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

 

[tecmatt]

does the equation still work for 3 phase? cant see 3 phase mentioned in the regs about the equation anywhere. just want to make sure im working it out right. thanks

 

[tony mc]

Same calculation making sure your K factor & PFC is correct x 2 ( 1.73) the highest line to neutral current.

 

[tony mc]

This topic might cause some debate re the size of the main Earth conductor re 3 phase.

EG Had one yesterday with a ZE of 0.05 ohms and a PFC of 3.2 kA on a TNS system 3P supply.
Main fuses BS 88 100A on each phase.

230/0.05 = 4600A

So a Adiabatic on this for the instalation would be. 4600 x 4600 x 0.01 square root/ 143 = 3.2

So S would be 3.2 csa. (As per calc)

However with it being 3 phase I did it this way and might be wrong.

PFC is 3.2 kA X 2 =6.4KA. Highest reading between phases and p-e

6400 x 6400 x 0.01 = 409600 Squre root = 640 /143 = 4.4

So S would be 4.4 csa. (As per calc)

I used 0.01 as t for instantaneous operation Fig 3.3B Appendix 3 and 143(k ) as per table 54.2

Any views welcome as tecmatt has quite rightly stated all calcs are based on single phase

Thanks !!!"!!

 

[Guest123]

Careful now chaps......dont forget you ae calculating the minimum size of the CPC so L-L & L-N KA values dont come into it as that isn't where EARTH fault current goes is it.

PEFC is the value you should use in the calc not PSCC, no need to multiply by 2 or 1.732 as it is a single phase calculation between a phase & earth.

 

[tony mc]

Thanks Lenny the example you have given makes total sense!!

 

[tecmatt]

Hi guys
needs some help. been a while since iv done this and seem to be having a blonde moment.
trying to work out the min csa of the cpc on a 3 phase 20A circuit. currently the cpc is only 2.5mm. Ze=0.05 R1+R2=0.23 k=115.
Id appreciate any advise/help, I keep getting odd results and other people iv spoke to are in the same position of not doing it for years

 

[Guest123]

Hi guys
needs some help. been a while since iv done this and seem to be having a blonde moment.
trying to work out the min csa of the cpc on a 3 phase 20A circuit. currently the cpc is only 2.5mm. Ze=0.05 R1+R2=0.23 k=115.
Id appreciate any advise/help, I keep getting odd results and other people iv spoke to are in the same position of not doing it for years

Ok so circuit Zs = 0.05 + 0.23 = 0.28ohms.

Fault current = 230 / 0.28 = 821.4A (I)

I assume the protective to be a 60898 'B', with a fault current of that value the device will operate in 0.1 so (T) = 0.1

Now we have (I) (T) & (K)

S = Sqroot of (I squared x T) / K

I squared = 674697.9

674697.9 x 0.1 = 67469.7

Sqroot of 67469.7 = 259.7

259.7 / 115 = 2.25, nearest available size = 2.5mm.

 

[redface]

so im trying to work out the minimum size main earth for main supply ze=0.11 its a tncs supply i would say k=143 and for the main consumerunit i would say its 5 second disconection time so
230/0.11=2091
so doing the caculation i get 32.9mm and this cant be right what am i doing wrong it must be T
a bs 1361 will go quicker than 0.1 s at that fault curent.
so taking pg 244 time/current characteristics for 1361 60 a fuse at 5 seconds is 330A so doing the calc again i get 5.16mm this is better but what is right or for a consumerunit disconection time is it 0.4
please help i done it on a pir with a 6mm main earth which i doubled up to 12mm but minimum size is 5.16mm now this is safe just not conforming to bs7671 2008

 

[andyb]

so im trying to work out the minimum size main earth for main supply ze=0.11 its a tncs supply i would say k=143 and for the main consumerunit i would say its 5 second disconection time so
230/0.11=2091
so doing the caculation i get 32.9mm and this cant be right what am i doing wrong it must be T
a bs 1361 will go quicker than 0.1 s at that fault curent.
so taking pg 244 time/current characteristics for 1361 60 a fuse at 5 seconds is 330A so doing the calc again i get 5.16mm this is better but what is right or for a consumerunit disconection time is it 0.4
please help i done it on a pir with a 6mm main earth which i doubled up to 12mm but minimum size is 5.16mm now this is safe just not conforming to bs7671 2008

I think it must be your maths

As you say, fault current of 2091
so S=sq rt ((2091x2091)0.1)/143
=sq rt(4372281 x 0.1)/143
=sq rt437228/143
=661.23/143

S=4.6mm

Don't forget though that the main bonding conductor will need to be 10mm.

 

[Mr CJ]

Many thanks Dave.

 

[sinemon]

However he did add it in his working out and not multiply (0.08+0.037)

 

[brizospark]

where do i find k values?

 

[topquark]

Table 43.1 in BRB/BGB (page 77 in BRB, page 85 in BGB).

 

[brizospark]

what do k values stand 4?

 

[topquark]

Resistivity of the material the conductor is made of, taking into account temperature and heat capacity of the conductor material.

 

[topquark]

Resistivity of the material the conductor is made of, taking into account temperature and heat capacity of the conductor material.

There is a more complete (and technically accurate) definition just above table 43.1.

 

[ian65]

Very good example thanks for sharig it i am due to take my exam next month, and i am told it is 50/50 if this question comes up

\thanks once again Ian.

 

[MacSparky]

Its been a while but I have my NAPIT inspection tuesday week and jumped on here for a refresher....lovely example of the adiabatic and easy to understand when put the way it was...thanks Dave....i'll keep reading....best sparky forum IMO.

 

[SIIVION]

I too have just joined AEI Cables. Thanks for the tip AEI has some very useful info.

 

[pactrol]

Hi If I may chip in what you say is correct but the driving force behind pme is to save the REC money on not having to provide an earth conductor .
my view on pme is its dangerous as it relys on the consumer having to install larger than normal earthing conductors in case the pme neutral fails otherwise you have a very dangerous situation give me TN-S any day

 

[tony mc]

If the PME neutral was to fail then no matter what size earthing conductor you had fitted the earth would still fail as the return path for the earth is the neutral. (Main bonding conductors are sized according to the neutral)

I was under the impression that if the DNO did not supply a earth connection then it was down to the customer who would then have to go down the TT route.

There have been a few posts recently re high ZE readings on TNS systems which turned out to be the suppliers sheath broken!

 

[jojerome12]

yes thanks for that,

 

[jojerome12]

that is a great help, always need to be aware, especially when you have not been doing it for a awhile

 

[danny26]

wqesasa

 

[pactrol]

Hi Tony mc

Im sorry I have to disagree with you pme installations have multiple earth spikes as back up for just such a senario they usually fit a earth rod at the transformer & another where the service enters the building so loss of a neutral will divert to the earthing system back to the transformer hence the need for large earth wires & bonding as every customer fed of that transformer has a neutral return via the earth. ( very dodgy but the rec don,t follow Bs7671 )