Cable selection project Factory office installation.

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) (B)• Circuit14 (3 double 10A outlets) (R)• Circuit15 (3 double 10A outlets) (W) • Circuit16 (2 double 1 single 10A outlet) (B)

MD Sub-mainsusing table C2

•

Circuit no

Load group

Load calculation R W B

1 A (4-250W)MV lamps 1.5A each 4 x 1.5 = 6 6

2 A (4-250W) MV lamps 1.5A each 4 x 1.5 = 6 6

3 A (4-18W) energy saver 0. 05A each

2x60w exhaust fans at 0.3A each0.8

4 A (3-18W) fluorescent 0.12A each 0.36

5 A 1-500w sodium vapour lamps 0.8 pf

2.72

6 B i 2-10A double 1Phase outlets = 4 14.14

7 B i 2-10A double 1Phase outlets =4 14.14

8 B i 2-10A double 1Phase outlets =4 14.14

36.312.0 =×

72.28.0230

500=

× pf

14.14230

75031000 =

×+

ww

14.14230

75031000 =

×+

ww

14.14230

75031000 =

×+

ww

8.0)23.0()405.0( =×+×

Factory

Circuit no

Load group

Load calculation R W B

9 B (iii) 20A three phase outlet Full load 20 20 20

10 B (iii) 20A three phase outlet 75% Full load 15 15 15

11 B (iii) 20A three phase outlet 75% Full load 15 15 15

12 G 15A three phase HWS Full load 15 15 15

office

13 A 8-twin 36w fluorescent 6.24

14 B (i) 3-10A double outlets single phase

19.56

15 B (i) 3-10A double outlets single phase

19.56

16 B (i) 2-10A double 1-10A single

Outlets single phase

16.3

Maximum D 104.7 104.7 105.6

56.19230

7506=

× w

56.19230

7506=

× w

16.3230

750w5=

×

24.678.08 =×

Maximum demand Sub-mains

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

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

Volts 6.521000

106381.62

1000

ILVcVd =

××=

××=

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

Load Group

load Calculation R W B

24 – 250W mercury vapour Lamps (8 per phase)

8 x 1.5 = 12 12 12 12

24 – 2x36W Fluorescent Luminaires (8 per phase)

6.24 6.24 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 phase0.12 x 3 =.36

0.12 x 2 = .24

0.36 0.36 0.24

3 – 500W 2.71 2.71 2.71

14 Double +1 single 10A single

outlets per phase (29 per phase)

87 outlets total

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

MD 302.8 302.8 302.7

24.678.08 =×

71.28.0230

500=

×

65.95230

750281000=

×+

Consumers Mains maximum demand

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

745.31000

30340309.0

1000=

××=

××=

ILVcVd

Progressive Voltage drop

Turret

MSB

DB unit3

DBUnit 1

DBUnit 2

Consumers main Volt drop 3.745 volts

Sub-main voltage drop 6.52 volts

3.745 Volts 6.52 Volts

40 metres

38 metres

20 metres

10 metres

Progressive Voltage drop distribution board unit 1

Volts 93.573.1

265.10

3

265.10==

Consumers mains

Sub-main Final sub-circuits

3.745 volts

3 phase Value

6.52 volts

3 phase value

3.745+6.52 = 10.265 volts 3 phase Therefore the 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

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

5.57 volts allowed in all single phase circuits 9.735 Volts allowed in all 3 phase circuits

Circuit arrangements

Cable Designation

Maximum demand

Installation Parameters AZ/NZS 3008/1/1

Table Nocolumn

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

Lamp Factory

2 circuits

6A TPS V90 installed with 3 other circuits

De rate 0.78

Table 2.1

Table 9

Table 24

4

2

8

10A outlets

Factory

16A TPS V90 installed with 3 other circuits spaced

De rate 0.87 item22

Table 2.1

Table 9

Table 24

4

2

8

Cable designation

Maximum demand

Installation

parameters

AS/NZS 3008

Table no

Column

no

3 phase outlets

20A 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

In Toilets

0.8A 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

4

16

Cable Size 20A 3Ø outlets Distribution board Unit 1

AmmV

IL

VdVc

/8.122038

735.91000

1000

=××

=

××

=

MSBUnit 1

DBU/G

turret

MainsSub-main

20A3Ø outlet

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 To satisfy voltage drop requirements

