Date post: | 20-Oct-2015 |
Category: |
Documents |
Upload: | kay-karthi |
View: | 36 times |
Download: | 2 times |
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
So how do we calculate MD?Single House Table C1Column
2LightingLoad groups Ai &
AiiChandelier or Multi-globe fitting =
Number of points =
= 3 points
Number of globes
So how do we calculate MD?Single House Table C1Column
2LightingLoad groups Ai &
AiiPower 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
2LightingLoad groups Ai &
AiiTrack 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
2LightingLoad groups Ai &
AiiIXL Tastics, Heat lamps etc
Not regarded as “Lighting”Fixed Space HeatingLoad Group “D”
So how do we calculate MD?Single House Table C1Column
2LightingLoad groups Ai &
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
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
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
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
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
CalculationsA house consists of the following:
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
CalculationsA house consists of the following:
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
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
43P 3 + 2 + 2 = 7A
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
7A
44P44 + 4 = 48P 10 + 5 + 5 = 20A
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
7A
20A
(10000/230) x 0.5 = 21.7A
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
7A
20A
13 x 0.75 =
21.7A9.75A
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
7A
20A
4600/230 =
21.7A9.75A20A
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
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
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 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
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?
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 = 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
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
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
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 =
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
(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
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
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