06 determining conductor cross sectional areas

Transcription

507
6. Determining conductor crosssectional areas
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Industrial electrical network design guide
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6. DETERMINING CONDUCTOR CROSS-SECTIONAL AREAS
Owing to the respective characteristics of LV and MV conductors, they have been dealt with in
separate paragraphs.
6.1.
Determining conductor cross-sectional areas and choosing protective
devices in low voltage
n definition of terms relating to low voltage wiring systems
(Insulated) cable
Assembly comprising:
- one or more insulated conductors
- their eventual individual screening
- any eventual assembly protection
- any eventual protective shielding
It may also comprise one or several bare conductors.
Multi-core cable
Cable comprising more than one conductor, which may eventually include bare conductors.
Note: the term three-core cable is used to designate the cable making up the phases of a three-phase
system.
Single-core cable
Cable comprising a single insulated conductor.
Note: the term single-core cable is especially used to designate a cable making up one of the phases of a
three-phase system.
Wiring system
Assembly made up of one or more electric conductors and the devices ensuring their fixation and, if
necessary, their mechanical protection.
Cable channel
Ventilated or enclosed duct located above or in the ground, having dimensions preventing persons from
moving around inside it but allowing access to the cables over their entire length during and after
installation.
Note: a cable channel may or may not form part of the building construction.
Cable tray
Holder made up of a base and sides but no cover.
Note: A cable tray may be perforated or unperforated.
Electrical circuit (of an installation)
All the electrical equipment of the installation fed from the same source and protected against
overcurrents by the same protective device(s).
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(Insulated) conductor
Assembly comprising the conductor, its insulating envelope and eventual screens.
(Circular) conduit
Enclosed envelope, having a circular cross-section, designed for the installation or the replacement of
insulated conductors or cables by capstan, in electrical installations.
Ducting
Assembly of closed envelopes having a non circular cross-sectional area, designed for the installation or
the replacement of insulated conductors or cables by capstan, in electrical installations.
Brackets
Horizontal cable supports fixed at one of their ends, arranged from point to point and on which the cables
rest.
Design current of a circuit
Current to be carried in a circuit in normal service
(Continuous) current carrying capacity of a conductor
Maximum value of the current that, in given conditions, can continuously flow in a conductor without its
steady-state operating temperature being higher than the specified value.
Cable ladder
Cable support made up of a series of non-touching elements firmly fixed to main vertical rods.
Sleeve (or tube)
Element surrounding wiring and providing it with extra protection in building passages (walls, partitions,
floor, ceiling) or in buried passages.
Sheath
Enclosure located above ground level having dimensions preventing persons from moving around inside
it but allowing access to the cables over their entire length. A sheath may or may not be built into the
masonry.
Trough
Assembly of envelopes closed by a cover and ensuring mechanical protection of insulated conductors or
cables not installed or removed by a capstan and which allow other electrical equipment to be added .
Building void
Space in a structure or building parts which is only accessible at certain places.
Note:
- spaces in walls, supported floors, ceilings and certain types of window or door frames and
jamb linings are examples of building voids.
- specially built building voids are also called "ducts".
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6.1.1.
Method principle
In compliance with the recommendations of IEC 364-4-43, the cross-sectional area of wiring
systems and the protective device must be chosen to meet several conditions necessary for
the security of the installation.
The wiring system must:
- carry the maximum design current and its normal transient peaks
- not generate voltage drops above the allowed values.
The protective device must:
- protect the wiring system against any overcurrents up to the short-circuit current
- ensure the protection of persons against indirect contact.
The logigram in figure 6-1 sums up the principle of the method which may be described by the
following stages:
1st stage:
- using the load power, the maximum design current I B is calculated and the rated current
I n of the protective device is deduced from this
- the maximum short-circuit current Isc at the origin of the circuit is calculated and the
breaking capacity of the protective device is deduced from this.
2nd stage:
- depending on the installation conditions (installation method, ambient temperature, etc.),
the overall correction factor f is determined
- the suitable conductor cross-sectional area is chosen in relation to I n and f .
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3rd stage:
- the maximum voltage drop is checked
- the thermal withstand of the conductors in the event of a short circuit is checked
- for TN and IT systems, the maximum length relating to the protection of persons against
indirect contact is checked.
The conductor cross-sectional area meeting all these conditions is then chosen.
Note:
an economic cross-sectional area larger than the cross-sectional area determined above may
be chosen if necessary (see § 6.3).
apparent power
to be carried
upstream or
downstream
network
short-circuit
power at the origin
of the circuit
short-circuit
current
design current
I sc
IB
rated current of
protective device
protective device
breaking capacity
In
choice of
protective device
choice of
protective device
installation
conditions
wiring system conductor
cross-sectional area
check of thermal
withstand in case
of short-circuit
IT or TN earthing system
maximum voltage
drop check
maximum wiring
system length chek
TT earthing
system
conductor crosssectional area
determination
confirmation of the choice of wiring system
cross-sectional area and its electrical protection
economic
cross-sectional
area possibly chosen
Figure 6-1: wiring system cross-sectional area and protective device choice logigram
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6.1.2.
Determining the maximum design current
The maximum design current ( I B ) is defined according to the type of installation fed by the
wiring system.
In the case of individual power supply to a device, the current I B will be equal to the rated
current of the device being fed. On the other hand, if the wiring system feeds several devices,
the current I B will be equal to the sum of currents absorbed, taking into account the
installation utilisation and coincidence factors.
In the case of motor starting or cyclical operating conditions of loads (spot welding station,
see § 3.4.2), current inrushes must be taken into account when their thermal effects are
cumulated.
Some installations are subject to future extensions. The current corresponding to this
extension will be added to the existing value.
In direct current:
I=
In alternating current: I =
S
U
P  power consumed (in W ) 


U  duty voltage (in V ) 
S
S
in three-phase.
in single-phase and I =
U
U 3
: apparent power consumed (VA)
: . voltage between the two conductors for a single-phase power supply
. phase-to-phase voltage for a three-phase power supply
When high harmonic currents circulate in the conductor, they must be taken into account. In
order to choose the cross-sectional area, the following must therefore be taken:
 ∞ 
I r .m.s. = 
I p2 


