Transcript powerpoint file

Electrical Contractor
Unit solution
Cable selection commentary
Quantity
Load
Load Group
Diversity
12
60W lights
A
3A 1-20 points + 2A
Next 20 or part there of
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
1
5KW Wall oven
C
1
6KW Hot plate
C
50% full load (11000/230)X 0.5
1
3KW continuous HW unit
F
Full load
3000/230
1
3.5KW Fixed Space Heater
D
3500/230 x 0.75
M.D.
Units 1to5
5
10A
23.92
13
M.D.
11.42
63.33A
Quantity
Load
Load Group
Diversity
12
60W lights
A
3A 1-20 points + 2A
Next 20 or part there of
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
1
5KW Wall oven
C
1
6KW Hot plate
C
50% full load (11000/230)X 0.5
1
3KW continuous HW unit
F
Full load
3000/230
1
Heat pump 8.7 A each
D
No contribution less than 10A
Consider as an additional outlet
M.D.
Units 6 to 9
5
10A
23.92
13
M.D.
51.92A
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
H
3x6.8 =
20.4
2
10A 230V outlets
I
2A per point
4.0
Maximum Demand
Communal load
25.26
Quantity
Load Group
Diversity
R
W
B
1,4,7
2,5,8
3,6,9
house
6A
6
6
6
10A + 5A x 3 =25A
25
25
25
15A
15
15
15
18
12
60W lights
A
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)
1
300W Tastic
B(i)
1
5KW Wall oven
C
1
6KW Hot plate
C
1
3KW continuous HW
unit
F
6A per unit
6A x 3 = 18 A
18
18
1
3.5KW Fixed Space
Heater
D
(7000/230) x 0.75
(3500/230 +)x0.75
22.82
22.82
Communal load
11.41
25.26
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:
Vc  L  I 0.878 10 131
Vd 

 1.1 5Volts
1000
1000
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.
voltage drop single phase 
1.15 1.2731

 0..6647 volts
1.73
3
M
S
B
U1
U2
10M
Car
Wash
bay
U3
20M
U4
25m
35M
Conduits in groups of 2
Driveway
Supply Mains
Conduits in groups of 2
20M
30M
U6
500KVA
Transformer
U7
40m
U8
40M
U9
U5
35M
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 . Derating 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.
Vd 
Vc  L  I 2.55  1.155  10  73

 2.15V
1000
1000
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
1000Vd 1000  5.29

 2.204m.V/A .m
L I
40  60
This Vc value is a single phase value.
To look up the cable for the sub - main this value must
be converted to a three phase value first.
2.204  0.866  1.908mV/Am
Vc 
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
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.
3  VL  IL
1000
KVA
500 ,000
IL 

 722.5 A
3  400 1.73  400
100
100
Iscc  Iflc 
 722.5 
 14450 A
5
5
KVA 
500KVA transformer
14450A
Fault Impedance at transformer Terminals
V
I short circuit current
230
Z(transfor mer) 
14450
Z (transformer )  0.0159ohms
MSB
Z(transfor mer) 
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
500KVA transformer
Reactance Table 30 = 0.0700 X 2 = 0.14 ohms/1000m.
Z  R 2  X2
Z  0.1052  0.14 2
Supply Mains
Z  0.175 ohms/1000m
mains 
0.175 29
  0.005075ohms
1000 1
Consumers
Mains
Fault level at the point of supply
I
230
Z TX  Z Distributors
230
I
0.0159  0.005075
I  10965A
POS
10965A
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
0.852
 10  0.00852 ohms
1000
To determine the fault level at the main switchboard
Fault level MSB

230
Ztx  Z dis  Zcm
230
0.0159  0.005075  0.00852
 7797A

MSB
Sub-board
Unit 1
7797A
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
1.26
 10  0.0126 ohms one conductor
1000
active  Neutral  0.0.0126  2  0.0252ohms
500KVA transformer
To determine the fault level at the unit switchboard
Fault level unit 1 DB
230

Ztx  Zsc  Zcm  Zsm
230

0.0159  0.005075  0.00852  0.0252
 4205A
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
Progressive volt drop
POS
Mains
Transformer
The volt drop in the mains is
Vc  L  I
Vd 
1000
.878 10 145
Vd 
1000
Vd  1.2731 Volts three phase
1.2731
 0.73589Volts single phase
1.73
Sub-mains
MSB
Unit 2
SB
Volt drop in the sub-main unit1
Vc  L  I
1000
2.43 1.155  24  63.33
Vd 
1000
Vd  4.26 Volts
Vd 
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 subcircuits of 2.2%
1000  Vd
1000  5.71
L max 

 32.13M
Vc  I
2.43 1.155  63.33
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 subcircuits of 2.2%
Vd 1000
5.711000
L max 

 34.8M
Vc  I
2.43 1.155  58.45
Units 6 and 7 can be supplied with 16mm² sub-mains
Units 8 and 9 require 25mm² sub-mains
Progressive volt drop
POS
Mains
Transformer
The volt drop in the mains is
Vc  L  I
Vd 
1000
.878 10 145
Vd 
1000
Vd  1.2731 Volts three phase
1.2731
 0.73589Volts single phase
1.73
Sub-mains
MSB
Unit 9
SB
Volt drop in the sub-main unit1
Vc  L  I
1000
1.55  1.155  40  58.45
Vd 
1000
Vd  4.185Volts
Vd 
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
Supply
Transformer
Supply
Earth
electrode
Main switchboard
Mains conductor 50mm² 10m.
Zcm Active + Neutral = 0.00852Ω
Sub-main Circuit breaker
Main earth
Electrode
Sub-main Earth
6mm²
Z = 0.15Ω
From table 34
Unit 9 SB
Sub-mains Active 40m
25mm² Zsm = 0.03536Ω
Table 34
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
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.
230
230
Iscc 

 401A
Ztotal 0.573439
• This value exceeds the required value 7.5 x 16= 120A
for a type C MCB