Transcript powerpoint file

Cable selection project
Factory office installation
maximum demand sub- mains
cable
• Each factory/warehouse consists of the
following loads
• 8- 250W mercury vapour lamps
• 4-60watt incandescent lamps
• 3-18watt fluorescent. External
• 1-500watt sodium vapour lamp. External
• 6-10A double single phase outlets
• 3-20A 3 phase outlets
• 1- 15A three phase storage hot water
Office
•
•
•
•
•
Lighting
8-double 36 watt fluorescent lights
Power
8- double10A single phase outlets
1- single 10A single phase outlet
Step 1
• Divide the installation into circuits and
distribute these circuits across the three
phases
• Calculate the maximum demand of the
installation
• The maximum demand of the sub- mains
is the load on the heaviest loaded phase
Arrange into circuits
• Factory
•
•
•
•
•
•
•
•
•
•
•
•
Circuit 1 (4-250W) lamps (R )
Circuit 2 (4-250W) lamps (W)
Circuit 3 (4-60watt) + 2 EF 60W (B)
Circuit 4 (3-18watt) F/lamps (W)
Circuit 5 (1x500watt SV lamps) (B)
Circuit 6 (2 double 10A) outlets (R)
Circuit 7 (2 double 10A) outlets (W)
Circuit 8 (2 double 10A) outlets (B)
Circuit 9 (20A 3 phase) outlet
Circuit 10 (20A 3 phase) outlet
Circuit 11(20A 3 phase) outlet
Circuit 12 (15A 3 phase) HWS
• Office
•
•
•
•
Circuit13 (8 double36 watt fluorescent)
Circuit14 (3 double 10A outlets)
Circuit15 (3 double 10A outlets)
Circuit16 (2 double 1 single 10A outlet)
(B)
(R)
(W)
(B)
MD Sub-mains
using table C2
Factory
Circuit
no
•
1
2
3
4
5
6
7
8
Load
group
Load
calculation
R
A
(4-250W)MV lamps 1.5A each
4 x 1.5 = 6
6
A
(4-250W) MV lamps 1.5A each
4 x 1.5 = 6
A
(4-18W) energy saver 0. 05A each
2x60w exhaust fans at 0.3A each
(0.05 4)  (0.3  2)  0.8
A
(3-18W) fluorescent 0.12A each
0.12  3  .36
0.36
A
1-500w sodium vapour lamps 0.8 pf
500
 2.72
230 0.8 pf
2.72
Bi
2-10A double 1Phase outlets = 4
1000 w 
3  750 w
 14.14
230
Bi
2-10A double 1Phase outlets =4
1000 w 
3  750 w
 14.14
230
Bi
2-10A double 1Phase outlets =4
1000 w 
3  750 w
 14.14
230
W
B
6
0.8
14.14
14.14
14.14
Maximum demand Sub-mains
Circuit
no
Load
group
Load
calculation
R
W
B
9
10
11
12
B (iii)
20A three phase outlet
Full load
20
20
20
B (iii)
20A three phase outlet
75% Full load
15
15
15
B (iii)
20A three phase outlet
75% Full load
15
15
15
G
15A three phase HWS
Full load
15
15
15
office
13
14
15
A
8-twin 36w fluorescent
8  0.78  6.24
B (i)
3-10A double outlets single
phase
6  750 w
 19.56
230
B (i)
3-10A double outlets single
phase
6  750 w
 19.56
230
16
B (i)
2-10A double 1-10A single
Outlets single phase
Maximum D
6.24
19.56
19.56
5  750w
 16.3
230
16.3
104.7
104.7
105.6
Cable size for sub-main to
factory/warehouse unit1
• The Maximum demand is 106A
Mains
Sub-mains
X90 SDI Cables double insulated buried in separate U/G conduit
Current carrying capacity T7/18 25mm² = 135A
Voltage drop T41 Vc =1.62mV/Am
Vd 
Vc  L  I 1.62  38 106

