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Everything posted by sethoflagos

  1. The mathematics topics that caught me under-prepared as a new undergrad (nearly 50 years ago!) were set theory and matrix algebra which weren't in our school curriculum. On the calculus side, the major missing item appears to be Numerical Solutions to ODEs, but as stated by several previously, such advanced topics will be covered during your course.
  2. To your excellent summary, I would stress the point that porous media - whatever the solid particle physical properties - make for first-class sound absorbers by their very nature. For a critical case scenario, google Hesco Bastions.
  3. Thanks! For a CoM reference frame with scattering into an arbitrary xy plane, I now have vx1 = +/- sqrt(1 - k^2) ux1; vy1 = k ux1; vx2 = -/+ sqrt(1 - k^2) ux1 / r; vy2 = - k ux1 / r for k = -1 ... +1 (0 = 'head-on') which looks quite sufficient to fill in any gaps in the particle velocity distribution without recourse to anything exotic.
  4. Quoting from https://en.wikipedia.org/wiki/Elastic_collision "Collisions of atoms are elastic, for example Rutherford backscattering. A useful special case of elastic collision is when the two bodies have equal mass, in which case they will simply exchange their momenta." Is this assertion mistaken? It does at least preserve both conserved quantities (momentum & energy), which your response appears to contradict.
  5. I've not really studied noble gases before, but on general principles, I understand that for 2 particles with mass ratio r, and initial scalar speeds u1,u2, the following equations should hold: u1 + ru2 = v1 + rv2 (conservation of momentum) u1^2 + ru2^2 = v1^2 + rv2^2 (conservation of energy) Leaving aside the trivial no collision solution (v1 = u1, v2 = u2), the quadratic formula gives the change in momenta during collision as: v1 - u1 = 2r (u2 - u1) / (r+1) ; v2 - u2 = 2r (u1 - u2) / (r + 1) ... which yields some rather puzzling consequences: 1) For a pure noble gas isotope (r = 1), v1 = u2, v2 = u1 - particle momenta are simply swapped. 2) If all particles have a uniform initial scalar speed (u2 = u1), they maintain their initial momenta indefinitely. In both cases, there is no apparent trend from a disequilibrium particle velocity distribution toward equilibrium. Is my reasoning correct so far? If so then where does the main mechanism for establishing an equilibrium particle velocity distribution come from: a) rare 3-particle interactions? b) quantum fluctuations? c) something else?
  6. Starting point without a doubt is Coulson and Richardson's Chemical Engineering: Volume 2A: Particulate Systems and Particle Technology, Sixth Edition. That's your broad brush introduction to the subject. But the range of application is extremely diverse - many, many specialist study areas with not so much common ground to unite them (or so it seems to me). If you have particular interest in some particular field, I may be able to give some relevant further references.
  7. Nearly. But it isn't the properties of the medium that are changing: it is whether or not the medium is flowing away from you toward the trains, which would delay your detection of the whistles; or flowing toward you from the trains, which would advance the detection. If you had no information on airspeed, say for instance you were observing via a remote video camera and microphone, your measurements of the (apparent) speed of sound would vary according to wind speed and direction wouldn't they? This is an example of Galilean non-invariance. Now substitute 'speed of light' & 'luminiferous aether' for 'speed of sound' & 'air'. What happened when Mickleson & Morley tried looking for these telltale Galilean non-invariances in their measurements of the speed of light through the medium of the luminiferous aether? They found none! This is why there is a fundamental difference between the non-invariant measured speed of sound in air and the invariant measured speed of light in vacuum. No worries. We've all had to work our way through this some time or other.
  8. In what circumstance would a medium be not at rest with respect to itself? Perhaps you intended "... at rest with respect to the observer", ie. a restatement of our 'still air' constraint. Could another restatement of this constraint be that observer and medium must be in the same Galilean reference frame? If so, then if the constraint was a necessary precondition, wouldn't this imply that the apparent speed of sound was not invariant under a Galilean transformation? I stress the word 'apparent' to distinguish between transmission over a measured distance in a measured time interval, as distinct from a physical property of the medium itself. These are two different quantities. Whether the latter is invariant to Galilean transformation is a moot point.
  9. Meteorites survive passage through the atmosphere, so perhaps you intend 'audible fireballs'. Even so, these propagate, initially at least, as supersonic bow shocks (sudden pressure discontinuities) rather than sound waves - at velocities dependent on their initial trajectories, and independent (for a while at least) of atmospheric properties. If one bolide were travelling towards the observer, and one away, they would be experienced as two blast wave fronts, the first perhaps considerably more powerful than the second. 'Pitch' is meaningless here. There may be some reflected aftershocks, but not necessarily in any ordered sequence. Perhaps your enquiries would be better served with a simpler less energetic case, such as two locomotives in still air, sounding their steam whistles as they passed each other in opposing directions. In the absence of relatavistic effects, and if the locomotives were perfectly streamlined (ie they weren't dragging a large envelope of air with them) an observer on the station platform would hear two perfectly simultaneous sounds pitched according to their respective Doppler shifts.
