to the 16th Edition IEE Regulations
   
   
   
 
 

chapter 5
Earthing

chapter 6
Circuits

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)
5.5 - Earth electrodes

5.10 - Combined functional and protective
---------earthing


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 mm 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 =
(Iat)
 
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 maximuminsulation 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 mm 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 mm 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 mm cable has a resistance of 7.4 mohms/m and 4 mm 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 =
(Iat)
=
(250 x 0.02)
mm = 0.97 mm
 
k
 
115
 

This result suggests that a 1.0 mm 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 mm conductor. Let us start again assuming a 1.5 mm 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 mm 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 =
(Iat)
=
(202 x 0.6)
mm = 1.36 mm
 
k
 
115
 

Thus, a 1.5 mm 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 mm, two sizes larger. A point to notice here is that the disconnection time with a 1.5mm 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 mm flat twin p.v.c. insulated and sheathed cable incorporating a 1.5 mm 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 mm 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 =
(Iat)
=
(426 x 0.10)
mm = 1.17 mm
   
k
 
115
 

Since this value is smaller than the intended value of 1.5 mm, 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 mm 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 mm copper conductor is 1.15 mOhms/m, whilst 10 mm and 6 mm 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 mm 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 =
(Iat)
=
(297 x 3.8)
mm = 5.03 mm
 
k
 
115
 

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

 

 

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Extracted from The Electricians Guide Fifth Edition
by John Whitfield
Published by EPA Press Click Here to order your Copy

Click here for list of abbreviations