02 Power System Protection Day 1 (114).pdf

7/26/2019 02 Power System Protection Day 1 (114).pdf

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Modern Power System Protective Relaying

Modern Power System Protective

Relaying

With Expert Course Faculty

Jelica Polimac

DAY 1

1




CONTENTS 1/2 INTRODUCTION TO THE TRAINING INTRODUCTION TO PROTECTION

Protection Role Protection Objectives Protection Requirements

Reliability Aspect Techno-Economical Aspects Protection in Power System

Protection Basic Principles Protection Principle Diagram Principle of a Unit Protection Principle of a Non-unit Protection Protection Types

Protection Types Protection Function Codes Relay Protection History

Electromechanical relays Static / Solid state relays Digital Numerical relays

Numerical Protective Relays Numerical Protection Concept Numerical Protection Signal Processing Numerical Protection Connections Numerical Protection Applications

POWER SYSTEM FAULT ANALYSIS

Power System Basics• What is a Power System?• Power Systems Types• Power System Parts• Power System Components• Terminology

Faults in Power Systems Type of Faults Balanced & Unbalanced Faults Fault Effects on the Power System Fault Current Factors Affecting a Fault Distorted Waveform

2




CONTENTS 2/2 POWER SYSTEM FAULT ANALYSIS (CONTINUES)

Power System Analysis Short Circuit Calculation Method

1. SC Calculations Basics2. Symmetrical Components

1. Positive Sequence System2. Negative Sequence System3. Zero Sequence System

3. Symmetrical Components for Faults1. Three-Phase Fault2. Earth Fault3. Two-Phase Fault4. Open Circuit

Symmetrical Components Example Modelling Components

• Generators Model• Transformers Model• Overhead Lines Model• Cables Model• Motors Model• Network Infeed• Load• Case Study: Applying Models

Short Circuit Calculation Procedure• Calculation Block Diagram• Standards for SC Calculations• Elements Affecting SC Calculation• SC Calculations by Computer Program• Short Circuit Calculation (Typical Data)• Short Cirdcuit Calculation (Typical Results)

Load Flow Calculations

Case Studies

3




Protection Role

Protection Objectives

Protection RequirementsProtection Basic Principles

Protection Types

Protection CodesRelay Protection History

Numerical Relays

INTRODUCTION TO PROTECTION

4



Modern Power System Protective Relaying5

Protection Role

What is protection?What is the role of protection?

400-1

L2

~GRID

~

M M

400-2

132-1

132-2

11-111-3

11-2

11-4

LV-M1 LV-M2

LV-L

L3L

Load2

Load1

L4

L1

Load

132L

400L

L5 L4

AT1 AT2

T1

T2

M1 M2

Load1

Load2

~




Objectives

Protect equipment from damage

Protect people from injury

Support uninterrupted power supply

6

Protection Objectives




Speed Sensitivity

Selectivity

Reliability Dependability (zone faults)

Security (no maloperation)

Specific requirements for protection types

7

Protection Requirements

F1F2F3F4

R

~

A B C

T




Principle of a Unit Protection Principle of a Non-unit Protection

8

Protection Basic Principles

Sensor

Circuit breaker

Measurement

Protection

Order




REF

Differential protection

Circulating current protection

Busbar protection

Restricted earth fault protection

Principle of Unit Protection

Protects a unit (selective element of the power system)Operates for internal faults within the setting range

Does not operate for any external fault outside the protective zone




OC Protection

Overcurrent protection

Earth fault protection

Voltage protection

Distance protection

Frequency protection

Principle of Non-Unit Protection

Protects more than one element of the power systemOperates for any fault (internal or external) within the setting range




Protection components: CT & VT Relay Trip relay Breaker trip coil DC supply Wiring

Protection errors and deterioration Errors in incorrect design / installation / setting Deterioration in service (component failure)

Components in series -> low reliability

Components in parallel -> high reliability

11

Reliability Aspect 1/4




Components in series (low reliability) Component failure causes system failure

12





Components in parallel (high reliability) Component failure doesn’t cause system failure

Duplicate: relays, CT, VT, Trip coils, dc supplies, wiring

13





Transmission circuits require high reliability Distribution HV circuits require medium reliability

LV circuits usually require low reliability

14


0

0.2

0.4

0.6

0.8

1

Cost

Reliability




Technical Aspects Protection requirements Reliability aspect New substation

Equipment selection (new technology)