 6.52 Volts
1000
1000
So 25mm² X90 SDI Cables in separate conduits will satisfy both current and
voltage drop requirements.
Unit 1 has the longest run (38 metres) so 25mm² will satisfy units 2 and 3
Installation load
•
•
•
•
•
•
•
•
The installation consists of the following loads.
Lighting
24 – 250W mercury vapour Lamps
24 – 2x36W Fluorescent Luminaires 0.78A each
12 – 18W fluorescent to replace 60W
8 – 18W Fluorescent
6 – 60W exhaust fans 0.3A each
3 – 500W
Installation load
•
•
•
•
•
Power
42 – double 10A single phase outlets
3 – single 10A single phase outlets
9 – 20A three phase outlets
3 – 3 phase HWS
Consumers Mains maximum demand
Load
Group
load
Calculation
R
W
B
24 – 250W mercury vapour
Lamps (8 per phase)
8 x 1.5 = 12
12
12
12
6.24
6.24
6.24
24 – 2x36W Fluorescent
Luminaires (8 per phase)
8  0.78  6.24
12 – 18W fluorescent (0.05A) to
replace 60W
4 x 0.05 = 0.2
0.2
0.2
0.2
6 – 60W exhaust fans (0.3A
each)
2 x 0.3 = 0.6
0.6
0.6
0.6
8 – 18W Fluorescent (0.12A)
3,3,2 per phase
0.12 x 3 =.36
0.12 x 2 = .24
0.36
0.36
0.24
2.71
2.71
2.71
1000  28  750
 95.65
230
95.65
95.65
95.65
9 – 20A three phase outlets
20 + 8 x 15 =140
140
140
140
3 - 15A 3phase HWS
3 x 15 = 45
45
45
45
302.8
302.8
302.7
3 – 500W
14 Double +1 single 10A single
outlets per phase (29 per phase)
87 outlets total
MD
500
 2.71
230  0.8
Cable Size consumers mains
• The consumer mains are X90 SDI cables installed in conduit U/G for a
length of 40 metres
• Determine the cable size to suit current and voltage drop requirements
• Table 2.4 item 2 refers to Table 7/16
• 150mm² = 330A The cable can carry the MD current
• Check for voltage drop.
Table 41 Vc for 150mm² conductors = 0.309mV/Am (60ºC) The cable
is rated at 90ºC and by choosing a Vc value at 60ºC this allows for
temperature rise under Short circuit conditions
Vd 
Vc  L  I 0.309  40  303

 3.745
1000
1000
Progressive Voltage drop
Consumers main
Volt drop 3.745
volts
40 metres
Turret
10 metres
20 metres
MSB
DB
unit3
DB
Unit 2
38 metres
Sub-main voltage drop 6.52 volts
3.745 Volts
6.52 Volts
DB
Unit 1
Progressive Voltage drop
distribution board unit 1
Consumers
mains
Sub-main
Final sub-circuits
5.57 volts allowed in all single phase
circuits
3 phase Value
3 phase value
9.735 Volts allowed in all 3 phase
3.745+6.52 = 10.265 volts 3 phase Therefore the circuits
3.745 volts
6.52 volts
3 phase voltage drop allowed in all 3 phase
circuits supplied from the DB Unit1 is 20 –
10.265 = 9.735 V
To determine the single phase voltage allowed in
final sub circuits
10.265 10.265

 5.93Volts
1
.
73
3
Therefore the single phase Voltage drop
allowed in all single phase circuits
supplied from distribution board Unit 1 is
11.5 – 5.93 = 5.57 Volts
Circuit arrangements
Cable
Maximum Installation Parameters
Designation demand
AZ/NZS
3008/1/1
Table No
column
Consumers
mains
302 A
XLPE (X90)
SDI enclosed UG
Table 2.4
Table 7
4
16
Sub-mains
106 A
X90 SDI enclosed U/G
Table 2.4
Table 7
4
18
4x250W MV 6A
Lamp Factory
2 circuits
TPS V90 installed with 3 other
circuits
De rate 0.78
Table 2.1
Table 9
Table 24
4
2
8
10A outlets
Factory
TPS V90 installed with 3 other
circuits spaced
De rate 0.87 item22
Table 2.1
Table 9
Table 24
4
2
8
16A
Cable
Maximum
designation demand
Installation
parameters
AS/NZS
3008
Table no
3 phase
outlets
TPS V90 installed
with 2 other circuits
On cable
De rate 0.82
Table 2.1
Table 12
Table 24
4
2
8
EF + Battens 0.8A
In Toilets
TPS unenclosed
3 other circuits on
cable tray up wall
from switchboard
Table 2.1
Table 9
Table 24
De-rate 0.88
4
2
8
HWS
15
TPS unenclosed
3 other circuits on
cable tray up wall
with 3 other circuits
Table 2.1
Table 12
Table 24
De-rate 0.88
4
2
8
500W SV
2.8A
TPS installed
enclosed U/G
Table 2.4
Table 9
20A
Column
no
4
16
Cable Size 20A 3Ø outlets
Distribution board Unit 1
Sub-main
Mains
U/G
turret
Determine the cable size for
the 20A 3 phase outlets 1
per circuit, longest run 38
metres The cable is 3core
TPS V 90 installed enclosed
in air. No de-rating for this
section. Unenclosed in air
spaced on perforated tray up
wall above switchboard 4
circuits
20
 24.4 A
0.82
T able12/4 a 4mm2 cable  29A
Unit 1
DB
MSB
20A
3Ø outlet
1000 Vd
Vc 
L I
1000 9.735

