11 kV Express Feeder: Tap Change or Tail-End Capacitor?
A 66/11 kV, 5 MVA transformer at a rural substation feeds an 8 km radial 11 kV express feeder serving an irrigation cluster of about 400 pump sets. For planning work the feeder is modelled as a single lumped load at the far end, with per-phase R = 0.27 Ω/km and X = 0.33 Ω/km. At the evening peak the tail-end load is 2.5 MW at 0.85 power factor lagging; the busbar voltage is held at 11.2 kV and the recorded tail-end voltage is 10.2 kV. Farmers report pump starters stalling and contactors chattering in the evening. A 3 × 2.5 MVAr capacitor bank at the substation has been out of service since a capacitor fuse failed last year. The Assistant Engineer proposes shifting the transformer's off-circuit tap from 0% to −5%; the Junior Engineer proposes re-commissioning 1.5 MVAr at the tail end of the feeder instead. The feeder is radial, with no sectionaliser, no interconnection and no possibility of load transfer this season.
1. Verify from the data whether the feeder drop explains the recorded 10.2 kV, and quantify the tail-end voltage and feeder loss if 1.5 MVAr is connected at the tail end.
Read the worked answer
Step 1 – load current. S = 2.5/0.85 = 2.94 MVA, so I = 2.94×10^6/(√3 × 11×10^3) = 154.4 A; cos φ = 0.85, sin φ = 0.527.
Step 2 – feeder constants. R = 0.27 × 8 = 2.16 Ω/ph; X = 0.33 × 8 = 2.64 Ω/ph.
Step 3 – drop. ΔV_ph ≈ I(R cos φ + X sin φ) = 154.4(2.16×0.85 + 2.64×0.527) = 154.4(1.836 + 1.391) = 498 V. Line-to-line drop = √3 × 498 ≈ 0.86 kV, giving a predicted tail end of 11.2 − 0.86 = 10.34 kV against a recorded 10.2 kV. The model is within about 1.5%, so the complaint is explained by feeder impedance and reactive flow - not by a transformer or metering defect.
Step 4 – loss check. 3I²R = 3 × 154.4² × 2.16 ≈ 154 kW. Warn that the lumped model overstates this; with a distributed load the true figure is roughly one-third, ≈ 50 kW.
Step 5 – with 1.5 MVAr at the tail end. Q falls from about 1.55 MVAr to 0.05 MVAr, so I ≈ 131 A at cos φ ≈ 1.0 and ΔV_ph = 131(2.16 + 2.64×0.02) ≈ 290 V, i.e. 0.50 kV line-to-line. Tail end becomes 10.70 kV, an improvement of about 0.36 kV (3.3%), matching the capacitor's own rise √3 × 78.7 × 2.64 = 0.36 kV. I²R loss falls by 3(154.4² − 131²) × 2.16 ≈ 43 kW on the lumped basis.
2. Compare the two proposals and recommend one, with reasons. Address the light-load condition explicitly.
Read the worked answer
Tap change to −5%. Raising secondary volts by 1/0.95 = 5.26% gives 11.79 kV at the busbar and about 10.93 kV at the tail end at peak. Three objections: (i) the busbar is already at 11.2 kV, so 11.79 kV leaves almost no margin to the 12 kV equipment limit; (ii) at light load (night, off-season) the tail end rises to roughly 11.6 kV - about 5.5% overvoltage - risking pump starter insulation and capacitor/PT failures; (iii) an off-circuit tap needs a shutdown, is only seasonally correct, and does nothing for loss or reactive flow.
Tail-end capacitor. It attacks the cause: reactive current and the IX drop. It improves voltage by 0.36 kV, cuts feeder losses by roughly 40 kW (lumped) and releases about 0.44 MVA of the 5 MVA transformer (loading falls from 59% to 50%). Its weakness is that the rise is load-independent, so at light load the tail end can reach about 11.2 + 0.36 = 11.56 kV.
Recommendation. Reject the tap change (keep 0%), install the capacitance at the tail end but in two switched steps of 0.75 MVAr with time/voltage control, and log tail-end voltage for a month before finalising the setting. If the recording shows the light-load rise is unmanageable, shift one step back to the substation bank instead.
3. The capacitor fuses failed last year. State the checks and protection features you would insist on before re-energising a capacitor bank, whether at the tail end or the substation.
Read the worked answer
Fuse failure is a symptom, so first establish the cause; re-fusing and re-energising without diagnosis usually repeats the failure.
Diagnostic tests
- Insulation resistance of each unit and of the rack insulators (2.5 kV megger); look for a value far below the nameplate/commissioning record.
- Capacitance of every unit against nameplate; a rise of more than about 5–10% indicates element failure, and a fall indicates a shorted element group.
- Discharge resistor continuity and discharge time check.
- Visual inspection for bulging, leakage, discoloured bushings, loose connections and corrosion.
- Verify the replaced fuse is correct: for 2.5 MVAr at 11 kV the rated current is 131 A, so an HRC fuse in the 1.4–1.8 × range (about 190–235 A) is right. An oversized fuse protects nothing and a grossly undersized one nuisance-fails.
System checks
- Harmonic background: irrigation VFDs and rectifier loads excite 5th/7th resonance. Measure THD-v; if it is high, add detuned reactors rather than plain capacitors.
- Check the switching device for restrike capability (vacuum breaker conditioned for capacitive duty) and that the bank is never re-energised before full discharge.
- Protection: unbalance/overcurrent relay, overvoltage relay, and step switching so the bank is not left connected at light load.