1. Electrical Contractor
Unit solution
This solution is an example of how an installation can be designed to meet the requirements of the AS/NZS 3000 Wiring Rules and The Service and Installation Rules.

2. Cable selection commentary
In this design solution it will be necessary to assess the consumers mains and at least sub-main 9 and sub-main 1.The consumers mains are very short, so negligible voltage drop is expected, and the fault level at the main switchboard will be similar to the fault level at the supply authority mains. The sub-mains to unit 9 are the longest, so the highest voltage drop will occur on these sub-mains. This will be the first sub-mains assessed. As the protective earthing cables for the sub-mains are long, it is expected that the fault-loop impedance may be a critical determinant for the earthing conductor size. Sub-mains 1 are the shortest set of cables and it is most likely that the fault level will be the highest on this sub-main.
The sub-mains will be installed in conduit under ground to a depth of 0.5m. The sub-main conduits will be installed in common trenches, so it will be necessary to apply de-rating factors for multiple circuits. The de-rating factor for a group of 4 circuits touching in a common trench is 0.79 (Table ASNZS ). This will have a significant effect on the cable selection and the use of the circuit protection, protecting the sub-mains. To minimise the effect of de-rating, the conduits are to be installed in groups of 2 at 300mm separation, and the de-rating factor from table 26.2 is 0.93.
The voltage drop has been assessed by allowing 2.2%for the final sub-circuits an then distributing the voltage across the consumer mains and sub-mains according to length. This assessment is project specific, and will vary according to the application. The cable lengths used include for vertical and horizontal components and terminations. The distribution board locations are dictated in part by the building designer as it is necessary to select suitable locations for the distribution boards. Separate earthing conductors will be run to each residence from the MSB. The water main and water meter are adjacent to the main switchboard. The earth electrode will be driven adjacent to the main switchboard.

3. AS/NZS 3000:2007 Wiring Rules Tables for maximum demand calculation.

4. Single domestic Maximum Demand column 2
Blocks of living units per phase e.g. 3 units per phase column 3

5. Foot notes to Table C1

6. Foot notes to Table C1 continued.

7. Quantity Load Load Group Diversity M.D. Units 1to512
60W lights A
3A 1-20 points + 2A
Next 20 or part there of 5 8
L.V down lights 0.4A each 1
100W Sensor Light 6
Double 10A 230V outlets
B(i)
12 points 2
Single 10A 230V outlets
2 points
300W Tastic
1 point = 15 points total
10A
5KW Wall oven C
6KW Hot plate
50% full load (11000/230)X 0.5
23.92
3KW continuous HW unit F
Full load
3000/230 13
3.5KW Fixed Space Heater D
3500/230 x 0.75
11.42
63.33A
Calculate the maximum demand of units 1,2,3,4 and 5
Units 1, 2, 3, 4 and 5 are supplied with a fixed space heater rated at 3.5Kw single phase.
Load Group A (i) Lighting 3amps for 1-20 points + 2A for each additional 20 points or part thereof
Load Group B (i) 10A single phase outlets. 10A 1-20 points + 5A for each additional 20 points or part thereof
Load Group C Wall Oven and Hot Plate 50% of the connected load.
Load Group D Heating 75% of the connected load.
Load group F Storage water heater full load current.

8. Quantity Load Load Group Diversity M.D. Units 6 to 912
60W lights A
3A 1-20 points + 2A
Next 20 or part there of 5 8
L.V down lights 0.4A each 1
100W Sensor Light 6
Double 10A 230V outlets
B(i)
12 points 2
Single 10A 230V outlets
2 points
300W Tastic
1 point = 15 points total
10A
5KW Wall oven C
6KW Hot plate
50% full load (11000/230)X 0.5
23.92
3KW continuous HW unit F
Full load
3000/230 13
Heat pump 8.7 A each D
No contribution less than 10A
Consider as an additional outlet
51.92A
Units 6, 7, 8 and 9 are supplied with a heat pump rated at 8.7A each (soft start). Technically the heat pump can be ignored from maximum demand as it is less than 10A. It could be included as an additional outlet and as such would not contribute to the MD.
Load Group A (i) Lighting 3amps for 1-20 points + 2A for each additional 20 points or part thereof
Load Group B (i) 10A single phase outlets. 10A 1-20 points + 5A for each additional 20 points or part thereof
Load Group C Wall Oven and Hot Plate 50% of the connected load.
Load Group D Heating 75% of the connected load.
Load group F Storage water heater, full load current.

