Fault current limiter heating - 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, Joule heat and the temperature rise be calculated?
Answer Typical applications Geometry
Given
Task Solution
Results
Fault current, Joule heating, and temperature rise in a resistive superconducting current limiter as the superconductor switches to a resistive state.
Engineering question
Calculate AC current and Joule losses in the superconducting cylinder, transfer the loss distribution to the coupled heat-transfer model, and calculate the resulting temperature rise.
fault current limiters, resistive protection devices, power system protection equipment
Download
Simulation problem
Problem Type
Axisymmetric multiphysics problem of AC Conduction coupled to Heat Transfer.
Superconductor conductivity in the normal state σ = 100 S/m.
Voltage drop across the superconductor ΔV = 1, frequency 50 Hz.
Coolant temperature is 77 K, convection coefficient is 10 W/K*m².
Calculate the fault current, Joule heat and the temperature rise.
To specify the voltage drop we assign zero electric potential to the left terminal and 1V electric potential to the right terminal.
It is convenient to use problems coupling to automatically pass the Joule heat distribution from the AC magnetic to the thermal problem.
The current is 0.93 A, Joule heat loss is 0.46 W, temperature rise is 1.1 K.
Video
Related examples