Extending a substation Organic growth (minimum discrepancy new & old

Economical consideration Single, dual or triple main protection Value for money

Combining technical and economical aspects

15

Techno-Economical Aspect




Power system behavior Sudden load loss Switching power transformers in Switching capacitors in Harmonics Unbalanced load Faults

Power system specific System earthing (solid, isolated, NER, Petersen) Transportation / Industrial systems connection

Power system configuration Double circuits Cable feeders, Long overhead lines Running arrangements with open points Outage planning

16

Protection in Power Systems




Question:

• What would be requirements and cost

implication for protection of:

a. LV feederb. Transmission feeder




Over-current Overload protection

Earth-fault protection

Differential protection

Distance protection Busbars protection

Voltage protection

Frequency protection

Reverse power

Unbalance protection

Mechanical protection

18

Protection Types




Protection Function CodeANSI code Function

2

3

11 Multifunction element

12 Overspeed

14 Underspeed

21 Distance protection

24 Volts / Hz Flux control

25 Synchronizing

26 Thermostat Winding temperature

27 Undervoltage32 Directional power 32P, 32Q

37 Undercurrent

37P Active under power

37Q Reactive under power




Protection Function Code

ANSI code Function38 Rotor bearing temperature

40 Field loss / Excitation loss

46 Negative sequence / unbalance

47 Negative sequence overvoltage

48 Excessive starting time supervision

49 Thermal overload

50 Instantaneous over-current 50, 50N (50G), 50BF

50BF Breaker fail protection 51, 51N

51 Delayed over-current

51N Delayed neutral earth fault

51LR Locked rotor

59 Overvoltage




Protection Function Code

ANSI code Function63 Pressure Buchholz relay

64 Earth fault Residual voltage

66 Excessive starting time

67 Directional over-current protection

67N Directional earth fault

78 Vector shift / Pole slip

79 Reclosing

81 Frequency (under / over)

86 Lockout relay/Trip circuit supervision

87 Differential protection 87, 87B, 87T, 87G




Power systems in 1880s: DC system (Thomas A Edison) AC three-phase system (Nikola Tesla)

Protection Electromechanical relays Static / Solid state relays

Digital Numerical relays

Relay Protection History

0

10

20

30

40

50

60

70

80

90

100

1 8 9 0

1 9 1 0

1 9 2 0

1 9 3 0

1 9 4 0

1 9 5 0

1 9 6 0

1 9 7 0

1 9 7 5

1 9 8 2

1 9 8 8

1 9 9 2

2 0 0 0

2 0 1 0

Electomech ptn Static ptn Digital ptn




•Moving parts – lower speed, longer reset• Robust, Reliable, accurate

• Significant wiring (logic & communication)

• Different relay names for same type (51: CDG, CDD, CAG)

• Connection to SCADA via transducers & I/P relays

• No requirements for aux supply• Deterioration due to ageing effect

• Several relays in protection

• High burden to CT & VT

• Easy plug set value

• Long service life• High maintenance

Electromechanical Relays




• No or few moving parts• Electronic components

• Connection to SCADA via interface relays &transducers

• Standard 19’’ rack design

• Low burden to CT / VT

• Requirement for aux supply

• Fast operation

• Quick reset

• Service life limited• Low maintenance

Static / Solid State Relays




• Occupy small space• Microprocessor

• Communication ports to SCADA

• Communication to other relays

• Standard relay for any application

• High functionality integration

• Different setting characteristics

• Self-monitoring

• Short lifetime due to continuesdevelopment of new technology

• Complicated setting files

• Specially trained staff for operation &maintenance

• Risk of hacking

Digital Numerical Relays



Modern Power System Protective Relaying 26

Numerical Protective Relays

• Numerical Protection Concept

• Numerical Protection Processing

• Numerical Protection Connections

• Numerical Protection Characteristics

• Numerical Protection Applications




Numerical Protection Concept• Analogue/Digital convertor• RAM – Random Access Memory• ROM – Read Only Memory• EPROM – Electrical Programmable ROM• HMI – Human Interface Machine (Local, PC, Web)

VT Inputs

A/D

Microprocessor HMI

Binary

Outputs

ROMRAM

Binary

Inputs

Communication

Port

EPROMCT Inputs




Numerical Protect ion Fron t & Back View

A – Aux supply & 4 outputs contacts

B1, B2 - CT inputs (I1,R , IY, IB )

