modul 4 cable
TRANSCRIPT
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Distribution Design System
A.Aman 1 BEKP 4783
UNIT 4: CABLE
1.0INTRODUCTION
1.1.
Cable is a mean or tool for which electrical energy is being distributed fromone place to others.
1.2. Cable also means to prevent any person from direct contact with the live
conductors.
1.3. The conductor has to be covered with a layer of insulating material, called
cable insulation.
1.4. To prevent any person from direct contact with the live conductors, a cable can
be definedas a length of insulated single conductor or two or more
conductors with its own insulation which are laid up together. The
insulated conductor or conductors may or may not be provided withoverall covering for mechanical protection.
1.5. Where mechanical protection is required, a second insulating layer known as
sheath is wrapped around the first layer ofinsulation. For stronger protection,
a layer ofsteel wire known as armour is wound around the first layer. A
sheath usually PVC is then placed over the armour.
2.0CONDUCTORS
2.1. The most common conductors used in cables are copper (Cu) and aluminum
(Al).
2.2. The resistance ofcopper at 70oC is 0.017/mm
2/m and aluminum
0.0283/mm2/m.
2.3. Aluminum has become a major alternative to copper as the price is cheaper.
However, it has poorer conductivity than cooper. Thus, the cross sectional
area of aluminum is larger. Example: for the same current rating, 300mm2
Al
cable is approximately equivalent to 185mm2.
2.4. A conductor and its immediate insulation known as core.
2.5. The conductor is identified by colour code:
2.5.1.1. Red, Yellow and Blue for phase conductors.
2.5.1.2. Black for Neutral.
2.5.1.3. Green/Yellow for CPC. Ground wire.
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2.6. Rated Voltage for cable is determined by Vo/V:
2.6.1.1. Vo is voltage between insulated conductor to earth and V is
between phase conductor.
Example: Low voltage power cable 600/1000V.
2.7. Conductors can be solid orstranded. Stranded conductors are more flexibleand easier to handle.
2.7.1.1. Solid conductor- single wire of the required cross-sectional area is
used.
2.7.1.2. Stranded conductor several wires of small cross-sectional are
twisted together to form a larger wire of the required sectional area.
Stranded conductors are more flexible and easier to handle.
Normally 3 to 7 wire twisted to form one conductor.
Figure 1- Cable conductor
2.8. Bare conductors are used in overhead HV transmission cable, earthing system,
lighting protection and busbars. Insulated conductors are used in domestic
and industrial installation, overhead LV (ABC), underground cable system and
submarine cable system.
2.9. Sizes of conductors are denoted by the cross-sectional area of the conductors for example 6mm
2cable means that the total cross-sectional area of all the
strands is 6mm2.
2.10. Copper conductor is widely used for the electrical earthing. The use of
aluminum as earth electrode or earthing conductor is prohibited.
3.0INSULATION MATERIALS
3.1.The selection of insulation materials is based on their physical and electrical
properties. Materials used as insulation for electrical cable must;
3.1.1.1. Provide safe condition of conductors at it rated voltage.
3.1.1.2. Electrical insulating properties should not deteriorate with time.
3.1.1.3. Have high mechanical strength during installation and operation.
3.1.1.4. Have thermal stability- chemical composition not effect by heat.
3.1.1.5. No environment or surrounding effect like weather or water.
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3.2.Types of insulation material used:
a) Butyl rubber can be used in ambient temperature of up to 80oC.
b) Silicone rubber suitable for temperature -60oC to 150
oC.
c) PVC poly-vinyl chloride or thermoplastic. Soften at 80oC.
d) Glass fiber Good heat resistance up to 155oC.
e) Paper paper impregnated with mineral oil mixed resin.
f) Mineral insulating material is magnesium oxide. Withstand very
high temperature up to copper melting point.
g) XLPE cross-linked polyethylene
h) FR fire retardant (specially formulated PVC compound)
4.0CABLE TYPES AND SPECIFICATIONS
4.1.Cables are named in accordance to:
a) Type ofinsulation
b) Mechanical protection
c) External sheath
Example: XLPE/SWA/PVC XLPE- (cross-linked polyethylene) insulation,
SWA steel wire armored (mechanical protection) and PVC external sheath.
