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7/29/2019 04 - Wk10 APM Handout Part 4 of 4
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PDS Emergency Configuration 1 (Cont)
APMDEPOT
11kV RMU
52S1 52S2
11kV 3 Supply
TR01 TR02
52F352F252F1
DS0
DSF1 DSF2
TR01 - Loss Power- Trains normally feeds by TR01
- 52S1, 52F1 & 52F2 closed
- DSF1, DSF2 closed
- DS0 close, DS1 open
- TR01 is de-energized
- 52S2 close automatically
- TR02 feeds the trains continuously
TractionCurrent
Flowdirection
DS1
East HallArrivalPlatform
West HallArrivalPlatform
Arrival Line
Power Rail
East HallDeparturePlatform
West HallDeparturePlatform
Departure Line
Power Rail
52FL252FL1
ZZ
40
PDS Emergency Configuration 2
APMDEPOT
11kV RMU
52S1 52S2
11kV 3 Supply
TR01 TR02
52F352F252F1
DS0
DSF1 DSF2
TR02 - Loss Power
- Trains normally feeds by TR02
- 52S2, 52F1 & 52F2 closed
- DSF1, DSF2 closed
- DS0 close, DS1 open
DS1
East HallArrivalPlatform
West HallArrivalPlatform
Arrival Line
Power Rail
East HallDeparturePlatform
West HallDeparturePlatform
Departure Line
Power Rail
52FL252FL1
ZZ
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PDS Emergency Configuration 2 (cont)
APMDEPOT
11kV RMU
52S1 52S2
11kV 3 Supply
TR01 TR02
52F352F252F1
DS0
DSF1 DSF2
TR02 - Loss Power- Trains normally feeds by TR02
- 52S2, 52F1 & 52F2 closed
- DSF1, DSF2 closed
- DS0 close, DS1 open
- TR02 is de-energized
- 52S1 close automatically
- TR01 feeds the trains continuously
TractionCurrent
Flowdirection
DS1
East HallArrivalPlatform
West HallArrivalPlatform
Arrival Line
Power Rail
East HallDeparturePlatform
West HallDeparturePlatform
Departure Line
Power Rail
52FL252FL1
ZZ
42
PDS Emergency Configuration 3
APMDEPOT
11kV RMU
52S1 52S2
11kV 3 Supply
TR01 TR02
52F352F252F1
DS0
DSF1 DSF2
ACB 52F2 fail to close
- 52S1, 52F1 & 52F2 closed
- DSF1, DSF2 closed
- DS0 close, DS1 open
TractionCurrent
Flowdirection
DS1
East HallArrivalPlatform
West HallArrivalPlatform
Arrival Line
Power Rail
East HallDeparturePlatform
West HallDeparturePlatform
Departure Line
Power Rail
52FL252FL1
ZZ
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PDS Emergency Configuration 3 (Cont)
APMDEPOT
11kV RMU
52S1 52S2
11kV 3 Supply
TR01 TR02
52F352F252F1
DS0
DSF1 DSF2
ACB 52F2 fail to close
- 52S1, 52F1 & 52F2 closed
- DSF1, DSF2 closed
- DS0 close, DS1 open
- 52F2 is tripped & cannot close
- Close DS1
- Departure Line feed continuously
- Traction current flow through 52F1is equal to total traction current inArrival and Departure Line
TractionCurrent
Flowdirection
DS1
East HallArrivalPlatform
West HallArrivalPlatform
Arrival Line
Power Rail
East HallDeparturePlatform
West HallDeparturePlatform
Departure Line
Power Rail
52FL252FL1
ZZ
44
PDS Emergency configuration 4
APMDEPOT
11kV RMU
52S1 52S2
11kV 3 Supply
TR01 TR02
52F352F252F1
DS0
DSF1 DSF2
ACB 52F1 fail to close
- 52S1, 52F1 & 52F2 closed
- DSF1, DSF2 closed
- DS0 close, DS1 open
TractionCurrent
Flowdirection
DS1
East HallArrivalPlatform
West HallArrivalPlatform
Arrival Line
Power Rail
East HallDeparturePlatform
West HallDeparturePlatform
Departure Line
Power Rail
52FL252FL1
ZZ
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PDS Emergency Configuration 4 (Cont)
APMDEPOT
11kV RMU
52S1 52S2
11kV 3 Supply
TR01 TR02
52F352F252F1
DS0
DSF1 DSF2
ACB 52F1 fail to close
- 52S1, 52F1 & 52F2 closed
- DSF1, DSF2 closed
