It is not necessary to know Iz if method use is5.1 and 5.2

Question

Basic cable calculation Checking Iz From BS7671 Method is used in items 5.1 and 5.2 Ib or In / Correction factors 
 
 It should be noted that the value of (It) appearing against the chosen cross-sectional area is not (Iz). 
 It is not necessary to know Iz where the size of conductor is chosen by this method. But it is possible to get incorrect coordination with cable calculations that seem acceptable? 
 
 Ib 28A 
 In 32A 
 L 40m 
 Ca 0.94 
 Cg 0.7 
 
 Cable T&E 
 Clipped direct 
 
 Calculation 
 
 It &ge; Ib &divide; correction factors so: It &ge; 28 &divide; 0.94 x 0.7 = 42.5A 
 
 Table 4D2A Method C 6mm2 cable - 46A It = 46 
 
 Volt drop 8.17V 
 
 So 6mm cabe looks OK 
 
 But if I do check Iz 
 
 Iz = It x Correction factors 46 x 0.94 x 0.7 = 30.26A 
 
 Ib &le; In &le; Iz 28 &le; 32 &le; 30.26

AJJewsbury · Accepted Answer

I think there are two different ideas getting mixed up here and making things look more complicated than they need be. 
 Firstly there's the idea of choosing "how much current you need the cable to carry" - if it only needs to carry what the load normally draws, then you can use Ib, if on the other hand it may need to carry more than that, up to the limit of the protective device (i.e. overload protection is needed) then you use the rating of the protective device - In. 
 Quite separately from that, there's the relationship between the cable's current carrying capacity in "standard" conditions (i.e. as tabulated, before applying correction factors) and that it will actually have under the conditions it's installed in (Iz). 
 The "obvious" way to calculate things might be to take the tabulated value, multiply by all the correction factors and compare that with "how much current you need the cable to carry". The trouble with that approach (especially in the days when electrician's didn't have a computer to hand) is that you end up having to do the same calculation umpteen times over to check a variety of cable sizes. 
 So they came up with an alternative approach of doing the calculation the "other way around" - i.e instead of calculating how much each size of cable could carry under the actual conditions, you'd instead calculate what the current carrying capacity of a just-the-right size cable would be if it was installed under standard conditions instead of the actual conditions - because you can then calculate one value and look it up in the tables once. 
 That might sound a bit odd, but mathematically it's quite simple - instead of multiplying two numbers to give a final answer, you take the final answer and divide by one number to give you the other. 
 So instead of calculating Iz for each cable size as It * C (and compare the result with "how much current you need the cable to carry"),you'd take "how much current you need the cable to carry" and divide by the correction factors to give you what would be the "tabulated" value of your ideal cable - you just then look up a size that's no smaller than that. 
 does that make things any clearer? 
 - Andy.

AJJewsbury · Answer

But I keep seeing Ib &ge; In &ge; Iz as the &lsquo;big thing&rsquo; 
 
 Hopefully the other way around: Ib &le; In &le; Iz (a small, but very significant difference!), but yes (where overload protection is required) that's the fundamental requirement. That doesn't mean you have to go about calculating it in that direct manner though - as long as the result you come out with satisfies that requirement, all is well. It's rather like saying you need to show that 1 &le; 2 &le; 3, but you've only got the figures 10, 20 and 30 instead - you could divide everything through by 10 and prove it, but you can also see directly that the same relationship will hold true. In our case things are multiplied (or divided) by all the correction factors, rather than 10, but the same logic holds. If it's true that "how much current you need the cable to carry" &le; Iz then it must also be true that ("how much current you need the cable to carry" divided by C) &le; (Iz divided by C) (for any positive value of C). 
 
 But it appears to me that (Iz) is not really established 
 
 Indeed - with the 'alternative method' you only establish your minimum acceptable figure for Iz (which is the same as "how much current you need the cable to carry") you don't need to calculate Iz of the actual cable size you select. Because the cable size you actually select can't be smaller than the theoretical "just the right size" cable calculated, the actual Iz of the chosen cable can't possibly be smaller than your minimum acceptable figure, so you can be sure the requirement is met even without calculating the exact figure. 
 - Andy.

mapj1 · Answer

It is worth noting that not all books use the same notation for the letters, although these days mostly they do, and I find it helpful to have a 'plain language' description of what is happening. Ib is the design current, which is the normal operating current of the circuit, while In is the rated current or current setting of the protective device (like a fuse or circuit breaker). The relationship between them is that In>=Ib In must be greater than or equal to Ib whis is the same as saying Ib must be less than In Ib<= In or the protective device might trip during normal operation, and at the same time Ib must be less than or equal to the cable's current-carrying capacity (Iz) so Ib<=Iz Iz>=Ib which is the cable's rating. Or the cable may be overloaded. Of course once exams are passed you then will realise its not really always true - the limiting of the load current could just as well be at the load end or the origin - Perhaps consider a single 13A fused spur, on 2.5mm2 fed by a 20A or even 32A MCB will be fine, as the 13A fuse means the cable cannot be overloaded even though the circuit protection appears to be 32A. More generally odd socket circuits are usually the ones where it is hard to be sure as you are often not really sure what might one day be plugged in and it maybe looks a if the breaker is too small, - perhaps a single 20A breaker feeding half a dozen BS4343 style 16A sockets for kit in a small workshop looks undersized, when the nature of the loads and the use pattern means it really isn't. Then there is the matter that short duration overloads like motor startup in-rush may trip the breaker, but assuming they don't, the cable is fine with it, as it takes several minutes to warm a large bundle of cables and the grouping derating can be pushed quite a lot, if cables are not all fully loaded at once for a long time. It is certainly worth keeping your wits about you above and beyond the algebra. Mike

AJJewsbury · Answer

Sorry, this must be painful for you. 
 
 Not at all!

But I think I&rsquo;ve got it.

It is basic transposition. 
 
 It is indeed! - plus realising that cables only come in certain fixed sizes, so you're likely to end up picking one that's a bit bigger than you actually need. Do the calculations (It and Iz) on the basis of what you actually need and you can guarantee that the corresponding values for the real cable won't be smaller. 
 - Andy.

It is not necessary to know Iz if method use is5.1 and 5.2

Community Rules & Guidelines