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If I wind a specific transformer with superconducting wire and take away temperature from the core by means of some way, how much power can I take from transformer? I should mention max power capacity of that transformer with normal copper wire is 100 W.

In normal usage:

  • Frequency: 50 Hz
  • Input 220V
  • Output 10V 10A
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  • \$\begingroup\$ You miss the core magnetic properties (saturation etc). The supercoductivity it is related with the external magnetic field applied to the cable \$\endgroup\$
    – GR Tech
    Commented Apr 12, 2014 at 19:40

2 Answers 2

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Assuming the core is at room temperature, the core will saturate at the same magnetic field with or without superconducting wire. The only thing you're saving is the copper losses.

However, if there was no issue with critical current density \$J_C\$ of the superconducting wire, the core would not be necessary and the transformer could operate as an air core transformer at unlimited power. It would draw enormous magnetizing current.

In practice, the critical current density of the wire would be exceeded, it would go normal, and the thing would go phhhhht.

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  • \$\begingroup\$ The LHC had some experience with the phenomena described in the last sentence. It was a nice bang. \$\endgroup\$ Commented Apr 12, 2014 at 19:49
  • \$\begingroup\$ I hear quenches can be dramatic. I assume LHC is recycling helium as it's in critical short supply (geopolitical is not helping) but I wonder if they captured that cloud. \$\endgroup\$ Commented Apr 12, 2014 at 19:58
  • \$\begingroup\$ "I wonder if they captured that cloud" - ironically they weren't quick enough!! \$\endgroup\$
    – Andy aka
    Commented Apr 12, 2014 at 20:03
  • \$\begingroup\$ suppose we will not reach critical current density and my aim of question is: I think we can take more power from the core at least 50 times or 100 times more.is it true?and core roles is to reduce magnetizing current and the only limit of power is copper loss that force us to select proper core size and core size is related to copper loss ,not any thing else. \$\endgroup\$
    – user40288
    Commented Apr 14, 2014 at 3:41
  • \$\begingroup\$ The point is that the core becomes irrelevant once it saturates, and that has nothing to do with the conductivity of the wire, only the current. \$\endgroup\$ Commented Apr 14, 2014 at 3:52
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Lets look at it from a purely electrical standpoint. I won't go into the Superconductor physics, as I'm no expert in that field. The equivalent circuit of a standard transformer is as follows. The values of the secondary are transformed to the primary side. (the \$R_{\sigma}\$ can be ommited completely as it is a representation of iron losses that we don't have here)

schematic

simulate this circuit – Schematic created using CircuitLab

This figure represents a classic transformer. You suggest that we remove \$R_{prim}\$, \$R_{\sigma}\$ and \$R_{sec}\$. If we do so we are left with a simpler circuit with only inductances and the resistance of the load.

schematic

simulate this circuit

Solving the circuit shouldn't be a problem for you I guess. Find the stationary values of the currents in this regime. You have now found your initial conditions for the circiut in the first schematic.

Explanation: You calculated the currents \$ I_{prim}, I_{sec} \dots \$ these currents will flow while the material is a superconductor. When it changes to a normal conductor, the currents will still be there. These currents are therefore your initial conditions for the network that has resistances(non superconducting superconductors).

The thing you need to do is calculate all of the parameters and solve these two networks. When you do that you will see if your transformer will evaporate from the transition strain :-)

EDIT: Another thing comes to mind. The exact point in time when they loose their superconductivity is important(sin goes from -1 to 1). There is probably even a optimal sequence for getting them to the non-superconduting state based on the phase difference.

I hope that my explanation makes sense!

Cheers

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