04 maximum demand

ELECTRICAL INSTALLATION PLANNING

LSEGG307A9080F

Lesson Content

• 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.

What is Maximum Demand?

Is the “Maximum continuous current that flows in a circuit for a period of 15 minutes or longer”

AS/NZS 30001.6.3

Why not just put a cable size in, that will handle the current rating for the maximum current rating

of the appliance?

MD

What is Maximum Demand?Domestic stove rated at 5kW

Max current = 22A

Min cable size = 6mm2

By calculation

Max current = 16A

AS/NZS 3000Table C4

Min cable size = 2.5mm2

MD can be determined by:

Calculation Assessment Measurement Limitation

AS/NZS 30002.2.2.

“If the measured current is larger than any of the listed methods, then the MD is considered the

measured value”

MD of Mains & Submains

“… may be the ….sum of the current settings of the circuit breakers protecting the

associated final sub-circuits….”

AS/NZS 3000Clause 2.2.2 (d)

MD of Mains & Submains

10AL2

16AP1

16AP2

32AStove

10AL1

MD = ?

10+10+16+16+32 =84 Amps

By Calculation

36 Amps

Domestic House

10mm2 2.5mm2

Most su

pply authorit

ies specif

y a minim

um

mains cable si

ze of 1

6 mm

2

20 metres 46 metres

16mm2

So how do we calculate MD?

Domestic Non-domestic Non- domestic Energy Demand

Table C1Table C2

Table C3

So how do we calculate MD?Single House Table C1Column

2LightingLoad groups Ai &

AiiBatten holder

=1 light point

25 lighting points = 1 -20 =

3 A

21 -25 =

2 A

5 A



AiiChandelier or Multi-globe fitting =

Number of points =

= 3 points

Number of globes



AiiPower points above 2.3 m for lighting =

1 point

Note “e”

Exhaust fans below 150W =

1 point



AiiTrack lights =

Every metre =

Note “d”2 points

2 x 2.6 metres track light =

5 points per track =

10 points



AiiIXL Tastics, Heat lamps etc

Not regarded as “Lighting”Fixed Space HeatingLoad Group “D”



AiiOutside Lighting

Floodlighting, Swimming Pool,

Tennis courts, etc

Not lighting around house walls

Load Group “Aii”

If total bigger than 1000W

If less than 1000W

Load Group “Ai”

CalculationsA house consists of the following:

36 light fittings, 3 consisting of 3 lamp combinations3 bathroom exhaust fans

2 x 750W heat lamp, fan, light combinations3 x 2metre track lights

12 wall lights around the veranda6 x 100W bollard lights on driveway

6 x 400W metal halide flood lights for pool

Ai

Ai

DAi

Ai

Aii

Aii

Assume to be <150W

6 x 100 = 600

6 x 400 = 2400

33 + (3 x 3) = 42P3P-3 x 2 x 2 = 12P12P

13.04A

3000 W

230V

Determine the lighting load in Amps






Ai

Ai

DAi

Ai

Aii

Aii

33 + (3 x 3) = 42P3P-3 x 2 x 2 = 12P12P

13.04A

42P3P12P12P

69P

1 – 20 =21 – 40 =41 – 60 =61 – 69 =

3A2A2A2A

9A + 13.04A =22.04A

3 Light Calculations

36 light fittings

Should indicate how it is to be broken up across the three phases

Broken up evenly

312 Fittings / 3A

NOT

A B C3A 3A

36 light fittings = 3A + 2A = 5A

31.6

A B C1.6 1.6






Ai

Ai

DAi

Ai

Aii

Aii

11 + (1 x 3) = 14P1P-1 x 2 x 2 = 4P4P

4.35A

14P1P4P4P

23P

1 – 20 =21 – 23 =

3A2A

5A + 4.35A = 9.35A /

Broken up evenly over 3 Phases

Divide by 3

Load Group Bi

Power Points

Double Power Points Notes h

36 Double Power Points6 Single power points

36 x 2 =72P6P

78P1 – 20 =

21 – 40= 41 – 60 =61 – 78 =

10A5A5A5A

25A

Load Group Bi

Power Points

Double Power Points Notes h

36 Double Power Points6 Single power points

36 x 2 =72P6P

78P

1 – 20 =21 – 40= 41 – 60 =61 – 78 =

10A5A5A5A

25AOther Direct Wired

Equipment that is <10A• Air conditioners• Cook tops• Water heaters

AND

Load Group Bii

Power Points15 A Power PointsNOT

Other Equipment that is Covered in Load Groups

below

C Ranges, Laundry equipmentD Fixed Space Heating or cooling

equipmentE & F Water HeatersG Spa & Swimming Pool Heaters15 A Power Points suppling this equipment is covered

under each load group


20 Double power points8 Single power points

8A swimming pool pump4 x 15A power points

12A Air-conditioner supplied via a 15A power point14A Pool heater direct wired

Bi

Bi

Bi

Bii

DG

20 x 2 = 40P8P1P10A--

3

Determine the Power point load in Amps

20 Double power points8 Single power points

8A swimming pool pump3 x 15A power points

12A Air-conditioner supplied via a 15A power point14A Pool heater direct wired


Bi

Bi

Bi

Bii

DG

20 x 2 = 40P8P1P10A--

40P8P1P

49P

1 – 20 =21 – 40 =41 – 49 =

10A5A5A

20A + 10A = 30A

Load Group C• 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 Bi, Bii or Biii

