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Generator and
Power Station Protection
Synopsis
Disclaimer
Barrie Moor : 2012 [email protected]
(07) 3298 5260
GENERATOR and
POWER STATION
PROTECTION
Barrie Moor, B Eng (Elec)
Slide [email protected]
DisclaimerSeminar Synopsis
[email protected]@powersystemprotection.com.auwww.powersystemprotection.com.au
Slide [email protected]
Disclaimer
The material presented in this module is for Educational purposes
only.
This module contains a summary of information for the protection of
various types of electrical equipment. Neither the author, nor
anyone acting on his behalf, makes any warranty or representation,
express or implied, as to the accuracy or completeness of theinformation contained herein, nor assumes any responsibility or
liability for the use, or consequences of the use, of any of this
information.
The practical application of any of the material contained herein
must be in accordance with legislative requirements and must give
due regard to the individual circumstances.
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Generator and
Power Station Protection
Synopsis
Disclaimer
Barrie Moor : 2012 [email protected]
(07) 3298 5260
Slide [email protected]
Seminar Synopsis
Over Current Protection
Differential Protection
High Impedance Differential Protection
Biased Differential Protection
Sequence Components
Motor Protection
Generator Protection
Generator Faults
Generator Events
Power System Events
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Seminar costs will vary depending on attendeenumbers and your individual circumstances.However, by organising your own in-houseseminar:
You can expect savings of between 40%and 65%.
Plus you eliminate travel andaccommodation expenses for all of yourattendees.
We provide:
2 or 3 day seminar presentation All seminar handout material, notes, folders,
CDs, etc Laptop Computer and Data Projector Note that we guarantee that all seminars will be presented personally by
our principal engineer and seminar author, Barrie Moor
Each attendee receives:
Two or three day seminar presentation Hard copy manual with all presentations, plus supporting technical
papers
CD with all printed material, plus considerable extra material and tools,including: pdf of seminar colour slides 2 per page additional technical papers tools for sequence component analysis of single, double and three
phase faults tools for grading of IDMT overcurrent relays tools for distance relay calculations, apparent impedance calculations,
fault resistance, mho and offset mho characteristic load limits
Certificate of attendance
You provide:
Seminar conference room (preferably on-site, within your own facilities) Whiteboard Any catering for lunch and tea breaks
To discuss your requirements, or to obtain a firm price quotation,please contact us at:
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Generator and
Power Station Protection
Over Current Protection
Fuses and Contactors
Barrie Moor : 2012 [email protected]
(07) 3298 5260
GENERATOR and
POWER STATION
PROTECTION
Barrie Moor, B Eng (Elec)
Slide [email protected]
OVER CURRENTPROTECTION
Overcurrent Relays
Fuses & Contactors
Directional Relays
Slide [email protected]
Over Current Protection
Over Load Protection
Operation to the thermal capability of plant
Over Current Protection
Primarily for clearance of faults
Some measure of over load protection may be provided
Slide [email protected]
Discrimination by Time
Setting chosen to ensure CB nearest to the fault opens
first
Often referred to as
Independent Definite Time Delay Relay
Timing intervals selected to ensure upstream relays do
not operate before CBs trip at fault location
Disadvantage
Longest fault clearing time occurs in section closest tothe power source where fault level is the highest
Slide [email protected]
Discrimination by Time
RELAY A RELAY B RELAY C
RELAY C
RELAY B
RELAY A
CURRENT
TIME
0.4 secs
0.4 secs
Slide [email protected]
Discrimination by Current
Apply where fault current varies with fault location due to
intermediate impedance
Set to operate at current values so that only relay
nearest to fault trips its CB
Difficulties
Same fault level at the end of one zone and the start of the
next
Fault levels vary with changing source impedance(eg. As generators come on and go off line)
Slide [email protected]
Discrimination by Current
RELAY A RELAY B
Relay A cannot distinguishbetween a fault here, forwhich it needs to operate
And a fault here forwhich it should notoperate
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Generator and
Power Station Protection
Over Current Protection
Fuses and Contactors
Barrie Moor : 2012 [email protected]
(07) 3298 5260
Slide [email protected]
Discrimination by Current
Significant difference betweencurrents seen for Faults A & B
Set HV OC to 1.3 x maximum
through current for LV Fault
HV OC
FDR OC
FDR OC
FDR OC
FDR OC
A
B
Slide [email protected]
Discrimination by Time & Current
Time and currentcoordination
RELAY A RELAY B RELAY C
RELAY C
RELAY B
RELAY A
CURRENT
TIME
ICmax IBmax IAmax
IAmax IBmax ICmax
Instantaneouselement
Slide [email protected]
Inverse Over Current Relays
Time of operation inversely proportional to fault current
Faster operating times at higher fault levels
Faster operating times for faults nearer to the source
Curves generally plotted in log - log or
log(current) linear(time) format
Slide [email protected]
Discrimination w ithInverse Time Over Current Relays
Inverse time andcurrent coordination
RELAY A RELAY B RELAY C
RELAY C
RELAY B
RELAY A
CURRENT
TIME
ICmax IBmax IAmax
IAmax IBmax ICmax
Instantaneouselement
Slide [email protected]
Relay Curves to IEC 60255(BS142)
I = Actual relay current
Relay Settings
TMS = Time Multiplier Setting
P = Plug (Current) pickup setting
Usual curve for t ransmission and distribution systems
1P
I
TMS14.0TIME
02.0Inverse_dardtanS
=
Slide [email protected]
Relay Curves to IEC 60255(BS142)
I = Actual relay current
Relay Settings
TMS = Time Multiplier Setting
P = Plug (Current) pickup setting
Systems where the fault level decreases significantlybetween relaying points
1P
I
TMS5.13TIME Inverse_Very
=
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Generator and
Power Station Protection
Over Current Protection
Fuses and Contactors
Barrie Moor : 2012 [email protected]
(07) 3298 5260
Slide [email protected]
Relay Curves to IEC 60255(BS142)
I = Actual relay current
Relay Settings
TMS = Time Multiplier Setting
P = Plug (Current) pickup setting
Grading with fuses
1P
I
TMS80TIME
2Inverse_Extremely
=
Slide [email protected]
Relay Curves to IEC 60255(BS142)
I = Actual relay current
Relay Settings
TMS = Time Multiplier Setting
P = Plug (Current) pickup setting
Long time thermal protection
Motor & Generator Protection
1P
I
TMS120TIME
Inverse_Time_Long
=
Slide [email protected]
Standard Characteristics to IEC 60255
Long Time (LTI)
Extremely Inverse (EI)
Very Inverse (VI)
Standard Inverse (SI)
Relay Characteristic
1I
TMS14.002.0
1I
TMS5.13
1I
TMS802
1I
TMS120
100 1.103
1.104
0.1
1
10
100
Standard Inverse
Very Inverse
Etremely Inverse
Long Time Inverse
IDMT Relay Grading Curves
Fault Current
Seconds
Slide [email protected]
US Characteristics to IEC 60255
U5 Short Time Inverse
U4 Extremely Inverse *
U3 Very Inverse
U2 Inverse
U1 Moderately Inverse
Relay Characteristic
+1M
0104.00226.0TD
02.0
+1M
95.5180.0TD
2
+1M
88.30963.0TD
2
+1M
64.502434.0TD
2
+1M
00342.000262.0TD
02.0
TD = Time dial (TMS)M = Multiple of pick-up current
Slide [email protected]
Electro Mechanical Relays
FLUX PRODUCED BY INPUT
TAPPEDCOIL
I
- L
SHADING LOOP
FLUX PRODUCED BY INPUT CURRENT
DISC DISC
FLUX PRODUCED BY SHADING LOOP
kI
L
(1-k) I
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Generator and
Power Station Protection
Over Current Protection
Fuses and Contactors
Barrie Moor : 2012 [email protected]
(07) 3298 5260
Slide [email protected]
Instantaneous Element
Reduces tripping time at high fault levels
Allows a the discriminating curves behind the high set
element to be lowered Grading of upstream relay now occurs at the instantaneous
setting and not at maximum fault level
Minimises fault damage in both cases
Beware Simple E/M instantaneous elements may have a substantial
transient overreach on fault currents that include DC offset
OC OC
100 1.103
1.104
0.5
1
1.5
2
2.5
3
3.5
4IDMT Relay Grading Curves
Fault Current
Seconds
100 1.103
1.104
0.5
1
1.5
2
2.5
3
3.5
4IDMT Relay Grading Curves
Fault Current
Seconds
100 1.103
1.104
0.5
1
1.5
2
2.5
3
3.5
4IDMT Relay Grading Curves
Fault Current
Seconds Set Tx HV inst
element and nowgrade here
OC
OC
GENERATOR and
POWER STATION
PROTECTION
Barrie Moor, B Eng (Elec)
Slide [email protected]
OVER CURRENTPROTECTION
Setting andCoordination
Procedures
Slide [email protected]
Relay Coordination ProcedureCurrent Setting
Start with selection of relay characteristic
As far as possible, use relays of the same characteristic
Choose current settings
Determine maximum load current limitations
Determine starting current requirements
As far as possible, select operating current of each
upstream relay greater than that of the successive
downstream relay
Slide [email protected]
Relay Current Pick-up Setting
Set above maximum load current
Allow for emergency loading conditions
Allow safety margin
Allow for relay reset ratio
Set below the current pickup level of the next upstream
relay
Allow for load pickup current
Slide [email protected]
Load Pickup Current
Motor starting current
Auxiliary heaters
Transformer magnetising inrush
Capacitor charging current
Lighting loads - 10s to 100s of msec
Filaments and electrodes heating
Arc lamps starting
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Generator and
Power Station Protection
Over Current Protection
Fuses and Contactors
Barrie Moor : 2012 [email protected]
(07) 3298 5260
Slide [email protected]
Load Pickup Current
Hot load pickup
Short term loss of supply and subsequent load pickup
currents on return of supply
Cold load pickup
Load pickup, but now with loss of diversity between cyclicloads
Voltage recovery pickup
Pickup currents not as severe as for complete loss of
supply and subsequent hot load pickup
But more motors may still be on-line as under voltage
releases may not have disconnected them
Slide [email protected]
Relay Coordination ProcedureTime Multip lier Setting
Coordinate relays via time multipliers to achieve
appropriate grading margins
Determine, under various system configurations, thevalues of short circuit current that will flow through eachprotective device
Set relays to give minimum operating time at maximum
fault currents
Check performance (discrimination) at lower fault levels
Plot and coordinate relay curves on log/log or log/linear
format
Plot to a common current base (across transformers)
Slide [email protected]
Relay TMS Grading
Must provide for
CB tripping time (0.1 sec ??)
