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I'm a science fiction author looking to describe what conditions might be like at the bottom of a cylindrical shaft ~ 30 meters across and ~ 35 km deep. It's located somewhere on the edge of Yellowstone national park (if that matters). The top is open to ambient, albeit controlled, conditions in a great big lab room. If you drop a coin down this well, there's nothing man-made in the way of it getting to the bottom.

The shaft itself uses advanced unobtanium wall liners, so the engineering challenge of keeping it from collapsing or otherwise self-destructing has been taken care of. I'm not sure how insulating the unobtanium is from the surrounding rock right now, could be "none" to "completely," depending on how interesting that makes what's happening at the bottom.

I'm pretty sure the atmosphere at the bottom will be amazingly hot and dense, probably quite a bit denser than several hundred feet of water, but I'm hoping someone more knowledgeable can help out with the details.

Contributors will definitely get a mention in the acknowledgements. Big contributions will likely result in a signed copy of the final book. This will be the third book in the Gemini Gambit series, which will likely hit the shelves in late 2017/early 2018.

UPDATE The hole is used as part of a super-massive geothermal power plant, so hot is good. Dense is good. I'm thinking the convection potential would perhaps make mounting ducted fan windmills somewhere also good for generating power.

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  • $\begingroup$ Where exactly are you drilling? The crustal structure in the vicinity of Yellowstone is complicated by the presence of a hotspot and associated magma chambers and the like. Hitting one of those would likely be a very bad idea. $\endgroup$
    – bon
    Jul 7, 2016 at 19:05
  • $\begingroup$ The hotspot and the magma chambers aren't a bug in my story, they're a feature. :) $\endgroup$ Jul 8, 2016 at 23:15
  • $\begingroup$ Still, where exactly you drill may have a significant effect on the outcome. $\endgroup$
    – bon
    Jul 9, 2016 at 7:42
  • $\begingroup$ You may find this and its references to be of interest, being the deepest hole ever drilled in real life: en.wikipedia.org/wiki/Kola_Superdeep_Borehole $\endgroup$ Jul 10, 2016 at 17:34

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If it is enclosed and undisturbed there would be a risk of the heavier gasses collecting at the base of the hole, with carbon dioxide, argon, neon and xenon displacing the nitrogen and oxygen. Hence the danger of asphyxiation for any organism at that depth. Mixing, to keep the air homogeneous would also have the effect of heat transfer to further up the hole, so you will need artificial mixing and cooling. There are likely to be traces of helium and hydrogen at such depth, but not enough for the upward flow of such light gasses to have any significant mixing effect. However, hydrogen will be a potentially explosive problem at the enclosed surface environment. At these temperatures water rock reaction at the exposed base of the hole would be rapid. It is not clear whether your hole would have punched through to the mantle. I'm assuming it would, in which case there will be water vapour - hartzburgite reaction to yield more hydrogen. Serpentinization of the rock will proceed, involving about a 17% rock expansion. The rate of such a reaction will be limited by the amount of water vapour in the hole. A by-product of the reaction will be moisture at pH12 which will rapidly react with the high partial pressure of CO2 to produce carbonate encrustation. So it will be a problem to keep the hole clean - unless you prevent any moisture entering the hole from the surface control area.

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  • $\begingroup$ Crustal thickness around Yellowstone is around 45 km, see eos.org/research-spotlights/… $\endgroup$ Jul 8, 2016 at 3:00
  • $\begingroup$ @WolfgangBangerth Actually in some places it seems to be thinner onlinelibrary.wiley.com/doi/10.1029/2009GC002787/abstract $\endgroup$
    – bon
    Jul 8, 2016 at 8:58
  • $\begingroup$ As I recall, I cited it on the northwest corner of the basin, up high so they wouldn't punch a hole all the way through the crust, but get very very close. $\endgroup$ Jul 8, 2016 at 23:16
  • $\begingroup$ If the air is well mixed you won't get any stratification from heavier gases separating; diffusion will keep them mixed. It would take a source of higher concentration of a heavier gas for it to settle in the bottom - which would slowly homogenize from diffusion. $\endgroup$
    – Ken Fabian
    Jul 12, 2023 at 22:12
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I found this while looking for something else (a Larry Niven story, with the condemnation of Mars as being "useless, at the bottom of a hole"), but meh.

