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Martoonsky

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Everything posted by Martoonsky

  1. I will first say that I am highly ignorant about how the food processing industry works, but here's how I envision it. A lab obtains a specimen of the desired species of bacteria from whatever source. They use whatever techniques are required to identify and isolate the species they wish to market. In a very controlled environment, they culture a large quantity of the target species with high purity. This high purity quantity is sold to a food manufacturer, which handles it carefully to avoid contamination and stores it in an environment that will preserve it effectively. Small quantities of this are used to grow large quantity cultures which are used to create vitamin B12 in marketable quantities. Eventually these cultures will spoil from contamination and need to be discarded. A new culture is then started from another small sample from the high purity quantity in storage. Eventually, the high purity quantity will be used up and a new quantity must be purchased from the lab. The lab, with its ability to test and isolate, should be able to keep the target species growing for as long as they like without having to bring more in from an external source. Is there anyone with some solid knowledge of food industry procedures who can confirm, deny, or correct any of this? Thanks.
  2. @CharonY Thank you for your response. Neither I nor my friend know much about industrial fermentation. In fairness to my friend, I have been doing a fair bit of searching recently and haven't found a definitive refutation of his idea. I find no mention anywhere of strains going bad or of a requirement for regular harvesting from animals, but as they say, absence of evidence is not evidence of absence. @zapatos I had already read the Wikipedia articles. They say that certain species and fermentation are used, but don't give enough information about the process to prove my friend wrong.
  3. 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.
  4. Sorry, I wasn't being specific enough. I was talking about the bacteria that produce the B12.
  5. He's looking. The problem is, I don't have any citations either.
  6. My friend claims that the strains used by industry for production by fermentation get "old" (contaminated, genetically mutated) and need to be replenished regularly from animal sources. He also claims that this means that the necessary quantity requires us to harvest significant quantities from animals that wouldn't be available in the same quantity from non-animal sources. I suspect that labs should be able to maintain their strains with minimal need to look for new sources. Some confirmation of this (that I can reference) is what I'm looking for.
  7. My friend says that the vitamin B12 produced as a food supplement is harvested from animals. I pointed out that online articles indicate that B12 is produced through fermentation. He claims that the strains used for fermentation get "old" and have to be replaced by new ones harvested from animals. I'm skeptical about that. Who's right? If you could provide sources so I could prove it to my friend (assuming I'm right), I would appreciate it. Thanks.
  8. 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.
  9. I found my math error so I'm feeling a little better. Maybe I won't give up so easily after all.
  10. 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.)
  11. 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.
  12. 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.
  13. 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.
  14. @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.
  15. 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.
  16. 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.
  17. 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.
  18. It is essential that the walls of the borehole be able to withstand the pressures exerted on them by the earth. There's no way the air pressure in an open tube would be able to do it. Also, the volume of the troposphere is 1E10 cubic kilometers whereas the volume of a one meter borehole would be 0.01 cubic kilometers, so the volume of air should not be an issue.
  19. I think it's reasonable to expect some plasma development in the earth's core. The temperatures there are hotter than the photospheres of some stars. It's a matter of degree, though, and I'm hoping that the ideal gas law may still be accurate. I don't think there would be a jet of hot air from the borehole. This would evacuate the air from the interior, creating a vacuum and eliminating the pressure which would drive the flow of air. I believe the air would settle into hydrostatic equilibrium, as it does in stars and the earth's atmosphere. It would likely be very rarefied near the center. I don't plan on considering the earth's rotation for now, but that's a good point. The plasma issue is looking a bit thorny to figure out, so I think I'll go ahead for now just assuming that the ideal gas law applies throughout.
  20. Yeah, if I decided the borehole was lined with a perfect insulator, that would make life easier. I decided to take temperature into account. I'm pretty sure it would make a difference - it is part of the ideal gas law.
  21. This would be true if the earth were of constant density. I was planning on modeling the earth as a number of layers of different densities, which would complicate the formula for g(r). I had a strategy all mapped out, but now I'm having a bit of trouble dealing with the air molecules dissociating and becoming a plasma near the center.
  22. @studiot Yes, that's right. I was also planning on having the temperature be variable as we go down into the earth. @joigus I've run into a snag. I was planning on using the ideal gas law but I now realize that deep inside the earth, the air would be heated to a plasma.
  23. For constant [math]a [/math] and variable [math] r [/math], [math]\int a dr = a \int dr = ar + c \\[/math] However, for a function [math]f\left(r\right), \int f\left(r\right) dr \neq f\left(r\right)\int dr [/math] Thus [math]\int \rho\left(R\right)g\left(R\right)dr = \rho\left(R\right)g\left(R\right)\int dr [/math] and [math]\int_{0}^{R} \rho\left(R\right)g\left(R\right)dr = \rho\left(R\right)g\left(R\right)\int_{0}^{R}dr = \rho\left(R\right)g\left(R\right)[r \vert_{0}^{R}] = \rho\left(R\right)g\left(R\right)R.[/math] However, when [math] \rho [/math] and [math] g [/math] are the functions [math] \rho\left(r\right) [/math] and [math] g\left(r\right)[/math] then [math]\rho[/math] and [math]g[/math] may not be removed from the integral and the functions will affect the result. For instance, if [math]\rho = 6 r^3[/math] and [math] g = \frac{1}{r^2}[/math] then [math]\rho g = 6r[/math] and [math]\int_{0}^{R} \rho g dr = \int_{0}^{R} 6r dr = 6\int_{0}^{R} r dr = 6[\frac{r^2}{2} \vert_{0}^{R}] = 3R^2.[/math] But this is not equal to [math]\rho\left(R\right)g\left(R\right)R = 6R^3 \left(\frac{1}{R^2}\right)R = 6R^2.[/math] Thus when [math]\rho[/math] and [math]g[/math] are functions of [math]r[/math] they make a difference in the value of the integral.
  24. Well, if anyone is following this, I've been working on it. I think I have a method worked out, but I'm still working on the details. More later.
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