Charging reactive power in sine form

For a lossless line of length $L$, phase constant $\beta = 2\pi/\lambda$ and surge impedance $Z_0$, the generalised constants are

$$A = D = \cos\beta L, \qquad B = jZ_0\sin\beta L, \qquad C = \frac{j\sin\beta L}{Z_0}$$

Sending-end voltage and current at no load

At no load $I_r = 0$:

$$E_s = AE_r + BI_r = E_r\cos\beta L$$ $$I_s = CE_r + DI_r = \frac{j\sin\beta L}{Z_0}E_r$$

So the sending-end current leads $E_r$ by $90^\circ$. Since $E_s$ is in phase with $E_r$ (because $\cos\beta L$ is real), $I_s$ also leads $E_s$ by $90^\circ$. The source therefore supplies purely capacitive (charging) current.

Charging reactive power

Complex power supplied by the source per phase:

$$\begin{aligned} S_s &= E_s I_s^* = E_s\left(\frac{j\sin\beta L}{Z_0}E_r\right)^* \\ &= -j\,\frac{E_sE_r}{Z_0}\sin\beta L \end{aligned}$$

The real power is zero, as expected for a lossless line at no load. The negative sign means the reactive power is capacitive (it is generated by the line and absorbed by the source). The magnitude of the charging reactive power is

$$Q_0 = \frac{E_sE_r}{Z_0}\sin\beta L$$

Substituting $\beta = 2\pi/\lambda$:

$$\boxed{Q_0 = \frac{E_sE_r}{Z_0}\sin\left(\frac{2\pi L}{\lambda}\right)}$$

Remarks

• With $E_s = E_r\cos\beta L$, this can also be written $Q_0 = \dfrac{E_r^2}{2Z_0}\sin 2\beta L$.
• For short lines, $\sin\beta L \approx \beta L$ and $E_s \approx E_r$, so $Q_0 \approx E^2\omega C_{total}$. This is the usual charging MVAr.
• $Q_0$ increases with line length and with the square of voltage. EHV lines generate large charging MVAr at light load, which causes the Ferranti rise and must be absorbed by shunt reactors.