No, the figures mentioned were clearly conductor CSAs in mm^2, so the answer was right, at least for DC.
An easier way would to use the tabulated values for 35mm^2 and halve them, giving 0.524 mohm/m for R1+R2 for 70+70 mm^2.
However the OP should be aware that for these large sizes used on AC the impedance will be higher then the DC resistance because (a) skin effect increases the resistance and (b) the inductive reactance is not necessarily negligible. The OSG is only relevant to installation work up to 100 A per phase, in which you would not usually need to use 70 mm^2 conductors!
The OP stated 50mm csa - csa = cross sectional area. For a direct current, resistance is inversely proportional to the cross sectional area of a material so doubling the area halves the resistance.
OK so far, assuming single-phase, which is not explicitly stated (although the words "consumer unit" do tend to suggest it's 1-ph).
Only applies if overload protection is required. Also if the six circuits of the group are not liable to simultaneous overload you can use Appendix 4, 6.2.2, equations (7) & (8) which will almost certainly help towards a smaller cable.
Six armoured cables clipped to a wall will not be bunched, they'll be in a single layer and spaced by more than one diameter. (Otherwise there won't be room to get the cleats in!) This gives Cg = 0.9. (If spaced by more than 2 diameters you can used Cg = 1.0 - no derating.)
Do all six cables go through the insulation, or just this one? The question isn't very clear. If only this cable, then Cg and Ci won't both apply at the same time. Then you have to consider the installation conditions for each section of the circuit in turn and take the highest It rating obtained.
This is getting silly - no-one would install like that, on grounds of cost if nothing else. You'd adjust the method of installation so that not all factors applied simultaneously, and not bunch the cables as I said above.
Since it's SWA Table 9C won't help with R2 (assuming armour as CPC). Or if a third core is used as CPC then R2 will equal R1.
The question seems to leave a lot of necessary detail out. Also what's the ultimate aim of the problem - to verify that Zs is low enough, or to verify compliance on all grounds?
The values were just picked out of the air by the lecturer, it was so that the class could work through correction factors, finding out Zs, fault currents and selecting cable sizes. I think he left out all the details so that our poor little brains don't overload :-), but the values he used cause more problem in the end.
Thanks for information, helps move class room knowledge to the real world (Even if the above example would never happen).
In my yoof, I was taught to calculate deflections on rolled steel beams.
Years later, I worked out thta the cost of calculating the size I required was greater in terms of my time than getting the 'big f*ck off beam that is no way ever going to bend' from the steel merchant..
..and anyway, my architect had a table of deflections for standard beams per ton per unit length squared...;-)
I just add this to make sure you get the course work in perspective...
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