4.2.The most common type of cables used in distribution systems are:
a) PVC PVC insulated.
b) PVC/PVC PVC insulated, PVC sheath
c) PVC/SWA/PVC PVC insulated, steel wire armored, PVC sheath.
d) FR(Fire Retardant) fire rate equipment e.g. emergency light,KELUAR sign, BOMBA lift.
e) MICC Mineral insulated copper covered.
f) MICS Mineral insulated copper sheath.
g) XLPE/SWA/PVC XLPE insulation, steel wire armored, PVC sheath
underground cables.
h) PILC Paper Insulated Lead Covered used in HV system.
4.3.The current capacity must be sufficient to cater for the maximum sustained
current which normally flows through the cable and the insulation must be
adequate with the system voltage.
4.4.In a single line diagram, the type and size of the cable is written as follows:
Example: 2x120mm2
4C PVC/SWA/PVC (Al) cable means that 2 set of
120mm2, 4 Core, PVC insulated, steel wire armored, PVC sheath and aluminum
conductor.
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4.5.According to BS 6004, all PVC insulated cables are suitable only as long as
the conductor temperature does not exceed 70oC and formineral insulated
cables to BS 6207 can be operated up to 135oC.
4.6.A sample of current carrying capacity (ampacity) is shown in TABLE 4D1A
(IEE Wiring Regulation BS 7671- Requirement for Electrical Installation)
for single core PVC insulated cables, non-armored with or without sheath.
Figure 2- Types of cables
5.0INSTALLATION
5.1. Small wiring PVC or FR cables can be installed in conduits or trunkings. The
conduits and trunkings are made of steel or PVC.
5.2. Schedule ofmethods of installation of cables is shown in TABLE 4A1 ofIEE
Wiring Regulation BS 7671.
5.3.
Table 4A1 in IEE Wiring Regulations BS 7671 shows the methods ofinstallation of cables and conductors.
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5.4. There are seven types of wiring systems and there are twenty methods of
installations in general wiring or in electrical installation. The using of method
of installation is to find the appropriate reference for determining current
carrying capacity of the cable. Below are the lists of seven types of wiring
system.
i. Open and clip directii. Cable embedded direct in building material.iii. In conduitiv. In trunkingv. In the free air, on cleats, brackets or a laddervi. Cable in building voidsvii.Cable in trenches
5.5. Fordomestic wiring installation only method 1, method 3 and method 4
have to consider.
5.6. The method of installation in final circuits wiring installation is importance,
where the methods of installations will influence ofcable size and selection ofrated current of protective devices.
5.7. Common used of LV cables:
a) Non-Armoured PVC cables installed in conduits & trunking for
internal wiring.
b) Non-Armoured PVC/PVC cables general indoor use in domestic &
commercial installation.
c) Armoured PVC cables for mains & sub-mains application.
d) MICS cables areas of extreme temperature or for circuit supplied to
firefighting equipment.
e) FR cables used in severe condition (firefighting / flammable
equipment)
6.0CABLE SIZING AND CURRENT RATING
6.1.The size of the cable to be used in a circuit depends on:
a) The full load current based on design current Ib.
b) Ambient temperature Ca factor.
c) Maximum allowable conductor temperature.
d) Conductor material current ampacity. - Table 4D1A and 4D2A in
BS7671 for copper conductor.
e) The method ofinstallation. Table 4A1 in BS7671
f) Insulation material. - Table 52B (Regulation 523-1) in BS7671
g) The number of circuits installed together Cg.
h) The voltage drop in the cable. - Table 4D1B and 4D2B in BS7671.
