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ELECTRICAL INSTALLATION PLANNING. LSEGG307A 9080F. 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. MD.

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lesson content
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.
slide3

MD

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?

slide4
What is Maximum Demand?

Domestic stove rated at 5kW

Max current = 22A

Min cable size = 6mm2

By calculation

AS/NZS 3000

Table C4

Max current = 16A

Min cable size = 2.5mm2

md can be determined by

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”

md of mains submains
MD of Mains & Submains

“… 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)

md of mains submains7
MD of Mains & Submains

10+10+16+16+32 =

By Calculation

16mm2

84 Amps

36 Amps

MD = ?

Most supply authorities specify a minimum mains cable size of 16 mm2

2.5mm2

10mm2

20 metres

46 metres

10A

L1

10A

L2

16A

P1

16A

P2

32A

Stove

Domestic House

so how do we calculate md
So how do we calculate MD?

Table C1

Table C2

Table C3

  • Domestic
  • Non-domestic
  • Non- domestic Energy Demand
so how do we calculate md9
So how do we calculate MD?

Table C1

Column 2

Lighting

Load groups Ai & Aii

Batten holder =

1 light point

25 lighting points =

1 -20 =

3 A

21 -25 =

2 A

5 A

Single House

so how do we calculate md10
So how do we calculate MD?

Table C1

Column 2

Lighting

Load groups Ai & Aii

Chandelier or Multi-globe fitting =

Number of points =

Number of globes

= 3 points

Single House

so how do we calculate md11
So how do we calculate MD?

Table C1

Column 2

Lighting

Load groups Ai & Aii

Power points above 2.3 m for lighting =

1 point

Note “e”

Exhaust fans below 150W =

1 point

Single House

so how do we calculate md12
So how do we calculate MD?

Table C1

Column 2

Lighting

Load groups Ai & Aii

Track lights =

Every metre =

2 points

Note “d”

2 x 2.6 metres track light =

5 points per track =

10 points

Single House

so how do we calculate md13
So how do we calculate MD?

Table C1

Column 2

Lighting

Load groups Ai & Aii

IXL Tastics, Heat lamps etc

Not regarded as “Lighting”

Fixed Space Heating

Load Group “D”

Single House

so how do we calculate md14
So how do we calculate MD?

Table C1

Column 2

Lighting

Load groups Ai & Aii

Outside Lighting

Floodlighting, Swimming Pool, Tennis courts, etc

Not lighting around house walls

If total bigger than 1000W

If less than 1000W

Load Group “Aii”

Single House

Load Group “Ai”

calculations
Calculations

A house consists of the following:

Assume to be <150W

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

Ai

Ai

D

Ai

Ai

Aii

Aii

33 + (3 x 3) = 42P

3P

-

3 x 2 x 2 = 12P

12P

13.04A

6 x 100 =

600

3000 W

6 x 400 =

2400

230V

Determine the lighting load in Amps

calculations16
Calculations

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

Ai

Ai

D

Ai

Ai

Aii

Aii

33 + (3 x 3) = 42P

3P

-

3 x 2 x 2 = 12P

12P

13.04A

42P

3P

12P

12P

1 – 20 =

21 – 40 =

41 – 60 =

61 – 69 =

3A

2A

2A

2A

22.04A

69P

9A

+ 13.04A =

3 light calculations
3 Light Calculations

Broken up evenly

ABC

36 light fittings

12 Fittings /

3A

3A

3A

3

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

NOT

ABC

36 light fittings

= 3A + 2A = 5A

1.6

1.6

1.6

3

calculations18
Calculations

A house consists of the following:

Broken up evenly over 3 Phases

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

Ai

Ai

D

Ai

Ai

Aii

Aii

11 + (1 x 3) = 14P

1P

-

1 x 2 x 2 = 4P

4P

4.35A

Divide by 3

14P

1P

4P

4P

1 – 20 =

21 – 23 =

3A

2A

9.35A /

23P

5A

+ 4.35A =

load group b i
Power Points

Notes

h

Double Power Points

Load Group Bi

36 Double Power Points

6 Single power points

36 x 2 =72P

6P

1 – 20 =

21 – 40=

41 – 60 =

61 – 78 =

10A

5A

5A

5A

78P

25A

load group b i20
Power Points

Notes

h

Double Power Points

Load Group Bi

36 Double Power Points

6 Single power points

36 x 2 =72P

6P

1 – 20 =

21 – 40=

41 – 60 =

61 – 78 =

10A

5A

5A

5A

AND

78P

Other Direct Wired Equipment that is <10A

25A

  • Air conditioners
  • Cook tops
  • Water heaters
load group b ii
Power Points

NOT

15 A Power Points

Other Equipment that is Covered in Load Groups below

Load Group Bii

C Ranges, 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

calculations22
Calculations

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

Bi

Bi

Bi

Bii

D

G

20 x 2 = 40P

8P

1P

10A

-

-

3

Determine the Power point load in Amps

calculations23
Calculations

A house consists of the following:

