# High current in a supercooled wire?

1. Apr 12, 2016

### PhiowPhi

I'm trying to understand heat-transferring process and the maximum amount of current and coolant required to apply and maintain(for a short-duration) high current in a certain conductor, before I start considering an experiment, I'd rather have my work correct in-terms of the calculations and predictions of what might happen.

The conductor is a copper plate, the dimensions: 250mm H x 5mm W x 10mm T
Weight: 111.58 grams

Can this conductor sustain 50kA in a duration of 100ms without fusing/melting?

Since it's a plate, I used this online calculator for a quicker computation of the resistance and confirmed it with my own work they are approximately the same, the resistance at room temperature is: 0.0000843 Ohms.

The idea here is to have the conductor initially cooled to -196°C using liquid nitrogen, using the same calculator above I changed the temperature to -196°C, and now R = 0.0000117 Ohms

Using Ohm's law I'm assuming that the applied voltage($V$) would be: $0.0000117\Omega \times 50kA$ = $0.585V$ The power is ≈ $30kW$

Now when working out the heat transfer and the rate of transfer I lose myself, it's like all the things I studied in Physics 101 and Chem. 101 faded away... here are some questions I couldn't figure out aside from the initial one:

1) How long would it take to cool the plate from RT(20 - 25°C) to (-196°C) to apply the 50kA?
2) How much(volume) liquid nitrogen would I need to sustain this process for 100ms?
3) How long would it take to cool the wire(or transfer all the dissipated power) for a re-run(somewhat relates to Q1)?
4) Would the induced magnetic field have any ramifications? I assume a large spike if disconnected quickly due to the induced EMF.

The reason I considered liquid nitrogen is to reduce the resistance greatly, and cool the system rapidly, however, I think water would be a good substitute? Or even air cooling? Or possibly both or all?
I'm pretty sure due to the rapid boiling of liquid nitrogen I'd require a lot of it.

2. Apr 12, 2016

### Nidum

This experiment is far too dangerous for anyone with limited technical knowledge to contemplate doing . If it had to be carried out for some real purpose then stringent safety precautions would be required .

3. Apr 12, 2016

### Jeff Rosenbury

Have you considered the skin effect?

To calculate this you will need to know your waveform and its bandwidth.

The heat of fusion of nitrogen is 5.56 kJ/mol. So to dissipate 3000 joules (30,000 kW over 0.1 sec), a bit over 4 mols of vaporized N2. This gets into the Leidenfrost area I think.

Last edited by a moderator: Apr 12, 2016
4. Apr 12, 2016

### Staff: Mentor

Closed pending moderation.

Edit: we will go ahead and leave this closed due to safety policy.

Last edited: Apr 13, 2016
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