9. Quantity load Load Group Diversity C W B
Communal load
1,4,7
house
2,5,8
3,6,9 2
2x36W 0.43A each H
Full load 2x.43
0.86 3
1000W metal halide flood lights
(6.8A) each
3x6.8 =
20.4
10A 230V outlets I
2A per point
4.0
Maximum Demand
25.26
The communal load consists of:
2 twin 36W fluorescent lights at 0.43A per fitting. Load group H Full connected load.
3 x 1000W Metal halide rated at 6.8A each. Load group H Full connected load.
2 x 10A socket outlets. Load group I 2A per point up to a maximum of 15A.

10. Quantity Load Group Diversity R W B
1,4,7
2,5,8
3,6,9
house 12
60W lights A 6A 6 8
L.V down lights 0.4A each 1
100W Sensor Light
Double 10A 230V outlets
B(i) 2
Single 10A 230V outlets
10A + 5A x 3 =25A 25
300W Tastic
5KW Wall oven C
15A 15
6KW Hot plate
3KW continuous HW unit F
6A per unit
6A x 3 = 18 A 18
3.5KW Fixed Space Heater D
(7000/230) x 0.75
(3500/230 +)x0.75
22.82
11.41
Communal load
25.26
M.D.
86.82
100.67
Maximum Demand (Consumer’s Mains)
Units 1,4,7 and the communal load are connected to the red phase. By moving the communal load to the blue phase a better balance can be achieved,
Unit 1 and 4 are supplied with a 3.5Kw space heater. Unit 7 is supplied with a heat pump, 8.7A input.
Units 2,5,and 8 are connected to the white phase. Units 2 and 5 are supplied with a 3.5Kw space heater. Unit 8 is supplied with a heat pump, 8.7A input
Units 3,6 and 9 are connected to the blue phase. Unit 3 is supplied with a 3.5Kw space heater. Units 6 and 9 are supplied with a heat pump 8.7A input.
30A is to be applied to the maximum demand for future extension.
The maximum demand is A
=130.67A say 131A

11. Consumers mains cable size
The mains cable is X90 SDI installed in one conduit U/G. Referring to Table 3.4 item 2
Table 8 column 24, a 50mm² Cable is rated at 163A
Referring to Table 41, the Vc value for a 50mm² conductor is 0.878mV/Am
The Actual voltage drop in the mains is:
The consumers mains are 10 metres long with a maximum demand of 86.82A (red and white phase). The blue phase maximum demand is Connect the communal load to the blue phase. 30A needs to be added to the maximum demand for future development. Therefore the mains rating is ( )= A. Add 30A for future development = round up to 131A
The voltage drop in the mains is 1.15V three phase. The single phase equivalent is 1.15/1.73 = Volts.
This Value is a three phase value and must be converted to a single phase value to determine the voltage drop allowed in the sub-mains to each unit.

12. U1 U2 U3 U4 U5 U6 U7 U8 U9 Supply Mains Car Wash bay MS B 10M 20M 25m
Conduits in groups of 2
Driveway
Supply Mains
Conduits in groups of 2
20M
30M
40m
40M
Units 1,2,3,4 and 5 have a maximum demand of A. Add 15% for future additions. 15% of =9.5A
Therefore the maximum demand needs to be increased to = (Say 73A)
Units 6,7,8 and 9 have a maximum demand of 51.92A. 15% = 7.788A = 59.7A. (Say 60A) U6 U7 U8 U9
500KVA
Transformer

