 GROUP

1- Ibrahim Kamal Mahmoud

2- Ahmed Gamal Ahmed Diab
3-Ahmed Said Taha
4-Ahmed Saber Moawad Morsi
5- Ahmed Abd-almagid Said Kamel
6- Aya Saber Abd-alrehim
7- Mahmoud Essam Mohamed mlige
8- Hend Mahmoud Mohamed
Sizing and protection of
conductors

1 GENERAL
The cabling and its protection at each level must satisfy several
conditions
at the same time it must:

 Carry the permanent full load current, and normal short-time overcurrents

 Not cause voltage drops likely to result in an inferior performance of certain loads

 Protect the cabling and busbars for all levels of overcurrent, up to and including short-
circuit currents.

 Ensure protection of persons against indirect contact hazards particularly in TN- and IT-
earthed systems, where the length of circuits may limit the magnitude of short-circuit
currents, thereby delaying automatic disconnection.

Flow-chart for the selection of cable size and protective device rating for a given circuit
 DEFINITIONS:
MAXIMUM LOAD CURRENT: IB

>> AT THE FINAL CIRCUITS LEVEL, THIS DESIGN CURRENT CORRESPONDS TO THE RATED KVA OF THE LOAD.

IN THE CASE OF A HIGH IN-RUSH CURRENT SUCH AS MOTORS STARTING THE CUMULATIVE THERMAL EFFECTS OF THE
OVERCURRENTS MUST BE TAKEN INTO ACCOUNT. BOTH CABLES AND THERMAL TYPE RELAYS ARE AFFECTED.

>>AT ALL UPSTREAM CIRCUIT LEVELS THIS CURRENT CORRESPONDS TO THE KVA TO BE SUPPLIED,

WHICH TAKES ACCOUNT OF THE DIVERSITY AND UTILIZATION FACTORS, KS AND KU RESPECTIVELY.

CALCULATION OF MAXIMUM LOAD CURRENT IB

MAXIMUM PERMISSIBLE CURRENT: IZ

CURRENT CARRYING CAPACITY IZ IS THE MAXIMUM PERMISSIBLE THAT THE CABLING FOR THE
CIRCUIT CAN CARRY WITHOUT REDUCING ITS NORMAL LIFE EXPECTANCY.
THE CURRENT DEPENDS ON SEVERAL PARAMETERS:
1- CONSTITUTION OF THE CABLE AND CABLE-WAY
2-AMBIENT TEMPERATURE
3- METHOD OF INSTALLATION
4- INFLUENCE OF NEIGHBOURING CIRCUITS.
OVERCURRENTS
OCCURS EACH TIME THE VALUE OF CURRENT EXCEEDS THE MAXIMUM LOAD CURRENT IB
IT MUST BE CUT OFF WITH A RAPIDITY THAT DEPENDS UPON ITS MAGNITUDE
OVERCURRENT CAN BE :
1- OVERLOADS
CAN OCCUR IN HEALTHY ELECTRIC CIRCUITS, FOR EXAMPLE MOTOR STARTING LOADS.
2-SHORT-CIRCUIT CURRENTS
FROM THE FAILURE OF INSULATION BETWEEN LIVE CONDUCTORS OR/AND
BETWEEN LIVE CONDUCTORS AND EARTH (ON SYSTEMS HAVING LOW-IMPEDANCE-EARTHED
NEUTRALS)
IN ANY COMBINATION, VIZ :
A- 3 PHASES SHORT-CIRCUITED (AND TO NEUTRAL AND/OR EARTH, OR NOT)
B- 2 PHASES SHORT-CIRCUITED (AND TO NEUTRAL AND/OR EARTH, OR NOT)
C- 1 PHASE SHORT-CIRCUITED TO NEUTRAL (AND/OR TO EARTH)
 Overcurrent protection principles

1-Acting to cut-off the current in a time shorter than that given by the CIRCUIT PROTECTION BY •
I2t characteristic of the circuit cabling CIRCUIT BREAKER
for periods up to 5 seconds following short-circuit initiation, be
determined
approximately by the formula:
I2t = k which shows that the allowable heat generated is proportional to
to the squared cross-sectional-area of the conductor.
t: Duration of short-circuit current (seconds)
S: Cross sectional area of insulated conductor (mm2)
I: Short-circuit current (A r.m.s.)
k: Insulated conductor constant
Note:
1- ISC: 3-phase short-circuit current
2- ISCB: rated 3-ph. short-circuit breaking current of the circuit breaker
3- Ir (or Irth).
 GENERAL RULES

