Time/current characteristics

[philp]

HI I'm trying to understand how to read a time/current characteristics graph. I was looking at the link http://www. ****************** /Book/3.6.5.htm where it talks about 32 Amp MCB i.e.
For example, consider a cable system rated at 30 A and protected by a miniature circuit breaker type 3, rated at 32 A.
Reference to (Fig 3.17 shows that a prolonged overload of about 38 A will open the breaker after about 104 seconds (about two and a half hours!). I dont get how you can read these values from the graph and how does 103 seconds = 2.5 hours?? Is this just plain wrong, or am I totally misundersting things? Sorry if this is a pretty basic question. Thanks

 

[ackbarthestar]

I can't find fig 3.17 in the regs, but then I suspect your 'removed link' is pointing to a commercial 3rd party product.

The straight line represents the electromagnetic part of the MCB where it is expected that instantaneous disconnection will take place when a level of fault current flows greater than the minimum for that size of breaker. ie. 320A

The curved part of the graph reflects the action of the electromechanical bimetallic strip. This requires an overload lasting for more than 5 secs to operate the device.

A 32A type C running at 40A will not, according to Appendix 3, disconnect. However, we know that these graphs only show the trailing edge so will reflect the worst possible case. Probably better to use manufacturer's data when looking at the time/current characteristics and then you might use a fusing factor of 1.13 as a rule of thumb.

 

[philp]

Thankyou very much ackbarthestar. I really would like to understand these graphs better. Just to help me understand more... "The straight line represents the electromagnetic part of the MCB where it is expected that instantaneous disconnection will take place when a level of fault current flows greater than the minimum for that size of breaker. ie. 320A" If I take for example Fig 3A4 page 301 in BC7671. Say 6A curve. I asume that 31 Amps would mean instant disconnection. 29 amps would disconnect in about 11/14 seconds (bimetel strip). 30 Amps could disconnect anywhere between instant and maybe 12/13 secs. Is this a correct interpretation? "Probably better to use manufacturer's data when looking at the time/current characteristics and then you might use a fusing factor of 1.13 as a rule of thumb." Does the 1.13 factor tell you that for example for a 6 A MCB it will happily take a current of 1.13*6A i.e. 6.78 amps without tripping. I have seen values of this like 1.45 to do much the same. Does different manufacturers give different factors or does it depend on the type B,C,D etc Thanks again.

 

[ackbarthestar]

fig 3A4 shows time/current characteristics where the Ia is assumed to be 5 In. So for a 6A breaker it would require a fault current of 30A theoretically. However, to get the precise values you would need to look ate the manufacturer's graphs. These will specify a range of fault currents between 3 -5 In. So you may find that your 6A breaker disconnects at 18A.
The straight line is the electromagnetic section of the graph and would be reasonable to expect a disconnect time as fast as the electromechanical parts could operate. That would equate to 0.1sec from fig 3A4 but in reality it is likely to be 10 times faster.
Generally, If the leading edge of the time current characteristic for a particular manufacturer's device is say 3 times 'In' then anything less than 18A will operate on the bimetallic part of the graph.

Take an example of a 10A breaker from fig. 3A4 Theoretically a 10A breaker will not disconnect the circuit if the current in the circuit is 1.13*10 = 11.3A
anything above this (13A), theoretically, will now put the disconnection time on the graph, starting from about 7 days for very small overloads down to about 12 secs fora fault current approaching Ia, which from the regs is 30A??? - edit 50A

To keep things simple by following the trailing edge as seen in BS7671 appendix 3, you will be providing extra protection to your circuit.