Benefits of using a superconductor to create a magnetic field at 77K

trini

Hello all, i'm in the process of designing my own electromagnets, and would like to know the benefits(if any) of using superconducting wire at 77K( liquid nitrogen cooled) rather than an equivalent length of copper wire cooled to those temperatures in creating a solenoid.

Because the critical current of commercially available High Temp Superconducting wire at 77K is only around 120 A, and the thinnest diameter i can find is about 3mm, am i better off going with regular copper wire and using higher currents and just cool it with liquid nitrogen(or some other system, I'm thinking liquid N2 might be unsafe due to a fast rate of evaporation)? Is the magnetic field produced by a superconductor at a given current much larger than that in a copper coil at the same current? Or is it the same?

Also, if the field produced is larger, how much larger is it and why is it producing a larger B field?

fleem

The field is the same--its still simply directly proportional to the current. The type of conductor doesn't matter--unless it has other magnetic properties like ferromagnetism, etc (and neither copper nor common superconductor material have such attributes). Maybe try a copper pipe and run the coolant through it-that way water might work fine. Not sure teeny copper pipes are easy to come by, though! I take it you need magnetic intensities above the flux density saturation of any known material, since you didn't mention the idea of a core? This thread has a couple of interesting points: http://www.physicsforums.com/showthread.php?t=238353

trini

ok i thought as much but wasn't thoroughly sure. i will revert to copper wire for practicality. i have attached a draft idea for my N2 refrigeration unit. I was wondering as well the length of time a strong magnetic field must interact with a ferromagnetic material before saturation occurs in the hysterisis curve, and how can you determine when this point is reached?

Oh i will be using NdFeB as my core material.

Attached Thumbnails

 

Bob S

Several things you should evaluate before embarking an a specific design:
1 What peak field do you need (want)?
2 What volume and shape of field do you want?
3 What magnet geometry to you want (C-magnet, H magnet, solenoid, etc,)?
(H-magnets usually give highest field)
4 What field uniformity do you want?
5 How many amp-turns do you need (want)?
6 Is a single layer coil adequate?
(multilayer may require hollow conductor unless superconducting)
7 What amps and volts do you need (want)?
8 Will you design or build power supply?
9 What regulation do you want?

trini

1. H peak should be around 50 KOe
2. rectangular (because i will me trying to make the curved surface of my object magnetised)
3. solenoid(but the poles would be on the curved surface and not the ends)
4. should be fairly uniform throughout(so i think my coils should extend beyond the magnetization region so that the field lines through the material will be as straight throughout as possible.)
5. A amp turn/m value of 3978873.5772971 would be req'd to produce a 50 KOe field
(so using 2mm wire at 40 turns per layer, each layer 8cm wide, and 10 layers high = 400 turns/8 cm = 5000 Turns/m. So the current to achieve the required field would be about 796 A.)
6. For symmetry reasons, no.
8. http://www.elect-spec.com/current_source.htm

Bob S

Your current is way above guidelines for water cooled copper conductor: 600 amps per sq. cm of copper. So only superconducting wire will do. Your superconducting coil needs quench protection. If at some current I, a small segment of your wire in the coil goes normal (resistance R_w per meter), you need to get the current out FAST so that portion of the coil does not melt. The heating rate per meter is I-squared R_w. The energy is integral [I-squared R_w exp(-t/tau)] dt, where tau is discharge time of coil current. Calculate this integral. You detect a quench beginning when the voltage drop across coil is a few millivolts above L dI/dt. If you discharge the coil too fast, you will overvoltage the coil (L dI/dt). If you discharge the coil too slow, you will exceed the MIIT (maga amp squared second) rating of your coil and melt it. Note that your amps squared is 0.6 mega amps squared, so as soon as a quench starts, you have maybe 1 second to get the current out before destroying the coil. The coil leads need to be switched from the power supply to a dump resistor R_dump, if the coil has insufficient copper to absorb the energy. The stored energy is (1/2) L I² joules. Your dump resistor time constant is L/R_dump so the ohms equals inductance approximately.

Your wire is roughly the same dia as 9 gauge coppwe wire. If it is 50% copper, 50% superconductor, it is probably 6 milli-ohms per meter. So for your wire, I-squared R is maybe 3300 watts per meter. How many joules per meter will melt your wire?. You need to check on what % of your wire is copper. Some wire is labelled "copper stabilized", meaning the wire resistance can absorb all of the stored energy if it quenches..

