Identifying csa of t&e

[zap]

Greetings.
I tested an rfc and recorded the following results:
r1 = 0.62
rN = 0.89
r2 = 1.45
The difference in r1 and rN indicate loose connections so I checked all the sockets (bar one) on the ring which revealed two loose connections.
The rfc now tests as follows:
r1 = 0.58
rN = 0.59
r2 = 1.45
The measurements indicate a 80 metre circuit of 2.5mm^2 csa.
I've checked every socket in the dwelling except one which feeds a waterbed and is inaccessible. So I told the customer that I have some homework to do before I can issue certification.
BUT, I hadn't realised that it could be 2.5/1 t&e in which case the measurement of 1.45 (remaining constant) is correct.
I don't have a micrometer to measure the cpc so how do I confirm the cable conductor's csa? ... and if it's 1mm^2 cpc is that okay?

The wiring is about 25 - 30 years old.

 

[R G]

what you could try is connect L to E on one side of the ring and leave the other open at the board, check the R1 R2 readings at the sockets nearest then move further along until you notice any major jump , if its a gradual increase on all then o/k ,

 

[Octopus]

those readings look like 2.5mm/1.0mm dating from late 70's to early 80's

 

[shanky887614]

what you could try is connect L to E on one side of the ring and leave the other open at the board, check the R1 R2 readings at the sockets nearest then move further along until you notice any major jump , if its a gradual increase on all then o/k ,

to make this quicker, wire a plug with a bit of flex from it going to connector strip.

this makes it quick to a quick check without removing fronts (very useful if there if there is a dead short somewhere)

or if you have a multimeter etc with non gs38 leads just shove them in the socket

 

[zap]

Thanks, R G,
The route the cable takes is not obvious but I sort of did what you're suggesting with expected results.
If the cpc is 1mm^2 is it acceptable (i.e. compliant with BS767)?

 

[shanky887614]

Thanks, R G,
The route the cable takes is not obvious but I sort of did what you're suggesting with expected results.
If the cpc is 1mm^2 is it acceptable (i.e. compliant with BS767)?

as long as earthloop is acceptable at furthest point its fine

 

[R G]

I would compare a bit of new cable with the old , for the earth wire size , if you have test leads to a plug then you won't need to remove the socket fronts ...

If you have done this then as shanky says if Zs is good at end of circuit then its o/k

 

[zap]

those readings look like 2.5mm/1.0mm dating from late 70's to early 80's

Thanks Murdock, that's what I suspect. But is it satisfactory?

 

[zap]

I would compare a bit of new cable with the old , for the earth wire size , if you have test leads to a socket then you won't need to remove the socket fronts ...

Thanks again R G,
I started to doubt myself and stopped thinking clearly.

Is it satisfactory to have a 1mm cpc on a rfc protected by a 32A type b breaker?

 

[R G]

Thanks again R G,
I started to doubt myself and stopped thinking clearly.

Is it satisfactory to have a 1mm cpc on a rfc protected by a 32A type b breaker?

No problem if Zs is good and RCD protected whats the problem

 

[zap]

Thanks shanky, this appears to have answered my question.

I've lost the 'thanks' button. Can anyone elucidate?

 

[Octopus]

Thanks Murdock, that's what I suspect. But is it satisfactory?

What were your test results?

 

[shanky887614]

Thanks shanky, this appears to have answered my question.

I've lost the 'thanks' button. Can anyone elucidate?

can you click on like/thanks on the posts?

it shows people who the long standing members are that help people out

 

[zap]

can you click on like/thanks on the posts?

it shows people who the long standing members are that help people out

Apparently, yes, and thanks again.

 

[zap]

What were your test results?

Preliminary result was too high i.e. 1.45 Ohms. But two loose connections have been rectified since so I'm returning to retest and this time I expect satisfactory results.

Thanks again for helping me understand.

 

[telectrix]

1.omm cpc is acceptable provided that you calculate the min. size of cpc from the pefc, using the adiabatic equation and the trip time of the ocpd. ( usually 0.1 sec. with a k value of 115).

 

[zap]

1.omm cpc is acceptable provided that you calculate the min. size of cpc from the pefc, using the adiabatic equation and the trip time of the ocpd. ( usually 0.1 sec. with a k value of 115).

Yep, I'll do this too! ... and thanks again.

