Electrical Contractor Unit solution. Cable selection commentary.
-
Upload
ami-wilcox -
Category
Documents
-
view
229 -
download
0
Transcript of 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