JKR Elektrik-Saiz Cable
JKR Elektrik-Saiz Cable
JKR Elektrik-Saiz Cable
0908
Introduction
The tabulated current-carrying capacity relates to a single circuit in the installation
methods shown in Table 4A (Refer to IEE Wiring Regulation Sixteenth Edition
page 181, 182, 183, 184 & 185), in an ambient air temperature of 30C. The
current-carrying capacities given in the tables for a.c operation apply only to
frequencies in the range 49 to 61 Hz.
The tabulated current-carrying capacity relates to continuous loading and is also
known as the full thermal current rating of the cable, corresponding to the
conductor operating temperature indicated in the headings to the tables
concerned. It is intended to provide for a satisfactory life of conductor and
insulation subject to the thermal effects of carrying current for sustained periods in
normal service. A cable may be seriously damaged, leading to early failure, or its
service life may be significantly reduced, if it is operated for any prolonged period
at a temperature higher than the indicated value.
In addition, there are other consideration affecting the choice of the crosssectional area of a conductor, such as the requirements for protection against
electric shock, protection against thermal effects, over current protection, voltage
drop and the limiting temperatures for terminals of equipment to which the
conductors are connected.
Other than that, the conductors will need to be sized in accordance with the
relevant circuit criteria. This sizing will also have to be taken into account the type
of cable and its insulation. Other requirements that will also affect the size
indirectly may include: availability, type of building, environment, security, life
expectancy, adaptability, aesthetics and cost.
2.0
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b)
For grouping
Tables 4B1, 4B2 and 4B3 (Refer to IEE Wiring Regulation Sixteenth
Edition page 186 & 187) give the correction factor to be applied to the
tabulated current-carrying capacity where cables or circuits are grouped.
c) For thermal insulation
For a cable installed in a thermally insulating wall or above a thermally
insulated ceiling the cable being in contact with a thermally conductive
surface on one side, the rating factor to applied may, in the absence of
more precise information, be taken as 0.75 times the current carrying
capacity for that a cable likely to be totally surrounded by thermally
insulation material. The applicable rating factor may be as low as 0.5
3.0
It
the value of current tabulated in the Table with referring to the IEE Wiring
Regulation Sixteenth Edition for the type of cable and installation method
concerned, for a single circuit in an ambient temperature of 30C
Ib
the design current of the circuit, i.e. the current intended to be carried by
the circuit in normal service
In
the nominal current or current setting of the device protecting the circuit
against over current
In all circumstances Iz must be not less than Ib and In also must be not less than
Ib. However CKE practice is Iz > I n > Ib
4.0
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voltage drop in mV, divided by Ib and by the length of run, will give the value of
voltage drop in mV/A/m which can be tolerated. A voltage drop not exceeding that
value is identified in the appropriate table and the corresponding cross-sectional
area of conductor needed on this account can be read off directly before any other
calculation are made.
The conductor size necessary from consideration of the conditions of normal load
and overload is then determined. All correction factors affecting Iz (i.e. the factors
for ambient temperature, grouping and thermal insulation) can, if desired, be
applied to the values of It as multipliers. This involves a process of trial and error
until a cross-sectional area is reached which ensures that I z is not less than I b and
not less than In of any protective device it is intended to select. In any event, if a
correction factor for protection by a semi-enclosed fuse is necessary, this has to
be applied to In as a divisor. It is therefore more convenient to apply all the
correction factors to In as divisors.
Once the cable type and other characteristics have been considered, the
conductor size can be determined. This size will be dependent upon the current
required by the circuit load
The following procedure enables the designer to determine the size of cable it will
be necessary to use in order to comply with the requirement for overland
protection.
1. For single circuits
Divide the nominal current of the protective device (In ) by any applicable
correction factor for ambient temperature (Ca) given in Table 4C1 and then
further divide by any applicable correction factor for thermal insulation (Ci).
