What fault can turn a TT into a TNCS and disable the RCD?

[m4tty]

I've been scratching my head on this one. No good at drawings but if Dave can't get it, I give up as I've got no chance lol. Is it anything to do with high resistance of rod, not enough current flow ? Prob not

 

[SLIPSHOD & SLAPDASH]

It stumped me a bit aswell. All i can think is due to the higher resistance on the earth the same current is flowing down N & E so the rcd does not trip

 

[remustered]

Thank F for that....don't mind so much being thick if the big boys are scratching their heads too...can logic either but not both together...watching the experts with great interest.... .

 

[NozSpark]

Well if it's open to anyone here's the answer

Treat it as a potential divider,,, say 0.2ohms to neutral and 200ohms to earth with a (N-E) fault of say 1kohms, this will give us a fault resistance of 1000.2ohms - 1000 + ((200x0.2)/(200+0.2)) and a fault current of 230mA.

Any fault current down the earth will travel down the earths until it reaches the N-E fault, at which point 99% of the fault current will go down the Neutral and only 1% down the Earth, so that will give 227mA down the Neutral and 3mA down the earth..... Therefore not enough L-N imbalance to trip the RCD

 

[m4tty]

Nice one Noz ... legend scoobed for that

 

[roys]

Like it Noz

 

[park1]

Well if it's open to anyone here's the answerTreat it as a potential divider,,, say 0.2ohms to neutral and 200ohms to earth with a (N-E) fault of say 1kohms, this will give us a fault resistance of 1000.2ohms - 1000 + ((200x0.2)/(200+0.2)) and a fault current of 230mA.

Any fault current down the earth will travel down the earths until it reaches the N-E fault, at which point 99% of the fault current will go down the Neutral and only 1% down the Earth, so that will give 227mA down the Neutral and 3mA down the earth..... Therefore not enough L-N imbalance to trip the RCD

Don't think that's quite right, that's just a simple neutral to earth fault, another fault exists.

 

[kme]

Nozzer is spot on mate - it is worth being aware that a N-E short somewhere upstream can render an RCD inoperative as a safety device, yet not cause it to trip.

One of the (myriad) reasons for doing RCD tests with NO loads connected - just switching off the MCBs doesn`t cut the mustard.

Had this about 10 yrs ago - I replaced THREE F`in RCD, cos they HAD to be faulty - wouldn`t trip on 1x or 5x, or even test button - all MCBs off.

Finally did individuated circuit IRs, and found a N-E "compression connection" ( someone fitting a faceplate had the neutral just outside the edges of the pattress; and tightening the front had crushed it against the metalwork. RCD never tripped.

I believe the newer "electronic" ones do not suffer the same problem ( it used to be a well known failing), but it can still occur.

KME

 

[Andy™]

One of the (myriad) reasons for doing RCD tests with NO loads connected - just switching off the MCBs doesn`t cut the mustard.

i would say even more reason for testing RCD with the circuit connected - there is no point testing RCD, finding it wont trip, then isolating everything else to get it to pass, only to then turn the circuits back on, which you know will disable it when needed

 

[sparkytim]

Well if it's open to anyone here's the answerTreat it as a potential divider,,, say 0.2ohms to neutral and 200ohms to earth with a (N-E) fault of say 1kohms, this will give us a fault resistance of 1000.2ohms - 1000 + ((200x0.2)/(200+0.2)) and a fault current of 230mA.

Any fault current down the earth will travel down the earths until it reaches the N-E fault, at which point 99% of the fault current will go down the Neutral and only 1% down the Earth, so that will give 227mA down the Neutral and 3mA down the earth..... Therefore not enough L-N imbalance to trip the RCD

Well Done thats exactly it.

So if a larger fault current flows then of course the MCB will trip as in a TN system which our N-E fault has virtually turned this into, but what is the resultant poptential DAMAGE caused?

Think size of "earthing" conductor fault current will take.

 

[SLIPSHOD & SLAPDASH]

Thank F for that....don't mind so much being thick if the big boys are scratching their heads too...can logic either but not both together...watching the experts with great interest.... .

Who are you calling a big boy X(

 

[kme]

I couldn`t possibly comment, Lardy :slap

 

[Sidewinder]

[quote name='Andy

 

[PC Electrics]

For those having difficulty following the "whys and wherefors", let me try and simplify it for you:

Lets say we have a neutral path back to the star point of 0.2ohm.

And lets say that the earth path back to the star point (via our rod) is 100ohm

And lets say our N-E fault is a dead short, so 0ohms.

We have a total load of 10A on our circuit.

Now, the ratio of resistances between the neutral and earth paths is 1:500 (0.2:100)

So, on its way back to the star point our 10A current reaches the N-E fault point. Here it can 'see' two paths back to the transformer. The current will split in the same ratio as the resistances, ie 1:500. Thus 0.02A will flow down the earth path, whilst the remaining 19.98A will flow down the neutral path.

