Lesson Content
• Describethe 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.
What is MaximumDemand?
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 MaximumDemand?
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
5.
MD can bedetermined 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.
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)
7.
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
8.
So how dowe calculate
MD?
Domestic
Non-domestic
Non- domestic Energy Demand
Table C1
Table C2
Table C3
9.
So how dowe 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
10.
So how dowe 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
11.
So how dowe 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
12.
So how dowe 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
13.
So how dowe 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”
14.
So how dowe 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”
15.
Calculations
A house consistsof 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
16.
Calculations
A house consistsof 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
17.
3 Light Calculations
36light 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
18.
Calculations
A house consistsof 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
19.
Load Group Bi
PowerPoints
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
20.
Load Group Bi
PowerPoints
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
21.
Load Group Bii
PowerPoints
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
22.
Calculations
A house consistsof 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
23.
20 Double powerpoints
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
24.
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
25.
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
26.
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
27.
Maximum Demand
41 x50 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
28.
Maximum Demand
41 x50 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
29.
Maximum Demand
41 x50 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
30.
Maximum Demand
41 x50 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
31.
Maximum Demand
41 x50 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
32.
Maximum Demand
41 x50 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
33.
Maximum Demand
41 x50 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
34.
Uneven 3 Loads
Notall 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.
3 Maximum Demand
30x 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
36.
3 Maximum Demand
30x 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 =
(4100230)x0.5 =
(4600230)x0.5 =
3600230 =
(5700x0.75)(√3 x400) =
39.2A 40.2A
Questions to be
Asked?
Howmany 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
39.
Maximum Demand
26 xLight 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
40.
Multiple Domestic
26 xLight 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
41.
Multiple Domestic
26 xLight 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 =
42.
Multiple Domestic
26 xLight 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 =
43.
What Happens IfThe
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
44.
Multiple Domestic
26 xLight 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
(2400230V) x 5 =
45.
Multiple Domestic
26 xLight 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
(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 =
46.
What Happens IfThe
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
47.
Communal Load
Loads thatare used by all the unit dwellers
• Driveway/ parking lighting
• Stairwell lighting
• Power points for cleaners
• Lifts
• Swimming pool pumps
48.
Multiple Domestic
20 x80 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
(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
49.
Socket outlets
20x 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
5x 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 x30A 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 x20A 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
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
850m2
50 VA/m2
25 VA/m2
75 VA/m2
x 850 = 63.750kVA
√3 x 400
= 92 Amps/Phase