Expression for receiving-end shunt compensation

Neglect the source impedance and losses. Let a shunt reactor $jX_p$ be connected at the receiving end of a line with surge impedance $Z_c$ and electrical length $\beta l$.

Line equation with the reactor

$$V_s = V_r\cos\beta l + jZ_c I_r\sin\beta l$$

At no load the only receiving-end current is the reactor current:

$$I_r = \frac{V_r}{jX_p}$$ $$V_s = V_r\left(\cos\beta l + \frac{Z_c}{X_p}\sin\beta l\right)$$

Condition $|V_s| = |V_r|$

$$\cos\beta l + \frac{Z_c}{X_p}\sin\beta l = 1 \qquad (1)$$

Define an angle $\theta$ by

$$\tan\theta = \frac{Z_c}{X_p}, \quad\text{so}\quad \cos\theta = \frac{X_p}{\sqrt{Z_c^2 + X_p^2}}, \quad \sin\theta = \frac{Z_c}{\sqrt{Z_c^2 + X_p^2}}$$

that is, $\theta = \cos^{-1}\!\left(\dfrac{X_p}{\sqrt{Z_c^2 + X_p^2}}\right)$.

Multiply (1) by $\cos\theta$:

$$\begin{aligned} \cos\beta l\cos\theta + \sin\beta l\sin\theta &= \cos\theta \\ \cos(\beta l - \theta) &= \cos\theta \\ \beta l - \theta &= \theta \end{aligned}$$

So $\tan(\beta l - \theta) = \tan\theta = \dfrac{Z_c}{X_p}$, which gives

$$\boxed{X_p = \frac{Z_c}{\tan\left(\beta l - \cos^{-1}\dfrac{X_p}{\sqrt{Z_c^2 + X_p^2}}\right)}}$$

This is the required expression.

Explicit form

From $\beta l - \theta = \theta$ we get $\theta = \beta l/2$, so

$$X_p = \frac{Z_c}{\tan(\beta l/2)} = Z_c\cot\frac{\beta l}{2}$$

Example: a 400 km line with $Z_c = 316\ \Omega$ and $\beta l = 0.397$ rad gives $X_p = 316/\tan(0.199) = 1571\ \Omega$.

Physically, the reactor absorbs the line's charging VArs at the receiving end. Less net charging current then flows through the series inductance, which removes the Ferranti rise.