16th Edition (reference only) – NOW superseded by the 17th Edition IEE Regulations.
 chapter 1 The IEE Regulations chapter 2 Installation Requirements and Characteristics chapter 3 Installation Control and Protection chapter 4 Cables, Conduits and Trunking chapter 5 Earthing chapter 6 Circuits chapter 7 Special Installations chapter 8 Testing and Inspection chapter 9 Data cabling and Networks
 Cables, conduits and trunking
 4.1 - Cable insulation materials 4.4 - Cable supports, joints and terminations 4.2 - Cables 4.5 - Cable enclosures 4.3 - Cable choice 4.6 - Conductor and cable identification
 4.3.1 - Cable types 4.3.8 - Protection by semi-enclosed (rewirable) fuses 4.3.2 - Current carrying capacity of conductors 4.3.9 - Cable rating calculation 4.3.3 - Methods of cable installation 4.3.10 - Special formulas - grouping factor calculation 4.3.4 - Ambient temperature correction factors 4.3.11 - Cable volt drop 4.3.5 - Cable grouping correction factors 4.3.12 - Harmonic currents and neutral conductors 4.3.6 - Thermal insulation correction factors 4.3.13 - Low smoke-emitting cables 4.3.7 - When a number of correction --------- factors applies 4.3.14 - The effects of animals, insects and plants

4.3.9 - Cable rating calculation

The Regulations indicate the following symbols for use when selecting cables:

 Iz is the current carrying capacity of the cable in the situation where it is installed It is the tabulated current for a single circuit at an ambient temperature of 30°C Ib is the design current, the actual current to be carried by the cable In is the rating of the protecting fuse or circuit breaker I2 is the operating current for the fuse or circuit breaker (the current at which the fuse blows or the circuit breaker opens) Ca is the correction factor for ambient temperature Cg is the correction factor for grouping Ci is the correction factor for thermal insulation.

The correction factor for protection by a semi-enclosed (rewirable) fuse is not given a symbol but has a fixed value of 0.725.

Under all circumstances, the cable current carrying capacity must be equal to or greater than the circuit design current and the rating of the fuse or circuit breaker must be at least as big as the circuit design current. These requirements are common sense, because otherwise the cable would be overloaded or the fuse would blow when the load is switched an.

To ensure correct protection from overload, it is important that the protective device operating current (I2) is not bigger than 1.45 times the current carrying capacity of the cable (Iz). Additionally, the rating of the fuse or circuit breaker (In) must not be greater than the the cable current carrying capacity (Iz) It is important to appreciate that the operating current of a protective device is always larger than its rated value. In the case of a back-up fuse, which is not intended to provide overload protection, neither of these requirements applies.

To select a cable for a particular application, take the following steps: (note that to save time it may be better first to ensure that the expected cable for the required length of circuit will] not result in the maximum permitted volt drop being exceeded

1. - Calculate the expected (design) current in the circuit (Ib)

2. - Choose the type and rating of protective device (fuse or circuit breaker) to be used (In)

3. - Divide the protective device rated current by the ambient temperature
----- correction factor (Ca) if ambient temperature differs from 30°C

4. - Further divide by the grouping correction factor (Cg)

5. - Divide again by the thermal insulation correction factor (CI)

6. - Divide by the semi-enclosed fuse factor of 0.725 where applicable

7. - The result is the rated current of the cable required, which must be chosen
----- from the appropriate tables {4.6 to 4.9}.

Observe that one should divide by the correction factors, whilst in the previous subsection we were multiplying them. The difference is that here we start with the design current of the circuit and adjust it to take account of factors which will derate the cable. Thus, the current carrying capacity of the cable will be equal to or greater than the design current. In {4.3.7} we were calculating by how much the current carrying capacity was reduced due to application of correction factors.

{Tables 4.6 to 4.9} give current ratings and volt drops for some of the more commonly used cables and sizes. The Tables assume that the conductors and the insulation are operating at their maximum rated temperatures. They are extracted from the Regulations Tables shown in square brackets e.g. [4D1A]

The examples below will illustrate the calculations, but do not take account of volt drop requirements (see {4.3.11}).

Example 4.1

An immersion heater rated at 240 V, 3 kW is to be installed using twin with protective conductor p.v.c. insulated and sheathed cable. The circuit will be fed from a 15 A miniature circuit breaker type 2, and will be run for much of its 14 m length in a roof space which is thermally insulated with glass fibre. The roof space temperature is expected to rise to 50°C in summer, and where it leaves the consumer unit and passes through a 50 mm insulation-filled cavity, the cable will be bunched with seven others. Calculate the cross-sectional area of the required cable.

First calculate the design current Ib

 Ib= P = 3000A = 12.5A -U 240

The ambient temperature correction factor is found from {Table 4.3} to be 0.71. The group correction factor is found from {Table 4.4} as 0.52. (The circuit in question is bunched with seven others, making eight in all).

The thermal insulation correction factor is already taken into account in the current rating table (4D2A ref. method 4] and need not be further considered. This is because we can assume that the cable in the roof space is in contact with the glass fibre but not enclosed by it. What we must consider is the point where the bunched cables pass through the insulated cavity. From {Table 4.5} we have a factor of 0.89.

