It and Iz

Question

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 &ge; 32 / 0.87 It &ge; 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 &ge; In &ge; 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 &ge; 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 
 electrical.theiet.org/.../appendix-4-of-bs-7671.pdf

AJJewsbury · Accepted Answer

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.

gkenyon · Answer

JohnE said: 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.

AJJewsbury · Answer

gkenyon said: AMK said: I have not been following the discussion in its entirety, but I would like to clarify whether ( Iz ) is simply the value of ( It ) after applying the correction factors. 
 NO. 
 The two are not directly related. 
 I'm not sure I agree. 
 Iz is It multiplied by all the relevant correction factors (which is what I think AMK was suggesting). 
 (unlike the usual calculation where we divide In or Ib by the correction factors - in effect converting our actual installation conditions to the conditions the tables are based on) 
 - Andy.

AJJewsbury · Answer

Thank you for your continued help, but, I got a bit lost again, with the last comment. 
 
 I think Graham was just saying the current carrying capacity of a cable (Iz &ge; Ib) is just one of several points that have to be considered when designing a circuit - other factors such as achieving a certain loop impedance, or required maximum voltage drop, may well require a larger cable than CCC consideration alone would suggest. 
 Say for instance you had a 1A lighting load 1000m metres away - to achieve 3% v.d. (6.9V) you'd need a cable with a v.d. of 6.9mV/A/m or lower - which would probably mean 10mm&sup2; cable, even though 1mm&sup2; or even 0.75mm&sup2; would probably have been fine from just a current carrying capacity perspective. 
 - Andy.

gkenyon · Answer

AJJewsbury said: I think Graham was just saying the current carrying capacity of a cable (Iz ≥ Ib) is just one of several points that have to be considered when designing a circuit - other factors such as achieving a certain loop impedance, or required maximum voltage drop, may well require a larger cable than CCC consideration alone would suggest. 
 Well, that, plus the important fact that it's not as simple as looking up the current-carrying capacity from a table in Appendix 4, given the cross-sectional area and cable type, and bingo, the value you look up (I t ) is equal to I z . because it may be subject to factors such as grouping etc. that are not accounted for in the Tables themselves. 
 Slightly confusingly though, it is NOT ALWAYS true that, when you go through the selection process in Appendix 4, at the point you choose a csa from the value of "minimum I t " you have calculated considering the grouping factors, the value of I t you choose from a table in Appendix 4, in order to be greater than or equal to the calculated minimum I t , is I z regardless of have to increase S (as AJJewsbury says). This is because, when you apply correction factors, the tabulated value (you look up) drops down to I Z . 
 More simply ... it's not a two-way process.

Chris Pearson · Answer

Not quite sure what is bothering you. Isn't the upper half of your image saying exactly what I did? 
 In practical terms, you look up the tables and select cable size accordingly. It might be that on the north side of a house up north you could reduce the ambient temperature, but why bother? 
 Even here on a hot day on the south coast, ambient temperature is not 30 &ordm;C, though it could be higher if the sun is shining on the electrical equipment. 
 Perhaps a bit of engineering judgement is required?

AJJewsbury · Answer

Using the 3 formula mentioned in EIDG Calculating Iz Calculating It using In Calculating It using Ib 
 
 At first glance, it all looks reasonable to me 
 - Andy.

gkenyon · Answer

JohnE said: There is no term as far as I am aware for the calculated value, just our 'It' has to be ≥ than that value. That has been a major part of my confusion. 
 But ... that is the point. I responded a couple of minutes ago with the explanation that, I hope, makes sense: 
 gkenyon said: NO. 
 The two are not directly related. 
 I z is used in Part 4 of BS 7671. 
 I t is introduced in Appendix 4, as a means of providing a "lookup value" that is at least I z for the relevant conditions. 
 I t is one of the values in the tables in Appendix 4. The minimum value of I z , when you combine all of the requirements in Part 4, must, by definition, be less than, or equal to, I t ... but since I t relates to discrete values of cross-sectional area S, it's quite unusual (statistically improbable) for the two to be identical, if you are mathematically pedantic.

gkenyon · Answer

gkenyon said: The two are not directly related. 
 I z is used in Part 4 of BS 7671. 
 I t is introduced in Appendix 4, as a means of providing a "lookup value" that is at least I z for the relevant conditions. 
 I t is one of the values in the tables in Appendix 4. The minimum value of I z , when you combine all of the requirements in Part 4, must, by definition, be less than, or equal to, I t ... but since I t relates to discrete values of cross-sectional area S, it's quite unusual (statistically improbable) for the two to be identical, if you are mathematically pedantic. 
 HOWEVER ... 
 I will also throw this in. 
 
 Because of the definition of the symbol in Part 2, once you have selected the cable cross-sectional area S, I can see that, if you have followed the selection criteria in Part 4, the actual selected cable I z exceeds the minimum value I z that is gleaned from the requirements of Part 4 (which is, of course, the purpose of the design exercise). 
 BUT ... 
 We then need to consider voltage drop, adiabatic, etc., and any of these may lead to a larger S. 
 Which means that, in the general case, I Z &ge; I t , which is what EIDG advises ?

gkenyon · Answer

JohnE said: 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 I z . I t is only used to in Appendix 4. 
 I t 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: 
 I z =I t &times;C g &times;C a &times;C s &times;C d &times;C i &times;C c (&times;C f ) ... 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). 
 
 AJJewsbury said: 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 I z (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 I t 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, I Z &ge; I n &ge; I b 
 So, we select an overcurent protective device rating I n &ge; I b (taking into account, where necessary Rated Diversity Factor RDF), and now we can say 
 I Z &ge; I n , so SECOND we can use the formula to find the value of I t I need to ensure the minimum value of I Z 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 C f comes before C c in BS 7671.

AJJewsbury · Answer

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 &ge; 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.

gkenyon · Answer

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 ?

gkenyon · Answer

JohnE said: 
 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: 
 I t &ge; 20 &divide; (0.87 &times; 0.88) = 20 &divide; 0.7656 
 or I t &ge;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 'I t at least 26.12 A' is 2.5 sq mm with I t = 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: 
 I t &ge; 20 &divide; (0.87 &times; 0.88 &times; 0.725) = 20 &divide; 0.5551 
 or I t &ge;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 'I t at least 36.02 A' is 4.0 sq mm with I t = 37 A.

AJJewsbury · Answer

This calculated value it seems has no name? we just know our 'It' has to be equal to or greater than this value. 
 
 I guess that's true. You could call it 'minimum required It' if you like. 
 
 It isn't necessarily greater than Iz - it all depends on how the actual conditions vary from those used for the tabulated values. If ambient temperature was guaranteed to be lower than 30 degrees for example, Iz could be larger than It. 
 
 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). 
 
 Iz is the current carrying capacity of the cable for continuous service under the particular installation conditions concerned 
 
 Correct - Iz is the current carrying capacity of your actual cable, as installed. 
 
 Don't get that because its not taking into account the installation conditions. 
 
 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. 
 - Andy.

Chris Pearson · Answer

JohnE said: 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&deg;C or the conductor operating temperature is not 70&deg;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.

AJJewsbury · Answer

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 2.5mm&sup2; 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.

It and Iz

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