Regulation and efficiency of 100 km, 110 kV line

A 100 km line is a medium line, so it is solved here with the nominal-T model: total series impedance split into two halves, with the total shunt admittance at the middle.

Line constants (per phase, total)

$$\begin{aligned} R &= 0.153 \times 100 = 15.3\ \Omega \\ X_L &= 2\pi (50)(1.21\times10^{-3})(100) = 38.01\ \Omega \\ Y &= 2\pi (50)(0.00958\times10^{-6})(100) = 3.010\times10^{-4}\ \text{S} \\ Z/2 &= 7.65 + j19.01\ \Omega \end{aligned}$$

Receiving-end quantities

$$\begin{aligned} V_R &= \frac{110\,000}{\sqrt3} = 63\,508.5\angle 0^\circ\ \text{V} \\ I_R &= \frac{20\times10^6}{3 \times 63\,508.5 \times 0.9} = 116.64\angle -25.84^\circ\ \text{A} \end{aligned}$$

Nominal-T solution

$$\begin{aligned} V_1 &= V_R + I_R \frac{Z}{2} = 65\,297.6\angle 1.41^\circ\ \text{V} \\ I_C &= jY V_1 = 19.65\angle 91.41^\circ\ \text{A} \\ I_S &= I_R + I_C = 109.05\angle -16.62^\circ\ \text{A} \\ V_S &= V_1 + I_S \frac{Z}{2} = 66\,754.4\angle 2.88^\circ\ \text{V} \end{aligned}$$

Sending-end line voltage $= \sqrt3 \times 66.754 = 115.62$ kV.

Regulation

At no load the receiving voltage is $V_S/|A|$, with $A = 1 + \frac{YZ}{2} = 0.9943\angle 0.13^\circ$:

$$\begin{aligned} V_{R0} &= \frac{66\,754.4}{0.9943} = 67\,138.3\ \text{V} \\ \%VR &= \frac{67\,138.3 - 63\,508.5}{63\,508.5}\times 100 = 5.72\% \end{aligned}$$

(If the no-load Ferranti rise is ignored, $(V_S - V_R)/V_R = 5.11\%$.)

Efficiency

$$\begin{aligned} P_{loss} &= 3\left(|I_R|^2 + |I_S|^2\right)\frac{R}{2} \\ &= 3\,(116.64^2 + 109.05^2)(7.65) = 0.585\ \text{MW} \\ \eta &= \frac{20}{20 + 0.585}\times 100 = 97.16\% \end{aligned}$$

Answer: Voltage regulation ≈ 5.72 % (sending end 115.62 kV), efficiency ≈ 97.16 %.