Impedance diagram: 25 MVA generator, two motors

Base: 25 MVA and 13.8 kV in the generator circuit. T2 is read as three single-phase units of 69/13.2 kV connected Y (line side) – Δ (motor side).

Base voltages

• Generator zone: $V_{b1} = 13.8$ kV
• Line zone (T1 13.2/115 kV): $V_{b2} = 13.8\times\frac{115}{13.2} = 120.23$ kV
• T2 line-to-line ratio: $69\sqrt{3}/13.2 = 119.51/13.2$ kV. Motor zone: $V_{b3} = 120.23\times\frac{13.2}{119.51} = 13.28$ kV

$$Z_{b2} = \frac{120.23^2}{25} = 578.18\ \Omega$$

Per-unit values (25 MVA base)

$$\begin{aligned} X_{G1} &= 0.15\ \text{pu}\\ X_{T1} &= 0.11\times\frac{25}{30}\times\left(\frac{13.2}{13.8}\right)^2 = 0.0839\ \text{pu}\\ Z_{line} &= \frac{20 + j65}{578.18} = 0.0346 + j0.1124\ \text{pu}\\ X_{T2} &= 0.11\times\frac{25}{30}\times\left(\frac{13.2}{13.28}\right)^2 = 0.0906\ \text{pu}\\ X_{M1} &= 0.15\times\frac{25}{15}\times\left(\frac{13}{13.28}\right)^2 = 0.2396\ \text{pu}\\ X_{M2} &= 0.15\times\frac{25}{10}\times\left(\frac{13}{13.28}\right)^2 = 0.3594\ \text{pu} \end{aligned}$$

Impedance diagram

    j0.0839  0.0346+j0.1124  j0.0906
 +--/\/\/-------/\/\/--------/\/\/--+-------+
 |   T1          line          T2   |       |
j0.15                          j0.2396  j0.3594
 |                                  |       |
(Eg)                             (Em1)   (Em2)
 |                                  |       |
 +----------------------------------+-------+
              reference bus

Answer: $X_{G1}=0.15$, $X_{T1}=0.0839$, $Z_{line}=0.0346+j0.1124$, $X_{T2}=0.0906$, $X_{M1}=0.2396$, $X_{M2}=0.3594$ pu (25 MVA; 13.8 / 120.23 / 13.28 kV).