Mariana Trench but without water?

Thread Starter

killivolt

Joined Jan 10, 2010
835
Air pressure doubles for every 5 km of depth, so if the Mariana Trench were dry, at 11 km deep the pressure would be about 4 atmospheres. 10 meters of water exerts an atmosphere of pressure, so 4 atmospheres corresponds to 40 meters or 130 feet. That's the pressure of air a scuba diver would be experiencing
kv
 

cmartinez

Joined Jan 17, 2007
8,178
I get 3 ATM:

View attachment 308447

But don't forget: temperature increases with pressure (adiabatic compression) -- which decreases density. I think you could expect the temperature at 11 km to be significantly high, therefore, the overall pressure would be much lower.
I haven't googled this yet ... but I remember that in a mine, temperature goes down as one goes deep, but it actually increases after a 700m depth is reached, or so ... I wonder what the temperature below ground in dry conditions would be at - 11,000 meters
 

WBahn

Joined Mar 31, 2012
29,865
A quick look at deep mines showed that the rock temperature at the bottom of the deepest S. African mine is about 150°F. Didn't see any indication of pressure.

Did find a calculation for pressure at the bottom of the Kidd mine in Canada, which has the deepest accessible non-marine point on Earth.

It's not stated in the Wikipedia article, but an answer on Stack Exchange used the formula from the Environmental Engineering in South African Mines and estimated the pressure to be a bit over 20 psi. I haven't tried to verify the data they used or the calculations.

On first blush, this seems on the low side, but consider that we are sitting under about 100 km of air to get it to 1 atm and about half of that is below 18,000 ft (~6000 m).

This page appears to be in pretty good agreement with the Stack Exchange calculations.

EDIT: Meant to point out -- the fact that all of the pages I looked at talking about the climate in deep caves mentioned temperature and humidity and some other things, but none of them (that I came across in the few minutes I searched) talked about air pressure would seem to indicate that it is a very minor factor and, for instance, does not come close to approaching 4 atm, which is past the point where concerns about nitrogen narcosis become significant considerations in a normal air mixture.
 
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wayneh

Joined Sep 9, 2010
17,493
I answered this on Quora once but quickly realized I’d have to fudge the answer - which I gave as 2-3 atm - because of the issues already identified here.

It’s not super complicated, just tedious to calculate a precise answer. You need to break the problem into intervals, maybe 10 feet each, calculate the weight of that interval assuming a constant temperature, pressure and density for that interval. Then add them up for the entire depth.

It’s the same approach you have to take to calculate pressure at the depth of the Titanic, say, because water compresses also at such great pressures. For water you could assume constant density over longer intervals, even 100 feet or more.
 

WBahn

Joined Mar 31, 2012
29,865
I answered this on Quora once but quickly realized I’d have to fudge the answer - which I gave as 2-3 atm - because of the issues already identified here.

It’s not super complicated, just tedious to calculate a precise answer. You need to break the problem into intervals, maybe 10 feet each, calculate the weight of that interval assuming a constant temperature, pressure and density for that interval. Then add them up for the entire depth.

It’s the same approach you have to take to calculate pressure at the depth of the Titanic, say, because water compresses also at such great pressures. For water you could assume constant density over longer intervals, even 100 feet or more.
Put you also have to include a model for how temperature changes with depth (for the air case, I don't know to what degree temperature affects compressibility of water under extreme depths), because the temperature of a gas has a significant impact on it's density. There are a couple of approximations that can be used for this.
 

BobTPH

Joined Jun 5, 2013
8,664
It’s not super complicated, just tedious to calculate a precise answer. You need to break the problem into intervals, maybe 10 feet each, calculate the weight of that interval assuming a constant temperature, pressure and density for that interval. Then add them up for the entire depth.
That is just numeric integration. You can also solve the integral analytically. As I recall we did that in high school physics.
 

wayneh

Joined Sep 9, 2010
17,493
Put you also have to include a model for how temperature changes with depth (for the air case, I don't know to what degree temperature affects compressibility of water under extreme depths), because the temperature of a gas has a significant impact on it's density. There are a couple of approximations that can be used for this.
I was thinking of seawater, where you have to worry about temperature and salinity changes with depth. Yes, for air you'd have to be more precise in your handling of temperature gradient due to the adiabatic lapse rate. But I think 10' intervals would allow for calculating a good estimate.
 

wayneh

Joined Sep 9, 2010
17,493
FWIW, the pressure in this hypothetical trench would be a serious health problem, just as scuba divers must contend with. Both the nitrogen and the oxygen could be toxic at that depth, and you'd have to equilibrate on the way back to sea level to avoid the bends.
 

sagor

Joined Mar 10, 2019
897
A quick look at deep mines showed that the rock temperature at the bottom of the deepest S. African mine is about 150°F. Didn't see any indication of pressure.

Did find a calculation for pressure at the bottom of the Kidd mine in Canada, which has the deepest accessible non-marine point on Earth.

It's not stated in the Wikipedia article, but an answer on Stack Exchange used the formula from the Environmental Engineering in South African Mines and estimated the pressure to be a bit over 20 psi. I haven't tried to verify the data they used or the calculations.

On first blush, this seems on the low side, but consider that we are sitting under about 100 km of air to get it to 1 atm and about half of that is below 18,000 ft (~6000 m).

This page appears to be in pretty good agreement with the Stack Exchange calculations.

