16th Edition (reference only) – NOW superseded by the 17th Edition IEE Regulations.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
 Earthing
 5.1 - The earthing principle 5.6 - Protective multiple earthing (PME) 5.2 - Earthing Systems 5.7 - Earthed concentric wiring 5.3 - Earth fault loop impedance 5.8 - Other protection methods 5.4 - Protective conductors 5.9 - Residual current devices (RCDs) 5.10 - Combined functional and protective ---------earthing
 5.4.1 - Earthing conductors 5.4.2 - Protective conductor types 5.4.3 - Bonding conductors 5.4.4 - Protective conductor cross-section assessment 5.4.5 - Protective conductor cross-section calculation 5.4.6 - Unearthed metalwork

5.4.5 - Protective conductor cross-section calculation

The c.s.a. of the circuit protective conductor (c.p.c.) is of great importance since

the level of possible shock in the event of a fault depends on it (as seen in {5.4.4}).

Safety could always be assured if we assessed the size using {Table 5.7} as a basis.

However, this would result in a more expensive installation than necessary because we would often use protective conductors which are larger than those found to be acceptable by calculation. For example. twin with cpc insulated and sheathed cables larger than 1 mm² would be ruled out because in all other sizes the CPC is smaller than required by {Table 5.7}.

In very many cases, calculation of the CPC size will show that a smaller size than that detailed in {5.4.4} is perfectly adequate. The formula to be used is:

 S = Ö(Ia²t) k where S  is the minimum protective conductor cross-sectional area (mm2) Ia  is the fault current (A) t   is the opening time of the protective device (s) k  is a factor depending on the conductor material and insulation, and the initial and maximum insulation temperatures.

This is the same formula as in {3.7.3}, the adiabatic equation, but with a change in the subject. To use it, we need to have three pieces of information, Ia, t and k.

 1) To find Ia Since Ia = Uo we need values for Ia = uo and Zs Zs Uo is simply the supply voltage, which in most cases will be 240V. Zs is the earth-fault loop impedance assuming that the fault has zero impedance.

Since we must assume that we are at the design stage, we cannot measure the loop impedance and must calculate it by adding the loop impedance external to the installation (Ze) to the resistance of the conductors to the furthest point in the circuit concerned. This technique was used in {5.3.6}.

Thus, Zs = Ze + R1 + R2 where R1 and R2 are the resistances of the phase and protective conductors respectively from {Table 5.5}.

2) To find t
We can find t from the time/current characteristics of {Figs 3.13 to 3.19} using the value of Ia already calculated above. For example, if the protective device is a 20 A miniature circuit breaker type I and the fault current is 1000 A, we shall need to consult {Fig 3.16}, when we can read off that operation will be in 0.01 s (10 ms). (It is of interest here to notice that if the fault current had been 80 A the opening time could have been anything from 0.04 s to 20 s,so the circuit would not have complied with the required opening times).

3) To find k
k is a constant, which we cannot calculate but must obtain from a suitable table of values. Some values of k for typical protective conductors are given in {Table 5.8}.

It is worth pointing out here that correctly installed steel conduit and trunking will always meet the requirements of the Regulations in terms of protective conductor impedance.

Although appearing a little complicated, calculation of acceptable protective conductor size is worth the trouble because it often allows smaller sizes than those shown in {Table 5.7}.

 Table 5.8 - Values of k for protective conductors Nature of protective conductor Initial temp. (°C) Final temp (°C) Conductor material K p.v.c. insulated, not in cable or bunched 30 160 Copper 143 - 30 160 Aluminium 95 - 30 160 Steel 52 - p.v.c. insulated, in cable or bunched 70 160 Copper 115 - 70 160 Aluminium 76 - Steel conduit or trunking 50 160 Steel 47 - Bare conductor 30 200 Copper 159 - 30 200 Aluminium 105 - 30 200 Steel 58

Example 5.2
A load takes 30 A from a 240 V single phase supply and is protected by a 32 A HBC fuse to BS 88. The wiring consists of 4 mm² single core p.v.c. insulated cables run in trunking, the length of run being 18 m. The earth-fault loop impedance external to the installation is assessed as 0.7 Ohms. Calculate the cross-sectional area of a suitable p.v.c. sheathed protective conductor.

This is one of those cases where we need to make an assumption of the answer to the problem before we can solve it. Assume that a 2.5 mm² protective conductor will be acceptable and calculate the combined resistance of the phase and protective conductors from the origin of the installation to the end of the circuit. From {Table 5.5}, 2.5 mm² cable has a resistance of 7.4 mohms/m and 4 mm² a resistance of 4.6 mOhms/m. Both values must be multiplied by 1.2 to allow for increased resistance as temperature rises due to fault current.

 Thus, R1 + R2 = (7.4 + 4.6) x 1.2 x 18 Ohms = 12.0 x 1.2 x 18 = 0.26 Ohms 1000 1000

This conductor resistance must be added to external loop impedance to give the total earth-fault loop impedance.

 Zs = Ze + Rl + R2 = 0.7+ 0.26 Ohms = 0.96 Ohms

We can now calculate the fault current:

 la = Uo = 240 = 250A Zs 0.96

Next we need to find the operating time for a 32 A BS 88 fuse carrying 250 A. Examination of {Fig 3.15} shows that operation will take place after 0.2 s.