A PROTECTIVE DEVICE (CIRCUIT BREAKER OR FUSE) FUNCTIONS CORRECTLY IF:

1- ITS NOMINAL CURRENT OR ITS SETTING CURRENT IN IS GREATER THAN THE MAXIMUM LOAD
CURRENT IB BUT LESS THAN THE MAXIMUM PERMISSIBLE CURRENT IZ
2- ITS TRIPPING CURRENT I2 “CONVENTIONAL” SETTING IS LESS THAN 1.45 IZ
3- ITS 3-PHASE SHORT-CIRCUIT FAULT-CURRENT BREAKING RATING IS GREATER THAN THE 3-PHASE
SHORT-CIRCUIT CURRENT EXISTING AT ITS POINT OF INSTALLATION.
APPLICATIONS
1-PROTECTION BY CIRCUIT BREAKER
BY VIRTUE OF ITS HIGH LEVEL OF PRECISION THE CURRENT I2 IS ALWAYS .
2- PROTECTION BY FUSES
THE CONDITION I2 Y 1.45 I I2 IS THE FUSING MELTING LEVEL) CURRENT EQUAL TO K2 X IN (K2 RANGES
FROM 1.6 TO 1.9)
2 Practical method for determining the
smallest allowable cross-sectional area of
circuit conductors
2.1 General method for cables

 Possible methods of installation for different types of conductors or cables

 Possible methods of installation for different situations:

 The number given in this table refer to the different wiring systems
considered
 Maximum operating temperature:

 The current-carrying capacities

given in the subsequent tables
have been determined so that the
maximum insulation temperature is
not exceeded for sustained periods
of time.

 Correction factors:

 In order to take environnement or special conditions of installation into account, correction

factors have been introduced. The cross sectional area of cables is determined using the rated
load current IB divided by different correction factors, k1, k2, ...

 I’B is the corrected load current .

  Ambient temperature :

 The current-carrying capacities of

cables in the air are based on an
average air temperature equal to 30
°C.
 The related correction factor is here
noted k1

 The current-carrying capacities of

cables in the ground are based on an
average ground temperature equal to
20 °C.
 The related correction factor is here
noted k2.
 Soil thermal resistivity:

 he current-carrying capacities of cables in the ground are based on a ground

resistivity equal to 2.5 K.m/W.
 The related correction factor is here noted k3.

 Correction factors for cables in buried ducts for soil thermal resistivities other than
2.5 K.m/W to be applied to the current-carrying capacities for reference
method D

 Based on experience, a relationship exist between

the soil nature and resistivity.
depending on the nature of soil.
 Grouping of conductors or cables

 the values of correction factor k4 for different configurations

 the values of correction
factor k4 for different
configurations of unburied
cables or conductors, for
groups of more than one
circuit of single-core cables in
free air.
 the values of correction factor k4 for different configurations of cables
or conductors laid directly in the ground.
 Harmonic current:

 if the 3rd harmonic percentage h3 is greater than 33 %, the neutral current is

greater than the phase current and the cable size selection is based on the
neutral current. The heating effect of harmonic currents in the phase conductors
has also to be taken into account.

 The values of k5 depending on the 3rd harmonic content

 Admissible current as a function of nominal cross-sectional area of
conductors
 2.2 Recommended simplified
approach for cables
 In order to facilitate the selection of cables, 2 simplified tables are proposed, for unburied
and buried cables.

 Unburied cables:
 Correction factors are given in Figure G21b for groups of several circuits or
multi- core cables:
 Buried cables:
 2.3 Sizing of busbar trunking systems
(busways)

 The selection of busbar trunking systems is very straightforward

 correction factors for grouping are not relevant parameters for this technology

 The cross section area of any given model has been determined by the
manufacturer based on:

 The rated current

 An ambient air temperature equal to 35 °C
 3 loaded conductors
 Rated current:

 The rated current can be calculated taking account of:

 The layout
 The current absorbed by the different loads connected along the trunking system.

 Ambient temperature

 A correction factor has to be applied for temperature higher than 35 °C. The
correction factor applicable is provided by the busway manufacturer.
 Neutral current:

 Where 3rd harmonic currents are circulating, the neutral conductor may be
carrying a significant current and the corresponding additional power losses
must be taken into account.

