Loading a 630 kVA transformer beyond its derated capacity
A 630 kVA, 11/0.433 kV, ONAN distribution transformer serves a small industrial feeder. The monthly maximum demand is 480 kW at a power factor of 0.82, and the ambient in the transformer bay reaches 46°C in summer. The transformer nameplate gives a 40°C maximum ambient and 65 K oil temperature rise. A new 90 kW motor is proposed, with an expected power factor of 0.85 and a demand factor of 0.8. The existing load includes two 6-pulse VFDs totaling 150 kW, and the neutral current on the LV side is 180 A although the phase currents are 620 A, 590 A and 605 A. The engineer is asked whether the transformer can take the new motor without replacement.
1. Compute the existing transformer loading and its derated capacity at 46°C. What margin remains?
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Existing apparent power = 480/0.82 = 585.4 kVA. Loading = 585.4/630 = 92.9%. Ambient is 6 K above the 40°C nameplate limit. For an ONAN oil-filled transformer, a common conservative derating is 1% of rating per K above 40°C, i.e. 6%: derated capacity is about 630 x 0.94 = 592 kVA. Margin = 592 - 585.4 = 6.6 kVA, about 1%. So the transformer is effectively at its summer limit before any addition; the 92.9% figure understates the problem because it ignores ambient. A proper calculation should use the actual temperature-rise test data and daily ambient profile, but the margin is clearly very small.
2. Include the new motor and decide whether replacement is needed.
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New motor input = 90 x 0.8 = 72 kW. At 0.85 pf, its demand = 72/0.85 = 84.7 kVA (assuming 90 kW is input; if it is shaft output, divide by motor efficiency too, making it larger). New total is about 585.4 + 84.7 = 670.1 kVA. Against the derated 592 kVA, that is 113%; against the nameplate 630 kVA it is 106%. The transformer cannot carry the addition. Replacement or a second transformer is required; a 1000 kVA unit would give margin, or an 800 kVA unit if load growth is limited and harmonic derating is checked. Also check motor starting: a 90 kW DOL motor has FLC about 160 A and starting current near 960 A, which will cause a noticeable dip on a 630 kVA transformer.
3. What do you make of the 180 A neutral current, and what actions follow?
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With phase currents of 620, 590 and 605 A, the fundamental neutral current should be small. A 180 A neutral current, about 30% of phase current, indicates significant triplen (3rd, 9th, 15th) harmonic currents from the 6-pulse VFDs and any single-phase loads. Consequences: extra neutral heating, transformer eddy-current and stray losses, and possible resonance with power-factor capacitors. Actions: measure true RMS phase and neutral currents and harmonic spectrum with a power-quality analyser; calculate a K-factor or derating per IEEE C57.110; check neutral conductor cross-section (it may need to be full-size or larger); specify a K-rated or more generously rated transformer; install harmonic filters or 12-pulse/active front-end drives; do not add plain capacitors. The transformer decision must use the harmonic-derated capacity, not just kVA.