 p =1 
1
∑
(see § 8)
I1 : current value at 50 Hz (or 60 Hz)
I p : value of harmonic current of order p
For example, for a speed variator
Ir .m.s.
≅ 1.7
I1
When there are compensation capacitors downstream of the wiring system, the design current
is determined as follows:
- assuming that compensation is in operation: in case of failure of the capacitors, the wiring
system is placed out of service
- assuming that compensation is out of service; in case of failure of the capacitors, the
conductor cross-sectional area is sufficient and availability is thus improved.
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n factor taking into account the power factor and efficiency:
a
The apparent power of a load is:
S=
P
η × Fp
in kVA
P : active power in kW
η : efficiency
Fp : power factor
We define the coefficient: a =
1
η × Fp
When a current stripped of harmonics flows through the conductor, Fp = cos ϕ .
n load utilisation factor: b
In an industrial installation, it is assumed that loads will never be used at their full power level.
A utilisation factor ( b ) is therefore introduced which generally varies from 0.3 to 1.
Without knowing the accurate values, we may take:
-
b = 0.75
for motors
-
b =1
for lighting and heating
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c
n coincidence factor:
In an industrial installation, the loads (of a workshop, for example) fed by the same wiring
system do not operate simultaneously in all cases. To take this phenomenon, which is linked to
the operating conditions of the installation, into account, the coincidence factor is applied to
the sum of the load powers in conductor sizing.
In the absence of precise indications resulting from experience of standard installations, the
values of tables 6-1 et 6-2 may be applied:
Use
Coincidence factor c
Lighting
1
Lighting and air conditioning
1
Power outlets
0.1 to 0.2 (for a number > 20)
Table 6-1: coincidence factor for an administrative building
Number of circuits having
similar nominal currents
Coincidence factor
2 and 3
0.9
4 and 5
0.8
5 to 9
0.7
10 and more
0.6
Table 6-2: coincidence factor for industrial distribution switchboards
n factor taking into account possible future extensions: d
The value of factor
installation.
d
must be estimated according to the foreseeable extensions of the
In the absence of precise indications, the value 1.2 is often used.
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n power conversion factor in current:
e
The power conversion factor in current is:
-
e = 8 in single-phase 127 V
e = 2.5 in three-phase 230 V
-
e = 4.35 in single-phase 230 V
e = 1.4 in three-phase 400 V
n maximum design current
The maximum design current is thus:
IB = P × a × b × c × d × e
P
: active power in kW
6.1.3.
Choosing the protective device
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n general rule
In compliance with IEC 364, a protective device (circuit-breaker or fuse) correctly fulfils its
function if the conditions outlined below are met.
o nominal or setting current
This must be between the design current and the current carrying capacity I a of the wiring
system:
I B ≤ I n ≤ I a , which corresponds to zone a in figure 6.2.
o conventional tripping current
This must meet the following relation:
I2 ≤ 1.45 I a , which corresponds to zone b in figure 6.2.
case of circuit-breakers
- For domestic circuit-breakers, standard IEC 898 specifies:
I2 = 1.45 In
- For industrial circuit-breakers, standard IEC 947-2 specifies:
I2 = 1.30 I set
we thus have I2 ≤ 1.45 I n (or I set )
while
I n ≤ I a (above condition)
The condition I2 ≤ 1.45 I a (zone b ) is thus automatically met.
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case of fuses
Standard IEC 269-1 specifies that I 2 is the current which ensures that the fuse fuses in the
conventional time (1 h or 2 h); I 2 is referred to as the conventional fusing current (see § 6.3.1
of the Protection guide).
I 2 = k2 × I n
where k2 = 1.6 to 1.9
depending on the fuses
Let us define the coefficient k3 such that:
k3 =
k2
1.45
Thus, the condition I2 ≤ 1.45 I a is met if:
I
In ≤ a
k3
For gG fuses:
-
I n ≤ 10 A
10 A < I n ≤ 25 A
I n > 25 A
à
à
à
k3 = 1.31
k3 = 1.21
k3 = 1.10
o breaking capacity
This must be higher than the three-phase maximum short-circuit current
( Isc 3 )
at its
installation point:
Breaking capacity ≥ Isc 3 , which corresponds to zone c in figure 6.2.
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o associating protective devices
The use of a protective device having a breaking capacity below the short-circuit current at the
point where it is installed is permitted by standard IEC 364 under the following conditions:
- there is another device upstream having at least the necessary breaking capacity
- the energy that the device placed upstream lets through is lower than the energy that the
downstream device and wiring systems protected by these devices can withstand without
being damaged.
This possibility is implemented:
.
in circuit-breaker/fuse associations
.
in the cascading technique which uses the high current limitation capacity of certain
circuit-breakers (e.g. the Compact).
The possible associations resulting from actual tests performed in a laboratory are given in
manufacturer catalogues.
6.1.4.
Current-carrying capacity of wiring systems
This is the maximum current that the wiring system can continuously carry without this being
prejudicial to its lifetime.
To determine this current, it is necessary to carry out the following:
- using tables 6-3 to 6-5, define the installation method, its associated selection number and
letter, and correction factors to be applied
- using the installation conditions, the correction factor values which must be applied are
determined (see tables 6-6 to 6-15)
- calculate the overall correction factor f equal to the product of the correction factors
- using table 6-16 for selection letters B, C, E, F and table 6-17 for selection letter D, the
maximum current
I 0 that the wiring system can carry under standard conditions
( f0 to f10 = 1 ) is determined
- calculate the maximum current that the wiring system can carry in relation to its installation
conditions: I a = f I 0 .
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n installation methods
Tables 6-3 to 6-5 give the main installation methods used in industrial networks.
For each installation method, the following is given:
- its associated selection number and letter
- the correction factors to be applied.
Factor f 0 corresponds to the installation method; factors f1 to f10 are explained below
(see tables 6-6 to 6-15).
Example
Description
N°
Selection
Correction factors
letter
f0
to be applied
11
C
1
f1
f4
f5
11A
C
0.95
f1
f4
f5
12
C
1
f1
f4
f5
Single or multi-core cables with or
without armour
- fixed on a wall
- fixed to a ceiling
- on unperforated trays
cables
multi-core
singlecore
- on perforated trays run
horizontally or vertically
13
E
F
1
f1
f4
f5
- on brackets
14
E
F
1
f1
f4
f5
- on ladders
16
E
F
1
f1
f4
f5
Table 6-3: installation methods for selection letters C, E and F
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Example
Description
Single or multi-core cables in
building voids
conduits in building voids
N°
Selection
Correction factors
letter
f0
21
B
0.95
f1
f4
f5
--
22A
B
0.865
f1
f4
f5
f6
23A
B
0.865
f1
f4
f5
f6
24A
B
0.865
f1
f4
f5
f7
25
B
0.95
f1
f4
f5
--
31A
B
0.9
f1
f4
f5
--
32A
B
to be applied
ducting in building voids
ducting built into the masonry
Single or multi-core
conductors :
- in false ceilings
- in suspended ceilings
troughs fixed to walls:
- run horizontally
- run vertically
0.9
f1
f4
f5
--
Table 6-4: installation methods for selection letter B
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Example
Description
N°
Selection
Correction factors
letter
f0
to be applied
troughs built into floors
33A
B
0.9
f1
f4
f5
--
suspended troughs
34A
B
0.9
f1
f4
f5
--
Multi-core cables in enclosed
channels run horizontally or
vertically
41
B
0.95
f1
f4
f5
--
open or ventilated channels
43
B
1
f1
f4
f5
--
Table 6-4 (cont.): installation methods for selection letter B
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Example
Description
N°
Selection
Correction factors
letter
f0
to be applied
conduits or in buried ducting
61
D
0.8
f2
f3
f8
f9
Single or multi-core cables
buried without any extra
mechanical protection
62
D
1
f2
f3
f10
--
Single or multi-core cables
buried with extra mechanical
protection
63
D
1
f2
f3
f10
--
Table 6-5: installation methods for selection letter D
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n correction factors for ambient temperatures other than 30 °C (wiring systems above
ground): f1
When electrical wiring systems are built into walls having heating elements, it is generally
necessary to reduce current-carrying capacities by applying the reduction factors in table 6-6.
This supposes that the distribution of temperatures inside the heated walls in contact with the
electrical wiring system is known.
When the air temperature is other than 30 °C, the correction coefficient to be applied is given
in the formula:
f1 =
θ p −θ0
θ p − 30o
θ p : maximum temperature permitted by the insulating material under steady-state conditions, °C
θ 0 : air temperature, °C
The value of f1 is given in table 6-6 for different values of θ p and θ 0 .
Insulation
Ambient
temperatures (°C)
PVC
XLPE and EPR
θ0
Elastomers
(rubber)
θ p = 60 °C
θ p = 70 °C
θ p = 90 °C
10
15
20
25
35
40
45
50
55
60
65
70
75
80
85
90
95
1.29
1.22
1.15
1.07
0.93
0.82
0.71
0.58
-
1.22
1.17
1.12
1.06
0.94
0.87
0.79
0.71
0.61
0.50
-
1.15
1.12
1.08
1.04
0.96
0.91
0.87
0.82
0.76
0.71
0.65
0.58
0.50
0.41
-
Table 6-6: correction factors for ambient temperatures other than 30 °C
(above ground wiring systems)
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n correction factors for ground temperatures other than 20 °C
(buried wiring systems): f 2
When the ground temperature is other than 20°C, the correction coefficient to be applied is
given in the formula:
f2 =
θ p −θ0
θ p − 20
θ p : maximum temperature permitted by the insulating material under steady-state conditions, °C
θ 0 : ground temperature, °C
The value of f 2 is given in table 6-7 for different values of θ p and θ 0 .
Ground temperature
θ 0 (°C)
Insulation
θ p = 70 °C
PVC
XLPE and EPR
θ p = 90 °C
10
1.10
1.07
15
1.05
1.04
25
0.95
0.96
30
0.89
0.93
35
0.84
0.89
40
0.77
0.85
45
0.71
0.80
50
0.63
0.76
55
0.55
0.71
60
0.45
0.65
65
-
0.60
70
-
0.53
75
-
0.46
80
-
0.38
Table 6-7: correction factor for ground temperatures other than 20 °C
(buried wiring systems)
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n correction factors for buried wiring systems, in relation to the soil thermal
resistivity: f 3
The soil thermal resistivity depends on the type and humidity of the ground. The correction
factor to be applied according to the soil resistivity is given in table 6-8.
Soil
thermal resistivity
K.m/W
Correction
factor
0.40
1.25
underwater installation
marshes
0.50
1.21
very moist soil
sand
0.70
1.13
moist soil
clay
0.85
1.05
normal soil
and
1.00
1.00
dry soil
chalk
1.20
1.50
0.94
0.86
very dry soil
2.00
0.76
and
2.50
0.70
clinker
3.00
0.65
Humidity
Observations
Type of soil
ash
Table 6-8: correction factors for buried wiring systems
in relation to the soil thermal resistivity
n correction factors for a group of several multi-core cables or groups of single-core
cables
The circuits or cables may be:
- touching; the correction factor f 4 must be applied
- arranged in several layers; the correction factor f 5 must be applied
- both touching and arranged in several layers (see fig. 6-3); correction factors
must then be applied.
f 4 and
f5
Figure 6-3: 6 multi-core cables - 2 layers of 3 touching cables
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o touching multi-core or groups of single-core cables: f 4
The factors in table 6-9 are to be applied to homogenous groups of cables, equally loaded, for
the given installation methods.
When the horizontal distance between neighbouring cables is greater than twice their external
diameter, no reduction factor is necessary.
The same correction factors are applicable:
- to groups of two or three single-core cables
- to multi-core cables.
N° of installation
methods
Number of touching multi-core cables or
groups of single-core cables
1
2
3
4
5
6
7
8
9
12
16
20
21, 22A, 23A, 24A,
25, 31, 31A, 32, 32A,
33A, 34A, 41, 43
1.00
0.80
0.70
0.65
0.60
0.55
0.55
0.50
0.50
0.45
0.40
0.40
11, 12
1.00
0.85
0.79
0.75
0.73
0.72
0.72
0.71
0.70
No extra
11A
1.00
0.85
0.76
0.72
0.69
0.67
0.66
0.65
0.64
reduction
13
1.00
0.88
0.82
0.77
0.75
0.73
0.73
0.72
0.72
factor for
14, 16
1.00
0.88
0.82
0.80
0.80
0.79
0.79
0.78
0.78
more than 9 cables
Table 6-9: correction factors for touching multi-core cables or
groups of single-core cables
o multi-core cables or groups of single-core cables arranged in several layers: f 5
When cables are arranged in several layers, the correction factors in table 6-10 must be
applied.
Number of layers
Correction factors f 5
2
3
4 or 5
6 to 8
9 plus
0.80
0.73
0.70
0.68
0.66
table 6-10: correction factors for a group of multi-core cables
or groups of single-core cables arranged in several layers
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n correction factors in relation to the number of conduits in air and their arrangement
(see table 6-11): f 6
Number of
Number of conduits arranged horizontally
conduits arranged
vertically
1
2
3
4
5
6
1
1
0.94
0.91
0.88
0.87
0.86
2
0.92
0.87
0.84
0.81
0.80
0.79
3
0.85
0.81
0.78
0.76
0.75
0.74
4
0.82
0.78
0.74
0.73
0.72
0.72
5
0.80
0.76
0.72
0.71
0.70
0.70
6
0.79
0.75
0.71
0.70
0.69
0.68
Table 6-11: correction factors in relation to the number of conduits in the air and their arrangement
n correction factors in relation to the number of conduits buried or built into concrete
