adiabatic equation reg 543.1.3

[tony mc]

Can any one help please
on a tncs system (PME)
100amp main fuse bs1361 100amp
I am trying to work out the maths for the earthing is there a simple way all looks a little confusing
S is the csa of the conductor 25mm
I is ???
t is page 244 the table says 630A
K = ??
so 630A squared x 5 (5sec disconection time
Now I am stuck
Cheers

Where do you get 630 A from on PME as this equates to a ZE of 0.36 ohms.

 

[Guest 004]

630A is the fault current
he has an Ze of 0.11, well below the max allowed of 0.36

 

[Martin Bradley]

Sorry be he has a fault current off 2090 amps

Uo/Z

 

[tony mc]

Im missing something here 230/.11 gives you 2090 A so thats what he should be using on the adibatic.

 

[Guest 004]

read the whole thread peeps
type 1361 fuse to trip in 5 seconds (page 244 regs) = 630 amps

 

[Martin Bradley]

Sorry i missed a Zero

 

[yellowvanman]

In which case S could be a lot smaller - about 6mm - but still the bonding needs to be 10mm!

 

[pushrod]

Fault current comes from appendix 3

 

[Guest 004]

from the regs page 128

I is the value in amperes, of fault current, which can flow through the protective device (in this case 1361)

also shows the adiabatic in full, page 128

 

[Martin Bradley]

Sorry chap you have it wrong.
as his impedance is 0.11
voltage is 230.
230/0.11=2090 If

 

[Martin Bradley]

as we have all said 10mm2 will do.LOL my nuts are hurting.
Have a real close look at my picture, and all will come clear.

 

[tony mc]

Im sorry but I just can't see where the 630 A comes from if his ZE is 0.11.

230/0.11

2090 x 2090 x 0.01 square root / k 115 or 143
s = 1.12mm or 1.5 mm

 

[pushrod]

Sorry chap you have it wrong.
as his impedance is 0.11
voltage is 230.
230/0.11=2090 If

A fault current of 2090A could not flow thru the protective device.

 

[pushrod]

Im sorry but I just can't see where the 630 A comes from if his ZE is 0.11.

230/0.11

2090 x 2090 x 0.01 square root / k 115 or 143
s = 1.12mm or 1.5 mm

For I you take the fault current that can flow when you take account of the impedance and the protective device. The impedance in this case does not limit the fault current so the PD does. App 3 for a BS1361 100A for 5s is 630 A

 

[Martin Bradley]

Thats the figures he has given, teach me somthing new, and i might shine your shoe's. 1361 100A HE SAY'S THE IMPEDANCE VALUE IS .011
SO I CALC V/Z =2090A
adiabatic equation. works out at 5.74mm2. so i would use 10mm2 as min.