My problem is the impedance, (Zc) value is adjusted by the 7 cores meaning a 4mm cable is needed as opposed to 2.5.

I'm not following this bit. Most of the above replies seem to be considering thermal (current carrying capacity) conditions, but faults are normally considered to be adiabatic. As long as you consider the energy let-though from the protective device and the withstand of the cable (based on c.s.a. and k (which in turn is based o thermal capacity, initial temperature and final temperature)) then that's it. I'm not following where the number of cores comes into it or how it affects Zs.

   - Andy.

My problem is the impedance, (Zc) value is adjusted by the 7 cores meaning a 4mm cable is needed as opposed to 2.5.

I'm not following this bit. Most of the above replies seem to be considering thermal (current carrying capacity) conditions, but faults are normally considered to be adiabatic. As long as you consider the energy let-though from the protective device and the withstand of the cable (based on c.s.a. and k (which in turn is based o thermal capacity, initial temperature and final temperature)) then that's it. I'm not following where the number of cores comes into it or how it affects Zs.

   - Andy.

Hello Andy, sorry for the late reply to this, but work has been hectic. I think I understand now that only one core impedance is required? To be a little clearer,  when I' m calculating the fault current I need the system impedances, source (Zs) and Cable impedance (Zc). The formula I have been given for Zc for a multi core cable is, ((Ohms per kilometre * Length) / number of cores) / 1000 and I assumed this gave a figure for each core. Its been so long since I've done this by hand its an interesting little project and I'm building an excel sheet but my results don't match the shop bought version and I think this is why? 

Well only copper in the resistance loop you are considering contributes  to resistance so if all cores are parallel, yes divide by no of cores but generally treat each core as a resistor and add the in series or parallel as per how they are  seen by the current paths.

As a very quick wet finger check assume 16 milliohms per 1m of length, per 1mm squared for cold copper. If the area goes up (as it does with larger cores and cores in parallel), divide the resistance according to the total area carrying the current. If the length goes up, then  multiply by no of metres.

The tables in the regs are based on something closer to 19 or 20 milliohms as it assumes hot copper and an allowance for the cable to be on the limit of the permitted manufacturing diameter which is a tad thinner than nominal. Also it assumes the 'there and back' path, whereas the rule of 16 is one way.

I have used this rule of 16 a few times when walking about, just to spot circuits that are so obviously over long and have volt drop issues, or are so obvious short that no further consideration is needed.

With a bit of mental arithmetic it  allows one to impress the fondle-slab generation by saying - "that one is a bit near to the volt drop limit we should probably check it" - and being right -before they have even got the right software fired up. In the older times it was also faster than the youngest member of the team being sent to go back to the truck for the book of tables..

Use it here to check your spreadsheet is not miles out.

Mike