What is the critical current of the wire you plan to use? Will it quench at 2 tesla and 77 kelvin? (How about 1000 Gauss at 77 kelvin?) This critical current is often a major limit on high temperature superconducting wire.

OK. Answer these questions before beginning construction.

trini

There is a point that I would like to have clarified:
Does this mean that through liquid nitrogen cooling, the maximum stable current may be higher? I can run a test current through 1 m of liquid N2 cooled wire to see if there is any improvement.

Also, the purpose of my experiment is to experiment with different arrangements of the final induced B field in my ferromagnetic material, so the final strength of the field is less important than my ability to control its direction, meaning i can reduce my current if necessary.


EDIT:

I did some more reading and found out the resistivity of copper at 77k is only 0.213 n Ohm/m as compared to room temperature which is 17 n ohm/m. This means i will be generating far less power as heat( for 1 m of 1.5 mm dia copper wire at 77K, R = (0.213 x10-9)*(1/pi * 0.0075*0.0075) = 1.205 u ohm/m
as compared to room temp where it is a whopping 3 m ohm/m.)

Bob S

Using liquid nitrogen for cooling in hollow copper conductor is a bad choice, because if there is any phase change from liquid to gas, the very large volume change will plug up the cooling tube.
As soon as any part if your superconducting wire goes normal, the temperature of that section will start rising. At that point, your only choice is to turn off the current quickly, subject to the constraints in my previous post. Suppose you had 1 cm of your wire go normal. At 796 amps, the voltage drop across that portion of wire would be 796 amps x .01 x 1.205 u ohm/m = 10 microvolts. Are you prepared to measure the voltage across your coil to an accuracy of several microvolts, in the environment where L dI/dt is much higher?

Read about the recent "accident" at the CERN LHC, where failure to properly monitor a portion of their superconducting bus led to a 1-year shutdown costing $ hundreds of millions.

trini

Oh i just realised that the attachment showing my cooling plan is still pending approval. My plan is not to use hollow wire, but thin regular wire, and essentially create a refrigerating box(with a pressure release valve) to house the setup in.

fleem

Is this low duty cycle? For example, those fat copper cables that lead to your automobile starter motor go way above that 600A/cm^2 guideline. My guess is that several seconds at that current would be fine. Maybe even more than several seconds.

Bob_s i think he was saying that the copper resistivity goes surprisingly low at those cool temps--not the superconductor. (And it surprised me, too).

trini

yeah that's correct. And yes its low duty, cus i have to be as efficient as possible with space management, so the number of turns i can fit into my system is a huge factor. That is why i am more concerned about being able to cool thin wire than using a thicker one that can naturally support the current.

I will be conducting some tests on the effects of temperature on tolerable current over the coming days. Will post results.

Bob S

I was aware of that. The drop in resistance is similar to the drop in heat capacity, so the two effects roughly cancel out. Depending on the fraction of copper in the 3-mm dia wire, the heating could be as high as 5000 watts per meter at 796 amps at room temp. Using a heat capacity of 0.385 joules per gram per kelvin at room temp, and a copper mass of about 30 grams per meter (for 50% copper content), the temperature rise would be about 5000/(0.385 x 30) = 400 degrees K per sec. Not much wiggle room.

For 100% solid copper conductor, it would be about 1600/(.385 x 64) = 65 degrees K per sec.

Phrak

I put a superconducting ingition coil in my car. It was a waste of time.

trini

I just realised that i only have to run the current for something like a few ms to achieve magnetization, so heat should not play a huge factor, my biggest worry now will be the breakdown current. if need be i can try to fit more coils and drop my current to safer levels.

Also, since I only have to run the current for a very short time, and since i'm doing this at home, i think that the most practical way to cool the wire would be to place the current generator, coil, and ferromagnetic material into the bottom of an open steel drum, pour some liquid n2 in, and run the current. It'd look pretty cool too cus there will be a huge cloud of vapour caused by the current evaporating the n2.

*note, the only reason I'm bothering with the cooling is so that my wires don't melt on me, do you guys think that i could forego the cooling since my time of operation is going to be so low(about half a second)*

Phrak

Why would you use a neodymium magnetic material as a core? Is the saturation flux density higher than soft iron? What's the goal of this project?

trini

I am sorry, I perhaps should have been slightly clearer( I'm new here!). The purpose of creating the electromagnet is to permanently magnetize the core material. Thus i will be using a neodymium core.

Phrak

That's interesting. What are the ferromagnetic properties of neodymium magnet material at 77K? Then you also have to consider what happens to the binder at 77K if there is any. Or is your material pressed and sintered?