Series capacitor to limit power angle to 30°

Assumptions: the generator internal emf is not given, so $|E| = 1.0$ pu is assumed, with infinite bus voltage $|V| = 1.0$ pu. The base is 50 MVA, 132 kV on the line side.

Data (pu on 50 MVA): $X_g = 0.3$, $X_T = 0.1$, line sections $0.3 + 0.3$, series capacitor $-jX_C$.

Step 1: Required total reactance

$$\begin{aligned} P &= \frac{EV}{X}\sin\delta \\ 1.0 &= \frac{1\times1}{X}\sin30^\circ \;\Rightarrow\; X = 0.5\ \text{pu} \end{aligned}$$

Step 2: Capacitor reactance

Uncompensated reactance: $0.3 + 0.1 + 0.3 + 0.3 = 1.0$ pu. So

$$X_C = 1.0 - 0.5 = 0.5\ \text{pu}$$

Without the capacitor, $\sin\delta = 1.0$, giving $\delta = 90^\circ$, which is right at the limit. The capacitor is therefore essential.

Step 3: Actual value

$$\begin{aligned} Z_{base} &= \frac{(132)^2}{50} = 348.48\ \Omega \\ X_C &= 0.5\times 348.48 = 174.24\ \Omega \\ C &= \frac{1}{2\pi f X_C} = \frac{1}{2\pi\times50\times174.24} = 18.27\ \mu\text{F} \end{aligned}$$

Answer: $X_C = 0.5$ pu = 174.24 Ω per phase, i.e. $C \approx 18.27\ \mu$F per phase (50% compensation of the total reactance).