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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-parallel problem of AC magnetics.

Geometry
Transmission line pole Tangent tower design, 110kV A B C Soil 2 m 3.5 m 2 m 3 m 14.5 m 3 m

Aluminum Conductor Steel Reinforced (ACSR)
Aluminum Conductor Steel Reinforced (ACSR) ACSR conductor for HV transmission line Steel Aluminum

Scheme of transposition, transmission line length l = 120 km:
Transmission line full transposition Transmission line transposition scheme A B C C A B B C A 40 km 40 km 40 km

Given
Line rated voltage Uline = 110 kV
Rload = 100 Ohm, Lload = 0.23 H.
Line to phase voltage Three phase system line and phase voltages A B C Uphase Uline Uphase =Uline / √3

Task
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:

transmission line simulation model

transmission line electric circuit

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

Impedance of a line can be presented as the sum of active resistance (Rline) and inductive reactance (Xline):
Zline = Rline + jXline.

Line inductance can be calculated as:

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

Results
Table of the measured currents and voltages for 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 conductor : Lphase = LA = LB = LC = 0.155 H (total length 120 km).

Transmission line magnetic field
3 phase transmission line magnetic field lines

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