Substation earthing
Earth potential rise from earth fault current (split / reduction factor)
- where
- earth-return current through the soil (kA in the box, shown in A)
- earth-return (split or reduction) factor
- total earth fault current 3I₀ (kA)
- earth potential rise (V)
- earthing system impedance (Ω)
Explain this equation
Gives the earth potential rise of a substation or electrode during an earth fault: the voltage of the whole earthing system relative to remote earth. It is the starting point for HOT-site, touch voltage and transfer assessments.
IF is the total earth fault current (3I₀) at the site. Only part of it returns through the soil: cable sheaths, overhead earth wires and local neutrals carry the rest back to the source. r is the fraction that does flow into the soil, so IE is the current that actually raises the electrode. ZE is the impedance of the earthing system to remote earth, including any parallel paths such as cable sheaths and tower lines.
It is Ohm's law applied to the earth electrode. The reduction factor sits in front because the metallic return paths are in parallel with the soil, and a well-coupled return conductor takes most of the current.
Overhead lines with an earth wire typically give r of 0.3 to 0.7; sheathed cables bonded at both ends often 0.05 to 0.3. IEEE 80 Example 1: 3180 A × 0.6 × 2.78 Ω ≈ 5.3 kV. Halving r halves the EPR.
Watch out: r and ZE are complex, and their magnitudes are used here. That is normally conservative, but a broken or single-point-bonded sheath can push r close to 1.
Works out the earth potential rise of a substation or earth electrode from the part of the earth fault current that actually returns through the soil.
Use the split or reduction factor to remove the share carried back by cable sheaths, overhead earth wires or other metallic paths.
Enter your values to see the result.
Good to know
ENA EREC S34 writes IE = rE × (IF − IN), where rE is the reduction factor of the line or cable and IN is any current returning through local neutrals; IEEE 80 writes IG = Df × Sf × 3I0 with split factor Sf and decrement factor Df. Typical earth-return fractions: overhead lines with an earth wire 0.3–0.7; cables with sheaths bonded at both ends often 0.05–0.3 (sheaths can carry 70–95 % of the fault current). EREC S34 Case study 4: IE = 1.5 kA, RE = 0.5 Ω gives EPR = 750 V.
Common mistake
Multiplying the full fault current by the electrode resistance. This overstates EPR, often by a factor of 3 to 10 on cable-fed sites. The opposite error is assuming the sheath return without checking that the sheath is continuous and bonded at both ends.
Limits
Simple scalar product. Reduction factors and ZE are really complex quantities; this uses magnitudes, which is normally conservative. To find r, use EREC S34 Appendix E (overhead lines), the S34 C-factor method (cables) or BS EN 50522 Annex I/J. Apply Df (IEEE 80) yourself if DC offset matters.
Sources
For competent high-voltage engineers. Results are calculations from the published methods shown and depend entirely on the data you enter. They are not a substitute for a system study, manufacturer data, the applicable standards and Engineering Recommendations, or your organisation’s safety rules. Never use a result to justify working on or near live HV equipment.