Charging reactive power of line at no load

Consider a lossless line of length $L$ with inductance $l$ and capacitance $c$ per unit length, open at the receiving end (no load). Let $E_s$ be the sending-end phase voltage and $E_r$ the receiving-end voltage.

Line equations

For a lossless line the propagation constant is $\gamma = j\beta$ with

$$\beta = \omega\sqrt{lc} = \frac{2\pi}{\lambda}, \qquad Z_c = \sqrt{\frac{l}{c}}$$

The long-line equations at distance $x$ from the receiving end are

$$\begin{aligned} E(x) &= E_r\cos\beta x + jZ_cI_r\sin\beta x \\ I(x) &= I_r\cos\beta x + j\frac{E_r}{Z_c}\sin\beta x \end{aligned}$$

At no load

$I_r = 0$, so at the sending end ($x = L$):

$$E_s = E_r\cos\beta L, \qquad I_s = j\frac{E_r}{Z_c}\sin\beta L$$

Hence $E_r = \dfrac{E_s}{\cos\beta L}$ (this is the Ferranti rise), and

$$I_s = j\frac{E_s}{Z_c}\frac{\sin\beta L}{\cos\beta L} = j\frac{E_s}{Z_c}\tan\beta L$$

The sending-end current leads $E_s$ by $90^\circ$: it is a pure charging current.

Reactive power supplied by the source

Taking $E_s$ as reference, the complex power per phase is

$$S_s = E_sI_s^* = E_s\left(-j\frac{E_s}{Z_c}\tan\beta L\right) = -j\frac{E_s^2}{Z_c}\tan\beta L$$

The real power is zero (lossless line). The magnitude of reactive power is

$$Q_0 = \frac{E_s^2}{Z_c}\tan\beta L$$

the negative sign meaning the source absorbs lagging VArs, i.e. it supplies leading (capacitive) charging VArs to the line.

Substitution

With $\frac{1}{Z_c} = \sqrt{\frac{c}{l}}$ and $\beta L = \frac{2\pi L}{\lambda}$:

$$\boxed{Q_0 = E_s^2\sqrt{\frac{c}{l}}\tan\frac{2\pi L}{\lambda}}$$

which is the required result.

Remarks

• For a short line, $\tan\beta L \approx \beta L$, giving $Q_0 \approx E_s^2\,\omega cL$, the usual $\omega CV^2$.
• $Q_0$ grows rapidly with length and becomes infinite at $L = \lambda/4$ (1500 km at 50 Hz), so long EHV lines need shunt reactors to absorb charging VArs and limit the Ferranti rise.