Unbalanced loading on a rice mill's 100 kVA transformer
A rice mill in Barpeta is fed by a dedicated 100 kVA, 11 kV/415 V three-phase distribution transformer. The mill runs a 15 kW and a 7.5 kW three-phase induction motor, single-phase lighting, and a 5 kW single-phase husk blower. During peak husking a clamp meter on the LT side reads 205 A, 96 A and 84 A on the three phases, and the transformer oil temperature gauge sits in the red. The 15 kW motor's thermal overload relay trips after about 20 minutes of running. Line voltages at the motor terminals are 415 V, 407 V and 390 V. The motor nameplate reads 15 kW, 415 V, 28 A, 50 Hz, 1450 rpm. The owner wants to raise the relay setting and to add another single-phase lighting branch to the phase already carrying 96 A.
1. Assess the transformer loading. Quantify the phase currents, the negative-sequence content and the neutral current, and say whether the oil temperature can be explained.
Read the worked answer
Rated current. I = 100000 / (√3 × 415) ≈ 139 A per phase.
What the meter shows. Phase A at 205 A is about 47% above rating; B and C at 96 A and 84 A are lightly loaded. The transformer is not overloaded in total kVA so much as grossly unbalanced.
Negative sequence and neutral current. Using the standard approximation for three currents displaced by 120°:
radicand = 205² + 96² + 84² − (205×96) − (96×84) − (84×205) = 42025 + 9216 + 7056 − 19680 − 8064 − 17220 = 13333
√13333 ≈ 115.5, so:
- neutral current ≈ 115 A
- negative-sequence current ≈ 115.5/√3 ≈ 67 A, around 48% of rated current
Interpretation. A distribution transformer is normally expected to tolerate only around 10% sustained negative-sequence current; 48% is far outside that. Negative-sequence current produces a reverse rotating field, double-frequency currents in the tank and core clamps, extra eddy-current loss and local heating. Phase A's winding is also carrying 47% above rated. Taken together, the red oil temperature is fully explained - this is a loading and balance problem, not a faulty gauge.
Neutral. 115 A in the neutral also means the neutral conductor and its terminations must be full-size and sound; a reduced neutral here would itself overheat.
2. Is the thermal overload relay the problem? Explain why the 15 kW motor trips, using the voltage readings.
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No - the relay is doing its job.
Voltage unbalance. Average voltage = (415 + 407 + 390)/3 = 404 V. Maximum deviation from the average = 404 − 390 = 14 V. Percentage unbalance = 14/404 × 100 ≈ 3.5%.
Effect on the motor. At around 3.5% unbalance the standard derating for a three-phase induction motor falls to roughly 0.85–0.88 of nameplate output - call it about 13 kW instead of 15 kW. The motor must draw extra current to produce the same torque, and the negative-sequence component heats the rotor at double frequency, so the rotor heats faster than the stator current alone suggests.
Plus undervoltage. 390 V is about 6% below 415 V, so for the same shaft load the current rises further - roughly 6–10%.
Conclusion. The relay trips after 20 minutes because the motor genuinely reaches its thermal limit. It is correctly protecting the winding; the winding would otherwise run above its insulation class temperature and lose life quickly, or fail outright.
What raising the setting would do. It would let the motor run hotter for longer, with insulation ageing roughly doubling for every 10 K rise. The motor would eventually fail at a time and place of the owner's choosing, and the relay would have been defeated.
Before concluding, rule out mechanical causes: belt tension, bearing condition, blocked husk path, and a single-phased motor circuit. A blown fuse or open contactor pole in the motor circuit would look similar, but the presence of voltage on all three lines makes supply unbalance the leading explanation.
3. Give a corrective plan for the installation and state what should be measured to prove the fix worked.
Read the worked answer
Corrective plan.
- Redistribute the single-phase load. Move lighting and the 5 kW husk blower across phases so that the three phase currents come within roughly 10% of each other. Do not add the new lighting branch to the phase already at 96 A - and certainly not to phase A.
- Investigate phase A first. 205 A may include a fault as well as imbalance: check for a partially shorted single-phase appliance, a leaking capacitor, or a loose connection drawing reactive current.
- Check the neutral. Confirm the neutral conductor is full size, its terminals are tight, and there is no shared or borrowed neutral.
- Check the transformer. Oil level and colour, breather silica gel, loose HV/LV terminals, tap position against the recorded incoming voltage, and temperature gauge calibration. Arrange an oil test if the over-temperature has been repeated.
- Protect the motor properly. Fit a motor protection relay with phase-failure and unbalance protection set for the actual nameplate, and keep the thermal relay setting at 28 A (or the class appropriate to the starting duty).
- Consider load limitation until balance is achieved: no phase should exceed 139 A continuously.
Measurements to prove the fix.
- Three phase currents and the neutral current with a true RMS clamp meter at peak load; expect all phases within about 10% of each other and a neutral current of a few tens of amps at most.
- Voltage unbalance recalculated from three line voltages; target under 1–2%.
- Motor current on all three phases and winding resistance (balanced within about 2%), plus insulation resistance at 500 V.
- Transformer oil temperature after a full production shift, trending down compared with the pre-fix reading.