Modern Power System Protective Relaying With Expert Course Faculty
Jelica Polimac
DAY 1 Modern Power System Protective Relaying
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CONTENTS 1/2
INTRODUCTION TO THE TRAINING INTRODUCTION TO PROTECTION
Protection Role Protection Objectives Protection Requirements
Protection Basic Principles
Electromechanical relays Static / Solid state relays Digital Numerical relays
Numerical Protective Relays
Protection Principle Diagram Principle of a Unit Protection Principle of a Non-unit Protection Protection Types
Protection Types Protection Function Codes Relay Protection History
Reliability Aspect Techno-Economical Aspects Protection in Power System
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
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CONTENTS 2/2
POWER SYSTEM FAULT ANALYSIS (CONTINUES)
Power System Analysis Short Circuit Calculation Method 1. SC Calculations Basics 2. Symmetrical Components 1. 2. 3.
3.
Symmetrical Components for Faults 1. 2. 3. 4.
Positive Sequence System Negative Sequence System Zero Sequence System Three-Phase Fault Earth Fault Two-Phase Fault 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
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INTRODUCTION TO PROTECTION Protection Role Protection Objectives
Protection Requirements Protection Basic Principles Protection Types Protection Codes Relay Protection History Numerical Relays Modern Power System Protective Relaying
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Protection Role What is protection? What is the role of protection? ~ GRID
400L
400-1
400-2 AT2
AT1
132-2
132L
132-1 T2
Load1
T1
~
L4
11-1
L2
Load2
11-3
Load1
L3
L L5
LV-M1
11-4
L4
LV-M2
M
M
M1
M2
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Load2
11-2
~
L1
LV-L Load
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Protection Objectives
Objectives
Protect equipment from damage
Protect people from injury
Support uninterrupted power supply
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Protection Requirements
Speed
Sensitivity
Selectivity
Reliability Dependability (zone faults)
Security (no maloperation)
Specific requirements for protection types A
B
R
C
~ F4
F3
F2
F1
T 7
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Protection Basic Principles Principle of a Unit Protection Principle of a Non-unit Protection
Order Circuit breaker
Measurement Sensor
Protection
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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
Differential protection
Circulating current protection Busbar protection
Restricted earth fault protection
REF
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Principle of Non-Unit Protection Protects more than one element of the power system Operates for any fault (internal or external) within the setting range
Overcurrent protection
Earth fault protection Voltage protection
Distance protection
Frequency protection
OC Protection
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Reliability Aspect 1/4
Protection components:
DC supply Wiring
Protection errors and deterioration
CT & VT Relay Trip relay Breaker trip coil
Errors in incorrect design / installation / setting Deterioration in service (component failure)
Components in series -> low reliability Components in parallel -> high reliability
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Reliability Aspect 2 /4
Components in series (low reliability)
Component failure causes system failure
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Reliability Aspect 3/4
Components in parallel (high reliability) Component failure doesn’t cause system failure
Duplicate: relays, CT, VT, Trip coils, dc supplies, wiring
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Reliability Aspect 4 /4
Transmission circuits require high reliability Distribution HV circuits require medium reliability LV circuits usually require low reliability Reliability 1 0.8 0.6 0.4 0.2 0
Cost 14
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Techno-Economical Aspect
Technical Aspects
Protection requirements Reliability aspect New substation
Organic growth (minimum discrepancy new & old
Economical consideration
(new technology)
Extending a substation
Equipment selection
Single, dual or triple main protection Value for money
Combining technical and economical aspects 15
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Protection in Power Systems
Power system behavior
Power system specific
Sudden load loss Switching power transformers in Switching capacitors in Harmonics Unbalanced load Faults 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
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Question: •
What would be requirements and cost implication for protection of: a. b.
