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Vector group compensation for the 87T
Choose the transformer vector group, its rating, its voltages and the CTs on each side. You get the TAPs, the compensation matrix of each winding, whether zero sequence must be removed, and a check that through load leaves no differential current.
IEC 60076-1vector group, clock notationIEEE C37.91phase and zero-sequence compensation
Results
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Charts
- HV (A at 12 o'clock)
- LV
- Compensated HV
- Compensated LV, inverted
Calculation and details
Compensation matrices
Step-by-step calculation
What it means for the settings
What it is for
The 87T differential compares what goes into the transformer with what comes out. Before subtracting, the relay must correct three things that are not faults: the magnitude difference between HV and LV currents, the phase shift of the vector group and the zero sequence that exists on only one side. This tool reproduces those three corrections so you can check relay settings, prepare a stability test or find out why an 87T tripped on an external fault.
How it is calculated
Rated current and TAP
In = S / (√3 · U) · TAP = S[MVA] · 1000 / (√3 · U[kV] · CTR) · CCTR is the CT ratio (primary / secondary) and C = 1 for wye-connected CTs or √3 for delta-connected CTs. This is the form used by SEL manuals, among others. The TAP is the secondary current the relay sees at rated power; it is also shown in per unit of the CT rated current. The factor In,CT / In,obj is the same idea expressed the way other manufacturers ask for it.
IEC 60076-1clock notationIEEE C37.91numerical compensation
Compensation matrix
Mij = (2/3) · cos(m · 30° + (j − i) · 120°), with i, j = 1, 2, 3The group number is the clock position of the LV voltage when the HV voltage is at 12 o'clock: LV lags by h · 30°. Matrix m rotates the positive sequence by m · 30° counterclockwise and removes zero sequence. For m = 1 it is 1/√3 · [1 −1 0; 0 1 −1; −1 0 1]; for m = 11, 1/√3 · [1 0 −1; −1 1 0; 0 −1 1]; for m = 12, 1/3 · [2 −1 −1; −1 2 −1; −1 −1 2], which does not rotate but removes I0. Matrix 0 is the identity and lets zero sequence through. IEEE C37.91 describes this compensation; the actual matrices are those of the manufacturers' manuals (SEL calls them CTC(m)).
Compensated side
If LV is compensated, its matrix is m = h and HV stays at 0; if HV is compensated, its matrix is m = 12 − h and LV stays at 0. A side that does not rotate but has a path to ground gets m = 12. In automatic mode the side with a path to ground is compensated, or the wye side if neither has one, which is the practice SEL recommends, because any matrix with m ≠ 0 also removes zero sequence.
Check
The tool simulates rated through load with positive and negative sequence, compensates it with the chosen matrices and checks that the HV plus LV sum is zero in all three phases. The TAP indicators turn yellow if the TAP ratio exceeds 7.5 or if a TAP falls outside 0.1 to 6.2 times the CT rated current: these are Tiza reference limits for an SEL-387 type relay; confirm the limits of your relay in its manual.
Limits of the calculation
- It assumes ABC phase rotation, CTs with polarity toward the transformer on both sides and a two-winding transformer. With ACB rotation the direction of the shifts is reversed.
- The TAP is calculated at rated voltage; with a tap changer, the error at the extreme tap is covered by the relay slope.
- With delta-connected CTs the wiring does the compensation and the relay must stay at 0: check that the delta (DAB or DAC) matches the vector group, because the wrong delta leaves a 60° error.
- The rotation convention is SEL's. GE and Siemens ask for the vector group directly, and some relays number clockwise: check the numbering of your relay.
- It does not replace the relay manual or the settings study, which sets the slopes, minimum pickup and harmonic blocking.
Frequently asked questions
Which compensation matrix does the 87T use for a Dyn1 transformer?
With the usual practice of compensating the grounded-wye side: the delta side has no compensation (identity matrix) and the wye LV side gets the matrix that advances 30° and removes zero sequence, which SEL calls CTC(1): 1/√3 · [1 −1 0; 0 1 −1; −1 0 1]. Other manufacturers ask for the vector group directly and build the equivalent matrix internally.
Why must zero sequence be removed in a transformer differential?
Because on the side with a grounded neutral, an external ground fault drives zero-sequence current that does not appear on the other side of the transformer (for example, it circulates inside the delta). The 87T would see that current as differential and trip for a fault outside its zone. That is why I0 is subtracted on the side with a path to ground.
How is the TAP of a transformer differential calculated?
It is the secondary current that reaches the relay at rated power: TAP = MVA · 1000 / (√3 · kV · CTR), multiplied by √3 if the CTs on that side are delta-connected. With the two TAPs the relay brings the currents of both sides to the same base; many relays calculate it themselves from the rating, the voltages and the CT ratios.
We set up your transformer differential
We review the vector group, the CTs and the 87T settings and load the configuration into the relay remotely. If you need it, we also do the stability test and commissioning on site, in Colombia or elsewhere on request.