Rewound 30 kW Pump Motor Tripping on Overload
A dairy plant's 30 kW, 415 V, 3-phase, 4-pole squirrel-cage motor (nameplate FLC 54 A, efficiency 89%, power factor 0.86) drives a centrifugal pump. After a local rewind, it trips the 63 A thermal overload relay after 6–8 minutes at a steady valve position. The MCC ammeter reads 61 A, 48 A and 51 A on R, Y and B; a calibrated clamp meter confirms the readings. Line voltages at the MCC are 417 V, 415 V and 416 V. Insulation resistance is 500 MΩ at 500 V DC. Winding resistance measured with a micro-ohmmeter at the terminal block: R–R' 0.372 Ω, Y–Y' 0.408 Ω, B–B' 0.404 Ω. The overload relay is set to 58 A, mounted in a 42 °C pump house, and has an ambient-compensation bellows. Pump delivery and bearing temperature are normal. The rewind workshop has not supplied test sheets.
1. Diagnose the likely cause of the trip using the readings. Show calculations for full-load current, voltage unbalance, current imbalance and winding-resistance spread, and state whether the 58 A setting or the supply is responsible.
Read the worked answer
Full-load current check
[I_{FL}=\frac{30000}{\sqrt{3}\times415\times0.89\times0.86}\approx54.4\text{ A}]
The nameplate FLC is 54 A, so a 58 A setting is about 1.07 × FLC. That is not a low setting, and at 48–51 A on two phases a healthy thermal relay would not trip in 6–8 minutes. The R phase at 61 A is 1.13 × FLC and above the setting, so the trip is consistent with a genuine R-phase overcurrent, not with an underset relay.
Supply unbalance
Average line voltage = (417 + 415 + 416)/3 = 416 V. Maximum deviation = 1 V, so voltage unbalance ≈ 1/416 × 100 = 0.24%. This is negligible and cannot explain the current spread.
Current imbalance
Average current = (61 + 48 + 51)/3 = 53.3 A. Maximum deviation = 61 − 53.3 = 7.7 A, so current unbalance ≈ 14.4%. The spread between highest and lowest phase is 13 A, or about 24% of average. This is far too large for a 0.24% voltage unbalance.
Winding resistance
Average resistance = (0.372 + 0.408 + 0.404)/3 = 0.3947 Ω. The R phase is 0.0227 Ω below average, about 5.8% low, while its current is about 14% above average. That inverse relationship points to a partial short-circuit or incorrect turns/connection in the R phase, not to a high-resistance joint. A turn-to-turn fault can give normal insulation resistance to earth because the fault is within the winding, not to the frame.
Conclusion: the most probable cause is a rewind defect in the R phase. The overload setting and the supply are not the cause. Increasing the setting would only allow the defective winding to overheat further.
2. What additional tests would you perform before returning the motor to the rewinder, and what result would confirm a turn-to-turn fault?
Read the worked answer
Carry out the tests with the motor isolated and the rotor at standstill, then repeat at no load if safe:
- Surge comparison test – apply a low-energy surge to each phase pair and compare the waveforms. A shorted turn reduces the inductance of that phase, shifting the waveform and producing a visible mismatch or damping. This is the most direct shop test for turn-to-turn faults.
- DC resistance at a controlled current – repeat the micro-ohmmeter readings at the same terminal points and correct to a common temperature. A repeatable 5–6% low reading on R, with the other two phases matched, supports the diagnosis.
- No-load run – run the motor uncoupled with three ammeters and a wattmeter. A no-load current imbalance greater than about 10%, or higher-than-normal no-load loss, supports a winding defect. Do not rely on a single ammeter.
- Insulation and polarity checks – 500 V or 1000 V insulation resistance to frame, and winding polarity/connection check. These may be normal in a turn-to-turn fault, so they cannot exclude it.
- Visual inspection – look for discoloured end windings, burnt sleeving, damaged turn insulation, and verify the coil pitch, number of turns and star/delta connection against the original data.
A confirmed result would be a surge waveform mismatch on R together with a repeatable low resistance and abnormal no-load current. The motor should then be returned to the rewinder with a written defect report, not simply reset and run.
3. State the immediate operating decision and the quality-assurance action you would take before the motor is put back into service.
Read the worked answer
Immediate decision: keep the motor isolated and locked off. Do not increase the overload setting, do not bypass the relay and do not run the pump to “see if it settles”. A turn-to-turn fault can progress rapidly to an earth fault or open circuit, with smoke, fire risk and damage to the MCC. The standby pump should be used, and the process should be told that the duty pump is unavailable pending rewind repair.
Quality-assurance action:
- Send the motor back to the rewinder with the resistance, current and surge-test results and a clear defect description.
- Require a test sheet at return: DC resistance of all phases, insulation resistance, surge comparison, no-load current and vibration.
- Verify the repair against the original winding data and the nameplate; confirm the connection is correct for the 415 V supply.
- After refitting, check the overload relay setting against the actual nameplate FLC, not the old setting. For a 54 A FLC, a setting near 54–57 A is reasonable, with the ambient-compensation feature correctly set for the 42 °C location.
- Run the motor uncoupled or at low load first, record all three line currents, then load gradually and compare phase currents and winding temperature.
The key judgement is that the overload relay did its job: it flagged a real winding problem. The repair quality, not the protection setting, is the issue.