 Figure G23b represents the maximum admissible phase and neutral currents
(per unit) in a high power busbar trunking system as functions of 3rd harmonic
level.
 The layout of the trunking system depends on the position of the current
consumers, the location of the power source and the possibilities for fixing the
system.

 One single distribution line serves a 4 to 6 meter area

 Protection devices for current consumers are placed in tap-off units, connected
directly to usage points.
 One single feeder supplies all current consumers of different powers.

 Once the trunking system layout is established, it is possible to calculate the

absorbed current In on the distribution line. In is equal to the sum of absorbed
currents by the current In consumers: In = Σ IB. The current consumers do not all
work at the same time and are not permanently on full load, so we have to use a
clustering coefficient kS : In = Σ (IB . kS).
3 Determination of
voltage drop
 Determination of voltage drop
 This section deals with methods of determining voltage drops, in order to check that
 They comply with the particular standards and regulations in force.
 They can be tolerated by the load.
 They satisfy the essential operational requirements.

Maximum voltage drop limit

The value of 8 % can lead to problems for motor loads , for example starting current is 5-7
times of rated so voltage drop will reach 40 % , in that condition motor will either
 Over-heating and trip-out
 Or accelerate very slowly, so that the heavy current loading during starting period
The maximum value of 8 % in steady operating conditions should not be
reached on circuits which are sensitive to under-voltage problems

Calculation of voltage drop in steady load conditions

Use of formulae
IB : The full load current in amps
L : Length of the cable in Ω/km
R : Resistance of the cable conductor in
Ω/km
Un : phase-to-phase voltage
Vn : phase-to-neutral voltage
Note
R is negligible above a c.s.a. of 500 mm2
X is negligible for conductors of c.s.a. less than 50 mm2.
In the absence of any other information, take X as
being equal to 0.08 Ω/km.

cos ϕ = 1 (Incandescent lighting)

cos ϕ > 0.9 (Led lighting or Fluorescent with electronic ballast)
cos ϕ = 0.35 (At start-up of motor)
cos ϕ = 0.8 (In normal service of motor)
Example
A three-phase motor taking as shown take 100 A at a cos ϕ = 0.8 on normal permanent load
and 500 A (5 In) at a cos ϕ = 0.35 during start-up , The voltage drop at the origin of the motor
cable in normal ( with the distribution board ) is 10 V phase-to phase.
solution
During starting conditions , Table G28 shows 0.54 V/A/km
the voltage drop at the distribution board will exceed 10 Volts.
Supposing that the infeed to the distribution board during motor starting is
900+500=1400 A then the voltage drop at the distribution board will increase to
(10 x 1400)/1000 = 14 V
ΔU cable = 0.54 x 500 x 0.05 = 13.5 V
ΔU total = 14 + 13.5 = 15 V
(27.5/400)x 100 = 6.9 % , is less than that authorized (8 %) and is satisfactory.
4 Short-circuit current
 A KNOWLEDGE OF 3-PHASE SYMMETRICAL SHORT-CIRCUIT CURRENT VALUES
(ISC) IS NECESSARY IN ORDER TO DETERMINE SWITCHGEAR (BREAKING
CAPACITY), CABLES (THERMAL WITHSTAND RATING), PROTECTIVE DEVICES
(DISCRIMINATIVE TRIP SETTINGS).
4.1 Short-circuit current at the secondary
terminals of a MV/LV distribution transformer

 In a simplified approach, the impedance of the MV system is assumed to be negligibly

small, so that:
Inx 100 Sx 10³
Isc= where In= and:
Usc U2√3
 S = kVA rating of the transformer
 U20 = phase-to-phase secondary volts on open circuit
 In = nominal current in amps
 Isc = short-circuit fault current in amps
 Usc = short-circuit impedance voltage of the transformer in %.
Typical values of Usc for different kVA ratings
of transformers with MV windings ≤ 20 kV
 The case of several transformers in parallel feeding a busbar The value of fault
current on an outgoing circuit immediately downstream of the busbars can be
estimated as the sum of the Isc from each transformer calculated separately.
 The value of fault current on an outgoing circuit immediately downstream of the
busbars can be estimated as the sum of the Isc from each transformer calculated
separately.
 Other factors which have not been taken into account are the impedance of the
busbars and of the cable between transformers and circuit breakers
3-phase short-circuit current (Isc) at
any point within a LV installation:
 In a 3-phase installation Isc at any point is given by:
U20
 Isc = where:
√3 ZT

