05: Cable Size and Protective Device Selection
The cable most commonly used in domestic electrical installations is the 70 °C thermoplastic PVC insulated and sheathed flat cable which is often referred to as ‘Twin and Earth’ or ‘T&E’ and is available in various sizes ranging from 1.0 mm² up to 16 mm².
As well as twin and earth a 1.0 mm² or 1.5 mm² 3-core and earth 70 °C thermoplastic PVC insulated and sheathed flat cable is also available and often used in 2-way lighting and central heating controls etc.
When selecting which conductor size to use for any circuit there are numerous things you need to consider.
Current carrying capacity of the conductors ‘Iz’…
As current flows through a conductor due to its electrical resistance it produces heat. If the current in the conductor increases, then the heat also continues to rise and eventually we could reach a temperature which will cause permanent damage to the cable or possibly even a fire within the installation.
In order to prevent this from happening we have a 70 °C maximum permitted temperature at the conductor for a standard thermoplastic PVC Twin and Earth cable. (See regulation 523.1 and Table 52.1 BS7671:2018) and we must ensure that our conductor does not go above this temperature.
We must also allow for anything that could restrict a cables ability to dissipate its heat as this too can cause a problem. In a domestic environment although not the biggest problem, this could be by running the cables with other cables or along runs with any central heating pipes etc.
But by far the biggest problem particularly in domestic installations is the amount of thermal insulation present, especially within studded walls and in loft spaces.
This thermal insulation is designed to prevent heat loss and as such we should avoid it with our cables whenever possible.
Unfortunately keeping cables away from thermal insulation is not always possible, so if we are passing through any thermal insulation one method of reducing the amount of heat present in our cable is to reduce the amount of current flowing through it.
Regulation 523.9 and Table 52.2 in BS7671:2018 provides guidance on how much the current should be reduced by dependent on the amount of thermal insulation the cable is passing through.
For example, a cable passing through 200 mm of thermal insulation should have its current carrying capacity reduced by multiplying it by 0.63 and a cable passing through 500 mm or more of thermal insulation should be reduced by 0.5 times of its current carrying capacity when clipped direct.
If we can’t reduce the amount of current present to a value below the maximum of the reduced current carrying capacity, then we need to increase the conductor size.
This larger size will now have a lower resistance and a lower overall conductor temperature with the same amount of current flowing.
Looking at BS7671:2018 Table 4D5 which is specifically for 70 °C Thermoplastic PVC cable (Twin & Earth) we have the maximum current caring capacities for various conductor sizes depending upon where they are installed and the amount of thermal insulation present, better known as ‘reference methods’.
The reference methods used are…
Reference method 100: Above a plasterboard ceiling covered by thermal insulation not exceeding 100mm in thickness.
Reference method 101: Above a plasterboard ceiling covered by thermal insulation exceeding 100mm in thickness.
Reference method 102: In a stud wall with thermal insulation with cable touching the inner surface wall.
Reference method 103: In a stud wall with thermal insulation with cable not touching the inner surface wall.
Reference method C: Clipped directly to the surface / buried below the surface on a plastered wall.
As you can see reference method C is the optimum installation method and allows for the highest amount of current to be carried by the conductor.
For example, a 2.5 mm² T&E can carry up to 27A when clipped direct. However, when passing through thermal insulation which restricts the conductor’s ability to dissipate its heat this value needs to be reduced as we’ll see…
When a 2.5 mm² T&E is installed to reference method 100 ‘above a plasterboard ceiling covered by thermal insulation not exceeding 100mm in thickness’ it has a maximum current allowed of 21A.
This is reduced even further when a cable is installed in the worst conditions, reference method 103 ‘in a stud wall with thermal insulation with cable not touching the inner surface wall’ which has a maximum current carrying capacity of 13.5A, which is half of what it could carry if it was clipped directly to the surface and this is the same for all of the T&E conductor sizes.
So, as we can see in order to prevent a conductor from becoming too hot and damaged, we need to limit the amount of current flowing through it to a maximum value based upon where and how it’s installed ‘its method of installation’.
