Hi
Can someone help out with this Electrical problem. (I'm doing a CG2330 course)
Table 9A of the IEE On site guide only goes upto 50mm csa cable for the values (R1+R2)
Eg
50mm Phase+CPC = 0.774 mOhms/mWhat would you do if the conductor was say 70mm ???
Is there a formula I could use ?
Thanks
Leigh
50/70 x 0.774 mOhms/m?
Nope, nothing like. Its cross sectional area you need to use (pi * r^2)
So for 50mm = approx 1962 sq mm, 70 mm = approx 3846 sq mm.
i.e. it will be arount HALF the resistrance or about 385milliOhms/metre.
Slurp
Cables are usually quoted in mm2. It correct for 50mm2 and 70mm2 is it not.
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!
Yes you are right. I was going by the OP's spec of 50mm/70mm, not thinking it was mm2!
Have a team point and gold star.
Slurp
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.
Mr F.
Hmm, looks like I've opened up a can of worms here !
Here's the full example question and the the bits I've got to.
Consumer unit, 190V A.C., with BS3036 fuse, Ze=3D0.49Ohms
45m PVC Armoured cable, clipped directly to wall with 5 other circuits 50 Degree C ambient temp, 9Kw load Cable runs through 100mm InsulationIB =3D 9*103/190 =3D 47.36Amps
In =3D 60A (From Page 195 Fig3.2A Next biggest to 47A)
Correction Factors
1). Cf - A BS3036 device 0.725 2). Cg - From Table 4A1, (Page 210) Method 1 used From Table 4B1, 1+5=3D6 Circuits, Bunched =3D 0.57 3). Ca - From Table 4C2 50=B0C, Armoured PVC =3D 0.87 4). Ci - From Table 52A, page 97 100mm =3D 0.81Iz =3D 60 =3D 206Amps (0.725*0.57*0.87*0.81)
70mm2 csa conductorThen the problem, table 9A O/S/G only goes upto 50mm
Table 9A (Page 158) of the O/S/G gives mO per meter values to calculate R1 + R2
R1+R2 =3D mO/m * Length (m) * Temperature multiplier (Table 9C) 1000
So I'm a bit confused as to who's right.
Leigh
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?
Hi Andy
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).
Leigh
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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