6.2.It is a normal practice that the design engineer relies on the tabulated current
carrying capacities as shown in Appendix 4 of the IEE Wiring Regulation
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BS 7671. However, it should be noted that the tabulated current ratings are
based upon a given set of condition.
a) Ambient temperature.
b) The heating effect of adjacent cable is not considered.
c) The cable is installed according to the rating table being used.
d) There is no surrounding thermal installation.
6.3.Therefore, any changes on the above said conditions, the cable rating has to be
adjusted according to the appropriate correction factor as follows.
a) Ambient temperature Correction Factor (Ca). - Table 4C1 in
BS7671
b) Grouping Correction Factor (Cg). - Table 4B1 in BS7671
c) Thermal Insulation Correction Factor (Ci).
6.4.For ambient temperature higher than specified temperature of 30oC, the rated of
the heat flow out of the conductor will be lower and thus will increase the
conductors operating temperature above permitted value.
6.5.Correction factor of temperature Ca in determining the current carrying
capacity of cable are provided in Table 4C1 ofIEE Wiring Regulation BS
7671.
6.6.Cables installed in the same enclosure or bunch together will get warm when
all are carrying current. Those cables close to the edges of enclosures will be
able to release the heat outward but will be restricted in losing heat inwards
toward other warm cable. Cable in the centre of the enclosure may find it
difficult to lose heat at all and as a result will increase the conductor
temperature.
6.7.The correction factor for groups Cg is shown in Table 4B1 ofIEE Wiring
Regulation BS 7671.
6.8.Many new building are now provided with better thermal insulating material
for roofs and cavity walls to reduce the heat lose. As thermal insulation is
design to limit heat flow, a cable in contact with it will tend become warmer
than the preferred operation conditions.
6.9.IEE Wiring Regulation 523-04 recommends that for a single cable, which is
likely to be surrounded by thermally insulated material over a length of more
than 0.5m, the thermal insulated Ci is 0.5 times the tabulated current carryingcapacity (method 1). Table 52A ofIEE Wiring RegulationBS 7671 shows
the de-rating factor for cable surrounded by thermal insulation for a length less
than 0.5m.
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7.0DETERMINE THE MINIMUM TABULATED CURRENT RATING
7.1.After considering all the above correction factors, the minimum tabulated
current rating for the cable can be determined as recommended in the IEE
regulation.
7.2.As an example the related table for copper conductor current carrying capacity
7.3.The formula used is as follows:
a) Where overload protection is required:
It min In
Ca x Cg x Ci
b) Where overload protection is not required:
It min Ib
Ca x Cg x Ci
8.0VOLTAGE DROP
8.1.The resistance of the conductor increases as the length increases and/or the
cross sectional area of cable decreases.
8.2.Associated with an increase resistance is a drop in voltage, which means that a
load at the end of long or thin cable will not have the full supply voltage
available.
8.3.To ensure the voltage drop from the supply intake to the terminal of any
appliance does not exceed the appliance operating tolerance, which is normally10%.
8.4.IEE Regulation 525-01 specifies that the voltage drop between the origins of
installation does not exceed 4% of the nominal voltage of the supply.
8.5.The utility supply regulation normally require that the voltage drop in the
circuit does not exceed 2.5% of the nominal voltage, i.e. 6V for single phase
240V supply and 10.4V for a three phase 415V supply. This is also a good
practice for a designer since the cable routing on the paper may differ from the
actual at site installation.
8.6.Values of voltage drop are tabulated for a current of 1 Amp for a 1 meter run,i.e. for a distance of 1 meter along the route taken by the cables. This
Tabulated Voltage Drop constant (TVD) is expressed in m per ampere per
meter run mV/A/m.
8.7.Voltage drop formula is as follows:
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( )1000
llengthITVDV brdrop
= for cable 16mm2
or less
( )1000
sincos lengthITVDTVDV bxrdrop
+
=
8.1.1 VOLTAGE DROP CALCULATION
8.1.1.1 TEMPERATURE CORRECTION ON RESISTIVE VALUE
8.1.1.2In relevant standards and manufacturer specifications, the value of
conductor resistance is usually given at 20oC of conductor
temperature. This conductor resistance value will increase
according to the resistance temperature coefficient when conductor
temperature increases due to load current.
