Static compensation to hold sending end at 140 kV

Data: $|A| = 0.983$, $\alpha = 3^\circ$, $|B| = 110\ \Omega$, $\beta = 75^\circ$, $V_R$ = 132 kV, load 50 MVA at 0.8 lag = 40 MW + j30 MVAr. With line kV, powers are three-phase.

Without compensation, $V_S = AV_R + B(P - jQ)/V_R = 165.6$ kV, which is too high. We need $|V_S| = 140$ kV with the same 40 MW load.

Receiving-end power-circle equations

$$\begin{aligned} P_R &= \frac{|V_S||V_R|}{|B|}\cos(\beta-\delta) - \frac{|A||V_R|^2}{|B|}\cos(\beta-\alpha) \\ Q_R &= \frac{|V_S||V_R|}{|B|}\sin(\beta-\delta) - \frac{|A||V_R|^2}{|B|}\sin(\beta-\alpha) \end{aligned}$$ $$\frac{140\times132}{110} = 168.0,\qquad \frac{0.983\times132^2}{110} = 155.71,\qquad \beta-\alpha = 72^\circ$$

Find $\delta$ from $P_R$ = 40 MW

$$\begin{aligned} 40 &= 168.0\cos(75^\circ-\delta) - 155.71\cos72^\circ \\ 40 &= 168.0\cos(75^\circ-\delta) - 48.12 \\ \cos(75^\circ-\delta) &= 0.5245 \Rightarrow 75^\circ-\delta = 58.37^\circ,\ \delta = 16.63^\circ \end{aligned}$$

Reactive power the line delivers

$$\begin{aligned} Q_R &= 168.0\sin58.37^\circ - 155.71\sin72^\circ \\ &= 143.04 - 148.09 = -5.05\ \text{MVAr} \end{aligned}$$

So the line can deliver no lagging VAr; it actually needs 5.05 MVAr from the receiving end.

Compensator rating

Load requires 30 MVAr lagging. The compensator must supply:

$$Q_C = 30 - (-5.05) = 35.05\ \text{MVAr}$$

Answer: Static (capacitive) compensator of about 35 MVAr at the receiving end.