Line power loss, 138.5∠8° kV and 132 kV

Use the power-circle equations (with $D = A$). Line voltages in kV give three-phase MW and MVAr.

Data: $|V_S| = 138.5$ kV, $|V_R| = 132$ kV, $\delta = 8^\circ$, $|A| = 0.983$, $\alpha = 0.68^\circ$, $|B| = 36.5\ \Omega$, $\beta = 71.5^\circ$.

Constants

$$\begin{aligned} \frac{|V_S||V_R|}{|B|} &= \frac{138.5\times132}{36.5} = 500.88 \\ \frac{|A||V_R|^2}{|B|} &= \frac{0.983\times132^2}{36.5} = 469.25 \\ \frac{|A||V_S|^2}{|B|} &= \frac{0.983\times138.5^2}{36.5} = 516.61 \end{aligned}$$

Angles: $\beta-\delta = 63.5^\circ$, $\beta-\alpha = 70.82^\circ$, $\beta+\delta = 79.5^\circ$.

Receiving end

$$\begin{aligned} P_R &= 500.88\cos63.5^\circ - 469.25\cos70.82^\circ = 223.49 - 154.17 = 69.32\ \text{MW} \\ Q_R &= 500.88\sin63.5^\circ - 469.25\sin70.82^\circ = 448.25 - 443.21 = 5.05\ \text{MVAr} \end{aligned}$$

Sending end

$$\begin{aligned} P_S &= 516.61\cos70.82^\circ - 500.88\cos79.5^\circ = 169.72 - 91.28 = 78.45\ \text{MW} \\ Q_S &= 516.61\sin70.82^\circ - 500.88\sin79.5^\circ = 487.93 - 492.49 = -4.56\ \text{MVAr} \end{aligned}$$

Losses

$$\begin{aligned} P_{loss} &= P_S - P_R = 78.45 - 69.32 = 9.12\ \text{MW} \\ Q_{loss} &= Q_S - Q_R = -4.56 - 5.05 = -9.60\ \text{MVAr} \end{aligned}$$

The negative reactive "loss" means the line generates about 9.6 MVAr net (its shunt charging exceeds the $I^2X$ absorbed in its series reactance).

Answer: Active power loss ≈ 9.12 MW; reactive power loss ≈ −9.60 MVAr (line is a net source of 9.60 MVAr).