Relay timing errors
Relay overshoot
CT errors (10% ??)
Safety margin (10% ??)
A typical figure of 0.3 - 0.4 seconds is usually OK
0.3 for numerical relays
0.4 for electromechanical relays
Alternatively calculate a margin
Only necessary for slow tripping times (> 1.0 sec)
Slide [email protected]
Relay TMS Grading
Relay TMS Grading
Hence for an E/M relay tr ipping in 0.5 seconds
t = (7.5 + 7.5 + 10)% x 0.5 + 0.1 + 0.05 + 0.1
t = 0.375 seconds
0.30.30.350.4Typical margin (s)
0.030.030.050.1Safety Margin (s)
0.020.020.030.05Overshoot Time (s)
5557.5Timing Error %
NumericalDigitalStaticElecto-
Mechanical
Relay Technology
CT Errors Slide [email protected]
Grading of Parallel Elements
Worst case for grading is with only 1 transformer in service But this will be an unusual operating condition
E/M & Electronic Relays Only a single relay setting is available
Hence, effectively no option but to set for the worst case, namely1 transformer case
And accept slower performance for system normal,namely when both transformers are in service
Microprocessor based relays
These relays have multiple setting groups
So, maybe set Group 1 for system normal : 2 transformers
And change to group 2 when one transformer is OOS Automatically ??
Via SCADA & operator intervention ??
OC OC
OC OC
OC
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Generator and
Power Station Protection
Over Current Protection
Fuses and Contactors
Barrie Moor : 2012 [email protected]
(07) 3298 5260
Grading of Parallel Elements
Maximum through fault level occurs when both transformers
are in service
But the maximum individual transformer current flows whenthe 2nd transformer is OOS
Need to consider both conditions when grading relays
HV OC
HV OCFdr_1 OC
3 Fault Levels 2 Tx IN : 16000A 1 Tx IN : 12000A
3 Fault Levels 2 Tx IN : 10000A 1 Tx IN : 7500A
33kV 11kV
20MVA
20MVA
300A FLC
800A FLC
Fdr_2 OCSI 400ATMS 0.2
Fdr1_TMS 0.28=Fdr1_TMS round Fdr1_TMS .003+ 2,( ):=Round Up
Fdr1_TMS 0.276=
Fdr1_TMS 1Fdr1_Tmin
Fdr1_TMS_1:=Hence we can calculate the required TMS to achieve the required tripping time
Fdr1_TMS_1 2.971=This would result in a tripping time of
Fdr1_TMS_1 SI Fdr1_Plug 1.0, Imax,( ):=Assume TMS = 1.0
Fdr1_Tmin 0.821=Fdr1_Tmin Fdr2_Tmin 0.4+:=Required tripping time
Imax
Fdr1_Plug10=Fdr1_Plug 1000:=So select settings for Feeder 1
Fdr2_Tmin 0.421=Fdr2_Tmin SI Fdr2_Plug Fdr2_TMS, Imax,( ):=Tripping time at maximum fault level
Fdr2_TMS 0.2:=
Imax
Fdr2_Plug25=
Fdr2_Plug 400:=Given data for Feeder 2
Imax 10000:=Grade Fdr_1 OC over Fdr_2 OC at the maximum through fault level of 10kASet Fdr_1 OC above maximum feeder load of 800A
and check against maximum fault level of 10kA
SI P TMS, I,( ) 0.14TMS
I
P
0.02
1
:=Relay Characteristic
Feeder 1 Relay_2n
SI Fdr2_Plug Fdr2_TMS, I2n
,( ):= SI Fdr2_Plug Fdr2_TMS, Imax,( ) 0.421=
Feeder 2 Relay_1n
SI Fdr1_Plug Fdr1_TMS, I1n
,( ):= SI Fdr1_Plug Fdr1_TMS, Imax,( ) 0.832= _T 0.411=
100 1 .103
1 .104
1 .105
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2
2.2
2.4
2.6
2.8
3
Fdr 2 OCFdr 1 OC
Tx OC Grading (11kV Base Currents)
Tx_HV_TMS 0.36=Tx_HV_TMS round Tx_HV_TMS .003+ 2,( ):=Round up
Tx_HV_TMS 0.355=
Tx_HV_TMS 1 Tx_HV_Tmin
Tx_HV_TMS_1:=Hence we can calculate the required TMS to achieve the required tripping time
Tx_HV_TMS_1 3.297=This would result in a tripping time of
Tx_HV_TMS_1 SI Tx_HV_Plug 1.0, Imax,( ):=Assume TMS = 1.0
Tx_HV_Tmin 1.169=Tx_HV_Tmin Fdr1_Tmin 0.4+:=Transformer HV OC
Fdr1_Tmin 0.769=Fdr1_Tmin SI Fdr1_Plug Fdr1_TMS, Imax,( ):=Fdr Tripping time at maximum fault level
Tx_HV_Plug 1500=
Tx_HV_Plug 3 Tx_HV_Plug:=Allow for 33/11kV ratio
Tx_HV_Plug 500:=Set130% FLC_33kV 455=FLC_33kV 350=FLC_33kV 20000000
3 33000:=
Imax 12000:=
Grade Transformer HV OC under the maximum current condition, namely with one transformer OOS
Feeder 1 Relay_1n
SI Fdr1_Plug Fdr1_TMS, I1n
,( ):= SI Fdr1_Plug Fdr1_TMS, Imax,( ) 0.769=
Tx HV Relay_3n
SI Tx_HV_Plug Tx_HV_TMS, I3
n
,
( ):= SI Tx_HV_Plug Tx_HV_TMS, Imax,( ) 1.187= _T 0.418=
100 1.10
3
1.10
4
1.10
50
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2
2.2
2.4
2.6
2.8
3
Fdr 2 OC
Fdr 1 OC
Tx HV OC
Tx OC Grading (11kV Base Currents)
Slide [email protected]
Sequential Operation of Over CurrentRelays
As CBs trip, fault current magnitudes and flows will change
We need to integrate how far each relay progresses towardstripping in each stage
To determine total tripping times
To ensure relays that should not trip, remain stable
Relay 1 operating time must have a suitable margin above thetotal of Relay 2 and the subsequent Relay 3 operations
Relay 3Relay 2
Relay 1
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Generator and
Power Station Protection
Over Current Protection
Fuses and Contactors
Barrie Moor : 2012 [email protected]
(07) 3298 5260
GENERATOR and
POWER STATION
PROTECTION
Barrie Moor, B Eng (Elec)
Slide [email protected]
OVER CURRENTPROTECTION
Directional Relays
Slide [email protected]
Directional Over Current Relays
Extra discrimination may be achieved by making the
response of the relay directional when current can flow in
both directions Achieved via voltage (polarising) connections to the relay
Digital and numeric relay achieve phase displacementsvia software
EM & Static relays require suitable connection of input
quantities to the relay
Slide [email protected]
Directional Over Current RelaysAppl ication to Paral lel Feeders
Apply directional relays at the feeder receiving ends
Typically set to 50% of FLC, TMS = 0.1
Grade below non-directional relays at the source end
Ensure DOC relay thermal rating is OK
OC OC
OCOC
Fdr 1
Fdr 2
BA
Slide [email protected]
0.90.9
0.
5
0
.5
0.1 0.1
1.3
1.
