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Minimum cross-section of protective conductor is determined by the adiabatic equation. S = √(I²·t) / k The equation is calculated using: fault current (I), tripping time (t), and a coefficient k depending on the conductive material/insulation. The equation is valid only for t ≤ 5 s.

S = √(I²·t) / k [mm²] (t ≤ 5 s)
Example calculation: Fault current 6,000 A, tripping time 0.4 s, copper-PVC (k=143) → S = √(6000²×0.4)/143 ≈ 26.5 mm² → nearest standard 35 mm².

Protection/Grounding Conductor Cross-Section (Adiabatic)

Calculate the minimum protective conductor cross-section required for a given fault current and tripping time using an adiabatic equation.

Calculated S_min
Proposed standard cross-section
Notes: S = √(I²·t)/k (TS EN 60364-5-54, Annex A). The adiabatic equation is just t ≤ 5 s This applies to; the k value depends on the conductor material and insulation. An alternative simple rule: protective conductor ≥ half of the phase cross-section. The final choice requires engineering verification.

How is it calculated?

  1. Enter the fault current (A) and tripping time (s).
  2. Select the k value according to the conductive material/insulation.
  3. Press Calculate; the minimum cross-section and the closest standard cross-section will be displayed.

Frequently Asked Questions

How to determine the cross-section of a grounding conductor?
The adiabatic equation S=√(I²·t)/k, or as a simple rule, using the approximation that the protective conductor cross-section is ≥ half of the phase cross-section.
What is the value of k?
It is a coefficient that depends on the conductor material and insulation (TS EN 60364-5-54 Appendix A): e.g., copper-PVC 143, copper-XLPE 176, aluminum-PVC 95.
Does the adiabatic equation always hold true?
No, this only applies when the opening time t ≤ 5 seconds; longer times require the full method.

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