Download - IET Lecture LV Wiring Design September 14
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Cable Sizing For Safe Power
System Based on IEC Standards
I r . H . P. L o o i ( m e k t r i c o n @ g m a i l . c o m )
B . E n g ( H o n s ) , F I E M , J u r u t e r a G a s
Cable Sizing Fundamentals and
E l e c t r i c a l L V S t a n d a r d
A r m a d a H o t e l , P e t a l i n g J a y a
2 3 r d A u g u s t 2 0 1 4
P a r t 2 C a b l e S i z i n g
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We had an overview of the following topics in Part 1
2 SYNOPOSIS
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1. Introduction
2. Scope
3. General Design Procedure (design road map)
4. Earthing system
5. Cable types & installation method
6. Circuit configuration
Part 2 will deal with Cable Sizing with the following:
1. Overview
2. Protection Device & Implication for cable sizing
3. Simplified Method 60364-5-52
4. Heat Flow Calculation 60297 series
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3 OVERVIEWTOCABLESIZING
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4 OVERVIEWTOCABLESIZING
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5 OVERVIEWNEHERMCGRATHAMPACITY
Cable sizing or Amps capacity calculation is basically a heat transfer problem. 1957 Neher
MacGrath equation makes the following simplification assumptions:
(a) Assumption of steady state conditions;
(b) The heat transfer media surrounding the cable system is homogeneous and the following
conditions apply:
(i) In case of cables in air, thermal transfer via convection is assumed to predominate
(direct solar radiation to be considered if relevant);
(ii) In case of cables in ground or embedded in solid media, conduction predominate;
(c) Cable diameter is assumed negligible compared to cable length, this reduces the model to2-dimensions.
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I = Ampacity (current carrying capacity), (kA)
Tc = Conductor temperature, (C)
Ta = Ambient temperature, (C)
TD= Conductor temp rise due to dielectric loss, (C)
Rdc = Conductor dc resistance, (/foot)Yc = Loss increment due to skin & proximity effects
Rca= Thermal resistance between conductor and
ambient (thermal feet)
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6 OVERVIEW2 METHODS
Simplified 60364-5-52 Annex B to Elook up tables
IEC 60287 series of equations,
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CircuitCapacity
CableAmpacity
VoltageDrop
I2
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8 IEC 60364-4-43 CIRCUITCAPACITY
(1) IB< In < Iz
(2) I2< 1.45 x Iz
Calculate
current
from load
Select
protective
devices,
In > IB
Calculate cable size
based on:
Iz > In
IBdecide current
rating of circuitIB> load current
Check I2< 1.45 Iz
Electrical load
Iz = current capacity of cable.
Cables
Protection devices
In = nominal
current rating
of protective
devices
I2= current ensuring effective
operation within time
prescribed for protective
device
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9 IEC 60364-4-43 CIRCUITCAPACITY
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10 IEC 60364-5-52 ANNEXB
Current capacities
Tables B52.2 to B52.13
Installation method
Reference methods Table B52.1
Temperature correction factor
Table B52.14 (cables in air);B52.15 (cables in ground) and
B52.16 (cables in ducts)
Correction factor for soil
thermal resistivityTable B52.16 for method D only
where soil resistivity is other
than 2.5 K-m/W.
Reduction factor for more than one (single or multi core) cable
per circuit.Table B52.17 for installation method E to F
Table B52.18 for cables laid direct in ground
Table B52.19 for cables laid in duct in ground
Table B52.20 for multi core cables in free air (method E only)
Table B52.21 for single core cables in free air (method F only)
Annex E reduction factor due to harmonic currents
Currentcapacity
required
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12 IEC 60364-5-52 TABLEB52.16
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13 IEC 60364-5-52EXAMPLE
EXAMPLE
1Estimate load current 0.9x25kW/(0.85x415Vx3
36A.25kW1
1 Choose rating of circuit IB= 50A.36A2
1 Choose protection device In= 63A (In>IB)60A3
1 Size cable IZ > In; choose 4x25mm PVC; IZ 90ATable B52-5
25mm4
1 Check I2 < 1.45Iz 130AFuse 63A blow in abour 200s at 130A (I2)
MCB 63A trip in about 240s at 130A (I2)
200s5
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14 IEC 60364-5-52EXAMPLE
Melt TimeCurrent Data2-125 Amperes
TimeinSecond
Current in Amperes
Example
gG Fuse 63A
Melt in 200s for 130A
current (over-load)
Fuse 63AMelt at about 0.01s for 2kA
current (short circuit
current)
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15 IEC 60898 BREAKERCLASS
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16 IEC 60898 FUSES
(a) The first letter indicates the breaking range of fuse-links:(i) g for full range breaking capacity; general purpose fuse link interrupts
all faults and is usually used for all overload and short circuit application.