C1, C2 – Communication ports

D1, D2 – Remote module connection ports

E – VT input, Residual voltage, Residual current

F – Communication port for old relays only

H1, H2, H3 – Input / output modules




Numerical Protect ion - HMI




Numerical Protect ion Diagram




Numerical Protect ion Schemat ic




Numerical Protection Signal Processing




Sampl ing• Sampling rate (fixed or adaptive)• Resolution• Simultaneous sampling for parallel channels

• (more channels, more precise protection)

A/D




Analogue/Digi tal Conversion

Quantify

0101 Digital

HoldSample Coding

Analogue

Word length [bit] Number of steps Resolution [%]

1 2 50

2 4 25

3 8 12.5

4 16 6.25

5 32 3.125

6 64 1.563

7 128 0.7818 256 0.391

9 512 0.195

10 1024 0.098

11 2048 0.049

12 4096 0.024




Digital Fil ter ing

20ms window length 10ms window length

• Filtered values are used to surpress transients• Longer window length, better harmonics elimination• Longer window length, slower processing




Fourier Transfo rmat ion

Fourier Transformation

Original curve

i(t)

Compute imaginary component

IS=2/N * [Ssin(w*n*Dt)*in]

Compute real component:

IS=2/N * [i0/2+iN/2+Ssin(w*n*Dt)*in]




Distance Relay - Algorithm




Slid ing Data Windows 1/2

• Longer window length, better harmonics elimination• Longer window length, slower processing




Sliding Data Window 2/2

• Placing data window for distance protection




Numerical Protection Connections

40

.

Relay

To relays

Remote

control

Bay

control

S/S control

Relay

BAY

S/S

SYSTEM

BAY BAY S/S2 S/S3

To bay

controls

To substations




Protection Connections Levels




Numerical Protection Characteristics

• Uniform design for all applications• Some variations for the type

• Self monitoring

• Multi-functionality

• Incorporates event/fault recorder• Extensive setting characteristics

• User configurable via keyboard, switches

• Accessibility (local & remote)

• Communications

• Unique IP address• Optional IEC 61850

• .




Numerical Protection in Transmission Concept of a separate relays for each main protection

Relays for 1st Main, 2nd Main, 3rd Main Protection

Provides higher reliability for protection systems

Independent aux supply for each relay

Independent trip circuits for each relay

Independent self-monitoring

1MA 2M

1 2

3M

3

+ + +

43

Numerical Protection Applications




• Concept of a relay per bay (Figure a & b)• Concept of a relay for several bays (Figure c)

Figure a

I>

+

Id AR Sy

Figure b

M

I>

+

Id In> U< t,63 t,3846

I>

+

I> I> AR

Figure c

Numerical Protection in Distribution

44




Numerical Distance & Diff Protection

45




Case Study: - Protection Requirements

Relay R is a distance protection, installed in substation A.Considering Figure above what is the right statements:

a. Relay R trips for fault F4

b. Relay R operates for fault F3

c. Relay R will not operate for short circuits in transformer T

d. Relay R protects feeder AB, transformer T and busbars B

e. Relay R is sensitive to earth faults within feeder AB

f. After F2 fault inception relay R operates within 3 secondsg. After F3 fault inception relay R operates within 50ms

h. After F4 fault inception relay R operates within 100 ms

i. Relay R trips for fault F2

j. Relay R trips for fault F1

F1F2F3F4

R

~A B C

T




Power System Basics

Faults in Power Systems

Power System Analysis Short Circuit Calculation Methods

Fault Calculation Procedure

Load Flow Calculations Case Study: LF & SC Calculations

POWER SYSTEM FAULT ANALYSIS




Power System Basics

• A Power system is a combination of electricalcomponents, which supply, transmit,

distribute and consume electrical energy

• AC Power systems:• AC 3-phase systems (Grid)

• AC 3-phase industrial / commercial systems

• AC 2-phase systems (25kV traction)

• AC 1-phase systems (Building services)

• DC systems• HVDC

• DC Traction systems




Generation Transmission Distribution

49

Power System Parts




• Generators

• Transformers

• Overhead lines (EHV, HV)

• Cables (EHV, HV, MV)• Power Quality equipment

• Rectifiers

• Motors

Power System Components




Type of Faults

Balanced & Unbalanced Faults

Fault Effects on the Power System Fault Current

Factors Affecting Fault Severity

Distorted Waveform

Faults In Power Systems




Three phase fault

Single phase fault

Two phase faultTwo phase to

earth fault

Discontinued

phaseDiscontinued phase

to earth fault

Type of Faults

Three phase to

earth fault

Discontinued

two phases




Balanced & Unbalanced Faults

5% of all faults are balanced faults80% of line faults are earth faults (unbalanced)

