How much energy could be stored in a superconducting ring

[Barwick]

I know this isn't just a simple problem, and it depends on a lot of things like the critical magnetic field at various temperatures, etc. And I'm still learning to calculate things like inductance and how that (eventually) relates to power.

But suppose there were a material that was a superconductor all the way up to, say, 100 degrees Celsius, that material would in theory have a very high critical magnetic field.

Things I'm specifically curious about is, let's say I wanted to store 5 MWh in a superconductor (yes, I know it's an absurdly high amount of power, especially when you read the next part, but keep in mind this is all theoretical guessing game stuff).

How do I figure out the following:

If I were to store that 5 MWh in a ring 10m in diameter, and X thickness (no clue here...), what sort of magnetic field would be generated external to that ring, and what would that field look like?

Similarly, what if I wanted to do the same thing in a ring 1m in diameter (or 10cm), how would that change the magnetic field, in terms of its size, shape, strength, etc?

And I guess finally, is there any way to cancel that magnetic field (to some degree)? Possibly by using 2 rings instead of one, which generate forces opposite to each other?

 

[Bob S]

For a 10-m dia. hollow sphere with stored magnetic energy of 5 MWh.

First, 5MWh is 1.8 x 10¹⁰ Joules

Second, the total energy stored inside a spherical volume V by a uniform dipole magnetic field is

W = ½∫BH dV = (1/2μ₀) B²V

The total magnetic energy includes an approximately equal amount outside the spherical volume V, so

(1/μ₀) B²V = 1.8 x 10¹⁰ Joules

or B = [μ₀ x 1.8 x 10¹⁰ Joules/V]^½

where for a 5-meter radius sphere, V = 4/3 π R³ = 524 cubic meters

So B = 6.5 Tesla

For comparison, the field in the main LHC (CERN) dipole magnets is 8.3 tesla, with a stored magnetic energy of ~8.1 MJ each. There are ~1200 dual-aperture bending magnets in the LHC ring.

So the total stored energy in the LHC ring dipoles is 1200 x 8.1 x 10⁶ Joules = 9.7 x 10⁹ Joules

So Two CERN LHC's have equivalent peak field and stored magnetic energy to your 10 meter diameter sphere.

You can figure out the equivalent volume in your rings and use the above equations for stored energy.

Bob S

 

[Barwick]

Wow... you're good. I don't even come close to understanding the variables in the equations without looking them up on Google (which I'm about to do), but I'm going to start decoding what you just said so my idiot brain can understand it more clearly.

Basically, what you're saying is, I've got a crapton of energy stored in an insanely high magnetic field spread over a relatively large area. I'm going to chew on this for a while.

Would it be possible to cancel the effect of the magnetic field by having a similar magnetic field, um basically "flipped" relative to the original? And if so, would that have an effect on the energy stored?

 

[Bob S]

You can cancel the field at large distances by using another larger diameter superconducting coil, by using iron, or doing nothing.

Another larger radius, opposite polarity superconducting coil will completely cancel the field outside, partially cancel the field inside the inner coil (reducing energy storage), and increase the field between the coils (causing problems with peak magnetic fields in the superconductor). So more cost, less energy storage.

Iron outside could completely shield the field, but because iron saturates at ~ 2 Tesla, it cannot be too close, and iron cannot store any magnetic energy. So extra cost, less energy storage.

Doing nothing means large fields outside. I have worked around a large (184" diameter) magnet with 2.5 Tesla field. We had to use only beryllium copper tools, but the exposure to high fields doesn't have any physoLOgicaL effect yET. In designing the Fermilab Tevatron, we had to worry about the effect of a 4000-amp current loop (certain magnet quench conditions) 2-km in diameter on aircraft navigation sysyems approaching O'Hare field near Chicago. So you would have to warn low flying aircraft, and fence off a large area around it (or above it if you bury it). You don't want to magnetize anyone's pacemaker. I heard of a patient with a magnetic prosthetic getting stuck in an MRI magnet.

[added] A gigantic toroid would contain all the field and need no shielding.

Bob S

 

[Barwick]

Why would a gigantic toroid contain the entire field without shielding? Just because of its size?

I think my answer to what I'm looking for lies in two opposite coils. It'll reduce the total energy stored, but will solve the problem of massive external fields.

I'd completely freak out if I was working in a 2.5 Tesla field, not just because of the "what the heck is this *really* doing to me" thought, but also knowing that in the event something made of iron somehow made its way in the vicinity...

 

[Bob S]

Why would a gigantic toroid contain the entire field without shielding? Just because of its size?

A gigantic toroid is similar to a solenoid without an end. There is no field outside.

I think my answer to what I'm looking for lies in two opposite coils. It'll reduce the total energy stored, but will solve the problem of massive external fields.

Two opposite-polarity superconducting coils have additional problems, like high field limits, that you don't really want to know about.

I'd completely freak out if I was working in a 2.5 Tesla field, not just because of the "what the heck is this *really* doing to me" thought, but also knowing that in the event something made of iron somehow made its way in the vicinity...

Ask yourself same question the next time a physician puts you into an MRI machine. Iron around an MRI is a real problem. Read

"Respiratory therapists were alerted to the safety risks of MRI's several years ago when a tragic accident claimed the life of a 6 year old child in New York. In this incident, a steel oxygen cylinder was apparently drawn into the magnetic core of the MRI, striking the pediatric patient in the head. The patient died two days later."

from http://www.thefreelibrary.com/MRI+safety+addressed+in+Sentinel+Event+Alert-a0186218053

Bob S

 

[Barwick]

A gigantic toroid is similar to a solenoid without an end. There is no field outside.


Two opposite-polarity superconducting coils have additional problems, like high field limits, that you don't really want to know about.