8.1.1.3The coefficient is approximately equal to 0.004 pero
C at 20o
C forboth copper and aluminium conductor. The value of conductor
resistance is zero mathematically at a conductor temperature of -
230oC.
8.1.1.4The conductor resistance value at other temperature such as 115can be calculated as:
115 = 20 230 + 115 = 20 x 1.38
230 + 20
Fi ure 3: Calculation of resistance value at various
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8.1.1.2CONDUCTOR TEMPERATURE ON VOLTAGE DROP (TVDr)
8.1.2.3 The TVDrvalue is based on the resistance of conductor at rated
temperature (i.e. 70oC for PVC-insulated Cu conductor)
corresponding to the conductor carrying its rated current.
8.1.2.4 If design current is significantly less than the rated current, the actualvalue of TVDrwill be lower due to lower conductor temperature.
Thus, the new value of TVDrat other temperatures is given by:
( TVDr)40o
C = 230 + 40 x ( TVDr)70o
C
230 70
8.1.2.5 The value of reactance (TVDx) is not influenced by temperature,
therefore the temperature correction does not apply to the reactance
values.
9.0PROTECTION AGAINST OVERLOAD
9.1 Overload currents may occur in various circuits in an installation where they are
carrying more current than their rated capacity. The consequences of overload
currents will increase the temperature of conductors and the effectiveness of the
insulation and its expected life may be reduced.
9.1.1REQUIRED CONDITION FOR OVERLOAD PROTECTION
9.1.1.1 To provide an adequate protection for overload, IEE Regulation 43
requires the following conditions to be satisfied:
1) IN IZ
2) I2 1.45 IZ
Where
IN : current rating of the protective device
IZ : current rating of the cable under particular installation methods.
I2 : current magnitude causing an effective operation of the protective
device.
Figure 4: Rated conductor temperature
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9.1.1.2 For MCB, the value I2 is 1.45 IN. For MCCB and ACB, the value I2 is
1.3 IN. For fuse, the value I2 is 1.6 IN.
9.1.1.3 The effective operating time is 2 hours (for MCB, MCCB and ACB),
except for breaker less than 63A where the effective operating time is
reduced to 1 hour. For fuse, the effective operating time is in the rangefrom 1 to 4 hours depending on the current rating.
9.1.1.4 In loading condition (I2 1.45 IZ), the cable is allowed to increase145% of its rated capacity even though conductor temperature is higher
than rated temperature but still below its critical temperature. Critical
temperature refers to the temperature that will cause insulation failure.
9.1.1.5 To assess whether there is an overload protection and how adequate it
is, we define the degree of overload protection as:
OL_P_Yes = 1.45 IZ - I2 x 100%
1.45Z
9.1.1.6 A positive value of (OL_P_Yes) means that the circuit is adequately
protected against overload current and a higher percentage indicate that
the circuit is more unlikely to be overloaded. The condition is vice
versa for negative value of (OL_P_Yes).
9.1.2SMALL OVERLOADS AND CABLE UTILISATION
9.1.2.1 There is possible to have a range of very small overload which will not
be detected by the protective device (i.e. at the range between IZ I I2). In general statement IEE Regulation 433-01-01, it states that every
circuit shall be designed so that a small overload of long duration is
unlikely to occur.
9.1.2.2 Therefore, we define the existence of small overload as:
Sm_OL_No = IZ - I2 x 100%
IZ
9.1.2.4 A positive value of (Sm_OL_No) indicates no undetected small
overload while a negative value means that small overload range
exists. Thus, designer should as far as possible minimize the negative
percentage (i.e. increase the conductor size will totally eliminate the
occurrence of such small overloads).
