70 km, 40 MW line: regulation and zero-regulation pf

Short line (70 km), shunt capacitance neglected.

Data

$$\begin{aligned} R &= 0.03\times70 = 2.1\ \Omega,\quad X = 0.15\times70 = 10.5\ \Omega \\ V_R &= 66\,000/\sqrt3 = 38\,105.1\ \text{V} \\ I &= \frac{40\times10^6}{\sqrt3\times66\,000\times0.8} = 437.39\angle-36.87^\circ\ \text{A} \end{aligned}$$

(i) Regulation and sending-end pf

$$\begin{aligned} IZ &= 437.39\angle-36.87^\circ\times10.708\angle78.69^\circ = 4683.5\angle41.82^\circ\ \text{V} \\ V_S &= 38\,105.1 + 4683.5\angle41.82^\circ = 41\,712.5\angle4.29^\circ\ \text{V} \end{aligned}$$

Line value $= \sqrt3\times41.713 = 72.25$ kV.

$$\%VR = \frac{72.25 - 66}{66}\times100 = 9.47\%$$

Angle between $V_S$ and $I$ $= 4.29^\circ + 36.87^\circ = 41.16^\circ$:

$$\text{pf}_S = \cos41.16^\circ = 0.753\ \text{lagging}$$

(ii) Receiving-end pf for zero regulation

Approximate condition (leading pf):

$$\begin{aligned} IR\cos\phi - IX\sin\phi &= 0 \\ \tan\phi &= \frac{R}{X} = \frac{2.1}{10.5} = 0.2 \\ \cos\phi &= 0.981\ \text{leading} \end{aligned}$$

(The exact condition $|V_S| = |V_R|$ with $P$ = 40 MW gives $Q = -10.13$ MVAr, i.e. pf 0.969 leading.)

Answer: (i) Regulation ≈ 9.47 %, sending-end pf ≈ 0.753 lag; (ii) receiving-end pf ≈ 0.981 leading (exact 0.969 leading).