The Adiabatic Equation and k Factor for Cable Thermal Withstand

When calculating the fault rating of a cable, it is generally assumed that the fault duration is short enough that no heat is dissipated from the cable to its surroundings. This adiabatic assumption simplifies the calculation and is normally conservative.

The adiabatic equation is used to check whether a cable conductor has enough cross-sectional area to withstand a short-circuit current for the protective device operating time. The factor k connects that equation to conductor material, insulation type and permissible temperature rise.

Adiabatic equation

For a fault current I lasting for time t, the minimum required conductor cross-sectional area is:

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

A	Nominal conductor cross-sectional area, mm2
I	Fault current, A
t	Duration of fault current, s
k	Factor dependent on conductor material, insulation type and temperature limits

Alternatively, if the cable cross-sectional area and fault current are known, the maximum allowable disconnection time is:

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

Typical k values

The factor k depends on the cable insulation, allowable temperature rise during the fault, conductor resistivity and conductor heat capacity. Typical values are shown below.

Insulation type	Initial temperature, °C	Final temperature, °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, EPR)	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

Calculating k from material properties

IEC 60364-5-54 gives the following formula for calculating k:

$$k=\sqrt{\frac{Q_c(\beta +20)}{{\rho }_{20}}\ln \left(\frac{\beta +{\theta }_f}{\beta +{\theta }_i}\right)}$$

Qc	Volumetric heat capacity of the conductor at 20 °C, J·K-1·mm-3
β	Reciprocal of the temperature coefficient of resistivity at 0 °C
ρ20	Electrical resistivity of conductor material at 20 °C, Ω·mm
θi	Initial conductor temperature, °C
θf	Final conductor temperature, °C

Strictly speaking, β is specified at 0 °C. In myCableEngineering calculations, β is determined by taking the reciprocal of the 20 °C temperature coefficient. This introduces a small but negligible error.

Material constants

Conductor material	β, °C	Qc, J·K-1·mm-3	ρ20, Ω·mm
Copper	234.5	3.45 × 10-3	17.241 × 10-6
Aluminium	228	2.5 × 10-3	28.267 × 10-6
Steel	202	3.8 × 10-3	138 × 10-6

Simplified conductor equations

Substituting the values above and rearranging the IEC equation gives the following practical forms.

Copper conductors:

$$k=226\sqrt{\ln \left(1+\frac{{\theta }_f-{\theta }_i}{234.5+{\theta }_i}\right)}$$

Aluminium conductors:

$$k=148\sqrt{\ln \left(1+\frac{{\theta }_f-{\theta }_i}{228+{\theta }_i}\right)}$$

Steel conductors:

$$k=78\sqrt{\ln \left(1+\frac{{\theta }_f-{\theta }_i}{202+{\theta }_i}\right)}$$

Example

Consider a maximum fault current of 13.6 kA where the protective device trips in 2.6 s. For a copper thermosetting 90 °C cable, use k = 143. The minimum safe cable cross-sectional area is:

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

Any selected conductor larger than this value will withstand the fault on the basis of the adiabatic calculation.

Derivation of the adiabatic equation

The term adiabatic describes a process with no heat transfer. For cable faults, this assumes that all heat generated during the fault is retained in the conductor rather than dissipated away. This is not completely true in practice, but it is generally conservative.

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

$$Q=cm\Delta T$$

Q	Heat added, J
c	Specific heat constant of the 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:

$$Q=I^{2}Rt$$

From the physical properties of the cable conductor:

$$m={\rho }_{c}Al$$

and

$$R=\frac{{\rho }_{r}l}{A}$$

ρc	Material density, g/mm3
ρr	Resistivity of the conductor, Ω·mm
l	Cable length, mm

Combining these terms:

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

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

Rearranging for area gives:

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

where

$$k=\sqrt{\frac{c{\rho }_c\Delta T}{{\rho }_r}}$$

The maximum allowable temperature rise is:

$$\Delta T={\theta }_f-{\theta }_i$$

θf	Final, or maximum, cable insulation temperature, °C
θi	Initial operating cable insulation temperature, °C

The equations above use grams and square millimetres rather than kilograms and metres. This convention is widely used by cable specifiers; the equations can be reformulated in SI base units if required.