From: neufeld@helios.physics.utoronto.ca (Christopher Neufeld)
Subject: Re: Superconductors
Organization: University of Toronto Physics/Astronomy/CITA
Date: Wed, 19 Aug 1992 06:59:41 GMT

In article <1992Aug19.061628.5126@ee.eng.ohio-state.edu> rob@guinness.eng.ohio-state.edu (Rob Carriere) writes:
>In article <Bt66ry.H2B@helios.physics.utoronto.ca>
>neufeld@helios.physics.utoronto.ca (Christopher Neufeld) writes: 
>>In article <1992Aug18.060748.2986@ee.eng.ohio-state.edu>
>rob@guinness.eng.ohio-state.edu (Rob Carriere) writes: 
>>>No matter what the critical current density, that [storing energy in a
>>>magnetic field] is your _only_ option.  The
>>>electrical energy stored in the loop is I^2*R, and since R=0...
>>>
>>   No, the electrical energy stored in the loop is I^2 * L / 2. L, the
>>inductance, is not zero. However, this expression for stored energy is
>
>Err, that's the _magnetic_ energy, thank you.  I'm happy to be corrected, but
>only when the correction is, well, correct.  Calling the energy stored in a
>magnetic field `electrical' just because there happens to be a handy current
>loop in the immediate vicinity is a bit Humpty-Dumpty-ish.
>
   Not at all. Look, I can take a superconducting inductor and apply a
voltage to it. It draws current by the familiar V = L dI/dt, and hence
absorbs power at the rate of V * I. Later I can extract the energy from
the loop by putting the inductor across a resistance. I've stored
electrical energy. I happen to have stored it in a magnetic field, but so
what? Does a rechargeable battery store electrical energy or chemical
energy? The energy stored in a superconducting loop is certainly entirely
different from I^2 * R, since that represents the instantaneous power
dropped in a load.

>>rigorously equal to the energy stored in the field, the integral over all
>>space of one half the dot product of the 'B' and 'H' vectors. They're two
>>different ways of expressing the same thing.
>
>Oh, so you knew that.  Then what was your point?  In effect, you just said:
>"You're wrong, it's not like what you said at all, it's like what you said."
>
   No, check your quoted section above. It says that the energy has to be
stored in the field since the electric energy stored in the loop is zero
for a superconductor. This isn't true. I pointed out that you can talk
about it either way, either from a second year electromagnetics viewpoint
dealing with the energy density expression, or from a second year
electronics viewpoint where you keep track of integrated values of V * I,
exactly the way you would consider energy storage in a capacitor. The two
approaches are rigorously equal.
   A person doing an electronics course at the Horowitz & Hill level will
happily calculate the 'electrical energy stored in a perfect inductor' as
the time integral of V * I without ever considering, or even being aware
of, the magnetic field of the system, and she will get the right answer.

>>A well designed superconducting coil doesn't dump the field in
>>seconds even during a quench. 
>
>Hmm.  I'd like to see what your coil does in a severe mechanical impact
>situation.  (Your car battery can be pretty much relied on to survive a
>crash.) 
>
   What does it do? Well, superconducting wires are embedded in a copper
matrix for mechanical, thermal, and safety reasons. The majority of the
material inside a 'superconducting' wire is actually normal metal copper.
This provides a convenient thermal mass which absorbs the energy safely.
A typical superconducting coil is further designed with multiple taps to
safety resistors. It can safely dissipate the energy from any two breaks
in the cable. Crushing the coil in many places at once is an invitation
to excitement if things happen to conspire against you. No more so than
crushing a gas tank (which is a better example, I think, than a car
battery). And who's going to hit a multi-hundred metre diameter buried
superconducting coil?

>>We have a superconducting magnet which
>>occasionally quenches, dropping 200 kilojoules into the helium vessel. We
>>can have the magnet up and running again in a couple of hours, at the
>>cost of a few dozen litres of liquid helium. Of course our helium
>>recovery system is incapable of taking the helium flow during a quench,
>>so we lose the evaporated gas during a quench, but even that is not an
>>inevitable cost of running a superconducting coil.
>
>200 kJ?  That's what my hot pot expends to make two cups of coffee.  As an
>energy storage device, this thing is a toy.  I wouldn't recommend
>extrapolating from it.
>
   The point is, protection of coils is well understood and doesn't
result in a catastrophic, near-instantaneous release of the stored energy
into the environment. The copper cladding alone provides a very low
resistance short around parts of the superconductor which have gone
normal. With a time consant of L/R the field can take a long time to
decay. We've had a segment of our coil go normal and it took twenty
minutes with a sensitive gauss-meter to convince ourselves that the field
was actually decaying. Hardly an armageddon of lightning and danger.


-- 
 Christopher Neufeld....Just a graduate student  | Confidence is the feeling
 neufeld@helios.physics.utoronto.ca    Ad astra  | you have before you
 utzoo.utoronto.ca!generic!cneufeld              | understand the situation.
 "Don't edit reality for the sake of simplicity" |