LV feeder Transmission feeder
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Protection Types Over-current Overload protection Earth-fault protection Differential protection
Distance protection Busbars protection Voltage protection Frequency protection Reverse power Unbalance protection Mechanical protection
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Protection Function Code ANSI code
Function
2
3 11
Multifunction element
12
Overspeed
14
Underspeed
21
Distance protection
24
Volts / Hz
25
Synchronizing
26
Thermostat
27
Undervoltage
32
Directional power
37
Undercurrent
37P
Active under power
37Q
Reactive under power
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Flux control
Winding temperature
32P, 32Q
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Protection Function Code ANSI code Function 38
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
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Protection Function Code ANSI code
Function
63
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
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87, 87B, 87T, 87G
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Relay Protection History
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
100
90 80 70 60 50 40 30 20 10 0
0 9 8 1
0 1 9 1
0 2 9 1
0 3 9 1
0 4 9 1
Electomech ptn Modern Power System Protective Relaying
0 5 9 1
0 6 9 1
0 7 9 1
5 7 9 1
Static ptn
2 8 9 1
8 8 9 1
2 9 9 1
0 0 0 2
0 1 0 2
Digital ptn 22
Electromechanical Relays • 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
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Static / Solid State 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
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Digital Numerical 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 continues
development of new technology • Complicated setting files • Specially trained staff for operation & maintenance • Risk of hacking Modern Power System Protective Relaying
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Numerical Protective Relays • Numerical Protection Concept • Numerical Protection Processing • Numerical Protection Connections • Numerical Protection Characteristics • Numerical Protection Applications
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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)
RAM
CT Inputs
ROM
EPROM Binary Outputs
A/D VT Inputs Binary Inputs
Microprocessor
HMI
Communication Port
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Num erica l P rotection Fron
t & B ack Vie w
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 Modern Power System Protective Relaying
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Num erical Protectio
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n - HMI
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Num erical Prote ctio n Dia g ra m
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Num erica l Prote ction Sch em atic
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Numerical Protection Signal Processing
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Sampling Sampling rate (fixed or adaptive) Resolution Simultaneous sampling for parallel channels
• • • •
(more channels, more precise protection)
A/D
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A nalogu e/Di gita l Con version Sample
Hold
Quantify
Analogue
Coding 0101
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
0.781
8
256
0.391
9
512
0.195
10
1024
0.098
11
2048
0.049
12
4096
0.024
Digital
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Digital Fi lte rin g • • •
Filtered values are used to surpress transients Longer window length, better harmonics elimination Longer window length, slower processing
20ms window length
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10ms window length
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Fo urier Tra n sfo rm ation 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]
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Distance Relay - Algorithm
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Sli d ing Da ta Win d o w s 1/ 2 • •
Longer window length, better harmonics elimination Longer window length, slower processing
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Sliding Data Window 2/2 • Placing data window for distance protection
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Numerical Protection Connections
Remote control To substations
SYSTEM S/S control
. To relays
S/S
Relay
To bay controls
Bay control
Relay BAY
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BAY BAY
S/S2
S/S3
40
Protection Connections Levels
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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 • .
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Modern Power System Protective Relaying
Numerical Protection Applications
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
A
+
+
+
1M
2M
3M
1
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2
3
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Numerical Protection in Distribution Concept of a relay per bay (Figure a & b) • Concept of a relay for several bays (Figure c) •
+
Id
+
I> AR Sy
I>
I>
I> A R
Figure a +
Id
I> In> U< 46
t,63
t,38
Figure c
M
Figure b
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N umer i cal D i stance & D i ff Prote ction
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Case Study: - Protection Requirements A
B
R
C
~ T
F4
F3
F2
F1
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 seconds g. After F3 fault inception relay R operates within 50ms h. After F4 fault inception relay R operates within 100 ms Relay R trips for fault F2 i. j. Relay R trips for fault F1 Modern Power System Protective Relaying
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POWER SYSTEM FAULT ANALYSIS
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
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Power System Basics • A Power system is a combination of electrical
components, 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 Modern Power System Protective Relaying
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Power System Parts Generation Transmission Distribution
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Power System Components • Generators • Transformers • Overhead lines (EHV, HV) • • • •
Cables (EHV, HV, MV) Power Quality equipment Rectifiers Motors
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Faults In Power Systems
Type of Faults
Balanced & Unbalanced Faults
Fault Effects on the Power System Fault Current
Factors Affecting Fault Severity
Distorted Waveform
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Type of Faults Three phase fault
Three phase to earth fault
Single phase fault
Discontinued phase
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Two phase fault
Discontinued two phases
Two phase to earth fault
Discontinued phase to earth fault
52
Balanced & Unbalanced Faults 5% of all faults are balanced faults 80% of line faults are earth faults (unbalanced) 5% two-phase faults (unbalanced) 5% two-phase with earth faults (unbalanced)
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Fault Effects on the Power System
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
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Fault Current 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
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Factors Affecting a Fault
Value of Short Circuit (MVA or kA)
Network electrical parameters
System Earthing Network configuration
DC Component
Voltage values
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Distorted Waveform Harmonics 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
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Question: • What type of faults is the most common fault
in overhead lines: a. b.
