High Ze causing headache !

[sythai]

Hi Guys

Any advice/ pointers would be appreciated

Have ‘pretty big’ new build house we’re wiring from scratch.

New supply in from DNO is PME, but we have Ze of 0.51

Are DNO obliged to keep this under 0.35?

Wouldn’t be that much of an issue normally if regular size house and meter was attached to house.

Before we came along on the scene Client had separate garage block built, meter permanently sited in here.

We’ve run a SWA sub-main across to house DB 25metres away, but struggling on suitable overcurrent protection. Even at 80amps BS 88…. Max EFLI 0.40!

Don’t really want to be putting new build on TT system, if it comes to it.

Thank you
Sy

 

[mattg4321]

100mA S Type RCD on sub main if it comes to it. Then make sure all RCD protection disconnects both line and neutral downstream

 

[Julie.]

Thanks Pete

How does that work with my 80amp sub-main supply though max EFLI is 0.40 for BS-88 ?

My thinking for TT as an option, was if so could then put all on an RCD (time delayed for sub / house db rcbo's) and use 1667 for max ZS values.

Maybe I'm missing something - has been long week!

Yes, time delayed rcd on distribution circuits, standard rcd on final circuits - rcds as fault protection.

No need to convert to TT

leave it as TN-C-S and you can still use 0.4s / 5s but add rcd as fault protection where required.

So if the cable from the meter location to cu fails to operate in 5s due to fault, add a time delayed rcd.

If any final circuit fails to operate in 0.4s due to fault add a standard rcd.

If a circuit will operate within these times, then a rcd is not required for fault protection, but may be required for additional protection.

 

[TJ Anderson]

You also need to consider the protection for L-N faults which won't be covered by adding RCD's. This often gets missed.

If you have a circuit that has cpc size equal or greater than the neutral and you have inadequate fault current to operate ocpd fast enough with regard to thermal constraints, then you will have same issue with L-N faults. You can add RCD to cover the L-E fault protection, but L-N will not be covered.

 

[Julie.]

You also need to consider the protection for L-N faults which won't be covered by adding RCD's. This often gets missed.

Yes, you still need overcurrent protection for fault protection, and where necessary overload protection, but there is no requirement to achieve the same trip time.

You also have to achieve voltage drop limits etc, etc - basically all the other requirements remain in place.

Using rcd for fault protection is merely a technique to achieve ads

 

[Pretty Mouth]

You also need to consider the protection for L-N faults which won't be covered by adding RCD's. This often gets missed.

The disconnection times only need to be met for L-E faults, as far as the regs are concerned anyway. Though interestingly there is something in the OSG that leads you to believe they do need to be met for L-N faults. I'll find a page number...

 

[Pretty Mouth]

7.2.4 on p77, note that it doesn't reference any regulation though:

 

[TJ Anderson]

7.2.4 on p77, note that it doesn't reference any regulation though:

View attachment 94443

It's also the thermal constraints part you need to comply with not just disconnection time. I'm not near my big blue book but it is in there. You have to use the 0.1s value of MCB and use it in the adiabatic to confirm.

 

[Pretty Mouth]

The disconnection times are to do with protection against electric shock, so I'm not sure why the OSG asks for them to be met for a L-N fault, as no exposed parts would go live.

Unless, perhaps on a TN-C-S, could a significant voltage appear at the MET due to the high voltage drop under such a fault??

 

[Pretty Mouth]

It's also the thermal constraints part you need to comply with not just disconnection time. I'm not near my big blue book but it is in there. You have to use the 0.1s value of MCB and use it in the adiabatic to confirm.

Indeed, I think I read your post pre-edit so got the wrong end of the stick

 

[Julie.]

It's also the thermal constraints part you need to comply with not just disconnection time. I'm not near my big blue book but it is in there. You have to use the 0.1s value of MCB and use it in the adiabatic to confirm.

Why would you use 0.1s for this calculation?

And of course all other aspects of the regs still apply, the use of an rcd to achieve one aspect of the regs, doesn't mean other aspects are no longer required.