The size of cable to be used is to be such that its tabulated currentcarrying capacity (It) is not less than the value of nominal current of the
protective device adjusted as below:
It In
CaCi
2. For groups
Divide the nominal current of the protective device (In) by the correction
factor for grouping (Cg) given in Tables 4B1, 4B2 or 4B3:
It In
Cg
Where a rewire able fuse to BS036 is to be used, an additional factor (0.725) must
be included, hence
It
In
0.725Ca CiCg
Thus, the factors to determining current carrying capacity of conductors will be
such as:
a.
b.
c.
d.
cross-sectional area
type of cable or conductor
method of installation
number of conductors grouped together
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5.0
Voltage Drop
Values of voltage drop are tabulated for a current of one ampere for a meter run
i.e. for a distance of 1 m along the route taken by the cables, and then present the
result of the voltage drops in the entire circuit conductor. For any given run the
values need to be multiplied by length of the run in meters and by the current the
cables are to carry, in amperes. The voltage drop for any particular cable run must
be such that the voltage drop in the circuit of which the cable forms a part does
not exceed 4% of the nominal voltage of the supply. As a guide, refer to the
following diagram. Volt drop from DB to final point need not be calculated. It is
assumed to be negligible since we are limiting our final circuit to not more than 80
meters from the DB.
DB
DB
1%
DB
1.5%
SSB
1.5%
SSB
1%
4%
4%
4%
2.5%
SSB
2%
MSB / SOURCE
Diagram 1
However for external lighting the volt drop from DB to final point must be
calculated. The total must not exceed 4%.
MSB
2%
SSB
DB CL
1.5%
0.5%
Final Point
4%
Diagram 2
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Where Vd
In
VD
L
= Vd L In
1000
It is pertinent to point out that the cable rating must always be higher than (or
equal to) that of the fuse or circuit breaker that is supposed to protect that part of
the installation.
6.0
ii)
iii)
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7.0
DB GAL
1.5%
L2
SSB G1
2.5%
SSB G3
L1
SSB G2
MSB NE
MD = 527.36 kW
IMD = 863.14A
Diagram 3
Allowable voltage drop:
2.5% = 10.375V
1.5% = 6.225V
3 phase
Iz > In > Ib
i)
Calculate cable size used and voltage drop from SSB G to DB GAL
a) Consider copper conductor used e.g. single core PVC insulated cable
and installation method is enclosed in conduit on a wall where L2 = 60
metre.
Iz > In > Ib
Refer to table 4D1A from IEE Wiring Regulation Sixteenth Edition, I z = 50A
Iz > 40 > 22.28
50 > 40 > 22.28
From value of Iz we get cable size of 10mm.
From table 4D1B voltage drop (per ampere per metre), V d = 4.4 mV/A/m
Voltage drop (VD)
= Vd x L 2 x I n / 1000
= 4.4 x 60 x 40 / 1000
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= 10.56V
Therefore cable to be used is 4 x 10mm PVC in conduit.
b) Say L 2 is longer i.e. L2 = 80 metre
Voltage drop (VD)
= Vd x L 2 x I n / 1000
= 4.4 x 80 x 40 / 1000
= 7.84V
(exceeding allowable voltage drop of 6.225V)
= Vd x L 2 x I n / 1000
= 2.8 x 80 x 22.28 / 1000
= 4.99V
Calculate cable size used and voltage drop from MSB NE to SSB G
a) Consider copper conductor XLPE/SWA/PVC cable, installed underground
where L1 = 110 metre
Iz > In > Ib
Refer to table 4E4A from IEE Wiring Regulation Sixteenth Edition, Iz =
251A
Iz > 250 > 137.09
251 > 250 > 137.09
From value of Iz we get cable size of 70mm.
From table 4E4B voltage drop (per ampere per metre), Vd = 0.60 mV/A/m
Voltage drop (VD)
= Vd x L 1x I n / 1000
= 0.60 x 110 x 137.09 / 1000
= 9.05V
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= Vd x L 1 x I n / 1000
= 0.60 x 155 x 137.09 / 1000
= 12.75V
(exceeding allowable voltage drop of 10.375V)
= Vd x L 1 x I n / 1000
= 0.45 x 155 x 137.09 / 1000
= 9.56V
= Vd x L 1 x I n / 1000
= 0.60 x 250 x 137.09 / 1000
= 20.56V
(exceeding allowable voltage drop of 10.375V)
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