Per unit circuit of three-generator system, 30 MVA base

Base: 30 MVA, 6.6 kV in the G1 circuit. Base voltages change by the line-to-line turns ratio of each transformer.

Base voltages

• G1 zone: 6.6 kV
• Line zone (T1 6.6/115 kV): 115 kV, $Z_b = \frac{115^2}{30} = 440.83\ \Omega$
• G2 zone (T2 115/6.6 kV): 6.6 kV
• G3 zone: T3 is a Y–Y bank of 6.9/69 kV single-phase units, so its line-to-line ratio is $\sqrt3\times69/\sqrt3\times6.9 = 119.5/11.95$ kV, i.e. 10 : 1:

$$V_{b,G3} = 115\times\frac{6.9}{69} = 11.5\ \text{kV}$$

Per-unit values (30 MVA base)

$$\begin{aligned} X_{G1} &= 0.20\times\frac{30}{25} = 0.24\ \text{pu}\\ X_{G2} &= 0.15\times\frac{30}{15} = 0.30\ \text{pu}\\ X_{G3} &= 0.15\times\frac{30}{30}\times\left(\frac{13.2}{11.5}\right)^2 = 0.1976\ \text{pu}\\ X_{T1} &= 0.10\ \text{pu}\\ X_{T2} &= 0.10\times\frac{30}{15} = 0.20\ \text{pu}\\ X_{T3} &= 0.10\times\frac{30}{3\times10}\times\left(\frac{11.95}{11.5}\right)^2 = 0.1080\ \text{pu}\\ X_{L1} &= \frac{120}{440.83} = 0.2722\ \text{pu}\\ X_{L2} &= \frac{90}{440.83} = 0.2042\ \text{pu} \end{aligned}$$

Per-unit circuit

    j0.10   j0.2722       j0.2042   j0.108
 +--/\/\/---/\/\/----+----/\/\/-----/\/\/---+
 |   T1     Line-1   |    Line-2     T3     |
j0.24              j0.20 (T2)            j0.1976
 |                   |                      |
(E1)               j0.30 (G2)             (E3)
 |                   |                      |
 |                 (E2)                     |
 +-------------------+----------------------+
            reference (neutral) bus

Answer: $X_{G1}=0.24$, $X_{G2}=0.30$, $X_{G3}=0.1976$, $X_{T1}=0.10$, $X_{T2}=0.20$, $X_{T3}=0.108$, $X_{L1}=0.2722$, $X_{L2}=0.2042$ pu.