300 km line: short line vs nominal-π

Given: $Z = 40 + j125 = 131.24\angle 72.26^\circ\ \Omega$, $Y = j10^{-3}$ S (totals), load 50 MW at 220 kV, 0.8 p.f. lagging. Per-phase working.

$$\begin{aligned} V_R &= \frac{220}{\sqrt3} = 127.02\angle 0^\circ\ \text{kV} \\ I_R &= \frac{50\times10^6}{\sqrt3\times220\times10^3\times0.8} = 164.02\angle -36.87^\circ = 131.22 - j98.41\ \text{A} \end{aligned}$$

$P_R = 50$ MW, $Q_R = 37.5$ Mvar.

(i) Short-line approximation ($A = D = 1$, $B = Z$, $C = 0$)

$$\begin{aligned} I_S &= I_R = 164.0\angle -36.87^\circ\ \text{A} \\ ZI_R &= (40 + j125)(131.22 - j98.41) = 17{,}550 + j12{,}466\ \text{V} \\ V_S &= V_R + ZI_R = 144{,}567 + j12{,}466 = 145.10\angle 4.93^\circ\ \text{kV/phase} \\ V_{S(L)} &= \sqrt3\times145.10 = 251.33\ \text{kV} \end{aligned}$$

p.f. $= \cos(4.93^\circ + 36.87^\circ) = \cos 41.80^\circ = 0.7455$ lagging

$$P_S = 3V_SI_S\cos\phi_S = 3\times145.10\times10^3\times164.02\times0.7455 = 53.23\ \text{MW}$$

(ii) Nominal-Π method

$$\begin{aligned} ZY &= (40 + j125)(j10^{-3}) = -0.125 + j0.04 \\ A = D &= 1 + \frac{ZY}{2} = 0.9375 + j0.02 = 0.9377\angle 1.22^\circ \\ B &= Z = 131.24\angle 72.26^\circ\ \Omega \\ C &= Y\left(1 + \frac{ZY}{4}\right) = -1.0\times10^{-5} + j9.6875\times10^{-4} = 9.688\times10^{-4}\angle 90.59^\circ\ \text{S} \end{aligned}$$ $$\begin{aligned} V_S &= AV_R + BI_R = (119{,}080 + j2540) + (17{,}550 + j12{,}466) \\ &= 136{,}630 + j15{,}006 = 137.45\angle 6.27^\circ\ \text{kV/phase} \\ V_{S(L)} &= 238.07\ \text{kV} \\ I_S &= CV_R + DI_R = (-1.27 + j123.05) + (124.98 - j89.64) \\ &= 123.71 + j33.41 = 128.1\angle 15.11^\circ\ \text{A} \end{aligned}$$

p.f. $= \cos(6.27^\circ - 15.11^\circ) = \cos 8.85^\circ = 0.9881$ leading (current leads voltage)

$$P_S = 3\times137.45\times10^3\times128.1\times0.9881 = 52.21\ \text{MW}$$

Comparison

The short-line model ignores about 123 A of charging current, so for a 300 km line it badly overestimates $V_S$ and $I_S$. The nominal-Π (or better, the exact long-line) model must be used.

Answer: (i) $V_S = 251.33$ kV, $I_S = 164.0$ A, p.f. 0.7455 lag, $P_S = 53.23$ MW. (ii) $V_S = 238.07$ kV, $I_S = 128.1$ A, p.f. 0.9881 lead, $P_S = 52.21$ MW.