# 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

Parents
• 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.

• 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.

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