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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 cross sectional area is given by:

$A= I 2 t k$

where: A - the nominal cross section area, mm2
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:

$t= k 2 A 2 I 2$

The factor k is dependant on the cable insulation, allowable temperature rise under fault conditions, conductor resistivity and heat capacity.  Typical values of 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 mm2
* 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 cross sectional area of a copper thermosetting 90°C cable (k=143) is:

$S= 13600 2 ×2.6 143 =154 mm 2$

Any selected cable larger than this will withstand the fault.

### Derivation  - Adiabatic Equation and 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).  Obviously this is not fully true, but it is on the safe side.

From physics, the heat Q, required to rise a material ΔT is given by:

$Q=c m ΔT$

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:

$Q= I 2 R t$

where R - the resistance of the cable, Ω

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

$m= ρ c A l$   and     $R= ρ r l A$

where ρc - material density in g.mm-3
ρr - resistivity of the conductor, Ω.mm
l - length of the cable, mm

Combining and substituting we have:

$I 2 Rt=cm ΔT$

$I 2 t ρ r l A =c ρ c A l ΔT$

and rearranging for A gives:

$S= I 2 t k$   by letting    $k= c ρ c ΔT ρ r$

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

$ΔT= θ f − θ i$

where θf - final (maximum) cable insulation temperature, °C
θi - initial (operating) cable insulation temperature, °C

Units:  are expressed in g (grams) and mm2, 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.