Maximum power with 138.5∠8° kV and 132 kV

For fixed voltage magnitudes, the receiving-end power

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

is maximum when $\delta = \beta$. (Line kV give three-phase MW/MVAr.)

Data: $|V_S| = 138.5$ kV, $|V_R| = 132$ kV, $|A| = 0.983$, $\alpha = 0.68^\circ$, $|B| = 36.5\ \Omega$, $\beta = 71.5^\circ$, so $\beta-\alpha = 70.82^\circ$.

Maximum active power

$$\begin{aligned} \frac{|V_S||V_R|}{|B|} &= \frac{138.5\times132}{36.5} = 500.88 \\ \frac{|A||V_R|^2}{|B|} &= \frac{0.983\times132^2}{36.5} = 469.25 \\ P_{R,max} &= 500.88 - 469.25\cos70.82^\circ \\ &= 500.88 - 154.17 = 346.71\ \text{MW} \end{aligned}$$

Reactive power at this condition

At $\delta = \beta$, $\sin(\beta-\delta) = 0$:

$$\begin{aligned} Q_R &= -\frac{|A||V_R|^2}{|B|}\sin(\beta-\alpha) \\ &= -469.25\sin70.82^\circ = -443.21\ \text{MVAr} \end{aligned}$$

The negative sign means the receiving end must supply about 443 MVAr to the line (the load must be strongly leading, pf ≈ 0.62 lead) for maximum power transfer. The load angle at this condition is $\delta = 71.5^\circ$.

Answer: $P_{R,max}$ ≈ 346.7 MW; reactive power at the receiving end ≈ −443.2 MVAr (443.2 MVAr must be supplied at the receiving end).