7
2.1 2.1
1.
7
1.31.3
Ring Mains Systems
Open the ring at thesupply point
And then gradeclockwise
And then anticlockwise
Source substation relaysdo not HAVE to be
directional Intermediate substation
slower relays do notHAVE to be directional
GENERATOR and
POWER STATIONPROTECTION
Barrie Moor, B Eng (Elec)
Slide [email protected]
OVER CURRENTPROTECTION
Earth Fault Relays
Slide [email protected]
Earth Fault Protection
Implement more sensitive protection responding only to
residual current of the system
Low settings are permissible and beneficial
Earth faults are the most frequent
Earth faults may be limited by earth fault resistance
Earth faults may be limited by neutral earth impedance
Typical settings 20 - 40% x FLC
Time grade in the same manner as for phase OC relays
Beware of the burden that electromechanical relays may
place on CTs at low current settings
Although burden does decrease at very high currents with
saturation of the relays magnetic circuits
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Generator and
Power Station Protection
Over Current Protection
Fuses and Contactors
Barrie Moor : 2012 [email protected]
(07) 3298 5260
Slide [email protected]
Expulsion Fuses
Used where expulsion gases cause no problem such as in overhead
circuits and equipment
Special materials (fiber, melamine, boric acid, liquids such as oil orcarbon tetrachloride ) located in close proximity to fuse element andarc rapidly create gases
These produce a high pressure turbulent medium surrounding the
arc
Expulsion process deionises gases them as well as removing themfrom arc area
In inductive circuits, transient recovery voltage (TRV) will be
maximum at current zero.
Slide [email protected]
Fuses & TRV Performance
0 0 .0 02 0 .0 04 0 .0 06 0 .0 08 0 .0 1 0 .0 12 0 .0 14 0 .0 16 0 .0 18 0 .0 2 0 .0 22 0 .0 242
1.5
1
0.5
0
0.5
1
1.5
2
Circuit VoltageFuse Voltage
Current
0 0 .0 02 0 .0 04 0 .0 06 0 .0 08 0 .0 1 0 .0 12 0 .0 14 0 .0 16 0 .0 18 0 .0 2 0 .0 22 0 .0 242
1.5
1
0.5
0
0.5
1
1.5
2
Circuit VoltageFuse Voltage
Current
0 0 .0 02 0 .0 04 0 .0 06 0 .0 08 0 .0 1 0 .0 12 0 .0 14 0 .0 16 0 .0 18 0 .0 2 0 .0 22 0 .0 242
1.5
1
0.5
0
0.5
1
1.5
2
Circuit VoltageFuse Voltage
Current
0 0 .0 02 0 .0 04 0 .0 06 0 .0 08 0 .0 1 0 .0 12 0 .0 14 0 .0 16 0 .0 18 0 .0 2 0 .0 22 0 .0 242
1.5
1
0.5
0
0.5
1
1.5
2
Circuit VoltageFuse Voltage
Current
0 0 .0 02 0 .0 04 0 .0 06 0 .0 08 0 .0 1 0 .0 12 0 .0 14 0 .0 16 0 .0 18 0 .0 2 0 .0 22 0 .0 242
1.5
1
0.5
0
0.5
1
1.5
2
Circuit VoltageFuse Voltage
Current
Current lagsVoltage by 90
deg
SystemVoltage
Currentinterrupted at
naturalcurrent zero
TRV acrossblown fuse
element
Fuse Voltage
Slide [email protected]
Current Limiting Fuses (HRC Fuses)
Fuse is designed to insert a large resistance
Hence, prospective level of fault current is reduced
And zero crossing of the current and voltage will be reasonably
in phase TRV significantly reduced
Fuse element is completely surrounded with f iller material, typically
silica sand
Arc energy melts the sand, thus inserting the required high
resistance
But this design may have difficulty interrupting low level overloads.Overcome by
M Effect designs
Spring assisted designs
Slide [email protected]
Current Limiting Fuses
Tin for M Effect lowoverload fuse performanceSee later
Slide [email protected]
Current Limiting Fuses
Slide [email protected]
Current Limiting FusesM Effect for low level overloads
M Effect : A.W. Metcalf - 1939
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Generator and
Power Station Protection
Over Current Protection
Fuses and Contactors
Barrie Moor : 2012 [email protected]
(07) 3298 5260
Slide [email protected]
Current Limiting Fuses
Slide [email protected]
Grading Relays wi th Fuses
Extremely Inverse curve
follows a similar I2t
characteristic Relay current setting should
be approximately 3 times the
fuse rating
Grading margin of not less
than 0.4 seconds
recommended
Or 15.04.0' + tT
1PI
TMS80TIME
2Inverse_Extremely
=
Slide [email protected]
Grading Relays with Fuses
First relay upstream of the fuse should be set to EI
characteristic
Now to coordinate further upstream relays
Option 1 : Also select EI characteristics
Option 2 : Check also for the possibility of setting
The next relay to a VI characteristic
And subsequent further upstream relays to SI characteristics
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Fuse Contactors
High fault level applications eg
40kA fault level
Contactor rated to only 10kA
Fuse operates for all faults above say 7 kA
Contactor and associated protection relay operate for
lower fault levels
Warning the fuse may also have a minimum breaking
capacity and the contactor must be set to operate abovethis point
Slide [email protected]
100 1.103
1.104
1.105
0.01
0.1
1
10
100
Fuse
Relay / ContactorFuse
Fuse Contactors
10kA Contactoroperates for faults
below 7kA
Fuse operates forfaults above 7kA
Fuse operationbelow 2kA is not
permissible
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Generator and
Power Station Protection High Impedance Differential Protection
Barrie Moor : 2012 [email protected]
(07) 3298 5260
GENERATOR and
POWER STATION
PROTECTION
Barrie Moor, B Eng (Elec)
Slide [email protected]
HIGH IMPEDANCEDIFFERENTIALPROTECTION
Busbar Protection and
Galvanically Connected
Plant
Slide [email protected]
Synopsis
HZ Differential Protection Principles
Determination of setting voltage
CT requirements
Current operated schemes and stabilising resistors
Limiting secondary system voltages to safe levels
Primary operating current and application of shuntresistors
CT supervision requirements
Application of high impedance differential protection
schemes to other galvanically connected plant
Slide [email protected]
SIMP
LE!!!
Bus Zone Protection Requirements
Dependability
Must trip for all in-zone faults
Discrimination
Must not trip for any out-of-zone faults
Security
Against all sources of mal-tripping
Speed of operation
As quickly as possible
Dependability & Security
Slide [email protected]
CT Connections & Polarity
S2
P2
S1
P1
I1
I2
P2
S2
P1
I1 S1
I2
Slide [email protected]
RELAY
Internal Fault
Slide [email protected]
RELAY
External Fault
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Generator and
Power Station Protection High Impedance Differential Protection
Barrie Moor : 2012 [email protected]
(07) 3298 5260
Slide [email protected]
3 Phase CT Connections
CT MARSHALLING
DIFFERENTIALRELAY
Slide [email protected]
Current Mismatch
CT Manufacturing Variations
Inequality of CT Burdens
CT Saturation
Highest Fault Current on CT exposed to through fault
Worst possible mismatch is
Total saturation of the CT on the faulted plant
All other CTs transform perfectly
Slide [email protected]
15000A 5000A
RELAY
5000A
External Fault
5000A
External Fault & CT Saturation
Slide [email protected]
RELAY
External Fault
Rlead
Rlead
Rct High
Impedance
Relay
CT Saturates :Magnetising branch
impedance becomes zero
LEADSCTFAULTRELAY RRIV
Slide [email protected]
Setting Voltage and Margins
Fault current comprises
AC Component
DC Component
Hence, employ a DC Stabilised Relay
No additional margin on the setting is required
And considering 0% / 100% CT saturation case
This in an unrealistically extreme case
100% safety margin is automatically built in
So, no additional safety margin on setting is required
Slide [email protected]
RELAY
Internal Fault
High
Impedance
Relay
RELAYKNEE V2V
CTs will saturate under internal fault conditions, butrelay operation is assured provided absolutely all CTs
meet the requirement
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Generator and
Power Station Protection High Impedance Differential Protection
Barrie Moor : 2012 [email protected]
(07) 3298 5260
Slide [email protected]
CT Selection
All CTs to be the same ratio
All CTs to have Vk 2.Vsetting
This is an absolute MUST
Preferably Vk 5.Vsetting
Need to know
Knee Point voltage
CT Resistance
Class Requirements
Not absolutely critical
But highly recommended to specify class PX CTs
0.1 PX 500 R3
Magnetisingcurrent at kneepoint voltage
Magnetisingcurrent at kneepoint voltage
CT knee
point voltage
CT knee
point voltage
CT internalresistance
CT internalresistance
Slide [email protected]
Summary
Ensure Stability under through faults
Ensure Operation for genuine in-zone faults
Beware of short cut methods
Do not simply set
RELAYKNEE V2V
2
VV KNEERELAY=
LEADSCTFAULTRELAY RRIV
Preferably 5 times to optimiserelay performance, but 2 is theabsolute minimum to ensure
reliable relay operation
Preferably 5 times to optimiserelay performance, but 2 is theabsolute minimum to ensure
reliable relay operation
Slide [email protected]
Current Operated Schemes
Voltage operated
Current operated, incl stabilising
resistor
Typical current settings
as low as possible, but
> 20% of plant rating
< 30% of fault current
20% setting is usually OK
Assuming the CT has been