- DS0 close, DS1 open
- 52F1 is tripped & cannot close
- Close DS1
- Departure Line feed continuously
- Traction current flow through 52F2is equal to total traction current inArrival and Departure Line
TractionCurrent
Flowdirection
DS1
East HallArrivalPlatform
West HallArrivalPlatform
Arrival Line
Power Rail
East HallDeparturePlatform
West HallDeparturePlatform
Departure Line
Power Rail
52FL252FL1
ZZ
46
APM Train System Train Operation Modes
Mode 1 - Dual Shuttle Mode
- APM train is running between East Hall and West Hall without crossing toother line
- Only two trains running
- PDS is operated at ~ 50% of installed capacity
East HallArrivalPlatform
West HallArrivalPlatformArrival Line
East Hall
DeparturePlatform
West Hall
DeparturePlatform
Departure Line
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APM Train System Train Operation Modes
Dual Shuttle Running Demonstration
East HallArrivalPlatform
West HallArrivalPlatformArrival Line
East HallDeparturePlatform
West HallDeparturePlatform
Departure Line
48
APM Train System - Train Operation Modes
Mode 2 - Normal Pinched Loop
- APM trains are running between East Hall and West Hall in a loop
- Maximum four trains are running in system
- PDS is operated at ~ 100% of installed capacity
East HallArrivalPlatform
West HallArrivalPlatformArrival Line
East Hall
DeparturePlatform
West Hall
DeparturePlatform
Departure Line
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APM Train System - Train Operation Modes
Normal Pinched Loop Demonstration
East HallArrivalPlatform
West HallArrivalPlatformArrival Line
East HallDeparturePlatform
West HallDeparturePlatform
Departure Line
50
APM Trains Operation Modes
Normal Pinched Loop Demonstration
East HallArrivalPlatform
West HallArrivalPlatformArrival Line
East Hall
DeparturePlatform
West Hall
DeparturePlatform
Departure Line
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PDS Design Requirement
- System design, Equipment Rating Selection (S, RAM, $)
- Fault level and protection coordination Safety
- Voltage Drop Train performance at minimum voltage
- Cable system temperature rise Thermal impact- Harmonic assessment - Interference impact
Design Deliverables (Safety, RAM, $)
52
PDS Design Flow Case Study
Assumptions
- The harmonic filter / power factor correction equipment ispurposely put out of service to simulate the maximum loadcurrent scenario.
- The fault level at source point 11kV side is 350MVA- Cable have same dimension will have equal impedance and will
share equal current when connected in parallel
Case Study
Voltage drop Calculation
Harmonic Assessment
Cable Temperature rise
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PDS - Design Interface with Train System
Train current profile- Different Service Modes Different Current profile, full load, train
start and coast
1. Dual Shuttle Mode (one train per line)
2. Normal Pinched Loop (two trains per line)
- Train Consist
- Maximum 4 nos. train running in system
1x 2 car train or 2 x2cartrain
1 x 4 car train
54
PDS - Design Interface with Train System
Train current profile Starting and Running
Peak 1178.6ARMS 321A
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PDS - Design Interface with Train System
Train current profile - Enlargement
Peak 1178.6A
56
PDS Design Interface with Train System
Train current and voltage characteristics
