studiot 2037 Posted August 15 (edited) There a problem with following the meteoroligical formula for the troposphere as this is based on measurements at many values of r. The procedure runs as follows. Using the general hydrostatic equation for a fluid under its own weight in a gravitational field [math]dp = - \rho gdr..............1[/math] and the equation of state with variable density for an ideal gas [math]\rho = \frac{P}{{RT}}.....................2[/math] We obtain the aerostatic equation [math]\frac{{dP}}{P} = - \frac{g}{{RT}}dr...............3[/math] Now direct measurement show that the temperature in the troposphere (where most of the mass of air resides) is a linear function of distance, with a slope coefficient, alpha, called the adiabatic lapse rate. [math]T = {T_R} - \alpha r....................4[/math] Measurements show that [math]{T_R} = {288^o}K[/math] and [math]\alpha = {6.5^o}K/km[/math] substituting equation 4 into equation 3 we find [math]\frac{{dP}}{P} = \frac{g}{{R\alpha }}\frac{{dT}}{T}...................5[/math] Integrating [math]\ln P = \frac{g}{{R\alpha }}\ln T + C............6[/math] and using the measured data to quantify the constants [math]\ln \frac{P}{{{P_R}}} = \frac{g}{{R\alpha }}\ln \frac{T}{{{T_R}}} = \frac{g}{{R\alpha }}\ln \left( {1 - \frac{{\alpha r}}{{{T_R}}}} \right).........7[/math] Now the points here are that this equation is experimentally observed over many data points. You only have the one single data point. Secondly this is an adiabatic calculation. So if you are going to follow it you must have adiabatic conditions in the tunnel. Thirdly the trposophere is about 11 km so this is the range of r. This allows this analysis to assume g is constant over that distance, which is reasonable. but the radius of the Earth is nearly 6 and a half thousand kilometers so such an assumption is unwarranted. Edited August 15 by studiot 0 Share this post Link to post Share on other sites

Martoonsky 1 Posted August 16 9 hours ago, studiot said: Now the points here are that this equation is experimentally observed over many data points. You only have the one single data point. Secondly this is an adiabatic calculation. So if you are going to follow it you must have adiabatic conditions in the tunnel. Thirdly the trposophere is about 11 km so this is the range of r. This allows this analysis to assume g is constant over that distance, which is reasonable. but the radius of the Earth is nearly 6 and a half thousand kilometers so such an assumption is unwarranted. There is a lot happening in the atmosphere - exchange of latent heat, absorption of sunlight, ionization by ultraviolet light. It's a different environment from the borehole. Adiabatic processes are those in which there is no exchange of heat or mass between a quantity of gas and its environment. There are both adiabatic and diabetic processes occurring in the atmosphere. However, the equation dP = (-) rho g dr describes how the weight of the air column changes as you change elevation. The weight of the air column does not depend on whether the processes occurring within it are adiabatic. In my analysis, g is not constant. 0 Share this post Link to post Share on other sites

studiot 2037 Posted August 16 On 7/23/2020 at 4:03 PM, Martoonsky said: I wonder if anyone has any ideas as to how one might figure this out. Since you continue to talk at cross purposes with me I see no point in my continuing the discussion. Thank you for wasting my time. 0 Share this post Link to post Share on other sites

Martoonsky 1 Posted August 16 17 hours ago, MigL said: Good Lord ! This is what you consider fun ??? For now, sure, but not usually. I usually hang out at cafes doing some reading. I also do a bit of bike riding and knitting. 0 Share this post Link to post Share on other sites

joigus 243 Posted August 17 On 8/16/2020 at 5:48 AM, Martoonsky said: diabetic Please, don't bring up diabetes. That would further complicate the problem. 0 Share this post Link to post Share on other sites

Martoonsky 1 Posted August 19 Well, I'm getting rather bummed out. At first, I seemed to be getting some kind of reasonable result, but then I discovered some math errors. One of my own (forgot a minus sign) plus I got an incorrect integration formula from a website. I think I have the right formula now, but when I do the integration manually, I get a negative value when I know the answer must be positive. When I plug the values into an online integration calculator, it comes up with a positive result. That doesn't make me feel good about my math skills. When I use the value from the online calculator, I get a final result of 3.49x10^19 bars, which is pretty hard to believe. The pressure at earth's center is 3.6 megabars. The "air" at the earth's center is likely to be a plasma. Also, at pressures higher than 34 bars, it would be a supercritical fluid rather than a gas. I still think it would be an interesting problem to solve, but it's starting to seem kind of intractable. I think I've put enough time and effort into it. I'm going to stop working on it. Thanks everyone for your help. (Just for reference, I'll post the math later that isn't working out.) 0 Share this post Link to post Share on other sites

Martoonsky 1 Posted August 21 I found my math error so I'm feeling a little better. Maybe I won't give up so easily after all. 0 Share this post Link to post Share on other sites

