Choosing an overcurrent protective device. | on ElectriciansForums

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Goody

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Where an overcurrent protective device is used the following condition shall be fulfilled:

Zs x Ia < or = Uo

Ia is the current in amperes (A) causing the automatic operation of the disconnecting device…

Can I assume Ia is the “PFC”? Thanks
 
Ia is the current in amperes (A) causing the automatic operation of the disconnecting device within the time specified in Table 41.1 of BS 7671.
 
Ia = the amount of current that is needed to operate the protective device (before Telectrix gets in ;)). Page 35 brb gives you all the symbols, very handy whilst learning. (IQ beat me to it.
 
Last edited:
Take the device type eg. B, C or D

The current multipliers are 5, 10 and 20 respectively

So a B 10 device will disconnect at 5 X 10 Amps so 50A is the value for Ia
 
Take the device type eg. B, C or D

The current multipliers are 5, 10 and 20 respectively

So a B 10 device will disconnect at 5 X 10 Amps so 50A is the value for Ia


You can also calculate the maximum Zs for a particular device-in the example above for a B 10 device:

230/50 = 4.6 Ohms

For a D 10 you would have 230/200 =1.15 Ohms
 
so mmax. zs is by ohms law, V/I. or , for a 32A type B, 230/160=1.4375 rounded to 1.44 as in brb.
 
...

Can I assume Ia is the “PFC”? Thanks

PFC (Prospective Fault Current) is measured at the supply side of the main switch.

It is whichever is the highest reading you get from measuring PEFC (Prospective Earth Fault Current) and PSCC (Prospective Short Circuit Current)
 
Hi


With the greatest deference to those on here who have a lot more knowledge and experience than I have, I wonder if I could explain it slightly differently.


Overcurrent means two different things.
Fault current - due to a fault.
Overload – due to overloading the circuit.


I think (I may be wrong) that the formulae that you quote arises in fault conditions where Ia is the current that will cause the protective device to cut out within the maximum time allow by the regulations i.e. 0.4 secs or 0.2 secs or 5 secs depending on what the circuit is.


Ia then depends on what protective device you are talking about.
1> Find out the maximum time to disconnection based on what circuit you have.
2> choose your protective device based on the load it is protecting.
3> look in appendix 3 for the current that will cause the device you chose to cut out in the time required.
4> That current is then Ia – for that device, in that circuit.
5> Use the Ohms law formulae to find Zs – for that circuit. This is the figure you have to measure and comply with. If Zs is too high then Ia will be too low and the time to disconnection may not be met.


If the device is a fuse then it straightforward. If it an MCB then there is one more twist;


The specification for a type B MCB says that it will cut off instantly (taken as being 0.1 sec) when the current is greater than 3 times the MCB rating and less than 5 times the rating. In other words for a 20amp type B MCB it will cut off in 0.1 second at anywhere between 60 and 100 amps.
For a type C it's between 5 and 10 times the rating and for a type D it's between 10 and 20 times the rating.
The reason the graphs figure 3.4, 3.5, 3.6 on page 249, 250 and 251 only show the upper figure (i.e. 5 times the rating for type B for example) is that this is the worst case you can meet in the real world. For example the figure 3.4 shows that for a type B 20 amp MCB it is possible that 99 amps could flow in your circuit for over 10 seconds before the thermal overload part of the MCB would trip, whereas 100amps would cause the magnetic part of the MCB to trip in 0.1 sec.


That's my understanding. If I am wrong in any aspect I hope someone will correct me. I'm trying to learn this stuff myself.
Laurie
 
i struggle reading long posts but at a glance it looks like your thoughts are correct, we tend to use the worst case scenario (ex 5x mcb rating and not 3x etc)
 

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