joigus 283 Posted August 13 2 minutes ago, studiot said: That cannot possibly be because heat must be transported down the temperature gradient. But heat transport can happen even though the situation does not depend on time. Heat transfer must have reached a stationary regime (I'm not saying heat doesn't flow). Also, they're assuming perfectly isolating walls... Maybe I got the premises wrong. 0 Share this post Link to post Share on other sites

studiot 2063 Posted August 13 4 minutes ago, joigus said: But heat transport can happen even though the situation does not depend on time. Heat transfer must have reached a stationary regime (I'm not saying heat doesn't flow). Also, they're assuming perfectly isolating walls... Maybe I got the premises wrong. Yes they are. My model is isothermal so doesn't depend on adiabatic walls. The North Atlantic Gyre does not depend on time either, but there is still a dynamic equilibrium with coldwater descending and warm water rising. 0 Share this post Link to post Share on other sites

joigus 283 Posted August 13 3 minutes ago, studiot said: My model is isothermal so doesn't depend on adiabatic walls. Mmmm. But temperature must go up as you go down the hole, irrespective. Gravitation always heats up any stuff as you go down towards the core. Never mind gravity field going down to zero. Pressure builds up --> temperature goes up. It's not temperature coming from Earth's core transferring it to the gas. It's the gas' own internal energy/volume that does it. Say, I may have misunderstood something. I must confess I'm a bit confused about this one. If you want to solve the problem, the only thing you can do is let the air in the hole acquire temperature from its own pressurization, so to speak, as it builds up weight on top. The Earth can't touch it, either thermally or pressure-wise. I've just done a lookup and the Van der Waals eq. is not good enough to deal with this either. 0 Share this post Link to post Share on other sites

studiot 2063 Posted August 13 (edited) 38 minutes ago, joigus said: If you want to solve the problem, the only thing you can do is let the air in the hole acquire temperature from its own pressurization, so to speak, as it builds up weight on top. The Earth can't touch it, either thermally or pressure-wise. I've just done a lookup and the Van der Waals eq. is not good enough to deal with this either. Yes you seem to have understood exactly what I am thinking. The gas in the tunnel gains internal energy from the pressure of the gas above it (at r greater than its own). You cannot change the internal energy of a gas without changing its temperature so it also heats up. Heating up is a very important point since the air entering at the surface is already nealy 100^{o} above critical. From then on the compressibility v pressure is fairly linear. Adiabatic, semiadiabatic insulating non insulating walls, the gas will also gain some heat from its surroundings -- more heating up. 38 minutes ago, joigus said: I must confess I'm a bit confused about this one. If you want to solve the problem, the only thing you can do is let the air in the hole acquire temperature from its own pressurization, so to speak, as it builds up weight on top. The Earth can't touch it, either thermally or pressure-wise. I've just done a lookup and the Van der Waals eq. is not good enough to deal with this either. I was not thinking virial or VDW when I mentioned power series solutions. I was thinking about the aerostatic differential equation with a differential radial element dr. I was thinking of a power series solution to this equation. It is this element which is confined round the tunnel perimeter, but subject to compressibility radially. We thus have to consider the radial compressibility and have a choice isothermic or adiabatic ? I choose isothermic because I consider this differential element to be all at one temperature. Edited August 13 by studiot spelling 1 Share this post Link to post Share on other sites

joigus 283 Posted August 13 14 minutes ago, studiot said: I choose isothermic because I consider this differential element to be all at one temperature. Aaaah. Now I understand much better what you're trying to do. Thanks for careful explanation. +1 0 Share this post Link to post Share on other sites