selected to match plant
rating
V = I.R = 0.2 x (200 + 10) = 42 volts
Relay0.2A
10 ohms
200 ohms
Slide [email protected]
Metrosils
In the case of a heavy internal fault, secondary system voltages maybecome excessive
Implications include damage to equipment and safety of personnel
Empirical Formula
VK = CT RMS knee point voltage
VF = Maximum RMS voltage that would occur if the CT did notsaturate
Install metrosils if this voltage become excessive(eg. >2.8kV peak)
KFKPEAK VVV22V
Slide [email protected]
Metrosil Parameters
Where V & I are PEAK
values
C = Metrosil constant
B = Metrosil constant
(0.2 0.25)
And, because of the
non-linearity
I52.0IRMS
ICV METROSIL NON-LINEAR RESISTORS
V = C . IB
10
100
1000
1104
0.001 0.01 0.1 1 10 100
C = 450 B = 0.25
C = 600 B = 0.25
C = 900 B = 0.25
Metrosil Current - Amps
Slide [email protected]
Metrosil Parameters
Based on the previous equations, the RMS current at the
relay setting voltage Vs(rms) is:-
52.0
I
2
CV RMS
RMSSetting
1
Setting
RMSC
V252.0I RMS
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Generator and
Power Station Protection High Impedance Differential Protection
Barrie Moor : 2012 [email protected]
(07) 3298 5260
Slide [email protected]
MAGMETROSILRELAYP INIICTI
RELAYImag
R
shunt
Ishunt
Irelay
Imet r
osil
Imag
N = Number of CTs in parallel
Primary Operating Current
Slide [email protected]
Shunt Resistors
Desensitise scheme
Prevent tripping on open CTs
Primary operating current > maximum plant loading
Effectively becomes a medium impedance scheme
Slide [email protected]
MAGSHUNTMETROSILRELAYP INIIICTI
RELAYImag
R
shunt
Ishunt
Irelay
Imet r
osil
Imag
N = Number of CTs in parallel
Primary Operating Current
Slide [email protected]
CT Supervision
Effect of CT problems when Shunt Resistors are installed
Scheme is part way to a trip condition
Effect of CT problems when BZ Check scheme is installed
Half of the scheme is already tripped
CT Supervision Setting Principles
Set to 50% of minimum load
Operation : initially (eg. < 3 secs)
Nil : allow for correct operation of BZ protection
Operation : short time (eg. > 3 secs)
Alarm
CT Shorting
GENERATOR and
POWER STATION
PROTECTION
Barrie Moor, B Eng (Elec)
Slide [email protected]
HIGH IMPEDANCEDIFFERENTIALPROTECTION
Application to other
Plant
Slide [email protected]
HZ Protn Application to Plant
Requires Galvanic Connection
All CT ratios the same
Can Apply To
Busbars
Transformers
Generators & Motors
Capacitors
Reactors
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Generator and
Power Station Protection High Impedance Differential Protection
Barrie Moor : 2012 [email protected]
(07) 3298 5260
Slide [email protected]
Relays
StabilisingResistors
Generators & Motors
Slide [email protected]
DIFF
Auto Transformers
All CT r atio s tobe the same
This CT will carrymaximum currentand hencedictates ALL CTratios
But this CT is internaland may have a singlefixed ratio.Thus, must be specifiedcorrectly at time ofpurchase !!
Slide [email protected]
REF
Restric ted Earth Fault Protection
CT terminalsaway fromprotected objectare connected
CT terminalsnear toprotected objectare connected
Slide [email protected]
DIFFDIFF
A
DIFF
B C
Reactors Earthed Neutral
Slide [email protected]
DIFFDIFF
A
DIFF
B C
Reactors Floating Neutral
Floating neutralbus is also protected
Slide [email protected]
DIFFDIFF
A
DIFF
B C
Capacitors Earthed Neutral
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Generator and
Power Station Protection High Impedance Differential Protection
Barrie Moor : 2012 [email protected]
(07) 3298 5260
Slide [email protected]
DIFFDIFF
A
DIFF
B C
Capacitors Floating Neutral
Floating neutralbus is also protected
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Generator andPower Station Protection Sequence Components
Barrie Moor : 2012 [email protected](07) 3298 5260
GENERATOR and
POWER STATION
PROTECTION
Barrie Moor, B Eng (Elec)
Slide [email protected]
SEQUENCECOMPONENTS
A Quick Review
Slide [email protected]
Sequence Components
Positive Sequence
A B C
Equal in magnitude 120 degrees apart
Negative Sequence
A C B
Equal in magnitude
120 degrees apart
Zero Sequence
A B C
Equal in magnitude
In phase
V1
V2
V0
I1
I2
I0
Slide [email protected]
Sequence Components
I1 I2 I0
I phase
IC
IB
IA
Slide [email protected]
Sequence Networks
Source
RelayLocation
FaultLocation
PositiveSequenceNetwork
Z1s
Z1f
I1
Source
RelayLocation
FaultLocation
NegativeSequenceNetwork
Z2f
Z2s
I2
Source
RelayLocation
FaultLocation
Zero
SequenceNetwork
Z0f
Z0s
I0
V1 = 1 / 0 V2 = 0 V0 = 0
Slide [email protected]
Sequence ComponentsThree phase conditions
Positive sequence only
Three phase load
Three phase fault
No neutral (earth fault) current
Slide [email protected]
In = 0
3 Phase Balanced Current
Balanced currents sum to zero
Positive sequence currents
Negative sequence currents
But zero sequence current will sum to 3.IoIn = 3.Io
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Generator andPower Station Protection Sequence Components
Barrie Moor : 2012 [email protected](07) 3298 5260
Slide [email protected]
Sequence Networks3 Phase Fault
Source
RelayLocation
FaultLocation
PositiveSequenceNetwork
Z1s
Z1f
I1
f1Zs1ZZ
Z
V1II
POS
POS
Fault_Phase_3
=
Slide [email protected]
Sequence ComponentsPhase Phase fault
Positive and Negative sequence components only
And consider the special case where A phase equal in magnitude but opposite in phase
B to CPhase to Phase fault
Slide [email protected]
Sequence Networks (A phase)Phase Phase fault
A phase
IA1 & IA2 antiphaseSum to zero
B phase
IB1 & IB2 at 60o
C phase IC1 & IC2 at 60o
IB = - IC
Source
RelayLocation
FaultLocation
Positive
SequenceNetwork
Z1s
Z1f
I1
Source
RelayLocation
FaultLocation
NegativeSequenceNetwork
Z2f
Z2s
I2
2I1I=
Slide [email protected]
Sequence Networks (A phase)Phase Phase fault
Source
RelayLocation
FaultLocation
Positive
SequenceNetwork
Z1s
Z1f
I1
Source
RelayLocation
FaultLocation
NegativeSequenceNetwork
Z2f
Z2s
I2
Since Z1 ~ Z2
|I1| = |I2| = 50%of 3 phase faultlevel
negpos ZZ
V2I1I
=
Slide [email protected]
Sequence ComponentsPhase Phase fault
|I1| = |I2| = 50% of 3 phase fault level
Thus |IB| = |IC| = 86.6% of 3 phase fault level(because of 60o angles)
Slide [email protected]
Sequence ComponentsEarth Fault
A phase positive sequencenegative sequencezero sequence
Equal in magnitudeand phase
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Generator andPower Station Protection Sequence Components
Barrie Moor : 2012 [email protected](07) 3298 5260
Slide [email protected]
Sequence ComponentsEarth Fault
Slide [email protected]
Sequence NetworksA Phase Earth Fault
Source
RelayLocation
FaultLocation
PositiveSequenceNetwork
Z1s
Z1f
I1
Source
RelayLocation
FaultLocation
NegativeSequenceNetwork
Z2f
Z2s
I2
Source
RelayLocation
FaultLocation
ZeroSequenceNetwork
Z0f
Z0s
I0
Slide [email protected]
Sequence NetworksA Phase Earth Faul t
Source
RelayLocation
FaultLocation
PositiveSequenceNetwork
Z1s
Z1f
I1
Source
RelayLocation
FaultLocation
NegativeSequenceNetwork
Z2f
Z2s
I2
Source
RelayLocation
FaultLocation
Ze
roSequenceNetwork
Z0f
Z0s
I0
I = 1.0 I = 1.0 I = 1.0
V1 = 1 / 0 V2 = 0
V0 = -0.15
V0 = -0.50
V2 = -0.10
V2 = -0.25
V0 = 0
V1 = 0.90
V1 = 0.75
= 0.10 = 0.10
= 0.15= 0.15
= 0.15
= 0.35
Slide [email protected]
Sequence NetworksA Phase Earth Fault
PositiveSequence
FaultLocation
NegativeSequence
FaultLocation
ZeroSequence
FaultLocation
RelayLocation
ZS1
Zl
ZS2
Zl
I2I1
RelayLocation
ZS0
Zl
I0
RelayLocation
zeronegpos ZZZ
V0I2I1I
=
0I2I1IIA
0IB=
0IC=
0I3INEUT
Slide [email protected]
Sequence ComponentsSummary
Positive Sequence
Balanced three phase load
Balanced three phase fault
No neutral (earth) current
Negative Sequence
Unbalanced load
Phase to phase fault
No neutral (earth) current Zero Sequence
Earth fault
Neutral current = 3 . Io
Cannot flow into or out of a delta
Can circulate around (within) the delta
Io = 0
Io = 0
Io = 0
Io
Io
Io
3Io
GENERATOR and
POWER STATION
PROTECTION
Barrie Moor, B Eng (Elec)
Slide [email protected]
TRANSFORMERSand
SEQUENCECOMPONENTS
DifferentialProtection
Requirements
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Generator andPower Station Protection Sequence Components
Barrie Moor : 2012 [email protected](07) 3298 5260
Slide [email protected]
Transformer Current Flows
Star / Star Transformer : LV Earth Fault Current flows in corresponding HV winding
Appears as EF on the HV system also
I1, I2 & I0I1, I2 & I0I1, I2 & I0I1, I2 & I0
Slide [email protected]
Transformer Current Flows
Star / Star Transformer : LV Earth Fault But, suppose we dont have an upstream power system earth However, consider the effect of adding a delta connected tertiary
winding HV line current flows in a 2:1:1 ratio No I0 on the HV system as there is no path for neutral current flow
I1, I2I1, I2
I0I0
I1, I2 & I0I1, I2 & I0
So where did the I0 go ??So where did the I0 go ??