- Minimum Operation Voltage
i.e. Terminal Voltage at pantograph = 440V
- Amperage Characteristic
4001601 x 4-car (no load)
200801 x 2-car (no load)
1,3402601 x 4-car (full load)
6701301 x 2-car (full load)
Peak StartingCurrent (A), 15S
RMS RunningCurrent (A)
Different Train ConsistConfiguration
- Power Factor = 0.97
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Traction Power System Design
4202102102 x 2 car
running
full load + no
load
2 x 2 carrunning
full load + no
load
6Normal
Pinched Loop
17408708702 x 2 car
starting
full load + no
load
2 x 2 carstarting
full load + no
load
5Normal
Pinched Loop
520260260
1 x 4 car
running full load
1 x 4 car
running full load4Dual Shuttle
2680134013401 x 4 car
starting full load
1 x 4 carstarting full load
3Dual Shuttle
2601301301 x 2 car
running
full load
1 x 2 carrunning
full load
2Dual Shuttle
13406706701 x 2 car
starting
full load
1 x 2 carstarting
full load
1Dual Shuttle
Transformercircuit
52S1 / 52S2
Departure TrackFeeder circuit
52F2
Arrival TrackFeeder circuit
52F1
Train onDeparture
Line
Train onArrival Line
ModeOperation
Summary of Train CurrentService Modes and Trains Consist Combination
58
Traction Power System Design Interface
PDS equipment rating
APMDepot
APMDEPOT
East HallArrivalPlatform
East Hall
DeparturePlatform
End ofLine
West HallArrivalPlatform
West Hall
DeparturePlatform
11kV RMU
East HallSubstation
11kV 3 Supply
Power Rail
Power Rail
Traction Tx
L5 : 7 x 240 mm2 cables
L1 : 4 x 240 mm2 cables
4 x 240mm2
cables
L3 : 4 x 240 mm2 cables
TR1 TR2
52S1 52S2
52F1 52F2
DSF1 DSF2
52FL252FL1
Z
Z
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PDS Design Equipment rating
Traction Supply System Equipment Capacity
30,2902,3302,330A600V Power Rail SystemPower Rail
27,7201,6821,682A600V Power Cables ( 4 nos. )L1, L3
46,0002,7882,790A600V Power Cables ( 7 nos. )L5
22,0002,5002,500A600V IsolatorDSF1, DSF2
22,0002,5002,500A600V ACB52F1, 52F2
22,0002,5002,500A600V ACB52S1, 52S2
6,3502,4002,500 kVA11/0.6kV Traction TransformerTR1, TR2
PeakAmperageCapacityin 15 sec.
(A)
RatedAmperageCapacity
(A)
InstalledRating
Equipment TypeEquipment ID
Equipment Short Circuit Rating = 50kA, 1 second
Manufacturersdata
60
PDS Design of Equipment Rating
Verification on Equipment Continuous CapacityAcceptance Criteria:The Installed Amperage Capacity shall higher than the Maximum Continuous Train
Current (in RMS value).
1116
19
10
10
10
22
Capacity utilized( %)
260260
520
260
260
260
520
Max. RMSContinuous
Train Current(A)
Yes2,330Power RailYes1,682L1, L3
Yes2,788L5
Yes2,500DSF1, DSF2
Yes2,50052F1, 52F2
Yes2,50052S1, 52S2
Yes2,400TR1, TR2
Fulfillacceptance
criteria
InstalledAmperageCapacity
(A)
Equipment ID
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Traction Power System Design
Verification on Equipment Peak Amperage CapacityAcceptance Criteria:The Installed Amperage Capacity shall higher than the Maximum Peak TrainCurrent.
5
5
6
6
6
6
42
Capacity utilized( %)
1340
1340
2680
1340
1340
1340
2680
PeakTrain Current
(A)
Yes30290Power Rail
Yes27720L1, L3
Yes46000L5
Yes22000DSF1, DSF2
Yes2200052F1, 52F2
Yes2200052S1, 52S2
Yes6350TR1, TR2
Fulfillacceptance
criteria
InstalledPeak Amperage
Capacity(A)
Equipment ID
62
Given Data
- 240mm2 3/C traction power cable Impedance: 0.175+j0.125 ohm/km
- Power Rail Impedance: 0.05+j0.086 ohm/km