38 20
 12.8m V / Am
To satisfy voltage drop requirements
Table 42 a 4 mm² Cable with a Vc value
of 9.71 mV/Am value is required
Cable Size 20A 3Ø outlets
Distribution board Unit 1
Sub-main
Mains
U/G
turret
MSB
Determine the cable size for the 20A 3 phase outlets 1 per circuit,
Route length 25 metres
Current carrying capacity is the limiting factor in this circuit
20A
3Ø outlet
To satisfy Current carrying capacity, a
4mm² TPS cable is required
Unit 1
DB
Cable Size 20A 3Ø outlets
Sub-main
Mains
U/G
turret
MSB
Unit 1
DB
Determine the cable size for the
20A 3 phase outlets 1 per circuit,
Route length 15 metres
20A
3Ø outlet
In this instance Voltage
drop is not the governing
factor. A 4mm² cable is
required for CCC
10A single phase outlets
• 6 outlets in warehouse 2 per phase, (3 circuits).
• Circuit 1 (38m route length) The outlet is at the end of the run
therefore use MD =10A
1000Vd 1000 5.57
Vc 

 14.65m V/A.m
L I
38 10
convert tothreephase
14.65 0.86  12.69m V/A.m
•
•
•
•
•
Therefore from Table 42, 4mm² cable is required for volt drop
Cable enclosed in conduit on wall with 3 0ther circuits spaced
( 4 circuits )
Table 9/6 4mm² cable = 26A
De-rating Table 22 (0.9) 26 x 0.87 = 23.4A
10A single phase socket outlets
• Circuit 2 (30m route length)
1000Vd 1000 5.57
Vc 

 18.56m V/A.m
L I
3010
convert tothree phase
18.56 0.86  16.07m V/A.m
• Therefore from Table 42 2.5mm² with a Vc value of
15.6mV/A.m is required
• Circuit 3 route length 20m route length can also be wired
in 2.5mm²
• Table 9 column 6 (2.5mm² cable)=20Ax0.9 =18A
So a 2.5mm² cable will satisfy both CCC and Vd
Light Circuits unit 1
Circuit 1 (4x250W mercury vapour)
• Route length 38m
Maximum Demand 4 x 1.5A = 6A.
• Circuit beaker rating 10A . Determine Cable size
1000Vd 1000 5.57
Vc 

 29.316m V / A.m
L I
38 5
convert tothreephase Vc  0.866
29.31 0.866 25.39m V / A.m
• So from Table 42 a 2.5mm² cable with a Vc value of
15.6mV/A.m three phase 15.6 x 1.155 =18mV/A.m is
required
Light Circuits unit 1
Circuit 2 (4x250W mercury vapour)
• Route length 44m
Maximum Demand 4 x 1.5A = 6A. See clause 3.5.2
50% of circuit protective device can be used
• Circuit beaker rating 10A . Determine Cable size
1000Vd 1000 5.57
Vc 

 25.328m V / A.m
L I
44  5
convert tothreephase Vc  0.866
25.318 0.866 21.925m V / A.m
• So from Table 42 a 2.5mm² cable with a Vc value of
15.6mV/A.m three phase 15.6 x 1.155 =18mV/A.m is
required
Circuit 3 four battens and 2 EF
Toilets
• Route length 28m
1000Vd 1000 5.57
Vc 

