electrical contractor unit solution. cable selection commentary

Electrical Contractor

Unit solution

Cable selection commentary

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 A

1 100W Sensor Light A

6 Double 10A 230V outlets B(i) 12 points

2 Single 10A 230V outlets B(i) 2 points

1 300W Tastic B(i) 1 point = 15 points total 10A

1 5KW Wall oven C

1 6KW Hot plate C 50% full load (11000/230)X 0.5 23.92

1 3KW continuous HW unit F Full load 3000/230

13

1 3.5KW Fixed Space Heater D 3500/230 x 0.75 11.42

M.D. 63.33A

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 A

1 100W Sensor Light A

6 Double 10A 230V outlets B(i) 12 points

2 Single 10A 230V outlets B(i) 2 points

1 300W Tastic B(i) 1 point = 15 points total 10A

1 5KW Wall oven C

1 6KW Hot plate C 50% full load (11000/230)X 0.5 23.92

1 3KW continuous HW unit F Full load 3000/230

13

1 Heat pump 8.7 A each D No contribution less than 10AConsider as an additional outlet

M.D. 51.92A

Quantity load Load Group

Diversity C W B

Communal load 1,4,7house

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

H 3x6.8 = 20.4

2 10A 230V outlets I 2A per point 4.0

Maximum DemandCommunal load

25.26

Quantity Load Group Diversity R W B

1,4,7 2,5,8 3,6,9house

12 60W lights A 6A 6 6 6

8 L.V down lights 0.4A each

A

1 100W Sensor Light A

6 Double 10A 230V outlets

B(i)

2 Single 10A 230V outlets

B(i) 10A + 5A x 3 =25A 25 25 25

1 300W Tastic B(i)

1 5KW Wall oven C 15A 15 15 15

1 6KW Hot plate C

1 3KW continuous HW unit

F 6A per unit6A x 3 = 18 A

18 18 18

1 3.5KW Fixed Space Heater

D (7000/230) x 0.75(3500/230 +)x0.75

22.82 22.8211.41

Communal load 25.26

M.D. 86.82 86.82 100.67

Consumers mains cable size

5Volts 1.11000

13110878.0

1000

ILVcVd

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 163AReferring to Table 41, the Vc value for a 50mm² conductor is 0.878mV/AmThe Actual voltage drop in the mains is:

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.

volts6647..073.1

2731.1

3

1.15 phase single drop voltage

CarWashbay

U1 U2 U3 U4 U5

U6 U7 U8 U9

10M 20M 25m 35M 35M

20M 30M 40m 40M

MSB

500KVATransformer

Conduits in groups of 2

Driveway

Conduits in groups of 2Supply Mains

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.

2.15V1000

7310 1.155 2.55

1000

ILVcVd

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 90ºCReferring 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

1.908mV/Am 0.866204.2

first. valuephase threea toconverted be

must valuemain this-sub for the cable theuplook To

value.phase single a is valueVc This

.m2.204m.V/A0640

5.291000

IL

1000VdVc

Fault level at the transformer terminals500KVA 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.

MSB

Sub-board Unit 1

A. IflcIscc

A..

,KVA

144505

1005722

5

100

5722400731

000500

4003 IL

1000

ILVL3KVA

500KVA

transformer

Fault Impedance at transformer Terminals

0.0159ohms)(14450

230mer)Z(transfor

currentcircuit short I

Vmer)Z(transfor

rtransformeZ

14450A

Fault level at the point of supply

MSB

Sub-board Unit 1

500KVA transformer

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.

ms0.005075oh1

29

1000

0.175 mains

ohms/1000m 0.175Z

0.140.105Z

XRZ

22

22

Fault level at the point of supply

10965AI0.0050750.0159

230I

Z Z

230I

rsDistributoTX

POS10965A

Supply Mains

Consumers Mains

Sub- mains

Fault level at the Main switch board

MSB

Sub-board Unit 1

50mm² SDI conductor a.c. resistance is 0.426 ohms per 1000m at 45°C Table 34Active + Neutral = 0. 426 x 2 = 0. 852 0hms/1000mConsumers mains are 10m so the phase to N resistance is

ohms 0.00852101000

0.852

To determine the fault level at the main switchboard

7797A0.008520.0050750.0159

230

ZcmZZtx

230

MSB levelFault

dis

500KVA transformer

7797A

Fault level at the Unit 1 switch board

MSB

Sub-board Unit 1

16mm² two core conductor. The a.c. Resistance is 1.26 ohms Per 1000m table 34Sub- mains are 20m so the phase to N resistance is

0.0252ohms 20.0.0126 Neutral active

conductor one ohms 0126.0101000

1.26

To determine the fault level at the unit switchboard

4205A 0252.000852.0005075.00159.0

230 ZsmZcmscZtx

230

DB 1unit levelFault

Z

500KVA transformer

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

Progressive volt drop

Transformer MSB Unit 2 SB

POS

Mains Sub-mains

The volt drop in the mains is

phase single Volts735890731

27311

phase threeVolts273111000

145108781000

..

.

.Vd

.Vd

ILVcVd

Volt drop in the sub-main unit1

Volts2641000

33632415514321000

.Vd

...Vd

ILVcVd

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%

MIVc

VdL 13.32

33.63155.143.2

71.510001000max

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%

MIVc

VdL 8.34

45.58155.143.2

100071.51000max

Units 6 and 7 can be supplied with 16mm² sub-mainsUnits 8 and 9 require 25mm² sub-mains

Progressive volt drop

Transformer MSB Unit 9 SB

POS

Mains Sub-mains

The volt drop in the mains is

phase single Volts735890731

27311

phase threeVolts273111000

145108781000

..

.

.Vd

.Vd

ILVcVd

Volt drop in the sub-main unit1

Volts185.41000

45.5840155155.11000

Vd

.Vd

ILVcVd

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

Mains conductor 50mm² 10m. Zcm Active + Neutral = 0.00852Ω

Sub-mains Active 40m25mm² Zsm = 0.03536ΩTable 34

Sub-main Circuit breaker

Main switchboard

Final sub-circuit 16A Circuit breaker

Final sub-circuit Active 2.5mm²Route length 20mZ = 0.18Ω From Table 35

Final Sub-circuit Protective earth 2.5mm² Z = 0.18Ω

Sub-main Earth6mm²Z = 0.15ΩFrom table 34 Unit 9 SB

Supply Transformer

Main earth Electrode

Supply Earth electrode

Sum of impedance values in theEarth 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 0.18

Sub-main earth 0.15

Total Impedance 0.573439

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

AZtotal

Iscc 401573439.0

230230