The correction factors must now be considered to see if more than one of them applies to the same part of the cable. The only place where this happens is in the insulated cavity behind the consumer unit. Factors of 0.52 (Cg) and 0.89 (CI) apply. The combined value of these (0.463), which is lower than the ambient temperature correction factor of 0.71, and will thus be the figure to be applied. Hence the required current rating is calculated:-

 Table 4.6 - Current ratings and volt drops for unsheathed single core p.v.c. insulated cables Cross sectional area In conduit in thermal insulation In conduit in thermal insulation In conduit on wall In conduit on wall Clipped direct Clipped direct Volt drop Volt drop (mm²) (A) (A) (A) (A) (A) (A) (mV/A/m) (mV/A/m) - 2 cables 3 or 4 cables 2 cables 3 or 4 cables 2 cables 3 or 4 cables 2 cables 3 or 4 cables 1.0 11.0 10.5 13.5 12.0 15.5 14.0 44.0 38.0 1.5 14.5 13.5 17.5 15.5 20.0 18.0 29.0 25.0 2.5 19.5 18.0 24.0 21.0 27.0 25.0 18..0 15.0 4.0 26.0 24.0 32.0 28.0 37.0 33.0 11.0 9.5 6.0 34.0 31.0 41.0 36.0 47.0 43.0 7.3 6.4 10.0 46.0 42.0 57.0 50.0 65.0 59.0 4.4 3.8 16.0 61.0 56.0 76.0 68.0 87.0 79.0 2.8 2.4

 Table 4.7 - Current ratings and volt drops for sheathed multi-core p.v.c.-insulated cables Cross sectional area In conduit in thermal insulation In conduit in thermal insulation In conduit on wall In conduit on wall Clipped direct Clipped direct Volt drop Volt drop (mm²) (A) (A) (A) (A) (A) (A) (mV/A/m) (mV/A/m) - 2 core 3 or 4 core 2 core 3 or 4 core 2 core 3 or 4 core 2 core 3 or 4 core 1.0 11.0 10.0 13.0 11.5 15.0 13.5 44.0 38.0 1.5 14.0 13.0 16.5 15.0 19.5 17.5 29.0 25.0 2.5 18.5 17.5 23.0 20.0 27.0 24.0 18.0 15.0 4.0 25.0 23.0 30.0 27.0 36.0 32.0 11.0 9.5 6.0 32.0 29.0 38.0 34.0 46.0 41.0 7.3 6.4 10.0 43.0 39.0 52.0 46.0 63.0 57.0 4.4 3.8 16.0 57.0 52.0 69.0 62.0 85.0 76.0 2.8 2.4

Iz =     in     =      15 A      = 32.4 A
Cg x Ci     0.52 x 0.89

From {Table 4.7}, 6 mm² p.v.c. twin with protective conductor has a current rating of 32 A. This is not quite large enough, so 10 mm²with a current rating of 43 A is indicated. Not only would this add considerably to the costs, but would also result in difficulties due to terminating such a large cable in the accessories.

A more sensible option would be to look for a method of reducing the required cable size. For example, if the eight cables left the consumer unit in two bunches of four, this would result in a grouping factor of 0.65 (from {Table 4.4}). Before applying this, we must check that the combined grouping and thermal insulation factors (0.65 x 0.89 = .0.58) are still less than the ambient temperature factor of 0.71, which is the case.

 Table 4.8 - Current ratings of mineral insulated cables clipped direct Cross-sectional area Volt p.v.c. sheath 2 x single or twin p.v.c. Sheath 3 core p.v.c. Sheath 3 x single or twin Bare sheath 2 x single Bare sheath 3 x single (mm²) (A) (A) (A) (A) (A) 1.0 500v 18.5 16.5 16.5 22.0 21.0 1.5 500v 24.0 21.0 21.0 28.0 27.0 2.5 500v 31.0 28.0 28.0 38.0 36.0 4.0 500v 42.0 37.0 37.0 51.0 47.0 1.0 750v 20.0 17.5 17.5 24.0 24.0 1.5 750v 25.0 22.0 22.0 31.0 30.0 2.5 750v 34.0 30.0 30.0 42.0 41.0 4.0 750v 45.0 40.0 40.0 55.0 53.0 6.0 750v 57.0 51.0 51.0 70.0 67.0 10.0 750v 78.0 69.0 69.0 96.0 91.0 16.0 750v 104.0 92.0 92.0 127.0 119.0

Note that in (Tables 4.8 and 4.9) 'P.V.C. Sheath means bare and exposed to touch or having an over-all covering of p.v.c. or LSF and 'Bare' means bare and neither exposed to touch nor in contact with combustible materials.

