Sending and receiving power, 140∠10° and 132 kV

Use the power-circle equations (taking $D = A$). Line kV give three-phase MW/MVAr.

Data: $|V_S| = 140$ kV, $|V_R| = 132$ kV, $\delta = 10^\circ$, $|A| = 0.99$, $\alpha = 0.1^\circ$, $|B| = 18\ \Omega$, $\beta = 68^\circ$.

Constants

$$\begin{aligned} \frac{|V_S||V_R|}{|B|} &= \frac{140\times132}{18} = 1026.67 \\ \frac{|A||V_R|^2}{|B|} &= \frac{0.99\times132^2}{18} = 958.32 \\ \frac{|A||V_S|^2}{|B|} &= \frac{0.99\times140^2}{18} = 1078.0 \end{aligned}$$

Angles: $\beta-\delta = 58^\circ$, $\beta-\alpha = 67.9^\circ$, $\beta+\delta = 78^\circ$.

Receiving-end power

$$\begin{aligned} P_R &= 1026.67\cos58^\circ - 958.32\cos67.9^\circ = 544.05 - 360.54 = 183.51\ \text{MW} \\ Q_R &= 1026.67\sin58^\circ - 958.32\sin67.9^\circ = 870.66 - 887.91 = -17.25\ \text{MVAr} \end{aligned}$$

Sending-end power

$$\begin{aligned} P_S &= 1078.0\cos67.9^\circ - 1026.67\cos78^\circ = 405.57 - 213.46 = 192.11\ \text{MW} \\ Q_S &= 1078.0\sin67.9^\circ - 1026.67\sin78^\circ = 998.80 - 1004.23 = -5.43\ \text{MVAr} \end{aligned}$$

Line loss $= 192.11 - 183.51 = 8.60$ MW.

Answer: Receiving end: 183.51 MW − j17.25 MVAr; sending end: 192.11 MW − j5.43 MVAr.