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Posers and Puzzles

Posers and Puzzles

  1. Subscriber sonhouse
    Fast and Curious
    20 Aug '06 20:25
    How do you calculate the power flow in a superconductor? If you have a normal conductor, it still has resistance so you can calculate the power, I^2*R But if R=0 this formula fails, so how do you do it in a superconductor?
  2. 20 Aug '06 22:24 / 1 edit
    what is superconductor
  3. Subscriber sonhouse
    Fast and Curious
    20 Aug '06 22:51
    Originally posted by bloodyboy
    what is superconductor
    He's the best one on the line, he gives kids lollipops.
    In physics, its an electrical conductor that shows zero resistance which means all the power that comes in one end gets to the other end with no loss. In regular wire, like copper, silver, aluminum, etc., there is some loss because although those three are very good conductors of electricity, they are not perfect and so loss some of the energy pumped into them and get a bit warmer as a result, some of the energy flowing though the wire gets dissapated as heat. In a superconductor there is no heating because the electrons are coupled in such a way as to make for perfect flow of this energy. The present day generation of superconductors have to be cooled down, way down, to operate. Above a certain tempurature which varies from near absolute zero to above 40 degrees K and higher, they stay superconductive. They all loss this superconductivity if the resultant magnetic field goes past a certain strength, also a differant level for each type of superconductor so there is a maximum magnetic field intensity that can be generated by these devices. There is some new work that suggests with the right development we might end up with superconductors that don't need cooling at all and may stay superconductive past the boiling point of water. That would be a great day for civilization for sure. I keep expecting these ones to surface but there is a lot of work to do before we ever get that one! Still doesn't answer my question about how you calculate the power flowing in a superconductor, however!
  4. Standard member XanthosNZ
    Cancerous Bus Crash
    20 Aug '06 23:20
    P=I^2*R calculates power loss in a resistor not power flow. And of course it's zero for a superconductor that's the point.
  5. Subscriber sonhouse
    Fast and Curious
    21 Aug '06 01:25 / 1 edit
    Originally posted by XanthosNZ
    P=I^2*R calculates power loss in a resistor not power flow. And of course it's zero for a superconductor that's the point.
    Sure, it calculates through differential resistance. But how do you do it in a superconductor. I can see a little electron counter counting each electron as it goes by and then adding up the 1.6 E-19 coolohms per electron and coming up with it that way but what about for real? Measuring the resultant magnetic field? What if it has a mu metal shield that hides the field?
    I am thinking you can only do it after the current gets injected into an ordinary conductor. What about if you have a superconducting generator, superconducting cables and a superconducting motor. What method would you use to measure power directly without resorting to some kind of a physical measure such as RPM against a load or some such.
  6. Standard member XanthosNZ
    Cancerous Bus Crash
    21 Aug '06 01:36
    Go read about maximum current flow and skin effect.
  7. Subscriber sonhouse
    Fast and Curious
    22 Aug '06 20:15
    Originally posted by XanthosNZ
    Go read about maximum current flow and skin effect.
    Well maximum current flow is a problem inherent to superconductors in that they lose superconductivity if the internal magnetic field goes past a certain limit, thats a problem specific to super's but skin effect is actually used in super's, there are micron level coatings that have superconductivity. There is also recent work (see New Scientist) about some new calculations based on the recent Magnesium Bromide type and those calcs indicate there may be a path to way beyond room temp super's, like 400 C! Which still doesn't anwer my q. In regular wires, we have a dc magnetic field detector (hall effect) that can measure the current indirectly by measuring the external field strength so that must be the only way to do it in a super. I think I am thinking about the fact that super's reject external magnetic fields but I guess they don't confine fields from internal current flow, so that must be the answer. For some reason I was thinking that field rejection was a two way thing but it must not be or else we could not make those 20 tesla magnets, eh. BTW, did you see the photo's of the levitated frogs? It seems if you put a frog in a 10 tesla field, it floats! I figure about 100 tesla and HUMANS would float too. Don't know what that would do to all the metal we carry though!
  8. Standard member XanthosNZ
    Cancerous Bus Crash
    23 Aug '06 00:35
    Originally posted by sonhouse
    Well maximum current flow is a problem inherent to superconductors in that they lose superconductivity if the internal magnetic field goes past a certain limit, thats a problem specific to super's but skin effect is actually used in super's, there are micron level coatings that have superconductivity. There is also recent work (see New Scientist) about some ...[text shortened]... HUMANS would float too. Don't know what that would do to all the metal we carry though!
    Read about paramagnetism and the levitating frog will become clear. And don't try to tell me what it is. I already know. Just like I already know what maximum current flow and the skin effect are. So why are you telling me about them?
  9. Standard member abejnood
    Independant Thinker
    24 Aug '06 22:11
    Originally posted by sonhouse
    Sure, it calculates through differential resistance. But how do you do it in a superconductor. I can see a little electron counter counting each electron as it goes by and then adding up the 1.6 E-19 coolohms per electron and coming up with it that way but what about for real? Measuring the resultant magnetic field? What if it has a mu metal shield that hid ...[text shortened]... ly without resorting to some kind of a physical measure such as RPM against a load or some such.
    A superconducter is not a normal "resistor", however. It is like an inductor or semiconductor of sorts where Ohm's law does not apply. In any event, a superconductor is not either a circuit nor does it do any work, and therefore it doesn't have any power.
  10. Standard member XanthosNZ
    Cancerous Bus Crash
    24 Aug '06 22:26
    Originally posted by abejnood
    A superconducter is not a normal "resistor", however. It is like an inductor or semiconductor of sorts where Ohm's law does not apply. In any event, a superconductor is not either a circuit nor does it do any work, and therefore it doesn't have any power.
    Ohm's law does apply to semiconductors except you have to use resistance at a given voltage difference (which granted isn't very useful) if you want to find I from V or V from I.