13. Sub-main to unit 1
Referring to Table 3.4 ASNZS3008 item 1, Table 8column 24, 16mm² X90 conductor = 86A. The conduits are installed in a trench in groups of 2 separated 300mm.apart. Table De-rating factor 0.93
(86 x 0.93 =79.98A)
Table 41, the Vc value for a 16mm² conductor is 2.55mV/Am. At 90ºC
Therefore:
A 16mm² Conductor will satisfy current requirements
Voltage drop allowed in the sub-mains is 2.3% = 5.29Volt single phase.
Units 1,2,3,4 and 5 have a maximum demand of A. Allow 15% for future addition
15% of 63.33=9.4995A, therefore the maximum demand is 72.8, say 73A
Unit 1 sub-main is 10 m to the main switchboard.
The meters are installed at the main switchboard position therefore the sub-mains require two active conductors and a common neutral conductor.
To calculate the volt drop in the sub-mains with a two tariff arrangement it will be necessary to determine the volt drop in a circuit with active and neutral the same size to determine the size of the neutral conductor. The voltage drop in the neutral is 50% of this value.
The neutral conductor will have a volt drop of 50% of the total volt drop of this value
The actives can then be sized to carry the proportion of the load that is connected to them.
From the above calculation the neutral conductor is 16mm². The active conductors are now sized according to the proportion of the maximum demand current they carry.
40A light and power a 10mm²conductor
33A hydro heat a 6mm² conductor
The voltage drop in the neutral conductor is therefore 50% of this value
2.15 x 0.5 = 1.075V

14. Sub-main to unit 9
Referring to Table 41 (ASNZS ) a 25mm² cable has a Vc value of 1.62mV/Am at 90ºC
Referring to Table 8 column 24, A 25mm² conductor can carry 113A.
The de-rating factor for the arrangement are 0.8 and 0.93
131 x 0.8 x 0.93 = 84 A. Current is not the determining factor. The conductor size will need to be determined from voltage drop requirements.
Therefore a 25 mm² cable is required for the sub-main to unit 9
Unit 9 is 40 metres from the main switchboard. The MD=51.92 allow 15% for future extension x .15 =59.7A say 60A
Table 5 column 24 can be used for current carrying capacity.

15. Fault level at the transformer terminals
500KVA transformer. Assume a 5% impedance value. This value refers to the value of the primary voltage required to cause full load current with a short circuit on the secondary. With 100% primary voltage applied the short circuit current would be 20 times the full load current.
500KVA transformer
14450A
Fault Impedance at transformer Terminals
The fault current is 14450A at the transformer terminals. The opposition to the fault current at the transformer terminals is ohms. See above for method of calculation.
MSB
Sub-board
Unit 1

16. Fault level at the point of supply
500mm² Supply Authority conductors run from the transformer. Length to the consumers mains point of supply is 29M.
a.c. Resistance Table 35 = ohms/1000m. At 75ºC 0ne conductor.
Active + Neutral = x 2 = ohms/ 1000m
Reactance Table 30 = X 2 = 0.14 ohms/1000m.
500KVA transformer
Supply Mains
POS
10965A
Consumers Mains
The point of supply is where the consumers mains connect to the supply conductors. The supply conductors run from the transformer. The supply conductors are 29m in length at the point of supply (pos).
Fault level at the point of supply
MSB
Sub- mains
Sub-board
Unit 1

17. Fault level at the Main switch board
50mm² SDI conductor a.c. resistance is ohms per 1000m at 45°C Table 34
Active + Neutral = x 2 = hms/1000m
Consumers mains are 10m so the phase to N resistance is
500KVA transformer
To determine the fault level at the main switchboard
The impedance of a cable can be calculated. And this will give a lower fault level. However by using the resistance of the cable a higher fault level will result and this will result in increased KA ratings of protection devices and equipment. Also the length of the consumers mains is short (10 metres) so negligible impedance is expected.
MSB
7797A
Sub-board
Unit 1

18. Fault level at the Unit 1 switch board
16mm² two core conductor. The a.c. Resistance is 1.26 ohms Per 1000m table 34
Sub- mains are 20m so the phase to N resistance is
500KVA transformer
To determine the fault level at the unit switchboard
The sub-mains to unit 1 are short so negligible impedance is expected. Determine the fault level using the resistance of the sub-main.
MSB
Sub-board
Unit 1
4205A

19. Actual progressive volt drop
Although this project specified an allowable voltage drop as follows:
Consumer mains 0.5%
Sub-mains 2.3%
Final sub-circuits 2.2%
The actual voltage drop in the mains and sub-mains is less than the allowed percent.
There fore the voltage drop in the final sub-circuit can be increased
Actual progressive volt drop

20. Progressive volt drop POS Mains Sub-mains Unit 2 MSB Transformer SB
Volt drop in the sub-main unit1
The volt drop in the mains is
The voltage drop allowed in final sub-circuits is 11.5 – ( ) 6.5 Volts

21. Maximum length of 16mm² conductor for sub-mains Units1 to 5
The next step is to determine the maximum length a 16mm² conductor can be run allowing for a voltage drop in the final sub-circuits of 2.2%
Units 1, 2, and 3 can be supplied with a 16mm² sub-main
Units 4 and 5 require a 25mm² sub-main.