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

Bi

Bi

Bi

Bii

D

G

20 x 2 = 40P

8P

1P

10A

-

-

40P

8P

1P

1 – 20 =

21 – 40 =

41 – 49 =

10A

5A

5A

30A

49P

20A

+ 10A =

load group c
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

50% of connected Load

An installation has:

2.3 kW Cook top

4 kW Wall oven

6.3 kW of Load Group C

(6300/230)

x 0.5 =

13.7A

load group d
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 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

load group e f
Load Group E &F

Water Heaters

Load Group E

15 litre heater

with a 2.4kW element

33.3% of connected load

2400/15 =

160 W/L

  • Instantaneous
  • Quick recovery

If greater than 100W/L

Load Group F

300 litre heater

with a 2.4kW element

100% of connected load

  • Storage
  • Off peak

2400/300 =

8 W/L

maximum demand
Maximum Demand

A house consists of the following:

Ai

Ai

Bi

Bi

C

D

F

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

maximum demand28
Maximum Demand

A house consists of the following:

Ai

Ai

Bi

Bi

C

D

F

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

43P

3 + 2 + 2 =

7A

maximum demand29
Maximum Demand

A house consists of the following:

Ai

Ai

Bi

Bi

C

D

F

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

7A

44 + 4 = 48P

10 + 5 + 5 =

20A

44P

maximum demand30
Maximum Demand

A house consists of the following:

Ai

Ai

Bi

Bi

C

D

F

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

7A

20A

(10000/230) x 0.5 =

21.7A

maximum demand31
Maximum Demand

A house consists of the following:

Ai

Ai

Bi

Bi

C

D

F

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

7A

20A

21.7A

9.75A

13 x 0.75 =

maximum demand32
Maximum Demand

A house consists of the following:

Ai

Ai

Bi

Bi

C

D

F

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

7A

20A

21.7A

9.75A

4600/230 =

20A

maximum demand33
Maximum Demand

A house consists of the following:

Ai

Ai

Bi

Bi

C

D

F

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

7A

20A

21.7A

9.75A

20A

78.45A

uneven 3 loads
Uneven 3Loads

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

3 maximum deman d
3Maximum Demand

A house consists of the following:

Ai

Ai

  • Aii

Bi

Bi

  • Bii

C

C

F

D

A

  • B
  • C
  • A
  • B
  • C
  • A
  • B
  • C
  • A
  • B
  • C

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 15ASingle 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
3 maximum deman d36
3Maximum Demand

A house consists of the following:

Ai

Ai

  • Aii

Bi

Bi

  • Bii

C

C

F

D

A

  • B
  • C
  • A
  • B
  • C
  • A
  • B
  • C
  • A
  • B
  • C

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 15ASingle 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

32P

3+2A =

5A

15A

8.91A

6.17A

3A

20A

10A

6.17A

18P

7.83A

10A

15.65A

6.17A

((6x400)230)x0.75=

30P

10+5A =

10+30P

10+5A =

(4100230)x0.5=

(4600230)x0.5=

3600230=

(5700x0.75)(√3 x400)=

35.1A

39.2A

40.2A

multiple domestic
Multiple Domestic

Unit 1

MSB

Unit 2

Unit 3

Columns 3, 4 or 5

Column 2

Submains to individual units

questions to be asked
Questions to be Asked?
  • How many units are there?
  • How are they spread across the 3 phases?

6 Units

= 2 Units / Phase

Use Colum

3

3

18 Units

= 6 Units / Phase

Use Colum

4

3

66 Units

= 22 Units / Phase

Use Colum

5

3

maximum demand39
Maximum Demand

A house consists of the following:

Ai

Ai

  • Aii

Bi

Bi

  • Bii

C

D

F

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 15ASingle socket outlet

1 x 5.6kW wall oven / cook top

1 x 3.6kW air conditioner

1 x 2.4kW storage HWS (Heat pump)

26 + 4 = 30P

3A + 2A =

5A

4.9A

20A

10A

12.2A

11.7A

10.4A

2 x 2 = 4P

1500W

(1500W/230V)x0.75

10A + 5A + 5A =

26 x 2 = 52P

52 + 5 = 57P

(5600W/230V)x0.5

(3600W/230V)x0.75

(2400W/230V)

74.2A

multiple domestic40
Multiple Domestic

9 Units consists of the following:

9 Units

= 3 Units / Phase

Use Colum

3

3

Ai

Ai

  • Aii

Bi

Bi

  • Bii

C

D

F

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 15ASingle socket outlet

1 x 5.6kW wall oven / cook top

1 x 3.6kW air conditioner

1 x 2.4kW storage HWS (Heat pump)

6A

25A

10A

15A

35.2A

18A

Not counted

10A + (5A x 3) =

(3600W/230V) x 0.75 x 3

6A x 3 =

109A

multiple domestic41
Multiple Domestic

30 Units consists of the following:

30 Units

= 10 Units / Phase

Use Colum

4

3

Ai

Ai

  • Aii

Bi

Bi

  • Bii

C

D

F

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 15ASingle socket outlet

1 x 5.6kW wall oven / cook top

1 x 3.6kW air conditioner

1 x 2.4kW storage HWS (Heat pump)