Elsewhere I'd collated a number of equations relating height and atmospheric pressure. It's not the simplest of topics, which is why I made a pile of notes then forgot the gory details. But tweaking my collection of models for a 35km (below sea level, not below ground level ; if you'd done geological work at the shore of the Caspian Sea and altitude -23m, you too would have got fed up with debugging other people's software over this point) deep borehole. My collection gives bottom-hole pressures of 2,095,904.36 ; 1,529,809.85 and 6,098,523.97 Pa, or 20, 15 and 61 atmospheres. Considerable variation.

From SCUBA training, any of those pressures would rapidly induce rapidly lethal oxygen toxicity if exposed to "air" at that pressure. The pressure itself is in the realm of "saturation diving", where you'd take several days to "descend" (in a pressure chamber) to that depth, and simultaneously change your breathing gas to one dominated by helium (or even hydrogen - really scary stuff!). Coming back "up", to surface, you'd take a week or two, on steadily changing gas mixtures.

As a geologist, your wallrock temperatures could easily vary by a factor of "several", from an upper range approaching 1000 K to low ranges of 500 or 600 K. The bottom ranges might, briefly, be survivable by unprotected human meat. Long enough that the oxygen poisoning would get you before you died of hyperthermia. But you wouldn't last long, and would come back up approximating "bil tong" or "jerkey". Filling your descent suit with barbecue sauce and adding ... "plumbing" might be considerate for people who want to grok your fullness.

My "Atmospheric Pressure" notes refers to Wiki (no article name), and "https://agupubs.pericles-prod.literatumonline.com/doi/10.1029/2002JA009430" but I honestly can't remember the gory details. That's what notes are for.

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Atmospherically speaking, it would be incredibly hot. Considering a dry adiabatic atmosphere, it would be nearly 343 degrees Celsius (617.4 degrees Fahrenheit) hotter at the bottom than at the top. If we were to assume that it was a lapse rate of 6.5 degrees celsius per km, than the temperature would be more like 227.5 degrees celsius (409.5 degrees faherenheit) hotter. Plus, you need to start thinking about geothermal heating at that depth too.

The other thing to keep in mind is that the pressure of the air would be very high too. Atmospheric pressure decreases exponentially with increasing height, so it would increase exponentially with decreasing height.

Assuming STP for outside the hole with a dry adiabatic lapse rate, the pressure at the bottom would be around 1.92*10^20 hPa. The density of the air at the bottom would also be very high.

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  • $\begingroup$ That number can't be right for the air pressure. 1.92*10^20 hPa is about 2*10^17 atmospheres. But in free air, air pressure only halves at about 5km, so a hole 35 km deep would have only 7 doublings -- 128 or so atmospheres at the bottom. $\endgroup$ Jul 8, 2016 at 3:02
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    $\begingroup$ You can check this math out: $$T(z)=T_{0}-\gamma z$$, $$\frac{dP}{dz}=-\rho g$$, $$P=\rho R T $$, $$ \rho=\frac{P}{RT} $$, $$\frac{dP}{dz}=-\frac{Pg}{RT}$$ $$\frac{d ln(P)}{dz}=-\frac{g}{RT}$$, $$dln(P)=-\frac{g}{R(T_{0}-\gamma z)} dz$$, $$ln(P/P_{0})=\frac{g ln(R * (T_{0}-\gamma z))}{R \gamma}$$, $$P(z)=P_0 (R (T_0-\gamma z))^{\frac{g}{R \gamma}}$$, subsitute the following values: $P_0=1000 hPa, R=287 \frac{J}{kg K}, T_0=273.15 K, \gamma=9.8 \frac{K}{km}= 9.8 * 10^{-3} \frac{K}{m} , g=9.8 \frac{m}{s^2}, z=-35 km = -35000 m$ $\endgroup$ Jul 8, 2016 at 3:37
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    $\begingroup$ This is all great stuff, thanks everyone! $\endgroup$ Jul 8, 2016 at 23:18

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