Martoonsky 1 Posted August 22 (edited) So, I'm feeling a bit better. I found the (algebraic) error in my math and I've had some time to think over my result for [math]P_0 [/math]. Here is the calculation: I ended with [math]P_0 = P_R e^-(\frac{\mu m_H}{K_B}\int_{R}^{0}\frac{g}{T}dr) [/math] First, a calculation of [math]\mu [/math], which is the mass of the average air particle in hydrogen mass units. The atmosphere is 78.08\% nitrogen gas, 20.95\% oxygen gas, 0.934\% argon gas, plus trace quantities of other gases. I will ignore the other gases. Let [math]\mu_N [/math] = 14.01 be the number of atomic mass units for a molecule of nitrogen gas, and similarly [math]\mu_O [/math] = 16.00 for oxygen gas and [math]\mu_{Ar} [/math] = 39.95 for argon gas. Then [math]\mu = 0.7808 \, \mu_N + 0.2095 \, \mu_O + 0.00934 \, \mu_{Ar} = 14.66 [/math] [math]m_H = 1.674 * 10^{-27} \, kg [/math] is the mass of a hydrogen atom. [math]K_B = 1.381 * 10^{-23} \, \frac{m^2kg}{s^2K} [/math] is the Bolzmann constant. We need to concern ourselves with the integral [math]\int_{R}^{0}\frac{g(r)}{T(r)}dr [/math] In my model, the earth has a constant density and the acceleration due to gravity changes linearly from the core to the surface as follows: [math]g(r) = K_\gamma r [/math] [math]K_\gamma = \frac{9.807 \, ms^{-2}}{6.371 * 10^6 \, m} = 1.539 * 10^{-6} \, s^{-2} [/math] The temperature in my model also varies linearly from the core to the surface as follows: [math]T(r) = K_T \, r + T_0 [/math] [math] K_T = \frac{T_R-T_0}{R} = \frac{288K - 5500K}{6.371*10^6\,m} = -8.181*10^{-4}\,\frac{K}{m} [/math] Thus [math] \frac{g}{T} = \frac{K_\gamma r}{K_T r + T_0} [/math] By polynomial long division, [math] \frac{K_\gamma r}{K_T r + T_0} = \frac{K_ \gamma}{K_T} - \frac{K_\gamma T_0}{K_T^2 r + K_TT_0} [/math] Thus we need to calculate [math]\int_R^0 \frac{g}{T}\, dr = \int_R^0 (\frac{K_ \gamma}{K_T} - \frac{K_\gamma T_0}{K_T^2 r + K_TT_0})\, dr [/math] \begin{math}= -\frac{K_\gamma}{K_T} R - K_\gamma T_0 \int_R^0 \frac{dr}{K_T^2 r + K_T T_0} [/math] We need to remove the units from the integral so they don't get lost in the math. The units of [math]K_T^2 r [/math] are [math]\{(\frac{K}{m})^2 * m\} = \{\frac{K^2}{m}\} [/math] and the same for [math]K_T T_0 [/math]. Thus the units of [math]\frac{dr}{K_T^2 r + K_T T_0} [/math] are [math]\{\frac{m^2}{K^2}\} [/math] [math]\int_R^0 \frac{dr}{K_T^2 r + K_T T_0} [/math] with the units included [math] = \{\frac{m^2}{K^2}\} \int_R^0 \frac{dr}{K_T^2 r + K_T T_0} [/math] where the integral is performed using dimensionless numbers. [math]\int_R^0 \frac{dr}{K_T^2 r + K_T T_0} = \frac{1}{K_T^2}\, (ln \vert K_T^2 r + K_T T_0\vert\,\vert_R^0) [/math] [math]= \frac{1}{K_T^2}\, (ln \vert K_T T_0\vert - ln \vert K_T^2 R + K_T T_0\vert) [/math] Back to our equation [math]P_0 = P_R e^-(\frac{\mu m_H}{K_B}\int_{R}^{0}\frac{g}{T}dr) [/math] [math]P_0 = P_R e^-(\frac{\mu m_H}{K_B} \frac{1}{K_T^2}\, (ln \vert K_T T_0\vert - ln \vert K_T^2 R + K_T T_0\vert)) [/math] Plug in all the values and we get [math]P_0 = 3.52*10^{19}\,P_R = 3.52*10^{19}\, [/math] bars [math] = 3.52*10^{24}\, [/math] Pascals. When first saw this value, I rejected it. The pressure at the center of the earth is only [math]3.64*10^{11} [/math] Pascals. The pressure at the center of the sun is [math]2.65*10^{16} [/math] Pascals! Then I began to warm up to it a little. I thought that the rock would be basically incompressible but the gas could be compressed, perhaps to densities that would require these pressures. And the center of the sun has a temperature of 16 million degrees. Maybe it would be okay. Then I had a look at the density equation, [math]\rho = \frac{P\mu m_H}{K_B T} [/math]. This gave a density of [math]1.14*10^{18}\, \frac{kg}{m^3} [/math] for the air at the center of the earth. The density of a neutron star is [math]5.9*10^{17}\, \frac{kg}{m^3} [/math]. Obviously, this wasn't a realistic model. So applying the equation [math]dP = -\rho g dr [/math] and making the assumption that the ideal gas law will apply all the way to the earth's center will not work, not even as an approximation. Perhaps the equation could still be used with some other description of how the density varies. Above pressures of about 34 bars, the air should become a supercritical fluid and begin to resist compression, thus slowing down the accumulation of weight as you go downward. Edited August 22 by Martoonsky 0 Share this post Link to post Share on other sites

joigus 243 Posted August 23 That's the direction I was going when I said that naive hydrodynamical calculations seem to produce ridiculous results. A back-of-the-envelope calculation in time saves a lot of work. Responsible for this could be the occurrence of exponentials. If you take a look at how spherically symmetric, self-gravitating fluids behave: https://en.wikipedia.org/wiki/Lane–Emden_equation.: Quote In astrophysics, the Lane–Emden equation is a dimensionless form of Poisson's equation for the gravitational potential of a Newtonian self-gravitating, spherically symmetric, polytropic fluid. You will realize that the dependence of density with the radius follows some kind of power law depending on the polytropic index n, which would have to be guessed. Your problem, I think, is still more complicated, because the fluid is not purely self-gravitating, but also is being subject to an external potential. 0 Share this post Link to post Share on other sites

Martoonsky 1 Posted August 27 I'm thinking of using a real gas equation of state and computing the integration numerically. I also want to take into account the supercritical state and the transition from gas-like to liquid-like fluids. Any info on the compressibility of supercritical fluids would be handy. 0 Share this post Link to post Share on other sites