Slide [email protected]
Transformer Current Flows
Star / Star Transformer : LV Earth Fault
Retain the delta connected tertiary winding
But, lets reinstate the power system earth
Power system and delta winding zero sequence current flowdistributions will depend on their relative Z0 impedances
I1, I2 & I0I1, I2 & I0I1, I2, I0I1, I2, I0
I0I0
Slide [email protected]
Transformers, Sequence Componentsand Differential Protection
Star/Star transformers, with a delta tertiary winding:
Will have a mismatch between zero sequence current flows onthe HV & the LV windings
It is thus necessary to exclude zero sequence current from thedifferential relay protection algorithms
Star/Star transformers, without a delta tertiary winding:
May still have a mismatch between zero sequence current flowson the HV & the LV windings
The transformer tank can act as a low quality tertiary delta winding It is thus still necessary to exclude zero sequence current from
the differential relay protection algorithms
Slide [email protected]
Transformer Current Flows
Delta / Star Transformer : LV Earth Fault Current in corresponding HV winding only
Appears as phase to phase fault from the perspective ofthe HV system
I1 & I2 onlyI1 & I2 only
So where did the I0 go ??So where did the I0 go ??
I1, I2 & I0I1, I2 & I0
Slide [email protected]
Positive Sequence Network
LVZS
ZS HVZ1HL
HV
LV
Zfdr
Zfdr
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Generator andPower Station Protection Sequence Components
Barrie Moor : 2012 [email protected](07) 3298 5260
Slide [email protected]
Negative Sequence Network
ZS
ZS Z1HL
LV
HV
HV
LV
Zfdr
Zfdr
Slide [email protected]
Zero Sequence Network
ZS
ZS Z1HL
LV
HV
HV
LV
Zfdr
Zfdr
Slide [email protected]
Transformer Current Flows
Delta / Star Transformer : LV phase to phase fault Current in 2 LV windings
Current in 2 HV windings
Appears as 2:1:1 fault on the HV system
I1 & I2I1 & I2I1 & I2I1 & I2
Slide [email protected]
Transformer Current Flows
Star / Delta Transformer : LV phase to phase fault Current in all 3 LV windings
Current in all 3 HV windings
Appears as 2:1:1 fault on the HV system
I1 & I2I1 & I2I1 & I2I1 & I2
Slide [email protected]
Sequence ComponentsTransformer LV ph-ph fault
30 deg
Consider B-C fault on the LV of either of Star Delta transformer Delta Star transformer
30 deg phase shift Positive seq components Negative seq components
will shift + 30 deg
will shift - 30 deg
Slide [email protected]
Sequence ComponentsTransformer LV ph-ph fault
LV phase to phase fault HV distribution is 1 : 2 : 1 LV distribution is 0 : 1 : 1
So be careful when grading HV & LV IDMT OC Relays |I1| = |I2| = 50% of 3 fault level But note that I1 & I2 are 60o apart on the LV, but are shifted 30o and
are thus, for the 2 phase, in phase on the HV LV current is 86.6% of 3 fault level in both phases HV current in the 2 phase is the same as for LV 3 fault !!
30 deg
LVHV
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Generator andPower Station Protection Sequence Components
Barrie Moor : 2012 [email protected](07) 3298 5260
Slide [email protected]
Transformers, Sequence Componentsand Differential Protection
Compensate for the transformer phase shift
Exclude zero sequence current from the differential relay
protection scheme Zero sequence current can flow into and out of earthed
star windings
Zero sequence current cannot flow into or out of deltawindings
Zero sequence current can circulate around delta windings(said to be trapped in the delta)
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Generator and
Power Station Protection Transformer Protection
Barrie Moor : 2012 [email protected]
(07) 3298 5260
GENERATOR and
POWER STATION
PROTECTION
Barrie Moor, B Eng (Elec)
Slide [email protected]
TRANSFORMERPROTECTION
Slide [email protected]
Types of Fault
Phase-ground faults - from winding to core or winding to tank
Phase-phase faults - between windings
Interturn faults - between single turns or adjacent layers of the samewinding (Buchholz)
Arcing contacts
Local hotspots caused by shorted laminations
Low level internal partial discharges (moisture ingress or designproblems)
Bushing faults (internal to the tank)
Tapchanger faults (often housed in a separate tank)
Terminal faults (external to the tank, but inside the transformer zone)
Slide [email protected]
Buchholz Protection
Two floats in the relay:
Upper float
Detects accumulation of gas
Detects loss of oil
Incipient faults
Partial discharge
Winding & core overheating
Bad contacts and joints
May alarm only or may be set to trip
Lower float
Detects surge in oil < 100ms
Although it does take a finite time for
pressure waves to initiate Buchholz
tripping
To Conservator
Gas Sample
Trip
Alarm
BUCHHOLZ RELAY
Float
To Tank
Float
Slide [email protected]
Pressure Relief Device (Qualitrol )
Spring assisted pressure relief devices
Relieves pressure impulses due to massive internal fault conditions.
Helps prevent the tank bursting or splitting
Relay contacts are also connected to trip the transformer.
Since pressure waves travel with a finite velocity, they may rupture
the tank locally before the pressure wave has reached the pressure
relief device, if it is some distance away. Several units may therefore
be required on larger transformers.
Slide [email protected]
Basic Transformer Protection
Fuses
Transformers without CBs
Perhaps to a few MVA
Overcurrent & Earth Fault Protection
Transformers with CBs
Perhaps 5 - 50MVA
Differential Protection
Transformers > 10MVA Fast
Can be sensitive
May detect terminal faults also
GENERATOR and
POWER STATION
PROTECTION
Barrie Moor, B Eng (Elec)
Slide [email protected]
TRANSFORMERPROTECTION
Biased Differential
Protection
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Generator and
Power Station Protection Transformer Protection
Barrie Moor : 2012 [email protected]
(07) 3298 5260
Slide [email protected]
Differential Protection
DifferentialRelay
P1
S1
P2
S2
SIDE OF CT AWAYFROM PROTECTEDPLANT CONNECTS
TO RELAY
CURRENT FLOWSINTO PLANT
CURRENT FLOWSOUT OF PLANT
CURRENT FLOWSINTO RELAY
CURRENT FLOWSOUT OF RELAY
SIDE OF CT AWAYFROM PROTECTEDPLANT CONNECTS
TO RELAY
TRIPELEMENT
IT IS NOT THE P1/S1 OR P2/S2ORIENTATIONS THAT ARE
IMPORTANT, BUT THEPREFERENCE FOR THE
AWAY SIDES OF THE CTsTO CONNECT TO THE RELAY
Slide [email protected]
Differential Protection of Transformers
11/132kV
2400/1 200/12400A 200A
1A 1A
BIAS orRESTRAINTELEMENT
BIAS orRESTRAINTELEMENT
TRIPPING ELEMENTDETECTS ONLY THE
MIS-MATCH CURRENT
TRIPELEMENT
Slide [email protected]
Transformer Differential Mismatch
Differential CT ratio selection
CT ratios selected to compensate for the transformer turnsratio
Mismatched CTs
CTs do not exactly compensate for transformer turns ratio
Transformer turns ratio changes with tap changing
Implement a biasing restraint system
Magnetizing current in the CTs, especially as somesaturation due to DC fault current sets in.
