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LSEGG307A 9080F. Describe the acceptable methods for determining the maximum demand on an installation’s consumer’s mains. Calculate the maximum demand.

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Presentation on theme: "LSEGG307A 9080F. Describe the acceptable methods for determining the maximum demand on an installation’s consumer’s mains. Calculate the maximum demand."— Presentation transcript:

1 LSEGG307A 9080F

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

3 Maximum Demand Non-Domestic Load are generally heavier Stay on for longer periods of time Heating & cooling has an effect on load

4 Residential Instillations Hotels Boarding Houses Hospitals Motels etc AS/NZS 3000 Table C2 Colum 2

5 Residential Instillations AS/NZS 3000 Table C2 Colum 2 Lighting Load group A 75% connected load Note B Incandescent Lamps = 60W Unless the fitting restricts the wattage to less Track lights = 0.5A/m

6 Residential Installations 10A Power Points Load group B If this section of building DOES NOT HAVE fixed heating/cooling Load Group B i 20 Double power points 20 x 2 = 40 Points1000W + (39 x 400W) = First Point Balance  230 = 72.2A

7 Residential Installations 10A Power Points Load group B If this section of building DOES NOT HAVE fixed heating/cooling Load Group B i 20 Double power points 20 x 2 = 40 Points1000W + (39 x 100W) = First Point Balance 4900  230 = 72.2A HAS Load Group B ii 21.3A 1000W + (39 x 400W) =  230 =

8 Residential Installations Load group B iii 1 st = full current of highest 2 x 20A power points 2 x 15A power points Power Points Larger than 10A 2 nd = 50% x full current 3 rd = 50% x full current 20A 0.5 x 20 = 10A 0.5 x 15A = 7.5A 45A

9 Residential Installations Power Points 3  Outlets Load Group B 10A 3  Outlet AA 1 Point BB CC 1 Point1 Point

10 Exercise 1 30 x Light points 40 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 10A 3  outlet An air conditioned motel consists of the following: A A A B ii B iii B ii ((30x60)  230)x0.75 = 5.87A 17.4A 23.3A AA BB CC AA BB CC AA BB CC 7.83A 21.7A 7.83A 34.3A ((6x400)  230)x0.75 = 31P (1000+(30x100))  230 = 11+30P (1000+(40x100))  230 = (1000  230)+15+((2x15)x0.5) = 29.5A42.1A ((40x60)  230)x0.75 = NSW S&IR Clause Less than 25A difference

11 Factories Shops Offices Schools Churches AS/NZS 3000 Table C2 Colum 3

12 Factories AS/NZS 3000 Table C2 Colum 3 Lighting Load group A 100% connected load Note B Incandescent Lamps = 60W Unless the fitting restricts the wattage to less Track lights = 0.5A/m

13 Factories 10A Power Points Load group B If this section of building DOES NOT HAVE fixed heating/cooling Load Group B i 20 Double power points 20 x 2 = 40 Points1000W + (39 x 750W) = First Point Balance  230 = 131.5A  Calculation  Assessment  Measurement  Limitation AS/NZS

14 Factories 10A Power Points Load group B If this section of building DOES NOT HAVE fixed heating/cooling Load Group B i 20 Double power points 20 x 2 = 40 Points1000W + (39 x 100W) = First Point Balance 4900  230 = 131.5A HAS Load Group B ii 21.3A 1000W + (39 x 750W) =  230 =

15 Factories Load group B iii 1 st = full current of highest 2 x 20A power points 2 x 15A power points Power Points Larger than 10A 2 nd = 75% x full current 3 rd = 75% x full current 20A 0.75 x 20 = 15A 0.75 x 15A = 11.25A 57.5A

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

17 Used to give a “Ball Park” figure of maximum demand when actual loads are not available in the initial design stages Values are given in VA per m 2 of floor area

18 10 metre x 10 metre air conditioned shop From C3 10 x 10 =100 m 2 Light & power (average value)=70 VA/m 2 Air conditioning=30 VA/m VA/m m 2 x100 VA/m 2 =10kVA 400 x √3 = 14.4A / 

19 Single items on circuit Multiple items on circuit Diversity may be applied Current rating of appliance Limit by circuit breaker Welders Domestic cooking appliances Final Sub-circuits

20 Arc Welding Machines Stick MIG TIG Types 1 st Machine 2 nd 3 rd 4 th 5 th + more 100% 85% 70% 60% Highest Lowest Rated Primary Current Actual Primary Current

21 Arc Welding Machines Resistance Welding Machines Stick MIG TIG Spot welders Types Rated Primary Current Actual Primary current Duty Cycle

22 Individual Machines Variable Operation Specific (fixed) Operation Seam welders Auto feed welders Manual welders Non-Auto Welders 70% 50% x Rated Primary Current Actual Primary current x Multiplier

23 Very few welders operate at 100% all of the time Changing the electrodes Repositioning the job Starting a new job Ratio of ON TIME/OFF TIME in 1 minute Weld for 30 seconds Off for 30 seconds 30/30 =50% Duty Cycle

24 Individual Machines Groups of machines Variable Operation Specific (fixed) Operation Seam welders Auto feed welders Manual welders Non-Auto Welders 70% 50% x Rated Primary Current Actual Primary current x Multiplier 1 st Machine 2 nd + More 100% 60% Highest Lowest

25 Table C4 4.8kW wall oven = 6.8kW wall oven = 9kW wall oven + 4.8kW hotplate = 16A 20A 13.8kW = 40A

26 TEST


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