in perspective, the United States’ current annual CO2 emissions are about 5,814 million metric tons per year.”

A quick calculation says that, if we were to capture 400 gigatonnes carbon and pile it 100 ft high, we’d need an area maybe a couple hundred miles on a side if it was packed densely as solid carbon.

This nice interview with Stephen Pacala, who chaired the aforementioned US National Academy of Science report on negative emissions, mentions approaches like injecting CO2 into saline aquifers:

“It looks like it’s not a problem. I would have bet a large amount of money 20 years ago, when I first started directing a group that works on this problem at Princeton, that there wouldn’t have been enough storage capacity, but now I think there is. It just turns out that there’s a lot, like injection of CO2 into saline aquifers, which is a kind of formation of salty water deep down under the ground. It’s where you get oil and gas from. Now there are some pretty strong indications that CO2 inside the salt reacts and turns into rocks really quickly.” Pascala also mentions that CO2 injection underground is already starting to happen already at a decent scale in the context of the natural gas industry: “Carbon capture and storage has gone from, “Well, maybe it’s possible to do,” to a big business. Sixty-one million tons of CO2 are going into reservoirs and staying there this year in the Lower 48 [U.S. states] alone. That’s a big number.”

It also seems like geological storage of captured CO2 could be robust and safe.

The conclusion to the NAS report summary puts it bluntly: “direct air capture and carbon mineralization have essentially unlimited capacity and are almost unexplored”.

The actual implementation of the burying surely adds cost, but things like the injection of CO2 underground for Enhanced Oil Recovery, or the natural gas industry use of saline aquifers just mentioned, show that the oil industry already has the general kind of technology needed. The NAS report has a chapter on this sort of thing. There are also ways of "turning captured CO2 into rock” for storage.

Cost

The Lawrence paper also gives a cost estimate for both biomass based and industrial chemical approaches to carbon sequestration:

“Published estimates cover a similar range to the biomass-based techniques, from about $20/t(CO2) to over $1000/t(CO2)27,28,60,65. Better estimates of the costs are particularly important for DACCS, since it essentially represents the cost ceiling for viability of any CDR measure due to its potential scalability and its likely constrainable environmental impacts.”

For enhanced weathering approaches (see below), Project Vesta estimated around $10 per tonne of CO2 removed, so also a trillion dollar scale project if done at the hundreds of gigatonne level. Likewise, if you want to plant a trillion trees (see below), and it costs perhaps on the order of $1 per tree, which is on the order of $5-$10 per tonne CO2, it is approaching the same scale of cost. Is there any way to break out of that roughly trillion dollar cost level?⊕Potentially with self-replicating solar-powered microorganismsWe’ll talk about that a bit later — but suffice it to say there are some major potential caveats and difficulties there too

Here are the cost reduction targets for the proposed 10 year R&D program from the Energy Futures Initiative, as you can see most approaches, especially technological ones, have heavy cost reduction targets.

“Whether the system is driven by water evaporation or by low grade heat, the cost of the thermodynamically-required energy can be as small as $1 to $2 per metric ton of carbon dioxide.”

Scale of cost

Quoting Stripe’s post again, for comparison:

“If there was scalable, verifiable negative emissions technology available in the vicinity of $100 per tonne of CO2 (tCO2) captured, it could be a trillion dollar industry by the end of the century and complement emissions reduction in halting anthropogenic climate change.”.

In practice, if someone can get it below $20 per tonne CO2 that would be amazing (hell, <$50/tonne CO2 would be amazing at this point), given all the other aspects than the thermodynamically necessary minimum energy for separation per se, like the need to get enough air flow or to ultimately store the captured CO2 somewhere. This means that, if we are ultimately going to suck hundreds of gigatonnes of CO2 out of the atmosphere, via an industrial direct air capture approach, it will probably have to cost trillions of dollars to do so, although from a pure thermodynamic and energy cost limit it could theoretically cost merely hundreds of billions.

⊕Though this might need to operate for longer than Manhattan or Apollo, and the structure of today’s economy and politics is of course very different from WWII or the early Cold War). Half of a percent of United States GDP over 10 years is a trillion dollars.As Michael Nielsen points out in reference to the NAS report, one could look at this in terms of fraction of GDP needed, and at historical precedents for significant fractions of GDP being spent on a single moonshot project — even at $50/tonne CO2, sucking out the US emissions rate would only be $235bn or.<2% of US GDP. At $20/tonne CO2, as Nielsen mentioned, we’d be at a $94bn or <1% GDP level, perhaps roughly comparable to previous major projects such as Manhattan or Apollo, and to implementation costs for programs like the Clean Air Act that Nielsen mentions.

Current funding

As a point of comparison this recent EFI report gives proposed breakdowns for proposed R&D funding for an interagency research initiative in the USA.