Reactance diagram: 20 MVA generator, 15 kV/25 MVA base

Single line diagram

  G         T1 (3 x 1-ph)     line       T2 (3 x 1-ph)
 (~)--||--[ D | Y ]----------j180----[ Y | Y ]----> Load
 20 MVA     5 MVA each                  5 MVA each   15 MVA
 11 kV      11/127.02 kV                127.02/      20 kV
 X"=10%     X=15%                       12.702 kV    0.9 lag
                                        X=20%

Transformer bank ratings

• T1 (Δ–Y): $3\times5 = 15$ MVA. Line voltages: 11 kV / $\sqrt3\times127.02 = 220$ kV.
• T2 (Y–Y): 15 MVA, 220 kV / $\sqrt3\times12.702 = 22$ kV.

Base values

Assumption: the 15 kV base is taken in the generator circuit, with $S_b = 25$ MVA.

Per-unit reactances

$$\begin{aligned} X_G &= 0.10\times\frac{25}{20}\left(\frac{11}{15}\right)^2 = 0.0672\\ X_{T1} &= 0.15\times\frac{25}{15}\left(\frac{11}{15}\right)^2 = 0.1344\\ X_{line} &= \frac{180}{3600} = 0.05\\ X_{T2} &= 0.20\times\frac{25}{15}\left(\frac{220}{300}\right)^2 = 0.1793 \end{aligned}$$

Load

15 MVA at 20 kV and 0.9 lagging: $S = 15/25 = 0.6$ p.u., $V = 20/30 = 0.6667$ p.u.

$$S = 0.6\angle25.84^\circ = 0.54 + j0.2615\ \text{p.u.}$$

As a series impedance:

$$Z_L = \frac{|V|^2}{S^*} = \frac{0.6667^2}{0.6}\angle25.84^\circ = 0.7407\angle25.84^\circ = 0.6667 + j0.3229\ \text{p.u.}$$

Reactance diagram

  j0.0672   j0.1344    j0.05    j0.1793
 +-/\/\------/\/\------/\/\------/\/\----+
 |                                       |
(Eg)                              0.6667 + j0.3229
 |                                     (load)
 +----------------- reference -----------+