Circuit breakers and fuses are used to prevent the current from going above the maximum current carrying capacity of the conductors and as such in accordance with regulation 433.1 BS7671:2018;
The rating of the circuit breaker or fuse ‘In’ should be chosen so that is equal to or less than the maximum current caring capacity of the conductors ‘Iz’ and also, the amount of current being used by the circuit, its design current ‘Ib’ should not be higher than the rating of the protective device.
This can be written in the following format…
Ib ≤ In ≤ Iz
“Design current ‘Ib’ should be less than or equal to the rating of the protective device ‘In’ which in turn should be equal to or lower than the current carrying capacity of the conductors ‘Iz’.”
Other things to consider which due to the short length of circuits within a domestic installation shouldn’t really be an issue, is exceeding the maximum loop impedance value (Zs) due to the resistance of the line and CPC conductors (R1+R2) of a circuit.
For example on a TN-S supply we have a maximum external earth loop impedance (Ze) of 0.8 Ω and as Zs = Ze + (R1+R2) this means that eventually if the circuit was long enough the resistance of the line and CPC conductors (R1+R2) when added to the external loop impedance (Ze) could exceed the maximum permitted Zs for that particular circuit (see Chapter 41 BS7671:2018).
However, as the combined resistance of the line (R1) and CPC (R2) conductors is very low i.e 36.2 mΩ per meter or R1+R2 = 0.0362 Ω per meter for a 1mm line and CPC cable it is unlikely that any circuits will be long enough for this to become a problem within a domestic installation.
Another thing that could only be a problem on larger installation and again not domestic installations is the length of the circuit and it’s effect on voltage drop.
The formula for calculating volt drop is (Ib x L x mV/a/m) ÷ 1000 which is design current, times the length of the circuit in metres, times the mV/a/m figure which is taken from Table 4D5 column 8 of BS7671. And then divide it all by 1000 to convert it from mΩ to Ω.
As you can see from the formula, the longer the circuit the higher the voltage drop is going to be and for our supply voltage (U0) of 230v there are maximum permitted values of 3% (6.9v) for lighting circuits and 5% (11.5v) for anything else (see table 4Ab BS7671:2018).
There is a very useful table within the current edition of the On-Site Guide see Table 7.1(i) and if you use this as your guide to the maximum length of a circuit after selecting the correct cable size and protective device rating should ensure that you don’t fall foul due to the length of your circuits within a domestic installation.
In summary.
When selecting the correct conductor size to use we need to know two things first…
The design current ‘Ib’ = watts ÷ voltage, and where the cable is being run ‘method of installation’
For example, if we had an appliance that consumed 6kW then our design current (Ib) would be 6000 ÷ 230 = 26.09A. If the cable was going to be installed to reference method 101 “above a plasterboard ceiling covered by thermal insulation exceeding 100mm in thickness”. Then looking at table 4D5 BS7671 we would need a cable size of 6mm2 as this could carry up to 27A (Iz).
We can now use the previously mentioned formula to find the correct protective device
Ib ≤ In ≤ Iz
Design current (Ib) should be less than or equal to the rating of the protective device (In) which in turn should be equal to or lower than the current carrying capacity of the conductors (Iz).
And this is where we sometimes can hit a problem as we have done here…
Our design current (Ib) is 26.09A and the closest available BS60898 type B circuit breaker rating (In) above or equal to this is 32A.
Unfortunately, the current caring capacity (Iz) of 6mm² ref method 101 is 27A which is less than the rating of the circuit breaker, meaning that the 6mm² cable will not be protected from overload / overheating so we will have to go bigger which will be 10mm² and this will now carry up to 36A.
So, we have a design current (Ib) of 26.09A, a protective device rating (In) of 32A and a current carrying capacity of the 10mm2 conductor (Iz) of 36A and this satisfies the formula…
Ib ≤ In ≤ Iz or 26.09A ≤ 32A ≤ 26A
and provided you don’t go above the maximum lengths in table 7.1(i) On-Site Guide which for our 10mm cable is…
On a TN-S system 105 metres with a 30mA RCD fitted or 74 metres with no RCD, and on a TN-C-S supply its 105 metres for both
As you can see these lengths are way above what is likely to be used within a standard domestic installation.
This is Part Five of our Electrical Foundation Series
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