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Martoonsky

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About Martoonsky

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  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.
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