- Distance between TR01 to 52S1 = 50m- Distance between DSF1 to Arrival Line Power Rail = 25m
- Distance between DSF2 to Departure Power Rail = 13m
- Distance between East Hall Substation and West Hall = 770m
- Peak Starting Current at full load scenario
IA = 1340A, IB = 1340A, IC = 2680A
PDS Design Voltage Drop Assessment
Power Rail
L1
DSF1 DSF2
Arrival
LineL3
Departure Line
52F1 52F2
52S1
TR01
L5
Power Rail
West
Hall
L2
L4
IC
IA
IB
IA
IB
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Transformer Circuit Impedance7 x 240 mm2 3/CCables in parallel
Impedance of cables = (Length x Unit Impedance) / no. of parallel cables
ohmj0.00090.001250.057
j0.1250.175ZL5 +=
+=
0.00125+j0.0009(7 cables in parallel)
Total Impedance (ohm)
0.0015435.5O0.175+j0.125L5 = 0.05Cables(TR01 to 52S1ZL5
UnitImpedance(ohm/km)
Length (km)Impedance ofItem
Traction Power System Design
52F1 52F2
52S1
TR01
L5
64
Arrival Line Impedance
West
Hall
East Hall
PowerRail
4 x 240 mm2 3/C
Cables in parallel
L2
L1
DSF1
Impedance of cables = (Length x Unit Impedance) / no. of parallel cables
ohmj0.00080.00110.0254
j0.1250.175ZL1 +=
+=
ohmj0.06620.03850.025j0.086)(0.05ZL2
+=+=
0.0385+j0.0662
0.0011+j0.0008(4 cables in parallel)
Total Impedance (ohm)
0.0765860O0.05+j0.086L4 = 0.77Power Rail(feeding train @ WestHall)
ZL2
0.0013636O0.175+j0.125L3 = 0.025Cables(DSF1 to Power Rail)
ZL1
UnitImpedance(ohm/km)
Length (km)Impedance ofItem
Traction Power System Design
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Departure Line Impedance
West
Hall
East Hall
Power
Rail
4 x 240 mm2 3/C
Cables in parallel
L4
L3
DSF2
0.0385+j0.0662
0.00055+j0.0004(4 cables in parallel)
Total Impedance (ohm)
0.0765860O0.05+j0.086L4 = 0.77Power Rail(feeding train @ WestHall)
ZL4
0.0006836O0.175+j0.125L3 = 0.013Cables(DSF2 to Power Rail)ZL3
UnitImpedance(ohm/km)
Length (km)Impedance ofItem
Impedance of cables = (Length x Unit Impedance) / no. of parallel cables
ohmj0.00040.000550.0134
j0.1250.175ZL3 +=
+=
ohmj0.06620.03850.025j0.086)(0.05ZL4
+=+=
Traction Power System Design
66
Formula
Transformer Circuit - Voltage Drop Calculation
( )5L5C
Z?ICircuitrTransforme@DropVoltageL=
Traction Power System Design
52F1 52F2
52S1
TR01
L5 IC
Calculation
Power Factor : cos = 0.97 then = cos-1 0.97 = 14O
Voltage drop at Transformer Circuit is
3 x 2680-14O x (0.00154 35.5O) = 7.1321.5O
Voltage appear at Switchboard terminal =600 - 7.1321.5O = 593-0.25O 593 Volt
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Formula
DropVoltage3-VVolatgeTerminalTrainAPM L =
East
Hall
West
Hall
Power Rail
Cables IA
DSF1
IA
IA
ZL2
ZL1
Full
Load
Arrival Line - Voltage Drop Calculation
( )2L21L1A
ZZ?IEndWest@DropVoltageLL +=
Traction Power System Design
Calculation
Voltage drop at West End is3 x 1340-14O x (0.00136 36O + 0.07658 36O ) = 180.6346O
APM Train Terminal Voltage = 593 180.6346O = 486 -15O
68
Departure Line Voltage Drop Calculation EastHall
WestHall
Power Rail
Cables IB
DSF2
IB
IB
ZL4
ZL3
FullLoad
Formula
DropVoltage3-VVolatgeTerminalTrainAPM L =
( )4L43L3B ZZ?IEndWest@DropVoltage LL +=
Traction Power System Design
Calculation
Voltage drop at West End is
3 x 1340-14O x (0.000686 36O + 0.07658 36O ) = 179.1946O
APM Train Terminal Voltage = 593 179.1946O = 485 -15O
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Traction Power System Design
Voltage drop - Summary
485-15O
486-15O