and their arrangement (see table 6-12): f 7
Number of conduits
arranged vertically
Number of conduits arranged horizontally
1
2
3
4
5
6
1
1
0.87
0.77
0.72
0.68
0.65
2
0.87
0.71
0.62
0.57
0.53
0.50
3
0.77
0.62
0.53
0.48
0.45
0.42
4
0.72
0.57
0.48
0.44
0.40
0.38
5
0.68
0.53
0.45
0.40
0.37
0.35
6
0.65
0.50
0.42
0.38
0.35
0.32
Table 6-12: correction factors in relation to the number of conduits buried or built into concrete and their
arrangement
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n correction factors for non-touching buried conduits run horizontally or vertically on
the basis of one cable or group of 3 single-core cables per conduit
(see table 6-13) : f 8
Distance between conduits (a)
Number of conduits
0.25 m
0.5 m
1.0 m
2
0.93
0.95
0.97
3
0.87
0.1
0.95
4
0.84
0.9
0.94
5
0.81
0.7
0.93
6
0.79
0.6
0.93
Table 6-13: correction factors for non-touching buried conduits run horizontally or vertically on the basis
of one cable or group of 3 single-core cables per conduit
The distances between conduits are measured as shown in figure 6-4.
a
multi-core cables
a
single-core cables
Figure 6-4: distance between conduits (a)
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n correction factors in the case of several circuits or cables in the same buried conduit
(see table 6-14): f 9
This is applicable to groups of cables with varying cross-sectional areas but having the same
allowable maximum temperature.
Correction factors
Arrangement of
touching circuits
or cables
Number of circuits or multi-core cables
Installed in a buried
conduit
1
2
3
4
5
6
7
8
9
12
16
20
1
0.71
0.58
0.5
0.45
0.41
0.38
0.35
0.33
0.29
0.25
0.22
Table 6-14: correction factors in the case of several circuits or cables
in the same buried conduit
n correction factors for a group of several cables installed directly in the ground - single
or multi-core cables arranged horizontally or vertically (see table 6-15): f10
Distance between cables or groups of 3 single-core cables (a)
Number of cables
or circuits
Zero
(touching
cables)
One cable
diameter
0.25 m
0.5 m
1.0 m
2
0.76
0.79
0.84
0.88
0.92
3
0.64
0.67
0.74
0.79
0.85
4
0.57
0.61
0.69
0.75
0.82
5
0.52
0.56
0.65
0.71
0.80
6
0.49
0.53
0.60
0.69
0.78
Table 6-15: correction factors for a group of several cables installed directly in the ground single or multi-core cables arranged horizontally or vertically
The distances between cables are measured as shown in figure 6-5.
a
a
a
multi-core cables
single-core cables
Figure 6-5: distance between cables (a)
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n current-carrying capacities (in amps) of wiring systems in standard installation
conditions for selection letters B, C, E, F
The current carrying capacities given in table 6-16 are valid for simple circuits made up of the
following number of conductors:
Selection letter B:
- two insulated conductors or two single-core cables or one two-core cable
- three insulated conductors or three single-core cables or one three-core cable
Selection letter C:
- two single-core cables or one two-core cable
- three single-core cables or one three-core cable
Selection letters E and F (see fig. 6-6):
- one two-core or three-core cable for letter E
- two or three single-core cables for letter F .
E
E
F
F
Figure 6-6: illustration of installation methods for selection letters E and F
The number of conductors to be considered in a circuit is that of the conductors through which
the current actually flows. When, in a three-phase circuit, the currents are assumed to be
balanced, it is not necessary to take into account the corresponding neutral conductor.
When the current value of the neutral conductor is close to that of the phases, a reduction
factor of 0.84 is to be applied. Such currents may, for example, be due to the presence of third
harmonic currents in the phase conductors (see § 6.2).
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Selection letter
B
Insulating material and number of loaded conductors
PVC 3
C
PVC 2
XLPE 3
PVC 3
PVC 2
E
PVC 3
F
XLPE 2
XLPE 3
PVC 2
PVC 3
XLPE 2
XLPE 3
XLPE 2
PVC 2
XLPE3
XLPE2
Copper crosssection (mm²)
1.5
2.5
4
6
10
16
25
35
50
70
95
120
150
185
240
300
400
500
630
15.5
21
28
36
50
68
89
110
134
171
207
239
17.5
24
32
41
57
76
96
119
144
184
223
259
299
341
403
464
18.5
25
34
43
60
80
101
126
153
196
238
276
319
364
430
497
19.5
27
36
48
63
85
112
138
168
213
258
299
344
392
461
530
22
30
40
51
70
94
119
147
179
229
278
322
371
424
500
576
656
749
855
23
31
42
54
75
100
127
158
192
246
298
346
395
450
538
621
754
868
1005
24
33
45
58
80
107
138
169
207
268
328
382
441
506
599
693
825
946
1088
26
36
49
63
86
115
149
185
225
289
352
410
473
542
641
741
16.5
22
28
39
53
70
86
104
133
161
186
18.5
25
32
44
59
73
90
110
140
170
197
227
259
305
351
19.5
26
33
46
61
78
96
117
150
183
212
245
280
330
381
21
28
36
49
66
83
103
125
160
195
226
261
298
352
406
23
31
39
54
73
90
112
136
174
211
245
283
323
382
440
526
610
711
24
32
42
58
77
97
120
146
187
227
263
304
347
409
471
600
694
808
26
35
45
62
84
101
126
154
198
241
280
324
371
439
508
663
770
899
28
38
49
67
91
108
135
164
211
257
300
346
397
470
543
161
200
242
310
377
437
504
575
679
783
940
1083
1254
Aluminium crosssection (mm²)
2.5
4
6
10
16
25
35
50
70
95
120
150
185
240
300
400
500
630
121
150
184
237
289
337
389
447
530
613
740
856
996
Table 6-16: current carrying capacities (in amps) of wiring systems in standard installation conditions
( f0 to f10 = 1) for selection letters B, C, E, F
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n current-carrying capacities (in amps) of wiring systems in standard installation conditions
for selection letter D (buried wiring systems) (see table 6-17)
The number of conductors to be considered in a circuit is that of the conductors through which
the current actually flows. When, in a three-phase circuit, the currents are assumed to be
balanced, it is not necessary to take into account the corresponding neutral conductor.
When the current value of the neutral conductor is close to that of the phases, a reduction
factor of 0.84 is to be applied. Such currents may, for example, be due to the presence of third
harmonic currents in the phase conductors (see § 6.2).
Selection letter
D
Insulating material and number of loaded conductors
PVC 3
PVC 2
XLPE 3
XLPE 2
26
34
44
56
74
96
123
147
174
216
256
290
328
367
424
480
32
42
54
67
90
116
148
178
211
261
308
351
397
445
514
581
31
41
53
66
87
113
144
174
206
254
301
343
387
434
501
565
37
48
63
80
104
136
173
208
247
304
360
410
463
518
598
677
57
74
94
114
134
167
197
224
254
285
328
371
68
88
114
137
161
200
237
270
304
343
396
447
67
87
111
134
160
197
234
266
300
337
388
440
80
104
133
160
188
233
275
314
359
398
458
520
Copper cross-sectional
area (mm²)
1.5
2.5
4
6
10
16
25
35
50
70
95
120
150
185
240
300
Aluminium cross-sectional
area (mm²)
10
16
25
35
50
70
95
120
150
185
240
300
Table 6-17: current carrying capacities (in amps) of wiring systems in standard installation conditions
( f0 to f10 = 1) for selection letter D (buried wiring systems)
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6.1.5.
Practical method for determining the minimum cross-sectional area of an LV
wiring system
conductor
installation
conditions
design current
IB
determination of the protective device rated current I n or
setting current I set taken to be just higher than the design
current: I n or I set I B
I n or I set
determination of the
selection letter and
overall correction
factor f
(see tab. 8-3 to 8-5)
determination of current I z of the wiring system to be
protected by the protective device
fuse
circuit-breaker
Iz
1.31 I n if I n
10 A
Iz
1.21 I n if I n
10 A
and I n
25 A
1.10 I n if I n
25 A
Iz
Iz
I n or I set
I z2
I z1
determination of the cross-sectional area S of the wiring system
conductors able to carry I z1 or I z2 :
I z1
I
or z 2 (1)
f
f
'
- determine the cross-sectional area able to carry I z in standard
installation conditions, depending on the insulating material, the
number of loaded conductors and the type of conductor (copper or
aluminium) (see tab. 8-16 and 8-17)
'
- calculate the equivalent current I z
S
check of other required conditions:
- maximum voltage drop
- maximum length for protection against indirect contact
(IT and TN earthing systems)
- check of thermal withstand in case of short circuit
(1) I z' is an equivalent current which, in standard installation
conditions, causes the same thermal effect as I z1 or I z2
in actual installation conditions
Figure 6-7: logigram for determining the cross-sectional area of a LV wiring system
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6.1.6.
Cross-sectional area of protective conductors (PE), equipotential bonding
conductors and neutral conductors (IEC 364)
In a low voltage installation, the protective conductors ensure that the exposed conductive
parts of loads are interconnected and insulation fault currents are evacuated to the ground.
The equipotential bonding conductors allow the exposed conductive parts and extraneous
conductive parts to be set at the same potential, or similar potentials.
In this chapter, we will limit ourselves to conductor sizing rules. Refer to paragraph 2 for the
protection and connection rules.
n cross-sectional area of protective conductors between MV/LV transformer and main
LV switchboard (see fig. 6-8)
main LV switchboard
PE
Figure 6-8: PE conductors between transformer and main switchboard
Table 6-18 gives the protective conductor cross-sectional areas (in mm²) in relation:
- to the nominal power of the MV/LV transformer
- to the operating time t (in seconds) of the MV protection. When protection is ensured by a
fuse, the cross-sectional area to be taken into account corresponds to t = 0.2 s
- to the insulating material and type of conductor metal.
In an IT earthing system, if an overvoltage limiter is inserted between the neutral and earth,
the same sizing is applied to its connecting conductors.
In the case where several transformers operate in parallel, the sum of their nominal powers will
be used to determine the cross-sectional area.
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Transformer power
(kVA)
LV voltage
127/220 V 230/400 V
Type of conductors
Copper
t (s)
Aluminium
Bare
conductors
0.2 s 0.5 s
-
PVC-insulated
conductors
-
0.2 s 0.5 s
0.2 s 0.5 s
-
-
0.2 s 0.5 s
XLPE-insulated
conductors
0.2 s 0.5 s
-
-
0.2 s 0.5 s
≤ 63
≤ 100
25
25
25
25
25
25
25
25
25
100
160
25
25
35
25
25
50
25
25
35
125
200
25
35
50
25
35
50
25
25
50
160
250
25
35
70
35
50
70
25
35
50
200
315
Potective conductor
35
50
70
35
50
95
35
50
70
250
400
cross-sectional area
50
70
95
50
70
95
35
50
95
315
500
S PE (mm²)
50
70
120
70
95
120
50
70
95
400
630
70
95
150
70
95
150
70
95
120
500
800
70
120
150
95
120
185
70
95
150
630
1 000
95
120
185
95
120
185
95
120
150
800
1 250
95
150
185
120
150
240
95
120
185
Table 6-18: cross-sectional area of protective conductors
between MV/LV transformer and main LV switchboard
n cross-sectional areas of low voltage exposed conductive part protective conductors: (PE)
The cross-sectional area of the PE conductor is defined in relation to the cross-sectional area
of the phases (for the same metal conductor) as follows:
- for S phase ≤ 16 mm² ,
S PE = S phase
- for 16 mm² < S phase ≤ 35 mm² ,
S PE = 16 mm²
- for S phase > 35 mm² ,
S PE =
(1)
S phase
2
(1) when the protective conductor is not part of the wiring system, it must have a cross-sectional area of
at least:
- 2.5 mm² if it comprises a mechanical protection
- 4 mm² if it does not comprise a mechanical protection
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In the TT earthing system, the protective conductor cross-sectional area may be limited to:
- 25 mm² for copper
- 35 mm² for aluminium
on condition that the neutral and exposed conductive part earth electrodes are separate,
otherwise the conditions of the TN earthing system are applicable (in a TT earthing
system, there may be an involuntary connection via the metal structure or other part between
the two earth electrodes; the earth fault current is then high).
n cross-sectional area of equipotential bonding conductors
o main equipotential bonding conductor
Its cross-sectional area must be at least equal to half the cross-sectional area of the
installation's largest protective conductor, with a minimum of 6 mm². However, it may be limited
to 25 mm² for copper or 35 mm² for aluminium.
o supplementary equipotential bonding conductor
If it connects two exposed conductive parts, its cross-sectional area must not be smaller than
the smallest of the protective conductors connected to these parts (see fig. 6-9-a).
If it connects an exposed conductive part to an extraneous conductive part, its cross-sectional
area must not be smaller than half the cross-sectional area of the protective conductor
connected to this exposed conductive part (see fig. 6-9-b).
S PE1 ≤ S PE 2
If
S
S LS = PE
2
S LS = S PE1
S PE1
S PE2
S PE
S LS
P1
(*)
S LS
P2
a) between two exposed conductive parts
P
b) between an exposed conductive part and a
structure
Figure 6-9: cross-sectional area of supplementary equipotential bonding conductors
(*)
with a minimum of:
- 2.5 mm² if the conductors are mechanically protected
- 4 mm² if the conductors are not mechanically protected
Conductors which are not incorporated in a cable are mechanically protected when they are
installed in conduits, troughs or casing or protected in a similar way.
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n cross-sectional area of PEN conductors
In the case of a TNC earthing system, the protective conductor also plays the role of the
neutral conductor.
In this case, the cross-sectional area of the PEN must be at least equal to the greatest value
resulting from the following requirements:
- S PEN ≥
− 10 mm 2 for copper

