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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.
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
What is Maximum Demand?
Domestic stove rated at 5kW
Max current = 22A
Min cable size = 6mm2
By calculation
Max current = 16A
AS/NZS 3000
Table C4
Min cable size = 2.5mm2
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
“… 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 & Submains
10A
L2
16A
P1
16A
P2
32A
Stove
10A
L1
MD = ?
10+10+16+16+32 =
84 Amps
By Calculation
36 Amps
Domestic House
10mm2 2.5mm2
M
ost supply authorities specify a m
inim
um
m
ains
cable size of 16 m
m
2
20 metres 46 metres
16mm2
So how do we calculate
MD?
 Domestic
 Non-domestic
 Non- domestic Energy Demand
Table C1
Table C2
Table C3
So how do we calculate MD?
Single House Table C1Column 2
Lighting
Load groups Ai & Aii
Batten holder
=
1 light point
25 lighting points = 1 -20
=
3 A
21 -25 = 2 A
5 A
So how do we calculate MD?
Single House Table C1Column 2
Lighting
Load groups Ai & Aii
Chandelier or Multi-globe
fitting =
Number of points
=
= 3 points
Number of globes
So how do we calculate MD?
Single House Table C1Column 2
Lighting
Load groups Ai & Aii
Power points above 2.3 m for
lighting =
1 point
Note “e”
Exhaust fans below 150W
=
1 point
So how do we calculate MD?
Single House Table C1Column 2
Lighting
Load groups Ai & Aii
Track lights = Every metre =
Note “d”
2 points
2 x 2.6 metres track
light =
5 points per track
=
10 points
So how do we calculate MD?
Single House Table C1Column 2
Lighting
Load groups Ai & Aii
IXL Tastics, Heat lamps
etc
Not regarded as
“Lighting”
Fixed Space
Heating
Load Group “D”
So how do we calculate MD?
Single House Table C1Column 2
Lighting
Load groups Ai & Aii
Outside 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”
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
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
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
69P
1 – 20 =
21 – 40 =
41 – 60 =
61 – 69 =
3A
2A
2A
2A
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
3
12 Fittings / 3A
NOT
A B
C
3A 3A
36 light fittings = 3A + 2A = 5A
3
1.6
A B C
1.6 1.6
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
11 + (1 x 3) = 14P
1P
-
1 x 2 x 2 = 4P
4P
4.35A
14P
1P
4P
4P
23P
1 – 20 =
21 – 23 =
3A
2A
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 Points
6 Single power points
36 x 2 =72P
6P
78P
1 – 20 =
21 – 40=
41 – 60 =
61 – 78 =
10A
5A
5A
5A
25A
Load Group Bi
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
5A
5A
25A
Other Direct Wired
Equipment that is <10A
• Air conditioners
• Cook tops
• Water heaters
AND
Load Group Bii
Power Points
15 A Power Points
NOT
Other Equipment that is
Covered in Load Groups
below
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
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
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
Calculations
A house consists of the
following:
Bi
Bi
Bi
Bii
D
G
20 x 2 = 40P
8P
1P
10A
-
-
40P
8P
1P
49P
1 – 20 =
21 – 40 =
41 – 49 =
10A
5A
5A
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 top
4 kW Wall oven 6.3 kW of Load Group C
(6300/230) 13.7A
x 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 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
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
Maximum Demand
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:
Ai
Ai
Bi
Bi
C
D
F
Maximum Demand
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:
Ai
Ai
Bi
Bi
C
D
F
43P 3 + 2 + 2 = 7A
Maximum Demand
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:
Ai
Ai
Bi
Bi
C
D
F
7A
44P
44 + 4 = 48P 10 + 5 + 5 = 20A
Maximum Demand
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:
Ai
Ai
Bi
Bi
C
D
F
7A
20A
(10000/230) x 0.5 = 21.7A
Maximum Demand
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:
Ai
Ai
Bi
Bi
C
D
F
7A
20A
13 x 0.75 =
21.7A
9.75A
Maximum Demand
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:
Ai
Ai
Bi
Bi
C
D
F
7A
20A
4600/230 =
21.7A
9.75A
20A
Maximum Demand
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:
Ai
Ai
Bi
Bi
C
D
F
7A
20A
21.7A
9.75A
20A
78.45A
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 Demand
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:
Ai
Ai
Aii
Bi
Bi
Bii
C
C
F
D
A
B
C
A
B
C
A
B
C
A
B
C
3 Maximum Demand
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:
Ai
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 =
(4100230)x0.5 =
(4600230)x0.5 =
3600230 =
(5700x0.75)(√3 x400) =
39.2A 40.2A
Multiple Domestic
Submains 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
3
= 6 Units / Phase Use Colum 4
66 Units
3
= 22 Units / Phase Use Colum 5
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 = 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
Multiple Domestic
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:
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
3
= 3 Units / Phase Use Colum 3
Not counted
Multiple Domestic
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:
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
3
= 10 Units / Phase Use Colum 4
Not counted
5A + (0.25A x 10) =
2.8A x 10 =
Multiple Domestic
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:
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
3
= 22 Units / Phase Use Colum 5
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
A

5 Units
Col 3
But the phases must not be unbalanced by more than 25A
B

C

5 Units 6 Units
Col 3 Col 4
NSW S&IR
Clause 1.10.3
Multiple Domestic
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:
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
(2400230V) x 5 =
Multiple Domestic
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:
Ai
Bi
Bi
Bii
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 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
A

5 Units
Col 3
B

C

5 Units 6 Units
Col 3 Col 4
176.6A 176.6A 177.2A
Communal Load
Loads 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
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
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 3000
C2.5.2.2 (b)
20A + 20A = 40 Amps
Per Phase
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
850m2
50 VA/m2
25 VA/m2
75 VA/m2
x 850 = 63.750kVA
√3 x 400
= 92 Amps/Phase