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Transmission line transposition

transmission line impedance calculation, wires transposition, electromagnetic FEA finite element analysis of the transmission line

The ACSR conductor 110 kV transmission line features complete transposition cycle at length of 120 kilometers.

Problem type:
Plane problem of AC magnetics.

Geometry:

Transmission line pole. All dimensions are in meters.

Aluminum Conductor Steel Reinforced (ACSR)

Scheme of transposition.
Transmission line length l = 120 km

Given:
Line rated voltage U_line = 110 kV
R_load = 100 Ohm, L_load = 0.23 H.

Problem:
Define transmission line phase inductance.

Solution:
At longer power lines wires are transposed according to the transposing scheme. Each of the three conductors must hang once at each position of the overhead line. On transmission lines of 110-500 kV rate one complete cycle of a transposition should be carried out if the line length exceeds 100 km. The transposition should be carried out so that total lengths of transmission line parts were approximately equal.
The length of our line is 120 km. The distance between transposition points is 40 km.

The simulation model includes field and circuit parts. There are three encapsulated field parts connected one with another by the external circuit. Each conductor consists of aluminum and steel wires:

Full impedance of a line develops of separate parts impedances and can be found as total voltage drop divided by a current:
Z_line = (U1 + U2 + U3) / I.

Impedance of a line can be presented as the sum of active resistance (R_line) and inductive reactance (X_line):
Z_line = R_line + jX_line.

Line inductance can be calculated as:

L = X_line / 2 π f,
where X_line - line phase inductive reactance;
f - frequency of the alternating current.

Results:
Table of the measured currents and voltages for a phase conductors

	Part 1	Part 2	Part 3	Total
Voltage UA, V	5676 + j2832	5943 + j2800	6142 + j2366	17761 + j7998
Voltage UB, V	-5078 + j4213	-5292 + j3501	-5438 + j3671	-15808 + j11385
Voltage UC, V	-547 - j6546	-1111 - j6494	-297 - j6331	-1955 + j19371
Current IA, A	263 - j270			263 - j270
Current IB, A	102 + j376			102 + j376
Current IC, A	-363 - j93			-365 - j93
Resistance ZA, Ohm				17.7 + j48.6
Resistance ZB, Ohm				17.7 + j48.6
Resistance ZC, Ohm				17.7 + j48.6

Inductance of a phase : L_phase = L_A = L_B = L_C = 0.155 H (total length 120 km).