  10. 'Robotic skin' is a highly active area of current research, and your OP seems to require a relatively low resolution, planar version of the same thing. I'm puzzled why you pick impedance as the sensed property for this. Most work in this field has been focused on traditional resistance strain gauges, piezoresistance, and capacitance methods though Fibre Bragg Gratings (FBGs) seem to have received a lot of attention recently. The EU Roboskin Project is funded to the tune of about 5 million Euros: details at https://cordis.europa.eu/project/id/231500
  11. I clearly remember my grandmother starting the fire like that as the family gathered for christmas dinner. The sudden updraught caught her pet budgie by surprise and... it didn't end well.
  12. Natural draught is used extensively as a power source, but not in an obvious way. It doesn't provide enough power to pull combustion air through a commercial power station, but it contributes significantly in reducing the power requirement of the forced- & induced-draught fans. Similarly, natural draught cooling towers (and domestic coal fires for those old enough to remember them) don't need forced ventilation in order to maintain air flow.
  13. Would I be reasonably correct in thinking that the product of mean particle momentum and mean particle separation would need to be oto Planck's constant?
  14. That's actually quite interesting. Thank you +1 Maybe pointing to the rarest naturally occurring element in the universe is stretching the limits of "commercially available" a tad though.
  15. Very few materials we commonly encounter have a structure that is stable under a pressure of 1 GPa. Most, much less. We do not have presses capable of exceeding 100GPa by a huge amount, because we do not have the commercially available materials to construct such presses. Give or take the odd colliding celestial body. And they are a one shot deal.
  16. Sorry, but I don't buy this explanation. Take this example from a previous post: If there was "no heat whatsoever" flowing into your cold sink, that would imply that the thermal efficiency of your machine was 100% (it can't be, but we'll let that pass for now). In other words, you were extracting every last milliJoule of work allowed by the 2nd Law of Thermodynamics. However, if your heat engine did indeed "utilize ALL the heat fed into it" from say 1 kg of hot cocoa, then you would not only have your 100% thermal efficiency, but you would also have attained 100% Carnot efficiency. Congratulations! Now explain to me what you did with that 1 kg lump of cocoa at absolute zero. Because that is what "ALL the heat" means. If on the other hand, you actually ended up with luke warm cocoa, the Carnot efficiency simply tells you what percentage of "ALL the heat" was actually available to you. It says nothing whatsoever about the virtues or quirks of your machine - it is an absolute limit for any machine that was fueled by 1 kg of hot cocoa and exhausted 1 kg of luke warm cocoa. And it's a simple function of those two temperatures. Please pause and reread the last few paragraphs. You've wasted a fair chunk of the last few years through not understanding the difference between the highlighted terms. How much more time can you afford to waste? Carnot's equation is a simple algebraic manipulation of the 2nd Law limit - delta S = 0 Experimental verification of the 2nd Law automatically verifies the Carnot limit. The first item to pop up on Google was: "Experiment to verify the second law of thermodynamics using a thermoelectric device", Gupta, V. K.; Shanker, Gauri; Saraf, B.; Sharma, N. K. American Journal of Physics, Volume 52, Issue 7, pp. 625-628 (1984). I've not read it, but I'm sure it's fine, and typical of many thousands of similar published papers. It's actually based on a Seeback-effect heat engine, but the underlying principles are just the same. If you want to see more, just Google "experimental verification of 2nd law of thermodynamics" as I did. There's over 6 million results for you to sift through. Good hunting.
  17. You're confusing Carnot efficiency, 1 - TC/TH ... with thermal efficiency, 1 - QC/QH It is entirely consistent with both classical thermodynamic theory, and centuries of detailed empirical observation by the most gifted of experimentalists, for a machine to have a low Carnot efficiency (the fraction of energy in an input stream that is available for conversion to work) and a high thermal efficiency (the fraction of that available work you manage to convert to actual work) The following demonstrates most clearly that you are totally unaware of this distinction: With an inescapable effect on the authority of your output:
  18. Please give me a credible mechanism for the creation of a slush (suspension of ice particles in water) in the system I have described. I currently regard the idea as a deus ex machina, but would be delighted to be shown otherwise.