Ken Fabian 135 Posted August 13 (edited) Studiot - I said - 12 hours ago, Ken Fabian said: I think it would reach pressure and temperature equilibrium and become effectively static. Air movement from convection can occur within the tunnel and changing weather based pressure differences at each end would generate air movements. You say - 8 hours ago, studiot said: That cannot possibly be because heat must be transported down the temperature gradient. I think pressure would approach equilibrium - and become effectively static, barring transient variations from pressure difference based air movements. And those would average out, even if they don't maintain a perfect equilibrium. With respect to the initial question I think heat is only relevant to pressure at the centre by changing the density of the air column and by that, the weight of the air column. Heat flow within a (not insulated) borehole/tunnel? Yes, it must, through conduction, convection and radiation but I think heat profile along the air column would still approach equilibrium over time as well, more slowly for a narrow borehole, more quickly for a large diameter tunnel that allows more convection - which I would expect to be the principal means of heat movement. 8 hours ago, studiot said: You don't need a source of air. I said circulation cell and countercurrent flow Thinking it further, convection would lead to hot air emerging from tunnel ends - but it would also draw in cooler surface air. Probably not a plume. And I would expect this outcome would still lead to an on-average, depth dependent, temperature equilibrium within the air column. With a through to other side tunnel - effectively two tunnels linked at the core - a continuous flow could be generated but off the top of my head (as most of this is) I would expect that the rate of air flow would have to be high enough that maximum air temperature is not reached before crossing through the centre - that it must continue warming (and losing density) on the way back up, in order to maintain the different densities to generate convection flow. For a continuing flow of air outwards you do need a source of colder air that undergoes heating, to generate expansion and lowered density. That has to come from convection within the tunnel or a continuous flow through from one end to the other. Some of this must depend on how the initial hypothetical conditions occur - a tunnel appearing magically, fully formed... but if it starts filled with air, is it surface temperature air? That would involve immediate expansion - and flow of air out both ends... until equilibrium. I wouldn't expect strong one way circulation - in one side, heating along the length, to emerge out the other without some significant other factor to get sufficient flow happening. Edited August 13 by Ken Fabian 0 Share this post Link to post Share on other sites

MigL 1388 Posted August 14 Since pressure is only dependent on the column of air above it, and ( pressure generated ) self heating, IOW we disregard any convection heating from the walls of the tube, why not just consider a gravitationally bound ball of gas ( say N2 ), the center of which is in equilibrium with pressure pushing outwards and gravity pushing inwards ? IIRC the Tolman-Oppenheimer--Volkoff equation is used for modelling star formation ( and compression of the fusible fuel on thermonuclear bombs ), and can be applied. The equation ( in spherical co-ordinates )for a static spherically symmetric star is derived by plugging in the energy momentum tensor for an ideal fluid into the Einstein field equations, along with conservation conditions. In the non-relativistic limit this reduces to simple hydrostatic equilibrium such that dP = -g(r) p(r) dr where P=pressure and p=density. I don't remember where I saw them, and perhaps someone more familiar with astronomy/astrophysics than I am might know, but I recall seeing tables, for the central pressures of various size stars, before the onset of fusion. It should not be a problem to extrapolate to a ball of nitrogen with the same radius as the Earth ( + atmosphere ). 1 Share this post Link to post Share on other sites

Martoonsky 1 Posted August 14 Interstellar gas clouds don't normally collapse unless some sort of compression or density wave triggers collapse by compressing the cloud beyond a critical value. I'm pretty sure that a gas cloud the size of the earth with a density similar to earth's atmosphere would not be gravitationally stable and would disperse. In any case, the gravity values and gradients would be different from that of the solid earth, so it's just a completely different situation. 0 Share this post Link to post Share on other sites

MigL 1388 Posted August 14 yeah, you're right. But the only difference would be in the -g(r) term due to the added mass of the solid part of the Earth. 0 Share this post Link to post Share on other sites

studiot 2063 Posted August 14 (edited) 14 hours ago, joigus said: Aaaah. Now I understand much better what you're trying to do. Thanks for careful explanation. +1 I wasn't going to post this before I had fin ished working on it but here is a summary. Please ask if there is anything you don't understand as it is not really beautiful enough for formal presentation. However this is not a competition so any suggestions as to the form of rho(r) is welcome. Edit I thought I posed this last night but obviously skipped the submit button. Note the correction to Martoonsky's original posting. It seems that we are all converging to a similar solution. My idea of a power series was to represent P(r) as a power series or inverse power series. This would then be multiplied through by the jr gravity term and allow easier term by term integration. The downside is that there is only one calibration point for the constants at r = R. Perhaps MigL's equation could help, I have not heard of this, cosmology/astrophysics is not my bag. +1 Have you any references MigL? Edited August 14 by studiot 1 Share this post Link to post Share on other sites

studiot 2063 Posted August 14 For those who are interested here is a chart of pertinent physical properties from Thermodynamics of the Earth and Planets by Douce (Cambridge University Press) 0 Share this post Link to post Share on other sites

