Regulation 434.5.2 Fault current!

[HappyHippyDad]

Cant quite get my head around how you use this reg?

Lets say S =16
I = 657
k = 115

You have to work out t first of all so

t = K[SUP]2[/SUP]S[SUP]2[/SUP]/I[SUP]2

[/SUP]so t = 7.843409 (to 6dp)

therfore K[SUP]2[/SUP]S[SUP]2[/SUP] = 3385600

The reg says that we need to make sure that K[SUP]2[/SUP]S[SUP]2 [/SUP]is greater than the value of the let through energy (I[SUP]2[/SUP]t),

But using the above equation I[SUP]2[/SUP]t also equals 3385600.

By using the above equation they will ALWAYS equal each other as your just messing around with the same equation.

I read the reg as K[SUP]2[/SUP]S[SUP]2 [/SUP]having to be greater than I[SUP]2[/SUP]t as it says:

'for current limiting devices K[SUP]2[/SUP]S[SUP]2 [/SUP]shall be greater than the value of the let through energy I[SUP]2[/SUP]t',

Bit confusing?

 

[D Skelton]

When working out your 'k[SUP]2[/SUP]S[SUP]2[/SUP]' you will be adjusting 'k' to take account of the actual conductor operating temperature, it isn't simply '115'. To work out the 'k' factor you will need to use more advanced formulae taking account of volumetric heat capacities of conductor materials, electrical resistivity of conductor materials, initial and final temperatures of conductors under fault and reciprocals of temperature coefficient of resistivity of conductors. If you want me to go in to it I will but if you're primarily working in a domestic environment it's never something your are ever going to need to know how to do.

 

[Richard Burns]

I²t as the energy let through of the device is a value given by the manufacturers (or an average value for the type) such as this from Crabtree.
View attachment Technical MCB data crabtree-2010-179.pdf
Because the device is designed to limit the current this may be lower than the calculated value. (and often makes required csa for cpc values much smaller when using the adiabatic equation)

 

[Geoffsd]

t = K[SUP]2[/SUP]S[SUP]2[/SUP]/I[SUP]2[/SUP] and I[SUP]2[/SUP]t

Are you not confused because you are taking 't' as the same value in both equations?

 

[HappyHippyDad]

When working out your 'k[SUP]2[/SUP]S[SUP]2[/SUP]' you will be adjusting 'k' to take account of the actual conductor operating temperature, it isn't simply '115'. To work out the 'k' factor you will need to use more advanced formulae taking account of volumetric heat capacities of conductor materials, electrical resistivity of conductor materials, initial and final temperatures of conductors under fault and reciprocals of temperature coefficient of resistivity of conductors. If you want me to go in to it I will but if you're primarily working in a domestic environment it's never something your are ever going to need to know how to do.

Thanks D S,

Sorry for not replying sooner but its all a bit busy at the moment! The reason why I was interested is because in the Best Practise Guide for domestic consumer unit change (number 6 I think) it says on page 5 (note 4 I think) that you can keep the 16mm tails even if there is a 100A cut out fuse..... as long as it meets the mentioned reg (434.5.2).

As for the value of K I got this by using the tables in BS7671 and it came out as 115

I'm just not entirely sure I'm understanding the reg though, is it saying that if K[SUP]2[/SUP]S[SUP]2[/SUP] is greater than I[SUP]2[/SUP]t then the protective device is still acceptable? Also.... I'll write this next bit on your's Richard...

I²t as the energy let through of the device is a value given by the manufacturers (or an average value for the type) such as this from Crabtree.
View attachment 19936
Because the device is designed to limit the current this may be lower than the calculated value. (and often makes required csa for cpc values much smaller when using the adiabatic equation)

It does make sense if I[SUP]2[/SUP]t is a value given by the maufacturer as you can compare it to the value you get for K[SUP]2[/SUP]S[SUP]2. [/SUP]I'm just not sure why we are given a formula to work out t. If I use the given formula to work out t and then multiply it by I[SUP]2 [/SUP]that result will always equal K[SUP]2[/SUP]S[SUP]2[/SUP] and if the idea is to compare K[SUP]2[/SUP]S[SUP]2 [/SUP]against I[SUP]2[/SUP]t then it doesn't make sense because they're always equal (unless we use the figure given by manufacturer - which makes much more sense)

 

[D Skelton]

Thanks D S,

Sorry for not replying sooner but its all a bit busy at the moment! The reason why I was interested is because in the Best Practise Guide for domestic consumer unit change (number 6 I think) it says on page 5 (note 4 I think) that you can keep the 16mm tails even if there is a 100A cut out fuse..... as long as it meets the mentioned reg (434.5.2).

As for the value of K I got this by using the tables in BS7671 and it came out as 115

Ok, I read your OP slightly wrong, although my answer applies and is open for designers to choose this option, my answer suits a slightly different question lol

I'm just not entirely sure I'm understanding the reg though, is it saying that if K[SUP]2[/SUP]S[SUP]2[/SUP] is greater than I[SUP]2[/SUP]t then the protective device is still acceptable?