  19. Average low night temperatures on mars are below 200 K throughout the year. There is next to no atmospheric blanketing and the (blackened) pipes are shielded from the ground by mirror finish parabolic troughs. They are in thermal communication with nothing but the empty vacuum of space. You may have noticed that I switched from an initial single 48" ND pipe to multiple 8" ND pipes. This was specifically directed at providing the necessary 'A' in sigma.A.T^4 to meet the night side heat shedding load. You may also have noticed that I've switched to extracting the 18 m^3/s via multiple tapping points (oto 10,000 spread over 1,000 km) so that it is drawn off at practically zero velocity. If there is any ice nucleation in the body of liquid (rather than at the annular interface), then it will float upwards as far as it can. There is insufficient fluid shear to overcome buoyancy forces. And certainly insufficient fluid shear to start ripping consolidated phase Ih ice away from the pipe wall. And yes, the project is impractical if not impossible. That was never in doubt. Of course, I'd like to carry everyone along with me on this. But it's as clear as day that there are a few individuals who will never concede as a matter of principle. And bottom line is, I don't see I'm under any real obligation to convince anybody. Except possibly myself. I'm sure your disbelief was exemplary. But now that you have updated information, do you agree that the interface pressures can be subject to operator control?
  20. ... So... Well you expressed disbelief that the interface pressures could be controlled by the operator. My second to last post sort of covered that, but maybe I could flesh out some details for you. It is relatively easy to set the pressure of a ringmain. A very simple example would be to fit it with an open header tank set at the appropriate elevation and sized to accommodate any expansion/contraction or temporary fluctuations in inventory. Something a little more sophisticated would be called for here. Maybe an underground reservoir with a substantial gas blanket to absorb the fluctuations within a tight pressure band. From a functional point of view, it's identical to the elevated tank, but without the open connection to the martian atmosphere. So we can create two ringmains with tightly controlled operating pressures. We now install tapping lines equipped with one-way flow valves (check valves) into the freeze/thaw system. Those connected to the high pressure ringmain would have their check valves so oriented to only allow flow into the ringmain. These will draw flow from the freeze/thaw system only when its pressure at that point exceeds that of the high pressure ringmain - ie in the vicinity of the freezing interface. Conversely, those tapping lines connected to the low pressure ringmain would be oriented in the opposite sense, feeding the freeze/thaw system only at those locations at a pressure below that of the low pressure ringmain - ie in the vicinity of the thawing interface. Obviously, the ringmains and tapping lines would be fully insulated and traced to prevent them from freezing up. Once the high pressure ringmain is pressurised, the turbine sluices can be opened, controlling the high pressure ringmain to a setpoint somewhere in the middle of its design operating band, this will automatically feed the low pressure ringmain with precisely the volume required to feed the low pressure injection tappings. Since it's a closed, very nearly constant volume system, fluctuations should be very small, and self-regulating. I trust this rather long and detailed post is sufficient to quell your disbelief.
  21. Not that interested in ad hominems. Guess we're done.
  22. I'm comfortable with a waterside delta P of ~100 bar (10^7 Pa). Over 10,000 km, this equates to a pressure gradient of 1 kPa/km. This gradient is compatible with a line velocity of 0.15 m/s in 8" double extra strong (XXS) linepipe. This is less than 1% of the flow generated by the freeze thaw cycles making ~ 18 m^3/s available to be tapped off into a high pressure ringmain operating at say 10 bar less than the high pressure (freezing) interface. Similarly, a low pressure ringmain operating at 10 bar above the pressure of the low pressure (thawing) interface will return the 'borrowed' 18 m^3/s back into the thawing zone. The residual 0.15 m/s water velocity in the freeze/thaw system will be naturally maintained as a consequence of the imposed delta P. Hydroelectric generators linking high and low pressure ringmains will utilise the 18 m^3/s flowing between them with a delta P of 80 bar (8 x 10^6 Pa) to yield: Est. Power Output = 0.9 x 18 x 8 x 10^6 = 129.6 MW (continuous)
  23. Why would any particular m^3 of ice need to be thawed in 1 second? So long as it's fully thawed by mid afternoon, say, before the heat input has reduced to the point where it starts to refreeze, then it's done its job. 6 hrs to thaw = ~5,000 km of the collection array doing the thawing. Actually, if your figure of 590 W/m^2 is good, a high efficiency collection strip 100 m wide will do the job over ~1,200 km or about an hour and a half. So there's a fair safety margin to play with. PS. Thinking about it, since I'm going to be reinjecting the somewhat warmish low pressure discharge from the water turbines back into the thawing zone to meet the contraction demand, that fact in itself should significantly accelerate the thawing process.
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