log tables for actual disconnect time

gmdaly · Jun 6, 2022

Hi,
could someone tell me how to calculate the actual disconnect time from a log table and where I can find this table for a 6A RCBO made by British General. I need this to get a meaningful CPC CSA using the adiabatic equation.

Many thanks,

g

Donkeybrains · Jun 6, 2022

gmdaly said:
Hi,
could someone tell me how to calculate the actual disconnect time from a log table and where I can find this table for a 6A RCBO made by British General. I need this to get a meaningful CPC CSA using the adiabatic equation.

Many thanks,

g

Define log table? Do you mean a table of logarithms?

Electro-tech · Jun 6, 2022

Do you have a copy of the 18th edition Regs book or onsite guides?

gmdaly · Jun 6, 2022

Electro-tech said:
Do you have a copy of the 18th edition Regs book or onsite guides?

hi yes i do have both and if you are saying i should use fig 3A4 that's ok but i have a fault current of 301A and if I take 0.4 seconds as t then I end up with a cross sectional area of >1.66mm. can I use the lower trip time of 0.1 seconds?
thanks

Electro-tech · Jun 6, 2022

Table 41.3 page 68 for Max earth loop impedances general RCBO's no specific manufacturers.

Useful video.

telectrix · Jun 6, 2022

use 0.1 sec. it's well in @ 301A.

Vortigern · Jun 6, 2022

I think the adiabatic applies to the main earth not cpc.

Electro-tech · Jun 6, 2022

[ElectriciansForums.net] log tables for actual disconnect time

gmdaly · Jun 6, 2022

telectrix said:
use 0.1 sec. it's well in @ 301A.

thanks telectrix but can you tell me how to justify its use? thanks,
g

Pretty Mouth · Jun 6, 2022

Vortigern said:
I think the adiabatic applies to the main earth not cpc.

It applies to all protective conductors, except bonding conductors

Pretty Mouth · Jun 6, 2022

gmdaly said:
thanks telectrix but can you tell me how to justify its use? thanks,
g

Take a look at the fault curves in appendix 3 of BS7671. For type B and C breakers/RCBOs, if it meets any permitted disconnection time, even 5s for distribution circuits, then it disconnects at 0.1 seconds (or faster). There is no middle ground, as above a certain current the magnetic trip will operate and will disconnect instantaneously. For example, you can't make it disconnect in exactly 1 second.

As you are presumably using a B or C type breaker/RCBO, use the fault current at the origin of the circuit for the calculation. For these devices, let through energy tends to increase with fault current, and fault current is greatest at the origin. This is the most onerous point in the circuit.

pc1966 · Jun 6, 2022

For MCB and RCBO in the thermal part of the trip curve you can use the graphs to get a meaningful measure of I2t for the adiabatic equation.

Once you hit the "instant" magnetic point though those plots are pretty meaningless. For an accurate figure you need to get the manufacturer's plot of let-through energy (i.e. I2t as a function of PFC) and some of the better makes provide this (e.g. Hager in their commercial catalogue). Otherwise you can refer to the generic specifications for that class of energy-liming breaker, for example Table B7 in the On-Site Guide.

Attached is an example page from Hager's catalogue.

pc1966 · Jun 6, 2022

Pretty Mouth said:
It applies to all protective conductors, except bonding conductors

Strictly speaking it applies to all conductors, it is just the CPC is often smaller than the live to save money as it has little to do normally. Of course if you meet the adiabatic for the CPC then the live conductors will be OK.

Vortigern · Jun 6, 2022

Pretty Mouth said:
It applies to all protective conductors,

Technically yes but practically I have never done an adiabatic for circuits issuing from the MCB as the fault current is automatically held to the level of the MCB. However, the main earth from the CU to the Neutral terminal where there is no protection, yes I can see the use of the adiabatic there as if the main earth goes meltdown there is a problem. But for 2.5 cable do you really do an adiabatic? What can you do if 1.5mm² is not enough? Thread a thicker cpc up the cable. Sorry If I am being a bit thick here but ready to learn something new.

pc1966 · Jun 6, 2022

Vortigern said:
Technically yes but practically I have never done an adiabatic for circuits issuing from the MCB as the fault current is automatically held to the level of the MCB. However, the main earth from the CU to the Neutral terminal where there is no protection, yes I can see the use of the adiabatic there as if the main earth goes meltdown there is a problem. But for 2.5 cable do you really do an adiabatic? What can you do if 1.5mm² is not enough? Thread a thicker cpc up the cable. Sorry If I am being a bit thick here but ready to learn something new.

You are quite right.

For any of the "standard circuits" you don't need to bother with the adiabatic as the cable size versus OCPD is already fine. That really is why using the usual suspects (standard cases, such as Table 7.1(i) from OSG) makes so much sense as it is one of many less things to worry about design-wise!