log tables for actual disconnect time

[gmdaly]

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]

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]

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

 

[gmdaly]

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]

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

Useful video.

 

[telectrix]

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

 

[Vortigern]

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

 

[Electro-tech]

 

[gmdaly]

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]

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

It applies to all protective conductors, except bonding conductors

 

[Pretty Mouth]

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]

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]

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]

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]

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!

 

[Pretty Mouth]

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!

Is this correct? The reason I query it is table B7 on p132 of the OSG. It shows minimum CPCs for class 3 B and C type MCBs, for different fault current levels.

For example, a B32 at a fault current of up to 3kA has a minimum CPC of 1.5mm. Above that and up to 6kA, it would be 2.5mm - thicker than the usual 1.5mm CPC that you'd typically find on a ring wired in 2.5mm T+E.

Is there something else at play here?

Calculating CPC size using manufacturer's data usually gives a much smaller min. CPC than those listed in table B7, but not always. BG and Contactum, when asked, gave identical data (which I suspect comes from the product standard rather than their own tests) which, when run through the adiabatic, gives the same results as table B7 for the higher 3-6kA fault.

 

[pc1966]

Is this correct? The reason I query it is table B7 on p132 of the OSG. It shows minimum CPCs for class 3 B and C type MCBs, for different fault current levels.

For example, a B32 at a fault current of up to 3kA has a minimum CPC of 1.5mm. Above that and up to 6kA, it would be 2.5mm - thicker than the usual 1.5mm CPC that you'd typically find on a ring wired in 2.5mm T+E.

Is there something else at play here?

I think it is down to the ideal case of DB PFC and max let-through of the MCB standard, and the real-world case of MCB generally being better, and the actual PFC being less.

For example, if you have a DB with 6kA PFC (assuming for now 230V) then your Ze is around 0.038 ohm. To bring that down to 3kA you need to double that, and with the R1+R2 of 2.5mm T&E being around 19.5mOhm/m that is 2m of cable.

So for most faults on any real-world installation using typical circuits you won't reach cable-damaging I2t, and if you do it will only be the first metre or so before a very close (and very hard) fault that needs replacing.

 

[Pretty Mouth]

I think it is down to the ideal case of DB PFC and max let-through of the MCB standard, and the real-world case of MCB generally being better, and the actual PFC being less.

For example, if you have a DB with 6kA PFC (assuming for now 230V) then your Ze is around 0.038 ohm. To bring that down to 3kA you need to double that, and with the R1+R2 of 2.5mm T&E being around 19.5mOhm/m that is 2m of cable.

So for most faults on any real-world installation using typical circuits you won't reach cable-damaging I2t, and if you do it will only be the first metre or so before a very close (and very hard) fault that needs replacing.

Ah, I see, that puts it in perspective!

 

[gmdaly]

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.

Thanks pretty mouth.