adiabatic equation reg 543.1.3

[yellowvanman]

If the supply is protected by a 100A fuse then the disconnection time required is 5 secs because its a distribution circuit and >32A. To get 5 secs disconnection the fuse needs to blow with 630A. If you get more - all well and good, fuse blows quicker.

Whilst Ze was 0.01 today what will it be in a few years when other houses are added to the road or other changes are made to the system - the only thing you can rely on is that the Ze will not be higher than 0.35 - because thats what the DNO has to provide!

 

[Martin Bradley]

Pushrod,i just noticed it dooh, that was lucky, i just got in before you could teach me somthing new, no need to shine them shoe's.LOL

 

[pushrod]

TBH although it is good practice on the adiabatic - as yellowvanman said main protective bonding is not calculated with it for a pme - you use table 54.8, or it is not less than 1/2 the size of the earthing conductor in other cases.

 

[tony mc]

Thanks pushrod I should of read the original post properly.

 

[pushrod]

Thanks pushrod I should of read the original post properly.

don't i know that feeling lol

 

[Chr!s]

If the system is TNCS, then the earthing conductor need to also be sized as a main bonding conductor.

1, Earthing conductor = Adiabatic or Table 54.7

2, Main bonding = 54.8

We then use the greater of the two requirements.

If the Ze = .11 then the figure used in the adiabatic is 2090 amps, if the fuse is a BS 1361 then the disconnection time will be .01, the K value = 143 At this level of pefc there will be no fault current limiting by the fuse.

So by my calc Adiabatic = 1.46 mm, but the bonding requirement will more than likely on a domestic be 10mm, so 10mm would be required.Though DNO may insist on a larger size.

 

[pushrod]

If the system is TNCS, then the earthing conductor need to also be sized as a main bonding conductor.

1, Earthing conductor = Adiabatic or Table 54.7

2, Main bonding = 54.8

We then use the greater of the two requirements.

If the Ze = .11 then the figure used in the adiabatic is 2090 amps, if the fuse is a BS 1361 then the disconnection time will be .01, the K value = 143 At this level of pefc there will be no fault current limiting by the fuse.

So by my calc Adiabatic = 1.46 mm, but the bonding requirement will more than likely on a domestic be 10mm, so 10mm would be required.Though DNO may insist on a larger size.

Think it is either - using the easier method of a table (54.7) will put a lot of slack in compared to the calculation.

Do you not think 1.46mm² for an earthing conductor seems too small?

 

[Chr!s]

Do you not think 1.46mm² for an earthing conductor seems too small?

well 1.462 mm is just telling us if our cable meets the adiabatic.

543.1.1 places some restrictions on size as does table 54.1, so the size of the cable is Dependant on various requirements, its just part of the overall selection process.

 

[Chr!s]

Think it is either - using the easier method of a table (54.7) will put a lot of slack in compared to the calculation.

Easier ?

On a large install, that could be a costly method though. Some time the adiabatic may produce a large size than that of table 54.7

 

[pushrod]

well 1.462 mm is just telling us if our cable meets the adiabatic.

543.1.1 places some restrictions on size as does table 54.1, so the size of the cable is Dependant on various requirements, its just part of the overall selection process.

I am no expert by any means but as i understand it the most difficult conditions for a protective conductor are when the fault current is lowest and the disconnection time is longest. In this case the fuse must disconnect in 5s (more than 32A) so this is the time to consider, not the easiest scenario. I think at the start of the thread it actually says to disconnect in 5s. So taking the 5s case you get If to be 630A and a csa for the earthing conductor around 10mm²

543.1.1 is only placing limits of 2.5mm² and 4.0mm² which to my mind are more likely to be applicable as min sizes when you have a TT earthing arrangement. Don't think that 54.1 would apply in most cases with a tn-c-s.

 

[pushrod]

Easier ?

On a large install, that could be a costly method though. Some time the adiabatic may produce a large size than that of table 54.7

"Easier" as in easier to work out

Agree - this is why i originally said either , as opposed to

We then use the greater of the two requirements.

Have never come across examples where calculation by the adiabatic will give you higher answers than table 54.7 IME you get smaller or the same size, but not saying it is a fact

 

[Chr!s]

I am no expert by any means but as i understand it the most difficult conditions for a protective conductor are when the fault current is lowest and the disconnection time is longest. In this case the fuse must disconnect in 5s (more than 32A) so this is the time to consider, not the easiest scenario. I think at the start of the thread it actually says to disconnect in 5s. So taking the 5s case you get If to be 630A and a csa for the earthing conductor around 10mm²

You use the I2T that the cable will be exposed to. This will basically be the product of 230/.11 = 2090, then we look how long that level of fault current will be maintained = t.

The 630a is the minimum current required for a disconnection of 5 sec.

So yes it would be prudent to use the worst case scenario, So in a domestic if we use the value of 16.5 ka(.014) for our front end fault and 690 amp(.33) for our far end.

 

[pushrod]

Hi Chris, i take it then (bearing in mind the original question- just actually had a read thru the whole thing lol) that you are in broad agreement with the 630A scenario. Just one other thing though, i noticed you used a K value of 143 in your calc would it not be 115 (using table 43.1) assuming the tails are 70*C thermoplastic?

 

[Chr!s]

Hi Chris, i take it then (bearing in mind the original question- just actually had a read thru the whole thing lol) that you are in broad agreement with the 630A scenario.

If i was inspecting the installation i would us the PEFC which would be 2090 amps
543.1.3 Note for I. Its a case of what the fault current level is, so I2T < equal K2S2

This is what the regulation asks.

If designing then you could you use the worst case scenario, like the circuits within the onsite guide. This allows standard circuits providing certain parameters are met.

Just one other thing though, i noticed you used a K value of 143 in your calc would it not be 115 (using table 43.1) assuming the tails are 70*C thermoplastic?

Are most earthing conductors incorporated in a cable or bunched with other cables?

I applied K 143, assuming its a single cable and not bunched with others.

And the table you are referring to is for are to be used for overcurrent calculations,Try table 54.2 Protective conductors

Regards Chris

 

[pushrod]

I applied K 143, assuming its a single cable and not bunched with others.

And the table you are referring to is for are to be used for overcurrent calculations,Try table 54.2 Protective conductors

Regards Chris

I looked at 54.3first, and then 54.2, but suppose i got sidetracked by all the chat about 5s so went for 43.1 in the end - when it was initially just a sizing question, not thermal withstand doh!.
Anyway certainly gets the old brain ticking over and good to get different points of view, cheers