15 km line: regulation and zero-regulation capacitor

Short line (15 km), so shunt capacitance is neglected.

Line constants

$$\begin{aligned} I &= \frac{5\times10^6}{\sqrt3 \times 11\,000 \times 0.8} = 328.04\ \text{A} \\ P_{loss} &= 0.12 \times 5 = 0.6\ \text{MW} = 3I^2R \\ R &= \frac{0.6\times10^6}{3\times 328.04^2} = 1.859\ \Omega \\ X &= 2\pi(50)(1.1\times10^{-3})(15) = 5.184\ \Omega \end{aligned}$$

(i) Sending-end voltage and regulation

$$\begin{aligned} V_R &= \frac{11\,000}{\sqrt3} = 6350.9\angle0^\circ\ \text{V} \\ I &= 328.04\angle-36.87^\circ\ \text{A} \\ V_S &= V_R + I(R + jX) \\ &= 6350.9 + 328.04\angle-36.87^\circ\,(1.859 + j5.184) \\ &= 7921.5\angle7.21^\circ\ \text{V} \end{aligned}$$

Line value $= \sqrt3 \times 7921.5 = 13.72$ kV.

$$\%VR = \frac{7921.5 - 6350.9}{6350.9}\times100 = 24.73\%$$

(ii) Load pf for zero regulation

Using the approximate drop $V_S - V_R \approx I(R\cos\phi \pm X\sin\phi)$, regulation is zero when the load is leading and

$$\begin{aligned} IR\cos\phi - IX\sin\phi &= 0 \\ \tan\phi &= \frac{R}{X} = \frac{1.859}{5.184} = 0.3585 \\ \phi &= 19.73^\circ,\quad \cos\phi = 0.941\ \text{leading} \end{aligned}$$

(Solving $|V_S| = |V_R|$ exactly with $P$ = 5 MW gives 0.890 leading.)

(iii) Capacitor at the receiving end

The 5 MW load stays at 0.8 lag (needs 3.75 MVAr). For zero regulation the net receiving-end reactive power must be

$$Q_{net} = -P\tan\phi = -5 \times 0.3585 = -1.793\ \text{MVAr}$$

Capacitor rating:

$$Q_C = 3.75 + 1.793 = 5.543\ \text{MVAr (3-phase)}$$

Capacitance per phase (star connected):

$$C = \frac{Q_C}{\omega V_L^2} = \frac{5.543\times10^6}{314.16\times 11\,000^2} = 145.8\ \mu\text{F}$$

Answer: (i) $V_S$ = 13.72 kV, regulation 24.73 %; (ii) pf ≈ 0.941 leading; (iii) 5.54 MVAr bank, ≈ 145.8 µF per phase (star).