Three phase faults Two-phase faults
c.
Earth faults
d.
Broken conductor
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Power System Analysis 1.Short Circuit Analysis (Fault Calculations) For control For specifying HV equipment For protection settings
2.Load Flow For control For specifying HV equipment For protection settings
3.Stability Studies For generators For transmission system
4.Harmonic Studies For power quality equipment
5.Other Analysis Modern Power System Protective Relaying
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Short Circuit Calculation Method • • • •
SC Calculations Basics Symmetrical Components Symmetrical Components for Faults Symmetrical Components - Example
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SC Calculations Basics 1/3 • 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 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
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SC Calculations Basics 2/3
Calculation of impedances from the fault point Voltage transformation (base values SB, VB ) Impedances in % -> ZS=Z400%+Z132%+Z11% Unbalanced faults (Symmetrical components)
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SC Calculations Basics 3/3 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: RL=0.021W/km, XL=0.16W/km, l=3km
•
ZL=RL+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
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Question: Line impedance is:
• •
ZL%=(5.2+j39.7)%
Calculate the line impedance in a vector form Z |j L
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Symmetrical Components 1/3 • 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 Modern Power System Protective Relaying
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Symmetrical Components 2/3 VB1
VB
VY2 VR1
VR VY
VY1
VB2
VR2
VR0 VY0 VB0
Voltage: VR=VR1+VR2+VR0=V1+V2+V0 VY=VY1+VY2+VY0= a2V1+aV2+V0 VB=VB1+VB2+VB0=aV1+a2V2+V0
V1= 1/3 * (VR+aVY+a2VB) V2= 1/3 * (VR+a2VY+aVB) V0= 1/3 * (VR+VY+VB)
Current: IR=IR1+IR2+IR0=I1+I2+I0 IY=IY1+IY2+IY0= a2I1+aI2+I0 IB=IB1+IB2+IB0=aI1+a2I2+I0
I1= 1/3 * (IR+aIY+a2IB) I2= 1/3 * (IR+a2IY+aIB) I0= 1/3 * (IR+IY+IB)
VIDEO: Symmetrical components Modern Power System Protective Relaying
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Symmetrical Components 3/3 • Symmetrical components of voltage, current and impedance 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
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a. Positi ve Sequence Syste m
Positive sequence short circuit impedance Z1
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b. Ne ga ti ve S equence S yste m
Negative sequence short circuit impedance Z2 (also associated with motors)
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c. Zero S eque nce Syste m
Zero sequence short circuit impedance Z0 (returns via earth) Modern Power System Protective Relaying
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Symmetrical Components For Faults Symmetrical Components for Faults a. 3-phase fault b. 2-phase fault c. Earth fault d. Open circuit
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Sym m etrica l C om po nents For 3 -ph Fault 1
ZN
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 I 1 = E / Z1 If3 =(U/√3)/Z1 If3-Three-phase fault current Modern Power System Protective Relaying
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Sym m etrica l C om po nents For 2 -ph Fault 2
ZN
IR = 0 IY = -IB VY = V B 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 Modern Power System Protective Relaying
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Sym m etrica l C om po nents For E arth Fa ult 1 ZA ZN
VR = 0 IY = 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+ZA ZN >> (isolated PS) or ZN=0 (solid earth) Modern Power System Protective Relaying
3Z
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Sym m etri ca l C om po nents F or Op en Circuit
IR = 0 IY = -IB VY = VB -I1 = I2 + I0
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Question: •
Where are the symmetrical components used and why?