joigus 283 Posted August 14 5 hours ago, studiot said: I wasn't going to post this before I had fin ished working on it but here is a summary. Please ask if there is anything you don't understand as it is not really beautiful enough for formal presentation. However this is not a competition so any suggestions as to the form of rho(r) is welcome. Edit I thought I posed this last night but obviously skipped the submit button. Note the correction to Martoonsky's original posting. It seems that we are all converging to a similar solution. My idea of a power series was to represent P(r) as a power series or inverse power series. This would then be multiplied through by the jr gravity term and allow easier term by term integration. The downside is that there is only one calibration point for the constants at r = R. Perhaps MigL's equation could help, I have not heard of this, cosmology/astrophysics is not my bag. +1 Have you any references MigL? Thank you for anticipating your ideas. +1 I'm in no hurry, in case you're wondering. I'm just interested. The problem is quite academic, but interesting nonetheless. Somewhat out of my scope. I think that's the route to solving it. Slicing the tube into infinitesimal slices of constant thermodynamic conditions. And let the gas equilibrate with itself. The only thing I don't see is the temperature. But maybe you've got that into account for later. 0 Share this post Link to post Share on other sites

Martoonsky 1 Posted August 14 (edited) People are going off on a number of tangents here, and that's cool. I like to see the different ideas. This is going to be my first really substantial post. I'm going to define the problem that I will be working on in a bit more detail. If others want to work on different variations, that's fine with me. I didn't intend for it to be an engineering problem. I'm not going to concern myself with how to construct or maintain the borehole. An alien race may have come along and punched a straw through the earth, or Harry Potter may have done it by magic, whatever. Here's the physical setup: there is a straight, narrow, cylindrical hole through the earth, passing through the center and open to the atmosphere on both sides. This will be known as "the borehole". The walls are strong enough to withstand the pressures exerted on it by the earth. In my model, the walls of the borehole will be sufficiently thermally conductive that the air in the borehole will be the same temperature as the earth at that radius. The problem is to determine what the air pressure would be at the center of the earth under these conditions. My original model was starting to get rather complicated, so I decided to start off with a very simple model, find a solution for that, and then perhaps work toward something more complicated. Here are the details of my first model: * The earth is spherical with radius [math] R = 6.371 * 10^6 m. [/math] * The earth has a constant density. * The earth's temperature varies linearly from the core to the surface. * The ends of the borehole have a temperature of 288 K and a pressure of one bar. * The temperature at the center is 5500 K. * The ideal gas law applies throughout the borehole. * The air in the borehole is in hydrostatic equilibrium. I will not consider effects due to the rotation of the earth. Distances are measured relative to the center of the earth, i.e. [math] r = 0 [/math] at the center and [math] r = R [/math] at the surface. We want to determine [math] P_0 [/math] which is the air pressure at the center of the earth. [math] P_R [/math] = 1 bar is the air pressure at the surface of the earth. I was originally considering using this equation: [math] P_0 = P_R + \int_{0}^{R}\rho g dr [/math] where [math] \rho [/math] is the density of the air and [math] g [/math] is the acceleration due to gravity. I thought this equation would give the result, but I wasn't sure how to express [math] \rho [/math] as a function of [math] r [/math], nor how to include the effect of the temperature. With a little help from Astronomy Education at the University of Nebraska-Lincoln (https://astro.unl.edu/naap/scaleheight/sh_bg1.html), I came up with the following: [math] dP = -\rho g dr [/math] We can then use the ideal gas law to come up with a substitute for [math] \rho [/math]: [math] PV = NK_BT [/math] where [math] V [/math] is the volume, [math] N [/math] is the number of gas molecules, [math] K_B [/math] is the Bolzmann constant, and [math] T [/math] is the temperature in Kelvin. For [math] N [/math] we can substitute [math] \frac{M}{\mu m_H} [/math] where [math] M [/math] is the total mass of the gas molecules, [math] \mu [/math] is the average mass of an air molecule in hydrogen atomic mass units, and [math] m_H [/math] is the mass of a hydrogen atom. Thus we have [math] PV = \frac{MK_BT}{\mu m_H} [/math] and [math] P = \frac{M}{V}\frac{K_BT}{\mu m_H} = \frac{\rho K_BT}{\mu m_H} [/math] We can solve for [math] \rho = \frac{P\mu m_H}{K_BT} [/math] Substituting this expression for [math] \rho [/math] in [math] dP = -\rho g dr [/math] we have [math] dP = -\frac{P\mu m_H g}{K_B T} dr [/math] and [math] \frac{dP}{P} = -\frac{\mu m_H g}{K_B T} dr[/math] Integrating downward from the surface to the earth's center, we have [math] \int_{P_R}^{P_0}\frac{dP}{P} = \int_{R}^{0}-\frac{\mu m_H g}{K_B T}dr = -\frac{\mu m_H}{K_B}\int_{R}^{0}\frac{g}{T}dr [/math] [math] \int_{P_R}^{P_0}\frac{dP}{P} = ln P \vert_{P_R}^{P_0} = ln P_0 - ln P_R = ln \frac{P_0}{P_R} [/math] Thus [math] ln \frac{P_0}{P_R} = -\frac{\mu m_H}{K_B}\int_{R}^{0}\frac{g}{T}dr [/math] and [math] \frac{P_0}{P_R} = e^-(\frac{\mu m_H}{K_B}\int_{R}^{0}\frac{g}{T}dr) [/math] and finally, we have an expression for the air pressure at the center of the earth: [math] P_0 = P_R e^-(\frac{\mu m_H}{K_B}\int_{R}^{0}\frac{g}{T}dr) [/math] It's getting late, so the completion of the calculation will wait for another day. Edited August 14 by Martoonsky 0 Share this post Link to post Share on other sites

