3 ms·

There is a non-negotiable entropy cost associated with removing CO2 from the atmosphere: you need to either pump it, so P0 = 0.0004 atm; if P1 = 5 atm (liquid C

by scythe 7y ago

There is a non-negotiable entropy cost associated with removing CO2 from the atmosphere: you need to either pump it, so P0 = 0.0004 atm; if P1 = 5 atm (liquid CO2) then E = M R T ln(5 / 0.0004) / 44 => E/M = about 500 J/kg, or you need to absorb it chemically, which requires a chemical with a vapor pressure of CO2 < 0.0004 atm, (e.g. NaOH) itself generally energy-intensive to produce. I expect that we will eventually develop technology for CO2 extraction, but it's not going to be easy or convenient. 500 joules per kilogram is a lot and I doubt chemical methods will beat it. Currently there are about 3.5 trillion tonnes of CO2 in the atmosphere, or about 3.5 * 10^15 kg. To decrease the concentration by 1/3, returning us to preindustrial CO2 levels, you'd need about 600 petajoules or 170 terawatt-hours of input energy assuming thermodynamically perfect isothermal compression (which is impossible to actually achieve). Global world electricity consumption per year is about 21000 TWh. No current method of CO2 extraction comes close to this limit and it's very unlikely that any ever will. Nonetheless, we have several orders of magnitude of wiggle room to let real-life methods work out okay. But the energy cost is not small and the negative effects of present-day CO2 emissions should not be considered easily curable. And just for the fun of it, a human being releases about 4 * 10^5 kg of CO2 during their lifetime, requiring about 2 * 10^8 J. So the most efficient CO2 pump possible would require at least 200 million watts of power (about a thousand Tesla Roadsters at maximum output) to confine this CO2 within a second.

avmich 7y ago
How did you calculate 500 J/kg? If you spend energy to pressurize gas, you can get it back when restoring the original pressure, right?
- scythe 7y ago
  I integrated the PV curve at constant temperature, so P = nRT/V and dW = P dV = nRT dV/V. Then we can write W = nRT log(V0 / V1) = nRT log(P1 / P0), and n = M / 44, the molar mass of CO2. Also, you can't restore the initial pressure because at the initial pressure the CO2 takes up a whole atmosphere of space!
  - avmich 7y ago
    Thanks. You've calculated the energy required to pressurize the gas. You can get that energy back if you make the gas doing work; if you pressurize it quasi-statically, you at least won't lose energy while compressing... I didn't get your last line though. I assume you compress "dirty" air.
    - scythe 7y ago
      You can't get the energy back because I assumed you have a magical compressor that only compresses carbon dioxide. I didn't include any terms for compressing air. You get a slightly cheaper term (ca 420 J/kg) if you set the upper pressure to atmospheric but then you're storing gaseous CO2 which is not practical. But recovering energy by restoring the original [partial] pressure of 0.0004 atm is not possible unless you somehow jettison the CO2 into outer space. To be clear: all real processes probably require many times more energy than this. There is no way I know of to selectively compress one component of a gas mixture. But if you did, this would be the energy cost.
      - avmich 7y ago
        Ok, let's take 500 J/kg of CO2. Do you think it's a lot? To get 1 kg of CO2 you'll need to process roughly a ton of air. There are ~5e15 tons of air in the atmosphere, so it's 3e18 J to process all the air. We produce annually 75e18 J annually on the planet - can we handle this expense? I'm afraid it will be more than that - but how much more exactly?
        scythe 7y ago
        That is roughly accurate for describing the operation of real compressors. The reason for calculating the bare minimum "magic compressor" cost is because it also applies to techniques like pressure-swing adsorption: https://en.wikipedia.org/wiki/Pressure_swing_adsorption https://en.wikipedia.org/wiki/Pressure_swing_adsorption Using PSA can reduce the energy cost of CO2 extraction significantly relative to direct compression technology. But because we know that the adsorbent is used in a cyclic process, we have dW[compression] = p dV + dE[adsorbent] and the long-time integral of dE[adsorbent] is effectively zero because the adsorbent cannot absorb or release energy forever, so ΔW[compression] = int[p dV] and thus any technology, compression-based or not, incorporating some finite amount of external components, is subject to the work limit. This is also why, for example, thermodynamic analyses of engine types are always concerned about the possible compression ratio. You could consider the partial pressure of atmospheric and stored CO2 as the compression ratio of a hypothetical compressed-air motor, which can then be run to extract thermal energy (i.e. violate the second law of thermodynamics) from any process more efficient than 500 J/kg CO2, since again W[motor] = int[p dV]. You can improve the compression ratio by taking advantage of the P-T curve and performing the whole compression at the South Pole, which saves you about 100 J/kg. Also, I was off by a factor of ten when I plugged in the vapor pressure of CO2. Correcting this is left as an exercise to the reader. Technologies not subject to the work limit include e.g. olivine crushing, somehow inhibiting volcanic activity (???), planting trees, etc. Another technology I just thought of would be to to form CO2@H2O clathrates in seawater starting with air, which would also lower the compression ratio, and extracting the (solid) clathrates to recover CO2 and desalinated water. Here the pressure required is much lower and additionally desalinated water is produced; this only works if we form the clathrates in a cold environment and decompose them in a warm one, which is kind of inconvenient. Then again, shipping gigatons of fresh water to LA might become necessary anyway!

