Maximum power with 132 kV at both ends

Data: $|A| = 0.98$, $\alpha = 3^\circ$, $|B| = 110\ \Omega$, $\beta = 75^\circ$, $|V_S| = |V_R| = 132$ kV. Take $D = A$. Line kV give three-phase MW/MVAr.

Maximum power

Maximum $P_R$ occurs at $\delta = \beta = 75^\circ$:

$$\begin{aligned} P_{R,max} &= \frac{|V_S||V_R|}{|B|} - \frac{|A||V_R|^2}{|B|}\cos(\beta-\alpha) \\ &= \frac{132^2}{110} - \frac{0.98\times132^2}{110}\cos72^\circ \\ &= 158.40 - 155.23\times0.3090 = 158.40 - 47.97 = 110.43\ \text{MW} \end{aligned}$$

Receiving-end power factor

$$Q_R = -\frac{|A||V_R|^2}{|B|}\sin(\beta-\alpha) = -155.23\sin72^\circ = -147.63\ \text{MVAr}$$ $$\text{pf}_R = \frac{110.43}{\sqrt{110.43^2 + 147.63^2}} = 0.599\ \text{leading}$$

Efficiency

Sending-end power at $\delta = 75^\circ$ ($\beta+\delta = 150^\circ$):

$$\begin{aligned} P_S &= \frac{|A||V_S|^2}{|B|}\cos(\beta-\alpha) - \frac{|V_S||V_R|}{|B|}\cos(\beta+\delta) \\ &= 155.23\cos72^\circ - 158.40\cos150^\circ \\ &= 47.97 + 137.18 = 185.15\ \text{MW} \end{aligned}$$ $$\eta = \frac{110.43}{185.15}\times100 = 59.64\%$$

The low efficiency shows that operating at the theoretical maximum (steady-state stability limit) is not practical.

Answer: $P_{max}$ ≈ 110.4 MW, receiving-end pf ≈ 0.599 leading, efficiency ≈ 59.6 %.