High Ze causing headache !

[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