5% two-phase faults (unbalanced)

5% two-phase with earth faults (unbalanced)




Damage at the point of fault

Depression of the voltage during the fault

Loss of load for generators close to the fault Generators stability

Induction motors slips

VIDEO: Faults

54

Fault Effects on the Power System




Fault Current

55

If =ISC Sub-transient: If =(5-10)*In, t<0.1s

Transient: If =(2-6)*In, t=0.1-1s

Steady state: If =(0.5-2)*In, t>1s




Value of Short Circuit (MVA or kA)

Network electrical parameters

System Earthing Network configuration

DC Component

Voltage values

56

Factors Affecting a Fault




Distorted WaveformHarmonics are components of current/voltage distorted waveform

Harmonics are generated by nonlinear load (not faults)

Welders, variable speed drives, static converters, rectifiers, FC lamps, PC

computers generate harmonics

Some harmonics are used in protection to distinguish faults & disturbances




Question:

• What type of faults is the most common fault

in overhead lines:

a. Three phase faultsb. Two-phase faults

c. Earth faults

d. Broken conductor




1.Short Circuit Analysis (Fault Calculations)For control

For specifying HV equipment

For protection settings

2.Load Flow

For controlFor specifying HV equipment

For protection settings

3.Stability StudiesFor generators

For transmission system4.Harmonic Studies

For power quality equipment

5.Other Analysis

Power System Analysis




• SC Calculations Basics

• Symmetrical Components

• Symmetrical Components for Faults

• Symmetrical Components - Example

Short Circuit Calculation Method




• Short Circuit (SC) calculations are carried out to define maximum &

minimum fault currents• The equipment is chosen for the maximum fault current

• Max & Min fault and load currents are used in protection settings

• For the maximum fault current all generation is in service

• For the minimum fault current minimum generation is in service

• Sub-transient calculations for the fault inception and transient

calculation (100ms) are carried out for the maximum and minimum

fault currents

• Voltage factors, applicable for fault calculations, are listed in the Table

SC Calculations Basics 1/3

Nominal voltage Voltage Factor for the SC calculations (IEC 60038)

Ifmax Ifmin

Vn < 1kV (Vn+6%)

Vn < 1kV (Vn+10%)

1.05

1.1

0.95

1kv < Vn <-35kV 1.1 0.95

Vn >36kV 1.1 1





Calculation of impedances from the fault point

Voltage transformation (base values SB, VB )

Impedances in % -> ZS=Z400%+Z132%+Z11%Unbalanced faults (Symmetrical components)





• Base values:• SB=100MVA

• VB = select the voltage where the fault is -> 11kV

• ZB=VB2/SB=112*106/(100*106)=1.21W

• Impedances in % (OR per unit):• Data for line L: R L=0.021W/km, XL=0.16W/km, l=3km

• ZL=R L+jXL=0.021*3+j0.16*3=(0.063+j0.48)W

ZL%=100%*ZL/ZB=(5.2+j39.7)%

Arc resistance: Rarc=28700*(a+2*vw*t)/I1.4 [W]a-Arc length, vw-wind speed, t-arc duration, I-current




Question:

• Line impedance is:

• ZL%=(5.2+j39.7)%

Calculate the line impedance in a vector form ZL|j




• Symmetrical Components, developed by Fortescue

in 1920s, consider a power system as superposition

of three independent symmetrical systems

a. positive sequence system (subscript 1)b. negative sequence system (subscript 2)

c. zero sequence system (subscript 0)

• Any fault can be calculated through the symmetrical

component method• Balanced fault through ‘a single phase system’ • Unbalanced faults through symmetrical components

Symmetrical Components 1/3





Voltage:VR=VR1+VR2+VR0=V1+V2+V0 V1= 1/3 * (VR+aVY+a2VB)

VY=VY1+VY2+VY0= a2V1+aV2+V0 V2= 1/3 * (VR+a2VY+aVB)

VB=VB1+VB2+VB0=aV1+a2V2+V0 V0= 1/3 * (VR+VY+VB)

Current:

IR=IR1+IR2+IR0=I1+I2+I0 I1= 1/3 * (IR+aIY+a2

IB)IY=IY1+IY2+IY0= a2I1+aI2+I0 I2= 1/3 * (IR+a2IY+aIB)