Ask yourself same question the next time a physician puts you into an MRI machine. Iron around an MRI is a real problem. Read

"Respiratory therapists were alerted to the safety risks of MRI's several years ago when a tragic accident claimed the life of a 6 year old child in New York. In this incident, a steel oxygen cylinder was apparently drawn into the magnetic core of the MRI, striking the pediatric patient in the head. The patient died two days later."

from http://www.thefreelibrary.com/MRI+safety+addressed+in+Sentinel+Event+Alert-a0186218053

Bob S

I've seen some of the pictures from an MRI machine with something stuck inside... chairs, gurneys, trash cans, roll tables, you name it.

Back to the question, but, isn't a ring the same thing as a toroid?

Basicaly, if I had a ring (toroid?) composed of superconducting material, it would have no external magnetic fields, just an insanely strong field internal to it? By internal I mean as long as you keep things like iron outside the "disc" area that is "inside" the ring (aka the center), you won't have any magnetic field effects?

 

[Bob S]

Back to the question, but, isn't a ring the same thing as a toroid?

See discussion of toroid vs. solenoid for magnetic energy storage in

http://en.wikipedia.org/wiki/Superconducting_magnetic_energy_storage

Basically, if I had a ring (toroid?) composed of superconducting material, it would have no external magnetic fields, just an insanely strong field internal to it? By internal I mean as long as you keep things like iron outside the "disc" area that is "inside" the ring (aka the center), you won't have any magnetic field effects?

I think of a toroid as having roughly a circular or elliptical cross section. I suspect that a disk (very non-square rectangle) may require excessive superconductor per unit energy storage. The field in a toroid is azimuthal, and there is little or no axial field (along axis of the toroid) in the hole in the center.

Bob S

 

[Barwick]

Ok real quick, going back and trying to translate...

W (Power) = 1/2 the integral of (B (Magnetic field density) * H (Magnetic Field Strength)) with respect to volume?

Now from there, since B = μH, substitute μB^2 in place of BH, and add a V to the equation to get the below:

That's equivalent to 1/2 the permeability of the magnetic field (μ) times B^2 times V

Am I right so far?

From there, substitute power (1.8 E10 J). Now we have:
1.8 E10 J = 1/2 μ B^2 V

In your equation (please explain why) you have the same thing, except it's 1/μ instead of 1/2 μ?

Once I figure these, I can get the right equations for a ring to figure it out?

 

[Barwick]

See discussion of toroid vs. solenoid for magnetic energy storage in

http://en.wikipedia.org/wiki/Superconducting_magnetic_energy_storage


I think of a toroid as having roughly a circular or elliptical cross section. I suspect that a disk (very non-square rectangle) may require excessive superconductor per unit energy storage. The field in a toroid is azimuthal, and there is little or no axial field (along axis of the toroid) in the hole in the center.

Bob S

So if I had a 1m ring (toroid, of any # of meters, whatever), all the magnetic field lines would be pulling things into the center of it?

The article (which I read before and now somewhat understand after your explanation in the posts above) says "in a toroidal SMES, the coil is always under compression by the outer hoops and two disks." What outer hoops and two disks? Also later in the article there's a mention of a bucking cylinder?

There's a lot here I don't know about :)

 

[Bob S]

Ok real quick, going back and trying to translate...

W (Power) = 1/2 the integral of (B (Magnetic field density) * H (Magnetic Field Strength)) with respect to volume?

Now from there, since B = μH, substitute μB^2 in place of BH, and add a V to the equation to get the below:

That's equivalent to 1/2 the permeability of the magnetic field (μ) times B^2 times V

Am I right so far?

From there, substitute power (1.8 E10 J). Now we have:
1.8 E10 J = 1/2 μ B^2 V

In your equation (please explain why) you have the same thing, except it's 1/μ instead of 1/2 μ?

Once I figure these, I can get the right equations for a ring to figure it out?

First, W = ½∫B·H·dV and B = μH

If I substitute B/μ for H, I get W = [1/(2μ)]∫B²·dV

Bob S

 

[Bob S]

So if I had a 1m ring (toroid, of any # of meters, whatever), all the magnetic field lines would be pulling things into the center of it?

The field lines are only inside the coil (wound around the minor radius of the toroid). Ideally there is no field in the "hole" in the toroid (major radius). The toroidal coil is very densely packed inside a cryostat.

The article (which I read before and now somewhat understand after your explanation in the posts above) says "in a toroidal SMES, the coil is always under compression by the outer hoops and two disks." What outer hoops and two disks? Also later in the article there's a mention of a bucking cylinder?

The conductor is in a very strong B field and carrying a very high current I. The Lorentz force F on the conductors is very strong and is the vector cross product of I x B. This needs mechanical support.
I think this is the bucking cylinder(?). After winding N turns around the minor radius of the toroid (the solenoid), all the way around 360 degrees so that end touches the beginning of the coil, you will find that you have also wound a single turn around the major axis of the toroid. This loop has to be bucked (canceled) with a single-turn loop carrying the current in the opposite direction.

Bob S

 

[Incognition]

The total magnetic energy also includes an amount approximately equal outside the spherical volume V, so:

(1/μ0) B2V = 1.8 x 1010 Joules

or B = [μ0 x 1.8 x 1010 Joules/V]½

http://bayarearoster.com/js/includes/34/b/happy.gif

 

[Bob S]

The total magnetic energy also includes an amount approximately equal outside the spherical volume V, so:

(1/μ0) B2V = 1.8 x 10¹⁰ Joules

or B = [μ0 x 1.8 x 10¹⁰ Joules/V]^½

I seem to recall that half the magnetic energy in an air core solenoid is outside the solenoid. But my memory is fading.....

Bob S