(ii) a for partial range breaking capacity,usually associated with another
protective device in cascade for discrimination between different
segments of a circuit. It is usually used for short circuit protection only.(b) The second letter indicates the utilisation category, which defines the
accuracy of the time-current characteristics,
(i) G indicates general application;
(ii) M indicates protection of motor;
(iii) R indicates semiconductor protection;
(iv) S indicates semiconductor protection;
(v) Tr indicates for transformer protection;
(vi) D indicates time-delay (UL 248); and
(vii) N indicates non-time-delay (UL 248).
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17 IEC 60898 FUSES
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18 IEC 60364-5-52VOLTAGEDROP
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19 NEUTRALCABLE
Neutral (TN and TT systems) shall have the same cross sectional area andnot less than that of the line conductor in the following cases (refer IEC
60364-5-52 Clause 524.2).
(i) In all cases of single-phase, 2-wire circuits.
(ii) In polyphase and single-phase 3-wire circuit, when the size of the line
conductors is less than or equal to 16mm2 copper or 25mm2
aluminium.
Note: Some national codes require that neutral cable should be the same size
as line conductor. This requirement is mandatory in Malaysia in all cases,
except between transformer and main switch board and at the discretion
of the designer who is expected to take into account issues relating to
neutral current listed below.. 23 rd August 2014
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20 HARMONICS, ANNEXE, 5-52
Where significant harmonic current occurs in three-phase circuits, reduction
factor for conductor sizing in accordance with Table 9 (Annex E of IEC 60364-5-
52) shall be applied.
When harmonics exceed 33%, neutral currents may exceed the line current in
which case neutral conductors may be oversized compared to line conductors.
Third harmonic content of
phase current (%)
Reduction factor
Size selection is based onphase current
Size selection is based onneutral current
0 -15 1.0 -
15 - 33 0.86 -
33 - 45 - 0.86
> 45 - 1.0
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R= Ro [1 + 20 ( 20)]
R = Rso [1 + 20 (sc 20)]X= 210-7Ln [2s/d]
1= (Rs/SR) [ 1 /(1 + (Rs/X))]
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23 VOLTAGESTANDARD
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IEC 60287 Electric cablesCalculation of the current rating
Part 1 Current rating equations (100 % load factor) and calculation oflosses:
Section 1: General;
Section 2: Sheath eddy current loss factors for two circuits in flat
formation;
Section 3: Current sharing between parallel single-core cables and -
calculation of circulating current losses;
Part 2Thermal resistance:Section 1: Calculation of thermal resistance;
Section 2: A method for calculating reduction factors for groups of cables
in free air, protected from solar radiation
Part 3Sections on operating conditions:Section 1: Reference operating conditions and selection of cable type;
Section 2: Economic optimization of power cable size;
Section 3: Cables crossing external heat sources
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25 EXAMPLE1
Circuit 1 Lighting 8 nos 80W each
Circuit 2 2x13A socket outlet in radial circuit
Circuit 3 1x15A water heater
Circuit 4 1x15A, air cool split unit air conditioning
Circuit 5 4x13A socket outlet in ring circuit
Circuit 6 6x13A socket outlet in ring circuitExample 1, domestic wiring diagram with typical circuits
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26 EXAMPLE1
Circuit 1 Lighting 8 nos 80W each
Circuit 2 2x13A socket outlet in radial circuit
Circuit 3 1x15A water heater
Circuit 4 1x15A, air cool split unit air conditioning
Circuit 5 4x13A socket outlet in ring circuit
Circuit 6 6x13A socket outlet in ring circuitExample 1, domestic wiring diagram with typical circuits
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27 EXAMPLE1
N
o Fuse rating Conductor
(mm)
Comments
1 6 Amps 1.5 mm Lighting circuit (8 nos x 80W). Clause 524,
Table 52.2 of 60364-5-52 requires minimum
cable size at 1.5mm.
2 15 Amps 2.5 mm 2x13A socket outlet in radial circuit (2x300W)
3 20 Amps 4 mm 1x15A socket outlet for water heater
(2,000W)
4 15 Amps 2.5 mm 1x15A socket outlet for (1x750W)
5 30 Amps 2.5 mm 4x13A socket outlet in ringcircuit(4x300W)
6 30 Amps 4 mm 6x13A socket outlet in ringcircuit(6x300W)
7 60 Amps 25 mm Main feeder cables to consumer unit (total
load = 6,690W (38A).