IB=IB1+IB2+IB0=aI1+a2I2+I0 I0= 1/3 * (IR+IY+IB)

VIDEO: Symmetrical components

VR1

VY1

VB1

VR2 VB2

VY2 VR0

VB0

VY0 VR

VY

VB




• Symmetrical components of voltage, current andimpedance correspond to physical phenomena (can be

measured)

• Generators produce positive sequence component

• Faults produce zero sequence components

• For motors – positive sequence component creates

moving force. negative sequence component creates

breaking force• For Transformers – for earth-faults zero sequence

component is closed via the transformer tank





Positive sequence short circuit impedance Z1

a. Pos i t ive Sequence System




b. Negat ive Sequence System

Negative sequence short circuit impedance Z2

(also associated with motors)




Zero sequence short circuit impedance Z0

(returns via earth)

c . Zero Sequence System




Symmetrical Components For Faults

Symmetrical Components for Faultsa. 3-phase fault

b. 2-phase faultc. Earth fault

d. Open circuit




Symmetr ical Components For 3-ph Faul t

VR = VY = VB = 0

IR + IY + IB = 0

IR+IY+IB=I1+I2+I0+a2I1+aI2+I0+aI1+a2I2+I=3I0=0, I0=0

VR = E – I1Z1 - I2Z2 VY = a2E –a2I1Z1-aI2Z2 ; when multiply with a: a*VY=E –I1Z1-a*aI2Z2

0 = E-I1Z1-I2Z2-E+I1Z1+a2I2Z2=(a2-1)I2Z2, therefore I2=0

VR = 0 = E – I1Z1 - I2Z2 ; E=I1Z1

I1 = E / Z1

If3 =(U/√3)/Z1 If3-Three-phase fault current

1

ZN




Symmetr ical Components For 2-ph Faul t

IR = 0IY = -IBVY = VB

I0=0

I1= -I2=E/(Z1+ Z2)

V1=E*(Z2)/(Z1+ Z2)V2= Z2*E/(Z1+ Z2)

V0=0

If2=(U)/(Z1+Z2)

If2- two-phase fault current

ZN

2




Symmetr ical Components For Earth Faul t

VR = 0IY = IB

I1= I2=I0=E/(Z1+Z2+Z0+3Z)

V1=E*(Z2+Z0 +3Z)/(Z1+ Z2+Z0+3Z)

V2= -Z2*E/(Z1+ Z2+Z0 +3Z)

V0= -Z0*E/(Z1+ Z2+Z0 +3Z)

If1 = √3*U/(Z1+Z2+Z0 +3Z)

Z=ZN+Z A

Z N >> (isolated PS) or Z N=0 (solid earth)

ZN

3Z

Z A

1




Symmetr ical Components For Open Circui t

IR = 0

IY = -IBVY = VB

-I1 = I2 + I0




Question:

• Where are the symmetrical components used

and why?




• Generators Model

• Transformers Model

• Overhead Lines Model

• Cables Model• Motors Model

• Network Infeed

• Load

• Applying Models

Modeling Components




• Synchronous generators are most complex equipment inthe power system

• Currents and voltages are calculated through differential

Laplace equations

Generators Model 1/2




• Sub transient period (80-120ms): Xd’’ (XST)• Transient period (up to 1s): Xd’ (XT)

• Steady state period: Xd (XS)

Generators Model 2/2




Transformers Model

• Model for two winding, 3-phase transformers

• Model for three winding, 3-phase transformers

• Model for 3-phase auto transformers• Model for single phase transformers




Two Wind ing Trans former Model

ZT= (uk/100% )* (UT2

/ ST), RT= (pk/100% )* (UT2

/ ST)uk (short circuit % voltage), pk (short circuit % transformer losses)




Three Wind ing Transfo rmer Model

a. Positive sequence b. Zero sequence

ZHL=(uHL/100)*(Ur 2

/SkHL), ZLT=(uLT/100)*(Ur 2

/SLT), ZHT=(uHT/100)*(Ur 2

/SHT)




Auto Trans former Model

a. Positive sequence b. Zero sequence

ZHL=(uHL/100)*(Ur 2/SkHL)

ZLT=(uLT/100)*(Ur 2/SLT)

ZHT=(uHT/100)*(Ur 2/SHT)




