Sending end from A = 0.98∠3°, B = 110∠75°

Given: 132 kV, 50 MVA at 0.8 p.f. lagging at the receiving end; $A = 0.98\angle 3^\circ$, $B = 110\angle 75^\circ\ \Omega$ per phase. Symmetrical line assumed, so $D = A$.

$$\begin{aligned} V_R &= \frac{132}{\sqrt3} = 76.21\angle 0^\circ\ \text{kV} \\ I_R &= \frac{50\times10^6}{\sqrt3\times132\times10^3} = 218.7\angle -36.87^\circ = 174.95 - j131.22\ \text{A} \end{aligned}$$

$P_R = 50\times0.8 = 40$ MW, $Q_R = 30$ Mvar.

(i) Sending-end voltage and power angle

$$\begin{aligned} AV_R &= 0.98\angle 3^\circ\times76{,}210 = 74{,}584 + j3909\ \text{V} \\ BI_R &= 110\angle 75^\circ\times218.7\angle -36.87^\circ = 24{,}056\angle 38.13^\circ = 18{,}923 + j14{,}854\ \text{V} \\ V_S &= AV_R + BI_R = 93{,}507 + j18{,}762 = 95.37\angle 11.35^\circ\ \text{kV/phase} \\ V_{S(L)} &= \sqrt3\times95.37 = 165.19\ \text{kV} \end{aligned}$$

Power angle $\delta = 11.35^\circ$.

(ii) Sending-end active and reactive power

Using the sending-end power equations with line values ($\beta = 75^\circ$, $\alpha = 3^\circ$, $D = A$):

$$\begin{aligned} P_S &= \frac{|D||V_S|^2}{|B|}\cos(\beta - \alpha) - \frac{|V_S||V_R|}{|B|}\cos(\beta + \delta) \\ Q_S &= \frac{|D||V_S|^2}{|B|}\sin(\beta - \alpha) - \frac{|V_S||V_R|}{|B|}\sin(\beta + \delta) \end{aligned}$$ $$\begin{aligned} \frac{|D||V_S|^2}{|B|} &= \frac{0.98\times165.19^2}{110} = 243.10\ \text{MVA} \\ \frac{|V_S||V_R|}{|B|} &= \frac{165.19\times132}{110} = 198.22\ \text{MVA} \\ P_S &= 243.10\cos 72^\circ - 198.22\cos 86.35^\circ = 75.12 - 12.63 = 62.49\ \text{MW} \\ Q_S &= 243.10\sin 72^\circ - 198.22\sin 86.35^\circ = 231.20 - 197.82 = 33.38\ \text{Mvar} \end{aligned}$$

(Check: the receiving-end form of the same equations gives back $P_R = 40$ MW and $Q_R = 30$ Mvar.)

(iii) Line losses

$$\begin{aligned} P_{loss} &= P_S - P_R = 62.49 - 40 = 22.49\ \text{MW} \\ Q_{loss} &= Q_S - Q_R = 33.38 - 30 = 3.38\ \text{Mvar} \end{aligned}$$

Answer: (i) $V_S = 165.19$ kV (line), $\delta = 11.35^\circ$; (ii) $P_S = 62.49$ MW, $Q_S = 33.38$ Mvar; (iii) losses $= 22.49$ MW and $3.38$ Mvar.

Note: the active loss is large because the 3° angle of $A$ implies a resistive (lossy) shunt branch: from $AD - BC = 1$, $C = 9.996\times10^{-4}\angle 39.1^\circ$ S. The series resistance alone ($110\cos 75^\circ = 28.5\ \Omega$) would account for only about 4.1 MW.