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 mm² 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 mm² 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 mm² 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 mm² 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 mm² in the first
case and 4 mm² 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 mm² 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 mm² 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 mm² 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.