− 16 mm 2 for aluminium
- meet the conditions relating to the PE conductor
- meet the conditions required for the neutral conductor cross-sectional area.
n cross-sectional area of the neutral conductor
- The neutral conductor must have the same cross-sectional area as the phase conductors in
the following cases:
.
single-phase circuit
.
three-phase circuit having phase cross-sectional areas smaller than or equal to 16 mm²
for copper or 25 mm² for aluminium.
- For three-phase circuits having a phase cross-sectional area greater than 16 mm² for
copper or 25 mm² for aluminium, the neutral cross-sectional area may be smaller than that
of the phases as long as the following conditions are met:
.
the maximum current likely to continuously circulate in the neutral is lower than the
current-carrying capacity of the chosen cross-sectional area. The unbalance of singlephase loads and third and multiples of third harmonics which may require the use of a
cross-sectional area greater than the phases must be taken into account (see § 8.2 neutral conductor heating).
.
the neutral conductor is protected against overcurrent by a fuse or a circuit-breaker trip
setting suitable to its cross-sectional area.
.
the cross-sectional area of the neutral conductor is at least equal to 16 mm² for copper or
25 mm² for aluminium.
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6.1.7.
Checking voltage drops
The voltage drop over a wiring system is calculated using the following formula:
L


∆V = b  ρ1 cos ϕ + λ L sin ϕ  × I B

 S
∆V
: voltage drop, in volts
= 1 for three - phase circuit

= 2 for single - phase circuit
b
: coefficient
ρ1
: conductor resistivity during normal service, i.e. 1.25 times that at 20 °C
ρ1 = 0.0225 Ω mm²/m for copper; ρ1 = 0.036 Ω mm²/m for aluminium
L
S
cos ϕ
IB
λ
: length of wiring system, in metres
: cross-sectional area of conductors, in mm²
: power factor, in the absence of specific indications we can take
cos ϕ = 0.8 ( sin ϕ = 0.6)
: maximum design current, in amps
: reactance per unit length of the conductors, in Ω/m
The values of λ in LV are:
-
0.08 × 10 −3 Ω / m for three-core cables
-
0.09 × 10 −3 Ω / m for single-core cables in a flat formation
-
0.15 × 10 −3 Ω / m for single-core cables spaced by d = 8 r
d
r
or triangular formation
: mean distance between conductor
: radius of conductor cores
The relative voltage drop is defined as:
∆V
Vn
for phase-to-neutral fed three-phase or single-phase circuits
∆V
Un
for phase-to-phase fed single-phase circuits (in this case,
∆V represents a phase-to-phase voltage drop)
Vn : nominal single-phase voltage
U n : nominal phase-to-phase voltage
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In accordance with IEC 364-5-52, in the absence of other considerations, it is recommended
that in practice the voltage between the origin of consumer's installation and the equipment
should not be greater than 4% of the nominal voltage of the installation.
n circuits feeding motors
The voltage drop is calculated by replacing the design current I B by the motor starting current.
Taking into account all the motors able to start simultaneously, the voltage drop must be lower
than 10% to ensure correct motor starting and not disturb the rest of the installation too much.
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6.1.8.
Maximum lengths of wiring systems for protection against indirect contact ( TN
and IT earthing system)
Standard IEC 364 specifies that the fault current for TN and IT earthing systems must be
cleared in a time compatible with the protection of persons.
This time is determined by a curve in relation to the prospective touch voltage; it is based on
the physiological effects of the electrical current on the human body. To simplify matters, using
this curve, it is possible to determine a maximum disconnecting time in relation to the nominal
voltage of the installation (see table 6-20 and 6-21).
Nominal a.c. voltage
Vn / U n
(Volts)
Disconnecting time
(seconds) (*)
non-distributed neutral
distributed neutral
120/240
0.8
5
230/400
0.4
0.8
400/690
0.2
0.4
580/1000
0.1
0.2
Table 6-20: maximum disconnecting times in the IT earthing system (second fault)
Nominal a.c. voltage Vn
(Volts) (**)
Disconnecting time
(seconds) (*)
120
0.8
230
0.4
277
0.4
400
0.2
> 400
0.1
Table 6-21: maximum disconnecting times in the TN earthing system
(*) these values are not valid in premises containing a bath or shower.
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Note 1: if the disconnecting time is more than the time t0 , but less than 5 seconds, protection is
allowed by IEC 364 (§ 413.1.3.5) in the following cases:
-
in distribution circuits when the protective conductor at the downstream end of the circuit is
directly connected to the main equipotential bonding.
-
in terminal circuits supplying stationary equipment only and having a protective conductor
that is connected to the main equipotential bonding and which is located in the area that is
influenced by the main equipotential bonding.
Note 2 : in the TT earthing system, protection is in general ensured by residual current devices which
are set to meet the following condition (see IEC 364, § 413.1.4.2):
RA IA ≤ 50 V
RA : resistance of the earth electrode of the exposed conductive parts
IA : rated residual current of the circuit-breaker
If selectivity is seen to be necessary, an operating time at the most equal to 1 second is
allowed in the distribution circuits without taking into account the touch voltage
Note 3 : in an IT earthing system, when the exposed conductive parts are earthed individually or in
groups, the conditions of the TT earthing system given in Note 2 must be met (see IEC 364,
§ 413.1.5.3).
n circuit-breaker protection
IEC 364 specifies that the magnetic tripping threshold of the circuit-breaker in TN and IT
earthing systems must be lower than the minimum short-circuit current. Furthermore, any
eventual circuit-breaker time delay must be shorter than the maximum disconnecting time
defined in tables 6-20 and 6-21.
For a given circuit-breaker and cross-sectional area, there is thus a maximum circuit length not
to be exceeded in order to comply with the requirements concerning the protection of persons
against indirect contact.
In the following part of the chapter, we will apply the conventional method for determining
maximum circuit lengths. This is more restrictive than the impedance method, but can be
applied by carrying out the calculations by hand.
In the conventional method, we neglect the influence of the reactance of the conductors for
cross-sectional areas smaller than 150 mm².
For large cross-sectional areas, we will take into account the influence of the reactance by dividing
Lmax by:
-
1.15 for a cross-sectional area of 150 mm²
1.30 for a cross-sectional area of 300 mm².
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Note:
for minimum short-circuit current calculations, refer to the "Industrial network protection guide"
§ 4.4.1.
o TN earthing system
The maximum length of a circuit in a TN earthing system is:
Lmax =
0.8 × Vn × S ph
ρ × (1 + m) × Im
Lmax : maximum length in m
Vn : single-phase voltage in volts
S ph : cross-sectional area of the phases in mm²
ρ
: resistivity of the conductors taken to be equal to 1.5 times that at 20 °C ( ρ
copper;
= 0.027 Ω mm 2 / m for
ρ = 0.043 Ω mm 2 / m for aluminium)
m
S ph : cross - sectional area of phases


=

 SPE : cross - sectional area of protective conductor 
Im
: circuit-breaker magnetic trip operating current
o IT earthing system
The maximum length of a circuit in an IT earthing system is:
- if the neutral conductor is not distributed:
Lmax =
0.8 × 3 × Vn × S ph
2 ρ × (1 + m) × I m
- if the neutral conductor is distributed:
Lmax =
= Sph
S1 : 
0.8 × Vn × S1
2 ρ × (1 + m) × Im
if the outgoing feeder considered does not have a neutral
= Sneutral if the outgoing feeder considered has a neutral
o TT earthing system
No condition on the wiring system length is specified since the protection of persons is
ensured by the residual current device.
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n fuse protection
Using the fuse fusing curve, we can determine the current I a ensuring fusion of the fuse in
the time t 0 specified in tables 6-20 and 6-21 (see fig. 6-10). We can then calculate the
maximum length of the wiring system in the same way as for the circuit-breaker replacing I m
by I a .
t
t0
I
Ia
Figure 6-10: fuse fusing curve
n application
In practice, checking the cross-sectional area of the wiring system in relation to the protection
of persons against indirect contact consists in making sure that the length of the wiring system
is less than Lmax for a given arrangement.
If the wiring system length is greater than Lmax , we can take the following measures:
- choose a circuit-breaker (or trip relay) with a lower magnetic threshold if the selectivity
requirements permit this
- install a residual current circuit-breaker for TNS
earthing system it is not possible to use a RCD)
and
IT
earthing system (in a TNC
- take larger phase and protective conductor cross-sectional areas meeting the maximum
length condition.
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6.1.9.
Checking the thermal withstand of conductors
When a short-circuit current flows through the conductors of a wiring system for a very short
time (up to five seconds), the heating of the conductors is considered to be adiabatic; this
means that the energy stored remains in the metal of the core and is not transmitted to the
insulating material. It is therefore necessary to check that the short-circuit thermal stress is
lower than the conductor thermal withstand:
2
tdis I sc
≤ k2 S2
tdis
: protective device disconnecting time in seconds
S
Isc
: cross-sectional area of conductors in mm²
: short-circuit current in A
The value of
k
(see table 6-22).
depends on the core metal and the type of insulating material
Insulating material
PVC
XLPE
Copper
115
135
Aluminium
74
87
Core
Table 6-22: value of factor k in accordance with IEC 364-4-43
If the disconnecting time is given, the cross-sectional area must comply with:
I
S ≥ sc × tdis
k
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n circuit-breaker protection
The check must be carried out for the maximum short-circuit current at the circuit-breaker
location.
The curves in manufacturers' catalogues give the maximum disconnecting time of the circuitbreaker. When circuit-breaker tripping is time delayed, the disconnecting time is taken to be
equal to the time delay.
To check the thermal withstand, the short-circuit current value must be calculated with a
resistivity ρ of the conductors taken to be equal to 1.5 times that at 20 °C :
-
ρ = 0.027 Ω mm 2 / m for copper
-
ρ = 0.043 Ω mm 2 / m for aluminium
o case of current-limiting circuit-breakers
On occurrence of a short circuit, current-limiting circuit-breakers only let a current below the
prospective fault current through (see fig. 6-11).
Isc
prospective peak I sc
prospective Isc
limited peak I sc
t
Figure 6-11: current limiting curve
The wiring system protected by this type of device is not therefore subjected to the
(prospective) calculated Isc thermal stress, but a much smaller stress defined by
manufacturers' limiting curves for each type of circuit-breaker.
2
The limiting curves give the thermal stress tdis I sc
expressed in A 2 × second .
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o example
We want to check the thermal withstand of a PVC-insulated 6 mm² copper conductor protected
by a Compact NS 80H-MA 380/415 V circuit-breaker fitted with an LR2-D33 63 thermal relay.
The themal withstand of the cable is:
k 2 S 2 = (115) 2 × 6 2 = 4.76 × 10 5 A2 × s .
The limiting curves in figure 6-12 give the maximum thermal stress of the circuit-breaker:
2 × 105 A2 × s .
The cable is thus protected up to the circuit-breaker breaking capacity.
The curves are in the table order
Figure 6-12: thermal stress limiting curves
for Compact NS 80H-MA-380/415V circuit-breakers
n fuse protection
The current causing the most stress is the minimum short-circuit current at the end of the
wiring system.
The fusing time t f of the fuse corresponding to I sc min must comply with the relation:
2
2 2
t f Isc
min ≤ k S
The method for calculating I sc min is given in paragraph 4.4.1 of the Protection guide.
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6.1.10.
Application example
n hypotheses
Let us consider the diagram in figure 6-13 the data of which is given below.
Since the installation feeds loads requiring good continuity of service the IT earthing system
without distributed neutral is chosen.
o W2 wiring system
This is made up of a PVC insulated copper three-core cable which is installed touching 3 other
multi-core cables on perforated trays in an ambient temperature of 40°C. It is protected by
fuses. It feeds a load having the following characteristics:
- active power P = 15 kW
- efficiency η = 0.89
-
cos ϕ = 0.85
- utilisation factor b = 0.9 .
o W1 wiring system
This is made up of 3 XLPE-insulated copper single-core cables in a triangular formation. The
cables are buried alone, without any extra mechanical protection, in soil which has a thermal
resistivity of 0.85 K.m/W and a temperature of 35 °C. They are protected by a circuit-breaker.
The wiring system feeds load L1 and 3 other outgoing feeders the I B current values of which
are given in figure 6-13.
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250 kVA
U sc = 4 %
unearthed neutral
400 V
W1
L1 = 100 m
cos
= 0.8
400 V
25 A
50 A
40 A
IB
W2
L2 = 15 m
R1
Figure 6-13: diagram of the installation
n determining the maximum design current
o W2 wiring system
-
P = 15 kW
1
= 1.32
η cos ϕ
- the utilisation factor b = 0.9
- the factor a =
- for a single load the coincidence factor is c = 1
- no extension is planned, thus d = 1
- for a 400 V three-phase network, the power conversion factor in current is e = 1.4 .
We then have:
I B = P × a × b × c × d × e = 15 × 1.32 × 0.9 × 1 × 1.4 = 24.9 A
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o W1 wiring system
The maximum design current of the W1 wiring system is obtained by calculating the sum of
currents ( I B ) of all the outgoing feeders fed by W1 and by applying a coincidence factor
estimated at 0.8 (see table 6-2):
I B = (25 + 50 + 40 + 24.9) × 0.8 = 115.9 A
n correction factors
o W2 wiring system
Table 6-3 gives the installation method N° 13 and the selection letter E .
The correction factors to be applied are:
-
ambient temperature
cable group
(see table 6-6)
(see tables 6-9 et 6-10)
: f1 = 0.87
: f 4 = 0.77 and f 5 = 1
The overall correction factor is:
f = 0.87 × 0.77 × 1 = 0.67
o W1 wiring system
Table 6-3 gives the installation method N° 62 and the selection letter D .
-
ground temperature
soil thermal resistivity
cable group
(see table 6-7)
(see table 6-8)
(see table 6-15)
: f 2 = 0.89
: f 3 = 1.05
: f10 = 1
The overall correction factor is:
f = 0.89 × 1.05 × 1 = 0.935
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n determining the cross-sectional area and choosing the protective device
o W2 wiring system
I B = 24.9 A
f = 0.67
The fuse nominal current must comply with the condition I n ≥ I B .
The fuse with a rating of I n = 25 A is chosen.
For 10 A < I n ≤ 25 A , the current I z of the wiring system protected by this fuse is:
I z = k3 In = 1.21 In = 30.3 A
The equivalent current that the wiring system must be able to carry in standard installation
I
. A
conditions is: I 'z = z = 451
f
Table 6-16 (selection letter E , PVC3, copper) gives a minimum cross-sectional area of
S = 10 mm2 which has a current-carrying capacity of I 0 = 60 A .
o W1 wiring system
I B = 115.9 A
f = 0.935
For an adjustable circuit-breaker, the setting current must comply with the condition Iset ≥ I B ;
I set = 120 A is chosen.
The current I z of the wiring system protected by this setting is:
I z = I n = 120 A
The equivalent current that the wiring system must be able to carry in standard installation
I
conditions is: I 'z = z = 128.3 A
f
Table 6-17 (selection letter D , XLPE3, copper) gives a minimum cross-sectional area of
S = 25 mm2 which has a current-carrying capacity of I 0 = 144 A .
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n maximum length of the wiring system
o W2 wiring system
For
S ph = 10 mm² , we have S PE = S ph = 10 mm²
whence
m=
S ph
S PE
=1
Table 6-20 gives a maximum disconnecting time of t = 0.4 s for a network with non-distributed neutral.
The time-current characteristic for a 25 A rated fuse gives us a current of
I a = 200 A for a
disconnecting time of 0.4 s.
The neutral is not distributed and we thus have:
Lmax =
0.8 × 3 × Vn × S ph 0.8 × 3 × 230 × 10
= 147.5 m
=
2 ρ (1 + m ) Ia
2 × 0.027 × 2 × 200
The length of the W2 wiring system (15 m) is far smaller than
Lmax and the protection of
persons against indirect contact is thus ensured.
o W1 wiring system
For 16 mm² < S ≤ 35 mm² , we have S PE = 16 mm²
whence
m=
S ph
SPE
=
25
= 1.56
16
The circuit-breaker chosen is a Compact NS 125E with an STR 22SE trip relay having a
magnetic tripping threshold set at I m = 1 250 A because of the selectivity.
The neutral is not distributed and we thus have:
Lmax =
0.8 × 3 × Vn × S ph
2 ρ (1 + m) I m
=
0.8 × 3 × 230 × 25
= 46.1 m
2 × 0.027 × 2.56 × 1 250
The length of the W1 wiring system (100 m) is greater than Lmax .
By taking cross-sectional areas greater, i.e. S ph = 35 mm² and S PE = 35 mm² ( m = 1) , we find
Lmax = 82.6 m < 100 m ; which is not sufficient.
So as not to oversize the conductors, it is decided that the outgoing feeder should be fitted
with a residual current device which ensures the protection of persons against indirect contact.
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n checking the voltage drop
o W2 wiring system
S = 10 mm² , L = 15 m , I B = 24.9 A
The cable is three-core and we thus have λ = 0.08 × 10 −3 Ω / m .
The power factor is cos ϕ = 0.85 , whence sin ϕ = 0.53 .
For a three-phase circuit b = 1 .
For copper ρ1 = 0.0225 Ω mm 2 / m .
We deduce from this that
15


