Reactance diagram: 90 MVA generator (25%), two motors

Base: 90 MVA and 11 kV in the generator circuit (generator rating). Base voltages in other zones follow the line-to-line turns ratios.

Base voltages

• Generator zone: $V_{b1} = 11$ kV
• Line zone (T1 is 10/132 kV): $V_{b2} = 11\times\frac{132}{10} = 145.2$ kV
• T2: three single-phase 66/10 kV units connected Y (line side) – Δ (motor side), so line-to-line ratio is $66\sqrt{3}/10 = 114.32/10$ kV. Motor zone:

$$V_{b3} = 145.2\times\frac{10}{114.32} = 12.70\ \text{kV}$$ $$Z_{b2} = \frac{145.2^2}{90} = 234.26\ \Omega$$

Per-unit reactances (on 90 MVA)

$$\begin{aligned} X_G &= 0.25\ \text{pu}\\ X_{T1} &= 0.06\times\frac{90}{100}\times\left(\frac{10}{11}\right)^2 = 0.0446\ \text{pu}\\ X_{line} &= \frac{100}{234.26} = 0.4269\ \text{pu}\\ X_{T2} &= 0.05\times\frac{90}{3\times30}\times\left(\frac{10}{12.70}\right)^2 = 0.0310\ \text{pu}\\ X_{M1} &= 0.2\times\frac{90}{50}\times\left(\frac{10}{12.70}\right)^2 = 0.2231\ \text{pu}\\ X_{M2} &= 0.2\times\frac{90}{40}\times\left(\frac{10}{12.70}\right)^2 = 0.2789\ \text{pu} \end{aligned}$$

Reactance diagram

     j0.0446   j0.4269   j0.0310
 +---/\/\/-----/\/\/-----/\/\/---+--------+
 |    T1        line       T2    |        |
j0.25                         j0.2231  j0.2789
 |                               |        |
(Eg)                           (Em1)    (Em2)
 |                               |        |
 +-------------------------------+--------+
            reference bus

Answer: On 90 MVA base: $X_G = 0.25$, $X_{T1} = 0.0446$, $X_{line} = 0.4269$, $X_{T2} = 0.0310$, $X_{M1} = 0.2231$, $X_{M2} = 0.2789$ pu; base voltages 11 kV, 145.2 kV and 12.70 kV.