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Brief Introduction for Automated Relay Testing (Relay Type: RET670) Kamin Dave Manager – Marketing & Protection Applications South Asia & Middle East
DOBLE ENGINEERING COMPANY, India
Protection Relay Testing:
¾Testing of Differential Protection Relay type RET670
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Relay Details: RET670 ¾ Three Wdgs Differential Protection relays ¾ Automatic calculation for Correction factor based on CTR ¾ Settings for Transformer & CT vector group compensations ¾ 2nd Harmonic Blocking during inrush period due to point on wave switching ¾ 5th Harmonic Blocking for Over excitation phenomena
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Relay Details: RET670 ¾ ¾ ¾ ¾ ¾
MAKE : ABB MODEL : RET670 LOCATION : GETCO, 220KV Dahej S/S Relay Settings: Differential Element: HV Voltage = 220KV LV Voltage = 66KV Transformer MVA = 100MVA Transformer VECTOR GROUP = YnYn0. CT CONNECTION = YY IDMin> = 0.2PU ID>> = 8.0PU Slope section2 = 30% Slope section3 = 70% End section1 = 1.25PU End section2 = 2.75PU In = 1A Knowledge Is Power SM
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Relay Details: RET670 ¾ MAKE : ABB ¾ MODEL : RET670 ¾ LOCATION: GETCO, 220KV Dahej S/S
¾ Relay Settings: ¾ Differential Element:
CT Ratio on HV Side = 300/1A Transformer full load current at HV side = 262.36A Calculated correction factor at HV side = 1.141965 CT Ratio on LV Side = 1200/1A Transformer full load current at LV side = 874.54A Calculated correction factor at LV side = 1.370358 Knowledge Is Power
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Relay Characteristic: Dual Slope Second slope is start from End section2
First slope is start from End section1
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Relay Characteristic: Dual Slope
SLP1=20%
SLP2=70%
Relay Characteristic plotted based on settings
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Relay Details: RET670 ¾ FLC on TR HV side1: IFLC = (100*1000) / (1.7325*220) = 262.36 Ampere ¾ FLC on TR LV side2: IFLC = (100*1000) / (1.7325*66) = 874.54 Ampere ¾ CF1 = (1*300) / (262.36*1) = 1.141965 ¾ CF2 = (1*1200) / (874.54*1) = 1.370358
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Relay Details: RET670 Start
Phase current for side1
Y/N
Zero sequence current side1
Zero sequence current reduction
Y
Phase current for side2
Deduct zero sequence current
N
Zero sequence current reduction
Y/N
Perform Angle compensation
Y
Deduct zero sequence current
N
Y/N
Perform magnitude compensation
Zero sequence current side2
Perform magnitude compensation Perform differential current
Perform Angle compensation
Perform differential Vs Restraint char.
Trip decision
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Y/N
Relay Details: RET670 The power transformer differential protection: 1.
Calculates three fundamental frequency differential currents, and one common bias current. The zero-sequence component can optionally be eliminated from each of the three fundamental frequency differential currents, and at the same time from the common bias current.
2.
Calculates three instantaneous differential currents. They are used for harmonic, and waveform analysis. Instantaneous differential currents are useful for post-fault analysis using disturbance recording
3.
Calculates negative-sequence differential current, Contributions to it from both (all three) power transformer sides are used by the internal/external fault discriminator to detect and classify a fault as internal or external.
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Relay Details: RET670
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Relay Details: RET670 The power transformer differential protection: 1.
Positive sequence components are presents during Phase to Phase faults, Phase to Ground faults & 3-phase faults.
2.
Negative sequence components are presents during Phase to Phase faults & Phase to Ground faults
3.
