2396 style question, 2400, short circuit, RCBO,

Richard Burns · Apr 11, 2018

I am certainly not an expert, just generally able to apply the requirements of the regs.
To expand on Strimas post, if you have an evenly distributed load on a long lighting circuit, say 100W every 20m then on average you can use the formula you describe as the volt drop will be high because all lights are taking current at the start of the circuit and drop down as you get to the end where only one light is taking current but it will also rise at the end because of the length of circuit, so on average your volt drop would be that for when the load is only half the total design current.
Not an equation I would usually use as you normally need to know the maximum volt drop. but for a rough estimate and a likely successful outcome it would work.

Additionally it is possible on long radial circuits to have a large csa at the start where the load current is high, so as to reduce volt drop and reduce the csa as you get further out and the load current, and so volt drop, is lower, in this case it may be necessary to know the fault current at the points where the csa changes in case the cable does not have the energy withstand of the fault current at origin.
I cannot see why you would ever need to know fault current at the end of the circuit though.

I attach a diagram from a previous post about calculating volt drop that is not totally in line with this post but may help explain the idea.

Daniel Jones · Apr 12, 2018

Sorry guys went off on a bit of a tangent asking about the volt drop. But thank you for your responses.

Sorry Strima, we must have posted those replies at the same time.

The reason I was asking about the fault current was in all the books you calculate the fault current at the extremity of the circuit then use that to calculate the minimum CPC. How ever that is not the largest current that will flow in a fault. A fault close to the source would be the largest. so shouldn’t we calculate the CPC size based on that?. Why do we need to calculate the fault current at the extremity? Is it just to check disconnection times are less than that of table 41.1.

Dan

Strima · Apr 12, 2018

Calculate the fault at the end of the circuit. If you did it at the source the current would be much higher therefore you would need a larger CPC which would in turn increase the fault current meaning you would need a larger CPC... And breathe...

Richard Burns · Apr 12, 2018

I am sorry, I was in error, I was still thinking energy withstand and maximum fault currents. You do not need to know the maximum fault current at the end of circuit but you do need to know the minimum.
You need to know the lowest fault current that can flow to ensure that the circuit will meet the disconnection times in all cases.
As the current will be at a minimum at the end of the circuit (or mid point of a ring) as the resistance of the cables will be highest. You also use Cmin to ensure that even if the supply voltage drops to (just above) the minimum permitted level you will still meet disconnection times.

Daniel Jones · Apr 12, 2018

ok,

a little confused now. sorry for throwing in the volt drop calculate that hasn't helped to flow.

I am trying the establish the minimum circuit protective conductor.
i have a 50mm² line conductor with a 25mm² CPC protected by a bs88 125A. i have determined the fault current at the extremity of the circuit.

Determine Earth Current,
Ief = (Uo × Cmin ∕ Zs)
therefore 230 × 0.95 ∕ 0.081 = 2698A
therefor the device will disconnect in less than 0.01 seconds

Determine Minimum Circuit Protective Conductor,
(S= √ (Ief² × t) ∕ K)
as t is less than 0.01 i use the energy let through = 78400(A²s)
therefore.
√(78400) ∕ 115 = 2.43mm²
as the CPC is 25mm² this is adequate

or does that equation not work and should i just be checking that the energy let through does not exceed the with stand of the cable. I.E
I²t ≤ K²S²
therefore
78400 ≤(115²25²)8'265'625
as the CPC is 25mm² this is adequate.

Dan

Richard Burns · Apr 13, 2018

Possibly it might be worth checking on your level of current training, you say you are studying for the 2396 and so have presumably got a good existing background in electrical compliance.
You started the thread with a fairly high level design query but are now getting confused about basics.

Now I do not profess to know lots about the subject but I tend to work from first principles and apply common sense, this is not always a good idea for the wiring regulations!
Always remember that I am posting my ideas in a similar manner to the way you are posting yours. Always check back with your notes and ensure I am not making errors.

If you have reference books that state that the fault current should be determined at the extremity of the circuit then they are more likely to be correct than I am. However do check the reasoning and intent behind the reason why they are determining the fault current.

In your last post you ask if you should use one of two equations but the equations are the same equation just mathematically rearranged.
S= √ (I² × t) ∕ k
multiply both sides by k
S * k = √ (I²t)
square both sides (or multiply by √I²t which you know equals Sk)
S²k² = I²t
(and if you divide both sides by I² you get t=k²S²/I²)
either way works.

My reasoning for using the PFC as fault current when determining energy withstand or cpc suitability, as opposed to ADS disconnection time, is hopefully explained by the attached diagram.

Richard Burns · Apr 13, 2018

OK so actually looking things up clarifies things.
Because of the nature of the tripping curve for fuses as the fault current decreases the time to disconnect increases.
For circuit breakers there is little difference in trip time once the fault current has reached the level of operating the magnetic trip.
Therefore when considering the energy withstand of the cable, even though the fault current is lower at the extremity of the circuit the increased time means that the cpc has to withstand a greater total energy so this is the worst case scenario.
I considered only circuit breakers where there is not much difference, but even then it is intended that the minimum fault current should be used. (not that I necessarily agree with this).
It would be necessary for checking the breaking capacity of the protective device to consider the maximum fault current at the origin if the circuit, but in other cases it is the minimum fault current (or maximum resistance).
A lack of attention to detail and just randomly answering lead me to the erroneous conclusion of thinking I²t included time. Sorry for causing confusion.