MigL 1388 Posted August 14 (edited) It seems a number of people are converging on the Newtonian hydrostatic equilibrium equation. dP = - g(r) p(r) dr where P=pressure p=radius dependent density and g=radius dependent gravity https://en.wikipedia.org/wiki/Hydrostatic_equilibrium I only gave the Toll/Opp/Volk root because I was intending to use data tables I'd seen for proto-stars before fusion begins. However as Martoonsky pointed out, this will not work because the weight of the gas is affected by the much denser solid portions of the Earth around the bore-hole. Further complicating matters, if we are not going to consider an insulated bore-hole, then p, the density, is dependent on pressure, which is dependent on temperature of the surroundings ( up to 5500 deg as Martoonsky stated ). Back to the drawing board ... Edited August 14 by MigL 0 Share this post Link to post Share on other sites

studiot 2063 Posted August 14 2 hours ago, Martoonsky said: For N we can substitute MμmH where M is the total mass of the gas molecules, μ is the average mass of an air molecule in hydrogen atomic mass units, and mH is the mass of a hydrogen atom. Surely N is not " the number of molecules",, but " the number of molecules between r and (r+dr) ? In other words surely N is a function of r and increases as the gas layers become more dense ? Haven't you treated N as a constant in your integration? I look forward with interest to your integration of g/T, both of which are also functions of r. I looked at this analysis which is a standard fluid mechanics/meteorology solution. I did not pursue it because there are direct measurements that support the assumptions it makes and provide many calibration points for the constants. As a matter of interest, this article is a bit old but contains estimates of the variation of gravity, temperature and pressure in the rocks as functions of depth. https://www.researchgate.net/figure/The-pressure-at-different-depths-in-the-Earths-interior_fig12_228377528 Look at pages 12 through 17 in particular 0 Share this post Link to post Share on other sites

Martoonsky 1 Posted August 15 (edited) 5 hours ago, studiot said: Surely N is not " the number of molecules",, but " the number of molecules between r and (r+dr) ? In other words surely N is a function of r and increases as the gas layers become more dense ? Haven't you treated N as a constant in your integration? The ideal gas law describes the behavior of a particular parcel of gas without any gas entering or leaving, so N would be constant in the equation. After the substitutions, the only things removed from the integral are [math] \mu, m_H [/math], and [math] K_B [/math], which are constants. Edit: I take that back. In the laboratory, one would normally be dealing with a fixed quantity of gas in a container, but you could also have an imaginary boundary within a volume of air through which gas could pass, so N could be variable. However, N is not in the integral and it's still true that [math] \mu, m_H [/math], and [math] K_B [/math] are constants, so removing them from the integral is valid. Further edit: The variability of N is implicit in the variability of P, rho, and T. Once the physical substitutions have been made, any constants in the integrand can be removed from the integral. Edited August 15 by Martoonsky 0 Share this post Link to post Share on other sites

joigus 283 Posted August 15 3 hours ago, studiot said: I look forward with interest to your integration of g/T, both of which are also functions of r. That's exactly what I was thinking. There's a circularity here. g(r) is no problem. It's T(r) what's a pain in the neck. Last thing I've tried is to use the energy equation for a diatomic ideal gas, which is, \[U_{\textrm{int}}=\frac{5}{2}Nk_{B}T\] and then to relate the internal energy of the gas with the gravitational potential energy to eliminate the temperature, but no luck. The density comes back. Maybe I was too tired. You still have the T(r) problem. Even if you use a virial expansion, you still have the T(r) problem. 0 Share this post Link to post Share on other sites