Zero sequence components are presnets during Phase to Ground faults & 2 phase to Ground faults (Generally on HV bushings)
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Relay Details: RET670 Normal Load Condition
100MVA 220/66kV
I1
Y
I2
Y
IB1 IR1
E
IR1
E
IY1 Positive seq. components
IY1
IB1 I1@0O
R
I2@180O
Positive seq. components
Negative sequence components are very less or almost zero Knowledge Is Power
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Relay Details: RET670 Power system fault data (External fault) simulated: HV SIDE: Phasor Data IR = 5A@65O IY = 5A@245O IB = 0.1A@130O Positive sequence components IR1 = 2.887A@35O IY1 = 2.887A@275O IB1 = 2.887A@155O Negative sequence components IR2 = 2.887A@95O IY2 = 2.887A@335O IB2 = 2.887A@215O Knowledge Is Power
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Relay Details: RET670 Power system fault data (External fault) simulated: LV SIDE: Phasor Data IR = 5A@245O IY = 5A@65O IB = 0.1A@310O Positive sequence components IR1 = 2.887A@215O IY1 = 2.887A@95O IB1 = 2.887A@335O Negative sequence components IR2 = 2.887A@-85O IY2 = 2.887A@155O IB2 = 2.887A@35O Knowledge Is Power
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Relay Details: RET670 External Fault Condition
100MVA 220/66kV
I1
Y
Y Positive seq. components
Positive seq. components
IB1
IR2
IB2
IR1
F1
I2
IY1
E
E
IR1
IB1
IY1 I1@65O IY2
R
I2@245O
IY2
IB2 IR2
Negative seq. components
Negative seq. components
Negative sequence components are significant but opposite in direction
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Relay Details: RET670 Power system fault data (Internal fault) simulated: HV SIDE: Phasor Data IR = 5A@73O IY = 5A@253O IB = 0.1A@130O Positive sequence components IR1 = 2.887A@43O IY1 = 2.887A@283O IB1 = 2.887A@163O Negative sequence components IR2 = 2.887A@103O IY2 = 2.887A@343O IB2 = 2.887A@223O Knowledge Is Power
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Relay Details: RET670 Power system fault data (Internal fault) simulated: LV SIDE: Phasor Data IR = 5A@68O IY = 5A@248O IB = 0.1A@100O Positive sequence components IR1 = 2.887A@38O IY1 = 2.887A@278O IB1 = 2.887A@158O Negative sequence components IR2 = 2.887A@98O IY2 = 2.887A@338O IB2 = 2.887A@218O Knowledge Is Power
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Relay Details: RET670 Internal Fault Condition
100MVA 220/66kV
I1
Y
F1
I2
Y
Positive seq. components
Positive seq. components
IB1
IR2
IB2
IR1
E
E
IY1 I1@73O IY2
R
IR1
IB1
IR2
I2@68O
IB2
Negative seq. components
IY1 IY2
Negative seq. components
Negative sequence components are significant & in same direction
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Relay Details: RET670 Figure from Manual
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Relay Details: RET670 100MVA 220/66kV
I1 Positive seq. components
IB1
IR2
IB2
Y
Side1
IR1
External Fault Condition & Side2 CT saturate for 3 cycles F1 I2
Y
Side2
Positive seq. components
IY1
E
E
IR1
IB1
IY1 I1@65O IY2
R
I2@245O
IY2
IB2 IR2
Negative seq. components
Negative seq. components
Negative sequence components flows through Side2 CT is higher than Side1 CT but Positive seq. components are lower than Side1 CT
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Relay Details: RET670 100MVA 220/66kV
I1 Positive seq. components
Internal Fault Condition & Side1 CT saturate
Y
Side1
F1
Y
I2
Side2
IB1 E
IR2
IB2
E
IY2
R
IR1
IB1
IY1 I1@73O
Positive seq. components
IR2
I2@68O
IB2
IY1 IY2
Negative seq. components
Negative seq. components
Negative sequence components flows through side1 CT is higher than side2 CT but same in direction
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Relay Details: RET670 Figure from Manual
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Relay Details: RET670 Inrush Detection by Adaptive Techniques: The combination of the 2nd harmonic and waveform analysis methods, allows the relay designer to optimize the detection of inrush currents while avoiding some of the potential drawbacks. One possible way, with good field experience, is to combine these methods as follows: ♦ Employ both the 2nd harmonic and the waveform criteria to detect the initial inrush condition ♦ One minute after power transformer energizing, the 2nd harmonic criterion can be disabled in order to avoid long clearance times for heavy internal faults and the waveform criterion alone can take care of the sympathetic and recovery inrush scenarios; and, ♦ Temporarily enable the 2nd harmonic criterion for six seconds when a heavy external fault has been detected, to gain additional security for external faults.
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Relay Details: RET670
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Relay Details: RET670 Setting Parameters
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Relay Details: RET670 Setting Parameters
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Relay Details: RET670 Setting Parameters
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Test Equipment •
Make
: DOBLE
•
Model : F6150
Software •
F6TesT Version 2.21-1156
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Test Circuit: Source Configurations: Six current sources Winding 1 currents Winding 2 currents
I1
IA IB
IC
RET
I2
I3
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Winding1 & Winding2 CT connections Knowledge Is Power
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F6150 Test set
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Tests Conducted: • Pickup test for Winding1 • Pickup test for winding2 • Slope Characteristic test • Stability test
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Winding Pickup Test Use “RAMP” Module
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Winding Pickup Test: ¾ Pickup Test for Winding1 (WDG1):Ipickup = (ID)*(1/CF1) = (0.2)*(1/1.141965)
= 0.1752A ¾ Pickup Test for Winding2 (WDG2):Ipickup = (ID)*(1/CF2) = (0.2)*(1/1.3701558)
= 0.1459A Not need to enter correction factor for CTR matching. Relay will Automatically adjust TAP value based on CTR Knowledge Is Power
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Winding Pickup Test:
¾How to Create Test Plan???
