150 km line, A = 0.96∠1°: regulation, max power

Data: $A = D = 0.96\angle1^\circ$, $B = 100\angle80^\circ\ \Omega$, load 30 MW at 0.8 pf lag, $V_R$ = 110 kV. $C$ from $AD - BC = 1$: $C = (A^2-1)/B = 8.53\times10^{-4}\angle77.84^\circ$ S.

(i) Sending-end voltage and regulation

$$\begin{aligned} V_R &= 110/\sqrt3 = 63.51\angle0^\circ\ \text{kV} \\ I_R &= \frac{30\times10^6}{\sqrt3\times110\times10^3\times0.8} = 196.82\angle-36.87^\circ\ \text{A} \\ AV_R &= 60.97\angle1^\circ\ \text{kV} \\ BI_R &= 19.68\angle43.13^\circ\ \text{kV} \\ V_S &= AV_R + BI_R = 76.71\angle10.91^\circ\ \text{kV} \end{aligned}$$

Line value $= \sqrt3\times76.71 = 132.87$ kV.

$$\%VR = \frac{|V_S|/|A| - |V_R|}{|V_R|}\times100 = \frac{76.71/0.96 - 63.51}{63.51}\times100 = 25.82\%$$

(ii) Reactive power supplied by the line

$$\begin{aligned} I_S &= CV_R + DI_R = 174.38\angle-19.35^\circ\ \text{A} \\ S_S &= 3V_SI_S^* = 34.66 + j20.22\ \text{MVA} \end{aligned}$$

Load reactive power $Q_R = 30\times0.75 = 22.5$ MVAr.

$$Q_S - Q_R = 20.22 - 22.5 = -2.28\ \text{MVAr}$$

The sending end supplies only 20.22 MVAr, so the line itself supplies about 2.28 MVAr net to the load (its charging MVAr exceeds its $I^2X$ consumption).

(iii) Maximum power transfer

With $|V_S|$ = 132.87 kV and $|V_R|$ = 110 kV, maximum $P_R$ occurs at $\delta = \beta$:

$$\begin{aligned} P_{R,max} &= \frac{|V_S||V_R|}{|B|} - \frac{|A||V_R|^2}{|B|}\cos(\beta-\alpha) \\ &= \frac{132.87\times110}{100} - \frac{0.96\times110^2}{100}\cos79^\circ \\ &= 146.15 - 22.16 = 123.99\ \text{MW} \end{aligned}$$

Answer: (i) $V_S$ = 132.87 kV, regulation 25.82 %; (ii) the line supplies ≈ 2.28 MVAr (the sending end gives 20.22 MVAr of the 22.5 MVAr load); (iii) $P_{max}$ ≈ 124.0 MW.