Transformer High-Set Trips and LV Earth-Fault Detection

Transformer full-load currents. HV: I_FL = 1,000,000 / (√3 × 11,000) = 52.5 A LV: I_FL = 1,000,000 / (√3 × 433) = 1,333 A

The setting in primary amperes. High-set = 6 × 52.5 = 315 A, which is 15.75 A secondary on the 100/5 CT - well within the relay's range, so the relay is operating exactly as set.

Inrush. A distribution transformer of this size draws a magnetising inrush typically in the range 8–12 times rated current for around 0.1 s, with a decaying DC component and a strong second-harmonic content. That is: 8–12 × 52.5 = 420–630 A, which is above the 315 A setting, so the relay trips in its instantaneous time of about 0.05 s. The magnetitude depends on the point on the voltage wave at which the breaker closes, so the trips can appear intermittent.

Consistency with the evidence. Trips only on energisation, no fault found, transformer tests passed, no trips in service - all consistent with inrush, not with a winding fault. A winding fault would show on insulation resistance, ratio or excitation current, and would generally not clear itself when the transformer is re-energised.

Could 6 times ever be correct? No. The setting has to be above the inrush, and 6 times rated current is below any plausible inrush for this transformer. It is also below the reflected LV fault current (next task), so even if inrush were unusually small, the setting would destroy discrimination with the LV air circuit breaker.

First action: record the actual inrush from the relay's event log or a transient recorder rather than working from a rule of thumb, then set the high-set from that measurement with margin.

LV three-phase fault level. Convert the source to the transformer base: transformer base 1,000 kVA; source fault level 200 MVA. Source impedance on 1,000 kVA base = 1,000 / 200,000 = 0.5% Total impedance to the LV busbar = 6% + 0.5% = 6.5% I_f,LV = 1,333 / 0.065 ≈ 20,500 A = 20.5 kA This is below the 50 kA switchboard rating, so the board itself is adequate.

Reflected to the 11 kV side. I_f,HV = 20,500 × 433 / 11,000 ≈ 807 A ≈ 15.4 × HV full-load current.

Why the high-set must respect this number. At 6 × (315 A), the relay would operate for any LV busbar fault (807 A) in about 0.05 s, before the LV air circuit breaker has cleared. A fault on the LV board would therefore trip the whole transformer instead of only the affected LV section - loss of discrimination and an unnecessary outage for every LV fault.

Recommended setting.

• High-set above both inrush and the maximum reflected LV fault. With measured inrush up to about 630 A and a reflected LV fault of 807 A, set the high-set at roughly 1.2 × 807 ≈ 970 A, i.e. about 18.5 × HV full-load current. Confirm this is above the measured inrush; if the measured inrush exceeds it, do not raise the high-set further - use a definite-time element instead.
• Definite-time overcurrent element at about 4 × FLC (210 A) with a 0.3–0.4 s delay, so that it grades above the LV ACB's instantaneous clearing time of about 0.05 s, with the usual 0.3 s grading margin.
• Verify the earth-fault element separately (next task).

Grading principle. Protection closest to the fault operates first. The downstream device (LV ACB) clears a LV fault in about 0.05 s; the upstream HV relay must therefore be slower than that by a defined margin across the whole range of fault currents where both can see the fault. A setting that is faster than the downstream device at any current in the overlapping range is a discrimination failure, regardless of how good the curve looks elsewhere.

Also check: the LV ACB's own short-time and instantaneous settings against the LV feeder devices, and the CT's composite error at the maximum through-fault current.

The physics. For an LV phase-to-earth fault, current flows from the faulted LV winding through the fault, through the general mass of earth and back to the transformer LV neutral star point. On the HV side, that current is a zero-sequence component circulating entirely inside the delta winding; it does not appear in the 11 kV line currents. The HV relay therefore sees only a small positive-sequence unbalance, typically a fraction of the LV current reflected through the ratio, and almost certainly below its 1.1 × FLC pickup.

Conclusion: the 11 kV overcurrent relay cannot detect an LV earth fault, no matter how it is set. This is a property of the Dyn11 vector group, not a setting error.

Consequence if nothing is added: an LV earth fault at the transformer terminals or on the LV board is cleared only by whatever LV protection sees it. If the LV ACB has no earth-fault element, a high-resistance arcing fault can persist indefinitely, which is a fire and touch-voltage risk.

Protection to add:

1. A dedicated earth-fault element on the LV ACB, fed either from a CT in the transformer LV neutral or from a core-balance CT around the outgoing conductors. Typical setting 10–20% of LV full-load current (133–267 A) with a delay of about 0.4 s, graded above the LV feeder earth-fault protections.
2. For the transformer LV winding itself, a restricted earth-fault scheme using the neutral CT and the phase CTs, which gives fast, sensitive detection of winding earth faults and is unaffected by downstream faults.

Verification: check the earth-fault loop impedance at the far end of the LV feeders against the disconnection times required by the installation standard (0.4 s for 230 V final circuits, up to 5 s for distribution circuits), and confirm the neutral earthing conductor and the main earthing terminal are correctly sized and continuous. If the LV system were TT rather than TN-S, an RCD would be the required solution instead.