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Location Level1
Right click on “Mouse” & click on “Append” to create new Location Knowledge Is Power
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Apparatus Maintenance and Power Management for Energy Delivery
Location Level2
Right click on “Mouse” & Click on “Append” to create location of feeder
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Relay Level
Right click on “Mouse” & Click on “Append” to create relay
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Test plan Level
Right click on Mouse & Press “Append” & create New Testplan Knowledge Is Power
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Double click on “New test plan (Say; 87T)”
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Right click on “Mouse” & Press on “Append” & Create new Test Module Knowledge Is Power
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Select “RAMP” Module from list
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Press on “OK” & Create New Test Definition Knowledge Is Power
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Select “6 Currents (Right bank)” Knowledge Is Power
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Press on “Yes” Knowledge Is Power
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Winding 1 Pick up Test Select “Trigger”
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Winding 1 Pick up Test
Ramp Graph Enter “Action” details
WDG1 current sources
Enter Expected Value = 0.175A
Result = 0.18A Knowledge Is Power
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Relay operated on this point
Enter CT ratio (300/1A) of winding1 current
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Select values in Primary
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Shows Primary values (300*0.175 = 52.5A) & (300*0.18=54A) of winding1 current Knowledge Is Power SM
Apparatus Maintenance and Power Management for Energy Delivery
Winding 2 Pick up Test
Ramp Graph Enter “Action” details
WDG1 current sources
Enter Expected Value = 0.145A
Result = 0.17A Knowledge Is Power
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Relay operated on this point
Enter CT ratio (1200/1A) of winding2 current
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Shows Primary values (1200*0.145 = 174A) & (1200*0.16=204A) of winding2 current Knowledge Is Power SM
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Winding 2 Pick up Test
Ramp Graph Enter “Action” details
WDG1 current sources
Enter Expected Value = 0.217A
Result = 0.23A Knowledge Is Power
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Relay operated on this point
Slope Characteristic Test for YnYn0 TR Configuration Use “DiffChar” Module
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Slope Characteristic Test: YnYn0
Vector Compensation
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How to Create Slope Characteristic???
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Option1: How To Draw Slope Characteristic in F6TEST software??
Click on “Add” button Knowledge Is Power
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Option1: How To Draw Slope Characteristic in F6TEST software??
Select “Differential” Knowledge Is Power
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Option1: How To Draw Slope Characteristic in F6TEST software??
Bias current Formula:
Ibias = MaxIIPI * IISI
Select Bias current (Ibias) formula from list Knowledge Is Power
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Apparatus Maintenance and Power Management for Energy Delivery
Option1: How To Draw Slope Characteristic in F6TEST software??
Bias Formula
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Software Configuration Star conn.
YnYn0
YdY0
Star conn.
YdY0
R Amplitude Compensation
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Option1: How To Draw Slope Characteristic in F6TEST software??
Click on “Pickup” Column & then after Click on “Modify” Knowledge Is Power
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Enter Pickup setting Say; 0.2
Option1: How To Draw Slope Characteristic in F6TEST software??
Select IBias Vs Idiff
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Option1: How To Draw Slope Characteristic in F6TEST software??
SLP2=70% SLP1=30%
Enter Slope 30% & 70% based on relay setting Knowledge Is Power
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Option1: How To Draw Slope Characteristic in F6TEST software??
Right click on “Mouse” & click on “Maximize Graph” Knowledge Is Power
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Maximize Graph on full screen
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Option1: How To Draw Slope Characteristic in F6TEST software??
Click on “Close” Knowledge Is Power
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Option2: How To Import RIO file from Excel tool??? Knowledge Is Power
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Option2: How To Import RIO file from Excel tool???
Click on “Import” button Knowledge Is Power
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Option2: How To Import RIO file from Excel tool???
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Option2: How To Import RIO file from Excel tool???
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Option2: How To Import RIO file from Excel tool???
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Select “Open to Close” Knowledge Is Power
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Select “Open to Close” Knowledge Is Power
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How to create New Test Module?
Right click on “Mouse” & click on “Append”
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How to create New Test Module?
Select “DiffChar” Module from list
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Double Click on Test Module & create New Test Definition Knowledge Is Power
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Select characteristic from list
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Select element
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Enter Max. Test current (i.e. 15A) Knowledge Is Power
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Select Trigger (i.e. Open to Close)
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Test results
Blue color shows Expected Curve & Pink color shows actual Curve Knowledge Is Power
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Stability Test Use “RAMP” Module
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Stability Test: IA@0O
IDiff = 0A
I1@180O
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Enter “Action” details
Ramp Graph
WDG1 current starts to reduce
Drop out value = 0.7A
Relay operated on this point
Pickup value = 0.6A Knowledge Is Power
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Stability Test Use “SSIMUL” Module
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Winding1 current Negative seq. current of winding1
Positive seq. current of winding1
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Winding2 current
Positive seq. current of winding2
Negative seq. current of winding2
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External Fault condition
Normal Load condition Knowledge Is Power
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Relay was not operated
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DOBLE ENGINEERING COMPANY 2nd Floor Suvidhi Pride, Gorwa Refinery Road VADODARA INDIA. Tel:- +91 98980 55956 EMAIL ID:-
[email protected] Website:- www.doble.com
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