Capacitor for zero regulation, 70 km 220 kV line

Short line: $Z = 8 + j22\ \Omega$/phase, $V_R$ = 220 kV, load 100 MW at 0.9 pf lag ($Q_L = 100\tan25.84^\circ = 48.43$ MVAr). Without compensation the regulation is 3.92 %.

Condition for zero regulation

With line kV and three-phase MW/MVAr, the net receiving-end $P + jQ$ must satisfy $|V_S| = |V_R|$:

$$\begin{aligned} V_S &= V_R + \frac{(R + jX)(P - jQ)}{V_R} \\ |V_R|^4 &= (V_R^2 + RP + XQ)^2 + (XP - RQ)^2 \end{aligned}$$

Substituting $V_R = 220$, $P = 100$, $R = 8$, $X = 22$:

$$48\,400^2 = (48\,400 + 800 + 22Q)^2 + (2200 - 8Q)^2$$ $$548Q^2 + 2\,129\,600Q + 82\,920\,000 = 0$$

Taking the root nearer zero:

$$Q = -39.34\ \text{MVAr}\quad(\text{net pf} = 0.931\ \text{leading})$$

Capacitor rating

$$Q_C = Q_L - Q = 48.43 + 39.34 = 87.77\ \text{MVAr (3-phase)}$$

Capacitance per phase (star connected)

$$C = \frac{Q_C}{\omega V_L^2} = \frac{87.77\times10^6}{314.16\times(220\times10^3)^2} = 5.77\ \mu\text{F}$$

(For a delta-connected bank, $C_\Delta = C/3 = 1.92\ \mu$F.)

Using the approximate rule $\tan\phi = R/X$ leading gives $Q_C = 48.43 + 100(8/22) = 84.80$ MVAr and $C$ ≈ 5.58 µF, close to the exact value.

Answer: About 87.8 MVAr of shunt capacitors, i.e. C ≈ 5.77 µF per phase (star).