∆ V =  0.0225 ×
× 0.85 + 0.08 × 10 −3 × 15 × 0.53 × 24.9


10
∆ V = 0.73 V
whence
∆ V 0.73
=
= 0.3 %
Vn
230
The total voltage drop is 4.2 % (the voltage drop in the W1 wiring system is 3.9 %, see below).
o W1 wiring system
S = 25 mm² , L = 100 m , I B = 115.9 A
The 3 single-core cables are in a flat formation and we thus have:
λ = 0.09 × 10 −3 Ω / m
The overall power factor of the installation is cos ϕ = 0.8 , whence sin ϕ = 0.6 .
100
We deduce from this that
∆ V =  0.0225 ×
× 0.8 + 0.09 × 10 −3 × 100 × 0.6 × 115.9


25
∆ V = 8.97 V
whence
∆ V 8.97
=
= 3.9 %
Vn
230
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n checking the thermal stress
o W2 wiring system
For fuse protection, the current to be taken into account is the minimum short-circuit current at
the end of the wiring system. For the IT earthing system, this is the short-circuit current for a
double phase-earth fault.
By applying the conventional method (see § 4.4.1.2 of the Protection guide), we can calculate:
Isc min =
3 × Vn × 0.8
3 × 230 × 0.8
=
= 1.97 kA
 1
1  2 × 15 × 0.027  1 + 1 

