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LSEGG307A 9080F

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Describe the acceptable methods for determining the maximum demand on an installations consumers mains. Calculate the maximum demand for the consumer's mains for given installations up to 400 A per phase.

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What is Maximum Demand? Is the Maximum continuous current that flows in a circuit for a period of 15 minutes or longer AS/NZS Why not just put a cable size in, that will handle the current rating for the maximum current rating of the appliance? MD

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What is Maximum Demand? Domestic stove rated at 5kW Max current = 22A Min cable size = 6mm 2 By calculation Max current = 16A AS/NZS 3000 Table C4 Min cable size = 2.5mm 2

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MD can be determined by: Calculation Assessment Measurement Limitation AS/NZS If the measured current is larger than any of the listed methods, then the MD is considered the measured value

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… may be the ….sum of the current settings of the circuit breakers protecting the associated final sub-circuits…. AS/NZS 3000 Clause (d)

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10A L2 16A P1 16A P2 32A Stove 10A L1 MD = ? = 84 Amps By Calculation 36 Amps Domestic House 10mm 2 2.5mm 2 M o s t s u p p l y a u t h o r i t i e s s p e c i f y a m i n i m u m m a i n s c a b l e s i z e o f 1 6 m m 2 20 metres46 metres 16mm 2

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So how do we calculate MD? Domestic Non-domestic Non- domestic Energy Demand Table C1 Table C2 Table C3

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So how do we calculate MD ? Single House Table C1Column 2 Lighting Load groups A i & A ii Batten holder = 1 light point 25 lighting points =1 -20 = 3 A = 2 A 5 A

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So how do we calculate MD ? Single House Table C1Column 2 Lighting Load groups A i & A ii Chandelier or Multi-globe fitting = Number of points = = 3 points Number of globes

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So how do we calculate MD ? Single House Table C1Column 2 Lighting Load groups A i & A ii Power points above 2.3 m for lighting = 1 point Note e Exhaust fans below 150W = 1 point

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So how do we calculate MD ? Single House Table C1Column 2 Lighting Load groups A i & A ii Track lights = Every metre = Note d 2 points 2 x 2.6 metres track light =5 points per track = 10 points

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So how do we calculate MD ? Single House Table C1Column 2 Lighting Load groups A i & A ii IXL Tastics, Heat lamps etc Not regarded as Lighting Fixed Space Heating Load Group D

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So how do we calculate MD ? Single House Table C1Column 2 Lighting Load groups A i & A ii Outside Lighting Floodlighting, Swimming Pool, Tennis courts, etc Not lighting around house walls Load Group A ii If total bigger than 1000W If less than 1000W Load Group A i

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A house consists of the following: 36 light fittings, 3 consisting of 3 lamp combinations 3 bathroom exhaust fans 2 x 750W heat lamp, fan, light combinations 3 x 2metre track lights 12 wall lights around the veranda 6 x 100W bollard lights on driveway 6 x 400W metal halide flood lights for pool AiAi AiAi D AiAi AiAi A ii Assume to be <150W 6 x 100 =600 6 x 400 = (3 x 3) = 42P 3P - 3 x 2 x 2 = 12P 12P 13.04A 3000 W 230V Determine the lighting load in Amps

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A house consists of the following: 36 light fittings, 3 consisting of 3 lamp combinations 3 bathroom exhaust fans 2 x 750W heat lamp, fan, light combinations 3 x 2metre track lights 12 wall lights around the veranda 6 x 100W bollard lights on driveway 6 x 400W metal halide flood lights for pool A i D A i A ii 33 + (3 x 3) = 42P 3P - 3 x 2 x 2 = 12P 12P 13.04A 42P 3P 12P 69P 1 – 20 = 21 – 40 = 41 – 60 = 61 – 69 = 3A 2A 9A A = 22.04A

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36 light fittings Should indicate how it is to be broken up across the three phases Broken up evenly 3 12 Fittings / 3A NOT A B C 3A 36 light fittings= 3A + 2A = 5A A B C 1.6

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A house consists of the following: 36 light fittings, 3 consisting of 3 lamp combinations 3 bathroom exhaust fans 2 x 750W heat lamp, fan, light combinations 3 x 2metre track lights 12 wall lights around the veranda 6 x 100W bollard lights on driveway 6 x 400W metal halide flood lights for pool AiAi AiAi D AiAi AiAi A ii 11 + (1 x 3) = 14P 1P - 1 x 2 x 2 = 4P 4P 4.35A 14P 1P 4P 23P 1 – 20 = 21 – 23 = 3A 2A 5A+ 4.35A = 9.35A / Broken up evenly over 3 Phases Divide by 3

