It is said that ACs are counterproductive in fight against global warming, in that while they may make the local environment temporarily livable, the greenhouse gases produced while making the electricity needed to operate them heat up the rest of the Earth by much more than the relief from the AC itself. By how much exactly is that? Note that here I am interested in the global impact of greenhouse gases specifically, not in the local heat island effect (given how ACs do not destroy heat but only move it from inside to outside, and add extra heat from running the compressor itself). Let’s also assume all electricity comes from fossil fuels (ACs might become a viable solution if 100% of AC electricity came from renewable solar, which is actually a reasonable goal to strive for given how both AC and solar are most active during the day, but at the moment most of electricity delivered to me specifically, for example, comes from natural gas.)

Here’s my estimate. Let me know if it is reasonable! Methane has energy density of 891 kJ/mol, burnt into CO2 at 1 mol : 1 mol. Gas turbines have efficiency up to 60%. The radiative forcing of CO2 can be calculated as: ln(new ppm/old ppm)/ln(2)*3.7 W/m**2. For example the 131 ppm increase in CO2 since 1750 up to 411 ppm has a radiative forcing of 2.05 W/m**2 (is that across the entire Earth’s surface? or only its crosssection?), and CO2 has persistence in atmosphere for at least 1000 years. The atmosphere composition is 78% nitrogen 21% oxygen 0.9% argon so its molar mass is:

.78 * 28 g/mol + .21*32 g/mol + .009*18 g/mol = 28.7 g/mol 

And total atmospheric mass:

4*3.14*(6.37e6 m)**2 * ~10000 kg/m**2 * 1000 g/kg / (28.7 g/mol) = 1.78e20 mol

Suppose 8 billion people each run 1kW AC for 1 year, with electricity from natural gas. (That’s similar to our total current global energy consumption of 20TW, though of course we use power for things other than just AC or electricity, but also most energy comes from coal and gasoline not just gas, and 80% comes from fossil fuels not renewables.)

8e9 people * 1000 W/person * 60*60*24*365 s / (891e3 J/mol * 0.6) = 472e12 mol

That’s 472 teramols of CO2 (20.8 gigatons) added to the atmosphere each year, or 472e12 / 1.78e20 * 1e6 = 2.65 ppm (parts per million). It is believable that having done so for a hundred years we have raised CO2 concentration from pre-industrial levels up to 411 ppm. The radiative forcing is:

ln((411 ppm + 2.65 ppm)/(411 ppm)) / ln(2) * 3.7 W/m**2 = 0.0343 W/m**2

Or for the whole earth:

4*3.14*(6.37e6 m)**2 * 0.0343 W/m**2 = 17.5 TW

What is my individual contribution for 1 hour?

17.5e12 W / 8e9 / (24*365) = 0.25 W

That is, if I run my 1kW air conditioner for 1 hour, the entire Earth will be solar heated by an extra 0.25 W for the next 1000 years. That doesn’t sound like much, but it adds up over time: I spent one kilowatt-hour in one hour on cooling, but the rest of the Earth will be heated by an extra 0.25 W * 24*365 hours = 2.2 kilowatt-hours in the next year, and again every year thereafter. Multiply that by 8 billion people or a hundred years and it adds up a lot, even considering the heat is distributed across entire planet surface not just areas where people live.

So my answer is 1 kWh of cooling = 2.2 kWh of heating per year for the next 1000 years. By same calculation in terms of mass, 1 kg of CO2 = 7.4 kWh of heating for every year thereafter. Is this accurate?

  • tburkhol@lemmy.world
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    1 year ago

    I won’t comment on the final accuracy, but I will note that this is an extremely roundabout path to your final answer, and some of the intermediate steps are…weird. Most notably, the speculation that every man, woman, and child on the planet might run a 1 kW appliance 24/7/365. This is 7e13 kWh or 70k TWh, about 3x current global energy use (not just electicity) before accounting for efficiency. The equation you cite for radiative forcing, specifically its ln(new/old) term is very non-linear, so you should get a much lower marginal effect from 70k TWh than from 1 kWh.

    A simpler approach is to calculate the CO2 required for your 1 kWh AC, i.e.: 1kWh * 3600 kJ/kWh / 0.6 efficiency / 890 kJ/mol = 6.7 mol CO2. Current atmospheric CO2 is 75 Pmol. From that, I get radiative forcing of ln((7.4e16 + 6.7)/7.4e16)/ln(2)3.7 * 4pi*(6.4e6^2). Numpy won’t tell me what ln(74000000000000006.7/74000000000000000). It will tell me the forcing from 10 kWh is ~2.5W, or the same 0.25W/kWh you got. I guess ln is not that nonlinear in the 1+1e-16 to 1+1e-4 range, after all.

    0.25W/kWh seems improbably high. 1 kWh is about 0.1 W running 24/365. At 60% efficiency, that’s burning 0.2W of natural gas and implies that the radiative forcing from CO2 is much greater than the energy to produce the CO2 in the first place. I get that the energy source for heating is different from the energy source for electricity, but it feels wrong, even without the 1000 year persistence. I don’t know where the radiative forcing equation came from nor its constraints, so I’m suspicious of its application in this context. There’s a lot of obscenely large numbers interacting with obscenely small numbers, and I don’t know enough to say whether those numbers are accurate enough for the results to be reasonable. Then there’s the question of converting the energy input to temperature change.

    • TauZeroOP
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      1 year ago

      Numpy won’t tell me what ln(74000000000000006.7/74000000000000000).

      Ran into exactly this problem for individual calculation 😆. Which is also why I multiplied by 8 billion and divided in the end - make the calculator behave. ln is linear enough around 1±epsilon to allow this.

      implies that the radiative forcing from CO2 is much greater than the energy to produce the CO2 in the first place

      That’s what I wanted to find out and it does appear to look exactly that way. Makes sense in retrospect since the radiative forcing is separate from the energy content of CO2 itself, same way as a greenhouse gets hot for no energy expended on its own.

      • tburkhol@lemmy.world
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        1 year ago

        Numpy won’t tell me what ln(74000000000000006.7/74000000000000000). Ran into exactly this problem for individual calculation

        Trouble is that 74000000000000006.7/74000000000000000 ~ 1.000 000 000 000 000 1 and double-float precision is 0.000 000 000 000 000 2. Needs a 96 or 128 bit float. The whole topic of estimating one’s personal contribution to global phenomena is loaded with computer precision risks, which is part of what makes me skeptical of the final result, without looking far more closely than my interest motivates. Like calculating the sea level rise from spitting in the ocean - I believe it happens, but I’m not sure I believe any numerical result.

        • TauZeroOP
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          1 year ago

          Your skepticism is excessively cautious 😁. You can work around precision limits perfectly fine as long as you are aware they exist there. Multiplying your epsilon and then dividing later is a legitimate strategy, since every function is linear on a small enough scale! You can even declare that ln(1+x) ~= x and skip the logarithm calculation entirely. Using some random full precision calculator I get:

          ln((74e15+6.7)/74e15) = 0.000000000000000090540540...
          

          Compare to the double-precision calculator with workaround:

          ln((74e15 + 6.7*10e9)/74e15) / 10e9 = 9.0540499...e-17
          

          Or even:

          ln(1+x) ~= x
          6.7/74e15 = 9.0540540...e-17
          

          You are worried about differences in the final answer of less than 1 part in a million! I try to do my example calculations in 3 significant figures, so that’s not even a blip in the intermediate roundoffs.