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Modeling Components • Generators Model • Transformers Model • Overhead Lines Model •• • • •
Cables Model Motors Model Network Infeed Load Applying Models
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Generators Model 1/2 • Synchronous generators are most complex equipment in the power system • Currents and voltages are calculated through differential Laplace equations
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Generators Model 2/2 • Sub transient period (80-120ms): Xd’’ (XST) • Transient period (up to 1s): Xd’ (XT) • Steady state period: Xd (XS)
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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
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Tw o Wi nd ing Tra ns form er Mod el ZT= (uk/100% )* (UT2 / ST),
RT= (pk/100% )* (UT2 / ST)
uk (short circuit % voltage), p k (short circuit % transformer losses)
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Th re e Wind ing Tra n sfo rm er M o d el ZHL=(uHL/100)*(Ur2/SkHL), ZLT=(uLT/100)*(Ur2/SLT), ZHT=(uHT/100)*(Ur2/SHT)
a. Positive sequence
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b. Zero sequence
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A uto T ra ns form er M od el ZHL=(uHL/100)*(Ur2/SkHL) ZLT=(uLT/100)*(Ur2/SLT) ZHT=(uHT/100)*(Ur2/SHT)
a. Positive sequence
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b. Zero sequence
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Overhead Lines Model Zl = Rl + jXl Values per unit length RL’= r / qn
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 rB = n (nrRn-1) where R is the bundle radius n – Number of bundled conductors, m0=4px10-7 H/m Modern Power System Protective Relaying
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Cables Model The equivalent model for OHL is applicable for cables Zl = Rl + jXl Rl and Xl dependent on the geometry Rl and Xl are measured and recorded in commissioning
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Motors 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 the sum 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
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Network Infeed
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Load 1.
2.
Load Type: 1.
Inductive load (L)
2. 3.
Capacitive load (C) Resistive load (R)
Load modeling: 1.
S[MVA] and power factor
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Applying Models 1/3
Source: 20kA, 132kV Transformer T: 132/33kV, 60MVA, uk=12%, pk=0.3%, Yd1, Z1=Z0 Feeder L: 3km, RL1=0.021 W/km, XL1=0.16 W/km RL0=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 Modern Power System Protective Relaying
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Applying Models 2/3 ZS1
ZT1
ZL1
Select SB=100MVA, UB=33kV ~
E
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)=10.9W ZS1=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=RL1+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 Modern Power System Protective Relaying
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Applying Models 3/3 b: 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
~
ZS1
ZT1
ZL1
ZS2
ZT2
ZL2
E
For the zero sequence system impedances are: ZB=10.9% Z Z Z ZS0=ZS1=34.95% ZT0=ZT1=0.5+j20% ZL0=RL0+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 S0
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L0
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Fault Calculation Procedure • Calculation Block Diagram • Standards for Fault Calculations • Elements Affecting SC Calculation • SC Calculations by Computer Program • Short Circuit Calculations
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Calculation Block Diagram
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Standards for Fault Calculation
IEC 60909 and 61393 ANSI / IEEE Standard C37 and UL 489
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IEC 609 09
Calculation is based on equivalent voltages at the point 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 calculating according to IEC 60909-0 (see 2.3.1);
the sub-transient behaviour of generators, powerstation units and motors.
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A NSI I EEE C37. 010- 1979
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 the exclusion of all 3-phase induction motors below 50 hp and all singlephase motors. Hence, no reactance adjustment is needed for these motors. The Chart at right clarifies the ANSI/IEEE procedure.
