The purpose of performing a cable calculation is to ensure that a cable is not overloaded. This can produce an excessive amount of heat which in turn can damage the cable and the installation and may even cause a fire.
So, here’s how that’s done…
Firstly how much current is your appliance going to use?
Let’s say it’s a 5kW appliance and we are going to be using a 70 deg C thermoplastic flat sheathed cable. So the design current (Ib) would be ‘5000/230 = 21.73 Amps’. Knowing this we can select the size of the protective device (In) and in this instance we are going to use a 32A circuit breaker.
We’ve not finished yet though as we need also to make allowances for anything that may restrict a conductors ability to disperse any heat that it may produce. These are known as rating factors. For example ‘ambient temperature (Ca)’, ‘buried cables (Cc)’, ‘Semi-enclosed fuse to BS3036 (Cf)’ etc (Take a look at Appendices 4 Section 3 of BS7671 for more rating factors that may apply). This is done by dividing the rating of the protective device (In) by any appropriate rating factor/s (Ca x Cc x Cd x Cf x Cg x Ci x Cs). If any of the rating factors do not apply to your cable then simply remove them from the equation.
In our example we have an ambient temperature of 35 deg C (Ca = 0.94 From Table 4B1) and some thermal insulation in a stud wall but with the cable touching the inner surface of the wall (aka reference method 102 in table 4D5). Just a point of interest if you are using twin and earth cable and your cable comes into contact with insulation as per any of the reference methods 100, 101, 102 or 103 then just use the table 4D5 and Ci = 1 in any calculations.
So, we have In/(Ca x Ci) or 32/(0.94 x 1) = 34.04A and looking at Table 4D5 column 4 ‘Method 102’ we can see that a 6mm cable gives us a current carrying capacity of 35A which is what we’ll be using providing that the Volt drop is okay?
More on Voltage Drop next time
Cheers Paul