Motor Trips on Every Start After a Rewind
A 500 hp, 4160 V, 4-pole, 60 Hz TEFC induction motor drives a boiler feedwater pump through a coupling. Nameplate: 63 A FLA, 6.0× locked-rotor current, thermal limit 60 seconds at locked rotor from a cold start. The motor was just reinstalled after a rewind. The 4160 V breaker relay has a 51 element at 75 A pickup on a very inverse curve and a 50 element at 340 A with a 2.0-second definite delay. Since the rewind, the motor trips on every start at about 1.8 to 2.0 seconds. Recorded starting current: 372 A, 368 A, 341 A, decaying to 63 A at 7.5 seconds when the pump reaches full load. Shop insulation resistance and surge comparison tests passed, and the discharge valve is confirmed open at start. The electrician proposes raising the 50 pickup to 600 A.
1. Diagnose the trip. Is the motor drawing abnormal current, and would you expect the 50 element to operate?
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The current is normal; the setting is not.
- Locked-rotor current from the nameplate = 6.0 × 63 A = 378 A
- Highest recorded start current = 372 A = 5.9 × FLA, below nameplate LRC. The motor is drawing exactly what it should on a start.
- The 50 element picks up at 340 A, which is below the 378 A the motor legitimately draws while accelerating. With a 2.0 s definite delay, it times out on every start. That matches the observed 1.8 to 2.0 s trip precisely.
So the trip is a coordination defect introduced or exposed by the outage, not a rewind failure. Do not accept the trip as evidence the motor is bad - the shop tests and the current magnitudes both point the other way.
Also flag the 51 as a latent second problem: at 372 A with a 75 A pickup the multiple is about 5×. On a very inverse curve at a typical time dial, the operate time can fall inside the 7.5 s acceleration. Even after the 50 is fixed, the 51 may trip the next start. Verify it against the actual curve before you leave.
2. Evaluate the proposal to set the 50 pickup at 600 A, and state what setting change you would actually make.
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600 A would stop the nuisance trip, but it is a number without a calculation and it leaves the 51 unfixed.
Why 600 A alone is not the answer:
- Any pickup above 378 A stops the trip; 600 A is just one of infinitely many values, chosen with no margin criterion and no check of what the element then protects.
- Once the 50 is set above locked-rotor current, it can no longer see a stalled start - the motor sits at 378 A and the 50 stays picked down. Stall protection must live somewhere else.
A defensible approach:
- Set the 50 pickup above maximum locked-rotor current with margin - roughly 1.2× × 378 A ≈ 455 A, so a round 480 A - with a short definite delay (0.1–0.25 s). Its job becomes short-circuit protection for the motor and leads, which is appropriate.
- Put stall protection on the 51 or a dedicated thermal/stall element. It must ride through the 7.5 s normal acceleration at about 5 to 6× FLA and still trip before the 60 s cold locked-rotor thermal limit. Choose the curve and time dial to satisfy both ends, then confirm against the manufacturer's thermal-limit curve, including hot-start limits for a restart.
- Verify the result against the upstream feeder relay so the new motor curve still coordinates.
3. The starting currents are 372 A, 368 A, and 341 A. What does the spread tell you, and what do you do with it?
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There is a real current unbalance and it needs a cause before it needs a setting.
Using the standard definition (maximum deviation from average, divided by average):
- Average = (372 + 368 + 341)/3 = 360.3 A
- Maximum deviation = 360.3 − 341 = 19.3 A
- Unbalance = 19.3/360.3 = 5.4%
That matters because current unbalance produces rotor heating far out of proportion to its size - on the order of 1% voltage unbalance producing roughly 6 to 10% current unbalance and a large increase in temperature rise. Next step is measurement, not adjustment:
- Measure voltage unbalance at the motor terminals during the start and at load. At about 1% voltage unbalance, the 5.4% current unbalance is consistent with the supply, and the motor should be derated or the supply imbalance corrected.
- If the supply is balanced (well under 0.5%), the asymmetry is inside the motor. That is a rewind defect - winding asymmetry - and the shop should be confronted with the voltage and current data, not with an opinion.
Either way this is also a plausible contributor to the 7.5 s acceleration time, so resolve it before finalizing the relay curve.