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9.1.2.5 If value of IZ is very close to IN, the un-detected small overloads may
occur (under-design) and if the value of IZ is very much higher than IN,
then it implies a larger size of cable has been selected which will
increase the installation cost and not fully utilized (over-design).
9.1.3OMISSION OF OVERLOAD PROTECTION
9.1.3.1 There are several circuits in which a break in current (tripping) by the
operation protective devices can cause danger. Example; breaking the
current of a lifting electromagnet could cause the load to drop, or
breaking the current in CT could induce a very high E.M.F
(electromagnetive force). So, it can be replaced by an overload alarm
instead of overload breaker.
9.1.3.2 For starting current of a motor, the designer has to ensure that the
protective device of the motor circuit will not trip during motor
starting. If starting current is large, the rated current IN may have to
be much higher than the design current IB.
9.1.3.3 Therefore, the current rating of cable, IZ has to be equal to or larger
than IN.
IN >> IB and IZ >> IB
9.1.3.4 However, if overload protection is not required, the designer can select
IZ to be independent of IN, and IZ >> IB.
9.1.3.5 Ifoverload protection is required, the minimum tabulated current
rating for the circuit is obtained by:
It,min = IN
Ca x Cg x Ci
9.1.3.6 Ifoverload protection is not required, the minimum tabulated
current rating for the circuit is obtained by:
It,min = IB
Ca x Cg x Ci
10.0 PROTECTION AGAINST SHORT CIRCUIT
10.1 It is necessary to ensure that the circuits conductor are protected adequately
against short-circuit current by interrupting breaker quickly enough to prevent
thermal damage to the cable.
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10.1 REQUIRED CONDICTIONS FOR SHORT CIRCUIT PROTECTION
10.1.1 During fault conditions up to duration of 5 seconds, the maximum time that
the cable can withstand the fault current can be approximately by the
following formula:
Tcable,max = k2S2
IF2
Where k : maximum thermal capacity of the conductor for the type of
insulation being used
S : cross-sectional area of the conductor in mm2
IF : prospective short-circuit current in amperes
(Note: For more than 5 seconds, more complicated formula has to be used)
10.1.2 (Tcable,max) also known as critical time, which is the time taken for temperature
of conductors to rise from rated value, Q1 to the critical value, QF. If QF is
exceed, the insulation material fails and the whole cable is thermally damage.
10.1.3 To assess whether short-circuit protection is provided and how adequate it is,
let us define:
SC_P_Yes = ( Tcable,max Tbk,3-phase F ) x 100%
Tcable,max
Where Tbk,3-phase F : operating time of the breaker corresponding to a current
during 3-phase fault at the cable.
10.1.4 A positive percentage (SC_P_Yes) indicates that the circuit meets the
requirement for shor-circuit protection and higher percentage implies a higher
margin of short-circuit protection. A negative percentage (SC_P_Yes) is vice
versa.
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10.2 FORMULA FOR SHORT-CIRCUIT CURRENT
10.2.1 Three Phase Fault: the current magnitude in each phase is identical except
that the angle is shifted by 120o
in each phase. Thus the current in each phase
can be calculated as:
IF,3 = VLL / 3
(RS + R1)2
+ (Xs + X1)2
10.2.2 Line-to-Neutral Fault :
IF,LN = VLL / 3
(RS + R1 + RN)2
+ (Xs + X1 + XN)2
10.2.3 Line-to-Line Fault :
IF,LL = VLL
(2RS + 2R1)2
+ (2Xs + 2X1)2
Where RS : Resistance value of supply source (per phase)
XS : Reactance value of supply source (per phase)
R1 : Resistance value of phase conductor (per phase)
Figure 5: Values of k for calculation of the effects of fault current
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X1 : Reactance value of phase conductor (per phase)
RN : Resistance value of the neutral conductor
RN : Reactance value of the neutral conductor
VLL : Line-to-Line voltage.
