Help with ring main calculations

[i=p/u]

hes a joiner

 

[mechelec]

Max voltage in uk should be 253V not 254V. For design purposes 230V should be used. If you have got 254 V you should get the DNO out.

 

[SirKit Breaker]

couldnt of put it better myself

And I couldn't have put it at all......

 

[pushrod]

why by 4. just curious

just to expand on Sintra's earlier answer - the resistance per metre is divided by 4 in a ring because the distance travelled by the current is effectively half the length of the ring and there are 2 paths, ie they are in parallel.

 

[Sparky Ninja]

Come on everybody, sintra and malcomsanford have bent over backwards on this one.. hit that 'Thanks' button...

If your using a conductor within the SWA as your CPC then you would just use line and CPC conductor as your R1 + R2 in the design stage.

If your designing a circuit where you feel you need to combine a conductor and the SWA as a CPC, though never seen this or done this myself then you would use a combined design to get that R1+R2

As you know though if you take Zs by measurement say on a PIR or extending a circuit then you will get, or should get a really lower than calcualted Zs because of the Parallel paths, which will included your SWA .

This is why unlike the nice people at NICEIC I like to do a Zs by measurement, it's ok doing Zs by calcualtion on a design, but when there is a fault on an installation it won't happen on a piece of paper it will happen when there is bonding. parallel paths etc etc and a truer reflection of what is going to happen to that fault.

I honestly think the NICEIC have advised the calculation method becasue they feel that some, lets call them electrical operatives, are not capable of live testing, and they maybe right!!!

Words can't describe it..

 

[i=p/u]

Originally Posted by i=p/u
couldnt of put it better myself
And I couldn't have put it at all......

but onlky thing you know

 

[Dave_]

If your designing a circuit where you feel you need to combine a conductor and the SWA as a CPC, though never seen this or done this myself then you would use a combined design to get that R1+R2

Hi malc,

I suppose you would calculate both resistances and use the lower value? how would you 'intermingle' them in a calc? not sure on that one, not something i've covered boss.

 

[Sintra]

To calculate R1 + R2 on an SWA using both the armour and a core.

Resistance of the core and the resistance of the armour have to be added together to get the R2. The figures i am going to use are purely for an example so don't shoot me down on it.

Resistance of core = 0.2 ohms (R1)
Resistance of armour = 0.5 ohms (R2)

We then need to add the 2 values of parallel resistances together.

1/Rt = 1/R1 + 1/R2

1/Rt = 1/0.2 + 1/0.5

1/Rt = 5 + 2

1/Rt = 1/7

Rt = 0.143

So your combined value of both resistances which is the R2 is 0.143 ohms.

This is then added to your R1 value which in this case will be 0.2 ohms.

R1 + R2 = 0.343 ohms

 

[pushrod]

Hey Sintra you need to change that signature
____________________________________________________________________________________

I'm a mathematician and an electrician !

 

[Sintra]

lol....

 

[Dave_]

To calculate R1 + R2 on an SWA using both the armour and a core.

Resistance of the core and the resistance of the armour have to be added together to get the R2. The figures i am going to use are purely for an example so don't shoot me down on it.

Resistance of core = 0.2 ohms (R1)
Resistance of armour = 0.5 ohms (R2)

We then need to add the 2 values of parallel resistances together.

1/Rt = 1/R1 + 1/R2

1/Rt = 1/0.2 + 1/0.5

1/Rt = 5 + 2

1/Rt = 1/7

Rt = 0.143

So your combined value of both resistances which is the R2 is 0.143 ohms.

This is then added to your R1 value which in this case will be 0.2 ohms.

R1 + R2 = 0.343 ohms

Cheers mate! Issue i have is with thermal constraints, The K value would be 46 for the armour and 100 for the core (5467 or 6724) do you think you could apply a similar formula to calculate this? Not sure....... but wouldn.t really matter cos im betting either would suffice

 

[pushrod]

Cheers mate! Issue i have is with thermal constraints, The K value would be 46 for the armour and 100 for the core (5467 or 6724) do you think you could apply a similar formula to calculate this? Not sure....... but wouldn.t really matter cos im betting either would suffice

When you look up your R2 value for the armour it has already taken account of the fact that it is not copper - is this what you mean? In normal circumstances if you have a spare core in the armour that will alway suffice as the cpc. The only situation i can think of where you might want to use the armour as well, is if you are wanting to use it as a main protective bond. You would however always earth the armour even if you have a spare core as it would then be an exposed-conductive-part.

 

[telectrix]

To calculate R1 + R2 on an SWA using both the armour and a core.

Resistance of the core and the resistance of the armour have to be added together to get the R2. The figures i am going to use are purely for an example so don't shoot me down on it.

Resistance of core = 0.2 ohms (R1)
Resistance of armour = 0.5 ohms (R2)

We then need to add the 2 values of parallel resistances together.

1/Rt = 1/R1 + 1/R2

1/Rt = 1/0.2 + 1/0.5

1/Rt = 5 + 2

1/Rt = 1/7

Rt = 0.143

So your combined value of both resistances which is the R2 is 0.143 ohms.

This is then added to your R1 value which in this case will be 0.2 ohms.

R1 + R2 = 0.343 ohms

very clearly put there sintra.

 

[Dave_]

When you look up your R2 value for the armour it has already taken account of the fact that it is not copper - is this what you mean? In normal circumstances if you have a spare core in the armour that will alway suffice as the cpc. The only situation i can think of where you might want to use the armour as well, is if you are wanting to use it as a main protective bond. You would however always earth the armour even if you have a spare core as it would then be an exposed-conductive-part.

No, im aware of that, what i'm getting at is what k value you would use to make a calculation regarding the thermal constraints of the cable, i would imagine you would make 2 calculations, one using 46 for the armour and one using 100 for the coppe core, then you could use either result to confirm compliance. I was asking if there is a way of making one adiabatic calculation but some how using both values, as the result obtained from making 2 calculations wouldn't necessarily reflect the true size of the earth required... if you know what i mean.... Kinda hard to explain!!!

 

[pushrod]

Got you now (i think) Sorry don't know any formal way of combining the 2 K values - suspect it would a ratio of the csa's of both conductors with the steel corrected (reduced) similar to table 54.7 (but reversed) ie a ratio of the 2 K values. Certainly not quick and simple to do, unless there is a table somewhere that has already calculated it. At least if you use the 46 value for steel you know you have a lot of leeway in the selected csa of the conductor. Which is is probs no bad thing as the quality of an armouring connection can deteriorate over time.

edit but as i said earlier it would only be exceptional circumstances where a similar sized core for a cpc would not be sufficient.