Induction Machine Motoring

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A 1 MVA, 480 V induction machine driving a constant 0.4 pu load torque — the motoring operating point.

An ideal 480 V source feeds a single-cage induction machine whose shaft drives a constant 0.4 pu load. The two torques fight over the rotor: the electrical torque $T_e$ the air gap develops against the braking $T_{\mathrm{load}}$. The rotor settles wherever slip $s=1-{\omega }_r$ makes them equal, which for these rotor parameters is about 1.1 %.

An ideal 480 V source is connected straight across a single-cage machine, and the shaft carries a constant braking torque. There is no Tmech wire, so the machine falls back to its tm0 parameter for shaft torque; NumaSim uses the generator convention throughout, so a load is a negative drive torque and tm0 = -0.4 pu is what makes this a motor. init_slip = 0 releases the rotor at synchronous speed with only its magnetizing flux seeded. The plots show rotor speed w above developed torque Te, plus the three phase currents; slip, Pmot and Qmot are published too if you want to add them.

The first three cycles are the price of switching a machine on directly. Full voltage lands on windings whose working flux has yet to build, and until it does the machine draws whatever the leakage reactances allow — Xls + Xlr = 0.2 pu, so about five times rated current before you count the DC offset. Phase current peaks at 16.5 kA around 6.8 ms, roughly fourteen times the 1.2 kA rated current, and Te spikes to +9.6 pu. That transient torque pulls the rotor down to w = 0.976 by 65 ms before the machine starts developing useful torque. From there it is a clean pull-in: by about 1 s the rotor holds w = 0.9890, slip = 0.0110, and Te = -0.400 pu exactly balancing the load, drawing Pmot = -0.409 MW and Qmot = -0.349 MVAr (negative in generator convention because a motor absorbs both). Steady-state phase current peaks near 900 A, comfortably inside the rating.

The slip is the interesting number. An induction machine develops torque only by slipping, and near the operating point torque is close to proportional to slip divided by rotor resistance — with Rr1 = 0.025 pu, 0.4 pu of torque needs roughly 1 % slip, which is what the run gives. Double Rr1 to 0.05 pu and the slip goes to 0.0220 for the same torque — almost exactly double, which is the trade a deep-bar rotor makes to buy starting torque. Make tm0 more negative and the rotor sinks further below synchronous; raise H and the pull-in takes proportionally longer. The run is 1.5 s at a 50 µs step, roughly five times the settling time. Starting this machine from a genuine standstill is not yet a supported operating region for the model, so for the starting story use this sample's inrush and then step the load torque.