220 kV double circuit efficiency and regulation

The line is 130 km long, so it is a medium line and the nominal-π model is used. The two circuits are in parallel.

Assumptions: power factor 0.9 lagging (not given in the question); receiving-end voltage held at 220 kV.

Line parameters (per phase)

Per circuit:

$$\begin{aligned} R &= 0.0544 \times 130 = 7.072\ \Omega \\ X &= 2\pi(50)(0.494\times10^{-3})(130) = 20.175\ \Omega \\ Y &= 2\pi(50)(23.1\times10^{-9})(130) = 9.434\times10^{-4}\ \text{S} \end{aligned}$$

Two circuits in parallel:

$$Z = \frac{7.072 + j20.175}{2} = 3.536 + j10.088\ \Omega, \qquad Y = j1.887\times10^{-3}\ \text{S}$$

Constants and currents

$$\begin{aligned} A &= 1 + \frac{ZY}{2} = 0.9905\angle 0.19° \\ V_R &= \frac{220}{\sqrt{3}} = 127.02\angle 0°\ \text{kV} \\ I_R &= \frac{144\times10^6}{\sqrt{3}\times220\times10^3\times0.9} = 419.9\angle -25.84°\ \text{A} \end{aligned}$$ $$\begin{aligned} V_S &= AV_R + ZI_R = 129.04\angle 1.59°\ \text{kV/phase} \\ &= 223.51\ \text{kV (line)} \\ I_S &= CV_R + AI_R = 379.1\angle 8.88°\ \text{A} \end{aligned}$$

a) Efficiency

$$\begin{aligned} P_S &= \text{Re}(3V_SI_S^*) = 145.56\ \text{MW} \\ \text{Loss} &= 1.56\ \text{MW} \\ \eta &= \frac{144}{145.56}\times100 = 98.93\% \end{aligned}$$

b) Voltage regulation

At no load, $V_{R0} = V_S/|A|$:

$$\begin{aligned} V_{R0} &= \frac{223.51}{0.9905} = 225.65\ \text{kV (line)} \\ \%VR &= \frac{225.65 - 220}{220}\times100 = 2.57\% \end{aligned}$$

c) Receiving-end voltage rise at no load (Ferranti effect)

With $V_S$ held at 223.51 kV and the load removed:

$$\Delta V = V_{R0} - V_S = 225.65 - 223.51 = 2.15\ \text{kV}$$

The no-load voltage is 5.65 kV above the rated 220 kV.

Answer: η = 98.93%, VR = 2.57%, no-load rise = 2.15 kV above $V_S$ (receiving end 225.65 kV).

The line already meets η > 94% and VR < 12% by a wide margin.

Options if the limits were not met

1. Larger conductor (lower $R$) for higher efficiency, or bundled conductors (lower $X$) for better regulation.
2. Raise the power factor with shunt capacitors or SVCs at the load end. At unity pf the regulation falls to about 1.1%.
3. Series capacitor compensation to cut $X$ and the voltage drop.
4. More circuits or a higher voltage (for example 400 kV), which reduces the current and therefore $I^2R$ and $IX$.
5. Shunt reactors at the receiving end to limit the no-load (Ferranti) rise.
6. On-load tap-changing transformers to correct the receiving-end voltage.