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Linear induction motor - QuickField simulation example

This example is prepared by the Metropolian Hyperloop Design Team participating in the Hyperloop Global competition in May 2024.
To be able to include the motor into the circuit simulator it is necessary to calculate electric motor equivalent circuit parameters. A no-load mode test and the fixed rotor test are carried out to calculate these parameters.

Engineering question

How to find linear induction motor thrust force versus speed?

Answer
Set up a plane-parallel QuickField AC Magnetics problem for a linear induction motor and evaluate thrust force versus speed from computed field results.

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Typical applications
linear induction motors, transport propulsion motors, electromagnetic drive systems

Linear induction motor

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Simulation problem

Problem Type
Plane-parallel problem of AC magnetics.

Geometry
Z-length is 70 mm. Slot dimensions are 20*85 mm. Number of slots is 25.
Linear induction motor equivalent circuit Simulate transients in the winding of a linear induction motor (LIM) when feeding with pulse-width modulated voltage. 722 mm 90 mm 8 mm 1 2 3 4 5 6 7 8 9 10 15 20 25 Core Rail A A A A A A A A A' A' A' A' A' A' A' A' B B B B B B B B B' B' B' B' B' B' B' B' C C C C C C C C C' C' C' C' C' C' C' C'

Given
Rail is made of Aluminum 6061-T6 with electrical conductivity of σ = 35 MS/m.
Number of conductors in the slot is 24, current is 20 A (r.m.s). Frequency f = 50 Hz.
Magnetic permeability of the stator is 1000.

Task
Calculate the equivalent electric circuit parameters.

Solution
The resistance of the stator windings can be calculated from the winding geometric dimensions as Rs = resistivity * Length / Conductor cross-section. We do not calculate it here.
In this example we only calculate the resistance caused by the losses in the rotor and the inductance.

We are going to reproduce 2 experiments that you can do with the real motor: a synchronous motion speed regime and a locked rotor regime.
First we simulate the synchronous speed regime (slip value s=0). In this regime there are no eddy currents in the rotor as the rotor speed is equal to the speed of the magnetic field produced by the stator winding. We model this by specifying zero conductivity of the rotor.
Then we simulate a locked rotor regime.
Req Leq

Unlike real experiments we have all field parameters readily available.
The inductance is the ratio of the magnetic field energy doubled to the electric current value squared: L = 2*W / I²
The resistance is the ratio of the active power to the electric current value squared: R = P / I²

Results
No load regime current I = 20 A (r.m.s). Magnetic field energy is 12.0 J. Inductance Leq = 2*12.0 / 20² = 0.06 H, reactance X = 2*π*f*Leq = 18.85 Ohm.
Core loss Pcore = 0.117 W. Resistance Req = 0.117/20² = 0.0003 Ohm.
Hyperloop linear induction motor

Fixed rotor regime current Is = 20 A (r.m.s).
Joule heat in the rotor is 2293 W, equivalent resistance Req = 2557/20² = 6.39 Ohm.
Magnetic field energy is 15.0 J. Inductance Leq = 2*15.0 / 20² = 0.075 H, reactance X = 23.6 Ohm.
Hyperloop linear induction motor

Reference: Induction motor, Wikipedia.

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