White Paper By the Author John Horak of Basler Electric

Abstract - This paper reviews the basic equations for the sequence impedance of transmission lines, including ground loop current flow, how neutral wires are included in the equations, and how this impedance is transformed from an ABC domain impedance to the 012 domain impedance. A side benefit of the approach taken is that the paper shows how one calculates the sequence impedances of untransposed power lines, including calculation of the off-diagonal (mutual) elements of the sequence component 012 domain impedances. The paper also addresses the calculation of mutual impedances between two parallel lines.

I. INTRODUCTION
The paper begins by analyzing system impedances in the ABC (physical or phase) domain, first without any overhead ground wires, and then shows how the overhead ground conductors are incorporated into the analysis. Then the paper shows the translation of these ABC domain impedances to the 012 (sequence) domain, which provides the zero sequence impedances, and other sequence impedances. Thereafter, the paper discusses the calculation of mutual impedance coupling between two parallel lines.

II. BASIC ABC DOMAIN IMPEDANCE CONCEPTS
In Fig. 1 there are three phase current loops, A, B, and C, each passing through a common neutral/ground. For this initial investigation, there is no independent metallic neutral/ground conductor, so the phase currents sum together and return through the earth in the current named IG. The phase A loop is shown as a dotted line.

Each phase loop in Fig. 1 has different impedance; each loop is defined by a different current path, a different loop cross section area and, when magnetic core material is involved, a different permeability of the material through which the flux passes. To keep the drawing from becoming exceedingly complex, only the main representative flux loops are shown and only for phase

A. One’s imagination should be used to fill in the blanks. An inspection of Fig. 1 shows that we have 3 current loops, but 4 currents are shown on the diagram. This means we can actually only set up 3 voltage drop equations, and we will need to take advantage of IG = IA + IB + IC to remove IG from directly being part of the voltage drop equations.