Superconducting fault current limiter - QuickField simulation example
This simulation example is prepared by Professor James R. Claycomb as a part of the webinar Fault Current Limiter Simulations using QuickField.
How can the fault current be calculated?
Answer Typical applications Geometry
Given
Task Solution
Results
Fault current in an inductive superconducting current limiter, showing current limitation when the superconducting element leaves its shielding state.
Engineering question
Solve the AC-magnetic limiter for the superconducting and normal states of the screen and compare the resulting winding fault current.
superconducting fault current limiters, inductive current limiters, magnetic shielding cores
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Simulation problem
Problem Type
Plane-parallel problem of AC magnetics.
Axial length is 30 mm
Mu-metal permeability is 10000.
Winding number of turns is 900, conductor cross-section area is 1 mm², copper conductivity σ = 56 MS/m.
Voltage source V = 220 V (r.m.s.), frequency 50 Hz.
Superconductor permeability in superconducting state is 10-7, in normal state 1.
Calculate the fault current.
The model includes left and right sides of the coil both that have the length of 30 mm each, but does not include front and back sides of the coil, that have the length of 55 mm each. To take into account the resistance of missing parts, electrical conductivity of the coil is attenuated: σ * (30 + 30) / (30 + 30 + 55 + 55) = 19.76 MS/m.
If a superconductor would remain in superconducting state, the fault current will be 51 A. But strong magnetic field causes the superconductor to go to a normal state. Then the fault current is limited to a value of 2.1 A.
Left picture shows the case, when the superconductor is in the superconducting state and fully screens off the core. Right picture shows a case when the superconductor is in a normal state.
Video
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