The adiabatic equation in Regulation 543.1.3 of BS 7671:2018+A4:2026
By: Michael Peace CEng MIET
The term ‘adiabatic’ means without heat transfer to the surroundings. The adiabatic equation assumes that all of the heat that is generated by a fault current is absorbed by the conductor and no heat is dissipated to its surroundings. This is because the duration of a fault is very short, typically less than one second, therefore no heat is dissipated.
The purpose of the adiabatic equation in Regulation 543.1.3 of BS 7671:2018+A4:2026 (BS 7671) is to size a conductor so that damage to the conductor due to the temperature rise will not occur.
The adiabatic equation can seem a bit daunting at first. However, when the values required are understood, it is simply a case of entering those values into a calculator to determine the cross-sectional area (CSA) of a protective conductor.
There are two methods of selecting the size of protective conductor, selection and calculation. However, this article focuses on calculation. The method of selection is often used as it is easier than performing calculations, however, determining the minimum CSA by calculation often results in a more economical design. This article explains step by step how to perform the calculation.
What is the adiabatic equation?
Regulation 543.1.3 of BS 7671 provides a method of calculating the CSA of a protective conductor using the adiabatic equation (Equation 1). The purpose of the adiabatic equation is to check the thermal withstand of the cable during fault conditions which takes into account factors such as fault current, operating time of the protective device and k values. It provides a minimum required CSA of the protective conductor.
How do I choose the CSA of a protective conductor?
There are two methods when it comes to choosing the CSA of a protective conductor provided in Regulation 543.1.1 of BS 7671. It is important to note that Regulation 543.1.1 of BS 7671 excludes sizing of protective bonding conductors by calculation. This is because protective bonding conductors are not designed to clear an earth fault.
The first method is selection, according to the size of the line conductor. Table 54.7 of BS 7671 shown below, provides a method of selecting the minimum CSA of the protective conductor according to the CSA of the line conductor. The first column is for the CSA of the line conductor, the second column is where the protective conductor is of the same material as the line conductor and the third column is where the protective conductor is not of the same material as the line conductor.
Figure 1: Table 54.7 of BS 7671
The other method of determining the required size of protective conductor for that part of the installation is by calculation using the adiabatic equation that is included in Regulation 543.1.3 of BS 7671 and shown below in Equation 1.
Equation 1: The adiabatic equation from Regulation 543.1.3 of BS 7671
where:
S is the nominal CSA of the conductor in mm2
I is the value in amperes (RMS for AC) of fault current for a fault of negligible impedance, which can flow through the associated protective device, due account being taken of the current limiting effect of the circuit impedances and the limiting capability (I2t) of that protective device
t is the operating time of the protective device in seconds corresponding to the fault current I amperes
k is a factor taking account of the resistivity, temperature coefficient and heat capacity of the conductor material, and the appropriate initial and final temperatures.
How do I determine I?
In the adiabatic equation, I represents the maximum fault current (If) that can flow through the associated protective device. The prospective fault current can be calculated or measured.
The value of I Is determined by dividing the voltage by the earth fault loop impedance value at the end of the circuit (Zs).
Regulation 533.3 of BS 7671 requires minimum and maximum fault current conditions to be taken into account, this is so that the highest energy let-through is taken into account.
Therefore, it may be necessary to evaluate the prospective earth fault current at the distribution board and at the end of the circuit because the impedance will be higher at the end of the circuit. The energy let-through is electrical and thermal energy passed through a protective device to a circuit during the time it takes to disconnect a fault.
Circuit-breakers feature a thermal component for overloads and a magnetic component to protect against short-circuits. The time current curve is made up of two parts. The magnetic part is represented by the vertical line and the curved line indicates the thermal operation as shown in Figure 3.
When calculating the maximum fault current, it is important to use the correct voltage factor. Generally, Cmin (0.95) is used for fuses and Cmax (1.1) is used for a circuit-breaker when the fault is interrupted in the magnetic part of the device.
The reason that Cmin is used for fuses is the disconnection time is greater, therefore the total energy let-through is higher. This is because within a range of fault currents, the magnetic part of a circuit-breaker disconnects within the same time, whereas the disconnection time of a fuse is variable throughout the time current curve.
Ohm’s law says that the lower the voltage, the lower the fault current which can mean a longer disconnection time. For example, where the earth fault path is 1 ohm and the voltage is 200 V, the earth fault current is 200 A. If the voltage is 100 V, the fault current is 100 A which will take longer to disconnect. For example, a 32 A BS 88-3 fuse will disconnect in 1 second for a 200 A fault current. For a 100 A fault current, it will take 30 seconds to disconnect.
The longer the disconnection time the greater the thermal stress to the cable because more energy will be delivered. Therefore, the minimum supply voltage must be considered and the voltage factor Cmin must be applied to determine the fault current and this value is used to determine t. The earth fault loop impedance in this example is 1.56 Ω.
How do I determine value of t?
The value of t is often confused with the maximum disconnection time. It is not. The value of t is the actual operating time of the protective device in seconds corresponding to the fault current I amperes.
The value of t can be determined from the time/current characteristics of overcurrent protective devices in Appendix 3 of BS 7671, using the value of I calculated.
Figure 2 shows that for a fault current of 140 amps, a 32 A BS 88-3 fuse will operate in 5 seconds. Therefore, the value of t in this example is 5 seconds.
Figure 2: Disconnection time for a 32 A BS 88-3 fuse (Fig 3A1 of BS 7671)
Disconnection time = 5 seconds
What are ‘k’ values?
