p.u. impedance diagram: G, T1, line, T2, M

Choose the generator rating as base: 20 MVA, 6.6 kV in the generator circuit. (The 6.6 kV generator feeds the 11 kV winding of T1, so the transformer and motor values need voltage correction.)

Base voltages

• Generator zone: $V_{b1} = 6.6$ kV
• Line zone (T1 11/132 kV): $V_{b2} = 6.6\times\frac{132}{11} = 79.2$ kV
• Motor zone (T2 132/11 kV): $V_{b3} = 79.2\times\frac{11}{132} = 6.6$ kV

$$Z_{b2} = \frac{79.2^2}{20} = 313.63\ \Omega$$

Per-unit values

$$\begin{aligned} X_G &= 0.15\ \text{pu}\\ X_{T1} &= 0.1\times\frac{20}{25}\times\left(\frac{11}{6.6}\right)^2 = 0.2222\ \text{pu}\\ Z_{line} &= \frac{10 + j60}{313.63} = 0.0319 + j0.1913\ \text{pu}\\ X_{T2} &= 0.1\times\frac{20}{20}\times\left(\frac{132}{79.2}\right)^2 = 0.2778\ \text{pu}\\ X_M &= 0.15\times\frac{20}{15}\times\left(\frac{11}{6.6}\right)^2 = 0.5556\ \text{pu} \end{aligned}$$

Impedance diagram

     j0.2222  0.0319+j0.1913  j0.2778
 +---/\/\/-------/\/\/--------/\/\/---+
 |    T1          line          T2    |
j0.15                              j0.5556
 |                                    |
(Eg)                                 (Em)
 |                                    |
 +------------------------------------+
            reference bus

Answer: On 20 MVA (base 6.6 / 79.2 / 6.6 kV): $X_G = 0.15$, $X_{T1} = 0.2222$, $Z_{line} = 0.0319 + j0.1913$, $X_{T2} = 0.2778$, $X_M = 0.5556$ pu.