An installation has:2.3 kW Cook top4 kW Wall oven 6.3 kW of Load Group C

(6300/230) 13.7Ax 0.5 =

50% of connected Load

Load Group D• 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 Bi, Bii or Biii

75% of connected LoadIf 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

Load Group E &FWater 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 heaterwith a 2.4kW element

2400/300 = 8 W/L

15 litre heaterwith a 2.4kW element

2400/15 = 160 W/L

Maximum Demand41 x 50 W down lights2 x 60 W exhaust fans4 x Single 10 A socket outlets22 x Double 10 A socket outlets1 x 10kW range1 x Permanently connected air conditioner (full load current of 13 A)1 x 4.6 kW continuous HWS

A house consists of the following:Ai

Ai

Bi

Bi

CDF



Ai

Bi

Bi

CDF

43P 3 + 2 + 2 = 7A



Ai

Bi

Bi

CDF

7A

44P44 + 4 = 48P 10 + 5 + 5 = 20A



Ai

Bi

Bi

CDF

7A

20A

(10000/230) x 0.5 = 21.7A



Ai

Bi

Bi

CDF

7A

20A

13 x 0.75 =

21.7A9.75A



Ai

Bi

Bi

CDF

7A

20A

4600/230 =

21.7A9.75A20A



Ai

Bi

Bi

CDF

7A

20A

21.7A9.75A20A

78.45A

Uneven 3 LoadsNot all single phase loads will break up evenly over

three phasesBut the phases must not be unbalanced by more than

25A

NSW S&IRClause 1.10.3

3 Maximum Demand30 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


Ai

Aii

Bi

Bi

Bii

C

C

F

D

A

B

C

A

B

C

A

B

C

A

B

C

3 Maximum Demand30 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


Ai

Aii

Bi

Bi

Bii

C

C

F

D

32P 3+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 =

30P 10+5A =

10+30P 10+5A =

(4100230)x0.5 =

(4600230)x0.5 =

3600230 =

(5700x0.75)(√3 x400) =

39.2A 40.2A

Multiple DomesticSubmains to individual units

Columns 3, 4 or 5

MSB

Unit 1

Unit 2

Unit 3

Column 2

Questions to be Asked?

How many units are there?

How are they spread across the 3 phases?

6 Units

3 = 2 Units / Phase Use Colum 3

18 Units


66 Units


Maximum Demand

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:

Ai

Ai

Aii

Bi

Bi

Bii

C

D

F

2 x 2 = 4P

1500W

26 + 4 = 30P 3A + 2A =

26 x 2 = 52P52 + 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

Multiple Domestic

26 x Light points









9 Units consists of the following:

Ai

Ai

Aii

Bi

Bi

Bii

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


Not counted

Multiple Domestic

26 x Light points










Ai

Ai

Aii

Bi

Bi

Bii

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


Not counted

5A + (0.25A x 10) =

2.8A x 10 =

Multiple Domestic

26 x Light points










Ai

Ai

Aii

Bi

Bi

Bii

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


Not counted

0.5A x 22 =

2.8A x 22 =

What Happens If The Number Of Units Is Not

Devisable By 3?16 Units

3 = 5.33 Units / Phase

A5 Units

Col 3

But the phases must not be unbalanced by more than 25A

B C5 Units 6 Units

Col 3 Col 4

NSW S&IRClause 1.10.3

Multiple Domestic

26 x Light points








Ai

Bi

Bi

Bii

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 Colum 3

(2400230V) x 5 =

Multiple Domestic

26 x Light points








Ai

Bi

Bi

Bii

C

D

F

15A + (3.75A x 6) =

6.5A

37.5A

10A

16.8A

70.4A

36A

(3600230) x 0.75 x 6

177.2A

6 Units / C Use Colum 4

6A x 6 =

5A + (0.25A x 6) =

2.8A x 6 =

What Happens If The Number Of Units Is Not

Devisable By 3?16 Units

3 = 5.33 Units / Phase

A5 Units

Col 3

B C5 Units 6 Units

Col 3 Col 4

176.6A 176.6A 177.2A

Communal LoadLoads that are used by all the unit dwellers

• Driveway/ parking lighting

• Stairwell lighting

• Power points for cleaners

• Lifts

• Swimming pool pumps

Multiple Domestic

20 x 80 W bollard Lights

30 x light points


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

(3600230) x 0.5 =

83.33A

15 x 1.25 =

Table C2 Colum 2

((20x80)230) + ((30x 60)230) =

(3600230) x 0.75 =

These circuits could also be divided across three phases

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

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

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

Welders 3 x 20A 3 Welders

Majority of welders only use 2 phases

W 1 W 2

W 3

A B C

Only 2 welders on each phase

AS/NZS 3000C2.5.2.2 (b)

20A + 20A = 40 Amps Per Phase

63.750kVA

Floor area =

Light/Power =Reverse cycle AC =

Table C3Use for a ‘Ball Park’ figure in early planning stages.

Table based on a temperate climate

Office 850m2

50 VA/m2

25 VA/m2

75 VA/m2 x 850 =

63.750kVA

√3 x 400= 92

Amps/Phase

04 maximum demand

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