Identifying csa of t&e

[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.