Adiabatic Equation help

timhoward said:

I only have time for a quick comment. There’s two concerns
1 - will the thing trip quickly enough. Clearly for a B16 it will, using min or max fault current
2 - how hot will the thing get during the time it takes to trip. This is where the higher fault current nearer the origin could be an issue, and using the equation in its other form to find how long it takes to get to 70 degrees would be required. This should be > answer 1.

I agree with your interpretation of the regs. Rightly or wrongly I suspect the point of the original question in this case is that whatever we do we end up back at 2.5 sq mm as the minimum mech. protected size.
I’m also on the fence regarding whether a trainee should focus on the min fault current at end of circuit which is what their tutor is probably expecting. Maybe!

(I’ve also seen one example that says a B16 will trip instantaneously at whatever the box on the right of the graph says (80A?) so 80 and 0.1 should be used in the equation as as soon as it reaches 80A it will trip. I’m still musing that one!)

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The whole point of the design project is to demonstrate you understand the process. Ask yourself why they'd wanting a specific manufacturer (as opposed to generic 7671) data ["....with the aid of manufacturer data...."] and the purpose and reasoning is clear; so can work the equations and extrapolate the required information, even if that ends up just back at the 2.5mm minimum.

You can, in theory, run the equation and steps, get them wrong and STILL get positive marks because while the end result may be wrong, you are demonstrating knowledge of the process, equations and expected results.

I assumed I'd do the calcs with the pfcs at end of circuit but what you say makes sense. Although they have stated to show it will satisfy requirements for ADS.

I did find on page 409 of BS7671 "the impedance values given in Tables 41.3 and 41.6 can be used for BS EN 60898 circuit-breakers. These values are far more onerous and in some cases may be difficult to achieve without installing larger sized cpcs.

Above this "Where ever possible designers should use the manufacturer's specific data"

so hager 16A 60898 type B will disconnect in 0.01s at 719A. Is this correct?

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

Lister1987 said:

The whole point of the design project is to demonstrate you understand the process. Ask yourself why they'd wanting a specific manufacturer (as opposed to generic 7671) data ["....with the aid of manufacturer data...."] and the purpose and reasoning is clear; so can work the equations and extrapolate the required information, even if that ends up just back at the 2.5mm minimum.

You can, in theory, run the equation and steps, get them wrong and STILL get positive marks because while the end result may be wrong, you are demonstrating knowledge of the process, equations and expected results.

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Yes to be fair i could have maybe avoided this as I haven't done the previous question which is selecting the devices. First time doing all this so didnt realise how important skipping that question would have been.
Thankyou!

On this graph would you multiply the breaker rating for example 16 by the bottom number of this graph?

Sam1995 said:

so hager 16A 60898 type B will disconnect in 0.01s at 719A. Is this correct?

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16x5=80amps
Square root (80x80x0.1)÷115=answer.

Sam1995 said:

On this graph would you multiply the breaker rating for example 16 by the bottom number of this graph?

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Yes, or prob easier to divide the PFC by 16 and use the numbers on the graph. Either works though.

Sam1995 said:

so hager 16A 60898 type B will disconnect in 0.01s at 719A. Is this correct?

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Yes, see attached pdf.

Sam1995 said:

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

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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.

Pretty Mouth said:

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.

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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!!

Sam1995 said:

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!!

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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.

Another small detail, but important: r1, rn, and r2 (lower case 'r') are the resistances of the conductors of a ring final, measured end-to-end. I assume in this case you're dealing with a radial circuit? for which you should use R1, Rn, and R2 (capital 'R') to indicate the conductor resistances.

The basic steps are usually easy to follow, but the 't' in your I2t formulae is the one that usually catches folks out as the curves in the regs only go down to 0.1s even though standard energy-limiting breakers operate in less than a cycle of the supply (so below 20ms).

Another oddity is once you get down much below 0.1s there is the issue of where in a cycle of the AC a fault occurs and what that means. Fortunately that answer is "very little" as energy limiting devices like MCB and fuses react to a degree to the fault energy and so laraely compensate for phase of fault occurrence. The axis of time is what is known as "virtual time" since it relates directly to the resulting I2t let-through.

TL;DR the time values are their precisely to compute I2t

So for some fuses (see below) you will see the time axis go down to 0.0001 seconds = 0.1ms = 100us but that does not mean they open in microseconds, what is happening is the fuse is current-limiting as the internal wire arcs so the fault clearing might be a couple of milliseconds but the I2t value is far less (as limited current is below the PFC) but if you use the PFC and virtual time values you get the right answer.

MCB also provide some limitation but unlike fuses they don't limit nearly as well, even though the I2t immediately in the "instant" magnetic region is less then the equivalent fuse. Breakers have a mechanical opening time that speeds up a bit as PFC rises, but not nearly as quickly as the I2t rises, hence the typical shape shown in the Hager graph in post #4 of this thread.

Pretty Mouth said:

@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.

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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.
Therfore we calculate based on the PFC at the end of the circuit as it will automatically comply at the origin if it complies at the end, but not necessarily the other way round.

davesparks said:

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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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.

However, for breakers that is not the case. While the equivalent disconnection time still drops it is not fast enough to compensate for the square-law of the I2 current factor rising.