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 to find temperature rise in resistive fault current limiter?
Answer Typical applications Geometry
Given
Task Solution
Results
Superconducting cylinder is surrounded by the coolant. When the fault occurs the high fault current causes the superconductor to become normal, and the current starts generating the heat.
Engineering question
Set up an axisymmetric QuickField Steady-state Heat Transfer problem for a resistive fault current limiter and evaluate temperature rise from computed field results.
fault current limiters, resistive protection devices, power system protection equipment
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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.
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