400 km long line, 60 MVA at 220 kV

Given: $l = 400$ km, $z = 0.105 + j0.3768\ \Omega$/km, $y = j2.822\times10^{-6}$ S/km, load 60 MVA at 0.9 p.f. lagging, 220 kV. A 400 km line is a long line, so the exact (distributed) model is used. Values are per phase.

Step 1: Total parameters

$$\begin{aligned} Z &= zl = 42 + j150.72 = 156.46\angle 74.43^\circ\ \Omega \\ Y &= yl = j1.1288\times10^{-3}\ \text{S} \end{aligned}$$

Step 2: Propagation constant and surge impedance

$$\begin{aligned} \gamma l &= \sqrt{ZY} = 0.4203\angle 82.21^\circ = 0.05693 + j0.41638 \\ Z_c &= \sqrt{Z/Y} = 372.3\angle -7.79^\circ\ \Omega \end{aligned}$$

Step 3: ABCD constants

$$\begin{aligned} A = D &= \cosh\gamma l = 0.91604 + j0.02304 = 0.9163\angle 1.44^\circ \\ \sinh\gamma l &= 0.05209 + j0.40511 = 0.4084\angle 82.67^\circ \\ B &= Z_c\sinh\gamma l = 152.1\angle 74.89^\circ\ \Omega \\ C &= \frac{\sinh\gamma l}{Z_c} = 1.097\times10^{-3}\angle 90.46^\circ\ \text{S} \end{aligned}$$

Step 4: Receiving-end quantities

$$\begin{aligned} V_R &= \frac{220}{\sqrt3} = 127.02\angle 0^\circ\ \text{kV} \\ I_R &= \frac{60\times10^6}{\sqrt3\times220\times10^3} = 157.46\angle -25.84^\circ = 141.71 - j68.64\ \text{A} \end{aligned}$$

$P_R = 60 \times 0.9 = 54$ MW, $Q_R = 60 \times 0.4359 = 26.15$ Mvar

Step 5: Sending-end voltage and current

$$\begin{aligned} V_S &= AV_R + BI_R = (116{,}350 + j2926) + (15{,}695 + j18{,}083) \\ &= 132{,}050 + j21{,}009 = 133.71\angle 9.04^\circ\ \text{kV/phase} \\ V_{S(L)} &= \sqrt3\times133.71 = 231.59\ \text{kV} \\ I_S &= CV_R + DI_R = (-1.11 + j139.34) + (131.40 - j59.61) \\ &= 130.28 + j79.74 = 152.7\angle 31.47^\circ\ \text{A} \end{aligned}$$

Step 6: Power factor and power

Angle between $V_S$ and $I_S$ $= 9.04^\circ - 31.47^\circ = -22.43^\circ$ (current leads).

Sending-end p.f. $= \cos 22.43^\circ = 0.9244$ leading

$$S_S = 3V_SI_S^* = 56.64 - j23.38\ \text{MVA}$$

Step 7: Losses

$$\begin{aligned} P_{loss} &= P_S - P_R = 56.64 - 54 = 2.64\ \text{MW} \\ Q_{loss} &= Q_S - Q_R = -23.38 - 26.15 = -49.53\ \text{Mvar} \end{aligned}$$

The negative reactive loss means the line's shunt capacitance generates 49.53 Mvar more than its series inductance absorbs. This is why the sending-end current leads.

Answer: $V_S = 231.59$ kV (line), $I_S = 152.7$ A, p.f. $= 0.9244$ leading, active power loss $= 2.64$ MW, reactive power loss $= -49.53$ Mvar (net 49.53 Mvar generated by the line). Efficiency $= 54/56.64 = 95.35\%$.