Iz not greater than In

[Shpark]

I'm doing some calcs for college and need a little help.

So, Iz has to be greater than In, but I'm not achieving this.

My totals watts for a lighting circuit is 354w

So 354/230=1.54

Ib=1.54A

In=6A for lighting CPD

Correction factors of:-

Ca - 0.91
Cg - 0.7

Iz = 1.54/(0.91x0.7) = 2.4A

What do I do?

Cheers

 

[Strima]

You have worked out the current, you now need to select the correct cable and protective device.

Looks like you're getting confused, Iz is current carrying capacity, In is the current setting of protective device.

You have calculated Ib which is the design current.

 

[Shpark]

I have picked the protective device at 6A MCB and 1.5mm cable

 

[JDE397]

Are you getting a little confused between Iz and It. Page 333 of BS 7671 equation 2 states to work out the value of It you divide In by the correction factors. In your calculation you divided Ib by the correction factors. By using In you will always satisfy the requirements of regulation 433.1.1 of BS 7671.

 

[davesparks]

I have picked the protective device at 6A MCB and 1.5mm cable

How did you select the cable size without completing the cable size calculation?

You’ve divided Ib by the de rating factors where you should be dividing In by the de rating factors.
So It=In/(Ca.Cg)
Once you have done that you can use Iz to select your cable size from the tables.

It’s also worth noting that It doesn’t necessarily have to be greater than In, the rating factor Ca for example can be greater than 1 for low temperatures so can result in a lower It than In.

 

[telectrix]

the Iz for 1.5mm cable, clipped direct is about 19A. that's where you done gone wrong.

 

[Shpark]

At college I've always been taught that Iz=Ib/Cg x Ca

This is an example of what we do...attached

 

[davesparks]

At college I've always been taught that Iz=Ib/Cg x Ca

This is an example of what we do...attached

Then I’m afraid to say you may have misunderstood, or your college are teaching you incorrectly.
It can be calculated by this method if overload protection is not required, but not Iz

Iz is the current carrying capacity of a cable, this can be calculated by multiplying the tabulated current carrying capacity by the rating factors. This is explained in the appendix to the regulations along with the reasoning for why we rarely carry out this calculation, normally It is calculated to establish the required conductor size.
It should be greater than or equal to In divided by the rating factors.

 

[Shpark]

Yeah someone somewhere has it messed up! Back to the drawing board! Cheers guys

 

[jailson]

hi there everyone this is my first post here so be patience whit me.. Im doing the electric course and im doing my assessment and in need of some help.
for my understand iz=in/CaxCgxCi......
then it is from the tables 4D2A Correct?
on the images I add I did some of the calculations but I start to think im doing something wrong. can you clarify it for my.. like do I need to fill the iz when im doing a ring final?
Thanks for help...