Example of use of Adiabatic Equation.

[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.