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Power Points Double Power Points Notes h 36 Double Power Points 6 Single power points 36 x 2 =72P 6P 78P 1 – 20 = 21 – 40= 41 – 60 = 61 – 78 = 10A 5A 25A

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Power Points Double Power Points Notes h 36 Double Power Points 6 Single power points 36 x 2 =72P 6P 78P 1 – 20 = 21 – 40= 41 – 60 = 61 – 78 = 10A 5A 25A Other Direct Wired Equipment that is <10A Air conditioners Cook tops Water heaters AND

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Power Points 15 A Power Points NOT Other Equipment that is Covered in Load Groups below CRanges, Laundry equipment D Fixed Space Heating or cooling equipment E & F Water Heaters G Spa & Swimming Pool Heaters 15 A Power Points suppling this equipment is covered under each load group

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A house consists of the following: 20 Double power points 8 Single power points 8A swimming pool pump 4 x 15A power points 12A Air-conditioner supplied via a 15A power point 14A Pool heater direct wired BiBi BiBi BiBi B ii D G 20 x 2 = 40P 8P 1P 10A Determine the Power point load in Amps

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20 Double power points 8 Single power points 8A swimming pool pump 3 x 15A power points 12A Air-conditioner supplied via a 15A power point 14A Pool heater direct wired A house consists of the following: B i B ii D G 20 x 2 = 40P 8P 1P 10A - 40P 8P 1P 49P 1 – 20 = 21 – 40 = 41 – 49 = 10A 5A 20A+ 10A = 30A

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Cook tops Wall ovens Washing machines Clothes driers Must be over 10A If supplied via a power point That power point is NOT included in B i, B ii or B iii An installation has: 2.3 kW Cook top 4 kW Wall oven 6.3 kW of Load Group C (6300/230)13.7Ax 0.5 = 50% of connected Load

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IXL Tastics Heat lamps Air conditioners Heaters Under floor heating Must be over 10A If supplied via a power point That power point is NOT included in B i, B ii or B iii 75% of connected Load If reverse cycle Air-conditioner is used only the highest system is to be taken into account Unless it is multi zone, when both heating and cooling could operate simultaneously

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Water Heaters If greater than 100W/L Load Group E 33.3% of connected load Load Group F 100% of connected load Instantaneous Quick recovery Storage Off peak 300 litre heater with a 2.4kW element 2400/300 =8 W/L 15 litre heater with a 2.4kW element 2400/15 =160 W/L

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41 x 50 W down lights 2 x 60 W exhaust fans 4 x Single 10 A socket outlets 22 x Double 10 A socket outlets 1 x 10kW range 1 x Permanently connected air conditioner (full load current of 13 A) 1 x 4.6 kW continuous HWS A house consists of the following: AiAi AiAi BiBi BiBi C D F

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41 x 50 W down lights 2 x 60 W exhaust fans 4 x Single 10 A socket outlets 22 x Double 10 A socket outlets 1 x 10kW range 1 x Permanently connected air conditioner (full load current of 13 A) 1 x 4.6 kW continuous HWS A house consists of the following: AiAiBiBiCDFAiAiBiBiCDF 43P =7A

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41 x 50 W down lights 2 x 60 W exhaust fans 4 x Single 10 A socket outlets 22 x Double 10 A socket outlets 1 x 10kW range 1 x Permanently connected air conditioner (full load current of 13 A) 1 x 4.6 kW continuous HWS A house consists of the following: AiAiBiBiCDFAiAiBiBiCDF 7A 44P = 48P =20A

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41 x 50 W down lights 2 x 60 W exhaust fans 4 x Single 10 A socket outlets 22 x Double 10 A socket outlets 1 x 10kW range 1 x Permanently connected air conditioner (full load current of 13 A) 1 x 4.6 kW continuous HWS A house consists of the following: AiAiBiBiCDFAiAiBiBiCDF 7A 20A (10000/230) x 0.5 =21.7A

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41 x 50 W down lights 2 x 60 W exhaust fans 4 x Single 10 A socket outlets 22 x Double 10 A socket outlets 1 x 10kW range 1 x Permanently connected air conditioner (full load current of 13 A) 1 x 4.6 kW continuous HWS A house consists of the following: AiAiBiBiCDFAiAiBiBiCDF 7A 20A 13 x 0.75 = 21.7A 9.75A