7.5A

52.5A

10A

28A

117A

60A

5A + (0.25A x 10) =

Not counted

15A + (3.75A x 10) =

2.8A x 10 =

(3600W/230V) x 0.75 x 10

6A x 10 =

275A

multiple domestic42
Multiple Domestic

66 Units consists of the following:

66 Units

= 22 Units / Phase

Use Colum

5

3

Ai

Ai

  • Aii

Bi

Bi

  • Bii

C

D

F

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 15ASingle socket outlet

1 x 5.6kW wall oven / cook top

1 x 3.6kW air conditioner

1 x 2.4kW storage HWS (Heat pump)

11A

91.8A

10A

61.6A

258A

118A

0.5A x 22 =

Not counted

50A + (1.9A x 22) =

2.8A x 22 =

(3600W/230V) x 0.75 x 22

100A + (0.8A x 22) =

550A

what happens if the number of units is not devisable by 3
What Happens If The Number Of Units Is Not Devisable By 3?

16 Units

= 5.33 Units / Phase

3

A

B

C

5 Units

5 Units

6 Units

Col 3

Col 3

Col 4

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

NSW S&IR

Clause 1.10.3

multiple domestic44
Multiple Domestic

16 Units consists of the following:

5Units / A & B

Use Colum

3

Ai

Bi

Bi

  • Bii

C

D

F

26 x Light points

26 x Double 10 A socket outlets

  • 5 x Single 10 A socket outlets

1 x 15ASingle socket outlet

1 x 5.6kW wall oven / cook top

1 x 3.6kW air conditioner

1 x 2.4kW storage HWS (Heat pump)

6A

35A

10A

15A

58.7A

52.2A

10A + (5A x 5) =

(3600W/230V) x 0.75 x 5

(2400230V) x 5 =

176.9A

multiple domestic45
Multiple Domestic

16 Units consists of the following:

6Units / C

Use Colum

4

Ai

Bi

Bi

  • Bii

C

D

F

26 x Light points

26 x Double 10 A socket outlets

  • 5 x Single 10 A socket outlets

1 x 15ASingle socket outlet

1 x 5.6kW wall oven / cook top

1 x 3.6kW air conditioner

1 x 2.4kW storage HWS (Heat pump)

5A + (0.25A x 6) =

6.5A

37.5A

10A

16.8A

70.4A

36A

15A + (3.75A x 6) =

2.8A x 6=

(3600230) x 0.75 x 6

6A x 6 =

177.2A

what happens if the number of units is not devisable by 346
What Happens If The Number Of Units Is Not Devisable By 3?

16 Units

= 5.33 Units / Phase

3

A

B

C

5 Units

5 Units

6 Units

Col 3

Col 3

Col 4

176.6A

176.6A

177.2A

communal load
Communal Load
  • Driveway/ parking lighting
  • Stairwell lighting
  • Power points for cleaners
  • Lifts
  • Swimming pool pumps

Loads that are used by all the unit dwellers

multiple domestic48
Multiple Domestic

16 Units have a communal load consists of the following:

H

H

I

  • Ji

Jii

D

E

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

((20x80)230) + ((30x 60)230)=

14.78A

15A

7.8A

15A

12A

18.75A

2A x15= 30 but maximum of 15A

(3600230) x 0.5 =

(3600230) x 0.75 =

15 x 1.25 =

Table C2 Colum 2

83.33A

These circuits could also be divided across three phases

socket outlets
Socket outlets

In a Factory

On each Phase

1000W +

19 x 750W =

15.25kW

15250 ÷ 230 =

66.3 Amps

In areas with air conditioning

  • 20 x 10A single power points

1000W +

19 x 100W =

2.9kW

2900 ÷ 230 =

12.6 Amps

  • 20 x 10A single power points
socket outlets50
Socket outlets
  • 5 x 10A 3 outlets

Treat as B(i)

1000W + 4 x 750W =

4kW

Per Phase

17.4 Amps

4000 ÷ 230 =

  • 2 x 32A 3 outlets
  • 5 x 20A 3 outlets

Per Phase

75% x 32A

75% x 20A x 5

131Amps

32A +

24A +

75A =

motors
Motors

In Factories

1 per phase

75% x 30A

50% x 3 x 15A

30A +

22.5A +

22.5A =

75 Amps

  • 2 x 30A 3 motors
  • 3 x 15A 1 motors
welders
Welders

AS/NZS 3000

C2.5.2.2 (b)

Majority of welders only use 2 phases

B

A

C

Only 2 welders on each phase

W 1

W 2

W 3

20A + 20A =

40 Amps

Per Phase

  • 3 x 20A 3 Welders
table c3
Table C3

Table based on a temperate climate

Office

Floor area =

Light/Power =

Reverse cycle AC =

850m2

50 VA/m2

25 VA/m2

x 850 =

63.750kVA

75 VA/m2

63.750kVA

92 Amps/Phase

=

√3 x 400

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