22. Maximum length of 16mm² conductor for sub-mains units 6 to 9
The next step is to determine the maximum length a 25mm² conductor can be run allowing for a voltage drop in the final sub-circuits of 2.2%
Units 6 and 7 can be supplied with 16mm² sub-mains
Units 8 and 9 require 25mm² sub-mains

23. Progressive volt drop POS Mains Sub-mains Unit 9 MSB Transformer SB
Volt drop in the sub-main unit1
The volt drop in the mains is
The voltage drop allowed in final sub-circuits is 11.5 – ( ) 6.58 Volts

24. Fault level
The fault level can be determined at each point in the installation as follows
Cable
Impedance ohms A + N
V/Z
Fault level
Transformer
0.0159
230/0.0159
14450A
Supply Mains
230/( )
10965A
Consumer Mains
230/ ( )
7797A
Sub-mains
0.0252
230/( )
4205A

25. Earth fault loop impedance.
The earth fault loop impedance can be determined as shown in the next slide.

26. Supply Mains Active + Neutral 500mm² conductor 29m Z = 0.003659
Transformer
Main switchboard
Mains conductor 50mm² 10m. Zcm Active + Neutral = Ω
Sub-main Circuit breaker
Supply Earth electrode
Main earth Electrode
Sub-main Earth
6mm²
Z = 0.15Ω
From table 34
Sub-mains Active 40m
25mm² Zsm = Ω
Table 34
The earth fault loop impedance can be determined by identifying the impedance of the active conductors in series in the fault loop and the value of the protective earthing conductor that carry the earth fault current.
In the event of a fault in a piece of equipment, the fault current will flow through the Protective earth of the faulty equipment back to the earth bar of the unit 9 switchboard. From there it will flow through the sub-main (laid up) earth to the main switchboard earth bar. Through the MEN link to the neutral link, through the main neutral of the consumers mains, the (PEN) conductor, to the supply distributor neutral and on to the star point of the Transformer. From there it will flow through the phase winding of the Transformer supplying the faulty circuit, through the active of the supply, through the mains active conductor, through the 16A type C MCB, out through the active conductor of the faulty circuit and back to the point of the fault.
Referring to Table 8.1 the maximum Earth Fault Loop Impedance for a 16A Type C circuit breaker is hms.This value can be measured with supply connected, however the installer must know that earth earth fault loop impedance is satisfactory before he finally selects cables for the installation.
Unit 9 SB
Final sub-circuit 16A Circuit breaker
Final Sub-circuit Protective earth
2.5mm² Z = 0.18Ω
Final sub-circuit Active 2.5mm²
Route length 20m
Z = 0.18Ω From Table 35

27. Sum of impedance values in the Earth fault-loop
Device/ cable9
Impedance Ω
Transformer
0.0159
Supply Mains A+N
Consumer Mains A+N
Sub-main Active
Final sub-circuit active
0.18
Protective earth
Sub-main earth
0.15
Total Impedance
The total impedance of the earth fault loop in the sum of all the impedance values in the path of the fault current. See above.

28. Fault current in the Final sub-circuit
Table 8.1 requires a maximum value of earth fault loop impedance of 1.91Ω. The actual value for this circuit ( ohms) is below the required maximum value.
The current flowing in this fault loop will be sufficient to operate the protective device as required.
This value exceeds the required value 7.5 x 16= 120A for a type C MCB
Table 8.2 requires a maximum value of earth fault loop impedance of 1.91Ω. The actual value ( ) is way below the maximum value allowed. The current flowing in this fault loop will be sufficient to operate the protective device as required. See above for calculation.