by scythe 7y ago

avmich 7y ago

How did you calculate 500 J/kg? If you spend energy to pressurize gas, you can get it back when restoring the original pressure, right?

scythe 7y ago
I integrated the PV curve at constant temperature, so P = nRT/V and dW = P dV = nRT dV/V. Then we can write W = nRT log(V0 / V1) = nRT log(P1 / P0), and n = M / 44, the molar mass of CO2. Also, you can't restore the initial pressure because at the initial pressure the CO2 takes up a whole atmosphere of space!
- avmich 7y ago
  Thanks. You've calculated the energy required to pressurize the gas. You can get that energy back if you make the gas doing work; if you pressurize it quasi-statically, you at least won't lose energy while compressing... I didn't get your last line though. I assume you compress "dirty" air.
  - scythe 7y ago
    You can't get the energy back because I assumed you have a magical compressor that only compresses carbon dioxide. I didn't include any terms for compressing air. You get a slightly cheaper term (ca 420 J/kg) if you set the upper pressure to atmospheric but then you're storing gaseous CO2 which is not practical. But recovering energy by restoring the original [partial] pressure of 0.0004 atm is not possible unless you somehow jettison the CO2 into outer space. To be clear: all real processes probably require many times more energy than this. There is no way I know of to selectively compress one component of a gas mixture. But if you did, this would be the energy cost.
    - avmich 7y ago
      Ok, let's take 500 J/kg of CO2. Do you think it's a lot? To get 1 kg of CO2 you'll need to process roughly a ton of air. There are ~5e15 tons of air in the atmosphere, so it's 3e18 J to process all the air. We produce annually 75e18 J annually on the planet - can we handle this expense? I'm afraid it will be more than that - but how much more exactly?
      - scythe 7y ago
        That is roughly accurate for describing the operation of real compressors. The reason for calculating the bare minimum "magic compressor" cost is because it also applies to techniques like pressure-swing adsorption: https://en.wikipedia.org/wiki/Pressure_swing_adsorption https://en.wikipedia.org/wiki/Pressure_swing_adsorption Using PSA can reduce the energy cost of CO2 extraction significantly relative to direct compression technology. But because we know that the adsorbent is used in a cyclic process, we have dW[compression] = p dV + dE[adsorbent] and the long-time integral of dE[adsorbent] is effectively zero because the adsorbent cannot absorb or release energy forever, so ΔW[compression] = int[p dV] and thus any technology, compression-based or not, incorporating some finite amount of external components, is subject to the work limit. This is also why, for example, thermodynamic analyses of engine types are always concerned about the possible compression ratio. You could consider the partial pressure of atmospheric and stored CO2 as the compression ratio of a hypothetical compressed-air motor, which can then be run to extract thermal energy (i.e. violate the second law of thermodynamics) from any process more efficient than 500 J/kg CO2, since again W[motor] = int[p dV]. You can improve the compression ratio by taking advantage of the P-T curve and performing the whole compression at the South Pole, which saves you about 100 J/kg. Also, I was off by a factor of ten when I plugged in the vapor pressure of CO2. Correcting this is left as an exercise to the reader. Technologies not subject to the work limit include e.g. olivine crushing, somehow inhibiting volcanic activity (???), planting trees, etc. Another technology I just thought of would be to to form CO2@H2O clathrates in seawater starting with air, which would also lower the compression ratio, and extracting the (solid) clathrates to recover CO2 and desalinated water. Here the pressure required is much lower and additionally desalinated water is produced; this only works if we form the clathrates in a cold environment and decompose them in a warm one, which is kind of inconvenient. Then again, shipping gigatons of fresh water to LA might become necessary anyway!