Sam Ayre said:

Hey guys I've been wondering when your wiring in parrallel do you divide the R1+R2 values by 4 when calculating thermal constraints and checking that the Zs value is okay. 

No, the (R1+R2) value of each separate cable in the pair is divided by 2, or alternatively if you measure (R1+R2) with the two cables connected together in parallel, you get is the real (R1+R2) value.

Don't be confused with RFC testing ... r1 and r2 are the end-to-end value along the whole length of the ring (and we use lowercase r). The resistance of each leg of the ring to the furthest point away from the DB is (r1+r2)/2 (half the value of the total ring), and the two legs in parallel therefore give you (r1+r2)/4.

Don't forget when you are looking at adiabatic, you need to take into account the current down each cable - each cable separately would have to meet adiabatic in case of a fault within the cable itself.

This highlights a danger in the use of multicore cables in parallel to meet disconnection times if the installation is not managed by a duty holder - if there is a fault in one cable, that blows the fault away, as well as operating the protective device, the protective device could simply be reset, and the circuit re-energized and no-one would be any the wiser. Less likely if the circuit has an RCD (but then, wouldn't that simply provide ADS disconnection times anyway).

Sam Ayre said:

Also when wiring in parrallel if you only use one of the earth connections would you divide by 3. 

No, just take the one value as R2 ... although if you ware wiring in SWA, the armour would usually be considered to require being part of the earth circuit, as usually it becomes an extraneous-conductive-part when using a gland (and if buried, both cables the SWA would need to be suitable for use as a cpc in its own right, regardless of the use of an internal copper core as supplementary cpc).

Are there some example adiabatic calculations for a ring final documented somewhere that explain the use of (r1+r2)/2 ?

Craig Nicholas said:

Are there some example adiabatic calculations for a ring final documented somewhere that explain the use of (r1+r2)/2 ?

Do you mean (r1+r2)/2, or (R1+R2)/2 ?

In a ring final circuit, r1 and r2 are the "end-to-end' resistance, so at the furthest point (mid-point) of the ring final, the resistance of each "leg" of the ring is (r1+r2)/2. Because they are in parallel, the effective (R1+R2) at the mid-point is:

(R1+R2) = (r1+r2)/4

Section 2.8 and 8.4 of the IET Electrical Installation Design Guide contain an explanation of the process for fuses, but when using circuit-breakers, as is the norm these days, because the disconnection time is so fast, we need to do the adiabatic calculation based on the let-through energy I²t and not using separate I and t from calculation and graphs. This is explained in Section 8.5 of the Electrical Installation Design Guide.

I did mean (r1+r2)/2 referring to your earlier reply where you mention the resistance of each leg and accounting for the current down each cable when using adiabatic.

I'm trying to determine the minimum CPC CSA for an earth fault in a ring final socket circuit with 2.5/1.5 cable and 32A In. It's the impedance to use in the fault current calculation that I'm unclear on. The R1+R2 value results in a high fault current where the CPC CSA is too small whereas if (r1+r2)/2 for a leg is used the CSA becomes acceptable.

I'm trying to determine the minimum CPC CSA for an earth fault in a ring final socket circuit with 2.5/1.5 cable and 32A In. It's the impedance to use in the fault current calculation that I'm unclear on. The R1+R2 value results in a high fault current where the CPC CSA is too small whereas if (r1+r2)/2 for a leg is used the CSA becomes acceptable.

Faults can occur anywhere - if you OPD is an MCB, worst case energy let-though is likely to be with the highest fault current - i.e. for a fault very close to the start of the circuit - so R1.0=0, R2=0.0 and so fault current just based on Zdb. (Yes there are known issues using reduced c.p.c.s with fault currents >> 3kA).

   - Andy.

I'm trying to determine the minimum CPC CSA for an earth fault in a ring final socket circuit with 2.5/1.5 cable and 32A In. It's the impedance to use in the fault current calculation that I'm unclear on. The R1+R2 value results in a high fault current where the CPC CSA is too small whereas if (r1+r2)/2 for a leg is used the CSA becomes acceptable.

Faults can occur anywhere - if you OPD is an MCB, worst case energy let-though is likely to be with the highest fault current - i.e. for a fault very close to the start of the circuit - so R1.0=0, R2=0.0 and so fault current just based on Zdb. (Yes there are known issues using reduced c.p.c.s with fault currents >> 3kA).

   - Andy.