Zl = Rl + jXl

Values per unit length

RL’= r / qn

84

Overhead Lines Model

d = 3 (dL1L2*(dL1L2*(dL2L3*dL3L1) – geometric mean distance

between conductors, or the centre of bundles

r – Radius of a single conductor or for conductor bundles

the radius is r B = n (nrRn-1) where R is the bundle radius

n – Number of bundled conductors, m0=4px10-7 H/m




The equivalent model for OHLis applicable for cables

Zl = Rl + jXl

Rl and Xl dependent on thegeometry

Rl and Xl are measured and

recorded in commissioning

85

Cables Model




Similar modeling as for the generators Motor characteristics are dependent on construction

Sub transient characteristic is related to the motor

inertia

Standard IEC 60909; take into account only if thesum of motor’s rated currents is greater than Ik/100

(Ik – short circuit current)

Pragmatic value 4 to 6 times rated current

Contribution to the fault level for smaller motors can

be often neglected

86

Motors Model




Network Infeed

87




1. Load Type:

1. Inductive load (L)

2. Capacitive load (C)

3. Resistive load (R)

2. Load modeling:

1. S[MVA] and power factor

Load




Applying Models 1/3

Source: 20kA, 132kV

Transformer T: 132/33kV, 60MVA, uk =12%, pk =0.3%, Yd1, Z1=Z0

Feeder L: 3km, R L1=0.021 W/km, XL1=0.16 W/km

R L0

=0.12 W/km, XL0

=0.04 W/km

Non-rotating load at A

Non-rotating load at F

For fault K3 calculate: a. 3-phase fault current b. Earth fault current




Applying Models 2/3

Select SB=100MVA, UB=33kV

a: Three Phase Fault

The equivalent diagram for the considered network is shown above

For positive sequence system impedances are:

ZB=VB2/SB=332*106/(100*106)=10.9WZS1=V/(√3*I)=132*103/(√3*20*103)=3.81W

ZS1%=100%*ZS1/ZB=100%*3.81/10.9=34.95%

XT1=12*100/60=20%, RT1=0.3*100/60=0.5%, ZT1=0.5+j20%

ZL1=R L1+jXL1=0.021*3+j0.16*3=(0.063+j0.48)

ZL1%=100%*ZL1/ZB=(0.58+j4.4)%ZS3%=ZS1+ZT1+ZL1=j34.95+0.5+j20+0.58+j4.4=1.08+j59.35=59.36|89

Sf3= (SB/ ZS3%)*100%=10000/59.36=168.46MVA

If3= Sf3/(√3*33*103)=2.95kA

ZS1

~ E

ZT1 ZL1




Applying Models 3/3b: Earth Fault

The equivalent diagram for the considered system

Positive sequence impedances are as calculated for

the 3-phase fault

Negative sequence impedances are equal to positive

For the zero sequence system impedances are:

ZB=10.9%

ZS0=ZS1=34.95%

ZT0=ZT1=0.5+j20%

ZL0=R L0+jXL0=0.12*3+j0.04*3=(0.36+j0.12)

ZL0

%=100%*ZL0

/ZB

=(3.3+j1.1)%

ZS1%=3*(ZS1+ZT1)+2*ZL1+ZL0=3*(j34.95+0.5+j20)+2*(0.58+j4.4)+3.3+j1.1=

=5.96+j174.75=174.85|88

Sf1= (SB/ ZS1%)*100%=10000/174.85=57.19MVA

If1= Sf1/(√3*33*103)=1.001kA

ZS1

~ E

ZT1 ZL1

ZS2 ZT2 ZL2

ZS0 ZT0 ZL0




• Calculation Block Diagram

• Standards for Fault Calculations

• Elements Affecting SC Calculation• SC Calculations by Computer Program

• Short Circuit Calculations

Fault Calculation Procedure




Calculation Block Diagram

9393




IEC 60909 and 61393 ANSI / IEEE Standard C37 and UL 489

94

Standards for Fault Calculation




Calculation is based on equivalent voltages at thepoint of the fault (LF is required prior to SC)

The introduction of a voltage factor c is necessary

for various reasons (IEC 60909-0, 1.3.15). These

are:

voltage variation depending on time and place;

changing of transformer taps;

neglecting loads and capacitances by calculatingaccording to IEC 60909-0 (see 2.3.1);

the sub-transient behaviour of generators, power-

station units and motors.

95

IEC 60909




The Method is described in IEEE Std.C37.010-1979 and its revision in 1999,

is used for high-voltage (above 1000V)

equipment

The IEEE standard permits theexclusion of all 3-phase induction

motors below 50 hp and all single-

phase motors. Hence, no reactance

adjustment is needed for these motors.The Chart at right clarifies the

ANSI/IEEE procedure.