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11.0 TABLES AND FIGURES (FROM IEE WIRING REGULATIONS)
11.1. Conductors Temperature
11.2. Cable Ampacity
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Table 4D1A Single-core Copper Conductor Current Capacity
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Table 4D2A Multi-core Copper Conductor Current Capacity
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11.3. Voltage Drop
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Table 4D1B Single-core Copper Conductor Voltage Drop
Table 4D2B Multi-core Copper Conductor Voltage Drop
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11.4. Ambient Temperature Correction Factor, Ca
Table 4C1 Correction factor for ambient temperature.
11.5. Grouping Correction Factor, Cg
Table 4B1 Correction factor for group
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11.6. Schedule of Methods of Installation
Table 4A1 An example of methods of insallation
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Example 1:
If four cores PVC insulated cable enclosed in trunking has tabulated current
capacity 80A. If the ambient temperature 50C, the current rating of cable
reduced 80A x 0.71 = 56.8A. The ambient temperature is refers to the
temperature of the cable surrounding media and does not temperature of
equipment.
* 0.71 refers to Table 4C1 temperature correction factor IEE Wiring Regulation.
* Tabulated current refers to Table 4D2A
Example 2:
A heater rated at 230V, 3kW is to be installed using twin-with-earth PVC
insulated and sheathed cable clip direct in a roof space which has an ambient
temperature of 40C. The circuit is protected by 15A MCB. The cable is
bunched with four other-twin-earth cables for a short distance as shown in
Figure 1. Determine the minimum tabulated current rating of the circuit and the
size of the conductor.
Figure Example1
Solution
The design current is
AIB
13230
3000==
and the current rating f the protective device is IN=15A.
From table 4C1 the temperature correction factor at 40C is:
Ca = 0.87
From table 4B1 the grouping correction factor of five circuits is:
Cg = 0.6
The minimum tabulated current rating Iz,min for the circuit is
ACC
II
ga
Nz 74.28
6.087.0
15min, =
=
=
Referring Table 4D2A of IEE Wiring Regulation, column 6, a 4mm2
which
tabulated current rating of 36A is selected. Although the design current is only
13A, a 4mm2cable rated 36A is selected. Without correction factor ofCa and
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Cg under same 13A operating conditions a 1mm2
cable rated at 15A is
sufficient.
Example 3:
The voltage drop on a circuit supplied from a 240V source by 16mm2
two-corecopper cable 23m long, clipped direct and carrying a design current 33A, will
be;
Solution:
Using formula( )
1000
llengthITVDV br
drop
= due to less than 16mm2
TVDr = 2.8mV - From Column 3 Table 4D2B, cable size 16mm2
two core
cable single phase.
Ib = 33A
Length = 23m
( )1000
llengthImVV bdrop
=
V125.21000
23338.2=
=
The maximum voltage drop for 240V installation less than 4% from the IEE
Regulation is 9.6V, so the voltage drop fulfills the IEE Regulation requirement.The maximum cable length of installation with these parameters could be
determined.
b
drop
ImV
VMaxLengh
=
1000
m149238.2
10006.9=
=
Example 4:
A 3 phase motor with full load current of 102A, and PF of 0.8 is to be fed
by single core, PVC insulated, copper conductor, non-armoured cables,
clipped direct on non metallic surface at 75 meter rub shown in Figure
Example 2. Determine the size of conductor if the permissible voltage
drop is 2%.
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Figure Example 2
Solution:
Let the design current Ib is the motor full load current. From table 4D1A,
column 7, a 25mm2
cable with rated current of 104A is initially selected
and the voltage drop is calculated as:
( )1000
sincos lengthITVDTVDV bxrdrop
+
=
( )
.983.9
1000
751026.0175.08.05.1
V=
+=
9.983 are 2.40% of 415V which exceeds 2% of 415V. Thus the next
higher size of 35mm2
is selected and Vdrop is re-calculated:
( )
.512.7
1000
751026.017.08.01.1
V=
+=
7.512V are 1.81% of 415V, thus in order to fulfill the requirement of Vdrop
less than 2% a 35mm2
cable is selected.