I am struggling a bit to visualise this - is the 7 core especially long so volt drop is a worry or is the concern heating ?

If you have 3 loaded cores and 4 cool ones, the multicore cable rating can be pushed relative to the '7 loaded cores' case, but how much is a question for the cable makers. Are you pulling more than about 20A phase ? for 2,5mm as part of a larger group, that is perhaps where thermal effects may start  to crimp the style but it will depend how the cable is installed and how it can cool itself.

Mike

Many thanks for the reply. This stems from the fact that there is a very limited amount of space for the cable installation and running two cables as opposed to one is far more convenient. I've checked the regulations and although not the prefered way this is an acceptable way. The voltage of the equipment is 600 volts and the calculations at 2.5mm prove to be acceptable for volt drop. Where it falls down is the short circuit calculation, I realise that using an appropriate MCCB is a mitigated but I'm curious, because I don't know and its not something I've ever come across, If using a multicore cable with only three cores used for power, and the remained for 24 volt switching and indication, does this allow the seven core to be calculated as a three or does it still require the seven core for the calculation of short circuit current?

step back one,
if this was 3 cables
1 x 3 core
2 x 2 core  (and at 24v would be light load anyway - could have been 1.0/0.5mm  lighting "flex" for the current needed ?)
How would you rate / calculate it ?

That they are all in one 'outer' case ("flexible conduit" with little air space...)
is kinda academic.
At a pragmatic level, one might even look to put 1 Hot  - 1 cool - 1 hot - 1 cool -  (if not for colours chosen) to allow any heat generated to be next to a cooler "spacer" - giving further practical allowances.
Then note in the installation cert / documentation that cable usage rated for 3 cores max only, and 4 cores for low current ELV auxiliary use. - so that later maintainers do not try to change use of the other cores, or negatively assess it at inspection time ?

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

In the ROI we effectively use Mike’s method. There are no voltage drop tables. If you want to know the voltage drop you have to use a given formula base on resistivity of copper at 70C and a set reactance value across all cross sectional areas then multiply by 2 for single phase or root 3 for three phase.

However, using Mikes method won’t take you a million miles from a more fastidious approach.

If you have a 4mm2 copper cable, for example, one core will have a resistance of around 20 divided by 4 milliohms per metre. Similarly a 10mm2 core will have 20 divided by 10 milliohms. 
Those can be multiplied by 2 or root 3 to five the single phase or three phase values per metre length of the circuit.

have a look at Table B1 in GN3 or the voltage drop values in say Table 4D2B in Appendix 4 of BS 7671 and you will see the results are not too far apart!