ANSI IEEE C37.010-1979




• Capacitors & non rotating load (not affecting calc)

• Static convertors (initial contribution to If ’’, but nocontribution on SC breaking current)

• Limiting reactors (taken as part of the current return)• Motors (to take it into account is: Sin > Ik/100):

• Synchronous motors begin to function like generators and feed

the fault (sub-transient reactance x’’d is applied to dissipate

energy stored in motors)• Asynchronous motors (neglected for some cases)

• The contribution of LV motors is negligible

Elements Affecting SC Calculation




• Power System Analysis and Studies:

• Use proven software

• Correct electrical parameters

• Verification of the model

Fault Calculations By Computer




Short Circuit Calculation (Typical Data)




Short Circuit Calculation (Typical Results)

S/S WINTER SUMMER

Sub-transient Transient Sub-transient Transient

3ph

[kA]

1ph

[kA]

3ph

[kA]

1ph

[kA]

3ph

[kA]

1ph

[kA]

3ph

[kA]

1ph

[kA]

A 30 34 22 28 19 16 12 14

B 45 38 41 37 40 35 27 30

C 23 23 22 20 20 20 12 16

D 23 21 19 24 20 21 13 16

E 46 39 42 36 41 34 29 33




• To calculate Load Flow (LF) the power system isrepresented through a model of its components

• The model for Load Flow Calculations is the

same as the model for short circuit calculations

• The load flow calculations show the flow of loadin the power system

• Maximum and minimum load flow is calculated,

which are used for equipment specification andoperations and for protection settings

• LF example is shown in the Case study

Load Flow Calculations

C St d LF&SC C l l ti




Case Study: LF&SC Calculations

400-1

L2

~

GRID

~

M M

400-2

132-1

132-2

11-111-3

11-2

11-4

LV-M1 LV-M2

LV-L

L3L

Load2

Load1

L4

L1

Load

132L

400L

L5 L4

AT1 AT2

T1

T2

M1 M2

Load1

Load2

~

Case Study: LF & SC Calculations Data




Case Study: LF & SC Calculations Data Grid Infeed: 400kV, 40kA, R/X=0.1

Generator: 21kV, 400MW, Pf=0.95, Xd=Xq=2.04pu, Xd’’=0.27pu, Rstator =0.0015pu , X0=0.1pu,

R0=0, X2=0.2pu, r 2=0,Generator transformer 21/400kV, 500MVA, uk=8.15%, ukr0=0.57%, Xm=0.29%, Rm-50,04kW,TC:-1,0,1, +1.25%

400L: 400kV, 20km, L12 tower, double lines A700mm2, earth wire Z400mm2, R1=0.54W,X1=0.48W,R0=4.78W, X0=31.6W

AT1, AT2: 400/132kV, 240MVA, uk=15%, ukr0=0.28%, Xm=0.82%, Rm-66kW, TC=1-12-15, 1.43%

132L: 132kV, 20km, L132 tower, double lines Z400mm2, earth wire L175mm2, R1=0.7W,X1=3.9W, R0=2.99W, X0=14.37W

Load 132kV: S=100MVA, Pf=0.95 ind

T1, T2: 132/11kV, 60MVA, uk=12%, ukr0=0.57%, Xm=0.29%, Rm-50.04kW, TC: 1-7-19, 1.67%

Generator 11kV, 1MVA, Pf=0.9, Xd=Xq=0.18pu, Xd’’=0.18pu, Rstator =0.027pu , X0=0.038pu,R0=0.00054pu, X2=0.18pu, r 2=0.0027pu

L, L2, L3: 11kV, XLPE 3-c 240mm2, R=0.098W/km, X=0.109W/km, R0=0.371W/km,X0=0.049W/km, B=132.3*10-6S/km

Load1, Load2: 11kV, 20MVA, Pf=0.95 indT1, T2: 11/0.4kV, 4MVA, uk=10%, ukr0=0.08%, Xm=0.05%, Rm-0.6kW, TC: 1-3-5, 2.5%

L1, L4, L5: 0.4kV, l=20m, XLPE 3-c 630mm2, R=0.06W/km, X=0.08W/km, R0=0.061W/km,X0=0.08W/km

M1, M2: Induction motor, 2.1625MVA, RS=0.008pu, XS=0.105pu, Xm=5.25pu, Rr =0.01pu,Xr =0.144pu, R/X=0.5

