Circuit Breakers

[DominicB]

Was just looking at circuit breaker types and I was a little bit confused so wondered if anyone could help clear it up.

Basically say If I determined my current would be 26a before taking into account any rating factors and chosen a 32a circuit breaker for my protective device. After taking into account my rating factors for length etc the actual current demand would be up to 43.7A (Which would exceed the 32a rating) so how does this work?

 

[Leesparkykent]

I think you've got things a little muddled here Dominic. These factors will decrease the current carrying capacity of the cable.

 

[DominicB]

Thanks for the reply I think i didn't explain my confusion very well basically how for example a type b 32a circuit breaker would actually trip at up to three to five times the rating (say 160a) but if the cable isn't rated that high then how comes this is okay?

 

[telectrix]

basically, the cable can withstand a massive overcurrent for the short duration that it takes the MCB to trip.

 

[DominicB]

Right that makes sense although if the sockets were protected by a 32a MCB and say a load of appliances were plugged in that drew 50-60a but the mcb would trip at a minimum of 96a (three times the 32a) wouldn't that be dangerous or is that okay for them to draw more then that because the cable should be able to handle more?

 

[telectrix]

a 32A type B breaker handling 50A would trip in approx 16 minutes.( see app.3 in BS76571).

 

[DominicB]

Right okay gotcha in my book it gives an example saying that the design current of a 6kw heater would be 26a (6000/230) so the nominal protective device would be 32a.

It then sase as the circuit is 21m in length installed via method C at 35 degress surrounded by 100mm of thermal insulation the cable must be capable of carrying a current of 43.7a as a minimum? If the cable is calculated to carry 43.7a then why is a 32a chosen?

 

[telectrix]

a 32 A breaker is selected because its rated current (In) is greater than the design current (Ib) so it will not trip in normal use. However, it's value must be less than the cable's max. current (Iz). the formula is Ib < In < Iz.

so we have here 26 < 32 < 43. i.e. it complies.obviously due to the installation method ( insulation) a 6mm or even 10mm cable would be appropriate, going by your 43A .

 

[DominicB]

So the 43.7a is the maximum current that would pass through the cable at normal conditions so if it did carry that much it would trip the circuit breaker in a certain amount of time? Thank you for your help

 

[davesparks]

Right okay gotcha in my book it gives an example saying that the design current of a 6kw heater would be 26a (6000/230) so the nominal protective device would be 32a.

It then sase as the circuit is 21m in length installed via method C at 35 degress surrounded by 100mm of thermal insulation the cable must be capable of carrying a current of 43.7a as a minimum? If the cable is calculated to carry 43.7a then why is a 32a chosen?

The 43.7A is not a real world value, it is a calculated value used to de-rate the cable.
What is confusing about the way the calculation works is that instead of actually reducing the current carrying capacity of the cable by calculation we instead multiply the nominal current (or design current in some cases) by factors which actually increase the value. This gives us the Iz value, which is not the current that will flow in the cable, but actually the current rating the cable is required to have to safely handle the nominal current (or design current) when subjected to the external factors such as temperature and grouping.

The example you have given is a pretty poor one in this case as an electric heater will almost certainly be able to be calculated based on its design current rather than the nominal current. So you would apply factors to the 26A rather than the 32A to get Iz.

 

[davesparks]

So the 43.7a is the maximum current that would pass through the cable at normal conditions so if it did carry that much it would trip the circuit breaker in a certain amount of time? Thank you for your help

No, see my above post, the 43.7A is a theoretical value only, it won't flow through the cable (except in a bizarre and very specific fault situation)

 

[DominicB]

Thanks for your replies, im just wondering why its only a theortical value if its based on external factors which have been calculated like the highest ambient temp and thermal insulation?

 

[Pat H]

It is an odd way of calculating things.
A cable that is in insulation has a de-rating factor. But what actually happens is the cable will heat up and its resistance increase.
So the current flowing will actually drop (voltage is nominal and fixed). However based on our de-rating value and subsequent thicker cable requirement the resistance will be lower. So less heat generated and less loss.
Its a sort of backwards way of achieving the same result.

 

[DominicB]

Sorry im still confused, whats the difference between the It and Iz because in the wiring regulations it says It is the current carrying capacity without correctional factors or installation conditons yet the calculation for It appears to take these into account

 

[gazdkw82]

As previously mentioned, these factors will decrease the amount of current the cable can safely carry. The de-rating factors and information on CCC’s can be found in Appendix 4 (BS7671)


The current carrying capacities can be found on page Appendix 4 (page 332 BS7671).


So first find the CCC of the chosen cable for the chosen circuit. Apply the factors, if the total CCC is now below the trip current of the protective device then the CSA will need to be increased.