+
2 L 2 ρ 
 10 10 
 S ph SPE 
The time-current characteristic of the 25 A rated fuse gives us a fusing time of t f = 5 ms for a
current of 1.97 kA.
The maximum thermal stress is thus:
(
2
3
Isc
min × t = 1.97 × 10
)
2
× 5 × 10 −3 = 19.4 × 10 3 A2 × s
2
The permitted cable thermal withstand is: k 2 S 2 = (115) × 102 = 1322 × 103 A2 × s .
The cross-sectional area of S = 10 mm2 is thus largely able to withstand to the fuse thermal
stress.
o W1 wiring system
The maximum short-circuit current of the circuit-breaker (neglecting the connection linking the
circuit-breaker to the transformer) is:
Isc =
Sn
1
250 × 103 100
×
=
×
= 9.02 kA
4
3 Un Usc
3 × 400
We assume that the circuit-breaker trip relay is delayed by 0.1 second, the maximum shortcircuit thermal stress is then:
(
2
Isc
t = 9.02 × 10 3
)
2
× 0.1 = 8.14 × 10 6 A2 × s
The permitted cable thermal withstand is: k 2 × S 2 = 1432 × 252 = 12.78 × 10 6 A2 × s
The cross-sectional area of
S = 25 mm2 is thus largely able to withstand the circuit-breaker
thermal stress.
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n conclusion
The cross-sectional areas to be chosen are:
- W1 wiring system: 3 × 35 mm 2 + 1 × 16 mm 2 copper
- W2 wiring system: 3 × 10 mm 2 + 1 × 10 mm 2 copper
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6.2.
Determining conductor cross-sectional areas in medium voltage
6.2.1.
Method principle
The method for determining the cross-sectional area of conductors in medium voltage consists
in:
- determining the maximum design current I B of the loads to be supplied
- determining the cross-sectional area S1 complying with the heating of the cable core under
normal operating conditions, which may be continuous or discontinuous. To do this, it is
necessary to know:
.
the actual installation conditions of the wiring system and consequently the overall
correction factor f
.
the current-carrying capacities of the different types of cable in standard installation
conditions.
- determining the cross-sectional area S 2 required for the thermal withstand of the cable in
the event of a three-phase short circuit
- determining the cross-sectional area
screen in the event of an earth fault
S3 required for the thermal withstand of the cable
- possibly checking the voltage drop in the wiring system for the chosen cross-sectional
area S. The technical cross-sectional area S to be selected is the maximum value among
cross-sectional areas S1 , S 2 and S3 .
- possibly calculating and choosing the economical cross-sectional area.
6.2.2.
Determining the maximum design current
The maximum design current I B is determined on the basis of the sum of powers of the loads
fed, applying if necessary utilisation and coincidence coefficients (see § 6.1.2.).
In medium voltage, a wiring system most often feeds a single load (transformer, motor,
furnace, steam generator), in this case I B is taken to be equal to the rated current of the
device.
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6.2.3.
Current-carrying capacities in wiring systems
n general rules
The current-carrying capacity is the maximum current that a wiring system can continuously
carry without this affecting its life span.
The current-carrying capacities of cables are given in standards or by manufacturers for
standard installation conditions.
To determine the current-carrying capacity of a wiring system in actual installation conditions,
the following must be carried out:
- using table 6-23, define the installation method, its associated table column number and
correction factors to be applied
- using the installation conditions, determine the correction factor values which must be
applied (see tables 6-24 to 6-28)
- calculate the overall correction factor f equal to the product of the correction factors
- using table 6-29 for impregnated paper-insulated cables and tables 6-30 to 6-34 for
synthetically-insulated cables, determine the maximum current that the wiring system can
carry in standard conditions ( f0 to f6 = 1)
- calculate the maximum current-carrying capacity of the wiring system in relation to its
installation conditions: I a = f I 0 .
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n installation methods
Table 6-23 gives, for each installation method, the current-carrying capacity table column to be
used for choosing the cross-sectional area of the conductors (see tables 6-29 to 6-34).
Factor
f 0 corresponds to the installation method; factors
f1 to
f 6 are explained below
Table
Correction factors
to be applied
f0
(see tables 6-24 to 6-28).
Installation methods
Example
column
A Conduits on wall
(3)
0.90
f1
f5
B Flush mounted
conduits
(3)
0.90
f1
f5
F Installed on cable
trays
(3)
1
f1
f5
G Installed on brackets
or cable ladders
(3)
1
f1
f6
H Troughs (enclosed)
(3)
0.90
f1
f5
(3)
1
f1
f6
(3)
0.80
f1
f5
J Ducts (open troughs)
L1 Conduits in open or
ventilated channels
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Installation method
Example
L3 Directly installed in
open or ventilated
channels
enclosed channels
channels filled with
sand
Table
Correction factors
to be applied
column
f0
(3)
0.90
f1
--
f5
(3)
0.80
f1
--
f5
(3)
0.80
f1
--
f5
(3)
0.90
f1
--
f5
(3)
0.90
f1
--
f5
1
f2
f3
f4
1
f2
f3
f4
N Troughs (in masonry)
P Manufactured blocks
S1 Directly buried
(armoured cables)
S2 Buried with
mechanical protection
P
: steady-state operating conditions
D
: discontinuous operating conditions
P D
____
___
(1) (2)
(1) (2)
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Installation method
Example
Table
column
f0
Correction factors
to be applied
P D
____
___
(1) (2)
S3 Buried in sleeves
0.8
f2
f3
f4
S4 Cables installed in
trefoil formation in a
prefabricated channel,
buried directly in the
ground, possibly with
extra backfill
(1) (2)
0.8
f2
f3
f4
S5 Single-core cables
installed in individual
channels, buried
directly in the
extra backfill
(1) (2)
0.8
f2
f3
f4
(1) (2)
0.8
f2
f3
f4
(3)
1,1
f1
--
--
Single-core cables in a
flat formation spaced
out in a prefabricated
channel, buried
directly in the
extra backfill
V Overhead lines
P
: steady-state operating conditions
D
: discontinuous operating conditions
Table 6-23: installation methods
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n correction factors for ambient temperatures other than 30 °C (cables installed in air): f1
Temperature
Type of insulating material
°C
PVC
PE
XLPE
EPR
10
1.22
1.15
15
1.17
1.12
20
1.12
1.08
25
1.06
1.04
30
1.00
1.00
35
0.94
0.96
40
0.87
0.91
45
0.79
0.87
50
0.71
0.82
55
0.61
0.76
Table 6-24: correction factors for ambient temperatures other than 30 °C (cables installed in air)
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n correction factors for ground temperatures other than 20 °C (buried cables): f 2
Temperature
Type of insulating material
°C
PVC
PE
XLPE
EPR
0
1.18
1.13
5
1.14
1.10
10
1.10
1.07
15
1.05
1.04
20
1.00
1.00
25
0.95
0.96
30
0.89
0.93
35
0.84
0.89
40
0.77
0.85
45
0.71
0.80
50
0.63
0.76
60
0.45
0.65
65
-
0.60
70
-
0.53
75
-
0.46
80
-
0.38
Table 6-25: correction factors for ground temperatures other than 20 °C (buried cables)
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n correction factors for soil thermal resistivities other than 1 K.m/W (buried cables): f 3
Soil thermal
resistivity
(K.m/W)
Humidity
0.5
Assembly of
three singlecore cables
Three-core
cables
Very moist soil
1.25
1.20
0.7
Moist soil
1.14
1.10
0.85
Normal soil
Clay
1.06
1.05
Dry soil
and
1.00
1.00
Chalk
0.93
0.95
Ash
0.85
0.88
and
0.75
0.79
Clinker
0.68
0.72
0.62
0.68
1
Type of soil
1.2
Sand
1.5
Very dry soil
2
2.5
3
Tableau 6-26: correction factors for soil thermal resistivities other than 1 K.m/W (buried cables)
n correction factors for a group of several wiring systems (buried cables): f 4
Number of
circuits
Distance between cables "a"
Zero (cables
touching)
One cable
diameter
0.125 m
0.25 m
0.5 m
2
0.75
0.80
0.85
0.90
0.90
3
0.65
0.70
0.75
0.80
0.85
4
0.60
0.60
0.70
0.75
0.80
5
0.55
0.55
0.65
0.70
0.80
6
0.50
0.55
0.60
0.70
0.80
Determination of the distance "a" in the case of single-core cables installed in a flat or trefloid
formation and three-core cables.
single-core cables
three-core cables
a
a
a
Table 6-27: correction factors for a group of several wiring systems (buried cables )
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n correction factors for a group of several circuits or several cables
(cables installed in air and away from direct sunlight): f 5 , f 6
Installation
method
Arrangement
f5
On unperforated horizontal trays.........................
f6
On perforated horizontal trays
or on brackets ....................................................
Number of circuits or
multi-core cables
2
3
4
6
>9
0.85
0.80
0.75
0.70
0.70
0.90
0.80
0.80
0.75
0.75
Table 6-28: correction factors for a group of several circuits or several cables (cables installed in air and
away from direct sunlight)
n current-carrying capacities of cables in standard installation conditions
( f0 to f6 = 1)
References (1), (2) and (3) of tables 6-29 to 6-34 correspond to the column number given in
table 6-23.
o impregnated paper-insulated cables
Impregnated paper-insulated cables have stopped being manufactured for several years.
However, for calculation purposes for existing installations, the current-carrying capacities may
be calculated to an approximate value of ± 5% using the following formula:
I = 10 B × S A
I
: current-carrying capacity, in A
S
: nominal cross-sectional area of the cable, in mm²
A and B : are coefficients given for each type of cable (see table 6-29)
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Wiring systems
Columns
Copper
Aluminium
A
B
A
B
Three-core
(1)
0.540
1.446
0.549
1.321
collectively
(2)
0.543
1.492
0.544
1.386
screened cable
(3)
0.588
1.371
0.598
1.293
3 single-core
(1)
0.556
1.269
0.571
1.130
cables
(2)
0.567
1.286
0.573
1.179
(3)
0.587
1.196
0.605
1.064
Three-core individually
(1)
0.581
1.215
0.594
1.089
screened cables
(2)
0.573
1.264
0.578
1.155
(3)
0.600
1.117
0.608
1.004
Table 6-29: values of coefficients A and B for impregnated paper-insulated cables
o synthetically-insulated cables
The detailed calculation method for current-carrying capacities of cables under steady-state
operating conditions is given in IEC publication 287.
The current-carrying capacities are given in tables 6-30 to 6-34, according to the type of
conductor, the type of insulating material and the rated voltage.
The rated voltage for which a cable is designed is expressed by a set of three values, in kV, as
U 0 / U (Um) , where:
-
U0 :
U :
Um :
voltage between the conductor core and a reference potential (screen or earth)
voltage between the cores of two phase conductors
maximum voltage which may occur between the network phases in normal operating
conditions
The expression of the rated voltage differs depending on whether the cable is an individually
screened type or not (see fig. 2.2.a and 2.2.b). For an individually screened cable, U 0 is
different from U , both values being generally in the ratio of
3 .
However, due to the way it is made, a collectively screened cable has an equivalent insulation
level between two phases and between one phase and the screen. This results in U 0 and U
having identical values.
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PVC-insulated
(*)
Nominal
crosssectional
area (mm²)*
EPR or XLPE-insulated
(1)
(2)
(3)
Copper
(1)
(2)
(3)
72
78
62
10
86
94
78
94
100
81
16
110
120
100
120
130
105
25
145
155
130
145
160
130
35
170
190
165
185
205
165
50
215
240
205
225
250
205
70
260
295
255
270
300
250
95
315
355
310
310
345
290
120
360
405
360
345
390
330
150
405
455
410
385
430
370
185
450
505
460
445
500
440
240
525
590
550
(1)
(2)
(3)
Aluminium
(1)
(2)
(3)
56
61
48
10
67
73
60
72
79
62
16
86
94
79
94
100
82
25
110
120
105
115
125
100
35
135
145
125
145
160
130
50
165
185
160
175
195
160
70
205
230
195
210
235
195
95
245
275
240
240
270
225
120
280
315
280
270
300
255
150
315
355
320
300
335
285
185
350
395
360
350
390
345
240
410
460
430
Above 50 mm², the values are calculated for sector conductors
Table 6-30: current-carrying capacities in three-core collectively screened cables having a rated voltage
lower than or equal to 6/6 (7.2) kV
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Copper
10
16
25
35
50
70
95
120
150
185
240
300
400
500
630
800
1 000
1 200
1 400
1 600
Aluminium
10
16
25
35
50
70
95
120
150
185
240
300
400
500
630
800
1 000
1 200
1 400
1 600
(*)
PE-insulated*
PVC-insulated
Nominal
crosssectional
area (mm²)
(1)
80
105
135
160
190
235
285
320
360
410
475
540
610
680
770
850
930
980
1 030
1 080
(1)
62
80
105
125
150
180
220
250
280
320
370
420
480
540
620
700
780
840
890
940
(2)
89
115
150
180
215
265
320
365
410
470
540
610
700
780
880
980
1 070
1 130
1 190
1 250
(2)
69
89
115
140
170
205
250
285
320
365
425
485
550
630
720
810
900
970
1 030
1 080
(3)
71
95
125
150
180
230
280
320
370
425
500
580
670
760
870
990
1 110
1 210
1 290
1 360
(3)
55
73
96
115
140
175
215
250
285
330
390
455
530
610
710
820
940
1 030
1 110
1 180
(1)
86
110
140
170
200
245
295
335
375
425
490
550
600
700
790
870
950
1 000
1 050
1 100
(1)
67
86
110
130
160
190
230
260
290
330
385
435
495
560
640
720
800
860
910
950
(2)
97
125
160
195
230
285
340
385
435
490
570
640
690
810
920
1 010
1 100
1 160
1 220
1 280
(2)
76
97
125
150
180
220
265
300
335
380
445
500
580
650
750
840
930
1 000
1 060