EDIT: Meant to point out -- the fact that all of the pages I looked at talking about the climate in deep caves mentioned temperature and humidity and some other things, but none of them (that I came across in the few minutes I searched) talked about air pressure would seem to indicate that it is a very minor factor and, for instance, does not come close to approaching 4 atm, which is past the point where concerns about nitrogen narcosis become significant considerations in a normal air mixture.
I’ve been in a mine just over 2km (6800 ft) down and the rock wall temperature was around 105F.
 

WBahn

Joined Mar 31, 2012
29,865
FWIW, the pressure in this hypothetical trench would be a serious health problem, just as scuba divers must contend with. Both the nitrogen and the oxygen could be toxic at that depth, and you'd have to equilibrate on the way back to sea level to avoid the bends.
Nitrogen would be an issue, but oxygen toxicity normally starts at a partial pressure of about 2 atm, which would require a total pressure of about 9.5 atm. Lots of variability, though. Assuming 3.1 atm is close, that would be a partial pressure for oxygen of about about 0.65 atm. I think that's going to be in the safe realm, since exposure to 100% oxygen at up to 0.5 atm can be tolerated indefinitely and the presence of other inert gases provides protection to somewhat higher pressures (don't know what that higher pressure would be).

As for the bends, we already live with that here, to some degree, when people from the coasts fly into Colorado and then rent a car and head to the top of Pikes Peak. This is usually altitude sickness and not truly the bends, but climbers of higher peaks and occupants of unpressurized aircraft are at risk for true decompression sickness.
 

BobTPH

Joined Jun 5, 2013
8,664
And what integral would that be?
Whatever formula f(h) based on height that you are using. You are numerically integrating it by applying it at every 10 feet. You can also write that as the integral of f(h)dh over the interval of 0 to the final depth, and it is then summed continuously. Sorry, don’t know how to use Tex to format it correctly.

Are you not familiar with the foundations of integral calculus? This is exactly the kind of problem it solves.
 

joeyd999

Joined Jun 6, 2011
5,171
Whatever formula f(h) based on height that you are using. You are numerically integrating it by applying it at every 10 feet. You can also write that as the integral of f(h)dh over the interval of 0 to the final depth, and it is then summed continuously. Sorry, don’t know how to use Tex to format it correctly.

Are you not familiar with the foundations of integral calculus? This is exactly the kind of problem it solves.
Air density and temperature are not linear when water vapor is present -- which is always the case on Earth.
 

BobTPH

Joined Jun 5, 2013
8,664
And do you think something needs to be linear to integrate it? If you can write a function for it, you can integrate it.

All I am saying is that what @wayneh was describing was numerical integration. Is that really controversial?
 

wayneh

Joined Sep 9, 2010
17,493
Whatever formula f(h) based on height that you are using. You are numerically integrating it by applying it at every 10 feet. You can also write that as the integral of f(h)dh over the interval of 0 to the final depth, and it is then summed continuously.
The properties of moist air are not well described by any such simple formula that would lend itself to integration. The component gases in air, including the water vapor do exhibit near ideal behavior, and so the ideal gas law is a decent approximation. But if you need more precision, you have to use more complex equations such as the one for the vapor pressure of water maintained by IAWPS. I don't recall exactly but I think it has over 60 terms. That's just the water, and you need additional equations for the properties of nitrogen and oxygen and the interactions between these components; fugacity and all that stuff that HVAC professionals look up in their charts.

I have two commercial apps that perform psychrometric calculations for the properties of moist air using these internationally accepted equations.

While I love any chance to put my integration chops to good use, this is one time I'm confident that the brute force numerical approach is orders of magnitude faster and easier.
 

wayneh

Joined Sep 9, 2010
17,493
Nitrogen would be an issue, but oxygen toxicity normally starts at a partial pressure of about 2 atm, which would require a total pressure of about 9.5 atm. Lots of variability, though. Assuming 3.1 atm is close, that would be a partial pressure for oxygen of about about 0.65 atm. I think that's going to be in the safe realm, since exposure to 100% oxygen at up to 0.5 atm can be tolerated indefinitely and the presence of other inert gases provides protection to somewhat higher pressures (don't know what that higher pressure would be).
Scuba divers use a value of 1.6 as a safe limit, where the factor is the product of atmospheres times composition. Ambient air is 1.0 x 0.21 = 0.21. Enhanced air for diving is typically 32% oxygen and will be labeled for 40m max depth where the factor is 5.0 x 0.32 = 1.6.

So, yeah, the factor at the bottom of the trench might be 3.0 x 0.21 = 0.63 and that's well below the 1.6 safety factor. However, the 1.6 limit is in the context of a relatively brief exposure during a dive, ie. 30-60 minutes. Long term exposure to elevated oxygen is know to be damaging (or therapeutic) and there's quite a bit of conflicting information on the topic. I wouldn't be afraid to visit such a place but I'd think twice before deciding to live there.
 

wayneh

Joined Sep 9, 2010
17,493
Just to wrap this up, I went ahead and used the simplified barometric formulas (derived from the ideal gas law) as described in the wiki link provided by @joeyd999. One assumes the temperature increases with the lapse rate, and the other assumes constant temperature. Both can be found implemented in many places on the internet but you need to be careful to make sure which one is being used.

Anyway, assuming a constant temperature gives you 3.68 atm at the bottom of the trench. Using a constant lapse rate as it is at sea level gives you a lower 3.20 atm due to the higher temperatures and lower densities of the air column above.

It's unlikely that the lapse rate would remain constant, so this is definitely a "ballpark" estimate, IMHO. And the ideal gas law becomes increasingly less precise as pressure increases. Four atmospheres is still a "low" pressure though and I wouldn't expect a huge error from assuming ideal behavior.

Screenshot 2023-11-30 at 10.56.03 AM.png
 
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