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41 x 50 W down lights 2 x 60 W exhaust fans 4 x Single 10 A socket outlets 22 x Double 10 A socket outlets 1 x 10kW range 1 x Permanently connected air conditioner (full load current of 13 A) 1 x 4.6 kW continuous HWS A house consists of the following: AiAiBiBiCDFAiAiBiBiCDF 7A 20A 4600/230 = 21.7A 9.75A 20A

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41 x 50 W down lights 2 x 60 W exhaust fans 4 x Single 10 A socket outlets 22 x Double 10 A socket outlets 1 x 10kW range 1 x Permanently connected air conditioner (full load current of 13 A) 1 x 4.6 kW continuous HWS A house consists of the following: AiAiBiBiCDFAiAiBiBiCDF 7A 20A 21.7A 9.75A 20A 78.45A

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Not all single phase loads will break up evenly over three phases But the phases must not be unbalanced by more than 25A NSW S&IR Clause

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30 x Light points + 2 x Exhaust fans 18 x Lights 6 x 400W mercury vapour flood lights 15 x Double power points (p/p) 10 x Single p/p & 15 x Double p/p 3 x 15A Single p/p 1 x 4.1kW wall oven 1 x 4.6kW cook top 1 x 3.6kW heat pump HWS 1 x 5.7kW 3 Air conditioner A house consists of the following: AiAi AiAi A ii BiBi BiBi B ii C C F D A B C A B C A B C A B C

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30 x Light points + 2 x Exhaust fans 18 x Lights 6 x 400W mercury vapour flood lights 15 x Double power points (p/p) 10 x Single p/p & 15 x Double p/p 3 x 15A Single p/p 1 x 4.1kW wall oven 1 x 4.6kW cook top 1 x 3.6kW heat pump HWS 1 x 5.7kW 3 Air conditioner A house consists of the following: A i A ii B i B ii C F D 32P3+2A = 5A 15A 8.91A 6.17A 35.1A A B C A B C A B C A B C 3A 20A 10A 6.17A 7.83A 10A 15.65A 6.17A 18P ((6x400) 230)x0.75 = 30P10+5A = 10+30P10+5A = ( )x0.5 = ( )x0.5 = = (5700x0.75) ( 3 x400) = 39.2A40.2A

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Multiple Domestic Submains to individual units Columns 3, 4 or 5 MSB Unit 1 Unit 2 Unit 3 Column 2

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How many units are there? How are they spread across the 3 phases? 6 Units 3 = 2 Units / Phase Use Colum3 18 Units 3 = 6 Units / Phase Use Colum4 66 Units 3 = 22 Units / Phase Use Colum5

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26 x Light points 2 metres of light track 2 x 750W flood lights 26 x Double 10 A socket outlets 5 x Single 10 A socket outlets 1 x 15A Single socket outlet 1 x 5.6kW wall oven / cook top 1 x 3.6kW air conditioner 1 x 2.4kW storage HWS (Heat pump) A house consists of the following: AiAi AiAi A ii BiBi BiBi B ii C D F 2 x 2 = 4P 1500W = 30P3A + 2A = 26 x 2 = 52P = 57P 10A + 5A + 5A = (5600W/230V)x0.5 5A 4.9A 20A 10A 12.2A 11.7A 10.4A (3600W/230V)x0.75 (2400W/230V) 74.2A (1500W/230V)x0.75

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26 x Light points 2 metres of light track 2 x 750W flood lights 26 x Double 10 A socket outlets 5 x Single 10 A socket outlets 1 x 15A Single socket outlet 1 x 5.6kW wall oven / cook top 1 x 3.6kW air conditioner 1 x 2.4kW storage HWS (Heat pump) 9 Units consists of the following: A i A ii B i B ii C D F 10A + (5A x 3) = 6A 25A 10A 15A 35.2A 18A (3600W/230V) x 0.75 x 3 6A x 3 = 109A 9 Units 3 = 3 Units / Phase Use Colum3 Not counted

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26 x Light points 2 metres of light track 2 x 750W flood lights 26 x Double 10 A socket outlets 5 x Single 10 A socket outlets 1 x 15A Single socket outlet 1 x 5.6kW wall oven / cook top 1 x 3.6kW air conditioner 1 x 2.4kW storage HWS (Heat pump) 30 Units consists of the following: A i A ii B i B ii C D F 15A + (3.75A x 10) = 7.5A 52.5A 10A 28A 117A 60A (3600W/230V) x 0.75 x 10 6A x 10 = 275A 30 Units 3 = 10 Units / Phase Use Colum4 Not counted 5A + (0.25A x 10) = 2.8A x 10 =

