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Yes, or prob easier to divide the PFC by 16 and use the numbers on the graph. Either works though.On this graph would you multiply the breaker rating for example 16 by the bottom number of this graph?
Discuss Adiabatic Equation help in the Electrician Courses : Electrical Quals area at ElectriciansForums.net
Yes, or prob easier to divide the PFC by 16 and use the numbers on the graph. Either works though.On this graph would you multiply the breaker rating for example 16 by the bottom number of this graph?
Yes, see attached pdf.so hager 16A 60898 type B will disconnect in 0.01s at 719A. Is this correct?
The calculation looks right. Is the value of R1+R2=0.21 correct? It was 0.22 in your OP.So with this I could go 2.5mm2 + 1mm2. r1+r2 = 0.21
Zs = .11 + .21 = .32
pfc = 230/.32 = 719A
so square root of 719 squared x 0.01/115 = 0.62 = S
I made a mistake with the initial post. r1+r2 was supposed to be .12, with 2.5mm2 + 2.5mm2.Yes, see attached pdf.
The calculation looks right. Is the value of R1+R2=0.21 correct? It was 0.22 in your OP.
As before, I'd have used the fault current from the origin (2091A) for the calc, but I think this is something you'll have to ask your tutor, to confirm what he/she is expecting.
FWIW, I discussed this with a fellow spark's apprentice a while back (generally, not related to any specific question he had). Prompted by me, next time he was at college, he posed the question to his tutor: where in the circuit should the fault current be taken from when calculating the adiabatic, for these types of breaker? His tutor admitted he didn't know.
The missing line on the pdf for b-curve is just a printing error. For all 3 curves, once a threshold current is exceeded, the breaker is not expected to operate any faster than 0.01 seconds.I made a mistake with the initial post. r1+r2 was supposed to be .12, with 2.5mm2 + 2.5mm2.
I've calculated the r1+r2 now using 2.5mm2 + 1mm2.
One last question hopefully... in the pdf you posted why does the line stop at 4 and 6 where as the type c graph the line continues upto 100?
My fault current would be at the 44 mark i think... 719/16 = 44.9 as timhoward said above?
I know this is a lot of questions however I have actually got a much better grasp on this now so thankyou all!!
@timhoward , I believe that, for B and C type breakers, the fault current used for the adiabatic equation should be taken at the point of the circuit that it would be highest, ie the origin. The reasons for this are in my above post, also the note to fig3A4 ('The application of the rules of Chapter 43 should take into account both minimum and maximum fault current conditions.').
To me, it seems that using the fault current at the far end of the circuit leads to a CPC being selected on a best case scenario, and a fault anywhere else in the circuit would raise the temperature of the conductors to a higher temperature than a fault here would.
For a fuse that is true as PFC rises the time decreases at a much higher rate (I^[3..4] sort of speed) so overall I2t drops.I raised this question with my tutor as an apprentice and was told that as the fault current is higher the speed of operation of the device will be faster at the origin.
As the device operates faster the temperature rise will be less, and not more, than at the end of the circuit.
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