Right, given slightly more info and your further example I understand what you're asking. With a 100 cutout fuse and 16mm tails:

Firstly, at 657A a 100A fuse will take approx 6s to blow, well within the time given of 7.8s until the fault current will raise the live conductors to the limiting temperature of operation. This shows that your 16mm tails are protected against fault current but not neccessarily overload current. However, your 5s max disconnection time (1s for a TT) is something that you have to bear in mind. All of this means that the second part of this regulation does not apply as it is not a fault of very short duration (less than 0.1s).

Using another set of example figures whereby the second part of this regulation applies:

S = 16mm
I[SUB]p[/SUB][SUB]f[/SUB] = 6.6kA (as opposed to 0.6kA)
k = 115
I[SUB]n[/SUB] = 100A (BS88 HRC)

The operating time of a 100 fuse at 6.6kA will be at most 0.01s. The I[SUP]2[/SUP]t of a 100A fuse with an I[SUB]pf[/SUB] of 6.6kA will be roughly between 250000* - 300000* depending on the manufacturer (*these are values that can be obtained from manufacturers information). The k[SUP]2[/SUP]S[SUP]2[/SUP] value is over three million so 434.5.2 is met. Reduce 'S' to 6mm and you can see that even then 434.5.2 is met, it is only when you drop the conductor csa to 4mm that in this example the I[SUP]2[/SUP]t value is higher than the k[SUP]2[/SUP]S[SUP]2[/SUP] value.

Essentially, what this further part of the regulation is explaining is that so long as the let through energy per second of the current limiting device for the highest fault current available is less than the amount of current per second that the conductor will be able to withstand then you have achieved effective fault current protection for that circuit. You must bear in mind that often your pssc and your pefc will give you different I[SUB]pf[/SUB] values and your calculations must take into account these variations.

It does make sense if I[SUP]2[/SUP]t is a value given by the maufacturer as you can compare it to the value you get for K[SUP]2[/SUP]S[SUP]2. [/SUP]I'm just not sure why we are given a formula to work out t. If I use the given formula to work out t and then multiply it by I[SUP]2 [/SUP]that result will always equal K[SUP]2[/SUP]S[SUP]2[/SUP] and if the idea is to compare K[SUP]2[/SUP]S[SUP]2 [/SUP]against I[SUP]2[/SUP]t then it doesn't make sense because they're always equal (unless we use the figure given by manufacturer - which makes much more sense)

As stated earlier, the formula given to work out 't' is to work out the time in which a given fault current will take to essentially destroy a conductor. You work out the time in which a given OCPD will take to disconnect a circuit under a certain fault current, then you work out whether this amount of time is lower that the amount of time it will take for that given current to burn out the conductors within that circuit.

I hope this helps explain things a little clearer

 

[HappyHippyDad]

Ok, I read your OP slightly wrong, although my answer applies and is open for designers to choose this option, my answer suits a slightly different question lol



Right, given slightly more info and your further example I understand what you're asking. With a 100 cutout fuse and 16mm tails:

Firstly, at 657A a 100A fuse will take approx 6s to blow, well within the time given of 7.8s until the fault current will raise the live conductors to the limiting temperature of operation. This shows that your 16mm tails are protected against fault current but not neccessarily overload current. However, your 5s max disconnection time (1s for a TT) is something that you have to bear in mind. All of this means that the second part of this regulation does not apply as it is not a fault of very short duration (less than 0.1s).

Using another set of example figures whereby the second part of this regulation applies:

S = 16mm
I[SUB]p[/SUB][SUB]f[/SUB] = 6.6kA (as opposed to 0.6kA)
k = 115
I[SUB]n[/SUB] = 100A (BS88 HRC)

The operating time of a 100 fuse at 6.6kA will be at most 0.01s. The I[SUP]2[/SUP]t of a 100A fuse with an I[SUB]pf[/SUB] of 6.6kA will be roughly between 250000* - 300000* depending on the manufacturer (*these are values that can be obtained from manufacturers information). The k[SUP]2[/SUP]S[SUP]2[/SUP] value is over three million so 434.5.2 is met. Reduce 'S' to 6mm and you can see that even then 434.5.2 is met, it is only when you drop the conductor csa to 4mm that in this example the I[SUP]2[/SUP]t value is higher than the k[SUP]2[/SUP]S[SUP]2[/SUP] value.

Essentially, what this further part of the regulation is explaining is that so long as the let through energy per second of the current limiting device for the highest fault current available is less than the amount of current per second that the conductor will be able to withstand then you have achieved effective fault current protection for that circuit. You must bear in mind that often your pssc and your pefc will give you different I[SUB]pf[/SUB] values and your calculations must take into account these variations.



As stated earlier, the formula given to work out 't' is to work out the time in which a given fault current will take to essentially destroy a conductor. You work out the time in which a given OCPD will take to disconnect a circuit under a certain fault current, then you work out whether this amount of time is lower that the amount of time it will take for that given current to burn out the conductors within that circuit.

I hope this helps explain things a little clearer

I really appreciate that answer D S, that makes everything MUCH more clear. I probably will have another question fairly soon as theres still one bit I'm not quite sure on but need to think how to word it first and wait until I get home to have a route through the BS7671 disconnection times.