Transformer Sc Test

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The classic short-circuit test: shorted secondary, reduced primary voltage, and a current reading that hands back the leakage impedance on the nameplate.

A 100 MVA 230/115 kV transformer with all three LV terminals shorted, fed from an ideal source at 23 kV — one tenth of rated, chosen so that rated current flows. The HV line current is the measurement.

With the LV winding shorted, the only impedance left between the source and the short is the leakage — the magnetizing branch is bypassed by a path a thousand times stiffer than itself. So the applied voltage needed to push rated current is directly the per-unit impedance, and this circuit applies exactly 0.1 pu (23 kV on a 230 kV winding) to a transformer whose nameplate says $x=0.10$ pu. The prediction is rated current, and after the switching transient dies the run settles on 250.8 A RMS against a rated HV current of $100\text{MVA}/(\sqrt{3}\times 230\text{kV})=251.0$ A. Read the test backwards, as an engineer with only the waveform would: $|Z|=0.1/0.999=0.1001$ pu, which is $\sqrt{r^2+x^2}$ for $r=0.005$ and $x=0.100$. The LV current settles at 501.3 A, exactly 1.999 times the HV current — the 230/115 turns ratio, recovered from two ammeters.

The magnetizing current is the reason the test works, and the run reports it: Imag_a settles at 0.217 A RMS, under a tenth of a percent of rated. Almost the entire applied voltage is dropped across the leakage reactance, so the core sees essentially nothing and the exciting branch draws essentially nothing. That is the assumption every short-circuit test rests on, made visible.

The first cycles are not the answer, though, and they are worth a look. The source closes at a voltage zero, and current through an inductive branch cannot jump, so the waveform starts with a DC offset that lifts the first peak to 658.4 A at 8.1 ms — 1.855 times the settled peak of 354.9 A. That number is not an accident: for a loop closed at voltage zero the first peak reaches $1+e^{-\pi R/X}$ times the symmetrical peak, and $1+e^{-\pi (0.005)/0.1}=1.855$. The offset then bleeds away with the loop's own time constant. It halves in 36.8 ms, i.e. $\tau =36.8/\ln 2=53$ ms, which is $X/(\omega R)=0.1/(377\times 0.005)$. So a single run yields $r$ and $x$ separately: the settled magnitude gives $|Z|$, and the decay rate of the offset gives $X/R=20$.

To run the other half of the classic pair, convert this into an open-circuit test: delete the LV short, terminate LV through a large resistance instead (1 MΩ per phase is ample), and raise the source to the full 230 kV. The magnetizing current then swings 3.55 A peak, i.e. 2.51 A RMS, which is 1.00% of rated — the i_mag_hv_pct = 1 on the nameplate, measured. It also arrives riding a 3.53 A DC offset that barely moves over the whole run (3.545 A at 0.1 s, 3.532 A at 0.5 s), because with the secondary open the flux loop is 140 H over 1.3 Ω and its time constant is near two minutes. That stubborn offset is the same mechanism that makes transformer inrush decay so slowly in the energization sample; here it is harmless, but read the amplitude, not the peak.

Other things worth trying: set x_hv_lv_pu = 0.06 and watch the settled current rise to 1.66 pu, since a fixed applied voltage across a smaller impedance is more current; or set the source phase = 90 so the circuit closes near a voltage crest, where the steady-state current is already passing through zero and no offset is needed — the first peak falls from 658 A to 366 A, within a few percent of the settled 355 A, and the waveform is symmetric from the start. Saturation is off here deliberately — at 0.1 pu applied voltage the core flux never approaches the knee, so it would change nothing, and leaving it off keeps the impedance the reader recovers unambiguously linear.