Short line 40 km, 381 MVA: regulation and efficiency

Short-line model: shunt capacitance neglected, $I_S = I_R$.

Line impedance

$$\begin{aligned} R &= 0.15\times40 = 6\ \Omega \\ X &= 2\pi(50)(1.3263\times10^{-3})(40) = 16.667\ \Omega \\ Z &= 6 + j16.667\ \Omega \end{aligned}$$

Receiving end

$$\begin{aligned} V_R &= \frac{220}{\sqrt3} = 127.02\angle0^\circ\ \text{kV} \\ I_R &= \frac{381\times10^6}{\sqrt3\times220\times10^3} = 999.87\angle-36.87^\circ\ \text{A} \end{aligned}$$

Sending-end voltage

$$\begin{aligned} V_S &= V_R + I_R Z = 127.02 + (0.99987\angle-36.87^\circ)(17.71\angle70.20^\circ) \\ &= 127.02 + 17.71\angle33.33^\circ \\ &= 142.15\angle3.93^\circ\ \text{kV (phase)} \end{aligned}$$

Line value $= \sqrt3\times142.15 = 246.21$ kV.

Sending-end power

$$\begin{aligned} S_S &= 3V_S I_S^* = 3(142.15\angle3.93^\circ)(0.99987\angle36.87^\circ) \\ &= 322.80 + j278.59\ \text{MVA} \end{aligned}$$

Sending-end pf $= \cos(40.80^\circ) = 0.757$ lagging.

Regulation

$$\%VR = \frac{246.21 - 220}{220}\times100 = 11.91\%$$

Efficiency

Receiving power $P_R = 381\times0.8 = 304.8$ MW.

$$\eta = \frac{304.8}{322.80}\times100 = 94.43\%$$

(Check: loss $= 3I^2R = 3(999.87)^2(6) = 18.0$ MW.)

Answer: $V_S$ = 246.2 kV (line), $S_S$ = 322.8 MW + j278.6 MVAr, regulation 11.91 %, efficiency 94.43 %.