imperial cable & adiabatic equation

[BES]

doing pir before cu change. property built 1968 imperial size cable all test results ok. just one circuit is bothering me and i maybe looking into this too much but radial circuit, which originally was for an electric old type two bar fire as informed by owner who lived there since it was built, circuit is now being used for 2g skt & is fed in t&e imperial 3/036 & 1/044 approx 1.3 & 1mm2. rated at 19a method c.so ok to go on 16a mcb . though id check the cpc size as per adiabatic equation as 1mm semed a bit small for thic type of circuit.
s=s/root I2t/k
If=230/0.34zs=676.4sqrd=457516.96x0.1=45751.696s/root=213.89/115=1.85mm2 cpc. cpc of cable too small.
but if i use 0.01 as t cpc =0.58mm2 so cpc ok.
i know we usually use 0.1 as time for type b mcb but if i use this cpc is to small but appx 3 time graphs go down to 0.01 and if this is used cpc is ok. i know if time is less than 0.1 then reg 434.5.2 applies but i cant find much info on manufactures let through energy and when i do i cant make head nor tails of their graphs.
always had good and very welcome help off you people before. hoping you can again
cheers
neil

 

[Deleted member 26818]

Your problem, is that you are using the PEFC value, rather that the value in Appendix 3.
A 16A Type B MCB will operate in 0.1s at an amperage of 80A.
As such you should carry out your Adiabatic equation using both the values from the time current characteristics in Appendix 3.
Using a disconnection time for a lower value of fault current will skew your calculations resulting in a higher value for CSA than is required.
Your calculation should be:
80A X 80A = 6400A
6400 X 0.1s = 640
√640 = 80
80 ÷ 115 = 0.7mm².

 

[BES]

cheers spin
i thought you used the ZS of the circuit (0.34) in this case, to get your If of the circuit (Uo/ZS=If) in this cicuit 676.4amp. the pefc of the installation is Uo/ZE in this case 230/0.21 (tncs)=1.1 KA
cheers
neil

 

[i=p/u]

like this thread!!! understand fomurlae but i dont understand what values you use if checking bonding conductors, even tho if you ment to use 10mm ??

 

[i=p/u]

do you not use 0.4 as this worst case??? just asking dont roar at me

Your problem, is that you are using the PEFC value, rather that the value in Appendix 3.
A 16A Type B MCB will operate in 0.1s at an amperage of 80A.
As such you should carry out your Adiabatic equation using both the values from the time current characteristics in Appendix 3.
Using a disconnection time for a lower value of fault current will skew your calculations resulting in a higher value for CSA than is required.
Your calculation should be:
80A X 80A = 6400A
6400 X 0.1s = 640
√640 = 80
80 ÷ 115 = 0.7mm².

 

[BES]

i=p/u
appex 3 brb bs60898 16a trips at 0.1s at 80a so 230/80=2.875 min zs for 16a bs60898. but graph goes down to 0.01s.

 

[i=p/u]

just tudor uses 0.4 when he showed us, so i asuumed worst case.. live and learn... what about checking your bonding in adibiatic what is I

 

[BES]

i=p/u
ref to reg 544.1.1/table 54.8 for main bonding conductors
adibiatic equation s=s/root I2t/k I is If the fault current determined by Uo/ZS OR ZE

 

[i=p/u]

so if had say 3ka fault current and k=115 with t=0.4 , a 16mm earthing conductor would be needed

and would i have a ma distance in that 16mm main bonding conductor?

 

[Deleted member 26818]

i=p/u.
You could use 0.4s, if you wanted to, but as the PEFC is known, and it is higher than that required for the device to operate in only 0.4s, why bother?
In many cases, it won't make any difference. The values for MCBs are the same from 0.1s through to 5s.
ccfc.
The term PEFC applies at whatever point in the installation you are assesing.
In this case the PEFC at the socket-outlet that used to be for the 2 bar fire, which is 676.4A.
The reason MCBs have the fastest operating time as 0.1s, is because they use two methods to operate
One is a bimetallic strip, which will bend as more current passe through it, and the other is an electro magnet which operates as soon as the current is above a certain value.
The electro magnet takes 0.1s to operate irrespective of the current, as long as it is above the threshold.
Of course, nothing is absolute, the values in Appendix 3 are those used for the British Standard applicable to whichever device.
They are the minimum that a test piece must meet in order for the rest of the devices to be marked as being compliant.
It may well be that many devices will operate faster than the minimum required by the Standard, and that individual devices manufactured in the same production run will operate at slightly different speeds.
The fastest that any device can be guaranteed to operate in this country, is 0.01s. That is the length of time it takes for a.c. current to alternate from zero voltage, to the peak voltage at 50Hz.

 

[BES]

in theory yeh apart from answer i got was 16.49 so 25mm cable

 

[BES]

cheers spin
thanks for reply
i bow to the more experianced man. :dunce2: i realise that pefc can be at any part of the installation but i have always used pfc/pefc at origin of installation and zs for the impedance of a circuit at its furthest point and i have always used this for the adiabatic equaition not the value in the appex 3 tables. my main concern was that the cpc of cable was ok as per reg 434.5.2 if time was under 0.1s
cheers
neil

 

[Chr!s]

The I2t refers to the energy let through of the device in a given time. So in your calculation your I2 is 676 Squared. Your "t" is the time in which the device operates. Now for times in excess of instantaneous we need to consult the manufacturers.
Now at that level of fault current, a 16 amp type b will disconnect in .01.

Regards Chris

 

[Guest123]

The I2T values from the manufacturer can also be used in the equation.