The amount of bias is increased under heavy through faultconditions to compensate for possible CT saturation
Slide [email protected]
Transformer Differential Mismatch
Inrush on energisation (2nd harmonic)
Over excitation (5th harmonic)
Transformer phase shifts
Earth fault (neutral zero sequence) currents
Slide [email protected]
Inrush Current onEnergisation of Transformer
TRIPELEMENT
Slide [email protected]
Second Harmonic on Inrush
Transformer inrush current on energization.
Inrush current produces a current from the energizing side
only, appearing as an internal fault.
Inrush current magnitude can be as great as a through 3
phase fault.
This current is characterized by the appearance of secondharmonics, so additional restraint can be based on this 2nd
harmonic signature
Relay setting below the 2nd harmonic level is required(Ratio of 2nd harmonic to fundamental)
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Generator and
Power Station Protection Transformer Protection
Barrie Moor : 2012 [email protected]
(07) 3298 5260
Slide [email protected]
Transformer Inrush Current
2
0
2
4
6
8
10
12
0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5
Transformer Inrush Current
Current
Seconds
Inrush current : with 2nd Harmonic
Slide [email protected]
Fifth Harmonic on over excitation
Overfluxing, caused by too high a voltage, or too low a frequency.
Increased magnetising current
This is characterized by third & fifth harmonics. Fifth harmonic restraint to retrain tripping of the differential
element
Typically no user calculations or settings are required
Sustained overfluxing may damage the transformer
Time delayed V/f tripping function (long time)
Especially applicable to generator transformers
Frequency can be anywhere from zero to nominal during run-up andrun down
Not so necessary for transmission or distribution applications
Frequency will not deviate significantly from nominal
Slide [email protected]
Bias Differential Protection
Allow for Transformer turns ratio
Allow for Transformer phase shifts
Eliminate Zero Sequence currents from the relaying system
OperatingWinding
P1
S1
Bias Windings
P1
S1
GENERATOR and
POWER STATION
PROTECTION
Barrie Moor, B Eng (Elec)
Slide [email protected]
SEQUENCECOMPONENTS
A Quick Review
GENERATOR and
POWER STATION
PROTECTION
Barrie Moor, B Eng (Elec)
Slide [email protected]
TRANSFORMERPROTECTION
Continued
Slide [email protected]
CT Connections and Ratios
Star/Delta and Delta/Star transformers have a 30 degree phase shift
Compensate with CTs connected opposite to the transformer
connections. ie:
Star connected CTs on the delta side of the transformer
Delta connected CTs on the star side of the transformer
Phase shift compensated
Zero sequence currents flowing in the transformer star windings
prevented from entering the relaying system
But how do we get the correct delta connection for our CTs ???
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Generator and
Power Station Protection Transformer Protection
Barrie Moor : 2012 [email protected]
(07) 3298 5260
Slide [email protected]
Determination of CT Connection
Diff Prot
Yd11
D11D11
CT Primary is star connectedCT secondary is D11 connectedOverall connection is thus YD11
Slide [email protected]
Determination of CT Connection
Dy11
D1D1
Diff Prot
CT Primary is star connectedCT secondary is D1 connectedOverall connection is thus YD1
Slide [email protected]
Star/Delta and Delta/Star TransformersCT Connection Summary
Transformer HV is STAR connected
HV CTs are delta connected (ie. phase shift ing)
HV CTs EQUAL to the transformer phase shift
LV CTs in star
Transformer LV is STAR connected
LV CTs are delta connected (ie. phase shifting) LV CTs OPPOSITE to the transfor mer phase shift
HV CTs in star
Slide [email protected]
CT Connection Summary
Compensates for the phase shift across a star-deltatransformer.
The correct vector group must be chosen for the CTs toensure that through currents balance.
Prevents any zero sequence currents flowing in the starwinding from entering the relay
Since they are not present in the line on the delta side.
And for Star / Star transformers ??
It is still necessary to eliminate Io from the relaying system
Connect CTs delta / delta
Or use the D12 / D12 feature of microprocessor relays
Slide [email protected]
A phase output is at "11 o'clock"
A phase "S1" connects to B phase "S2"B phase "S1" connects to C phase "S2"C phase "S1" connects to A phase "S2"
D11
S2
A
S1
S1
C
S2
S1
S2
B
C
B
A
CT YD11 Connections
D11D11
Slide [email protected]
A phase output is at "1 o'clock"
A phase "S2" connects to B phase "S1"B phase "S2" connects to C phase "S1"C phase "S2" connects to A phase "S1"
A phase output is at "11 o'clock"
A phase "S1" connects to B phase "S2"B phase "S1" connects to C phase "S2"C phase "S1" connects to A phase "S2"
D11
S2
A
S1
S1
C
S2
S1
S2
B
D1
C
B
A
S2B
S2
S1
S1
A
S1C
S2
C
B
A
CT YD1 Connections
D1D1
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Generator and
Power Station Protection Transformer Protection
Barrie Moor : 2012 [email protected]
(07) 3298 5260
A B
Bias Windings
P1 P2
S1 S2
A1 A2
OperatingWindings
C
a2 a1P1P2
S2 S1
Notice that theconnections forthe Delta windingsare the same !!
Away side of CTs connected to relay.Hence, transformer current in or outcorresponds to relay current in or out.
Away side of CTs connected to relay.Hence, transformer current in or outcorresponds to relay current in or out
Slide [email protected]
Transformer Current Flows
There must be a path for the current to flow
There must be an Ampere Turns balance
If there is current flowing in one winding
There must be current in the coupled winding
If there is no current flowing in one winding
There can be no current in the coupled winding
A B
Bias Windings
P1 P2
S1 S2
A1 A2
OperatingWindings
C
a2 a1P1P2
S2 S1
External Phase Earth Fault
Protection Scheme remains balanced
HV 0:1:1 (HV looks like a phase phase fault)
LV 0:0:1 (LV is actually a single phase fault)
A B
Bias Windings
P1 P2
S1 S2
A1 A2
OperatingWindings
C
a2 a1P1P2
S2 S1
Protection Scheme remains balanced
HV 1:2:1 (HV has a 2:1:1 current distribution)
LV 0:1:1 (LV is actually a phase phase fault)
External Phase Phase Fault
Slide [email protected]
Delta CTs and Ratio Selection
CT ratios must allow for the fact
that current flowing into therelay from the delta connected
CTs is 3 times the CTsecondary current
Hence, a standard 1A CT will
result in relay current of 3times the CT secondary current
Thus, CTs with ratios such as
1000/0.577 are, for this reason,quite common.
1/0
1 / -120
1/-
240
= 1.732 /-30
1 / 0 1/-240
Slide [email protected]
Delta CTs and Ratio Select ion
Ia - Ic = 1.732 /-30
CT ratios must allow for the fact
that current flowing into therelay from the delta connected
CTs is 3 times the CTsecondary current
Hence, a standard 1A CT will
result in relay current of 3times the CT secondary current
Thus, CTs with ratios such as
1000/0.577 are, for this reason,quite common.
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Generator and
Power Station Protection Transformer Protection
Barrie Moor : 2012 [email protected]
(07) 3298 5260
Slide [email protected]
CT Ratio Selection
The CT ratios must be opposite to the transformer ratio
Choose one CT ratio
Base all other CT ratios on this selection and transformerturns ratio
Not on winding MVA !!!
