# 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
• Sorry but I am just not getting Iz

I understand the cable calculation procedure and getting a value from dividing design current or In by correction factors

This calculated value it seems has no name? we just know our 'It' has to be equal to or greater than this value.

This calculated value we look up  in appendix 4 and find a cable that is again equal too or greater than our calculated value.

So this value from the tables will give us our  final cable size. This is the tabulated value. so will finally be our 'It'

Iz has not come into it.

But we could put our 'It' into formula 2.4.0 ( EIDG) See calculations

It gives us an Iz of 36.97 A
which is greater than 'In,'  and 'In' is greater than 'Ib' so coordination is OK

Iz is the current carrying capacity of the cable for continuous service under the particular installation conditions concerned
Don't get that because its not taking into account the installation conditions.

• Isn't it normally In rather than Ib, where overload protection is required (i.e. the actual current could exceed Ib and is limited only by the protective device, not necessarily the load).

I'm not following you now. Iz should take account of all the actual installation conditions (or rather it includes adjustments for all the conditions that differ from those assumed by the tables - which should amount to the same thing). So if the table assumes 30 degree ambient and you have 40 in reality, you have 0.87 (rather than 1.0) for Ca.

This is the thing.  What is Iz...
2.4.0 (formula multiplies It by correction factors)  What is actually happening here?

Installaion conditions generally derate the cable.

In my examples the calculate Iz is 36.97 A
The It is 85A
So how is Iz taking into account installation conditions.
I know I have a misunderstanding somewhere, which is frustrating me no end

• This is the thing.  What is Iz.

I think that you are making things too complicated.

It and Iz are defined in Part 2 (page 46). The definitions are repeated in Appendix 4, paragraph 3 (page 423).

If the installation conditions match one of the tables in Appendix 4, let's say 4D1A for singles in conduit on a wall, RM B, just use It, which is the same as Iz.

However, Iz could be higher or lower if the ambient temperature is not 30°C or the conductor operating temperature is not 70°C; or for grouping or other installation methods.

Paragraph 5 points out that you do not need to know Iz if the cable size is obtained from the tables, with corrections as appropriate.

• What is Iz..

It's the cable rating as it's actually installed - the cable's "real" rating if you like.

The tables only give you ratings for a few specific circumstances - so where your circumstances differ, you need to pick one and adjust accordingly.

Take a bit of 2.5mm² T&E for example. If you installed it according to method C (single cable clipped direct onto a wooden or masonry wall, 30 degree ambient) it would have a rating of 27A. (Iz = It).

If however you installed it somewhere with a maximum 25 degree ambient temperature, it would be able to carry 1.03x27A = 27.81A (It = 27A, Ca=1.03 :. Iz = 27.81A).

If two such cables were clipped direct together (still at 30 degree ambient) you'd be looking at 27A x 0.85 =  22.95A each.

The same two cables but in a 25 degree environment, then 23.6385A.

- Andy.

• Oooh, I'm getting a tingle of understanding.  Thank you

• we just know our 'It' has to be equal to or greater than this value.

And by doing this, you will have an Iz that meets the normative requirements of BS 7671 ... it's like a puzzle starting with the answer, not the question.

• 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).

Am I missing something? Surely that is not true if using a rewireable fuse. The current carrying capacity, Iz, is not affected by the type of fuse.

• The current carrying capacity, Iz, is not affected by the type of fuse.

Sort of - BS 3036 fuses are slower to respond to overloads so where overload protection is required, you may need a larger cable to give the same margin of safety as you'd have with other devices (which normally have a fusing factor of 1.45 or less) - so yes it's strictly true that Iz shouldn't alter, it is a convenient way of preserving the margin when matching a cable to the fuse rating. Like a while back when we had separate tables for 'course' and 'close' protection - with lower tabulated values for course (i.e. rewireable fuses).

- Andy.

• Surely that is not true if using a rewireable fuse. The current carrying capacity, Iz, is not affected by the type of fuse.

YEs ... that is what Cf is for .. and why it's in brackets because it can be ignored for cases where rewireable fuses are not used.

Iz=It×Cg×Ca×Cs×Cd×Ci×Cc(×Cf)

• Iz=It×Cg×Ca×Cs×Cd×Ci×Cc(×Cf

With Cf at the end, I deem the above to be technically incorrect. It may well assist in facilitating  how we arrive at;

but something technically incorrect is not a good starting point for an explanation. If it were correct, then we would need to use Cf for circuits that did not need overload protection but were protected by a 3036 fuse requiring only fault current protection.

Correctly, the EIDG does not include Cf as shown in the top equation as the type of overcurrent protective device does not affect Iz.

Whilst we ultimately end up at the same place, the strict application of regulation 433.1.202  gives the maximum rating of fuse to be used when the current carrying capacity (Iz) of the smallest conductor of the circuit is multiplied by 0.725 for a 3036 device with a fusing factor of two.

Since that would be kind of impractical, Cf is applied as a divisor to In to give the minimum value of Iz (or It if referring to the ccc in a table)

Suppose we had a 2,5mm2 single-phase cable clipped direct as Table 4D2A column 6. No correction factors apply other than we are using a 20A 3036 fuse for overload protection. It=27A and by dint Iz = 27A

(a) what is the value of Iz?

Since Cg, Ca, Ci etc are all 1 then Iz =It * 1 = 27A  (Cf is ignored as it does not alter the actual value of Iz). Iz is definitely not 27 * 0.725 as would  be the reasoning of the topmost equation.

(b) what is the maximum rating of 3036 fuse that could be employed?

In must be less than or equal to;  0.725 *27 = 19.575A .

That is hardly a practical method when we are looking for the minimum current carrying capacity to be used, although In could be dropped to 15A depending on the value of Ib.

(c) What is the minimum value of Iz that can be used; Iz minimum =  In/0.725 = 20/0.725 = 27.6A. Since the 2.5 can only carry 27A, strict application would indicate that when using a 20A 3036 fuse for the circuit, the arrangement would not comply.

In that regard, a 4mm2 cable would need to be used with an It of 36A.

Now of course, it would have been infinitely easier just to apply the second formula from the top to find the minimum value of It;

It  must be greater than or equal to In / 1*0.725  = 27.6A

(d) Minimum size of cable if only fault protection is required and Ib =20A;

Ib must be less than or equal to In

and Ib must be less than or equal to Iz

It must be greater than or equal to Ib/applicable correction factors.

In this case Cf is not applied so; 20/1 =20. So the minimum value of Iz is 20 and since 2.5mm2 cable is used, It is 27A. Thus, the conditions are satisfied (although further calculations for thermal protection will be required).

• but something technically incorrect is not a good starting point for an explanation. If it were correct, then we would need to use Cf for circuits that did not need overload protection but were protected by a 3036 fuse requiring only fault current protection.

But you can use Cf for all circuits?

433.1.202 gives you the value of Cf for a BS 3036 fuse different to that in general for 433.1.1 (iii) where Cf for other OCPDs is 1, if you take the view that Cf=1 applies to 433.1.1 (iii) an this is clarified for BS 3036 fuses by application of Cf = 0.725 ?

This is the view that is portrayed in 5.1.1 (a)  of Appendix 4 to BS 7671?