Surge Impedance Loading

The relationship between current and voltage a long one phase of the line in term of the distributed parameters.

R = Series resistance.
G = Shunt conductance.
L = Series inductance.
C = Shunt capacitance.

z = R + jωl = series impedence per unit length/phase.
y = G + jωl = shunt admittance per unit length/phase.
l = length of the line.

Since G is negligible and R is small, high voltage lines are assumed to be lossless when we are dealing with lightning and switching surges. Hence the characteristic impedance Zc with losses neglected is commonly referred to as the surge impedance.

It is equal to (L/C)^1/2 and has dimension of a pure resistance. The power delivered by a transmission line when it is terminated by its surge impedance is known as the natural load or surge impedance load (SIL)

SIL = [(Vo)^2/Zc] Watt

Where Vo is the rated voltage of the line. If Vo is the line to neutral voltage, SIL given by the above equation is the per-phase value. If Vo is the line to line value, the SIL is the three-phase value.


Voltage and current along the length of a lossless line at SIL are given by

V = VR.(e)^ϒx

I = IR.(e)^ϒx

Where ϒ = jβ = jω.(L.C)^0.5

At SIL, transmission line (lossless) exhibit the following special characteristic
1. V and I have constant amplitude a long the line.
2. V and I are in phase throughout the length line.
3. The phase angle between the sending end and receiving end voltages (currents) is equal to βl.

At the natural load, the reactive power generated by C is equal to reactive power absorbed by L, for each incremental length of the line.

Hence, no reactive power is absorbed or generated at either end of the line, and the voltage and current profiles are flat. This is an optimum condition with respect to control of voltage and reactive power.

As we will see the natural or surge impedance loading of a line serve as a convinient reference quantity for evaluating and expressing its capability.

Table below gives typical parameters of overhead lines of nominal voltage.
















For easy to remember appoximate equivalent circuit applicable to an overhead line of 160km (100mi) length and of any voltage rating is shown. For the 500kV line whose parameter are listed in table above with a line length of 160km, we have














XL = 160 x 0.325 = 52 Ω
BC = 160 x 5.20x10^-6 = 8.32 x 10^-4 siemens

Expressed in per unit of Zc (250 Ω)
XL = 52/250 = 0.208 pu
BC = 8.32x10^-4 x 250 = 0.208 pu

Two type of cables are included; direct-buried paper insulated lead covered (PILC) and high pressure pipe type (PIPE).
















For example the transmission line between Power Plant to Sub Station is 134.84 km; Rating 2.07 kA; R= 4.24 ohm; XL = 38.81 ohm; Bc = 554.2 x 10^-6 S.

XL/km = 38.81/134.82 = 0.273 ohm/km
XL = ω.L
L = (0.273 ohm/km) / (2.π.50Hz) = 8.69 x 10^-4 ohm/km

Bc/km = 4.11 x 10^-6 s/km
Bc = ω.C
C = 0.013 x 10^-6 s/km

Zc = (L/C)^0.5
Zc = 257.66 ohm

SIL = V^2 / Zc
SIL = 500^2 MV / 257.66 ohm
SIL = 970.27 MW

Safety factor 5%
SIL = 970 - (970x0.05) = 922 MW

So 922MW electrical power can be supliied as long 134.84 km with 500kV without voltage losses.
V = VR.(e)^Yx
V = VR.(e)^jω(L.C)^0.5
V = VR.(e)^ 2 x 2.14 x 50Hz x (8.69.10^-4 x 0.013.10^-6)^0.5
V = VR.1
V = VR