220 kV line, 200 MW: sending end from ABCD

Assumption: $B$ is not printed, so it is found from the reciprocity condition $AD - BC = 1$:

$$\begin{aligned} B &= \frac{AD - 1}{C} = \frac{0.81\angle0.002^\circ - 1}{1.14\times10^{-3}\angle90^\circ} \\ &= \frac{-0.19}{1.14\times10^{-3}\angle90^\circ} = 166.67\angle89.99^\circ\ \Omega \approx j166.67\ \Omega \end{aligned}$$

Receiving-end quantities

$$\begin{aligned} V_R &= 220/\sqrt3 = 127.02\angle0^\circ\ \text{kV} \\ I_R &= \frac{200\times10^6}{\sqrt3\times220\times10^3\times0.8} = 656.08\angle-36.87^\circ\ \text{A} \end{aligned}$$

(i) Sending-end voltage and current

$$\begin{aligned} AV_R &= 114.32\angle0^\circ\ \text{kV},\quad BI_R = 109.35\angle53.12^\circ\ \text{kV} \\ V_S &= AV_R + BI_R = 200.07\angle25.93^\circ\ \text{kV (phase)} \\ CV_R &= 144.80\angle90^\circ\ \text{A},\quad DI_R = 590.47\angle-36.87^\circ\ \text{A} \\ I_S &= CV_R + DI_R = 516.75\angle-23.92^\circ\ \text{A} \end{aligned}$$

Sending-end line voltage $= \sqrt3\times200.07 = 346.53$ kV.

(ii) Sending-end power

$$S_S = 3V_SI_S^* = 3(200.07\angle25.93^\circ)(0.51675\angle23.92^\circ) = 200.03 + j237.03\ \text{MVA}$$

$P_S$ = 200.03 MW, $Q_S$ = 237.03 MVAr (pf 0.645 lag).

(iii) Losses

Receiving end: $S_R = 200 + j150$ MVA.

$$\begin{aligned} P_{loss} &= 200.03 - 200 = 0.03\ \text{MW} \\ Q_{loss} &= 237.03 - 150 = 87.03\ \text{MVAr} \end{aligned}$$

The active loss is almost zero because the given $A$, $D$ have nearly zero angle, i.e. the line is practically lossless.

(iv) Regulation and efficiency

$$\begin{aligned} \%VR &= \frac{|V_S|/|A| - |V_R|}{|V_R|}\times100 = \frac{200.07/0.9 - 127.02}{127.02}\times100 = 75.02\% \\ \eta &= \frac{200}{200.03}\times100 = 99.99\% \end{aligned}$$

Answer: $V_S$ = 346.5 kV (line), $I_S$ = 516.8 A; $P_S$ = 200.03 MW, $Q_S$ = 237.03 MVAr; losses 0.03 MW and 87.03 MVAr; regulation ≈ 75 %, efficiency ≈ 99.99 %. (The very high regulation shows the line is loaded far above its surge impedance loading of about 127 MW.)