It and Iz

Im getting a bit confused with It and Iz, and which is out final value when calculating cable size

It  - the value of current tabulated in this appendix for the type of cable and installation method concerned, for a single circuit in the ambient temperature stated in the current-carrying capacity tables

Iz -  the current-carrying capacity of a cable for continuous service, under the particular installation conditions concerned.

In =32  Ca 0.87

so    It ≥  32 / 0.87

It ≥ 36.78A

Look up tables in BS 7671  clipped direct (c) one  2 core cable

4mm=36A

6mm= 46A

So we go with 46A

So our It (Tabulated value we look up) is 46A and we will use a 6mm2 cable

So what is Iz?

Iz ≥ In ≥ Ib  understand the relationship

Now Iz = It x correction factors.
Iz = 46 x 0.87 =  40.02A  which is greater than In (32A) so thats OK

But we got a value of 36.78 A when applying correction factors to In

We also have this formula...

It ≥ Iz / correction factors

40.02 / 0.87 = 46

We end up with the same number.

We have a calculated value of 36.78A  in the book the cable choice is 6mm which is rated at 46A

I'm struggling to see what we would call our final cable choice? Its not Iz , but the description (continuous service) makes it sound like it is

What is Iz

Hope that make some sense

Its based around this IET article

• OK, I understand the confusion. What's done is a little trick to make the procedure easier, but it doesn't help understanding - and it's often not explained at well either.

In principle what you should do is take It of all the various cable sizes, apply all the correction factors to each It as multipliers, to get the true Iz for each, then select the cable based on Ib (or In) against those values.

But that requires a lot of working out - you might have to go through the whole calculation umpteen times over, once for each candidate cable size, which especially in the days before electronic calculation (when most of these processes were first invented), was a bit of a pain in the butt to say the least.

So rather the textbook procedure is to take Ib (or In) and divide by the correction factors to get an It equivalent needed for your design current - then look down the table to pick a cable size large enough. Which saves a lot of calculation.

It doesn't directly give you a value for Iz though - you need to take It of your chosen cable size and multiply by the correction factors to get that. But mostly you don't need to know the exact value of Iz, since you've picked the correct cable size already ... which is the main point of the process.

- Andy.

• It ≥ Iz / correction factors

40.02 / 0.87 = 46

We end up with the same number.

Not always.

The first thing to understand, is that the normative requirements in Part 4 of BS 7671 do NOT refer to It at all, only Iz. It is only used to in Appendix 4.

It is the value you look up in a table in Appendix 4. It for the stated installation conditions (method) in the Tables doesn't  contain many of the factors required (grouping etc.), and for this reason, but they do contain things like thermal factors:

Iz=It×Cg×Ca×Cs×Cd×Ci×Cc(×Cf) ... see equation (2.4.0) in EIDG (page 19 in the 5th Ed) with the addition of see equation (2.4.3) in EIDG (page 21 in 5th Ed).

What's done is a little trick to make the procedure easier,

That is the point. If I want to look up a value in Appendix 4, to take into account the relevant factors, I need to take the desired Iz (the one I want to find to meet the normative requirements of BS 7671), and apply the factors not already taken into account in the Tables in Appendix 4..

The following are the two important steps ... To do this, I FIRST take the equation above (2.4.0 in EIDG) and re-arrange it to make It the subject, by swapping left for right, and dividing both sides by the product of the grouping factors, so that I get the following:

Now, we also know that where overload protection is required,  IZ ≥ In ≥ Ib

So, we select an overcurent protective device rating In ≥ Ib (taking into account, where necessary Rated Diversity Factor RDF), and now we can say

IZ ≥ In, so SECOND we can use the formula to find the value of It I need to ensure the minimum value of IZ required, as follows:

And ... this is the equation we see in Equation 1 of 5.11 and Equation 2 of 5.1.2 (i) of Appendix 4 to BS 7671 (page 425 of BS 7671:2018+A2:2022), except that Cf comes before Cc in BS 7671.

• Thank you both for two really helpful replies.

I have tried to follow the process you mentioned as in the EIDG

This is a really simple example I have chosen. probably too simple

We have the information we know.
Then we have the desired  Iz

Is this right?

• That doesn't look quite right to me. Fundamentally we need the cable as installed to carry the required current (in this case Ib = In) so I would have thought that Iz ≥ In would do. The correction factors convert It to Iz where the installation conditions differ from what's assumed by the tables.

I'd probably say your 26.3A is the minimum required It (rather than Iz).

Yes, 20.67A is the Iz of the cable as installed - which meets your required CCC  of 20A.

- Andy.

• Cf only applies for BS 3036 semi-encloses rewireable fuses.

Regards Ca, Your example doesn't say what the ambient temperature is ... but Reference Method C includes for that if the ambient is 30 degrees.

Similarly, what is your assumption for Ci ?

• Ci 50mm 0.88 Ca 40oC

• 2.4.3 in EiDG

• 2.4.3 in EiDG

Humm, I won't have the latest version to check against, but I wonder if you're spotted a typo there. To my mind it should be It ≥ In/C (or Iz  ≥ In).

- Andy.

• 2.4.3 in EiDG

This ONLY applies for rewireable fuses to BS 3036. For other OCPDs to standards quoted in BS 7671, Cf = 1

So, to use your example: In = 20 A.

We then use the following formula to see what value we need to look up in the Tables in Appendix 4:

(a) If this circuit is NOT protected by a 20 A rewireable fuse, but by say a 20 A circuit-breaker, then (using your example) we have Ca = 0.87, Ci = 0.88, and all the other C factors = 1, so:

It ≥ 20 ÷ (0.87 × 0.88) = 20 ÷ 0.7656

or  It ≥26.12 A.

If I am using 'clipped direct' and want to select 2c+E 'flat' UK cable, I look in Table 4D5, column 8, and see the first cable that meets the requirement 'It at least 26.12 A' is 2.5 sq mm with It = 27 A.

(b) If this is protected by a 20 A rewireable fuse, then (using your example) we have Ca = 0.87, Ci = 0.88, Cf = 0.725, and all the other C factors = 1, so:

It ≥ 20 ÷ (0.87 × 0.88 × 0.725) = 20 ÷ 0.5551

or It ≥36.02 A

If I am using 'clipped direct' and want to select 2c+E 'flat' UK cable, I look in Table 4D5, column 8, and see the first cable that meets the requirement  'It at least 36.02 A' is 4.0 sq mm with It = 37 A.

• Ci 50mm 0.88 Ca 40oC

OK ... makes sense it just wasn't stated and I didn't want to assume.