Reactance diagram: 90 MVA generator (15%), two motors

Choice of bases

• $S_b = 90$ MVA and $V_b = 11$ kV in the generator circuit.
• Line: through T1 (10/132 kV), $V_b = 11\times\dfrac{132}{10} = 145.2$ kV.
• T2 is a Y–Δ bank of three single-phase 66/10 kV units. Its line-voltage ratio is $\sqrt3\times66 = 114.32$ kV / 10 kV, and its three-phase rating is $3\times30 = 90$ MVA.
• Motor circuit: $V_b = 145.2\times\dfrac{10}{114.32} = 12.70$ kV.

Per-unit reactances

Use $X_{new} = X_{old}\,\dfrac{S_{b}}{S_{rated}}\left(\dfrac{V_{rated}}{V_b}\right)^2$.

$$\begin{aligned} G&: 0.15\ \text{(same base)}\\ T_1&: 0.06\times\frac{90}{100}\times\left(\frac{10}{11}\right)^2 = 0.0446\\ \text{Line}&: Z_b = \frac{145.2^2}{90} = 234.26\ \Omega,\quad X = \frac{100}{234.26} = 0.4269\\ T_2&: 0.05\times\frac{90}{90}\times\left(\frac{10}{12.70}\right)^2 = 0.0310\\ M_1&: 0.2\times\frac{90}{50}\times\left(\frac{10}{12.70}\right)^2 = 0.2231\\ M_2&: 0.2\times\frac{90}{40}\times\left(\frac{10}{12.70}\right)^2 = 0.2789 \end{aligned}$$

(T2 gives the same value on its HV side: $0.05\times(114.32/145.2)^2 = 0.0310$.)

Reactance diagram

  j0.15    j0.0446   j0.4269   j0.031
 +-/\/\-----/\/\------/\/\-----/\/\---+-----+
 |                                     |     |
(Eg)                              j0.2231 j0.2789
 |                                     |     |
 |                                   (Em1) (Em2)
 +------------- reference --------------+-----+

The transformer winding connections (Y, Δ, earthing) do not change the positive-sequence p.u. reactances. They only introduce a phase shift, which is ignored in the reactance diagram.