Power angle and series compensation

For a lossless line, $P = \dfrac{V_SV_R}{X}\sin\delta$. Series capacitors reduce the net reactance $X$. For the same power, this reduces $\delta$ and raises the stability margin.

Data: $X_L = 0.8$ pu, $V_S = V_R = 1$ pu, $P = 1$ pu.

(i) Power angle without compensation

$$\begin{aligned} \sin\delta &= \frac{PX_L}{V_SV_R} = \frac{1\times0.8}{1\times1} = 0.8 \\ \delta &= \sin^{-1}0.8 = 53.13^\circ \end{aligned}$$

(ii) Series compensation for $\delta = 30^\circ$

The net reactance needed is

$$X_{net} = \frac{V_SV_R\sin30^\circ}{P} = \frac{1\times1\times0.5}{1} = 0.5\ \text{pu}$$

So the series capacitor must have

$$X_C = X_L - X_{net} = 0.8 - 0.5 = 0.3\ \text{pu}$$ $$\text{Degree of compensation} = \frac{X_C}{X_L}\times100 = \frac{0.3}{0.8}\times100 = 37.5\%$$

Effect on stability

Answer: (i) $\delta = 53.13^\circ$; (ii) series capacitive reactance $X_C = 0.3$ pu, i.e. 37.5% series compensation.