Load: 0.4kV, 3.5MVA, Pf=0.95 ind

Case Study: LF Calculation




Case Study: LF Calculation

Case Study: LF Calculation Results




Case Study: LF Calculation Results Object V [kV] P [MW] Q [MVAR] IL [kA]

Grid 400 84.73 -10.17

AT1 400 115.96 43.44 0.18

132 -115.72 -32 0.54

GT 400 -149.51 -43.92 0.22

AT2 400 118.28 45.37 0.18

132 -118.03 -33.51 0.55

400L 400 -31.23 -53.61 0.09

400 31.23 -1.45 0.05

132L 132 -1.93 9.05 0.04

132 1.93 -5.71 0.03

T1 132 22.65 9.82 0.11

11 -22.53 -8.43 1.3

T2 132 21.1 7.99 0.1

11 -20.99 -6.8 1.19

L 11 1.43 0.54 0.08

11 -1.42 -0.59 0.08

Case Study: SC Calculation










Case Study: SC Calculation Results




Case Study: SC Calculation Results

S/S - Busbars Max Fault current [kA]

400-1 42.44

400-2 20.12

132-1 12.72

132-2 12.53

11-1 28.62

11-2 27.7

11-3 17.75

11-4 15.2

LV-M1 44.6

LV-M2 44.6

LV-L 10.47

C St d / Q ti i




Case Study / Questionnaire

C St d 1 1




For the circuit shown on the figure below, calculate the following:

a. Two-phase fault current at beginning of L2 line

Data:Grid Infeed: 400kV, 40kA, R/X=0.1

T1: 400/132kV, 240MVA, uk=15%, pk=0.3%, Yy0

T2: 132/11kV, 30MVA, uk=12%, pk=0.6%, Yd11

L1: R1=0.035W/km, x1=0.195W/km, R0=0.15W/km, x0=0.72W/km, 8km


Case Study 1.1: – SC Calculation

L2

400/132kV 132/11kV

400kV T1

L1

T2

C St d 1 2




For the circuit shown on the figure below, calculate the following:a. Three-phase fault current when all transformers are in service b. Three phase fault current when transformer T2 is out of service


T1: 400/132kV, 240MVA, uk=12%, pk=0.4%, Yy0

T2: 132/11kV, 50MVA, uk=10%, pk=0.6%, Yd5T3: 132/11kV, 50MVA, uk=10%, pk=0.6%, Yd5L1: R1=0.035W/km, x1=0.195W/km, R0=0.15W/km, x0=0.72W/km, 10km


400/132kV

L1 11kV

A

T1

T3

T2,

C St d 1 3




For the circuit shown below, calculate the following:a. Two-phase fault current at the end of L3 line

DataGrid Infeed: 400kV, 40kA, R/X=0.1

T1: 400/132kV, 240MVA, uk=12%, pk=0.4%, Yy0T2: 132/11kV, 50MVA, uk=10%, pk=0.6%, Yd5L1: R1=0.035W/km, x1=0.195W/km, R0=0.15W/km, x0=0.72W/km, 12km

L2: R1=0.035W/km, x1=0.195W/km, R0=0.15W/km, x0=0.72W/km, 12kmL3: R1=0.035W/km, x1=0.195W/km, R0=0.15W/km, x0=0.72W/km, 6km


400/132kV

400kV

132/11kV

T1

L1

T2

L3

L2

C St d 1 4




For the circuit shown on the figure below, calculate the following:

a. Three-phase fault current in s/s D when L2 and L3 are in service

b. Three-phase fault current in s/s D when L3 is out of service


T1: 400/132kV, 200MVA, uk=12%, pk=0.3%





L2

400/132kV

A T1

L1

L3 D

C St d P t ti R i t



Case Study: Protection Requirements

F1F2F3F4

R

~

A B C

T

Relay R is a distance protection, installed in substation A.Considering Figure above what is the right statements:

a. Relay R trips for fault F4 Yesb. Relay R operates for fault F3 Yesc. Relay R will not operate for short circuits in transformer T Yesd. Relay R protects feeder AB, transformer T and busbars B Noe. Relay R is sensitive to earth faults within feeder AB Yesf. After F2 fault inception relay R operates within 3 seconds Nog. After F3 fault inception relay R operates within 50ms Noh. After F4 fault inception relay R operates within 100 ms Yesi. Relay R trips for fault F2 Yesj Relay R trips for fault F1 No

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