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.

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 26.2 ASNZS 3008.1.1 ). 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.

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

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

Foot notes to Table C1

Foot notes to Table C1 continued.

Quantity Load Load Group Diversity M.D. Units 1to5 12 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.

Quantity Load Load Group Diversity M.D. Units 6 to 9 12 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.

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.

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 100.67A 100.67 + 30 =130.67A say 131A

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 75.41 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 (88.5 + 25.26)= 100.67A. Add 30A for future development 100.67 + 30 =130.67 round up to 131A The voltage drop in the mains is 1.15V three phase. The single phase equivalent is 1.15/1.73 = 0.6647 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.

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 63 .33A. Add 15% for future additions. 15% of 63.33 =9.5A Therefore the maximum demand needs to be increased to 63.33 + 9.5 = 72.8295. (Say 73A) Units 6,7,8 and 9 have a maximum demand of 51.92A. 15% = 7.788A. 51.92 +7.788 = 59.7A. (Say 60A) U6 U7 U8 U9 500KVA Transformer

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 26.2 . 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 63 .33A. 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

Sub-main to unit 9 Referring to Table 41 (ASNZS3008.1.1) 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. 51.92 x .15 =59.7A say 60A Table 5 column 24 can be used for current carrying capacity.

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 0.0159 ohms. See above for method of calculation. MSB Sub-board Unit 1

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 = 0.0525 ohms/1000m. At 75ºC 0ne conductor. Active + Neutral = 0.0525 x 2 = 0.105 ohms/ 1000m Reactance Table 30 = 0.0700 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

Fault level at the Main switch board 50mm² SDI conductor a.c. resistance is 0.426 ohms per 1000m at 45°C Table 34 Active + Neutral = 0. 426 x 2 = 0. 852 0hms/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

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

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

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 – (0.73589+4.26) 6.5 Volts

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.

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

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 – (0.73589+4.185) 6.58 Volts

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 0.005075 230/(0.159 + 0.005075) 10965A Consumer Mains 0.00852 230/ (0.0159 + 0.005075 +0.00852) 7797A Sub-mains 0.0252 230/(0.0159 + 0.005075 +0.00852 +0.0252) 4205A

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

Supply Mains Active + Neutral 500mm² conductor 29m Z = 0.003659 Transformer Main switchboard Mains conductor 50mm² 10m. Zcm Active + Neutral = 0.00852Ω 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 = 0.03536Ω 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 1.92 0hms.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

Sum of impedance values in the Earth fault-loop Device/ cable9 Impedance Ω Transformer 0.0159 Supply Mains A+N 0.003659 Consumer Mains A+N 0.00852 Sub-main Active 0.03536 Final sub-circuit active 0.18 Protective earth Sub-main earth 0.15 Total Impedance 0.573439 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.

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 (0.573439 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 (0.573439) 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.