studiot 2063 Posted August 15 (edited) 2 hours ago, joigus said: g(r) is no problem. Has no one looked at the paper I linked to it is a free pdf download. ? But too long (and against the rules) to fully reproduce. However I was suprised how much difference the different densities of mantle and core make to the variation of g. Edited August 15 by studiot 0 Share this post Link to post Share on other sites

joigus 283 Posted August 15 3 minutes ago, studiot said: Has no one looked at the paper I linked to it is a free pdf download. ? But too long (and against the rules) to fully reproduce. I may have missed it. I did take a look at your suggestion: On 8/14/2020 at 12:24 PM, studiot said: For those who are interested here is a chart of pertinent physical properties from Thermodynamics of the Earth and Planets by Douce (Cambridge University Press) and it got me linking back to my previous observation of how far away from an ideal gas we probably are. Very close to electron degeneracy regime at the deep core. I think @MigL's suggestion of thinking in terms of Oppenheimer-Volkoff equations of state is not very far-fetched, to say the least. I know next to nothing about that though. 0 Share this post Link to post Share on other sites

Martoonsky 1 Posted August 15 @joigus I believe I have the temperature problem solved. First of all, in my case, I decided to make the borehole walls thermally conductive so that the air in the borehole will have the same temperature as the earth at the same radius. But let's go the other way and let the walls of the borehole be perfect thermal insulators. Let's say the ends are sealed and the interior is a vacuum. You open the seals and let air flow in. It falls to the center and compresses, causing it to heat up. Now it's hot in the middle and normal atmospheric temperature at the ends. The only way you can really approach the problem is to let the system be in equilibrium. When there's a temperature gradient, heat will flow. Eventually (and it may take a very long time), the temperature throughout the borehole will constant and equal to the temperature at the ends. 0 Share this post Link to post Share on other sites

MigL 1388 Posted August 15 1 hour ago, joigus said: Very close to electron degeneracy regime at the deep core. I wouldn't think so, but it would be an ionized plasma at temps of 5500 deg. 0 Share this post Link to post Share on other sites

joigus 283 Posted August 15 49 minutes ago, Martoonsky said: I decided to make the borehole walls thermally conductive so that the air in the borehole will have the same temperature as the earth at the same radius. Oh, oh. Much of the heat in the Earth's interior comes from furious radioactive decay. Full of heavy elements there. That's why Kelvin got the age of the Earth wrong by quite a long shot. If you assume air temperature to come from there you may be faced with no sensible hydrodynamic regime being able to account for it. I think that's why, even though I was sloppy, sheer substitution in hydrodynamic formulas with T at centre = T_{core} gave me crazy numbers. And believe me, "crazy" is an understatement. It's true that I tend to do these calculations at 3 in the morning, so... I do believe you are deep in the regime of "hard balls" there. Ionization may play a part too, as MigL says. May I ask where this problem came from? Did somebody give it to you as an assignment or is it just for the fun of it? 23 minutes ago, MigL said: I wouldn't think so, but it would be an ionized plasma at temps of 5500 deg. Yep. I think it's a plasma down there even if you impose isolating walls. But even so, the "stellar" approximation would be less off than assuming an ideal gas. Nothing ideal in that situation. 0 Share this post Link to post Share on other sites

Martoonsky 1 Posted August 15 7 minutes ago, joigus said: May I ask where this problem came from? Did somebody give it to you as an assignment or is it just for the fun of it? It popped into my head one day. It's just for fun. I realize that there will be ionization and plasma development near the core. I started looking into it, but it was making life too difficult. I decided to go with some simplifying assumptions for starters. I have the first problem worked out. I'm working on the posting now. Later, I may take the earth's layers into account or have another look at plasma. 0 Share this post Link to post Share on other sites

joigus 283 Posted August 15 4 minutes ago, Martoonsky said: It popped into my head one day. It's just for fun. I realize that there will be ionization and plasma development near the core. I started looking into it, but it was making life too difficult. I decided to go with some simplifying assumptions for starters. I have the first problem worked out. I'm working on the posting now. Later, I may take the earth's layers into account or have another look at plasma. LOL. Pray you, insulate the walls!!! 0 Share this post Link to post Share on other sites

MigL 1388 Posted August 15 (edited) 1 hour ago, Martoonsky said: It's just for fun. Good Lord ! This is what you consider fun ??? Edited August 15 by MigL 0 Share this post Link to post Share on other sites