to the 16th Edition IEE Regulations
   
   
   
 
 

chapter 5
Earthing

chapter 6
Circuits

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
------------------------------------------------------------------
4.3.11 - Cable volt drop


4.3.11 - Cable volt drop

All cables have resistance, and when current flows in them this results in a volt drop. Hence, the voltage at the load is lower than the supply voltage by the amount of this volt drop.

The volt drop may be calculated using the basic Ohm's law formula
U = I x R
where U is the cable volt drop (V
  I is the circuit current (A), and
  R is the circuit resistance W(Ohms)

Unfortunately, this simple formula is seldom of use in this case, because the cable resistance under load conditions is not easy to calculate.

[525-01-03] indicates that the voltage at any load must never fall so low as to impair the safe working of that load, or fall below the level indicated by the relevant British Standard where one applies.

[525-01-02] indicates that these requirements will he met if the voltage drop does not exceed 4% of the declared supply voltage. If the supply is single-phase at the usual level of 240 V, this means a maximum volt drop of 4% of 240 V which is 9.6 V, giving (in simple terms) a load voltage as low as 230.4 V. For a 415 V three-phase system, allowable volt drop will be 16.6 V with a line load voltage as low as 398.4 V.

It should be borne in mind that European Agreement RD 472 S2 allows the declared supply voltage of 230 V to vary by +10% or -6%. Assuming that the supply voltage of 240 V is 6% low, and allowing a 4% volt drop, this gives permissible load voltages of 216.6 V for a single-phase supply, or 374.5 V (line) for a 415 V three-phase supply.

To calculate the volt drop for a particular cable we use {Tables 4.6, 4.7 and 4.9}. Each current rating table has an associated volt drop column or table. For example, multicore sheathed non-armoured P.V.C. insulated cables are covered by {Table 4.7} for current ratings, and volt drops. The exception in the Regulations to this layout is for mineral insulated cables where there are separate volt drop tables for single- and three-phase operation, which are combined here as {Table 4.9}.

Each cable rating in the Tables of [Appendix 4] has a corresponding volt drop figure in millivolts per ampere per metre of run (mV/A/m). Strictly this should be mV/(A m), but here we shall follow the pattern adopted by BS 7671: 1992. To calculate the cable volt drop:

1. - take the value from the volt drop table (mV/A/m)
2. - multiply by the actual current in the cable (NOT the current rating)
3. - multiply by the length of run in metres
4. - divide the result by one thousand (to convert millivolts to volts).

For example, if a 4 mm˛ p.v.c. sheathed circuit feeds a 6 kW shower and has a length of run of 16 m, we can find the volt drop thus:

From {Table 4.7}, the volt drop figure for 4 mm˛ two-core cable is 11 mV/A/m.


Cable current is calculated from
I =
P = 6000 A
= 25 A
   
U     240-----
 


Volt drop is then
11 x 25 x 16 V
= 4.4 V
 
1000
 



Since the permissible volt drop is 4% of 240 V, which is 9.6 V, the cable in question meets volt drop requirements. The following examples will make the method clear.

Example 4.5
Calculate the volt drop for the case of Example 4.1. What maximum length of cable would allow the installation to comply with the volt drop regulations?

The table concerned here is {4.7}, which shows a figure of 7.3 mV/A/m for 6 mm˛ twin with protective conductor pvc insulated and sheathed cable. The actual circuit current is 12.5 A, and the length of run is 14 m.

Volt drop =
7.3 x 12.5 x 14 V
= 1.28 V
 
1000
 


Maximum permissible volt drop is
4% of 240 V =
4 of 240 V
= 9.6 V
   
100
 


If a 14 m run gives a volt drop of 1.28 V,
the length of run for a 9.6 V drop will be:
9.6 x 14m
= 105m
1.28
 



Example 4.6
Calculate the volt drop for the case of {Example 4.2}. What maximum length of cable would allow the installation to comply with the volt drop regulations?

The Table concerned here is {4.7} which shows a volt drop figure for 4.0 mm˛ cable of 11mV/A/m, with the current and the length of run remaining at 12.5 A. and 14 m respectively.

Volt drop =
11 x 12.5 x 14 V
= 1.93 V
 
1000
 


Maximum permissible volt drop is
4% of 240 V =
4 of 240 V
= 9.6 V
   
100
 


If a 14 m run gives a volt drop of 1.93 V,
the length of run for a 9.6 V drop will be:
9.6 x 14m
= 70m
1.93
 



Example 4.7
Calculate the volt drop for the cases of {Example 4.3} for each of the alternative installations. What maximum length of cable would allow the installation to comply with the volt drop regulations in each case?

In neither case is there any change in cable sizes, the selected cables being 6 mm˛ in the first case and 4 mm˛ in the second. Solutions are thus the same as those in {Examples 4.5 and 4.6} respectively.

Example 4.8
Calculate the volt drop and maximum length of run for the motor circuit of {Example 4.4}.

This time we have a mineral insulated p.v.c. sheathed cable, so volt drop figures will come from {Table 4.9}. This shows 9.1 mV/A/m for the 4 mm˛ cable selected, which must be used with the circuit current of 15.3 A and the length of run which is 20 m.

Volt drop =
9.1 x 15.3 x 20 V
= 2.78 V
 
1000
 


Maximum permissible volt drop is
4% of 415 V =
4 of 415V
= 16.6 V
   
100
 


Maximum length of run for this circuit
with the same cable size and type will be:
  16.6 x 20m
= 119m
2.78
 



The 'length of run' calculations carried out in these examples are often useful to the electrician when installing equipment at greater distances from the mains position.

It is important to appreciate that the allowable volt drop of 4% of the supply voltage applies to the whole of an installation. If an installation has mains, sub-mains and final circuits, for instance, the volt drop in each must be calculated and added to give the total volt drop as indicated in {Fig 4.10}.

All of our work in this sub-section so far has assumed that cable resistance is the only factor responsible for volt drop. In fact, larger cables have significant self inductance as well as resistance. As we shall see in Chapter 5 there is also an effect called impedance which is made up of resistance and inductive reactance (see {Fig 5.8(a)}).

Inductive reactance
XL
=
2(pi)fL
where
XL
=
inductive reactance in Wohms
  (pi) =
the mathematical constant 3.142
  f =
the system frequency in hertz (Hz)
  L =
circuit self inductance in henrys (H)

It is clear that inductive reactance increases with frequency, and for this reason the volt drop tables apply only to systems with a frequency lying between 49 Hz and 61 Hz.

Fig 4.10 Total volt drop in large installations

For small cables, the self inductance is such that the inductive reactance, is small compared with the resistance. Only with cables of cross-sectional area 25 mm˛ and greater need reactance be considered. Since cables as large as this are seldom used on work which has not been designed by a qualified engineer, the subject of reactive volt drop component will not be further considered here.

If the actual current carried by the cable (the design current) is less than the rated value, the cable will not become as warm as the calculations used to produce the volt drop tables have assumed, The Regulations include (in [Appendix 4]) a very complicated formula to be applied to cables of cross-sectional area 16 mm˛ and less which may show that the actual volt drop is less than that obtained from the tables. This possibility is again seldom of interest to the electrician, and is not considered here.

 

 

 

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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