1 110
(3)
76
100
130
160
190
240
295
340
385
445
530
600
700
790
920
1 040
1 160
1 260
1 350
1 420
(1)
99
125
165
195
230
285
340
385
430
485
560
630
720
800
910
1 000
1 100
1 160
1 220
1 280
(3)
59
78
100
125
150
185
230
265
300
345
410
470
550
640
750
860
980
1 080
1 160
1 230
(1)
77
98
125
150
180
220
260
300
335
380
440
500
570
640
740
830
920
990
1 050
1 100
(2)
(3)
110
145
185
225
265
325
390
445
500
560
650
730
840
940
1 060
1 170
1 270
1 350
1 420
1 480
(2)
93
120
160
200
235
295
360
420
475
550
650
740
860
990
1 140
1 300
1 450
1 570
1 680
1 770
(3)
87
110
145
175
205
250
300
345
385
440
510
580
660
750
860
970
1 070
1 150
1 230
1 290
72
95
125
150
185
230
280
325
370
425
510
580
680
790
920
1 070
1 220
1 340
1 450
1 530
For cables having high density polythene insulation, the values are to be multiplied by:
1.05 for columns (1) and (2)
1.06 for column (3)
Table 6-31: current-carrying capacities in cables made up of three single-core cables having a rated
voltage lower than or equal to 6/10 (12) kV
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PE-insulated*
Nominal
crosssectional
area (mm²)
(1)
(2)
(3)*
110
140
170
200
250
295
335
375
425
490
550
630
700
790
870
960
1 010
1 070
1 110
125
160
195
230
280
335
385
430
490
560
640
720
810
920
1 010
1 100
1 170
1 240
1 290
105
135
165
200
250
300
350
395
455
530
610
710
810
930
1 050
1 180
1 270
1 360
1 430
(1)
(2)
(3)
86
110
130
155
190
230
260
290
330
385
435
495
560
640
720
800
860
920
960
96
125
150
180
220
260
300
335
380
445
500
570
650
740
830
930
1 000
1 060
1 110
81
105
130
155
190
235
270
305
355
420
480
560
650
750
860
990
1 090
1 170
1 240
(*)
Copper
16
25
35
50
70
95
120
150
185
240
300
400
500
630
800
1 000
1 200
1 400
1 600
Aluminium
16
25
35
50
70
95
120
150
185
240
300
400
500
630
800
1 000
1 200
1 400
1 600
(1)
(2)
(3)
125
165
195
230
280
335
385
430
490
560
640
720
810
910
1 010
1 110
1 180
1 240
1 290
140
185
220
260
320
385
440
495
560
650
730
830
940
1 060
1 170
1 280
1 360
1 440
1 500
130
170
200
245
305
375
425
485
560
660
750
870
1 000
1 150
1 300
1 470
1 590
1 700
1 790
(1)
(2)
(3)
98
125
150
180
220
260
300
335
380
440
500
570
640
740
830
930
1 000
1 060
1 110
110
140
170
205
250
300
340
385
435
510
570
660
740
850
960
1 070
1 160
1 230
1 290
99
130
160
190
235
290
330
375
430
510
590
680
790
930
1 060
1 230
1 350
1 450
1 540
1.06 for column (3)
Table 6-32: current-carrying capacities in cables made up of three single-core cables having a rated
voltage greater than 6/6 (7.2) kV and lower than or equal to 18/30 (36) kV
T&D
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568
Copper
PE-insulated*
PVC-insulated
Nominal
crosssectional
area (mm²)
(1)
(2)
(3)
(1)
(2)
(3)
(1)
(2)
(3)
10
80
87
71
85
94
75
97
110
92
16
100
115
90
110
120
98
125
140
120
25
130
145
120
140
155
125
160
180
155
35
160
175
145
165
190
155
190
215
190
50
185
205
175
195
220
185
225
250
225
70
230
255
215
240
270
230
275
310
280
95
275
305
260
285
320
275
330
370
340
120
310
345
300
325
365
315
370
420
385
150
345
385
340
365
415
365
420
475
445
185
390
435
385
410
465
410
470
535
510
240
450
500
450
475
530
485
540
610
590
300
500
560
520
530
605
560
610
690
680
Aluminium
(1)
(2)
(3)
(1)
(2)
(3)
(1)
(2)
(3)
(*)
10
62
68
55
66
73
58
75
84
71
16
79
87
71
84
94
76
96
110
92
25
100
115
93
110
120
99
125
140
120
35
120
135
115
130
145
120
150
165
145
50
145
160
135
150
170
140
175
195
175
70
180
195
165
185
210
175
215
240
215
95
210
235
205
220
250
215
255
285
260
120
240
270
235
250
285
245
290
325
300
150
270
300
265
285
325
280
325
370
345
185
305
340
300
320
360
320
365
415
395
240
350
390
355
370
420
380
425
480
465
300
395
440
405
420
475
435
480
540
530
1.06 for column (3)
Table 6-33: current-carrying capacities in three-core individually screened cables having a rated voltage
lower than or equal to 6/10 (12) kV
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Nominal crosssectional area (mm²)
Copper
(1)
(2)
(3)
16
125
140
125
25
160
175
160
35
190
210
195
50
225
250
230
70
270
305
280
95
330
370
345
120
370
420
395
150
415
465
450
185
465
525
510
240
540
610
600
Aluminium
(1)
(2)
(3)
16
96
105
95
25
125
135
125
35
145
165
150
50
175
195
175
70
210
235
220
95
255
285
265
120
290
325
305
150
320
360
345
185
360
410
395
240
420
475
470
Table 6-34: current-carrying capacities in three-core individually screened cables having a rated voltage
greater than 6/6 (7,2) kV and lower than or equal to 18/30 (36) kV
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6.2.4.
Thermal withstand of conductors in the event of a short circuit and
determination of the cross-sectional area S 2
The thermal withstand of live conductors must be checked for the maximum short-circuit
current at the origin of the cable. It is calculated using the impedance method taking into
account the participation of all the network elements (motors, generators, etc., see Protection
guide § 4.2).
In the case of an installation with an internal generator set, the thermal withstand is
established on the basis of the short-circuit current during the transient period, this
approximately corresponding to the short-circuit clearance time (see Protection guide § 4.1.2).
For a short-circuit time less than 5 seconds, cable heating is considered to be adiabatic; this
means that the energy stored stays in the core and is not transmitted to the insulating material.
The thermal calculations are then simplified. They are given below.
Note:
to check the thermal withstand of protective and equipotential bonding conductors, the earth
fault current must be taken into account (see § 4.2.2 of the Protection guide)
n general method
The heating calculation results are shown by the curves in figure 6-14. They give the current
density withstands δ 0 in different types of cable for a short-circuit time of one second, in
relation to the cable temperature before the short circuit.
The minimum conductor cross-sectional area complying with heating in the case of a short
circuit is determined by the fomula:
I
S = sc
δ
Isc : maximum short-circuit current, in A
δ : current density withstand, in A / mm²
for a short-circuit time other than one second, we have:
δ=
t
δ0
t
: short-circuit time
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Figure 6-14: short circuit in the core
T&D
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n simplified method
This assumes that the cable temperature before the short circuit is equal to the temperature
allowed in steady-state operating conditions.
In this case, the conductor cross-sectional area must meet the following condition:
I
S ≥ sc
k
t
Isc : maximum short-circuit current
t
: short-circuit time
k : coefficient the value of which is given in table 6-35
For protective conductors, the current to be taken into account is the earth fault current I f .
Insulating material
PVC
PE
XLPE
EPR
115
143
74
94
Live conductors
- in copper
- in aluminium
Protective conductors
a
b
a
b
143
115
176
143
- in aluminium
95
75
116
94
- in steel
52
_
64
_
- in copper
a
b
protective conductors not incorporated in cables
protective conductors incorporated in cables
Table 6-35: coefficient k values
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6.2.5.
Short-time withstand currents in cable screens with extruded synthetic
insulation (determination of S3 )
In the event of a phase-to-screen short circuit, the thermal withstand resulting from the
passage of the fault current I f for a time t , must not exceed the thermal withstand of the
cable screen.
If
is the earth fault current and the method for determining its value is
described in the Protection guide, paragraph 4-2.
The calculation of the overcurrent permitted in the cable screens depends on what the screen
is made of and the type of cable.
In the absence of precise indications, the values of tables 6-37, 6-38 and 6-39 can be used.
These values correspond to a screen made up of a copper band 0.1 mm thick wrapped around
the insulating material with an overlap of 15 %.
Table 6-36 gives, for each type of insulating material, the temperatures during normal service
and at the end of overcurrents used for calculating cable screen heating.
Type of insulating
material
Temperature on the screen during
service (°C)
Final temperature following
overcurrent
(°C)
XLPE
70
250
EPR
70
250
PE
60
150
PVC
60
160
Table 6-36: temperature conditions used for the calculation
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o overcurrent values permitted in cable screens with extruded synthetic insulation
See tables 6-37, 6-38 and 6-39.
Rated voltage
Short-circuit time
6/10 (12) kV
0.5 s
1s
8.7/15 (17.5) kV
2s
0.5 s
1s
2s
12/20 (24) kV
0.5 s
1s
2s
18/30 (36) kV
0.5 s
1s
2s
Conductor cross-sectional
area in mm²
16
1 100
900
650 1 350 1 000
800 1 800 1 400 1 100
25
1 200
950
700 1 400 1 050
800 1 800 1 400 1 100
35
1 400 1 000
50
1 600 1 150 1 000 1 750 1 350 1 050 1 950 1 450 1 150 2 500 1 950 1 550
70
1 750 1 250 1 050 1 900 1 450 1 150 2 100 1 600 1 250 2 700 2 050 1 650
95
1 850 1 350 1 100 2 050 1 550 1 200 2 200 1 700 1 300 2 800 2 150 1 700
120
1 900 1 400 1 150 2 150 1 650 1 300 2 500 1 950 1 550 3 100 2 400 1 900
150
2 150 1 650 1 300 2 400 1 850 1 500 2 600 2 000 1 600 3 150 2 450 1 950
185
2 400 1 850 1 450 2 600 2 000 1 600 2 750 2 150 1 700 3 350 2 600 2 100
240
2 700 2 050 1 650 2 800 2 150 1 700 3 100 2 400 1 950 3 600 2 750 2 200
300
2 800 2 150 1 750 3 150 2 450 1 950 3 300 2 550 2 050 3 800 2 950 2 350
400
3 050 2 350 1 800 3 450 2 650 2 150 3 650 2 800 2 250 4 200 3 300 2 650
500
3 400 2 550 1 950 3 800 2 950 2 350 4 100 3 200 2 550 4 550 3 550 2 850
630
3 750 3 000 2 300 4 250 3 300 2 650 4 450 3 450 2 800 4 950 3 850 3 100
800
4 400 3 400 2 600 4 650 3 600 2 900 4 850 3 750 3 000 5 300 4 150 3 300
1 000
5 100 3 900 3 050 5 200 4 050 3 250 5 350 4 200 3 350 5 850 4 550 3 650
1 200
5 350 4 100 3 300 5 450 4 250 3 400 5 650 4 400 3 550 6 150 4 800 3 850
1 400
5 600 4 400 3 550 5 900 4 550 3 650 6 050 4 700 3 800 6 550 5 100 4 100
1 600
6 000 4 700 3 800 6 200 4 850 3 900 6 400 5 000 4 000 6 900 5 350 4 300
900 1 650 1 250 1 000 1 850 1 400 1 100
Table 6-37: single-core or three-core individually screened cables with XLPE or EPR insulation short-circuit current permitted in the screen (A)
T&D
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Rated voltage
Short-circuit time
6/10 (12) kV
0.5 s
1s
8.7/15 (17.5) kV
2s
0.5 s
1s
2s
12/20 (24) kV
0.5 s
1s
2s
18/30 (36) kV
0.5 s
1s
2s
Conductor cross-sectional
area in mm²
16
800
650
490 1 000
740
560 1 200
870
660
25
900
700
510 1 000
750
570 1 200
870
660
35
1 000
750
540 1 100
800
600 1 200
880
660
50
1 100
800
580 1 150
840
640 1 250 1 000
770 1 750 1 300
70
1 300
920
700 1 350
990
760 1 450 1 100
820 1 750 1 300 1 000
95
1 350 1 000
750 1 450 1 050
820 1 550 1 150
880 2 050 1 550 1 200
120
1 450 1 050
800 1 500 1 150
860 1 650 1 200
930 2 150 1 650 1 230
150
1 550 1 100
840 1 600 1 200
910 1 700 1 300 1 000 2 250 1 700 1 300
185
1 650 1 150
900 1 700 1 250
970 2 000 1 500 1 200 2 350 1 800 1 400
240
1 800 1 450 1 100 2 000 1 550 1 200 2 150 1 650 1 250 2 650 2 050 1 600
300
2 000 1 550 1 200 2 150 1 650 1 300 2 300 1 750 1 350 2 800 2 150 1 700
400
2 300 1 750 1 400 2 600 2 000 1 550 2 650 2 050 1 600 3 000 2 300 1 800
500
2 550 1 900 1 500 2 900 2 200 1 750 3 050 2 350 1 850 3 400 2 600 2 050
630
2 750 2 050 1 550 3 000 2 300 1 800 3 150 2 400 1 900 3 500 2 650 2 050
800
3 000 2 250 1 700 3 300 2 500 2 000 3 450 2 600 2 100 3 700 2 800 2 200
1 000
3 300 2 400 1 800 3 500 2 700 2 100 3 650 2 800 2 200 3 950 3 000 2 400
1 200
3 550 2 550 1 900 3 700 2 850 2 200 3 850 2 950 2 300 4 200 3 200 2 550
1 400
3 650 2 750 2 000 3 900 3 000 2 350 4 050 3 100 2 450 4 350 3 350 2 650
1 600
3 750 2 850 2 100 4 000 3 100 2 400 4 150 3 200 2 500 4 500 3 400 2 700
990
Table 6-38: single-core or three-core individually screened cables with PE insulation short-circuit current permitted in the screen (A)
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Conductor crosssectional area
Short-cicuit time
mm²
0.5 s
1s
2s
10
1 550
1 200
980
16
1 700
1 300
1 050
25
1 950
1 450
1 200
35
2 050
1 550
1 250
50
2 150
1 600
1 300
70
2 300
1 700
1 400
95
2 550
1 900
1 550
120
2 750
2 100
1 650
150
2 900
2 200
1 750
185
3 350
2 450
2 050
240
3 500
2 650
2 200
Table 6-39: PVC-insulated three-core collectively screened cables with a rated voltage of 6/6 (7.2 kV) short-circuit current permitted in the screen (A)
o example
Let us consider a PE-insulated single-core cable in a 10 kV network having an earth fault
current I f limited to 1 000 A.
According to table 6-38, the minimum cross-sectional area of the conductor depends on the
short-circuit time:
- for
- for
- for
t = 0.5 s
t= 1s
t= 2s
,
,
,
S min = 35 mm²
S min = 95 mm²
S min = 240 mm² .
The cross-sectional area S3 is selected in relation to I f and the short-circuit time, which is
taken to be equal to the longest time needed to clear the fault (e.g., the back-up protection
time delay).
T&D
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6.2.6.
Checking voltage drops
Voltage drops in medium voltage cables in industrial networks are in general negligible.
However, it seems useful to give the calculation method able to be applied notably for very
long wiring systems.
For a three-phase circuit, the voltage drop (single-phase voltage) is calculated by the fomula:
L


