electrical contractor unit solution. cable selection commentary

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Electrical Contractor Unit solution

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Page 1: Electrical Contractor Unit solution. Cable selection commentary

Electrical Contractor

Unit solution

Page 2: Electrical Contractor Unit solution. Cable selection commentary

Cable selection commentary

Page 3: Electrical Contractor Unit solution. Cable selection commentary
Page 4: Electrical Contractor Unit solution. Cable selection commentary
Page 5: Electrical Contractor Unit solution. Cable selection commentary
Page 6: Electrical Contractor Unit solution. Cable selection commentary
Page 7: 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

Page 8: Electrical Contractor Unit solution. Cable selection commentary

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

Page 9: Electrical Contractor Unit solution. Cable selection commentary

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

Page 10: Electrical Contractor Unit solution. Cable selection commentary

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

Page 11: Electrical Contractor Unit solution. Cable selection commentary

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

Page 12: Electrical Contractor Unit solution. Cable selection commentary

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

Page 13: Electrical Contractor Unit solution. Cable selection commentary

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

Page 14: Electrical Contractor Unit solution. Cable selection commentary

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

Page 15: Electrical Contractor Unit solution. Cable selection commentary

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

Page 16: Electrical Contractor Unit solution. Cable selection commentary

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

Page 17: Electrical Contractor Unit solution. Cable selection commentary

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

Page 18: Electrical Contractor Unit solution. Cable selection commentary

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

Page 19: Electrical Contractor Unit solution. Cable selection commentary

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

Page 20: Electrical Contractor Unit solution. Cable selection commentary

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

Page 21: Electrical Contractor Unit solution. Cable selection commentary

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.

Page 22: Electrical Contractor Unit solution. Cable selection commentary

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

Page 23: Electrical Contractor Unit solution. Cable selection commentary

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

Page 24: Electrical Contractor Unit solution. Cable selection commentary

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

Page 25: Electrical Contractor Unit solution. Cable selection commentary

Earth fault loop impedance.

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

Page 26: Electrical Contractor Unit solution. Cable selection commentary

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

Page 27: Electrical Contractor Unit solution. Cable selection commentary

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

Page 28: Electrical Contractor Unit solution. Cable selection commentary

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