A line-compensation arrangement may use three phase reactors and an additional reactor in their neutral path. Its behavior depends on the phase connections and the connected line. The neutral branch is not simply a fourth phase reactor with an interchangeable rating. [1]
Draw the phase bank, neutral branch and earth connection. For the basic star-connected arrangement, define each phase impedance Zp and neutral impedance Zn. Under zero-sequence current, the neutral carries three times the per-phase zero-sequence current.
Separate the sequence paths
| Sequence or state | Basic path | Required extension |
|---|---|---|
| Balanced positive sequence | Phase impedances; ideal neutral current zero | Actual phase asymmetry and mutual effects |
| Zero sequence | Z₀ = Zp + 3Zn in the simple star model | Grounding, coupling and line capacitance |
| Single pole open | Unbalanced connected network | Phase-domain or complete sequence interconnection |
| Fault recovery | Time-dependent voltage/current | Switching, arc and controls if relevant |
The Z₀ expression uses impedances referred to each phase and assumes an uncoupled symmetric bank. A coupled magnetic construction or more complex neutral network needs the appropriate impedance matrix. Do not use the simple expression without its connection assumptions.
Include the line in the compensation problem
During a single-pole-open interval, healthy phases can feed the disconnected phase through electric and magnetic coupling. Reactor compensation changes this network but cannot be assessed from the bank alone. Line length, transposition, capacitances and mutual impedances belong in the model.
In the simple star model, KCL gives In = 3I₀. Each phase therefore sees I₀Zp + 3I₀Zn, giving Z₀ = Zp + 3Zn. [2] This bank relation does not select optimum line compensation.
Apply the neutral factor on one sequence basis
Assumptions: an uncoupled symmetric star bank at 50 Hz has Xp = 100 Ω and Xn = 20 Ω. At a prescribed 16 kV RMS per-phase zero-sequence voltage, X₀ = 160 Ω gives I₀ = 100 A and neutral current 300 A, exceeding an assumed 250 A neutral rating by 50 A. A project bank target of X₀ = 200 Ω would require Xn = (200−100)/3 = 33.33 Ω, giving I₀ = 80 A and neutral current 240 A. The 20 Ω selection fails this screen. The 33.33 Ω candidate still needs the complete open-pole line, recovery-voltage and arc study. [2]
References
- X. Zhang et al., Compensation scheme for secondary arc current on four-circuit parallel transmission lines, IET Generation, Transmission & Distribution 10, 2079–2086, 2016. DOI: 10.1049/iet-gtd.2015.1102.
- U.S. Department of Energy, DOE-HDBK-1011/3-92, Electrical Science, Volume 3, June 1992.