When calculating the fault ratings of a cable, it is generally assumed that the duration so short enough that no heat is dissipated by the cable to the surrounding. Adopting this approach simplifies the calculation and errs on the safe side.

The normally used equation is the so-called adiabatic equation. For a given fault of *I*, which lasts for time *t*, the minimum required cable

$A=\frac{\sqrt{{I}^{2}t}}{k}$

*where:* A - the nominal cross section area, mm^{2}

*I* - the fault current in, A

t - duration of fault current, s

k - a factor dependant on cable type (see below)

Alternatively, given the cable cross section and fault current, the maximum time allowable for the protective device can be found from:

$t=\frac{{k}^{2}{A}^{2}}{{I}^{2}}$

The factor *k* is *k* are:

Temperature | Conductor Material | ||||
---|---|---|---|---|---|

Initial °C | Final°C | Copper | Aluminium | Steel | |

Thermoplastic 70°C (PVC) |
70 |
160/140 |
115/103 |
76/78 |
42/37 |

Thermoplastic 90°C (PVC) |
90 |
160/140 |
100/86 |
66/57 |
36/31 |

Thermosetting, 90°C (XLPE, EDR) |
90 |
250 |
143 |
94 |
52 |

Thermosetting, 60°C (rubber) |
60 |
200 |
141 |
93 |
51 |

Thermosetting, 85°C (rubber) |
85 |
220 |
134 |
89 |
48 |

Thermosetting, 185°C (silicone rubber) |
180 |
350 |
132 |
87 |
47 |

*where two values; lower value applied to conductor CSA > 300 mm^{2 }

* these values are suitable for durations up to 5 seconds, source: BS 7671, IEC 60364-5-54

### Example

Consider a maximum fault current of 13.6 kA and the protective device trips in 2.6 s. The minimum safe cable *k*=143) is:

$S=\frac{\sqrt{{13600}^{2}\times 2.6}}{143}=154{\text{mm}}^{2}$

Any selected cable larger than this will withstand the fault.

**Derivation - Adiabatic Equation and ***k*

*k*

The term adiabatic applies to a process where there is no heat transfer. For cable faults, we are assuming that all the heat generated during the fault is contained within the cable (and not transmitted away).

From physics, the heat *Q*, required to *ΔT* is given by:

$Q=c\text{}m\text{}\Delta T$

*where * *Q* - heat added, J

*c* - specific heat constant of material, J.g^{-1}.K^{-1}

*m* - mass of the material, g

*ΔT* - temperature rise, K

The energy into the cable during a fault is given by:

$Q={I}^{2}R\text{}t$

*where* *R* - the resistance of the cable, Ω

From the physical cable properties we can calculate *m* and R as:

$m={\rho}_{c}A\text{}l$ and $R=\frac{{\rho}_{r}l}{A}$

*where ρ _{c}* - material density in g.mm

^{-3}

*ρ*- resistivity of the conductor, Ω.mm

_{r}*l*- length of the cable, mm

Combining and substituting we have:

${I}^{2}Rt=cm\text{}\Delta T$

${I}^{2}t\frac{{\rho}_{r}l}{A}=c\text{}{\rho}_{c}A\text{}l\text{}\Delta T$

and rearranging for A gives:

$S=\frac{\sqrt{{I}^{2}t}}{k}$ by letting $k=\sqrt{\frac{c{\rho}_{c}\Delta T}{{\rho}_{r}}}$

Note: *ΔT is the maximum allowable temperature rise for the cable:*

* $\Delta T={\theta}_{f}-{\theta}_{i}$*

*where θ _{f}* - final (maximum) cable insulation temperature, °C

*θ*- initial (operating) cable insulation temperature, °C

_{i}**Units:** are expressed in g (grams) and mm^{2}, as opposed to kG and m. This is widely adopted by cable specifiers. The equations can easily be redone in kG and m if required.