Receiving end power in ABCD terms and maximum power

Start from the ABCD relation of a two-port line:

$$V_S = A V_R + B I_R$$

Take $V_R$ as reference and let

$$V_R = |V_R|\angle0^\circ,\quad V_S = |V_S|\angle\delta,\quad A = |A|\angle\alpha,\quad B = |B|\angle\beta$$

Receiving-end current

$$I_R = \frac{V_S - A V_R}{B} = \frac{|V_S|}{|B|}\angle(\delta-\beta) - \frac{|A||V_R|}{|B|}\angle(\alpha-\beta)$$

Receiving-end complex power

$$\begin{aligned} S_R &= P_R + jQ_R = V_R I_R^* \\ &= \frac{|V_S||V_R|}{|B|}\angle(\beta-\delta) - \frac{|A||V_R|^2}{|B|}\angle(\beta-\alpha) \end{aligned}$$

Separating real and imaginary parts:

$$\begin{aligned} P_R &= \frac{|V_S||V_R|}{|B|}\cos(\beta-\delta) - \frac{|A||V_R|^2}{|B|}\cos(\beta-\alpha) \\ Q_R &= \frac{|V_S||V_R|}{|B|}\sin(\beta-\delta) - \frac{|A||V_R|^2}{|B|}\sin(\beta-\alpha) \end{aligned}$$

With phase voltages these give power per phase; with line voltages they give total three-phase power.

Maximum receiving-end power

For fixed $|V_S|$, $|V_R|$ and line constants, only $\delta$ can vary. The second term is constant, so $P_R$ is maximum when $\cos(\beta-\delta) = 1$, i.e.

$$\delta = \beta$$ $$P_{R,max} = \frac{|V_S||V_R|}{|B|} - \frac{|A||V_R|^2}{|B|}\cos(\beta-\alpha)$$

At this point the reactive power at the receiving end is

$$Q_R = -\frac{|A||V_R|^2}{|B|}\sin(\beta-\alpha)$$

which is negative, so the load must supply (leading) reactive power to receive maximum active power.

For a short line ($A = 1\angle0^\circ$, $B = Z\angle\theta$):

$$P_{R,max} = \frac{|V_S||V_R|}{|Z|} - \frac{|V_R|^2 R}{|Z|^2}$$