Adiabatic Calculation

Hi All,


Been looking into the Adiabatic equation, as its been bothering me for a while that im not 100% on the subject. But after reading a lot of the old forums posts and the Amtech handbook i have collated the below information. Hopefully it is correct, and can be used by others. If not i would be very appreciative of any feedback.


Of course, you could just look at table 54g, but where’s the fun in that ?. I would be interested in understanding how that table has been derived, Id assume the line conductor is associated from the selection of an sensible MCB size as there seems to be no input of the variables of the adiabatic equation


Adiabatic Calculation

This calculation is a check to ensure that the cross-section of the CPC is sufficient to allow it to withstand the energy let-through of the Circuit Protective Device (CPD), i.e., fuse or circuit-breaker under earth fault conditions.

S = √(I2t)/ k


where:

S = the minimum section of the conductor in mm2

I = the earth fault current in amperes*

t =the CPD disconnection time at the earth fault I*

k = a factor (k) which ‘takes account of the resistivity, temperature coefficient and heat capacity of the conductor material, and the appropriate initial and final temperatures’.

*Where the disconnection time is greater than 0.1 s, I2t is the earth fault current squared times the disconnection time. S = √(I2t)/ k

*where the CPD disconnection time is less than 0.1 s, I2t is the CPD energy let-through in ampere squared seconds (A2s).  The Calculation must be transposed as I2t<=K2S2 and manufacturers data/ EN 60898 must be consulted as BS 7671 only gives 0.1s minimum from the tables so the I2t figure cannot be ascertained.

Fuse Fault Rules

For fuses the worst-case is a fault at the far end of the circuit, so you can use the same values for checking both worst-case disconnection time for shock protection and worst-case energy let-through for conductor protection.

MCB Fault Rules

MCBs have a different characteristic, the energy let-through typically increases with fault current, so the worst case as far as conductor protection is concerned is right after the protective device. 

For a fault above 0.1 seconds Example

Assuming a Zs of 1.5 ohms on a circuit at supply distribution board with a line conductor size of 4mm2 (90 Degrees Thermosetting (LSF) where the protective device is a 32ABS60898 type B MCB. 

S = √(I2t)/ k

I = 230/1.5 = 154A

t =From BS7671, 15 seconds

k = is 176 for thermosetting cable from table 54.2

S = √(154 x 154 x 15)/ 176

S= 3.38mm2 (4mm2 CPC required)

MCBs (EN 60898) which stipulates the maximum allowable energy let-through for various fault currents and rating and type of device (B-type, C-type etc). E.g. for a B-type MCB rated 16A or less (and energy limiting class 3) the max permitted is 15,000 A²s for a 3kA fault. For devices over 16A but 32A or under, it's 18,000 A²s and so forth.  As previously stated the energy let-through typically increases with fault current for MCBs, as more energy is essentially “let through”, so ultimately a larger cpc will be required.

For a fault below 0.1 seconds Example:

Assuming a Zs of 0.06 ohms on a circuit at supply distribution board with a line conductor size of 16mm2 (90 Degrees Thermosetting (LSF) where the protective device is a 63ABS60898 type B MCB. 

I2t<=K2S2

First calculate Earth Fault Current to establish the let through current or energy from the breaker.:

230/0.06 = 3883A Earth Fault Current.

From there, the I2t figure can be found from EN 60898.

It can be seen from a 63A Type B Breaker that 48000 A2s (I2t) units of energy are let through at this fault for this specific breaker and fault level. Therefore, we can summarise that for a cable to be safe it has to withstand 48000  energy units otherwise it will get too hot and there will be a breakdown in insulation.

I2t<=K2S2

Try for a 10mm2 CPC:

S = 10mm2

k = is 176 for thermosetting cable from table 54.2

48000 A2s <= (10 x 10) x (176 x 176)

48000A2s <= 3097600 A2s

So, we now know the cable can with withstand 3097600 energy units before it gets damaged, as the (I²t) of 48000 energy units is less than this there is no problem. 

A 6mm2 cable can be tried:

It is also worth noting that the Earth Fault Loop impedance calculation is a completely different calculation to ensure that the breaker disconnects fast enough depending on the fault current and specified disconection times.

Thanks