 U20 = phase-to-phase voltage of the open circuited secondary windings of the

power supply transformer(s).
 ZT = total impedance per phase of the installation upstream of the fault location
(in Ω).
Method of calculating ZT

 The method consists in dividing the network into convenient sections, and to calculate the
R and X values for each.
 Where sections are connected in series in the network, all the resistive elements in the
section are added arithmetically; likewise for the reactances, to give RT and XT. The
impedance (ZT) for the combined sections concerned is then calculated from:
ZT = √RT² + XT²
 Any two sections of the network which are connected in parallel, can, if predominantly
both resistive (or both inductive) be combined to give a single equivalent resistance (or
reactance) as follows:
 Let R1 and R2 be the two resistances connected in parallel, then the equivalent resistance
R3 will be given by:
𝑅1 𝑋𝑅2 𝑿𝟏 𝑋 𝐗𝟐
R3= 𝑅1+𝑅2 or for reactances X3= 𝐗𝟏+𝐗𝟐
Determination of the impedance of
each component
1) Network upstream of the MV/LV transformer:
 The 3-phase short-circuit fault level Psc, in kA or in MVA is given by the power supply
authority concerned, from which an equivalent impedance can be deduced .

 A formula which makes this deduction and converts the impedance

to an equivalent value at LV is given, as follows:
𝑼𝒐²
Zs=
𝐏𝐬𝐜
Where :

Zs = impedance of the MV voltage network, expressed in milli-ohms

Uo = phase-to-phase no-load LV voltage, expressed in volts
Psc = MV 3-phase short-circuit fault level, expressed in Kva

 The upstream (MV) resistance Ra is generally found to be negligible compared with the
corresponding Xa
 If more accurate calculations are necessary, Xa may be taken to be equal to 0.995 Za
and Ra equal to 0.1 Xa.
2) Transformers:

for an approximate calculation, in the absence of more precise information on

transformer characteristics, Cenelec 50480 suggests to use the following guidelines:
 if U20 is not known, it may be assumed to be 1.05 Un
 in the absence of more precise information, the following values may be used:
Rtr = 0.31 Ztr and Xtr = 0.95 Ztr where:
𝑼𝟐𝟎²
Ztr= where:
𝐏𝐬𝐜∗𝑼𝒔𝒄

U20: open-circuit secondary phase-to-phase voltage expressed in volts

Psc = MV 3-phase short-circuit fault level, expressed in VA
Usc = the short-circuit impedance voltage of the transformer expressed in %
3) Busbars :

 The resistance of busbars is generally negligible, so that the impedance is practically all
reactive, and amounts to approximately 0.15 mΩ/metre length for LV busbars.
 In practice, it's almost never possible to estimate the busbar length concerned by a
short-circuit downstream a switchboard.

4)Circuit conductors:

The resistance of a conductor is given by the formula:

ℓ
Rc=ρS
Where
ρ = the resistivity of the conductor material at the normal operating temperature
ℓ = length of the conductor in m
S = c.s.a. of conductor in mm²
 Cable reactance values can be obtained from the manufacturers. For c.s.a.
of less than 50 mm2 reactance may be ignored. In the absence of other
information, a value of 0.08 mΩ/metre may be used (for 50 Hz systems) or
0.096 mΩ/metre (for 60 Hz systems). For busways (busbar trunking systems)
and similar pre-wired ducting systems, the manufacturer should be consulted.