593.37-0.25O
600
TerminalVoltage
(V)
179.1946OPoint C
180.6346OPoint B
7.1321.5OPoint A
0Transformer
Voltage Drop(V)
Location
Power Rail
DSF1 DSF2
Arrival
Line
Departure Line
52F1 52F2
52S1
TR01
West
Hall
A
B
Power RailC
Minimum System Voltage > Minimum Train Operation Voltage
70
Traction Power System Design
3 Phase Fault Level Calculation
Power Rail
DSF1 DSF2
Arrival
Line
Departure Line
52F1 52F2
52S1
TR01
West
Hall
F1
F2
Power RailF3
Given Data
- Transformer Impedance = 6.5%
- Base MVA = 2.5MVA
- Base Voltage = 600V 50Hz
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Traction Power System Design
Per Unit Impedance Conversion
3.85+j6.62%(0.05+j0.086) x (770/1000) x 100%Power RailF
0.057+j0.04%{(0.175+j0.125)/4} x (13/1000) x 100%240mm2 Power Cables(52F2-Power Rail)
E
0.109+j0.078%
{(0.175+j0.125)/4} x (25/1000) x 100%240mm2 Power Cables(52F1-Power Rail)
D
0.125+j0.089%
{(0.175+j0.125)/7} x (50/1000) x 100%240mm2 Power Cables(Tx-52S1)
C
j6.5%(2.5 MVA / 2.5 MVA) x 6.5%Traction TransformerB
j0.714%(2.5 MVA / 350 MVA) x 100%SourceA
ResultSequence ImpedanceDescriptionItem
Power Rail
DSF1 DSF2
Arrival
Line
Departure Line
52F1 52F2
52S1
TR01
West
Hall
F1
F2
Power RailF3
SOURCE A
B
C
D
EF
F
72
Traction Power System Design
3 Phase Fault Level Calculation
1000int6003
e Fault Polt Z at thTotal Fau
Base MVAent (kA)Fault Curr
=
4.0320+j13.963%A + B + C +
E + F
F3
4.0840+j14.0010%A + B + C +D + F
F2
0.125+j7.3030%A + B + CF1
Resultant ImpedanceTotal FaultImpedanceat FaultPoint
FaultPoint
Power Rail
DSF1 DSF2
Arrival
Line
Departure Line
52F1 52F2
52S1
TR01
West
Hall
F1
F2
Power RailF3
SOURCE A
B
C
D
EF
F
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Traction Power System Design
3 Phase Fault Level Calculation
Power Rail
DSF1 DSF2
Arrival
Line
Departure Line
52F1 52F2
52S1
TR01
West
Hall
F1
F2
Power RailF3
SOURCE A
B
C
D
EF
F
kA32.9345
1000100
3030.7125.06003
105.2F1,pointAt
6
=
+
=
j
entFault Curr
kA16.5
1000100
0010.14084.46003
105.2F2,pointAt
6
=
+
=
j
entFault Curr
kA16.55
1000100
963.13032.46003
105.2F3,pointAt
6
=
+
=
j
entFault Curr
Maximum System Fault Level 33kA
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Cable Temperature Rise Calculation
Formula:Since no heat energy is lost during the process, it is assumed:
Electrical heat generated = heat gained by the conductor
This can be expressed thus:-
( )
=+)(
0
1
0
1
2 12
1
ttP
dSWdtJ
RI
where:-J = Joules constant ( 4.18 Joules/calorie )I = Momentary Peak Load current ( A )
R1 = Resistance at temperature t1 ( ohm / cm )1 = Temperature coefficient of resistance of the conductor referred toa temperature t1 ( OC-1 )
t2 = Maximum conductor temperature ( OC )= conductor temperature rise ( OC )
W = mass of the conductor ( g / cm )S = specific heat of the conductor ( calories / g / OC )P = duration of momentary Peak Load ( seconds )
76
Cable Temperature Rise Calculation
By re-arranging the formula:
+=
)(
0 10
1
2 12
1
ttPd
dPJSW
RI
( )[ ]t-1lnP
1211
2
1 tJSW
RI +=
Temperature RiseT, i.e. (t2-t1)
1
1-
12
1
JSW
PRI
e
T
=
then
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Cable Temperature Rise Calculation
Calculations:N.B. 1: All cable parameter shall be referred to Chapter 13 of
Electrical Engineers Reference Book (14th edition) by M G Say and M ALaughton.
N.B. 2: The cable is operating at ambient temperature of 30OC initially.
