How does the pressure vary as we go deep into the mines

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Pressure decreases as altitude increases in the atmosphere, while in mines, pressure increases with depth. The relationship for liquids is expressed as P(h) = P_{atm} + ρgh, assuming constant density. An exponential formula, P_h ≈ P_0e^{-mgh/kT}, is discussed for estimating pressure at height, though it may overestimate due to temperature changes. The title of the thread, "Heat," is questioned, with suggestions for more descriptive titles to improve clarity and response relevance. The applicability of the pressure equations below Earth's surface is limited, as actual values diverge significantly from predictions at greater depths.
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Sir,
How does the pressure vary as we go deep into the mines and as move up into the atmosphere? I think as we move up into the atmosphere the pressure decreases. I am not sure.
 
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You can derive this yourself by considering the forces on a column of air in equilibrium.

For a liquid, the relationship is too familiar, P(h) = P_{atm} + \rho g h where h is the depth below the free surface at which the pressure is atmospheric. Note that this equation holds only if the density \rho is constant. This too can be derived as above.

Show your work here if you need more help.
 
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One can estimate the pressure at a certain height above the ground using the exponential relationship;

P_{h} \approx P_{0}e^{-\frac{mg}{kT}}

This is not entirely accurate as it does not take into account the changes in temperature which will occur. Therefore, this equation will give an overestimate of the pressure.

Also, I would like to inquire as why you title all your post Heat?

~H
 
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You are right about the atmosphere. In general how pressure varies with height (or depth) in a fluid depends on how density varies. I can't say much about a mine. Are you asking how pressure changes as you below the Earth's surface?

EDIT: Sorry this post was written before I saw Hootenanny's reply.
 
Sir,
This expression given by you i.e. P(h) = P(o)e^(-mg/kT) doesn't have a direct dependence on h. Is the pressure indirectly related to h through g(acceleration due to gravity)?
 
Amith2006 said:
Sir,
This expression given by you i.e. P(h) = P(o)e^(-mg/kT) doesn't have a direct dependence on h. Is the pressure indirectly related to h through g(acceleration due to gravity)?

My apologies Sir, it was a typo mistake on my part, the formula should read;

P_{h} \approx P_{0}e^{-\frac{mgh}{kT}}

Apologies for any inconvience. I am still interested however as to why this thread is entiled 'Heat'

~H
 
Sir,
I feel that Pressure, Volume and temperature are terms closely related to Heat. So whenever my doubts involve these terms I give them the title heat. I think its troubling you a lot. Next time I will try to give a different title.
 
Is the above mentioned expression applicable to pressure variation below the Earth's surface also?
 
Amith2006 said:
Sir,
I feel that Pressure, Volume and temperature are terms closely related to Heat. So whenever my doubts involve these terms I give them the title heat. I think its troubling you a lot. Next time I will try to give a different title.

You are right, they are inseparable. The title doesn't bother me it was just I noticed that you were posting a lot of Heat titled threads. I would just like to point out that you may get faster and more applicable responses if you had a more informative title. For example, one could entitle this thread 'Pressure dependence and height'. It is not a critisism and please do not take it as such, it is only a suggestion.

~H
 
  • #10
Amith2006 said:
Is the above mentioned expression applicable to pressure variation below the Earth's surface also?

I'm afraid not, the difference between predicted values and actual values become significant below the Earth's surface.

~H
 
  • #11
Sir,
Actually I never thought about it. I will surely take your suggestions in the right spirit. Next time I will try to be more specific in giving titles.
 
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