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
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[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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[Dylboy]

I was then doing more reading on Touch Voltage and came across the link from many moons ago which once more throws a spanner.... hahaha

https://www.theforumsa.co.za/forums/showthread.php/9042-Touch-Voltage

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[GCE]

Hi

I keep reading this thread and think you are confusing me or you are confusing yourselves.

You are taking 2 different regulations and test readings and combining them as one.

8.6.3 - Resistance of earth continuity conductor - Is the resistance of the earth wire used after the point of control and has been calculated to prevent touch voltages etc - The resistance you measure may not be higher than what is given in the table.
To make our lives easier all the clever people have completed the calculations and given us table 6.28 to ensure that earth cables are pulled in first time correctly

The values of the resistance given in Table 8.1 has become lower since they brought in the regulation that the phase conductor and earth conductor need to be the same size which is why it has changed from Ed2 to Ed 3 - Prior to that ED 2 is the same as ED1.1
You cannot use R= V/( 2*I) you need to stick to the tables otherwise your COC is incorrect

8.6.5 - Earth fault loop impedance at main switch - is measuring/calculating the supply authority earthing and ensuring that the neutral earth is bonded in the sub station - without the neutral bonded to earth there will be no reference to earth and so it cannot trip on an earth fault as there will be no current flow or very little.
Since they never ( hardly ever) run an earth wire from the sub to the premises we need to check that the earthing will be good enough to create a current flow in the event of an earth fault

Impedance is calculated using resistance , inductive reactance and capacitive reactance

[Dylboy]

Hey GCE

Yes ya I have been and still a tad confused, but what you said i agree with and makes sense as well.

Where is the regulation that the earth must be same size as lhase conductor ? Would that then not make FTE cable wrong?

One note for 8.6.3 it states "

In instances where it is not
practical to measure the resistance, suitability of the ECC can be confirmed
by doing an earth loop impedance test from the point of consumption as
prescribed in 8.6.5.1."

And in 8.6.5.1 it states

"8.6.5.1 At the main switch, the earth loop impedance shall be such that an
earth fault current double the rated current (or higher) of the main protective
device automatically disconnects the supply to the installation. Table 8.2
indicates the earth fault loop circuits for different distribution systems."

So that adds a bit more confusion now too as then the double the current formula is there but it does not meet TV?

Maybe I am now way over thinking all of this but I want to understand it properly rather than just follow a table for circuit design and CoC etc. For example a very old hosue is not to ED. 3 so does that mean if the Recc is less than the table it's a fail but Ed.2 had a higher value etc...

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[Derlyn]

Very interesting thread.

I have always been under the impression that the ECC can be downgraded one size from the phase conductor.

If this has been changed then it's not only FTE that does not comply, but also AIRDAC.

Should one pull in seperate ECC conductors when using the abovementioned cables ?

Peace out .. Derek Stuart.

[GCE]

Very interesting thread.

I have always been under the impression that the ECC can be downgraded one size from the phase conductor.

If this has been changed then it's not only FTE that does not comply, but also AIRDAC.

Should one pull in seperate ECC conductors when using the abovementioned cables ?

Peace out .. Derek Stuart.

The regs state

can carry earth fault currents, protective conductor currents and leakage currents to
earth without danger from thermal, thermomechanical and electromechanical
stresses and from electric shock that arise from these currents,

Ecc etc are designed to deal with thermal rise and electro mechanical stress and the calculation to work out the earth resistance and ability to get rid of a a fault current is complex hence the reason they refer us to table 6.28 - table D.2(b) and manufacturers spec

The regs allow for cables to be treated differently to bare copper

6.12.1.1 An earth continuity conductor shall
a) consist of compatible conductors,
b) if it forms part of a cable other than a flexible cable, comply with the relevant
requirements of the standard for the cable,
c) if it forms part of a flexible cable, be of the same material as, and have a
nominal cross-sectional area at least equal to, that of the largest phase
conductor,
d) be able to carry the prospective fault current without excessive temperature
rise of the conducor, within the disconnecting time
e) if it does not form part of a cable or flexible cable, have a nominal crosssectional
area at least equal to that determined in accordance with
table 6.28(a), as follows: