150 mile, 60 Hz line: ABCD and sending end

Given: $l = 150$ miles, 60 Hz; $R = 0.1858\ \Omega$/mi, $L = 2.6$ mH/mi, $C = 0.012\ \mu$F/mi; load 50 MVA at 0.85 p.f. lagging, 138 kV. 150 miles (241 km) is treated with the exact long-line model.

Step 1: Per-mile and total constants

$$\begin{aligned} z &= 0.1858 + j(2\pi\times60\times2.6\times10^{-3}) = 0.1858 + j0.9802\ \Omega/\text{mi} \\ y &= j2\pi\times60\times0.012\times10^{-6} = j4.524\times10^{-6}\ \text{S/mi} \\ Z &= zl = 27.87 + j147.03 = 149.6\angle 79.27^\circ\ \Omega \\ Y &= yl = j6.786\times10^{-4}\ \text{S} \end{aligned}$$

Step 2: $\gamma l$ and $Z_c$

$$\begin{aligned} \gamma l &= \sqrt{ZY} = 0.3187\angle 84.63^\circ = 0.02981 + j0.31727 \\ Z_c &= \sqrt{Z/Y} = 469.6\angle -5.37^\circ\ \Omega \end{aligned}$$

Step 3: ABCD constants

$$\begin{aligned} A = D &= \cosh\gamma l = 0.95051 + j0.00930 = 0.9506\angle 0.56^\circ \\ \sinh\gamma l &= 0.02832 + j0.31211 = 0.3134\angle 84.82^\circ \\ B &= Z_c\sinh\gamma l = 147.2\angle 79.45^\circ\ \Omega \\ C &= \frac{\sinh\gamma l}{Z_c} = 6.674\times10^{-4}\angle 90.18^\circ\ \text{S} \end{aligned}$$

Step 4: Receiving-end values

$$\begin{aligned} V_R &= \frac{138}{\sqrt3} = 79.67\angle 0^\circ\ \text{kV} \\ I_R &= \frac{50\times10^6}{\sqrt3\times138\times10^3} = 209.2\angle -31.79^\circ = 177.81 - j110.19\ \text{A} \end{aligned}$$

Step 5: Sending-end voltage and current

$$\begin{aligned} V_S &= AV_R + BI_R = (75{,}732 + j741) + (20{,}735 + j22{,}755) \\ &= 96{,}466 + j23{,}496 = 99.29\angle 13.69^\circ\ \text{kV/phase} \\ V_{S(L)} &= \sqrt3\times99.29 = 171.97\ \text{kV} \\ I_S &= CV_R + DI_R = (-0.17 + j53.17) + (170.03 - j103.09) \\ &= 169.86 - j49.92 = 177.0\angle -16.38^\circ\ \text{A} \end{aligned}$$

Step 6: Sending-end power

p.f. $= \cos(13.69^\circ + 16.38^\circ) = \cos 30.07^\circ = 0.8655$ lagging

$$\begin{aligned} P_S &= 3|V_S||I_S|\cos\phi_S = 3\times99.29\times10^3\times177.0\times0.8655 = 45.64\ \text{MW} \\ Q_S &= 26.42\ \text{Mvar} \end{aligned}$$

($P_R = 42.5$ MW, so the line loss is 3.14 MW and efficiency 93.12 %.)

Answer: $A = D = 0.9506\angle 0.56^\circ$, $B = 147.2\angle 79.45^\circ\ \Omega$, $C = 6.674\times10^{-4}\angle 90.18^\circ$ S; $V_S = 171.97$ kV (line), $I_S = 177.0$ A, $P_S = 45.64$ MW at 0.8655 p.f. lagging.