- 2071 Magh · 8 marks
Pilot current with tap change (10 MVA)
Short answer
LV CT ratio is N = 75 x (132/33)/√3 = 173.2 : 1, star connected. At 128 kV tap and 5 times full load, LV pilot current is 874.77/173.2 = 5.05 A, HV pilot current is √3 x 225.53/75 = 5.21 A, giving 0.158 A relay unbalance.
A 10 MVA, 132/33 kV, star-delta, three-phase, 50 Hz transformer is connected in delta on the low voltage side and in star with star point earthed on the high voltage side. If the CTs on the high voltage have a ratio of 75/1 A, determine the CT ratio on the low voltage side. What would be the current circulating through pilots for a through fault due to which a current of 5 times the full load occurs if the voltage tapping is set to 128 kV at the time of occurrence of fault?
HV is star connected, so the HV CTs are connected in delta; LV is delta connected, so the LV CTs are connected in star. The LV CT ratio is chosen so that pilot currents balance at the nominal ratio 132/33 kV. A tap change upsets this balance and a spill current flows.
Data: 10 MVA, 132/33 kV; HV CTs 75/1 (in delta); through fault = 5 × full load; tap set at 128 kV.
Step 1: LV CT ratio
For HV current I_H, the delta of HV CTs gives a pilot current √3 I_H / 75. The LV CTs (ratio N:1, star) give I_L / N, with I_L = 4 I_H (132/33).
I_L / N = √3 I_H / 75
N = 75 × (I_L/I_H) / √3 = 75 × 4 / √3 = 173.2
LV CT ratio = 173.2 : 1
Step 2: Fault currents with 128 kV tap
Full-load current on the LV (33 kV) side:
I_FL(LV) = 10 × 10⁶ / (√3 × 33 × 10³) = 174.95 A
Through-fault LV current = 5 × 174.95 = 874.77 A
HV current with tap at 128 kV = 874.77 × 33/128 = 225.53 A
Step 3: Pilot currents
From LV CTs: 874.77 / 173.2 = 5.051 A
From HV CTs: √3 × 225.53 / 75 = 5.208 A
Out-of-balance (relay) current = 5.208 − 5.051 = 0.158 A
Answer: LV CT ratio = 173.2 : 1 (star connected). For the through fault with the tap at 128 kV, the pilot currents are about 5.05 A (LV side) and 5.21 A (HV side), so an unbalanced current of 0.158 A circulates through the relay. A percentage-biased relay is needed so that this does not cause tripping. (If "5 times full load" is taken on the HV side, the corresponding values are 5.05 A, 4.90 A and 0.153 A.)
✓ Answer checked by Claude Opus 5.5 (high effort)
Related questions
- CT ratio for 33/6.6 kV star/delta (400/5) · Switchgear and Protection, chapter 7
- CT ratio for 33/6.6 kV star/delta (300:5) · Switchgear and Protection, chapter 7
- IDMT relays across 132/33 kV transformer (2000 A) · Switchgear and Protection, chapter 7
- IDMT relays across 132/33 kV transformer (2700 A) · Switchgear and Protection, chapter 7
- IDMT relays across 132/11 kV transformer (3500 A) · Switchgear and Protection, chapter 7
- Percentage differential relay: 125 MVA transformer · Switchgear and Protection, chapter 7
- 500 kVA delta/star transformer efficiency · Electrical Machine, chapter 2
- Nominal-T 100 km line, 50 MVA at 132 kV · Power System Analysis I, chapter 6
- Currents and efficiency of 132/11 kV transformer at 5 MW · Electrical Machines I, chapter 2 · asked 3×
- Charging current of 60 Hz horizontal 132 kV line · Power System Analysis I, chapter 4