Calculating csa of protective conductors

[RMS]

hi all,

when calculating the cross sectional area of protective conductors using the adiabatic equation, i am struggling to understand the operating time to apply when using a BS60898 overcurrent device.

When i refer to the time/current graph in BS7671 it shows that the circuit breakers will operate between 0.1-5 seconds. All my calculations for prospective fault current fall the 0.1-5 seconds on the graph.

Does anyone know the figure i should be using or should i be using 0.4seconds for sockets and 5 seconds for fixed equipment. ie. worst case scenario. My calculations using this method gives me very larged size protective conductors. My Zs values are well with limits.

any help appreciated.

 

[ricicle]

What figures do you have, what size/type of cable etc.What size are you using to calculate the Zs?

 

[RMS]

ok heres an example, please correct me if i am wrong,

using the equation S = (square root) I² *t / k

I = Uo / Zs with a Zs =1.32ohms 240v / 1.32ohms = 181.8A

When i refer to the time/current graph for BS EN 60898 for a 6A device it falls over the straight line which is equal to 0.1 - 5 seconds. Its what figure do i use for t.

The cable is PVC twin and earth(1.5mm² with 1.0mm² cpc copper with operating temp. of 70 degreesC) so k=115.

so,S = (square root) 181.8²A * 0.1 / 115 = 0.50mm² which is acceptable.

or using 5 seconds,

S = (square root) 181.8²A * 5 / 115 = 3.5mm² which is unacceptable

 

[Adam_151]

If you are meeting ELFI requirements the breaker will trip in 0.1sec or faster, the 0.1 sec value is the one to use* Do remember to check the calculation for the start of the circuit as well, the fault current will be greater there, so you are basically plugging Ze into the equation instead of Zs because faults don't only occur at the most distant point


*actually once you get to 0.1 you should be looking up the I²t from the makers data, but if your calculation works with I² x 0.1 then that will be ok, because at least for a breaker the makers figure will be lower

For breakers the figure will be on a graph listing fault current against I²t See here for the one for MK sentry B curve breakers

For a fuse the value is going to be a constant value

 

[ricicle]

Good lad Adam.I go and sort out my daughters bedtime come back and its sorted

 

[RMS]

Thanks for that Adam.

Good point about calculation at the closest point of the circuit. Although when calculating this proves to be a lot different.

Say I have 1.5/1.0mm² cable feeding lights protected by a BS EN 60898 MCB and the Ze = 0.40ohms.

I = 240V / 0.40ohms = 600A

S = sqrt 600A² * 0.1 / 115 = 1.65mm² and this would be unacceptable for my circuit.

I would have to go with a 2.5mm² earth.

Is this correct? and how would this work when the Ze is even lower which is very common, as the csa would be even larger?

 

[RMS]

Or would this be a time to get hold of the maker's data to get and exact time of disconnection as suggested?

 

[Adam_151]

We then have to go to the makers data, and compare I²t of the CPD to the K²S² of the cable, an analogy I picked up from somewhere is that of a glass and a bottle of wine, pour the glass into the bottle and it goes in nicely, likewise if I²t is less than K²S² the breaker shuts off before damage occurs, pour the bottle into the glass (K²S² less than I²t) and you end up with split wine (a shame) and melted cables (somewhat more of a problem!)

Right then, K²S² for 1mm² T&cpc is 115² * 1² giving 13,225

look at my data above for MK breakers (its available for all makes... or should be, but the MK is a nice big graph), to make reading it easier, lets say we want a max I²t of 10,000, so look up the side for 10,000 find the line and go across to the red line labeled for 6A breakers then when you hit the red line, read it off the bottom axis... which gives a fault current of a bit over 4kA before we start to see I²t values of that range

 

[RMS]

thanks once again Adam,

You should consider taking up teaching