sythai

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Apr 29, 2010
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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
 
Are you taking the Ze from source (at the new supply)
 
Listened to a podcast from the E5 group earlier today and DNO guidance states PME ZEs can actually be as high as 0.8. I will try to dig out the pdf
 
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New supply in from DNO is PME, but we have Ze of 0.51
Have you spoken to the DNO about it.
They might sort it out.
 
DNO is not obliged to supply Ze lower than 0.35 ohm; although all DNO strive to achieve it.

It appears to be a common myth!

Edit:

And why on earth (sic) would you consider changing it to TT?
 
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Two 60A / 63A sub-mains, to two separate DBs in the house, circuits suitably divided between them! 😆
 
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Is it the Zs of the sub main that is high or the Ze in the garage block? What size is the SWA?
 
if disconnection times cannot be achieved by the Breakers then residual current devices can be used.

no need to TT. a PME with a slightly high Ze gives a far better EFLI than any TT could achieve.



although a combination of earth rod and pme may be in order in case of dropped neutrals, but that’s a whole different thread on an abandoned proposed reg change.
 
Listened to a podcast from the E5 group earlier today and DNO guidance states PME ZEs can actually be as high as 0.8. I will try to dig out the pdf
I did hear that a while back.... that they wouldn't be concerned unless above 0.8! But always hoped they wouldn't go by that
 
Have you spoken to the DNO about it.
They might sort it out.
Did raise my concerns with Client before Xmas.... DNO apparently came out straight away (I wasn't on site) said no problem all fine !

Re-checked last week still 0.51😠

I'm going to raise it myself again directly with them, making sure I'm onsite when (if) they re-visit.
 
And why on earth (sic) would you consider changing it to TT?
Open to any other suggestions, any advice would be appreciated 😉.... definitely not my preferred method by far
 
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if disconnection times cannot be achieved by the Breakers then residual current devices can be used.

no need to TT. a PME with a slightly high Ze gives a far better EFLI than any TT could achieve.
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!
 
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100mA S Type RCD on sub main if it comes to it. Then make sure all RCD protection disconnects both line and neutral downstream
 
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.
 
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.
 
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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
 
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...
 
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7.2.4 on p77, note that it doesn't reference any regulation though:

S C.png
 
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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.
 
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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??
 
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
 
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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.
 
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.
 
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.
 
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.
 
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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.
 
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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
 
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If you meet ADS by OCPD then you meet thermal constraints. If you meet ADS by relying on an RCD then you may well not meet thermal constraints for L-N by the OCPD which is still covering that.
 
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If you meet ADS by OCPD then you meet thermal constraints. If you meetr ADS by relying on an RCD then you may well not meet thermal constraints for L-N by the OCPD which is still covering that.
Correct, which is why I said earlier on that all aspects of the regs still need to be adhered to.

I grant you that some people may not check various aspects of the regs, usually because "it's always been OK in the past" but I think that happens all the time, I come across so much work where the person installing hasn't thought about it one bit - basically it's never designed, just installed.
 
Correct, which is why I said earlier on that all aspects of the regs still need to be adhered to.

I grant you that some people may not check various aspects of the regs, usually because "it's always been OK in the past" but I think that happens all the time, I come across so much work where the person installing hasn't thought about it one bit - basically it's never designed, just installed.

Incorrect. You only stated that after I had mentioned it. It often gets missed.

Applying RCD to a circuit to cover ADS goes hand in hand with checking Line-Line or Line-Neutral thermal constraints.
 
If you meet ADS by OCPD then you meet thermal constraints.
Not always. With MCBs, the let through energy tends to increase with increasing fault current. So the most onerous part of the circuit for thermal constraint is close to the origin, where ADS is assured
 
Not always. With MCBs, the let through energy tends to increase with increasing fault current. So the most onerous part of the circuit for thermal constraint is close to the origin, where ADS is assured

Generally it does work out with MCB's if you are operating with pfc's withing the breaking capacity of the device. At least it always has in the real world scenarios I have applied it. There is also the other variable of manufacturers data where you can go with faster times than the 0.1s in the regs which I believe has additional considerations attached that I can't remember what they are?
 
Generally it does work out with MCB's if you are operating with pfc's withing the breaking capacity of the device. At least it always has in the real world scenarios I have applied it. There is also the other variable of manufacturers data where you can go with faster times than the 0.1s in the regs which I believe has additional considerations attached that I can't remember what they are?
I don't think there any additional considerations for using manufacturer's data, it just replaces the I2t in the equation AFAIK.

Using, for example, data for a 32A Hager B curve, and using a 1mm2 CPC, it is possible to fall foul for fault currents of ~3kA or greater, and for higher fault currents for lower rated MCBs. So potentially a problem for old ring finals wired with 1mm CPCs, or other such circuits.

Interesting to note that the adiabatic equation only seems concerned with protective conductors, as far as the regs are concerned, as far as I can tell anyway
 
I don't think there any additional considerations for using manufacturer's data, it just replaces the I2t in the equation AFAIK.

Using, for example, data for a 32A Hager B curve, and using a 1mm2 CPC, it is possible to fall foul for fault currents of ~3kA or greater, and for higher fault currents for lower rated MCBs. So potentially a problem for old ring finals wired with 1mm CPCs, or other such circuits.

Interesting to note that the adiabatic equation only seems concerned with protective conductors, as far as the regs are concerned, as far as I can tell anyway
I think the adiabatic appears only to be related to protective conductors in the regs as it's included in that section and the CPC is usually the smallest (or at least the same size).

It is interesting that it is assumed to be limited to disconnection times of less than 5s but essentially the same equation is used by most cable design books for up to 30 secs.

(And they have different k values as the temperature change is included within the calculation rather than within the k value - so you get exactly the same result)
 
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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.
Thanks Julie
 
I don't think there any additional considerations for using manufacturer's data, it just replaces the I2t in the equation AFAIK.

Using, for example, data for a 32A Hager B curve, and using a 1mm2 CPC, it is possible to fall foul for fault currents of ~3kA or greater, and for higher fault currents for lower rated MCBs. So potentially a problem for old ring finals wired with 1mm CPCs, or other such circuits.

Interesting to note that the adiabatic equation only seems concerned with protective conductors, as far as the regs are concerned, as far as I can tell anyway
It was on the IET forum about why we should not use adiabatic for fault currents operating devices <0.1S or >5S. I couldn't remember but just found it again on there.

Going over 5S was more straightforward. Because the adiabatic is simplified it does not allow for heat loss, this results in artificially high csa's

Going under 0.1s was more complex. 0.1S is 5 cycles of the supply waveform and going below that could give rise to greater currents than calculated due to the waveform being asymmetric and becoming distorted.
 
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Interesting to note that the adiabatic equation only seems concerned with protective conductors, as far as the regs are concerned, as far as I can tell anyway

I think the adiabatic appears only to be related to protective conductors in the regs as it's included in that section and the CPC is usually the smallest (or at least the same size).
My bad, it isn't only concerned with protective conductors:

434.5.2
A fault occurring at any point in a circuit shall be interrupted within a time such that the fault current does not cause the permitted limiting temperature of any conductor or cable to be exceeded.
 

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sythai

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