Sending end from A = 0.98∠0.3°, B = 110∠75°

Data: $A = 0.98\angle0.3^\circ$, $B = 110\angle75^\circ\ \Omega$, load 50 MVA at 0.8 lag at 132 kV, i.e. $S_R = 40 + j30$ MVA. $D = A$ is assumed (symmetrical line).

(a) Sending-end voltage and power angle

$$\begin{aligned} V_R &= 132/\sqrt3 = 76.21\angle0^\circ\ \text{kV} \\ I_R &= \frac{50\times10^6}{\sqrt3\times132\times10^3} = 218.69\angle-36.87^\circ\ \text{A} \\ AV_R &= 74.69\angle0.3^\circ\ \text{kV} \\ BI_R &= 110\angle75^\circ\times0.21869\angle-36.87^\circ = 24.06\angle38.13^\circ\ \text{kV} \\ V_S &= AV_R + BI_R = 94.84\angle9.25^\circ\ \text{kV (phase)} \end{aligned}$$

$|V_S|$ (line) $= \sqrt3\times94.84 = 164.27$ kV, power angle $\delta = 9.25^\circ$.

(b) Sending-end active and reactive power

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

With $\frac{0.98\times164.27^2}{110} = 240.41$, $\frac{164.27\times132}{110} = 197.12$, $\beta-\alpha = 74.7^\circ$, $\beta+\delta = 84.25^\circ$:

$$\begin{aligned} P_S &= 240.41\cos74.7^\circ - 197.12\cos84.25^\circ = 43.69\ \text{MW} \\ Q_S &= 240.41\sin74.7^\circ - 197.12\sin84.25^\circ = 35.76\ \text{MVAr} \end{aligned}$$

(Check with the receiving-end equations at $\delta = 9.25^\circ$: $P_R = 40.00$ MW, $Q_R = 30.00$ MVAr.)

(c) Line losses and VARs absorbed

$$\begin{aligned} P_{loss} &= 43.69 - 40 = 3.69\ \text{MW} \\ Q_{absorbed} &= 35.76 - 30 = 5.76\ \text{MVAr} \end{aligned}$$

Answer: (a) $V_S$ = 164.27 kV, δ = 9.25°; (b) $P_S$ = 43.69 MW, $Q_S$ = 35.76 MVAr; (c) loss 3.69 MW, line absorbs 5.76 MVAr.