 39.78m V / A.m
L I
28 5
convert t othreephase Vc  0.866
39.78 0.866 34.45m V / A.m
• So from Table 42 1.5mm² cable is required
• Table 9 column 6 (1.5mm² cable = 14A)
Circuit 4 (3x 18W) Fluorescent
• Route length 50m
1000Vd 1000 5.57
Vc 

 318.28V / A.m
L I
50 .36
convert tothreephase Vc  0.866
318.28 0.866 275.63m V / A.m
• Voltage drop is not a factor for this circuit
• Either 1mm² or 1.5mm² can be used
Circuit 5 (500W) sodium Vapour
• Route length 16m
• TPS Cable V90 Installed enclosed U/G
Distribution
Board
500W Sodium vapour
1000Vd 1000 5.57
Vc 

 127.98m V / A.m
L I
16 2.72
convert tothreephase Vc  0.866
127.98 0.866 110.8m V / A.m
Voltage drop is not a factor for this circuit
Either 1mm² or 1.5mm² can be used
Hot Water Cylinder
3 phase 15A
•
•
•
•
Route length 28m
Cable 3core + E enclosed TPS V90
Table 12 column 2 (2.5mm² = 23A)
Check Voltage drop
1000 Vd 1000  9.735
Vc 

 23.17mV/Am
L I
28 15
• Table 41 (2.5mm² = 15.6mm²)Therefore 2.5mm² Cable
Office
• 8-2x36W(0.78A) fluorescent one circuit
• Route length 40m Use 10A MCB
• Rule 3.6.2 50.% circuit protective device for voltage drop
1000Vd 1000 5.57
Vc 

 27.85m V / A.m
L I
40 5
convert tothreephase Vc  0.866
27.85 0.866 24.118m V / A.m
Table 42 1.5mm² = 28.6mV/Am.
Use 1.5mm² cable
Office10A socket outlets
• Circuit 14 (three double outlets)
• Route length 22m TPS cable installed unenclosed in air Table 9
column 4 (2.5mm² = 26A)Use 20MCB
• Rule 3.6.2 (50.% circuit protective device for voltage drop)
1000Vd 1000 5.57
Vc 

 25.318m V / A.m
L I
22 10
convert tothreephase Vc  0.866
25.318 0.866 21.925m V / A.m
Table 42
2.5mm² = 15.6mV/Am
2.5mm Cable for all socket outlet circuits in the office.
Fault loop impedance
• The earth fault-loop impedance in an MEN
system comprises the following parts, starting
and ending at the point of the fault.
• a) The protective earthing conductor, (PE),
including the main earthing terminal/connection
or bar and MEN.
• b) The neutral return path, consisting of the
neutral conductor, (N), between the main neutral
terminal or bar and the point at the transformer
(the earth return path RG to RB has a relatively
high resistance and may be ignored for an
individual installation in an MEN system)
Fault loop impedance
• c) The path through the neutral point of the
transformer and the transformer winding.
• d) The active conductors as far as the point of
the fault.
• The earth fault-loop is normally regarded as
consisting of the following two parts• i) conductors upstream or external to the
reference point; and
• ii) conductors down stream or internal to the
reference point.
• Refer to appendix B for detail
Path taken by an earth fault current
Earth fault-loop impedance
Fault current IA
Distributor's network
A
H
POS
MEN
Neutral
Bar
Main Earth
Faulty equipment
Soil resistance high between
electrodes
Determine maximum route length
to satisfy fault loop impedance.
• The maximum length of a circuit can be
determined using Table B1
• (Exceptions include circuits wired in 4mm²
cable protected by a 16A or 20A Type C
MCB)
• The maximum length for this example will
need to be calculated
Calculation 16A MCB
0.8Uo  Sph Spe
LMax 
Ia    ( Sph  Spe)
0.8  230 4  2.5
LMax 
3
16 7.5  22.5 10  (4  2.5)
L Max  109metres
A 4mm² Cable protected by a 16A MCB can be run 109m and
not exceed the earth fault loop impedance requirements
Switchboards Units 1-3
120A Main switch
MCB’S
L L L L L P P P P P
P HW L P P P
10 10 10 10 10 16 16 16 20 20 20 16 10 16 16 16
Main Switch Board