2365 design drawing level 3 Project

32A Radials in 2.5?
As a ring is fine but Radials should be 16/20 depending on installation method & factors

Eddy Currents said:

32A Radials in 2.5?
As a ring is fine but Radials should be 16/20 depending on installation method & factors

Click to expand...

But the stated load is 7kW so a 20A radial would be overloaded, it needs to be 32A for that load.
But yes it is unusual for 2.5mm to be suitable for a 32A radial circuit.

Radoslav said:

1. It with the aid of the regs book - formula It = In/(Ca*Cg)
2. Iz this is where I'm stuck as I know this should be a tabulated value but I'm struggling with the fact that it's value is smaller than In

Click to expand...

It is the tabulated value taken from the appropriate table in the regs before any correction factors are applied.
Iz is the current carrying capacity of the cable after the correction factors have been applied.
See appendix 4 section 3 "relationship of current carrying capacity to other circuit parameters".

I would think that the 32A radials should be 4mm minimum in insulation & the deeper the insulation even a 6mm.
Why 2 radials for a Hotel, is this just one room or do you have to cook your own dinner & plug in your kettle?

Thank you everyone for all the replies. The decision over the cable was made in college and can't remember why we decided to go with 2.5mm. I was thinking at first 6mm or even 10mm based on the length and installation method and my experience, but we were told 2.5mm for some reason and therefore I'm using 2.5mm. I will correct that and maybe go with my first thought of 6mm? The main thing in all this is to show I'm able to use the Regs Book and as it is really the very first time I'm doing this I wouldn't expect my project to be perfect, however I want to know when I'm wrong and try and fully understand why so I can correct myself.
Electro-tech there are 2 radials because columns 1-4 is ground floor and 5-8 is 1st floor.
On the reply from loz2754 I went through the information in the book few times and my understanding is the It value is in table 4D5, and Iz is calculated using the values for Ca and Cg and the formula Iz = Ib/(Ca*Cg)?

PS: English is not my first language and perhaps this could be why at first I couldn’t fully understand why It is the tabulated value from table 4D5 as in the table clearly says “current-carrying capacity” which is marked with Iz.

Radoslav said:

1. Conductor size chosen based on In - (not sure if my calcs are correct for this so a bit help will be appreciated)

Click to expand...

This may the start of your problem, you don't select conductor size based on In, you select your conductor size after doing the next step.

Radoslav said:

1. It with the aid of the regs book - formula It = In/(Ca*Cg)

Click to expand...

You've misread the equation, it is It > In/(Ca*Cg) (actually its greater than or equal to but I don't know how to type that on the phone)
It is the tabulated CCC of the cable, literally the number printed in the table.

So what you do is calculate In/(Ca*Cg) and then look up the cable size which has an It which is the next highest, or equal to, the product of that calculation (which I don't think actually has a symbol)

The use of the rating factors can be a bit confusing at first and is often not taught very well. We take the nominal current of the circuit and apply factors to that which will either increase it or reduce it, and then use this modified value to select the cable size (unless we can calculate based on design current)
So if we have an In of 6A and have a combined rating factor of 0.5 we select a cable that can carry a current of 12A (6A/0.5=12A)

Radoslav said:

1. Iz this is where I'm stuck as I know this should be a tabulated value but I'm struggling with the fact that it's value is smaller than In - (some help to get this right will be needed.

Click to expand...

Iz is the CCC of the cable after rating factors have been applied, you can calculate this seperately to the cable selection process by applying the rating factors to It.

Radoslav said:

View attachment 100377

Click to expand...

For circuit 1:
In = 6A
Ca = 0.89
Cg = 0.65

So we calculate:
=In/(Ca*Cg)
=6/(0.89*0.65)
=6/0.5785
=10.37
But we need a whole number so we round it to 10

Now we go to table 4D5 and look down the column for method 103 and find a cable which has a CCC greater than or equal to 10A
Thus we select 1.5mm as it has an It equal to 10A

Radoslav said:

View attachment 100377

Click to expand...

Next piece of the puzzle, I can see where your calculation has gone wrong.

You've got a figure of 4.38 in your It row for circuit 1 which suggests you've made a mathematical error in applying the formula.

You've tried to calculate 6/(0.89*0.65) but have actually calculated 6/0.89*0.65

Where an equation contains brackets you need to do the calculation within the brackets first and then continue with the rest of the calculation.

I used to struggle with this kins of maths and found it helped me to write out calculations the way I did in the post above. So wrote the equation, then write it with the symbols replaced by the actual values on the next line then continue with a new line for each step of the calculation.

Spreadsheets are good and certainly have their uses, but when you are learning you will find a lot of value in putting pen to paper and doing the calculations long-hand.

Radoslav said:

but we were told 2.5mm for some reason and therefore I'm using 2.5mm. I will correct that and maybe go with my first thought of 6mm?

Click to expand...

Who told you 2.5mm? If it was the college tutor then I'd be a bit concerned about their knowledge. Even if it was a genuine mistake, which we all make, then 2.5mm for a 32A radial would make anyone stop and double check (except in particular circumstances maybe)

Don't go with guesses or what just sounds right, go through the calculations as I did above and work out the right answer.
College projects will often be deliberately designed to give some unusual/unexpected cable sizes as a way of making sure you are actually doing the calculations and not just filling in boxes with what sounds right for a circuit.

@davesparks thanks a lot for the very detailed help and information. I will go step by step through your last replies.

1. I certainly did misunderstand the meanings of It & Iz, but I have a much better understanding and knowledge on this now.
2. With regards to the calculations I make them both on paper and in excel as I need to have them written down for future reference just to be able to explain how I got the results and it also helps remember the formulas when I write them down. I did it right in my notebook, but for some reason forgot to put the brackets in excel and therefore the values were incorrect - In my defense I will say I do this mainly after work before bed time and also having a baby certainly does not make things easy Also not very helpful attaching a screenshot of the wrong table.
3. We choose the 2.5mm cable in classroom during a discussion and from what I can remember was based on the installation method and design current, but all I don't have much in my notes as to why we thought the 2.5mm will be ok.

After taking into consideration all I've quickly learned in this forum I decided to do some changes in my project to try and simplify my detailed explanation after I'm done with the calculations.
I'm uploading my current calculations so whoever want's to take a quick look at it and if something is wrong please let me know and I will correct it. I'm not trying to make it perfect, just want to try and understand as much as I can in this project as I have 4 more to go and they won't be easier.
Thanks in advance for all the help.

Calcs file

Radoslav said:

1. We choose the 2.5mm cable in classroom during a discussion and from what I can remember was based on the installation method and design current, but all I don't have much in my notes as to why we thought the 2.5mm will be ok.

Click to expand...

Could you be confusing the towel rail circuit with the socket radial?

A towel rail is likely to be a fixed load which does not require overload protection, so you can use Ib to calculate cable size instead of In. I suspect that the project has been intentionally designed to create this situation.

Definitely the radial, we were talking about and discussing different size cables for the different circuits.
I have made some amendments to my calcs now so I think it looks a lot better. I'm thinking of using installation method 102 for the radial circuit as well? I was just thinking as it's a radial circuit doesn't that mean I'm technically doubling the cable? For example if I go for 4mm with CCC 18.5A this will technically be 37A as I have two cables connected into the socket? Having said that, my Iz value is still 55A which is higher than the 37A CCC so not sure if this will work ?