5) Motors:
 In general, this fault-current contribution may be ignored. However, if the total power of
motors running simultaneously is higher than 25 % of the total power of transformers, the
influence of motors must be taken into account. Their total contribution can be estimated
from the formula:
Iscm = 3.5 In from each motor
 The motors concerned will be the 3-phase motors only; single-phase-motor contribution
being insignificant.
6)Fault-arc resistance:
 Short-circuit faults generally form an arc which has the properties of a resistance.
 resistance is sufficient to reduce the fault-current to some extent. Experience has
shown that a reduction of the order of 20 % may be expected. This phenomenon
will effectively ease the current-breaking duty of a CB, but affords no relief for its
fault current making duty
Method:
 Select the c.s.a. of the conductor in the column for copper conductors (in this example
the c.s.a. is 47.5 mm2).
 Search along the row corresponding to 47.5 mm2 for the length of conductor equal to
that of the circuit concerned (or the nearest possible on the low side).
 Descend vertically the column in which the length is located, and stop at a row in the
middle section (of the 3 sections of the Figure) corresponding to the known fault-
current level (or the nearest to it on the high side).
 In this case 30 kA is the nearest to 28 kA on the high side.
 The value of short-circuit current at the downstream end of the 20 metre circuit is given
at the intersection of the vertical column in which the length is located, and the
horizontal row corresponding to the upstream Isc (or nearest to it on the high side).
 This value in the example is seen to be 14.7 kA.
 The procedure for aluminium conductors is similar, but the vertical column must be
ascended into the middle section of the table.
5 Particular cases of short-
circuit Current
5.1 Calculation of minimum levels of
short-circuit Current

 on LV circuits
 a single protective device protects against all levels of current ) overload
threshold- the maximum rated short-circuit current breaking capability(
 The protection device should be able to operate in a maximum time to
ensure people and circuit safety
 To check that behavior
calculation of minimal short-circuit current or fault current is mandatory
 The protective device must therefore satisfy the two following conditions:
 Its breaking capacity must be greater than Isc.

 Elimination of the minimum short-circuit current possible in the circuit, in a time tc

compatible with the thermal constraints of the circuit conductors

(valid for tc < 5 seconds)

Where
S is the cross section area of the cable,
k is a factor depending of the cable
conductor material, the insulation material and initial temperature

The protective device must fulfill:

b instantaneous trip setting Im < Iscmin for
a circuit breaker
b fusion current Ia < Iscmin for a fuse
The limiting effect of the impedance of long circuit conductors on the value of short-circuit currents must be
checked and the length of a circuit must be restricted.

1 - Calculation of Lmax for a 3-phase 3-wire circuit

Using the “conventional method ” the voltage at the point of protection P is assumed to be 80 % of the nominal
voltage during a short-circuit fault
so that 0.8 U = Isc Zd,
where:
Zd = impedance of the fault loop
Isc = short-circuit current (ph/ph)
U = phase-to-phase nominal voltage
so that
where
ρ = resistivity of conductor material at the average temperature during a short-circuit,
S ph = c.s.a. of a phase conductor in mm2
L = length in metres
calculated maximum length
calculation similar to that of example 1 above is required, but for a single-phase
fault (230V)

 If Sn (neutral cross-section) = Sph

Lmax = k Sph / Im

 If Sn (neutral cross-section) < Sph, then (for cable cross-section y 120mm

Lmax = 6.666 Sph / Im (1+m)
 5.2 Verification of the withstand capabilities
of cables under short-circuit conditions

Thermal constraints:
 When the duration of short-circuit current is brief all of the heat produced is assumed to remain in the
conductor, causing its temperature to rise. an assumption that simplifies the calculation and gives a
pessimistic result.
 a higher conductor temperature than that which would actually occur, in practice, some heat would
leave the conductor and pass into the insulation.
 For a period of 5 seconds or less, the relationship I2t = k2S2 characterizes the time in seconds during which
a conductor of c.s.a. S (in mm2) can be allowed to carry a current I, before its temperature reaches a
level which would damage the surrounding insulation
The method of verification consists in checking that the thermal energy I2t per ohm of conductor material,
allowed to pass by the protecting circuit breaker (from manufacturers catalogues) is less than that permitted
for the particular conductor
 Electrodynamic constraints:

 For all type of circuit (conductors or bus-trunking), it is necessary to take electro dynamic effects into
account.
 To withstand the electro dynamic constraints, the conductors must be solidly fixed and the
connection must be strongly tightened.
For bus- trunking, rails, etc.
 it is also necessary to verify that the electro dynamic withstand performance is satisfactory when
carrying short-circuit currents.
The peak value of current, limited by the circuit breaker or fuse, must be less than the bus bar system
rating.
Tables of coordination ensuring adequate protection of their products are generally published by the
manufacturers and provide a major advantage of such systems.
6 Protective earthing
conductor (PE)
 Connection and choice

Connection
PE conductors must:
1)Not include any means of breaking the continuity of the
circuit (such as a switch, removable links, etc.)
2) Connect exposed conductive parts individually to the
main PE conductor, i.e. in parallel, not in series, as shown in
Figure

3) Have an individual terminal on common earthing bars in distribution boards

 Conductor sizing

 This table provides two methods of determining the appropriate c.s.a. for
both PE or PEN conductors.

