Reactance diagram: 50 MVA, 13.8 kV base, three generators

All reactances are converted to a common base of 50 MVA; base voltages follow the transformer line-to-line ratios starting from 13.8 kV at G1.

Base values

• G1 zone: $V_b = 13.8$ kV
• Line zone (T1 13.8/220 kV): $V_b = 220$ kV, $Z_b = \frac{220^2}{50} = 968\ \Omega$
• G2 zone: T2 is a Y–Δ bank of 127/18 kV units, line-to-line ratio $\sqrt{3}\times127/18 = 220/18$ kV, so $V_b = 220\times\frac{18}{220} = 18$ kV
• G3 zone (T3 220/22 kV): $V_b = 22$ kV

Formula: $X_{new} = X_{old}\,\frac{S_{b,new}}{S_{old}}\left(\frac{V_{old}}{V_{b,new}}\right)^2$

Per-unit values

$$\begin{aligned} X_{G1} &= 0.2\times\frac{50}{20} = 0.5\ \text{pu}\\ X_{G2} &= 0.2\times\frac{50}{30}\times\left(\frac{18}{18}\right)^2 = 0.3333\ \text{pu}\\ X_{G3} &= 0.2\times\frac{50}{30}\times\left(\frac{20}{22}\right)^2 = 0.2755\ \text{pu}\\ X_{T1} &= 0.1\times\frac{50}{25} = 0.2\ \text{pu}\\ X_{T2} &= 0.1\times\frac{50}{3\times10} = 0.1667\ \text{pu}\\ X_{T3} &= 0.1\times\frac{50}{35} = 0.1429\ \text{pu}\\ X_{L1} &= \frac{80}{968} = 0.0826\ \text{pu}\\ X_{L2} &= \frac{100}{968} = 0.1033\ \text{pu} \end{aligned}$$

Reactance diagram

   j0.2   j0.0826        j0.1033   j0.1429
 +-/\/\/--/\/\/---+------/\/\/------/\/\/--+
 |  T1     L1     |        L2         T3   |
j0.5           j0.1667 (T2)              j0.2755
 |                |                        |
(E1)           j0.3333                   (E3)
 |                |                        |
 |              (E2)                       |
 |                |                        |
 +----------------+------------------------+
            reference (neutral) bus

Answer: $X_{G1}=0.5$, $X_{G2}=0.333$, $X_{G3}=0.276$, $X_{T1}=0.2$, $X_{T2}=0.167$, $X_{T3}=0.143$, $X_{L1}=0.0826$ and $X_{L2}=0.1033$ pu on 50 MVA, with base voltages 13.8, 220, 18 and 22 kV.