LV voltage of Δ–Y bank of 100 kVA transformers

In a Δ–Y bank each HV winding gets the full 2400 V line voltage, and each LV winding gives 120 V phase, so LV line voltage is √3 × phase voltage. The SC test was done on the HV side (LV shorted), so it gives impedance referred to HV. Turns ratio per transformer $a = 2400/120 = 20$.

Equivalent impedance (HV side, per transformer)

$$\begin{aligned} Z_{eq} &= \frac{53.4}{41.7} = 1.2806\ \Omega \\ R_{eq} &= \frac{850}{41.7^2} = 0.4888\ \Omega \\ X_{eq} &= \sqrt{1.2806^2 - 0.4888^2} = 1.1836\ \Omega \end{aligned}$$

Rated HV winding current $I = 100000/2400 = 41.67$ A.

LV voltage at rated current, 0.8 pf lag

Using $V_1 = V_2' + I(R_{eq}\cos\phi + X_{eq}\sin\phi)$ in phasor form with $V_1 = 2400$ V:

$$\begin{aligned} I(R\cos\phi + X\sin\phi) &= 41.67(0.4888\times0.8 + 1.1836\times0.6) = 45.88\ \text{V} \\ I(X\cos\phi - R\sin\phi) &= 41.67(1.1836\times0.8 - 0.4888\times0.6) = 27.23\ \text{V} \\ V_2' &= \sqrt{2400^2 - 27.23^2} - 45.88 = 2353.96\ \text{V} \end{aligned}$$

Referred to LV and changed to line value:

$$\begin{aligned} V_{2,ph} &= \frac{2353.96}{20} = 117.70\ \text{V} \\ V_{2,LL} &= \sqrt3\times117.70 = 203.86\ \text{V} \end{aligned}$$

(Regulation = (2400 − 2353.96)/2400 = 1.92%.)

Currents at rated load

Answer: LV line-to-line voltage ≈ 203.9 V; HV winding current 41.67 A, LV winding current 833.3 A.