 

[TJ Anderson]

l

Why would you use 0.1s for this calculation?

And of course all other aspects of the regs still apply, the use of an rcd to achieve one aspect of the regs, doesn't mean other aspects are no longer required.

Because you would be trying to meet the current to operate the magnetic part of the MCB which disconnects in 0.1s-5 on the tables. They show the values for 0.1s for you to use in calcs and state 5s (for exactly the same fault current) to show the ADS met incase someone can't work out that 0.1s is less time than 5 seconds. Funny they do that really.

 

[Julie.]

l

Because you would be trying to meet the current to operate the magnetic part of the MCB which disconnects in 0.1s-5 on the tables. They show the values for 0.1s for you to use in calcs and state 5s (for exactly the same fault current) to show the ADS met incase someone can't work out that 0.1s is less time than 5 seconds. Funny they do that really.

But we know the trip times in the example by the op is not within the instantaneous portion of the characteristics, so using that would be totally inappropriate.

If we use his 25m x 25mm^2 cable and say the original design (assuming Ze = 0.35) then the current would be around 560A, which would trip in 2 secs or so. That time and current should be used for the adiabatic calculations, which I guess would be around 4-6mm^2 minimum

With the 0.51 ohm Ze this would reduce the current to around 400A , and 20s trip time, again these figures in the adiabatic would give around 12-15mm^2 min cable.

 

[TJ Anderson]

But we know the trip times in the example by the op is not within the instantaneous portion of the characteristics, so using that would be totally inappropriate.

If we use his 25m x 25mm^2 cable and say the original design (assuming Ze = 0.35) then the current would be around 560A, which would trip in 2 secs or so. That time and current should be used for the adiabatic calculations, which I guess would be around 4-6mm^2 minimum

With the 0.51 ohm Ze this would reduce the current to around 400A , and 20s trip time, again these figures in the adiabatic would give around 12-15mm^2 min cable.

But we know the trip times in the example by the op is not within the instantaneous portion of the characteristics, so using that would be totally inappropriate.

If we use his 25m x 25mm^2 cable and say the original design (assuming Ze = 0.35) then the current would be around 560A, which would trip in 2 secs or so. That time and current should be used for the adiabatic calculations, which I guess would be around 4-6mm^2 minimum

With the 0.51 ohm Ze this would reduce the current to around 400A , and 20s trip time, again these figures in the adiabatic would give around 12-15mm^2 min cable.

But we know the trip times in the example by the op is not within the instantaneous portion of the characteristics, so using that would be totally inappropriate.

If we use his 25m x 25mm^2 cable and say the original design (assuming Ze = 0.35) then the current would be around 560A, which would trip in 2 secs or so. That time and current should be used for the adiabatic calculations, which I guess would be around 4-6mm^2 minimum

With the 0.51 ohm Ze this would reduce the current to around 400A , and 20s trip time, again these figures in the adiabatic would give around 12-15mm^2 min cable.

Sorry, yes you are absolutely right. I was generalising regarding final circuits and not his particular circumstance with a BS88. I had skimmed down through the posts.

The point I was getting at is that is that L-N often does not get considered when people are applying RCD's to cover fault protection when they can't meet max Zs by OCPD.

 

[Julie.]

Sorry, yes you are absolutely right. I was generalising regarding final circuits and not his particular circumstance with a BS88. I had skimmed down through the posts.

The point I was getting at is that is that L-N often does not get considered when people are applying RCD's to cover fault protection then they can't meet max Zs by OCPD.

Zs is only applicable to ads it isn't applicable to fault protection for L-N or L-L faults.

 

[TJ Anderson]

Zs is only applicable to ads it isn't applicable to fault protection for L-N or

Zs is only applicable to ads it isn't applicable to fault protection for L-N or L-L faults.

Thats the problem. The regs only require Zs and don't get people to take Zn because the cpc =/< than the neutral so the fault current will =/> too and still disconnect