# Adiabatic Short Circuit Temperature Rise

Adiabatic short circuit temperature rise normally refers to the temperature rise in a cable due to a short circuit current. During a short circuit, a high amount of current can flow through a cable for a short time. This surge in current flow causes a temperature rise within the cable.

## Derivation

An “adiabatic” process is a thermodynamic process in which there is no heat transfer. In the context of cables experiencing a short circuit, this means that the energy from the short circuit current contributes only to raising the temperature of the cable conductor (e.g. copper) without any heat loss through the cable (i.e. through resistive effects). This is obviously a simplifying assumption as in reality, there would be heat lost during a short circuit, but it is a conservative one and yields a theoretical temperature rise higher than that found in practice.

The derivation of the short circuit temperature rise is based on a simple application of specific heat capacity. The specific heat capacity of a body (for instance the solid conductor in a cable) is the amount of energy required to change the temperature of the body, and is given by the following basic formula:

$c_{p} = \frac{E}{m \Delta T} \,$

Where $E \,$ is the energy dissipated by the body (in Joules), $m \,$ is the mass of the body (in grams), and $\Delta T \,$ is the change in temperature (in Kelvins).

The energy from a current flowing through a cable is based on the SI definition for electrical energy:

$E = QV = i^{2}Rt \,$

Where $i \,$ is the current (in Amps), $R \,$ is the resistance of the body which the current is flowing through (in Ω) and $t \,$ is the duration of the current flow (in seconds).

The mass and resistance of an arbitrary conductive body is proportional to the dimensions of the body and can be described in general terms by the following pair of equations:

$m = \rho_{d} Al \,$
$R = \frac{\rho_{r} l}{A} \,$

Where $\rho_{d} \,$ is the density of the body (in g $mm^{-3}$), $\rho_{r} \,$ is the resistivity of the body (in Ω mm), $A \,$ is the cross-sectional area of the body (in $mm^{2}$) and $l \,$ is the length of the body (in mm).

Putting all of these equations together and re-arranging, we get the final result for adiabatic short circuit temperature rise:

$\Delta T = \frac{i^{2}t \rho_{r}}{A^{2} c_{p} \rho_{d}} \,$

Alternatively, we can re-write this to find the cable conductor cross-sectional area required to dissipate a short circuit current for a given temperature rise is:

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

With the constant term $k = \sqrt{\frac{c_{p} \rho_{d} \Delta T}{\rho_{r}}} \,$

Where $A \,$ is the minimum cross-sectional area of the cable conductor ($mm^{2}$)

$i^{2}t \,$ is the energy of the short circuit ($A^{2}s$)
$c_{p} \,$ is the specific heat capacity of the cable conductor ($J g^{-1}K^{-1}$)
$\rho_{d} \,$ is the density of the cable conductor material ($g mm_{-3}$)
$\rho_{r} \,$ is the resistivity of the cable conductor material ($\Omega mm$)
$\Delta T \,$ is the maximum temperature rise allowed (°$K$)

In practice, it is common to use an $i^{2}t$ value that corresponds to the let-through energy of the cable's upstream protective device (i.e. circuit breaker or fuse). The manufacturer of the protective device will provide let-through energies for different prospective fault currents.

### Worked Example

This example is illustrative and is only intended to show how the equations derived above are applied. In practice, the IEC method outlined below should be used.

Suppose a short circuit with let-through energy of $1.6 x 10^{7}$ occurs on a cable with a copper conductor and PVC insulation. Prior to the short circuit, the cable was operating a temperature of 75°C. The temperature limit for PVC insulation is 160°C and therefore the maximum temperature rise is 85°K.

The specific heat capacity of copper at 25°C is $c_{p}$ = 0.385 $J g^{-1}K^{-1}$. The density of copper is $\rho_{d}$ = 0.00894 $g mm^{-3}$ and the resistivity of copper at 75°C is $\rho_{r}$ = 0.0000204 $\Omega mm$.

The constant is calculated as k = 119.74. The minimum cable conductor cross-sectional area is calculated as 33.4 $mm^{2}$. it should be stressed that the calculated value of k is probably inaccurate because the specific heat capacity of copper is subject to change at different temperatures.

## Effects of Short Circuit Temperature Rise

High temperatures can trigger unwanted reactions in the cable insulation, sheath materials and other components, which can prematurely degrade the condition of the cable. Cables with larger cross-sectional areas are less sensitive to short circuit temperature rises as the larger amount of conductor material prevents excessive temperature rises.

The maximum allowable short circuit temperature rise depends on the type of insulation (and other materials) used in the construction of the cable. The cable manufacturer will provide specific details on the maximum temperature of the cable for different types of insulation materials, but typically, the maximum temperatures are 160°C for PVC and 250°C for EPR and XLPE insulated cables.

## Treatment by International Standards

IEC 60364 (Low voltage electrical installations) contains guidance on the sizing of cables with respect to adiabatic short circuit temperature rise. The minimum cable conductor cross-sectional area is given by the following equation:

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

Where $A \,$ is the minimum cross-sectional area of the cable conductor ($mm^{2}$)

$i^{2}t \,$ is the energy of the short circuit ($A^{2}s$)
$k \,$ is a constant that can be calculated from IEC 60364-5-54 Annex A. For example, for copper conductors:
$k = 226 \sqrt{\ln{\left(1 + \frac{\theta_{f}-\theta_{i}}{234.5+\theta_{i}}\right)}} \,$

Where $\theta_{i} \,$ and $\theta_{f} \,$ are the initial and final conductor temperatures respectively.

The National Electricity Code (NEC) does not have any specific provisions for short circuit temperature rise.