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Elements Affecting SC Calculation • Capacitors & non rotating load (not affecting calc) • Static convertors (initial contribution to If’’, but no contribution 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
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Fault Calculations By Computer
• Power System Analysis and Studies: • Use proven software • Correct electrical parameters • Verification of the model
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Short Circuit Calculation (Typical Data)
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Short Circuit Calculation (Typical Results) S/S
WINTER
Sub-transient
SUMMER
Transient
Sub-transient
Transient
3ph
1ph
3ph
1ph
3ph
1ph
3ph
1ph
A
[kA] 30
[kA] 34
[kA] 22
[kA] 28
[kA] 19
[kA] 16
[kA] 12
[kA] 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
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Load Flow Calculations • To calculate Load Flow (LF) the power system is represented 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 load in the power system
• Maximum and minimum load flow is calculated, which are used for equipment specification and operations and for protection settings
• LF example is shown in the Case study Modern Power System Protective Relaying
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Case Study: LF&SC Calculations ~ GRID
400L
400-1
400-2 AT2
AT1
132-2
132L
132-1 T2
Load1
T1
~
L4
Load2
11-2 L2
Load2
11-1 11-3 Load1
L3
L L5
11-4
L4
LV-M1
LV-M2
M
M
M1
M2
~
L1
LV-L Load
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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, r2=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=0.54 W, 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% W, 132L: 132kV, 20km, L132 tower, double lines Z400mm2, earth wire L175mm2, R1=0.7 W W W X1=3.9 , R0=2.99 , X0=14.37 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, r2=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 ind T1, 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=0.06 W/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 Modern Power System Protective Relaying
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Case Study: LF Calculation
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Case Study: LF Calculation Results Object
V [kV]
P [MW]
Q [MVAR]
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
400
-31.23
-53.61
0.09
400
31.23
-1.45
0.05
132
-1.93
9.05
0.04
132
1.93
-5.71
0.03
400L
132L
T1
T2
L
IL [kA]
132
22.65
9.82
0.11
11
-22.53
-8.43
1.3
132
21.1
7.99
0.1
11
-20.99
-6.8
1.19
11
1.43
0.54
0.08
11
-1.42
-0.59
0.08
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Case Study: SC Calculation
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Case Study: SC Calculation
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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
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Case Study / Questionnaire
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Case Study 1.1: – SC Calculation 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 L2: R1=0.035W/km, x1=0.195W/km, R0=0.15W/km, x0=0.72W/km, 6km 400/132kV
132/11kV L1
400kV
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T1
L2 T2
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Case Study 1.2: – SC Calculation 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 Data: Grid Infeed: 400kV, 40kA, R/X=0.1
T1: 400/132kV, 240MVA, uk=12%, pk=0.4%, Yy0 T2: 132/11kV, 50MVA, uk=10%, pk=0.6%, Yd5 T3: 132/11kV, 50MVA, uk=10%, pk=0.6%, Yd5 L1: R1=0.035W/km, x1=0.195W/km, R0=0.15W/km, x0=0.72W/km, 10km
A
400/132kV L1
T2,
11kV
T1 T3
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Case Study 1.3: – SC Calculation For the circuit shown below, calculate the following: a. Two-phase fault current at the end of L3 line Data Grid Infeed: 400kV, 40kA, R/X=0.1
T1: 400/132kV, 240MVA, uk=12%, pk=0.4%, Yy0 T2: 132/11kV, 50MVA, uk=10%, pk=0.6%, Yd5
W/km, W/km, W/km, W/km, L1: L2: R1=0.035 R1=0.035W /km, x1=0.195 x1=0.195W /km, R0=0.15 R0=0.15W /km, x0=0.72 x0=0.72W /km, 12km 12km L3: R1=0.035W/km, x1=0.195W/km, R0=0.15W/km, x0=0.72W/km, 6km 400/132kV
L1
132/11kV L3
400kV
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L2
T2
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Case Study 1.4: – SC Calculation 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 Data: Grid Infeed: 400kV, 40kA, R/X=0.1
T1: 400/132kV, 200MVA, uk=12%, pk=0.3% L1: R1=0.035W/km, x1=0.195W/km, R0=0.15W/km, x0=0.72W/km, 15km L2: R1=0.035W/km, x1=0.195W/km, R0=0.15W/km, x0=0.72W/km, 10km L3: R1=0.035W/km, x1=0.195W/km, R0=0.15W/km, x0=0.72W/km, 10km 400/132kV L1 A
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L2
L3
D
113
Case Study: Protection Requirements A
B
R
C
~ T
F4
F3
F2
F1
Relay R is a distance protection, installed in substation A. Considering Figure above what is the right statements: Relay R trips for fault F4 a. b. Relay R operates for fault F3 Relay R will not operate for short circuits in transformer T c. Relay R protects feeder AB, transformer T and busbars B d. e. Relay R is sensitive to earth faults within feeder AB f. After F2 fault inception relay R operates within 3 seconds g. 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 Relay R trips for fault F1 j. Modern Power System Protective Relaying
Yes Yes Yes No Yes No No Yes Yes No 114