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26 x Light points 2 metres of light track 2 x 750W flood lights 26 x Double 10 A socket outlets 5 x Single 10 A socket outlets 1 x 15A Single socket outlet 1 x 5.6kW wall oven / cook top 1 x 3.6kW air conditioner 1 x 2.4kW storage HWS (Heat pump) 66 Units consists of the following: A i A ii B i B ii C D F 50A + (1.9A x 22) = 11A 91.8A 10A 61.6A 258A 118A (3600W/230V) x 0.75 x A + (0.8A x 22) = 550A 66 Units 3 = 22 Units / Phase Use Colum5 Not counted 0.5A x 22 = 2.8A x 22 =

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16 Units 3 = 5.33 Units / Phase A 5 Units Col 3 But the phases must not be unbalanced by more than 25A B C 5 Units6 Units Col 3Col 4 NSW S&IR Clause

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26 x Light points 26 x Double 10 A socket outlets 5 x Single 10 A socket outlets 1 x 15A Single socket outlet 1 x 5.6kW wall oven / cook top 1 x 3.6kW air conditioner 1 x 2.4kW storage HWS (Heat pump) 16 Units consists of the following: A i B i B ii C D F 10A + (5A x 5) = 6A 35A 10A 15A 58.7A 52.2A (3600W/230V) x 0.75 x A 5 Units / A & B Use Colum3 ( V) x 5 =

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26 x Light points 26 x Double 10 A socket outlets 5 x Single 10 A socket outlets 1 x 15A Single socket outlet 1 x 5.6kW wall oven / cook top 1 x 3.6kW air conditioner 1 x 2.4kW storage HWS (Heat pump) 16 Units consists of the following: A i B i B ii C D F 15A + (3.75A x 6) = 6.5A 37.5A 10A 16.8A 70.4A 36A ( ) x 0.75 x A 6 Units / C Use Colum4 6A x 6 = 5A + (0.25A x 6) = 2.8A x 6 =

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16 Units 3 = 5.33 Units / Phase A 5 Units Col 3 B C 5 Units6 Units Col 3Col A 177.2A

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Communal Load Loads that are used by all the unit dwellers Driveway/ parking lighting Stairwell lighting Power points for cleaners Lifts Swimming pool pumps

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20 x 80 W bollard Lights 30 x light points 15 x Single 10 A socket outlets 1 x 3.6kW HWS for cleaners room 1 x 3.6kW Air conditioner for lobby 1 x 12A pool pump/filter 1 x 15A lift motor 16 Units have a communal load consists of the following: H H I Ji Jii D E 2A x15= 30 but maximum of 15A 14.78A 15A 7.8A 15A 12A 18.75A ( ) x 0.5 = 83.33A 15 x 1.25 = Table C2 Colum 2 ((20x80) 230) + ((30x 60) 230) = ( ) x 0.75 = These circuits could also be divided across three phases

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Socket outlets 20 x 10A single power points On each Phase 1000W +19 x 750W = In a Factory 15.25kW ÷ 230 = 66.3 Amps 20 x 10A single power points 1000W +19 x 100W = In areas with air conditioning 2.9kW 2900 ÷ 230 = 12.6 Amps

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Socket outlets 5 x 10A 3 outlets 4kW Per Phase Treat as B(i) 1000W + 4 x 750W = 17.4 Amps 4000 ÷ 230 = 2 x 32A 3 outlets 5 x 20A 3 outlets 32A + 75% x 32A 24A + 75% x 20A x 5 75A = 131Amps Per Phase

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Motors 2 x 30A 3 motors 3 x 15A 1 motors 1 per phase In Factories 30A + 75 Amps 75% x 30A 22.5A +22.5A = 50% x 3 x 15A

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Welders 3 x 20A 3 Welders Majority of welders only use 2 phases W 1W 2 W 3 A B C Only 2 welders on each phase AS/NZS 3000 C (b) 20A + 20A = 40 Amps Per Phase

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63.750kVA Floor area = Light/Power = Reverse cycle AC = Table C3 Use for a Ball Park figure in early planning stages. Table based on a temperate climate Office 850m 2 50 VA/m 2 25 VA/m 2 75 VA/m 2 x 850 = kVA 3 x 400 = 92 Amps/Phase

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