Cable size

[Wilson12]

I'm struggling to work out cable size.
a lighting circuit in a office with the distribution board next door,having three cables chased up.to the suspended office room next door then run in basket tray for two meters before separating to each circuit.
In=6
Ca.30 degrees seems a high ambient temperature
I'd of thought room temp is more like 20 so I don't get that

I put 25degrees so x1.03

Cg three circuits bunched in air or embedded or enclosed
So I put x0.7

Then the protective devise won't be a bs3036 fuse
So x1

6/1.03x0.7x1 =8.32 now I don't get how to use table f5 to get cable size

I then tried with cooker
9kw
/230 =39a
Diversity =18.73 a which sounds low for a cooker
In =20
Ca1 cg1 cf1
So again I have 20 as current carrying capacity but how do I get the cable size from the table
And am I right so far thanks

 

[Simonslimline]

What is the Ib (design current) for the lighting circuit?

 

[Simonslimline]

What are the distances of these final circuits as well?

 

[Wilson12]

Ib is 5.6
And length of lighting is
12 meters
Is that for the voltage drop as I first want to get my head around cable size and these tables thanks

 

[Simonslimline]

Ib is 5.6
And length of lighting is
12 meters
Is that for the voltage drop as I first want to get my head around cable size and these tables thanks

1. Work out Ib, design current.
2. select protective device with In greater than or equal to Ib.
3.The tabulated current carrying capacity of the cable (It) is given by
______In______
Ca x Cg x Ci x Cf

4. Select suitable cable size using the tables in BS7671.


5. Calculate Voltage drop to make sure it is within the required permitted limits. mV/A/M x Ib x L. Divide by 1000.


Been a long time since i have done this so hopefully its all there. I am sure i will pulled up if it is not.

 

[Wilson12]

Yes that's right. I know what the process is... Just not sure how to use table f5 to get the cable size

 

[telectrix]

just select the cable with the Iz> than the Ib that you have calculated

 

[Richard Burns]

If you perform the calculations as above to determine the tabulated current carrying capacity (It).

You also need to know the cable type you are intending to use as this will be dictated by the area in which you are installing it.
e.g if you are using singles in trunking then choose table F4, if you are using flat twin and earth then use table F6, etc.

Then you need to know how the cable is to be installed and whether it is single or three phase.
Note if the cable is T&E installed in insulating material you should not use the Ci rating factor when calculating It as this is accounted for by the installation methods 100-103 in the table (similarly for underground in a duct etc.)

On the appropriate table choose the correct column for installation method and number of phases and look down that column until you reach a current that is higher than the It you calculated, then move left along the row to read off the cross sectional area of the cable that is required.

 

[Wilson12]

For a 32amp ring final 2.5mm is used
Why here is 2.5mm only able to carry around 20amps

 

[Wilson12]

Why In all methods 2.5mm2 twin and earth can't carry more than 27Amps
When it's normal to use 2.5 on a ring final 32amp circuit?

 

[davesparks]

Why In all methods 2.5mm2 twin and earth can't carry more than 27Amps
When it's normal to use 2.5 on a ring final 32amp circuit?

Because in a ring circuit you have both ends of the circuit connected to the ocpd.
So the load current will be shared proportionally between the two possible paths of current flow. This means that the 2.5mm cable won't ever see more than 2/3 of the possible 32A