Adiabatic Equation
S = √(I²t) / k
Calculate the minimum cross-sectional area a protective conductor needs to survive a fault, from the fault current, the disconnection time and the material/insulation k factor.
Reference
BS 7671 Regulation 543.1.3; k from Table 54.3
How this works
Regulation 543.1.3 lets you calculate a protective conductor's minimum cross-sectional area rather than take it from Table 54.7. The adiabatic equation is:
S = √(I²t) / k
- S — minimum cross-sectional area, in mm²
- I — the fault current, in amperes
- t — the protective device's operating time at that fault current, in seconds
- k — a factor for the conductor material and its insulation, from Table 54.2 to 54.6
"Adiabatic" means the calculation assumes all the heat generated during the fault stays in the conductor, with none escaping to the surroundings. That is pessimistic, and deliberately so — it holds for the short durations involved and errs on the side of a larger conductor.
The equation is only valid for disconnection times up to 5 seconds. Beyond that the assumption breaks down and the calculation no longer applies.
k values used here
| Conductor and insulation | k |
|---|---|
| Copper, 70°C thermoplastic | 115 |
| Copper, 90°C thermoplastic | 100 |
| Copper, 90°C thermosetting | 143 |
| Aluminium, 70°C thermoplastic | 76 |
| Aluminium, 90°C thermoplastic | 66 |
| Aluminium, 90°C thermosetting | 94 |
Values for a protective conductor incorporated in a cable or bunched with cables, BS 7671 Table 54.3. Other tables apply where the conductor is separate from, or forms a sheath around, the cable.
Limitations
This calculator is a revision and cross-checking aid. It does not replace the published tables in BS 7671, the specific requirements of the installation in front of you, or the judgement of a competent person. Always verify against the current edition of the standard before relying on a result.