Finally, we need a value for k. From {Table 5.8} we can read this off as 115, because the protective conductor will be bunched with others in the trunking.

We now have values for Ia, t and k so we can calculate conductor size.

 S = Ö(Ia²t) = Ö(250² x 0.02) mm² = 0.97 mm² k 115

This result suggests that a 1.0 mm² protective conductor will suffice. However, it may he dangerous to make this assumption because the whole calculation has been based on the resistance of a 2.5 mm² conductor. Let us start again assuming a 1.5 mm² protective conductor and work the whole thing through again.

The new size protective conductor has a resistance of 18.1 mOhms/m, see {Table 5.5}, and with the 4 mm² phase conductor gives a total conductor resistance, allowing for increased temperature, of 0.491 Ohms. When added to external loop impedance this gives a total earth-fault loop impedance of 1.191 Ohms and a fault current at 240 V of 202 A. From {Fig 3.15} operating time will be 0.6 s. The value of k will be unchanged at 115.

 S = Ö(Ia²t) = Ö(202² x 0.6) mm² = 1.36 mm² k 115

Thus, a 1.5 mm² protective conductor can be used in this case. Note that if the size had been assessed rather than calculated, the required size would be 4 mm², two sizes larger. A point to notice here is that the disconnection time with a 1.5mm² protective conductor is 0.6 s, which is too long for socket outlet circuits (0.4 s max.).

Example 5.3
A 240 V, 30 A ring circuit for socket outlets is 45 m long and is to be wired in 2.5 mm² flat twin p.v.c. insulated and sheathed cable incorporating a 1.5 mm² cpc. The circuit is to be protected by a semi-enclosed (rewirable) fuse to BS 3036, and the earth-fault loop impedance external to the installation has been ascertained to be 0.3 Ohms. Verify that the 1.5 mm² cpc enclosed in the sheath is adequate.

First use {Table 5.5} to find the resistance of the phase and cpc conductors. These are 7.4 mOhms/m and 12.1 mOhms/m respectively, so for a 45 m length and allowing for the resistance increase with temperature factor of 1.2.

 R1 + R2 = (7.4 + 12.1) x 1.2 x 45 Ohms = 19.5 x 1.2 x 45 Ohms = 1.05 Ohms 1000 1000

 Zs = Ze + R1+ R2 = 0.3 + 1.05 Ohms = 0.3 + 0.263 Ohms = 0.563 Ohms 4 4

The division by 4 is to allow for the ring nature of the circuit.

 Ia = Uo = 240 A = 426A Zs 0.563

We must then use the time/current characteristic of {Fig 3.13} to ascertain an operating time of 0.10 s.

From {Table 5.8} the value of k is 115.

 Then S = Ö(Ia²t) = Ö(426² x 0.10) mm² = 1.17 mm² k 115

Since this value is smaller than the intended value of 1.5 mm², this latter value will be satisfactory.

Example 5.4
A 240 V single-phase circuit is to be wired in p.v.c. insulated single core cables enclosed in plastic conduit. The circuit length is 45 m and the live conductors are 16 mm² in cross-sectional area. The circuit will supply fixed equipment, and is to be protected by a 63 A HBC fuse to BS 88. The earth-fault loop impedance external to the installation has been ascertained to be 0.58 Ohms. Calculate a suitable size for the circuit protective conductor.

With the information given this time the approach is somewhat different. We know that the maximum disconnection time for fixed equipment is 5 s, so from the time/current characteristic for the 63 A fuse {Fig 3.15} we can see that the fault current for disconnection will have a minimum value of 280 A.

 Thus,  Zs = Uo = 240 Ohms  = 0.857 Ohms Ia 280

If we deduct the external loop impedance, we come to the resistance of phase and protective conductors.

 R1 +R2 = Zs-Ze = 0.857-0.58Ohms = 0.277 Ohms

Converting this resistance to the combined value of R1 and R2 per metre,

 (R1 + R2) per metre = 0.277 x 1000 mOhms/m = 5.13 mOhms/m 45 x 1.2

Consulting {Table 5.5} we find that the resistance of 16 mm² copper conductor is 1.15 mOhms/m, whilst 10 mm² and 6 mm² are 1.83 and 3.08 mOhms/m respectively. Since 1.15 and 3.08 add to 4.23, which is less than 5.13, it would seem that a 6 mm² protective conductor will be large enough. However, to be sure we must check with the adiabatic equation.

 R1 + R2 = (1.15+3.08) x 1.2 x 45 Ohms = 0.228 Ohms 1000

 Zs = Ze+(R1 +R2) = 0.58+0.228 Ohms = 0.808 Ohms

 Ia = Uo = 240 A = 297 A Zs 0.808

From {Fig 3.15} the disconnection time for a 63 A fuse carrying 297 A is found to he 3.8s.
From {Table 5.8} the value of k is 115.

 Then S = Ö(Ia²t) = Ö(297² x 3.8) mm² = 5.03 mm² k 115

Since 5.03 is less than 6 then a 6 mm² protective conductor will be large enough to satisfy the requirements