 Table 4.9 - Volt drops for mineral insulated cables Cross-sectional area Single-phase p.v.c. Sheath Single-phase bare Three-phase p.v.c. Sheath Three-phase bare (mm²) (mV/A/m) (mV/A/m) (mV/A/m) (mV/A/m) 1.0 42.0 47.0 36.0 40.0 1.5 28.0 31.0 24.0 27.0 2.5 17.0 19.0 14.0 16.0 4.0 10.0 12.0 9.1 10.0 6.0 7.0 7.8 6.0 6.8 10.0 4.2 4.7 3.6 4.1 16.0 2.6 3.0 2.3 2.6

This leads to a cable current rating of         15           A = 25.9 A
0.65 x 0.89

This is well below the rating for 6 mm² of 32 A, so a cable of this size could be selected.

Example 4.2
The same installation as in Example 4.1 is proposed. To attempt to make the cable size smaller, the run in the roof space is to be kept clear of the glass fibre insulation. Does this make any difference to the selected cable size?

There is no correction factor for the presence of the glass fibre, so the calculation of Iz will be exactly the same as Example 4.1 at 32.4 A.

This time reference method I (clipped direct) will apply to the current rating {Table 4.7}. For a two core cable, 4.0 mm², two core has a rating of 36 A, so this Will be the selected size.

It is of interest to notice how quite a minor change in the method of installation, in this case clipping the cable to the joists or battens clear of the glass fibre, has reduced the acceptable cable size.

Example 4.3
Assume that the immersion heater indicated in the two previous examples is to be installed, but this time with the protection of a 15 A rewirable (semi-enclosed) fuse. Calculate the correct cable size for each of the alternatives, that is where firstly the cable is in contact with glass fibre insulation, and secondly where it is held clear of it.

This time the value of the acceptable current carrying capacity Iz will be different because of the need to include a factor for the rewirable fuse as well as the new ambient temperature and grouping factors for the rewirable fuse from {Tables 4.3 and

Iz =           in           =              15 A             = 45.7 A
Cg x Ca x 0.725      0.52 x 0.87 x 0.725

In this case, the cable is in contact with the glass fibre, so the first column of {Table 4.7} of current ratings will apply. The acceptable cable size is 16 mm² which has a current rating of 57 A.

This cable size is not acceptable on the grounds of high cost and because the conductors are likely to be too large to enter the connection tunnels of the immersion heater and its associated switch. If the cables leaving the consumer unit are re-arranged in two groups of four, this will reduce the grouping factor to 0.65, so that the newly calculated value of Iz is 36.6 A. This means using 10 mm² cable with a current rating of 43A (from {Table 4.7}), since 6 mm² cable is shown to have a current rating in these circumstances of only 32 A. By further rearranging the cables leaving the consumer unit to be part of a group of only two, Cg is increased to 0.8, which reduces Iz to 29.7 A which enables selection of a 6 mm² cable.

Should it be possible to bring the immersion heater cable out of the consumer unit on its own, no grouping factor would apply and Iz would fall to 23.8 A, allowing a 4 mm² cable to be selected.

Where the cable is not in contact with glass fibre there will be no need to repeat the calculation of Iz, which still has a value of 29.7 A provided that it is possible to group the immersion heater cable with only one other where it leaves the consumer unit. This time we use the 'Reference Method 1 (clipped direct)' column of the current rating {Table 4.7}, which shows that 4 mm² cable with a current rating of 36 A will be satisfactory.

Examples 4.1, 4.2 and 4.3 show clearly how forward planning will enable a more economical and practical cable size to be used than would appear necessary at first. It is, of course, important that the design calculations are recorded and retained in the installation manual.

Example 4.4
A 415 V 50 Hz three-phase motor with an output of 7,5 kW, power factor 0.8 and efficiency 85% is the be wired using 500 V light duty three-core mineral insulated p.v.c. sheathed cable. The length of run from the HBC protecting fuses is 20 m, and for about half this run the cable is clipped to wall surfaces. For the remainder it shares a cable tray, touching two similar cables across the top of a boiler room where the ambient temperature is 50°C. Calculate the rating and size of the correct cable.

The first step is to calculate the line current of the motor.

Input =   output   = 7.5 x 100 kW = 8.82 kW
efficiency         85

Line current Ib =             P                 =       8.82 x 103 A  = 15.3 A
------------------------- Ö3 x UL x  cosÆ             Ö3 x 415 x 0.8

We must now select a suitable fuse. {Fig 3.15} for BS 88 fuses shows the 16 A size to be the most suitable. Part of the run is subject to an ambient temperature of 50°C, where the cable is also part of a group of three, so the appropriate correction factors must be applied from {Tables 4.3 and

Iz =   In   =----------  16A    = 34.2A
Cg x Ca  ------ 0.70x 0.67

Note that the grouping factor of 0.70 has been selected because where the cable is grouped it is clipped to a metallic cable tray, and not to a non-metallic surface. Next the cable must be chosen from {Table 4.8}. Whilst the current rating would be 15.3 A if all of the cable run were clipped to the wall, part of the run is subject to the two correction factors, so a rating of 34.2 A must be used. For the clipped section of the cable (15.3 A), reference method I could be used which gives a size of 1.0 mm² (current rating 16.5 A). However, since part of the cable is on the tray (method 3) the correct size for 34.2 A will be 4.0 mm², with a rating of 37 A.