∆V =  ρ1 cos ϕ + λ L sin ϕ  I B
 S

ρ1
: conductor resistivity during normal service, i.e. 1.25 times that at 20 °C
ρ1 = 0.0225 Ω mm 2 / m for copper; ρ1 = 0.036 Ω mm 2 / m for aluminium
L
S
cos ϕ
: length of wiring system, in metres
: conductor cross-sectional areas, in mm²
IB
λ
: maximum design current in
: power factor; in the absence of precise indications, we may take
cos ϕ = 0.8 (sin ϕ = 0.6)
A
: reactance per unit length of the wiring system, in
Ω/m .
The values of λ in MV are:
-
0.08 × 10 −3 Ω / m for three-core cables
-
0.15 × 10 −3 Ω / m for single-core cables
We define the relative voltage drop as:
∆V
Vn
Vn
: nominal single-phase voltage
T&D
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6.2.7.
Practical determination of the minimum cross-sectional area of a medium
voltage cable (see fig. 6-15)
cable installation
conditions
determination of
maximum design
current I B
determination of the
cable column and
overall correction
factor (see tab. 8-23)
equivalent current (1)
Iz
determination of the cross-sectional area S1 of the cable
able to carry I z in standard installation conditions in
relation to the type of cable, its insulation and rated
voltage (see tab. 8-29 to 8-34)
IB
f
thermal withstand
S2
I sc max
S3 function I f , t
(see tab. 8-37 to 8-39)
screen thermal
withstand: If
voltage drop
check
Isc max
t
k
S
max ( S1 , S 2 , S3 )
economic cross-sectional
area possibly chosen
(1) I z is an equivalent current which, in standard installation conditions,
causes the same thermal effect as I B in actual installation
conditions
Figure 6-15: logigram for determining the minimum cross-sectional area of a medium voltage cable
T&D
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6.2.8.
Cable screen earthing conditions
n single-core cables
The passage of a current in the cable core produces an induced voltage in the screen. This
voltage depends on the geometrical arrangement of the cables, the length and the current
carried:
2a
E0 = 0.145 × log10   × l × I
 d
a
d
l
I
:
:
:
:
distance between cable axes (mm)
mean diameter of the screen (mm)
connection length (km)
current carried in the core (A).
For very long cables,
E0
may reach dangerous values for persons. The standard
recommends screen earthing at both ends when E0 is likely to exceed the limit of 50 V
under steady-state operating conditions.
However, screen earthing at both ends produces currents continuously circulating in the
screen.
For screen earthing at one end only, on occurrence of a short circuit, the potential induced on
the second end may be high and cause a breakdown of the screen insulation where it is
connected. The necessary precautions must therefore be taken.
o calculation of the current circulating in screens earthed at both ends
In balanced steady-state operating conditions (or during a three-phase short circuit), the
induced voltage in screens earthed at both ends causes a three-phase current to circulate.
This current is given by the formula:
I0 =
where
Rs
Xs
l
E0
Rs2
+ Xs2
2a
Xs = 0.145 × log10   × l
 d 
: screen resistance (Ω)
: screen reactance (Ω)
: length of cable or line
T&D
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o example
Let us consider a 20 kV aluminium single-core cable with a cross-sectional area of 300 mm² ,
with PE insulation and a length of l = 3 km , buried in soil having a resistivity of ρ = 100 Ω ⋅ m ,
the characteristics of which are as follows:
-
Icapacity = 500 A
-
d = 33.5 mm
a = 38.5 mm
Rs = 0.45 Ω / km
It is installed in a network such that:
-
I B = 400 A
Isc = 8 kA
The induced voltage under steady-state operating conditions is:
 2a 
E0 = 0.145 × log10   × I B × l = 63 V
 d
The 50 V limit is exceeded and the screen must therefore be earthed at both ends.
The circulation current in the screen is in this case:
I0 =
E0
Rs2 + Xs2
Rs = 1.35 Ω
2a
X s = 0.145 × log10   × l = 0.157 Ω
 d
whence
Note:
I0 = 46.4 A
the circulation current in the screen is independent of the cable length.
T&D
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581
The induced voltage in the event of a short circuit is:
 2a 
E0 sc = 0.145 × log10   × I sc × l = 1 260 V
 d
The circulation current in the screen is then:
I0 sc = 927 A
This current must be withstood by the screen for the maximum short-circuit time. This is the
case since it can withstand 1 350 A for 2 s (see table 6-38).
Note:
if the cable length was 2 km, the screen would be earthed at one end only. The voltage induced
in the screen on occurrence of the short circuit will then be equal to 840 V. In this case it is
necessary to check that the screen insulation at the point where the terminal box is located is
sufficient.
Evaluation of Ws losses in the screen
Ws = Rs I02
for
Rs = 0.45 Ω / km , l = 3 km
and
I0 = 46.4 A
Ws = 0.45 × 3 × (46.4 )2 = 2.9 kW
The losses in the core are:
Wc = Rc × I B2
Rc : core resistance
For an aluminium conductor with a cross-sectional area of S = 300 mm² , Rc = 0.1 Ω / km
whence
Wc = 0.1 × 3 × ( 400)2 = 48 kW
We determine the ratio
Ws
=6%
Wc
The screen losses represent 6 % of the core losses. They must therefore be taken into
account when determining the maximum current-carrying capacity of the cable.
T&D
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582
o thermal effect in the cable screens
As we saw in the previous example, when the screen is earthed at both ends, the continuous
circulation of induced current in the screen causes extra heating in the cable and consequently
reduces its current-carrying capacity.
Generally, this phenomenon is only to be taken into account for cables with a cross-sectional
area greater than 240 mm².
We can apply the following rule:
- thin screen without armour, for S > 1 000 mm² the current-carrying capacity is reduced by 5 %
- non-thin screen without armour, the current-carrying capacity is to be reduced by:
. 5 % for
. 10 % for
240 mm² ≤ S ≤ 800 mm²
S > 800 mm²
- cables with screen and armour, the current-carrying capacity is to be reduced by:
. 5 % for
. 10 % for
. 15 % for
240 mm² ≤ S ≤ 400 mm²
500 mm² ≤ S ≤ 800 mm²
S > 800 mm²
n three-core cables
For three-core collectively screened cables, the electromagnetic field is zero in balanced
operating conditions.
In normal operating conditions, there is no circulation current in the screen.
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6.2.9.
Application example
Let us determine the conductor cross-sectional area of the W1 wiring system inserted into the
network illustrated in figure 6-17.
The W1 wiring system is made up of three single-core three-phase 6/10 (12) kV aluminium
cables with XLPE insulation, directly installed in a enclosed channel in a temperature of 35 °C.
The time delay of the protection against phase-to-phase short circuits is: t = 0.2 s .
20 kV
T1
1000 A
S n = 10 MVA
U sc = 8 %
U n = 5.5 kV
W1 wiring system
l = 1200 m
T2
S n = 630 kVA
U sc = 4 %
400 V
Figure 6-17: diagram of the installation
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n determining the maximum design current I B
The W1 wiring system only feeds the 630 kVA power transformer T2 .
The current I B is thus taken to be equal to the nominal transformer current:
I B = In =
Sn
630 × 10 3
= 66 A
=
3 Un
3 × 5.5 × 10 3
n correction factors and choice of S1
The direct installation in an enclosed channel corresponds to installation type L4
table 6-23). Column (3) in the current-carrying capacity tables must be used.
(see
- installation method: f0 = 0.8
- ambient temperature (see table 6-24): f1 = 0.96
- group of several cables (see table 6-28): f 5 = 1
The overall correction factor is: f = 0.8 × 0.96 = 0.77
The equivalent current that the cable must be able to carry in standard installation conditions
is:
I
I z = B = 86 A
f
Table 6-31 (column (3), XLPE, aluminium) gives a minimum cross-sectional area of
S1 = 16 mm2 which has a current-carrying capacity of I 0 = 95 A .
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(S 2 )
n checking thermal withstand
Neglecting the impedance upstream of the transformer and the impedance of the transformerbusbar connection, the maximum short-circuit current at the origin of the cable is equal to the
short-circuit current of the transformer.
The impedance of the transformer T1 is:
(
)
5.5 × 10 3
Un2 U sc
×
=
ZT 1 =
Sn 100
10 × 10 6
2
×
8
= 0.242 Ω
100
The maximum short-circuit current is thus:
Isc = 11
.
Un
5.5 × 10 3
= 11
= 14.4 kA
. ×
3 ZT 1
3 × 0.242
(see Protection guide § 4.2.1)
The cross-sectional area complying with the short-circuit requirement is:
I
S2 ≥ sc
k
t
k = 94 : value of the coefficient corresponding to a XLPE-insulated aluminium conductor (see table 6-35)
t = 0.2 s : short-circuit time equal to the protection time delay
whence
S 2 ≥ 69 mm2
The minimum cross-sectional area to be chosen is thus S 2 = 70 mm2
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n checking the cable screen thermal withstand ( S3 )
The 5.5 kV distribution network has an earthing system with a 1 000 A current limiting resistor.
The fault current is then:
If =
Vn
+ IC
RN
(see Protection guide § 4.3.2)
Vn : single-phase network voltage
RN : limiting resistance
I C : 5.5 kV network capacitive current ( I C = 3 j C ω Vn )
The capacitive current of an industrial network is of the order of several amps to several dozen
amps and it can thus be neglected in relation to the 1 000 A limiting current.
We thus have I f = 1 000 A
We assume that the screen must be able to withstand the fault current for 2 seconds, in order
to take into account the maximum time delay of the protection against phase-earth faults and
eventual reclosing.
The cross-sectional area of the conductor complying with the thermal withstand of the cable
screen is then:
S3 = 50 mm2
(see table 6-37)
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n checking voltage drops
The voltage drop is given by the formula:
l


∆V =  ρ1 cos ϕ + λl sin ϕ  I B

 S
l = 1 200 m ; S = 70 mm² ; λ = 0.15 × 10 −3 Ω / m ; I B = 66 A ; ρ1 = 0.036 Ω ⋅ mm 2 / m
We assume that the cable load has a cos ϕ = 0.6 (sin ϕ = 0.8)
whence
1 200


∆V =  0.036 ×
0.8 + 0.15 × 10 −3 × 1 200 × 0.6 × 66


50
∆V = 53 V
The relative voltage drop is:
53
∆V
=
= 1.7 %
Vn  5500 
 3 
In spite of a very long cable length for an industrial network, the voltage drop is acceptable.
n choosing the technical cross-sectional area
The calculations carried out give the following cross-sectional areas:
S1 = 16 mm²
S 2 = 70 mm²
S3 = 50 mm²
The technical cross-sectional area to be chosen is thus:
S = 70 mm²
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6.3.
Calculating the economic cross-sectional area
The methods described in chapters 6-1 and 6-2 lead to the choice of technical cross-sectional
areas of wiring systems, complying with the different thermal withstands, voltage drops and
protection of persons.
But it may be useful to take into account the economic criterion, based on the cost of
investment and the operating costs, when looking for the optimum cross-sectional area.
The investment cost is essentially composed of:
- the cable cost, linear function of the cross-sectional area S and length L ,
i.e. K1 L + K2 L S
- the cost of civil engineering and installation, depending on the length and regardless of the
cross-sectional area in a limited interval, i.e. K3 L .
The operating costs comprise:
- the Joule losses in the cable
- the maintenance costs.
To calculate the economic cross-sectional area, only the cost of the Joule losses w relative to
the wiring system is taken into account:
w=n×ρ
n
ρ
L
HC
× I2
S
1 000
Euros.
: number of live conductors
: resistivity of the live conductor during normal service, i.e. 1.25 times that at 20 °C.
ρ = 0.0225 Ω mm ²/ m for copper; ρ = 0,036 × Ω mm² / m for aluminium
L
S
I
H
C
:
:
:
:
:
cable length
cross-sectional area of conductors
current carried, assumed to be constant, in A
number of cable operating hours (for a year H = 8 760)
cost of kWh, Euro/kWh.
The cost of investment and the cost of losses w do not have the same term of payment. It is
necessary to change in order to carry out the sum of their values. This can be done by
converting the operating costs paid at the end of consecutive years to current value, i.e. by
converting them to the period in which the cable is purchased.
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If N (years) is the amortizement time forecast for the cable, and if the price of energy and
the cable load are assumed to be constant for the entire period, the sum of converted values
of Joule losses is:
 1
1
1
W =w
+
+ ....... +
2
1 + t (1 + t )
(1 + t ) N


(1 + t ) N − 1
 = w×
N

t (1 + t )

t being the forecast conversion to current value rate.
t (1 + t )
w
We can write W =
, where A =
A
(1 + t ) N − 1
N
The total cost is therefore:
P ( S) = K3 L + K1 L + K2 L S + n ρ
L 2 HC
I
S
1 000 × A
∂ P

The function P ( S ) goes via a minimum 
= 0
∂ S

for a cross-sectional area of S0 = I
nρHC
K2 A × 1 000
For an approximate calculation we can use the following formula:
S0 =
KI
100
HC
, mm²
A
where K = 2.56 for copper and 4.61 for aluminium.
The value of the economic cross-sectional area to be chosen is the closest standard
value to S 0 .
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n example
Taking the elements from the application example of § 6.2.9:
- design current I B = 66 A
-
energy cost: C = 0.061 Euro/kWh
- aluminium conductor, K = 4.61
- conversion to current value rate of 8 %
- amortizement time N = 20 years
- number of operating hours H = 3 800 hours.
A=
0.08 × (1 + 0.08)
S0 =
20
(1 + 0.08)20 − 1
= 0.102
66 × 4.61 3 800 × 0.061
= 145 mm ²
100
0.102
The economic cross-sectional area is the closest standard value to S 0 , i.e. S = 150 mm2 .
In practice, the economic cross-sectional area is often greater than the technical crosssectional area.
n advantages of cable oversizing
- Improved voltage quality under normal operating conditions and reduced amplitude of
voltage surges during motor or other machine starting.
- Presence of reserve power offering the possibility of future extensions.
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06 determining conductor cross sectional areas

Transcription

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