# Minimum size of CPC

When using the Adiabatic equation, only to calculate the minimum size CPC, I am thinking that using the fault currents given in Appendix 3 of BS7671 rather than that calculated (UO/ZS) would give a more realistic outcome.

For example: A D6 circuit breaker is protecting a circuit where the calculated fault current is 328A. The table associated with Fig 3A6 in Appendix 3 of BS7671 shows that an overcurrent of 120A will operate the 6A breaker in 0.4s. Using these values and a K value of 115 results in a minimum CPC of 0.66mm2 (1.0mm2), whereas if you use the 328A the calculation results in a minimum CPC of 1.80mm2 (2.5mm2).

I cannot see a problem with using the lower value, but would appreciate confirmation and guidance from others.

Parents
• I cannot see a problem with using the lower value, but would appreciate confirmation and guidance from others.

There absolutely is a key issue with using the lower value, because that does not represent the actual prospective fault current at either end of the cable.

However, potentially, also the 328 A value if that is calculated at the furthest point in the circuit is also a problem, because a fault may well occur at the "near end" of the cable.

This is why designers ought to use an "adiabatic plot" over the characteristics of the protective device ... discussed in detail in Section 8 of the IET Electrical Installation Design Guide (EIDG).

BUT

There is another issue with the fact you're using circuit-breakers... as discussed in section 8.5 of the EIDG, if it's in the "instantaneous" portion of the curve, as 328 A or more would be, you need to use either:

(a) The manufacturer's stated value of let-through energy I2t; or

(b) The let-through energy I2t from the circuit-breaker product standard.

You plug this let-through energy value in place of the product I2t, in the adiabatic formulas in 543.1.3 and 434.5.2, rather than calculating I and looking up t. THe let-through energy does depend on the actual fault current, so you need to know the range of prospective fault currents (near end and far end of the cable) and look up the let-through energy for both, and the largest calculated S is the one to use.

The reason for this, is that, for a cpc, as well as 541.1,3, you also need to meet Regulation 434.5.2 (which relates to the selection of the protective device in relation to fault currents ... fault currents includes prospective fault currents for all of L-L, L-N and L-PE faults.

If you want to use values from the product standards BS EN 60898 and BS EN 61009, these lead to the following minimum csa (copper conductors) for Type B and C circuit-breakers (see Table 8.6 of the EIDG) ... the use of manufacturer's data often leads to lower csa requirement:

• I cannot see a problem with using the lower value, but would appreciate confirmation and guidance from others.

There absolutely is a key issue with using the lower value, because that does not represent the actual prospective fault current at either end of the cable.

However, potentially, also the 328 A value if that is calculated at the furthest point in the circuit is also a problem, because a fault may well occur at the "near end" of the cable.

This is why designers ought to use an "adiabatic plot" over the characteristics of the protective device ... discussed in detail in Section 8 of the IET Electrical Installation Design Guide (EIDG).

BUT

There is another issue with the fact you're using circuit-breakers... as discussed in section 8.5 of the EIDG, if it's in the "instantaneous" portion of the curve, as 328 A or more would be, you need to use either:

(a) The manufacturer's stated value of let-through energy I2t; or

(b) The let-through energy I2t from the circuit-breaker product standard.

You plug this let-through energy value in place of the product I2t, in the adiabatic formulas in 543.1.3 and 434.5.2, rather than calculating I and looking up t. THe let-through energy does depend on the actual fault current, so you need to know the range of prospective fault currents (near end and far end of the cable) and look up the let-through energy for both, and the largest calculated S is the one to use.

The reason for this, is that, for a cpc, as well as 541.1,3, you also need to meet Regulation 434.5.2 (which relates to the selection of the protective device in relation to fault currents ... fault currents includes prospective fault currents for all of L-L, L-N and L-PE faults.

If you want to use values from the product standards BS EN 60898 and BS EN 61009, these lead to the following minimum csa (copper conductors) for Type B and C circuit-breakers (see Table 8.6 of the EIDG) ... the use of manufacturer's data often leads to lower csa requirement:

Children
No Data