Table 13A; Example 1, schedule of circuits
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28 EXAMPLE1
Load EstimateSelect Earthing
System
Determine
Installation
Method
Select & Size
Wiring / Cables
Table 13A
(previous slide)TN-S
PVC single core
in conduit;
method A1
IEC60364-5-52,
method A1; Table
B52.2
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29 EXAMPLE1
Cct
No
Load/ current(3)
IB(A)
MCB (type B) Cable Comments
Nom.rating IN
Effectivetrip I2
(1)Size mm Iz (A)
(2)
230V single phase
1 8x80W ~ 3.3A 6A 9A 1.5 14.5 Lighting circuit
2 2x300W ~ 3.3A 15A 22.5A 2.5 19.5 Socket outlet radial
circuit
3 2000W ~ 9.7A 20A 30A 4 26 Water heater
4 750W ~ 3.8A 15A 22.5A 2.5 19.5 15A outlet with airconditioning unit
5 4x300W ~ 6.5A 30A 45A 2.5 2x19.5 Ring circuit subject to
nationalconditions.6 6x300W ~ 9.8A 30A 45A 4 2x26
7 6,690W ~ 36A 60A 90A 25 80 Main power intake
(1) Assume effective trip current to be 1.5 x IN.(2) Cable installation assumes PVC insulated cables, 2 single core in conduit embedded in walls.
Current capacity referenced from IEC 60364, table B52.2, installation method A1.
(3) Load estimate are presented here without consideration of diversity as an illustration only.
The subject of load diversity can be a detail subject which may require national statistics on
average usage of electrical appliances.
Check: IB < IN< IZ I2 < 1.45 x IZ
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Installation Config- Conductor Current capacity Group Corrected
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31 EXAMPLE2LARGECABLESCase
Installation Config-
uration(1); XLPE cable
Conductor
size(1) (mm)
Current capacity
of cables
Group
rating factor(2) (3)
Corrected
current
capacity
Multi core cable (B52.12, column 3, method E) B52.20
(a) 3 x 4 core, 1 cable
diameter apart
3x300mm / 4
core
3x621A = 1,853A 1 1,853A
(b) 3 x 4 core, 1 cable
diameter apart
2x240mm / 4
core
3x538A = 1,614A 1 1,614A
Single core cables in trefoil (B52.12, column 5, method F) B52.21
(a) Trefoil, 2 circuit @ 2 cable
diameter apart on 1 ladder
7x400mm / 1
core (5)
2x823A = 1,646A 1 1,646A
(b) Trefoil, 3 circuit touching 11x240mm / 1core (5)
3x607A = 1,821A 0.82(4)
1,493A
(b) Trefoil, 3 circuit @ 2 cable
diameter apart on 1 ladder
11x240mm / 1
core (5)
3x607A = 1,821A 1) 1,821A
Single core cables in flat formation (B52.12, column 6 or 7,
method F or G)
B52.21
(a) Cable touching (methodF), 2 circuits, 1 ladder.
7x400mm / 1core (5)
2x868A = 1,736A 0.97 1,684A
(b) 1 cable diameter apart,
(method G), 2 circuits, 1
ladder.
7x300mm / 1
core (5)
2x902A = 1,804A 0.97 1,750A
(b) 1 cable diameter apart,
(method G), 2 circuits, 1ladder.
7x240mm / 1
core(5)
2x781A = 1,562A 0.97 1,515A
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32 EXAMPLE2LARGECABLES
Notes:
(1) Installation configuration and conductor sizes are based on table B52.12 forXLPE insulated copper cables.
(2) Group rating factor where only 1 cable ladder is used is referenced from table
B52.17 row 5. Finer graduation in group factor can be obtained from tables
B52.20 (for multi core cables and B52.21 (for single core cables).
(3) In all cases, we assume only one cable ladder used.(4) Group rating for case 2(b). Table B52.21 for single core cables on ladder in
trefoil only considers case of cable-circuits laid 2 Deapart. To consider case of
cable touching,table B52.17 is referred for reduction factor.
(5) For single core cables, we assume neutrals to be sizes. The odd cable is
therefore used as neutral e.g. 11 core means 3x1 core for each line conductorand 2x1 core for neutral. In some national jurisdiction, size neutral may not
be allowed.
(6) Ambient temperature is assumed to be at 30C, for temperature correction
factor at 1 (from table B52.14).
Table 14A; Example 2, schedule of cable sizes and configurations
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33 EXAMPLE2LARGECABLES
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34 EXAMPLE2LARGECABLES
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Cable Sizing For Safe Power System Based
o n I E C S t a n d a r d s
I r . H . P. L o o i ( m e k t r i c o n @ g m a i l . c o m )
B . E n g ( H o n s ) , F I E M , J u r u t e r a G a s
Cable Sizing Fundamentals and
E l e c t r i c a l L V S t a n d a r d
A r m a d a H o t e l , P e t a l i n g J a y a
2 3 r d A u g u s t 2 0 1 4
P a r t 2 C a b l e S i z i n g
ThankYou for your At tent ion !