 

[zap]

1.omm cpc is acceptable provided that you calculate the min. size of cpc from the pefc, using the adiabatic equation and the trip time of the ocpd. ( usually 0.1 sec. with a k value of 115).

This is too good an opportunity to check my understanding of the adiabatic equation.

Type B breaker

I = 5 x 32A = 160A
t = 0.1
k = 115
S = minimum csa

S = SQR(I² x t) ÷ k

50.6 ÷ 115 = 0.44 mm²

Therefore 1mm² is good to go.

 

[Risteard]

Not quite.

You need to calculate the ACTUAL disconnection time with the ACTUAL fault current at that point. Typically a BS EN 60898 circuit breaker has an instantaneous operating time quoted as 0.1 seconds, however if you consult the manufacturer's data then there may be a definite minimum operating time of around 0.01 seconds at your value of prospective fault current at the extremity of the circuit. This could help achieve a smaller cpc csa.

Then so long as the result, i.e. S, is no greater than the cpc cross-sectional area that you have then it is adequate. This is an iterative calculation, so when designing you would have to recalculate as you increase the cpc csa as this will affect the Zs and prospective earth fault current, which will subsequently affect the size of cpc given as the minimum (i.e. it will increase - so you need to ensure that your new csa is adequate after calculation).

 

[davesparks]

Thanks again R G,
I started to doubt myself and stopped thinking clearly.

Is it satisfactory to have a 1mm cpc on a rfc protected by a 32A type b breaker?

Sometimes it is and other times it isn't. Just calculate it using the adiabatic equation the same as you do for the cpc of all the circuits you install.

 

[electroyd]

This is too good an opportunity to check my understanding of the adiabatic equation.

Type B breaker

I = 5 x 32A = 160A
t = 0.1
k = 115
S = minimum csa

S = SQR(I² x t) ÷ k

50.6 ÷ 115 = 0.44 mm²

Therefore 1mm² is good to go.

the value for I is incorrect mate. You need to measure the zs of the circuit, to work out the actual fault current. Higher reading the better! In your situation

 

[HandySparks]

Not quite.

You need to calculate the ACTUAL disconnection time with the ACTUAL fault current at that point. Typically a BS EN 60898 circuit breaker has an instantaneous operating time quoted as 0.1 seconds, however if you consult the manufacturer's data then there may be a definite minimum operating time of around 0.01 seconds at your value of prospective fault current at the extremity of the circuit. This could help achieve a smaller cpc csa.

Then so long as the result, i.e. S, is no greater than the cpc cross-sectional area that you have then it is adequate. This is an iterative calculation, so when designing you would have to recalculate as you increase the cpc csa as this will affect the Zs and prospective earth fault current, which will subsequently affect the size of cpc given as the minimum (i.e. it will increase - so you need to ensure that your new csa is adequate after calculation).

I agree about using the actual prospective earth fault current.

However, shouldn't it be the PEFC at the origin of the circuit? After all, the fault might occur at the nearest socket to the origin, or even in the cable before the first socket.

You might have to refer to the manufacturer's spec for I²t let-through energy and put that figure into the adiabatic equation instead of separate figures for I and t to arrive at a sensible answer.

 

[Risteard]

However, shouldn't it be the PEFC at the origin of the circuit? After all, the fault might occur at the nearest socket to the origin, or even in the cable before the first socket.

No, because the worst case will be with the slowest disconnection time (i.e. at the extremity of the circuit). The lower prospective current at that point will cause the longest disconnection time which will thermally stress the conductors more.

In reality it can be advisable to carry out the calculation for both near end and far end faults to find the worst case, although it is generally assumed to be far end.

 

[HandySparks]

No, because the worst case will be with the slowest disconnection time (i.e. at the extremity of the circuit). The lower prospective current at that point will cause the longest disconnection time which will thermally stress the conductors more.

In reality it can be advisable to carry out the calculation for both near end and far end faults to find the worst case, although it is generally assumed to be far end.

Looks like I need to try plugging in a few figures and see how it works out.

 

[zap]

Thanks guys. Juggling figures is easy! knowing what figures to juggle is the essence and this forum undoubtedly helps me understand.
I'm going back to the job this afternoon to perform the relevant tests.
I'll also look up manufacturer's specifications and respond later too.