How did they calculate this ?

[Dylboy]

Happy Friday all!

Anyone know the formula used for the attached screenshot ? (The table )

The numbers in Ed 2 are different to Ed 3.0

Also I was taught to do the touch voltage formula to get the Recc

Just want to dive deeper into this


Cheers
Dyl

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[Derlyn]

Happy Friday all!

Anyone know the formula used for the attached screenshot ? (The table )

The numbers in Ed 2 are different to Ed 3.0

Also I was taught to do the touch voltage formula to get the Recc

Just want to dive deeper into this


Cheers
Dyl

Sent from my SM-N960F using Tapatalk

The readings shown in table 8.1 above are just plain wrong.

They were also incorrect in edition 1.8 and edition 2.
In both the above versions the comma needs to move one numeral to the right.

Formula is R(ohms) = V divided by 2 times I.

example for 63A R = 240/126 = 1,9 Ohms.

The reason 2 x I is used is that in the event of a short, the resistance must be such that at least twice the rated current of the protection must be able to flow in the circuit ensuring that the overcurrent protection will do it's job and trip.

Peace out .. Derek Stuart

[Dylboy]

Now that makes much more sense hahaha, that is what I call the Rmax formula or Loop formula. it even states above the table that if the ling lead test can't be done then a loop test is done and thus that formula is used.

Very intresting stuff here....

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[Dylboy]

It appears I need some work on this and what I was taught in my apprentice days may not be correct.... anyways,

So here is my process... back to basics so here is a every day sorta situation...

A 2.5mm flat twin and earth cable which has a 1.5mm CPC (Earth) on a 20A CB. How long can I run this cable? ( assume a 10A load at the end)

If i do the Rmax = v / 2 x I I get 5.75 ohms.
If I then apply it to Recc = p x L / A
(A being 1.5mm² as doing line to E fault) I get a length of 386 meters...

If I do the Touch Vtage formula so as to keep it to 30V ( in book says do calcs at 30V )

TV= 2 x I x R
30 = 2 x 20 x R
30= 40 x R
R = 0.75 ohm

Then again the Recc formula to calc length
Recc = P x L / A (A being 1.5mm²)
0.75 = 0.0223 x L / 1.5
Legth = 50meters

Then on page 168 of Ed 3.0 has the formula for the max length of the Ecc (CPC, Earth)
And when i do that with the 0.8 factor out i get 48meters (closer to the TV side of 50 meters)


Then the bloody table above also has very low Ohm ranges haha.

So anyone wiser than me on this know the best answer or method to determine that question ?

Also in section 8 it says that if the Recc can not be checked (long wonder lead) then the method for the loop test must be done which then included the Rmax formula which is 5.75ohm, I know it then has a loop so "double distance) and the phase and Earth are diffenrft CSA so not sure how to get that to a single Earth restaince number to try do the math...

I am down a rabbit hole at the moment with all this as need to prove to current boss that there may be an issue in the design but I want to make sure I am not to far gone and lost...

Edit: Vd excluded at this time, as could be a 5A load but this is just for some math and getting out the rabbit hole.


Edit again: had a brain wave at 1 this morning when small was awake that the TV formula I use is the same as the Rmax formula.... so the diffenrce then being the voktage of either 230v or 30v
Sent from my SM-N960F using Tapatalk

[Derlyn]

It appears I need some work on this and what I was taught in my apprentice days may not be correct.... anyways,

So here is my process... back to basics so here is a every day sorta situation...

A 2.5mm flat twin and earth cable which has a 1.5mm CPC (Earth) on a 20A CB. How long can I run this cable? ( assume a 10A load at the end)

If i do the Rmax = v / 2 x I I get 5.75 ohms.
If I then apply it to Recc = p x L / A
(A being 1.5mm² as doing line to E fault) I get a length of 386 meters...

If I do the Touch Vtage formula so as to keep it to 30V ( in book says do calcs at 30V )

TV= 2 x I x R
30 = 2 x 20 x R
30= 40 x R
R = 0.75 ohm

Then again the Recc formula to calc length
Recc = P x L / A (A being 1.5mm²)
0.75 = 0.0223 x L / 1.5
Legth = 50meters

Then on page 168 of Ed 3.0 has the formula for the max length of the Ecc (CPC, Earth)
And when i do that with the 0.8 factor out i get 48meters (closer to the TV side of 50 meters)


Then the bloody table above also has very low Ohm ranges haha.

So anyone wiser than me on this know the best answer or method to determine that question ?

Also in section 8 it says that if the Recc can not be checked (long wonder lead) then the method for the loop test must be done which then included the Rmax formula which is 5.75ohm, I know it then has a loop so "double distance) and the phase and Earth are diffenrft CSA so not sure how to get that to a single Earth restaince number to try do the math...

I am down a rabbit hole at the moment with all this as need to prove to current boss that there may be an issue in the design but I want to make sure I am not to far gone and lost...

Edit: Vd excluded at this time, as could be a 5A load but this is just for some math and getting out the rabbit hole.


Edit again: had a brain wave at 1 this morning when small was awake that the TV formula I use is the same as the Rmax formula.... so the diffenrce then being the voktage of either 230v or 30v
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Ok, just to confuse the issue I get 61,1 metres.

My method is as follows. I always first calculate for a 1A load.

Max volt drop allowed = 5% ( 11v )
2,5 mm cable volt drop per Amp per meter is 18mv.
Therefore length allowed at 1 Amp = 11000mv / 18mv = 611 meters.
Therefore length allowed at 10 Amp = 611/10 = 61,1 meters.

Peace out .. Derek Stuart

[Dylboy]

Hahaha I think I am far gone and confusing myself, if I work with the assumed 10A then yes I get same answer.


Bascially I am trying to work out that if we do the 61 meters to stay in the VD but now what happens if at that 61 meter point (the furthest point) the line and Earth had to touch.... would there be enough current to trip the CB? And what is the calculation to to tell this or get the current?

I may need to gets ducks in a row and actually figure out exactly what I need to know as it may be confusing myself with the UK stuff of the adiabatic equation and all sorts



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[Derlyn]

Resistance of cable = 18 milli ohms per meter. That is why there's a volt drop of 18mV per Amp per Meter.

Therefore total resistance of cable = 61,1 x 2 x 0,018 ohms = 2,2 ohms.

I = V/R = 230/2,2 = 104,55 A.

That should be enough to trip the breaker.

Peace out .. Derek Stuart

[Dylboy]

Perfect, I feel I will use this method and my own and use the Rmax or the double the amps to trip CB as it makes the most sense to me and that table is just whack...

Thanks so much as well for some light in this tunnel I have been in habah.

Also bud must that total resistance not be divided by 2 as the mVAm table includes the return path ?



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[Derlyn]

@ Dylboy.

You are correct. Just checked conductor specs of voltex.

2,5mm. 7,41 ohms/km.

So that table does take the return path into account.

That means it's gonna be 209 Amps. Even better.

Peace out .. Derek Stuart

[Dylboy]

Ah lekker ! Then ya even better!



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