(1) Data valid if the prospective conductor is of the same material as the line conductor. Otherwise, a
correction factor must be applied.
 (2) When the PE conductor is separated from the circuit phase conductors, the
following minimum values must be respected:
- 2.5 mm2 if the PE is mechanically protected
- 4 mm2 if the PE is not mechanically protected.
(3) For mechanical reasons, a PEN conductor, shall have a cross-sectional area not
less than 10 mm2 in copper or 16 mm2 in aluminium

Values of factor k to be used in the formulae

Protective conductor between MV/LV
transformer and the main general distribution
board (MGDB)
Recommended conductor sizes for bare and insulated PE conductors from the
transformer neutral point, are indicated below in Figure A

Fig.A
7 The neutral conductor
THE C.S.A. AND THE PROTECTION OF THE NEUTRAL CONDUCTOR, APART
FROM ITS CURRENT-CARRYING REQUIREMENT, DEPEND ON SEVERAL
FACTORS, NAMELY:

 THE TYPE OF EARTHING SYSTEM, TT, TN, ETC.

 THE HARMONIC CURRENTS.

 THE METHOD OF PROTECTION AGAINST INDIRECT CONTACT HAZARDS.

7.1 Sizing the neutral conductor
2- Influence of harmonic currentson Sizing the neutral conductor .
Protection against overload
 If the neutral conductor is correctly sized (including harmonics), no specific protection of
the neutral conductor is required because it is protected by the phase protection.

Protection against overload

 If the c.s.a. of the neutral conductor is lower than the c.s.a. of the phase
conductor, the neutral conductor must be protected against short-circuit.
 If the c.s.a. of the neutral conductor is equal or greater than the c.s.a. of the
phase conductor, no specific protection of the neutral conductor is required
because it is protected by the phase protection.
8- Worked example of
cable calculation
 The installation is supplied through a 630 kVA transformer. The process requires a high
degree of supply continuity and part of the installation can be supplied by a 250 kVA
standby generator. The global earthing system is TN-S, except for the most critical loads
supplied by an isolation transformer with a downstream IT configuration
THE SAME CALCULATION USING THE SIMPLIFIED METHOD
RECOMMENDED IN THIS GUIDE
1- DIMENSIONING CIRCUIT C1 •

TWO SINGLE-CORE PVC-INSULATED COPPER CABLES IN PARALLEL WILL BE USED FOR EACH PHASETHESE CABLES •
WILL BE LAID ON CABLE TRAYS ACCORDING TO METHOD F. EACH CONDUCTOR WILL THEREFORE CARRY 433 A.
FIGURE G21A INDICATES THAT FOR 3 LOADED .CONDUCTORS WITH PVC ISOLATION, THE REQUIRED C.S.A. IS 240
MM²

2-DIMENSIONING CIRCUIT C3
3-DIMENSIONING CIRCUIT C7
4-CALCULATION OF SHORT-CIRCUIT
CURRENTS FOR THE SELECTION OF CIRCUIT
BREAKERS Q1, Q3, Q7 (SEE FIG. G67)
5-THE PROTECTIVE CONDUCTOR •
WHEN USING THE ADIABATIC METHOD, THE MINIMUM C.S.A. FOR THE PROTECTIVE

EARTH CONDUCTOR (PE) •

CAN BE CALCULATED BY THE FORMULA GIVEN •

GENERALLY, FOR CIRCUITS WITH PHASE CONDUCTOR C.S.A. SPH ≥50 MM2 THE PE CONDUCTOR MINIMUM C.S.A. •

WILL BE SPH / 2. THEN, FOR CIRCUIT C3, THE PE CONDUCTOR WILL BE 95 MM2 AND FOR CIRCUIT C7, THE PE •
CONDUCTOR WILL BE 50 MM

6-PROTECTION AGAINST INDIRECT-CONTACT HAZARDS •

7-VOLTAGE DROP •

THE VOLTAGE DROP IS CALCULATED USING THE DATA GIVEN IN FIGURE G28, FOR BALANCED •

THREE-PHASE CIRCUITS, MOTOR POWER NORMAL SERVICE (COS Φ= 0.8). •

THE RESULTS ARE SUMMARIZED ON FIGURE G68: •