We are familiar with using values such as current and time in calculations but the k value can cause confusion. The ‘k’ value “is a factor taking account of the resistivity, temperature coefficient and heat capacity of the conductor material, and the appropriate initial and final temperatures’”.
The construction and material used for the protective conductor, will decide the k factor. Table 54. 2, Table 54.3, Table 54.4, Table 54.5 and Table 54.6 of BS 7671 provide k values for different types of protective conductors and materials.
What value of k should be used?
For protective conductors, Table 54.2, Table 54.3, Table 54.4, Table 54.5 and Table 54.6 of BS 7671 provide values of k for different types of protective conductors. For live conductors, Table 43.1 provides values of k for common materials, for calculation of the effects of fault current for disconnection times up to 5 seconds.
Where thermosetting 90 °C (XLPE) armoured cables are used and the armour is used as a protective conductor, the question is often asked which value of k should be used? This is a good question because when selecting a cable for current carrying capacity, Regulation 512.1.5 of BS 7671 states that switchgear, protective devices, accessories and other types of equipment being shall not be subjected to conductor temperatures exceeding 70 °C. This previous IET Wiring Matters article, “Thermoplastic (PVC) or Thermosetting (XLPE) SWA cable?” provides further information on thermoplastic and thermosetting cables. Ultimately, the current ratings in Table 4D4A for 70 °C thermoplastic cables must be used.
It would be easy to think that it is the same for the adiabatic equation, but it is not the case. The limiting factor of 70 °C conductor operating temperature for selection of cables for current carrying capacity is related to deterioration of equipment or accessories caused by constant loads of long duration.
When carrying out fault current calculations, and where thermosetting insulation is used, due to the short duration, the value of k, for calculation of the effects of fault current, may be taken to correspond to the final temperature for 90 °C thermosetting insulation. One limitation is that 90 °C thermosetting cables are not in contact with 70 °C thermoplastic cables.
How do I calculate the CSA of a protective conductor using the adiabatic equation?
Now that I, t and k have been determined, these values can be used to calculate S, the CSA of a protective conductor. It is just a matter of inputting the data into the adiabatic equation for calculation.
Therefore:
I = 140 A
t = 5 s (Fig 3A1 of BS 7671)
k = 115 (Table 54.3 of BS 7671)
Example 1 shows a distribution circuit protected by a 32 A BS 88-3 fuse system.,The maximum earth fault loop impedance is 1.1 Ω and the protective conductor is incorporated in a thermoplastic 70 ºC cable. Applying Cmin as it is a fuse, the fault current is determined to be 140 A.
Where cables are used as protective conductors and the application of the formula produces a non-standard size, a conductor having a larger standard CSA shall be used. Therefore, the next standard size of protective conductor to 1.93 mm² in this example is 2.5 mm².
Example calculation 1: 32 A BS 88-3 fuse
Next standard size cable = 4 mm².
What if the disconnection time is less than 0.1 seconds?
The shortest disconnection times provided for circuit-breakers in Appendix 3 of
BS 7671 is 0.1 seconds. This is because the disconnection time is characterised by the mechanical and physical design of the device and varies according to the manufacturer.
Where the disconnection time of the protective device is less than 0.1 seconds, manufacturer’s data must be consulted to ascertain I²t values. In such cases, I²t is the energy let-through in ampere squared seconds (A²s). The calculation is transposed as I²t < k²S².
For further information, see thermal withstand Wiring Matters article by Stuart Kershaw.
How do I check if live conductors can withstand short-circuit current without thermal damage?
A similar process can be carried out to determine any conductors that are not thermally damaged by short-circuit currents. [SG2.1]Regulation 434.5.2 of BS 7671 applies the same underlying logic to verify that the limiting temperature of any conductor is not exceeded by short-circuit current. The relationship is usually expressed as I²t ≤ k²S², where I²t is the thermal stress imposed by the fault and k²S² is the thermal withstand of the conductor. Table 43.1 of BS 7671 provides values of k for common materials, for calculation of the effects of fault current for disconnection times up to 5 seconds.
Example 2 shows a circuit using a 32 A B type circuit-breaker to BS EN 60898 and a fault current of 160 A. The conductor selected to check is 2.5 mm².
Because the prospective fault current causes instantaneous operation of the device as shown in Figure 3, that is, operation in 0.1 seconds or less, the manufacturers let-through energy I²t data, or the energy let-through (I²t) quoted for the class of device, must be used. The calculation is transposed as I²t < K²S².
For example, according to BS EN 60898 and BS EN 61009, the energy let-through (I²t) for a 32A Type B 6 kA circuit-breaker or RCBO is 45,000 A²s.
This calculation shown in Example 2 demonstrates that a 2.5 mm² conductor is adequate.
Example calculation 2: 32 A Type B circuit-breaker to BS EN 60898
Figure 3: Disconnection time for a 32 A BS EN 80898 Type B (Fig 3A4 of BS 7671)
Disconnection time = <0.1 seconds
Summary
The adiabatic equation can be used to determine the size of protective conductors regardless of construction and type.
The calculation method of determining the size of the protective conductor results in a more economical design compared to the selection method.
Where disconnection times of protective devices are less than 0.1 seconds, manufacturers’ data should be consulted.
Acknowledgments
- Craig O’Neill
- Gary Gundry
- Graham Kenyon
- Joe Cannon
- John Peckham
- Mark Coles
- Stuart Kershaw
Further reading