Table 42 a 4 mm² Cable with a Vc value of 9.71 mV/Am value is required

29A cable4mm a 12/4 Table

4.2482.0

20

2 =

= A

Cable Size 20A 3Ø outlets Distribution board Unit 1

MSBUnit 1

DBU/G

turret

MainsSub-main

20A3Ø outlet

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

To satisfy Current carrying capacity, a 4mm² TPS cable is required

Cable Size 20A 3Ø outlets

MSBUnit 1

DBU/G

turret

MainsSub-main

20A3Ø outlet

Determine the cable size for the 20A 3 phase outlets 1 per circuit,

Route length 15 metres

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

• 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

mV/A.m...

mV/A.m..

IL

VdVc

69128606514

phase threeconvert to

65141038

57510001000

=×

=××

=×

=

10A single phase socket outlets

• Circuit 2 (30m route length)

• 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

mV/A.m.

mV/A.m.

IL

VdVc

07.1686056.18

phase threeconvert to

56.181030

57510001000

=×

=××

=×

=

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

• 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

mAmV

mAmVIL

VdVc

./25.390.86629.31

0.866Vc phase threeconvert to

./316.29538

57.510001000

=××

=××

=×

=

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

• 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

mAmV

mAmVIL

VdVc

./21.9250.866318.25


./328.25544

57.510001000

=××

=××

=×

=

Circuit 3 four battens and 2 EFToilets

• Route length 28m

• So from Table 42 1.5mm² cable is required• Table 9 column 6 (1.5mm² cable = 14A)

mAmV

mAmVIL

VdVc

./45.340.86678.39


./78.39528

57.510001000

=××

=××

=×

=

Circuit 4 (3x 18W) Fluorescent

• Route length 50m

• Voltage drop is not a factor for this circuit• Either 1mm² or 1.5mm² can be used

mAmV

mAVIL

VdVc

./63.2750.86628.318


./28.31836.50

57.510001000

=××

=××

=×

=

Circuit 5 (500W) sodium Vapour

• Route length 16m• TPS Cable V90 Installed enclosed U/G

DistributionBoard

500W Sodium vapour

mAmV

mAmVIL

VdVc

./8.1100.86698.127


./98.12772.216

57.510001000

=××

=××

=×

=

Voltage drop is not a factor for this circuitEither 1mm² or 1.5mm² can be used

Hot Water Cylinder3 phase 15A

• Route length 28m • Cable 3core + E enclosed TPS V90• Table 12 column 2 (2.5mm² = 23A)• Check Voltage drop

• Table 41 (2.5mm² = 15.6mm²)Therefore 2.5mm² Cable

23.17mV/Am1528

9.7351000

IL

Vd 1000Vc =

××

=×

=

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

mAmV

mAmVIL

VdVc

./118.240.86685.27


./85.27540

57.51000 1000

=××

=××

=×

=

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)

mAmV

mAmVIL

VdVc

./925.210.866318.25


./318.251022

57.510001000

=××

=××

=×

=

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 impedanceDistributor's networkFault current IA

Main Earth

MEN NeutralBar

Faulty equipmentSoil resistance high between electrodes

A

H POS

Determine maximum route lengthto 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

metres109

)5.24(105.225.716

5.242308.0

)(

8.0

3

=+××××

×××=

+××××

=

−

MaxL

L

SpeSphIaSpeSphUo

L

Max

Max ρ

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

L L L L

L P P P P P P HW L P P P

120A Main switch

MCB’S

10 10 10 10 10 16 16 16 20 20 20 16 10 16 16 16

Main Switch Board

Cable selection project Factory office installation.

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