    And remember resistors don't do much. They just sit there, the current travelling through them however does work travelling through the resistor and because of this heats it up.
  11. Standard member abejnood
    Independant Thinker
    24 Aug '06 22:33
    Originally posted by XanthosNZ
    Ohm's law does apply to semiconductors except you have to use resistance at a given voltage difference (which granted isn't very useful) if you want to find I from V or V from I.

    And remember resistors don't do much. They just sit there, the current travelling through them however does work travelling through the resistor and because of this heats it up.
    Your last point illustrates the uselessness of superconductors today. The resistance does just sit there, but at least generates power. In any event, it takes much more work to keep superconductors at their level of temperature than anything can be gotten from it.
  12. 24 Aug '06 22:34
    Originally posted by XanthosNZ
    Ohm's law does apply to semiconductors except you have to use resistance at a given voltage difference (which granted isn't very useful) if you want to find I from V or V from I.

    And remember resistors don't do much. They just sit there, the current travelling through them however does work travelling through the resistor and because of this heats it up.
    Hm, interesting stuff. I didn't know any of this, I almost feel like looking this stuff up, but I'll probably hear about it in physics class pretty soon anyways...
  13. Standard member XanthosNZ
    Cancerous Bus Crash
    24 Aug '06 23:33
    Originally posted by abejnood
    Your last point illustrates the uselessness of superconductors today. The resistance does just sit there, but at least generates power. In any event, it takes much more work to keep superconductors at their level of temperature than anything can be gotten from it.
    Jesus Christ, resistors do not generate power they absorb it in the form of heat. Your bar heater? That's basically a big resistor. It's hot and glowing because it's giving off heat energy that it gained due to current passing through the resistor.

    Superconductors have a number of uses right now. Anything that involves large magnetic or electric fields? Probably uses superconductors so that's MRI machines use them (if you ever get to take a tour of one of these take it up, it's cool to watch them trace field lines in the entire room with a pair of scissors on a string), all high speed particle collision tests (CERN) use them, you can use them as giant batteries storing MegaWatts of energy. A few guys in the Electrical Engineering department build a rather impressive superconducting alternating transformer.

    There are limits that many people don't know about. I mentioned the skin effect. Basically it causes current to flow nearer the outside of conductors and is noticeable if the conductor is big enough (there's a reason power transmission lines don't just get fatter and fatter, the middle becomes useless). However in superconductors thanks to the zero resistance (until we reach saturation current) we get all the current flowing on the very outside of the conductor (maybe a few atoms deep). That severely reduces max current flow from what it could be.

    There's also the fact that superconductors will stop working above a certain magnetic field or a certain electrical field. And given that they generate both you have to be careful.
  14. Subscriber sonhouse
    Fast and Curious
    25 Aug '06 01:24
    Originally posted by XanthosNZ
    Jesus Christ, resistors do not generate power they absorb it in the form of heat. Your bar heater? That's basically a big resistor. It's hot and glowing because it's giving off heat energy that it gained due to current passing through the resistor.

    Superconductors have a number of uses right now. Anything that involves large magnetic or electric fields? ...[text shortened]... r a certain electrical field. And given that they generate both you have to be careful.
    I guess you heard about the latest result, 1500 amps/cm width, a new record for super's. Those were on about 10 micron thick layers. I think the skin effect goes a bit deeper than just a few atoms though, if that were true, these latest versions would not need to be even a micron deep. (10,000 angstroms, room for several thousand molecules of the stuff).
  15. Standard member XanthosNZ
    Cancerous Bus Crash
    25 Aug '06 01:49
    Originally posted by sonhouse
    I guess you heard about the latest result, 1500 amps/cm width, a new record for super's. Those were on about 10 micron thick layers. I think the skin effect goes a bit deeper than just a few atoms though, if that were true, these latest versions would not need to be even a micron deep. (10,000 angstroms, room for several thousand molecules of the stuff).
    How do you measure skin depth? Introducing a metre of any type will disrupt it.