Example of use of Adiabatic Equation.

[Guest123]

Many thanks to 'Dave_' for posting this example originally in a thread.


Minimum Size of CPC

Here I will calculate the minimum size of CPC in accordance with regulation 541.1.3 of BS7671.
Primarily I will need to calculate the minimum size of CPC (or confirm the armouring is adequate) of the distribution cables. I already know that the external loop impedance value is 0.08Ω, so I will now focus on the distribution cable to the distribution board, and secondly the supply cable between the suppliers cut-out and the CCU.
I have used data from AEI cables (this can be found in the index) to ascertain the values for conductor and armour resistance.
Distribution circuit to DB3
I will use the equation: Zs = Ze(R1+R2)
Ze = 0.08
R1 = 0.342 mΩ/m (at 90°C, i will not need to apply a correction factor of 1.28)
R2 = 1.2 mΩ/m (armour resistance)
Circuit length = 24m
Therefore: 0.342 + 1.2 = 1.542 mΩ/m × 24 = 37mΩ or 0.037Ω
0.037 + 0.08 = 0.117Ω
I can now calculate the fault current using the following equation:

If= Uo/Zs
Where If = fault current, Uo = line voltage to earth and Zs = total loop impedance
If= 230/0.117=1966A
Using Amtech software at work I have sourced the relevant time current graphs for SQUARE-D BSEN 60947-3 MCCB’s which I will use in the installation (I will go into more detail in part 7, time current graphs can be found in the index)
Using time current graph number 1: t = 0.1s @ 1966A
This is perfectly acceptable, maximum disconnection time = 5 seconds (over 32A)

I will now use the adiabatic equation below to confirm that the armouring of the cable can withstand the level of fault current:
S= √(I²×t)/k

Values for k can be found on p129 BS7671 (in this case table 54.4, the armour of a cable)

S= √(1966²×0.1)/46=13.5mm²

Using the table provided by AEI I can see that a 70mm² four core cable has a 131mm² armour CSA, therefore this is perfectly acceptable.

 

[i=p/u]

i learnt them last week, thanks and il read again

 

[kmcgold]

good example

 

[Sparky Ninja]

would be better if reference to BS7671 could have been used throughout IMO

 

[Amp David]

Found that the other day, really usefull for me and cleared up 99% of any confusion I had.

Registered with AEI cables website, loads of info on data sheets, very handy

 

[ian.settle1]

Even better still would be if he got the formula right

Zs = Ze + (R1+R2)

not the one shown which means Zs = Ze x (R1+R2)

 

[Amp David]

Even better still would be if he got the formula right

Zs = Ze + (R1+R2)

not the one shown which means Zs = Ze x (R1+R2)

Did notice that bit, but knew what he meant

 

[SimpleSimon]

Im a bit confused when trying this out to work out an earthing conductor?
I take it that the fault current would be the PFC so if say trying to work it out for a TNS system

230/0.8= 287.5a, seems a bit low, what am I doing wrong?

 

[johnboy6083]

its right. bear in mind the higher the resistance, the lower the current that wuill flow. 0.8 is the abolsute max Ze that you will have if you are supplied with an earth from the REC. so 287a should be the least amount of fault current you will have also, (assuming 230v).

i remeber when i was doing 2360, that they were telling us about the debate for PME, and one of the issues was that the lower Ze, wopuld cause higher fault currents, and increase the risk of fire, in poorly maintained installations.

The flip side is that along with increased fault currents come quicker disconnection times, which is i believe the driving force behind PME.

 

[SimpleSimon]

Thanks Johnboy6083, I just couldn't believe how low that was, cheers for the reassurance.

 

[B6 7BU]

I will now use the adiabatic equation below to confirm that the armouring of the cable can withstand the level of fault current:
S= √(I²×t)/k

Adriatic
[video=youtube;wHidlxvO0pg]http://www.youtube.com/watch?v=wHidlxvO0pg[/video]

 

[Wilbur]

I note you use the resistance value of R2 = 1.2 mOhms per metre for the armour, does the armour have no reactance? or if so is it so small as to be ignored in calculating earth fault loop impedance values. Ignoring the reactive component would give a lower earth fault loop impedance and hence a higher earth fault current than would occur if reactance was taken into account.

 

[Chr!s]

I note you use the resistance value of R2 = 1.2 mOhms per metre for the armour, does the armour have no reactance? or if so is it so small as to be ignored in calculating earth fault loop impedance values. Ignoring the reactive component would give a lower earth fault loop impedance and hence a higher earth fault current than would occur if reactance was taken into account.

You should take reactance into account in the armour

 

[High Tower]

Top job Len thanks for the re-post!

 

[Wilbur]

Chr!s, It is also my view that armour reactance should be taken into account. The example posted by Lenny only mentions resistance.
Do you know the reactance values for the armour of all BS standard armoured cables for both copper and aluminium conductors, 2 core, 3 core and 4 core?
I have searched the internet unsuccessfuly many times.

Regards

Wilbur