04 - Wk10 APM Handout Part 4 of 4

7/29/2019 04 - Wk10 APM Handout Part 4 of 4

1/2720

39

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


DS1



Arrival Line

Power Rail



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


- TR02 is de-energized

- 52S1 close automatically

- TR01 feeds the trains continuously

TractionCurrent

Flowdirection

DS1



Arrival Line

Power Rail



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


TractionCurrent

Flowdirection

DS1



Arrival Line

Power Rail



Departure Line

Power Rail

52FL252FL1

ZZ


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APMDEPOT

11kV RMU

52S1 52S2

11kV 3 Supply

TR01 TR02

52F352F252F1

DS0

DSF1 DSF2


- 52S1, 52F1 & 52F2 closed

- DSF1, DSF2 closed


- 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



Arrival Line

Power Rail



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


- 52S1, 52F1 & 52F2 closed

- DSF1, DSF2 closed


TractionCurrent

Flowdirection

DS1



Arrival Line

Power Rail



Departure Line

Power Rail

52FL252FL1

ZZ


4/2723

45


APMDEPOT

11kV RMU

52S1 52S2

11kV 3 Supply

TR01 TR02

52F352F252F1

DS0

DSF1 DSF2


- 52S1, 52F1 & 52F2 closed

- DSF1, DSF2 closed


- 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



Arrival Line

Power Rail



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


West HallArrivalPlatformArrival Line

East Hall

DeparturePlatform

West Hall

DeparturePlatform

Departure Line


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47

APM Train System Train Operation Modes

Dual Shuttle Running Demonstration





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 Hall

DeparturePlatform

West Hall

DeparturePlatform

Departure Line


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49

APM Train System - Train Operation Modes

Normal Pinched Loop Demonstration





Departure Line

50

APM Trains Operation Modes

Normal Pinched Loop Demonstration



East Hall

DeparturePlatform

West Hall

DeparturePlatform

Departure Line


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51

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


8/2727

53

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


Train current profile Starting and Running

Peak 1178.6ARMS 321A


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55


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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57

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 Hall

DeparturePlatform

End ofLine


West Hall

DeparturePlatform

11kV RMU

East HallSubstation

11kV 3 Supply

Power Rail

Power Rail

Traction Tx

L5 : 7 x 240 mm2 cables


4 x 240mm2

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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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


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


ohmj0.00080.00110.0254

j0.1250.175ZL1 +=

+=

ohmj0.06620.03850.025j0.086)(0.05ZL2

+=+=

0.0385+j0.0662



0.0765860O0.05+j0.086L4 = 0.77Power Rail(feeding train @ WestHall)

ZL2

0.0013636O0.175+j0.125L3 = 0.025Cables(DSF1 to Power Rail)

ZL1





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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



0.0765860O0.05+j0.086L4 = 0.77Power Rail(feeding train @ WestHall)

ZL4

0.0006836O0.175+j0.125L3 = 0.013Cables(DSF2 to Power Rail)ZL3




ohmj0.00040.000550.0134

j0.1250.175ZL3 +=

+=

ohmj0.06620.03850.025j0.086)(0.05ZL4

+=+=


66

Formula

Transformer Circuit - Voltage Drop Calculation

( )5L5C

Z?ICircuitrTransforme@DropVoltageL=


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 +=


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 +=


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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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


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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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



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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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


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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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



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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


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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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

QQ

1-)(cos?

1-1-

)?(cos

1?cosPQ

=

=

86


kVAr750

0.65

482

K

QQ

kVAr482

1-0.9

1-1-

0.37

10.37642.8Q

O

"

22

"

=

=

=

=

Harmonic Filter capacity Q :


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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

88


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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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

90


750 kVArTotal

150 kVArCapacitor bank

150 kVAr11th

200 kVAr7th

250 kVAr5thHarmonic

Filter

CalculatedCapacity


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END

04 - Wk10 APM Handout Part 4 of 4

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