And, as noted before CT ratios must allow for the fact
that current flowing into the relay from the delta connected
CTs is 3 times the CT secondary current
Slide [email protected]
Modern Microprocessor Relays
All CTs connected in Star
Relay has to process phase shifts
Relay has to remove neutral current
P1
P1
P1
S1
S1
S1
S1
S1
S1
P1
P1
P1
Slide [email protected]
Modern Microprocessor Relays
P1
P1
P1
S1
S1
S1
S1
S1
S1
P1
P1
P1
ICIAIARELAY
IAIBIBRELAY
IBICICRELAY
IC
IB
IA
110
011
101
3
1
IC
IB
IA
RELAY
RELAY
RELAY
IC
IB
IA
110
011
101
3
1
IC
IB
IA
RELAY
RELAY
RELAY
Slide [email protected]
D1
D11
A phase output is at "1 o'clock"
A phase "S2" connects to B phase "S1"B phase "S2" connects to C phase "S1"C phase "S2" connects to A phase "S1"
A phase output is at "11 o'clock"
A phase "S1" connects to B phase "S2"B phase "S1" connects to C phase "S2"C phase "S1" connects to A phase "S2"
D11
S2
A
S1
S1C
S2
S1
S2
B
D1
C
B
A
S2B
S2
S1
S1
A
S1C
S2
C
B
A
Modern Microprocessor Relays
=
IC
IB
IA
110
011
101
3
1
IC
IB
IA
RELAY
RELAY
RELAY
=
IC
IB
IA
101
110
011
3
1
IC
IB
IA
RELAY
RELAY
RELAY
Slide [email protected]
Modern Microprocessor Relays
=
IC
IB
IA
211
121
112
3
1
IC
IB
IA
RELAY
RELAY
RELAY
D12
D1
D11
=
IC
IB
IA
110
011
101
3
1
IC
IB
IA
RELAY
RELAY
RELAY
=
IC
IB
IA
101
110
011
3
1
IC
IB
IA
RELAY
RELAY
RELAY
Slide [email protected]
Modern Microprocessor RelaysD12 Zero sequence current elimination
=
IC
IB
IA
211
121
112
3
1
IC
IB
IA
RELAY
RELAY
RELAY
( )ICIBIAIA33
1=
( )( )ICIBIAIA33
1++=
( )0I3IA33
1=
0IIA=
( )ICIBIA23
1IARELAY =
8/10/2019 Generator and Power Station Protection_2012.pdf
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Generator and
Power Station Protection Transformer Protection
Barrie Moor : 2012 [email protected]
(07) 3298 5260
Slide [email protected]
Modern Microprocessor Relays
All CTs can now be connected in Star
Relay internal processing adjusts for phase angle
Relay internal processing rejects zero sequencecomponents
CT ratios mismatches can also now be accommodated
Internal processing within relay then adjusts CT current to
match transformer turns ratio
CTs can be fine tuned to match middle tap position
Allows for more sensitive relay settings
Slide [email protected]
CT Phase and Ratio Adjustment
Dyn120MVA 33/11kV
1500/1400/1
DifferentialElement
350A 1050A
0.7A
-300
0.875A
00
00
1A
00
1A
Transformer Microprocessor Differential Protection Relay
Magnitudes normalised to transformer FLC Phase angles compensated Zero sequence current eliminated
Yy0
Software CTx 1.143
Software CTx 1.429
Yd11
00 -300
TAP POSITION
SoftwareCT Ratio
Adju stm ent
Slide [email protected]
Differential Relay Bias Settings
CT ratio selection
Select one ratio to meet load requirements
Base all other ratios on the first (not on load !)
Allow Margin (perhaps 10-15%) allowing
the worst mismatch of transformer ratio
and CT ratios (remember 3 for delta CTs !!!)
To decide worst case - consider the overall scheme
At the top tap position .......... & then
At the bottom tap position.
And both should be the same for a microprocessor based
relay that is correctly mid tap balanced
Slide [email protected]
Differential Relay Bias Settings
Use the relay equations to determined the worstmismatch
Typically
I1 & I2 are the currents into the two sides of thetransformer
2I1IDiff
2
2I1IBias
50%Not Average !!
Slide [email protected]
Tap Changer Position
For any setting of tap changer and through current, andgiven the CT ratios, the values of bias current anddifferential current can easily be calculated.
Base calculation on the relay algorithm not on anarbitrary or simple mismatch calculation Note that the previous equations are not the only options
available to and used by relay manufacturers
Hence use THE ACTUAL RELAY equations !!
Allow the recommended safety margin Probably 10 15%
And check your e lbow !!
Slide [email protected]
Transformer Bias Differential Protection
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 60
0.5
1
1.5
2
2.5
3
3.5
4
15% Differential Setting
25% Differential Setting
35% Differential Setting
Differential
Current
Diff I1 I2+:=
Bias Current BiasI1 I2+
2:=
OPERATEOPERATE
TransformerInternal Fault
Protection Trips
TransformerInternal Fault
Protection Trips
Through Fault withCT Saturation
Through Fault withCT Saturation
Through FaultMismatch due to CT Ratios &Transformer Tap Changing
Through FaultMismatch due to CT Ratios &Transformer Tap Changing
RESTRAINRESTRAIN
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35/62
Generator and
Power Station Protection Transformer Protection
Barrie Moor : 2012 [email protected]
(07) 3298 5260
Slide [email protected]
CT Requirements
Some CT saturation is permissible for through faults
Such CT saturation occurs mainly due to the DC
component of the fault current
Most manufacturers provide simple equations to
determine CT class
No complex calculations required
GENERATOR and
POWER STATION
PROTECTION
Barrie Moor, B Eng (Elec)
Slide [email protected]
TRANSFORMERPROTECTION
Winding Neutral
End Faults
Slide [email protected]
Protection for neutral end earth faults
Small current Many turns
Large current Few turns
Differential Protection
Line current mismatchis too small to trip thedifferential relay
We need to monitorneutral current flow
Slide [email protected]
Transformer Neutral Fault Current & Primary Current
0 10 20 30 40 50 60 70 80 90 1000
2
4
6
8
10
12
14
16
Fault Current in Shorted TurnsPrimary Current
FAULT
CURRENT
multiples
of
rated
current
DISTANCE OF FAULT
FROM NEUTRAL
(Percentage of winding)
Significant current flows in theactual fault near to the neutralend of the transformer winding
But very little linecurrent flows at theHV terminals
Slide [email protected]
REF
Restric ted Earth Fault Protection
CT terminalsaway fromprotected objectare connected
CT terminalsnear toprotected objectare connected
Slide [email protected]
DIFF
Auto TransformerHigh Imp Phase Segregated Protection
All CT r atio s tobe the same
This CT will carrymaximum currentand hencedictates ALL CTratios
But this CT is internaland may only have asingle fixed ratio.Therefore, this must bespecified correctly atthe time of purchase !!
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Generator and
Power Station Protection Transformer Protection
Barrie Moor : 2012 [email protected]
(07) 3298 5260
Slide [email protected]
High Impedance Protection
Especially sensitive in detecting earth faults in the
bottom 25% of the winding
Application to
Auto Transformers
REF earth fault schemes
Tertiary windings not specifically protected
Setting principles same as for bus zone
Select voltage setting to achieve through fault stability
CT Vk > 2.Vset to ensure scheme operation
HighImpedanceRelays
Tertiary
HV CTs LV CTsNeutral CTs One per phase at neutral end But above the star point
Restric ted Earth Fault Protection
Restricted EF ProtRestricted EF Prot
Transformer delta winding and star winding REF protection schemes
LV scheme remainsstable for external EF
HV scheme remainsstable for external EF
HV scheme also remainsstable for all LV EFs
GENERATOR and
POWER STATION
PROTECTION
Barrie Moor, B Eng (Elec)
Slide [email protected]
EARTHINGTRANSFORMERS
Operationand
Protection
Slide [email protected]
Earthing Transformer
LOAD
Earthing Transformer Fault Currents
Slide [email protected]
B
A
C
Earthing Transformer
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Generator and
Power Station Protection Transformer Protection
Barrie Moor : 2012 [email protected]
(07) 3298 5260
Slide [email protected]
Earthing Transformer
LOAD
Earthing Transformer Fault Currents
Slide [email protected]
Earthing Transformers
Provide a good earth reference for a delta winding duringearth faults
Restrict the voltage rise on the healthy phase duringearth faults inoperative during balanced voltageconditions
Carry significant current only during earth faultsie. 3 x I0 only
Earthing transformer and associated power transformerconsidered as a single unit and tripped together
Slide [email protected]
Effectively Earthed System
R0/X1 < 1
X0/X1 < 3
Limit voltage rise on unfaulted phases
80% of rated phase - phase voltage
Otherwise healthy phases can reach 100% of rated phase
- phase voltage
May even exceed this on transients
Slide [email protected]
Protection of Earthing Transformer
Two types of faults we need to consider:
Internal faults - faults inside the earthing transformer, the
result of insulation breakdown.
