SI
SI
discoversearch

We've detected that you're using an ad content blocking browser plug-in or feature. Ads provide a critical source of revenue to the continued operation of Silicon Investor.  We ask that you disable ad blocking while on Silicon Investor in the best interests of our community.  If you are not using an ad blocker but are still receiving this message, make sure your browser's tracking protection is set to the 'standard' level.
Strategies & Market Trends : ahhaha's ahs -- Ignore unavailable to you. Want to Upgrade?


To: Bilow who wrote (2321)5/19/2001 5:27:14 AM
From: frankw1900Read Replies (1) | Respond to of 24758
 
From May 1998 IEEE Spectrum. article on Super Conducting Fault Current Limiter (SCFCL). [Power grid stuff]

"Superconductors, because of their sharp transition from zero resistance at normal currents to finite resistance at higher current densities, are tailor-made for use in FCLs [Fault Current Limiters].

"Super conductors lose their electrical resistance below certain critical values of temperature, magnetic field, and current density. A simplified phase diagram of a superconductor defines three regions [Fig.1}. In the innermost, where the values for temperature, field, and current density are low enough, the material is in its true superconducting state and has zero resistance. In a region surrounding that area, resistivity rises steeply as values for the three variables go higher. Outside that area, resistivity is in essence independent of field and current density, as with ordinary conductors."

Current density, I believe, is measured in Ampere per square meter - symbol Acm^-2. Do I have that right? It's been a while.

"All the same, the superconductor's impedance is truly zero only for DC currents. The more common ac applications are affected by two factors. First, the finite length of the conductor produces a finite reactance, which, however, can be kept low by a special conductor architecture. Second, a superconductor is not loss free in ac operation; the magnetic ac field generated by the current produces so called ac losses -- basically just eddy current losses. .... They barely contribute to the total SCFCL impedance but dissipate energy in the superconductor thus raising cooling costs."

"It was earlier assumed that while a fault current is being limited, the voltage drop is uniform throughout the conductor. But in practice superconductors tend to develop thermal instabilities, called hot spots, connected with the strong current and temperature dependence of their resistivity in the transitional state. If a part of the superconductor sees a greater voltage drop than the rest, as a result of an inhomogeneity, this part will heat up even faster, leading to an even greater voltage drop at that point and further accelerated heating. Burn-through can result."

I learn from this:

Superconductors are not uniformly superconducting in cross section.
One parameter of superconductance is a critical value of current density.
Localized voltage drops cause heatng. They burn through.

Things might get exciting in high current superconducting labs.

You gave as examples of working superconducting devices things which are low temperature super conducting technology, are relatively low powered, and use materials which are normally conducting anyway, I think. Are there MRIs which use high temperature (cooled by nitrogen) superconductors?

I have a feeling that superconductors, especially hte high temperature kind will be used either in relatively massive applications (power grid mangement not transmission, MRI) where the great expense can be justified by great economic benefits or very small things which don't use much material.



To: Bilow who wrote (2321)5/19/2001 9:19:32 PM
From: E. DaviesRead Replies (1) | Respond to of 24758
 
Apparently from the other posts my logic was appropriate for AC but not DC.

Personally I don't believe in the concept of zero resistance. That goes beyond reason. Nothing physical is *ever* zero (or infinite). But thats another topic.

BTW- just to continue our arguement purely for the sake of arguing:
Power dissipation in a wire of resistance R carrying current I is R*I^2.

True, but since I=V/R Power dissipation is also V^2/R. Does that mean it's independent of current too?

And this too..
If a part of the superconductor sees a greater voltage drop than the rest, as a result of an inhomogeneity, this part will heat up even faster

Voltage? Hmmm..

Eric