As can be seen, the fault current of 0.02A (20mA) won't be sufficiently high to trip an RCD.

If however the system were a TN, we would have essentially the same resistance back to the transformer down both neutral and earth paths. thus our 10A would have split roughly in half. With a 5A imbalance an RCD would trip.

Does that help?

 

[park1]

Nozzer is spot on mate - it is worth being aware that a N-E short somewhere upstream can render an RCD inoperative as a safety device, yet not cause it to trip.One of the (myriad) reasons for doing RCD tests with NO loads connected - just switching off the MCBs doesn`t cut the mustard.

Had this about 10 yrs ago - I replaced THREE F`in RCD, cos they HAD to be faulty - wouldn`t trip on 1x or 5x, or even test button - all MCBs off.

Finally did individuated circuit IRs, and found a N-E "compression connection" ( someone fitting a faceplate had the neutral just outside the edges of the pattress; and tightening the front had crushed it against the metalwork. RCD never tripped.

I believe the newer "electronic" ones do not suffer the same problem ( it used to be a well known failing), but it can still occur.

KME

Yes my bad, i was looking at it in a in service position, where there would need to be a double fault, L-E and then a N-E. Then realised the the RCD tester is in fact acting as the second fault, so current may divide down the neutral and earth .

As mentioned, the RCD need's to be tested in an isolated manor.

 

[steptoe]

[quote name='Andy

 

[sparkytim]

For those having difficulty following the "whys and wherefors", let me try and simplify it for you:Lets say we have a neutral path back to the star point of 0.2ohm.

And lets say that the earth path back to the star point (via our rod) is 100ohm

And lets say our N-E fault is a dead short, so 0ohms.

We have a total load of 10A on our circuit.

Now, the ratio of resistances between the neutral and earth paths is 1:500 (0.2:100)

So, on its way back to the star point our 10A current reaches the N-E fault point. Here it can 'see' two paths back to the transformer. The current will split in the same ratio as the resistances, ie 1:500. Thus 0.02A will flow down the earth path, whilst the remaining 19.98A will flow down the neutral path.

As can be seen, the fault current of 0.02A (20mA) won't be sufficiently high to trip an RCD.

If however the system were a TN, we would have essentially the same resistance back to the transformer down both neutral and earth paths. thus our 10A would have split roughly in half. With a 5A imbalance an RCD would trip.

Does that help?

Thats a good summary of it, thanks for that PC.

So still waiting on Potential DAMAGE this situation can result in?

Thinking along the line of size of the effective Earthing conductor under secound fault condition, i.e. a L-E Short on the cooker circuit sharing the same RCD.

 

[ProDave]

For those having difficulty following the "whys and wherefors", let me try and simplify it for you:Lets say we have a neutral path back to the star point of 0.2ohm.

And lets say that the earth path back to the star point (via our rod) is 100ohm

And lets say our N-E fault is a dead short, so 0ohms.

We have a total load of 10A on our circuit.

Now, the ratio of resistances between the neutral and earth paths is 1:500 (0.2:100)

So, on its way back to the star point our 10A current reaches the N-E fault point. Here it can 'see' two paths back to the transformer. The current will split in the same ratio as the resistances, ie 1:500. Thus 0.02A will flow down the earth path, whilst the remaining 19.98A will flow down the neutral path.

As can be seen, the fault current of 0.02A (20mA) won't be sufficiently high to trip an RCD.

If however the system were a TN, we would have essentially the same resistance back to the transformer down both neutral and earth paths. thus our 10A would have split roughly in half. With a 5A imbalance an RCD would trip.

Does that help?

That explains why the RCD does not trip.

Now the second part. Why has it "disabled" the RCD?

Well lets assume we now get a second fault, a faulty appliance plugged in that has a 1Kohm fault from L to E.

So that 1Kohm fault will cause a fault current of 0.23A to flow.

Without the N-E fault, that 0.23A fault would be seen as an imbalance by the RCD and it would trip.

But now, thanks to our N-E fault, the second L-E fault current will share the same two paths back to the star point, so will be split again in the same 1:500 ratio. So of that 0.23A fault current, only 0.46mA will flow down the earth lead, the rest will flow via the neutral, back through the RCD, maintaining an almost balanced load as seen by the RCD.

Hence on a second real L-E fault, the RCD won't trip.

 

[OnOff]

"So still waiting on Potential DAMAGE this situation can result in?"

In the case of a cooker the N is possibly a bigger CSA than that of the cable from the MET back to the rod.....could be as low as 2.5mm sq.....am I on the right track? Thinking again path of least resistance.......back into the load?

 

[sparkytim]

"So still waiting on Potential DAMAGE this situation can result in?"In the case of a cooker the N is possibly a bigger CSA than that of the cable from the MET back to the rod.....could be as low as 2.5mm sq.....am I on the right track? Thinking again path of least resistance.......back into the load?

On the right track.but just think of the route MOST OF the fault current will take NOW with the existing N-E short at a WALL LIGHT fitting wired in 1mm T&E.