Download presentation

Presentation is loading. Please wait.

Published byLukas Burton Modified over 3 years ago

1
LSEGG307A 9080F

2
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.

3
What is Maximum Demand? Is the Maximum continuous current that flows in a circuit for a period of 15 minutes or longer AS/NZS 3000 1.6.3 Why not just put a cable size in, that will handle the current rating for the maximum current rating of the appliance? MD

4
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

5
MD can be determined by: Calculation Assessment Measurement Limitation AS/NZS 3000 2.2.2. If the measured current is larger than any of the listed methods, then the MD is considered the measured value

6
… may be the ….sum of the current settings of the circuit breakers protecting the associated final sub-circuits…. AS/NZS 3000 Clause 2.2.2 (d)

7
10A L2 16A P1 16A P2 32A Stove 10A L1 MD = ? 10+10+16+16+32 = 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

8
So how do we calculate MD? Domestic Non-domestic Non- domestic Energy Demand Table C1 Table C2 Table C3

9
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 21 -25 = 2 A 5 A

10
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

11
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

12
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

13
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

14
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

15
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 =2400 33 + (3 x 3) = 42P 3P - 3 x 2 x 2 = 12P 12P 13.04A 3000 W 230V Determine the lighting load in Amps

16
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+ 13.04A = 22.04A

17
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 3 1.6 A B C 1.6

18
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

19
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

20
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

21
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

22
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 - - 3 Determine the Power point load in Amps

23
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

24
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

25
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

26
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

27
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

28
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 43P3 + 2 + 2 =7A

29
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 44 + 4 = 48P10 + 5 + 5 =20A

30
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

31
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

32
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

33
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

34
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 1.10.3

35
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

36
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 = (4100 230)x0.5 = (4600 230)x0.5 = 3600 230 = (5700x0.75) ( 3 x400) = 39.2A40.2A

37
Multiple Domestic Submains to individual units Columns 3, 4 or 5 MSB Unit 1 Unit 2 Unit 3 Column 2

38
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

39
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 26 + 4 = 30P3A + 2A = 26 x 2 = 52P 52 + 5 = 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

40
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

41
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 =

42
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 22 100A + (0.8A x 22) = 550A 66 Units 3 = 22 Units / Phase Use Colum5 Not counted 0.5A x 22 = 2.8A x 22 =

43
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 1.10.3

44
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 5 176.9A 5 Units / A & B Use Colum3 (2400 230V) x 5 =

45
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 (3600 230) x 0.75 x 6 177.2A 6 Units / C Use Colum4 6A x 6 = 5A + (0.25A x 6) = 2.8A x 6 =

46
16 Units 3 = 5.33 Units / Phase A 5 Units Col 3 B C 5 Units6 Units Col 3Col 4 176.6A 177.2A

47
Communal Load Loads that are used by all the unit dwellers Driveway/ parking lighting Stairwell lighting Power points for cleaners Lifts Swimming pool pumps

48
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 (3600 230) x 0.5 = 83.33A 15 x 1.25 = Table C2 Colum 2 ((20x80) 230) + ((30x 60) 230) = (3600 230) x 0.75 = These circuits could also be divided across three phases

49
Socket outlets 20 x 10A single power points On each Phase 1000W +19 x 750W = In a Factory 15.25kW 15250 ÷ 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

50
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

51
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

52
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 C2.5.2.2 (b) 20A + 20A = 40 Amps Per Phase

53
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 = 63.750kVA 3 x 400 = 92 Amps/Phase

Similar presentations

OK

Demand Resource Operable Capacity Analysis – Assumptions for FCA 5.

Demand Resource Operable Capacity Analysis – Assumptions for FCA 5.

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on bond length and bond Ppt on poverty and hunger in india Ppt on water activity for kids Ppt on op amp circuits examples Unified communications microsoft ppt online Ppt on limits and continuity pdf Ppt on internet as a post office Ppt on tourism industry in india Disaster management ppt on uttarakhand Ppt on air pollution control act