External faults - faults on the system outside the earthing
transformer. These can cause overheating of the earthing
transformer
Slide [email protected]
Over Current Protection
Over Current Protection
Interturn, interwinding or winding-to-core faults
Fed from delta-connected current transformers, so thatearth faults on the system, which generate only zero
sequence current, are not seen
Hence, O/C setting can be very low
Slide [email protected]
Earth Fault Protection
Earth Fault Protection
Long term low level unbalance may cause thermal
damage
Need to consider and set earth fault below
continuous rating capability
short-time rating capability
Combination of IDMT and definite time functions used
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Generator and
Power Station Protection Transformer Protection
Barrie Moor : 2012 [email protected]
(07) 3298 5260
Slide [email protected]
Earthing Transformer
O/C relay does not operate for externalearth faultsDef Time and IDMT E/F relays operate forexternal earth faults IDMT E/F relay
Def Time E/F relay
O/C relay
LOAD
Slide [email protected]
Overcurrent Setting Must be:
Greater than the magnetising current
Greater than the maximum inrush current. This depends on
Earthing transformers B-H characteristics The point-on-wave of the energisation
The remanence of the core
One common estimate of upper bound is 50x the magnetising
current
Tolerant of Earthing Transformer manufacturing variations
Impedances variations between the individual phases of the
earthing transformer may result in some spill current from the
delta CTs under through EFs
Slide [email protected]
Earth Fault Protection
Affected by long term residual voltage, which may cause
thermal damage
remember - no over temperature sensor is provided
Need to consider and set earth fault below
continuous rating capability
short-time rating capability
Combination of IDMT and definite time relays used to do
this
Thermal Protection
10 100 1 103
1 1040.1
1
10
100
1 103
1 104
1 105
EARTHING TRANSF THERMAL PROTECTION
EARTH FAULT CURRENT - AMPS
TIME-SECONDS
30
2300
contrating30A
max E/Fcurrent2300A
adiabatic thermal limit
actual thermal limit
earthing transformer E/F relay - Definite Time
downstream E/F relay
earthing transformerE/F relay - IDMT
Slide [email protected]
Biased Differential Protection
Earthing transformers are included inside the biased differential
zone of their power transformer
Current transformer connections important
LV CTs will be star connected
Thus Io can flow to the differential relay
And earthing transformer now supplies Io
Thus stability for external earth faults becomes an issue
To eliminate Io from the relaying system
YD12 CT connection with microprocessor differential relay
Or utilise additional Earthing Transformer CT connections
0
C
B
A
400/0.577
132 / 33 kV
externalearthfault
1600/1
0
c
b
0a
Earthing Transformer CTsDifferential Trip due to LV I0 Currents
8/10/2019 Generator and Power Station Protection_2012.pdf
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Generator and
Power Station Protection Transformer Protection
Barrie Moor : 2012 [email protected]
(07) 3298 5260
all 1600/0.333
0
C
B
A
400/0.577
132 / 33 kV
Nexternal
earthfault
1600/1
0
c
b
0a
Earthing Transformer CTsEliminate I0 from the relaying system
Slide [email protected]
Zero Sequence Current Trap
Star connected interposingCTs with neutral connection
Interposing CT secondarywithout neutral connection
CT includes a delta tertiary
winding to trap zerosequence currents
3 wire connection to the relayI1 & I2 only
Slide [email protected]
D1
D11
Modern Microprocessor Relays
IC
IB
IA
110
011
101
3
1
IC
IB
IA
RELAY
RELAY
RELAY
IC
IB
IA
101
110
011
3
1
IC
IB
IA
RELAY
RELAY
RELAY
IC
IB
IA
211
121
112
3
1
IC
IB
IA
RELAY
RELAY
RELAY
D12
No phase change
But, removes neutral current
GENERATOR and
POWER STATION
PROTECTION
Barrie Moor, B Eng (Elec)
Slide [email protected]
TRANSFORMERPROTECTION
Delta Windings withEarthed Corner
Slide [email protected]
Delta windings with earthed corner
Typical for transformer tertiary windings
To maintain stability on through faults
Need to include CT in the earthed corner
CT to be the same ratio as the main CT
Connect in parallel with the earthed phase CT
This cannot be overcome with phase shifting and Io
facilities of microprocessor based relays
Earthed Delta Corners inside theDifferential Zone
CT installed on the earthed corner.Same ratio as phase CTsIn parallel with the phase CT onthe earthed phase
This is also required formicroprocessor relay schemes
8/10/2019 Generator and Power Station Protection_2012.pdf
40/62
Generator and
Power Station Protection Transformer Protection
Barrie Moor : 2012 [email protected]
(07) 3298 5260
GENERATOR and
POWER STATION
PROTECTION
Barrie Moor, B Eng (Elec)
Slide [email protected]
TRANSFORMERPROTECTION
Neutral Displacement
Protection
Slide [email protected]
Neutral Displacement Protection
Applicable to delta windings with no earth reference
Connected to open delta of VT secondary
To provide a zero sequence flux path
3 x 1 phase VTs or
5 limb VT
Relay must be immune to 3rd harmonics
Tripping time can be relatively slow
Problems can occur with resonance on energisation
Slide [email protected]
Neutral Displacement Protection
VoltageDisplacement
Relay
VT with open deltasecondary winding
Slide [email protected]
Neutral Displacement Protection
Measures 3.Vo for a solid E/F
Healthy phases rise to full phase-phase potential
Healthy phases are now only 60 deg apart
Thus, for standard 63.5V VT, output voltage is 190.5V
Set relay to 10% say 20V
NeutralDisplacement
Protection
0 V0 V
VB = 3VC = 3
VA = 0
-150+150
63.5 V63.5 V110 V110 V190.5 V190.5 V
8/10/2019 Generator and Power Station Protection_2012.pdf
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Generator and
Power Station Protection Motor Protection
Barrie Moor : 2012 [email protected]
(07) 3298 5260
GENERATOR and
POWER STATION
PROTECTION
Barrie Moor, B Eng (Elec)
Slide [email protected]
MOTORPROTECTION
Slide [email protected]
Motor Protection
To detect faults within the motor and supply system
Faults on the supply system, cables, etc
Internal faults
Often as a result of previous thermal events
Thermal Withstand
Motor overload
Starting and Stalling currents & times
Unbalanced voltage supply (NPS)
System events
Loss of load, under voltage, etc.
Slide [email protected]
Motor Protection
Thermal Protection
Extended Start Protection
Stalling Protection
Number of Starts Limitation
Negative Sequence Current Detection
Short Circuit Protection
Earth Fault Protection Under Voltage Protection
Loss of Load Protection
Winding RTD measurement and trip
THERMALPROTECTION
GENERATOR and
POWER STATION
PROTECTION
Barrie Moor, B Eng (Elec)
Slide [email protected]
MOTORPROTECTION
Thermal Protection
Slide [email protected]
Thermal Overload Protection
Winding failures are often due to previous heating
Because of the motors thermal mass
Infrequent short term overloads may have no affect
But sustained overloads, of even just a few percent, may
cause aging and premature insulation failure
Insulation life may be halved for operation every 10 deg Cabove rated maximum.
Slide [email protected]
Starting Current
High starting current reducesto full load current as the motor
gains speed
Positive sequence impedance
of the motor at any slip is :-
SYNC
ACTSYNC
N
NNSlip
=
2
1R1S
2
1R1SPOS XX
S
RRZ
0 250 500 750 1000 1250 15000
0.2
0.4
0.6
0.8
1
Z s( )
s
0 250 500 750 1000 1250 15000
1
2
3
4
5
6
I s( )
s
8/10/2019 Generator and Power Station Protection_2012.pdf
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Generator and
Power Station Protection Motor Protection
Barrie Moor : 2012 [email protected]
(07) 3298 5260
Slide [email protected]
Negative Sequence Equivalent Circuitof Induction Motor
Positive Sequence Equivalent Circuitof Induction Motor
(Rs2 + Rr2)
(Rs1 + Rr1)
j(Xs2 + Xr2)
(S - 1) Rr2
(2 - S)
j(Xs1 + Xr1)
S
(1 - S) Rr1
Motor Impedances
Slide [email protected]
Positive Sequence Impedance
At any slip
At standstill with slip = 1.0
This is a small impedance, hence high starting current
eg. up to 6 pu
21R1S
2
1R1SPOS XX
SRRZ
2
1R1S
2
1R1SPOS XXRRZ
Slide [email protected]
Negative Sequence Impedance
At any slip
At normal running with slip 0
2
2R2S
2
2R2SNEG XX
S2
RRZ
2
2R2S
22R
2SNEG XX2
RRZ
Slide [email protected]
Motor Impedances
Positive Sequence : Standstill
Negative Sequence : Running
R is small compared with X:
These will be approximately the same !!
2
1R1S
2
1R1SPOS XXRRZ
2
2R2S
2
2R2SNEG XX
2
RRZ
RUNNINGSTANDSTILL NEGPOSZZ =
Slide [email protected]
Motor Impedances Consider an example where ISTART/IFLC = 6
Hence, 1pu positive sequence supply voltage results in:
1 pu current when running at full load
6 pu current at start-up
But the negative sequence impedance at running is thesame as the positive sequence impedance at startup
So, when running, 1 pu NPS supply voltage wouldresult in:
6 pu NPS current
Or, as a typical example, when running, 1% NPSsupply voltage would result in:
6% NPS current
Slide [email protected]
CurrenttartingS
CurrentLoadFull
Z
Z
RUNNING
STARTING
POS
POS=
Motor Impedances
Positive sequence impedances at Start-up & Running
And since:
CurrenttartingS
CurrentLoadullF
Z
Z
RUNNING
RUNNING
POS
NEG=
RUNNINGSTARTING NEGPOSZZ =
eg. 1/6 th
eg. ZNEG = ZPOS / 6
So, what does this
mean for NPS current
8/10/2019 Generator and Power Station Protection_2012.pdf
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Generator and
Power Station Protection Motor Protection
Barrie Moor : 2012 [email protected]
(07) 3298 5260
Slide [email protected]
Negative Sequence Current
Negative sequence current will equal:
% of NPS in the supply voltage
Multiplied by the ratio of starting / full load current
Multiplied by full load current
=NPSI
POS
NPS
V
V
FLC
CurrentStarting[ ]
FLCI