Calculating Cable Fault Ratings

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When selecting a cable, the performance of the cable under fault conditions is an important consideration. It is important that calculations be carried out to ensure that any cable is able to withstand the effects of any potential fault or short circuit.  This note looks at how to do this.

The primary concern with cables under a fault condition is the heat generated, and any potential negative effect this may have on the cable insulation.

Calculation of fault rating is based on the principle that the protective device will isolate the fault in a time limit such that the permitted temperature rise within the cable will not be exceeded.

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:

$A= I 2 t k$

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:

Value of k
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

Tip: for a better understanding of cable insulation and how it is categorised, refer to our Cable Insulation Properties note.

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:

$Q=c m ΔT$

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:

$Q= I 2 R t$

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     $R= ρ r l A$

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:

$ΔT= θ f − θ i$

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.

Obtaining Values of k

The constant k can be calculated from the above equation.

A more common approach for is to use the tabulated values for k, for example, from BS 7671[1].

IEC 60364-5-54[2] also allows a more direct calculation of k, using:

$k= Q c ( β+20 ) ρ 20 ln( β+ θ f β+ θ i )$

Qc - volumetric heat capacity of conductor at 20°C, J.K-1.mm-3

β - reciprocal of temperature coefficient of resistivity at 0°C, °C

ρ20 - resistivity of conductor as 20°C, Ω.mm

θi - initial conductor temperature, °C

θf - final conductor temperature, °C

β [°C]    Qc [J.K-1.mm-3 ρ20 [Ω.mm]
Copper 234.5 3.45 x 10-3
17.241 x 10-6
Aluminium 228 2.5 x 10-3 28.267 x 10-6
Steel 202 3.8 x 10-3 138 x 10-6

Substituting the above values and rearranging the IEC equation slightly, gives:

$k=226 ln( 1+ θ f − θ i 234.5+ θ i )$- copper conductors

$k=148 ln( 1+ θ f − θ i 228+ θ i )$- aluminium conductors

$k=78 ln( 1+ θ f − θ i 202+ θ i )$- steel

As mentioned, the adiabatic equation assumes no heat is dissipated from the cable during a fault.  While putting the calculation on the safe side, in some situations, particularly for longer fault duration there is the potential to be able to get away with a smaller cross section.  In these instances, it is possible to do a more accurate calculation.

Considering non-adiabatic effects is more complex.  Unless there is some driver, using the adiabatic equations is just easier.  Software is available to consider non adiabatic effects, however, there is a cost, time and complexity associated with this.

The IEC also publish a standard which deals with non-adiabatic equations:

• IEC 60949 "Calculation of thermally permissible short-circuit current, taking into account non-adiabatic heating effects".

The method adopted by IEC 60949 is to use the adiabatic equation and apply a factor to cater for the non-adiabatic effects:

$I=ε I AD$

I - permissible short circuit current, A (or kA)

ε - factor to allow for heat dissipation from cable

The bulk of the IEC 60949 standard is concerned with the calculation of ε.

Other Cable Fault Issues

In addition to the direct heating effect of fault currents, other considerations include:

• electro-mechanical stress and fault levels large enough to cause cable failure
• performance of joint and terminations under fault conditions

While in most cases the none heating effects are not serious, there may be situations where these could pose a risk to the cable or equipment/personnel in the vicinity.

References

• [1]. BS 7671 - Requirements for Electrical Installations. 17th ed. United Kingdom: IEE; 2008.
• [2] IEC 60364-5-54 Low-voltage electrical installations - Part 5-54: Selection and erection of electrical equipment - Earthing arrangements and protective conductors. 3rd ed. IEC; 2011.

More interesting Notes:

Steven has over twenty five years experience working on some of the largest construction projects. He has a deep technical understanding of electrical engineering and is keen to share this knowledge. About the author

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