Shunt reactors for 750 kV line ABCD

The line is a two-port with $A = 0.9$, $B = j130\ \Omega$. Shunt reactors $jX_L$ are connected at both ends.

Relation at no load

At no load the only "load" on the receiving end of the two-port is the receiving-end reactor:

$$I_R = \frac{V_R}{jX_L}$$

The sending-end reactor is directly across the source. It changes the current drawn from the source but not the relation between $V_s$ and $V_R$. Hence

$$\begin{aligned} V_s &= AV_R + BI_R = 0.9V_R + j130\cdot\frac{V_R}{jX_L} \\ V_s &= V_R\left(0.9 + \frac{130}{X_L}\right) \end{aligned}$$

Without compensation

$$\frac{V_R}{V_s} = \frac{1}{0.9} = 1.111\ \text{pu}$$

That is an 11% Ferranti rise.

Required reactor for $V_R = 1.05V_s$

$$\begin{aligned} 0.9 + \frac{130}{X_L} &= \frac{1}{1.05} = 0.95238 \\ \frac{130}{X_L} &= 0.05238 \\ X_L &= \frac{130}{0.05238} = 2481.8\ \Omega \end{aligned}$$

Check: $1/(0.9 + 130/2481.8) = 1/0.95238 = 1.05$ ✓

Three-phase reactor rating at 750 kV:

$$Q = \frac{V^2}{X_L} = \frac{750^2}{2481.8} = 226.6\ \text{MVAr}$$

 Vs o----+----[ A=0.9, B=j130 ]----+----o VR
         |                         |
        jXL                       jXL
         |                         |
 --------+-------------------------+----- gnd

Answer: $X_L \approx$ 2482 Ω per phase (about 227 MVAr at 750 kV) at the receiving end. The same value is used at the sending end for a symmetrical scheme.