Per unit values: 100 MVA generator, two motors

Common base: 100 MVA, 6 kV in the generator circuit. The reactance of T2 is not printed; it is taken equal to T1, X = 4 % on 120 MVA, 6.6/132 kV (an identical transformer).

Base voltages

• Generator zone: $V_{b1} = 6$ kV
• Line zone (T1 6.6/132 kV): $V_{b2} = 6\times\frac{132}{6.6} = 120$ kV, $Z_{b2} = \frac{120^2}{100} = 144\ \Omega$
• Motor zone (T2 132/6.6 kV): $V_{b3} = 120\times\frac{6.6}{132} = 6$ kV

Per-unit values (100 MVA)

$$\begin{aligned} X_G &= 0.06\times\frac{100}{100} = 0.06\ \text{pu}\\ X_{T1} &= 0.04\times\frac{100}{120}\times\left(\frac{6.6}{6}\right)^2 = 0.0403\ \text{pu}\\ Z_{line} &= \frac{10 + j50}{144} = 0.0694 + j0.3472\ \text{pu}\\ X_{T2} &= 0.04\times\frac{100}{120}\times\left(\frac{6.6}{6}\right)^2 = 0.0403\ \text{pu}\\ X_{M1} &= 0.06\times\frac{100}{50}\times\left(\frac{6}{6}\right)^2 = 0.12\ \text{pu}\\ X_{M2} &= 0.06\times\frac{100}{40} = 0.15\ \text{pu} \end{aligned}$$

Impedance diagram

    j0.0403  0.0694+j0.3472  j0.0403
 +--/\/\/-------/\/\/--------/\/\/--+-------+
 |   T1          line          T2   |       |
j0.06                             j0.12   j0.15
 |                                  |       |
(Eg)                             (Em1)   (Em2)
 |                                  |       |
 +----------------------------------+-------+
              reference bus

Answer: $X_G = 0.06$, $X_{T1} = X_{T2} = 0.0403$, $Z_{line} = 0.0694 + j0.3472$, $X_{M1} = 0.12$, $X_{M2} = 0.15$ pu on 100 MVA (6 / 120 / 6 kV).