400 km line at UPF: Zc, propagation, regulation

Given: 50 Hz, $l = 400$ km, 400 kV, 400 MW at unity p.f.; $r = 0.032\ \Omega$/km, $L = 1.066$ mH/km, $C = 0.0109\ \mu$F/km (converted from per-metre values). Receiving voltage 400 kV.

Step 1: Series impedance and shunt admittance per km

$$\begin{aligned} z &= 0.032 + j(2\pi\times50\times1.066\times10^{-3}) = 0.032 + j0.3349 = 0.3364\angle 84.54^\circ\ \Omega/\text{km} \\ y &= j2\pi\times50\times1.09\times10^{-8} = j3.424\times10^{-6}\ \text{S/km} \end{aligned}$$

Step 2: Characteristic impedance

$$Z_c = \sqrt{\frac{z}{y}} = \sqrt{\frac{0.3364\angle 84.54^\circ}{3.424\times10^{-6}\angle 90^\circ}} = 313.4\angle -2.73^\circ\ \Omega$$

Step 3: Propagation and phase-shift constants

$$\begin{aligned} \gamma &= \sqrt{zy} = 1.073\times10^{-3}\angle 87.27^\circ = 5.11\times10^{-5} + j1.0721\times10^{-3}\ /\text{km} \\ \beta &= 1.0721\times10^{-3}\ \text{rad/km}, \quad \alpha = 5.11\times10^{-5}\ \text{Np/km} \\ \gamma l &= 0.02044 + j0.42884 \end{aligned}$$

Step 4: ABCD constants

$$\begin{aligned} A &= \cosh\gamma l = 0.90964 + j0.00850 = 0.9097\angle 0.54^\circ \\ B &= Z_c\sinh\gamma l = 130.5\angle 84.71^\circ\ \Omega \end{aligned}$$

Step 5: Sending-end voltage

$$\begin{aligned} V_R &= 230.94\angle 0^\circ\ \text{kV}, \quad I_R = \frac{400\times10^6}{\sqrt3\times400\times10^3} = 577.35\angle 0^\circ\ \text{A} \\ V_S &= AV_R + BI_R = 230.3\angle 19.53^\circ\ \text{kV/phase} = 398.83\ \text{kV (line)} \end{aligned}$$

Step 6: Voltage regulation

$$\begin{aligned} V_{R(NL)} &= \frac{|V_S|}{|A|} = \frac{398.83}{0.9097} = 438.43\ \text{kV} \\ \%\text{VR} &= \frac{438.43 - 400}{400}\times100 = 9.61\ \% \end{aligned}$$

The load (400 MW) is below the SIL ($400^2/313.4 \approx 511$ MW), so the line generates net reactive power and $V_S$ is slightly below $V_R$ at full load.

Answer: $Z_c = 313.4\angle -2.73^\circ\ \Omega$, $\gamma = 1.073\times10^{-3}\angle 87.27^\circ$ per km, $\beta = 1.072\times10^{-3}$ rad/km, voltage regulation $= 9.61\%$.