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magnoliafan89

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Hi guys

Have a circuit its OCPD is a C type 80A 10ka 60898 MCB with a pfc of 1.11ka. On the log log graph for this device I’m struggling to understand the time for it. It only goes to 800a?

I’ve spoken to the NICEIC who say it would operate withing 0.1 seconds but I’m struggling to see how they’ve got to this.??
 
For an 80A C-curve the "instant" magnetic trip is 5-10*In so in this case 400-800A. Above that and your operating time is typically well under 20ms (i.e. energy limiting). Some plots do not show this region, or they show it as a vague band below 20ms time.

Sometimes they list the let-through I2t value and you can work backwards from the PFC 'I' and the 'I2t' to get the virtual operating time 't'. This "virtual time" is what is normally plotted, and for fuses that energy limit very well you might see tens of microseconds equivalent, but the fuse takes longer to burn, it just limits I to way below PFC so the equivalent 't' with the prospective current is much smaller.

Below that region, but above about 1.45*In, it will go eventually on the thermal trip side, but that is not below 5s so not listed. For D-curve the 5s time is before the 10-20*In "instant" region.

If you have a PFC of 1.1kA then your MCB will go in the < 20ms sort of time.
 
For an 80A C-curve the "instant" magnetic trip is 5-10*In so in this case 400-800A. Above that and your operating time is typically well under 20ms (i.e. energy limiting). Some plots do not show this region, or they show it as a vague band below 20ms time.

Sometimes they list the let-through I2t value and you can work backwards from the PFC 'I' and the 'I2t' to get the virtual operating time 't'. This "virtual time" is what is normally plotted, and for fuses that energy limit very well you might see tens of microseconds equivalent, but the fuse takes longer to burn, it just limits I to way below PFC so the equivalent 't' with the prospective current is much smaller.

Below that region, but above about 1.45*In, it will go eventually on the thermal trip side, but that is not below 5s so not listed. For D-curve the 5s time is before the 10-20*In "instant" region.

If you have a PFC of 1.1kA then your MCB will go in the < 20ms sort of time.
I just want to be clear you mean less than 20ms?? Thanks for your reply mate
 
For an 80A C-curve the "instant" magnetic trip is 5-10*In so in this case 400-800A. Above that and your operating time is typically well under 20ms (i.e. energy limiting). Some plots do not show this region, or they show it as a vague band below 20ms time.

Sometimes they list the let-through I2t value and you can work backwards from the PFC 'I' and the 'I2t' to get the virtual operating time 't'. This "virtual time" is what is normally plotted, and for fuses that energy limit very well you might see tens of microseconds equivalent, but the fuse takes longer to burn, it just limits I to way below PFC so the equivalent 't' with the prospective current is much smaller.

Below that region, but above about 1.45*In, it will go eventually on the thermal trip side, but that is not below 5s so not listed. For D-curve the 5s time is before the 10-20*In "instant" region.

If you have a PFC of 1.1kA then your MCB will go in the < 20ms sort of time.
I did call CPN earlier as rhat was the brand and I asked the girl on the phone for the let through energy of this particular MCB and her reply didn’t of “the what?” Didn’t fill me with much confidence lol
 
What's the model? There's gotta be manufacturers data out there to answer your question.
I did try looking it’s one of them cudis CPN ones.

Here’s a link to one

 
I just want to be clear you mean less than 20ms?? Thanks for your reply mate
Usually, yes. Here is the curve from the Hager commercial catalogue for let-through energy:
[ElectriciansForums.net] C type MCB time and Log logs

They only seem to list the common ones to 63A, but taking that as an example at 1,000A fault current the let-through is about 4 kA2s so the virtual time is:
t = (I2t) / I^2 = 4,000 / (1,000 * 1,000) = 0.004 = 4ms
Repeating for 10kA upper limit and we see the let-through is about 70 kA2s so we get:
t = 70,000 / (10,000 * 10,000) = 0.7ms
So the MCB opens faster at higher currents which is not really a surprise. The Hager curves seem to be for 'typical' values, so you might find some are a bit slower, and some a bit faster. But all energy-limiting breakers really have to open under one cycle of the supply.
 
Usually, yes. Here is the curve from the Hager commercial catalogue for let-through energy:
View attachment 110556
They only seem to list the common ones to 63A, but taking that as an example at 1,000A fault current the let-through is about 4 kA2s so the virtual time is:
t = (I2t) / I^2 = 4,000 / (1,000 * 1,000) = 0.004 = 4ms
Repeating for 10kA upper limit and we see the let-through is about 70 kA2s so we get:
t = 70,000 / (10,000 * 10,000) = 0.7ms
So the MCB opens faster at higher currents which is not really a surprise. The Hager curves seem to be for 'typical' values, so you might find some are a bit slower, and some a bit faster. But all energy-limiting breakers really have to open under one cycle of the supply.
Thanks for the replies will save me some work now then. The breaker was a small 2 way board which had the 16mm T and E feeding a second floor flats consumer unit and I was trying to figure if the 6mm CPC was an adequate Earth
 
Thanks for the replies will save me some work now then. The breaker was a small 2 way board which had the 16mm T and E feeding a second floor flats consumer unit and I was trying to figure if the 6mm CPC was an adequate Earth
If you take the adiabatic k as 115 (Table 54.3) and 6mm then max I2t is:

S = sqrt(I2t) / k -> sqrt(I2t) = S * k -> I2t = (S * k)^2 = (6 * 115)^2 = 476.1 kA2s

From the Hager plot, and assuming 80A is proportionally bigger then 63A for let-through, then it is safe for any fault current 800A ->10kA (i.e. any in safe instant-trip region).
 
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