Mass of a single 240mm2 conductor ( W ) Density of copper = 8890 kg / m3
Area of conductor core = 240 mm2 = 0.24x103 m2
Mass of a conductor ( W ) = 8890 x 0.24 x 103 kg / m= 21.336 g / cm
78
Temperature coefficient of resistance of the conductor at 30OC ( 1 ) 1 = 0.0039 OC 1
Resistance of the conductor ( R1 )
Resistance of the conductor at 20O
C = 0.0754 ohm/km
Resistance of the conductor at 30OC = 0.0754 [ 1 + 0.0039 x 10 ]= 0.0783x10-5 (ohm/cm)
Specific Heat of the conductor ( S ) S = 0.385 (calories / g / OC )
Joules Constant ( J ) J = 4.18 Joules/calories
Cable Temperature Rise Calculation
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Temperature Rise Calculation:
1
1-
12
1
JSW
PRI
eT
=
0.050.04Calculated Temperature Rise T ( OC )
3030Initial cable temperature ( OC )
383(= 2680 7)
335(= 1340 4)
Peak Current per cable (A)
1515Duration of peak current (seconds) - P
26801340Momentary Peak Current (A)
74No. of cables installed
240240Cable size (mm2)
L5L1 & L3
SectionSectionItem
Cable Temperature Rise Calculation
80
Harmonic Problems in AC Traction System
Impact of Harmonics Overheating of conductors and electrical equipment Mechanical oscillation of electric machine Telecommunication interference
Inaccurate meter reading Disturbances to sensitive electronic equipments Nuisance operation of protection equipments
Solution Harmonic Filter
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PDS - Harmonic Problem and Solution
Elimination Method By Power Factor Correction Equipment Harmonic Filter
600V Busbar
CapacitorBank
5th FL 7th FL 11th FL
82
Harmonic Filter Calculation
System Current Characteristics Always contain 5th, 7th, 11th , 13th & 17th harmonics due
to power electronics traction control
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Harmonic Filter Capacity Calculation
Target : System Power Factor shall maintain at 0.9 lagging
Given data
Maximum Train Current IMAX = 1383.1A
RMS of Train Current IRMS = 618.5A
Total active power consumptionof trains = 240kW
84
Example Harmonic Filter Value Calculation
1. Calculate capacity of power factor improvement equipment value
System Active power = 240kW
System Apparent power without LC filter and capacitor bank
= 3 ? IRMS x V= 3 x 618.5 x 600
= 642.8 kVA
System power factor without LC filter and capacitor bank
= 240 642.8= 0.37
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Example Harmonic Filter Value Calculation
Required capacity of harmonic filter for p.f. improvement
Here
P = Apparent Power = 642.8 kVA
cos ?1 = power factor without capacitor bank = 0.37
cos ?2 = Target system power factor = 0.9 KO = Utilization factor = 0.65
O
"
2
2
2
1
1
"
K
1-)(cos?
1-1-
)?(cos
1?cosPQ
=
=
86
Example Harmonic Filter Value Calculation
kVAr750
0.65
482
K
kVAr482
1-0.9
1-1-
0.37
10.37642.8Q
O
"
22
"
=
=
=
=
Harmonic Filter capacity Q :
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Example Harmonic Filter Value Calculation
2. Calculate capacity of tuned filter capacity
The LC filter shall compensate the current of each harmonic order based on the
system maximum current.
Harmonic current Ia at each harmonic order
= IMAX x K1
The value of K1 (in %) of individual harmonic order ampere is in accordance with thereference data of Uk Electrotechnical Committee.
Required LC filter capacity at each order of harmonic
= ( 3 ? Ia x V) / K2
The value of K2 is utilization factor of individual harmonic order
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Example Harmonic Filter Value Calculation
LC value of 5th harmonicK1 = 18.5%, K2 = 1.2,
Ia 5th = 1383.1 x 18.5% = 256A
Required LC filter capacity
= ( 3 ?256 x 600) / 1.2= 221.7 kVAr
250 kVAr
LC value for 7th harmonicK1 = 12%, K2 = 1.0,
Ia 7th = 1383.1 x 12% = 166A
Required LC filter capacity= ( 3 ?166 x 600) / 1.0
= 172.5 kVAr 200 kVAr
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Example Harmonic Filter Value Calculation
LC value of 11th harmonicK1 = 6%, K2 = 0.8,
Ia 11th = 1383.1 x 6% = 83A
Required LC filter capacity= ( 3 ? 83 x 600) / 0.8
= 107.8 kVAr 150 kVAr
LC value of tuned filter 5th harmonic filter = 250kVAr 7th harmonic filter = 200kVAr
11th harmonic filter = 150kVAr
Capacity of capacitor bank
= 750 250 200 150
= 150 kVar
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Example Harmonic Filter Value Calculation
750 kVArTotal
150 kVArCapacitor bank
150 kVAr11th
200 kVAr7th
250 kVAr5thHarmonic
Filter
CalculatedCapacity
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