Transcript
Relay Coordination
-BY JAIKISHAN GIANANI
WHY PROTECTION IS REQUIRED ?
TO PROTECT COMPONENTS OF POWER SYSTEM
(GENERATORS , TRANSFORMERS , OH LINES
CABLES , MOTORS , CIRCUIT BREAKERS ETC)
TO PROTECT THE POWER SYSTEM
COMPONENTS OF PROTECTION SYSTEM
CURRENT TRANSFORMERS & POTENTIAL
TRANSFORMERS
PROTECTIVE RELAYS
CIRCUIT BREAKERS
RELAY CO-ORDINATION
WHY RELAY CO-ORDINATION ?
TO ACHIEVE SELECTIVITY/ DISCRIMINATION
ONLY FAULTY SECTION OF POWER SYSTEM
TO BE ISOLATED
GENERALLY USED FOR OVERCURRENT AND
EARTH FAULT RELAYS
FUSES
• SIMPLEST AND MOST COMMONLY USED
OVERCURRENT PROTECTION DEVICE
• PROVIDES OVER LOAD PROTECTION TO A
CERTAIN DEGREE
• IT CONSISTS OF A FUSIBLE METAL (
TIN, LEAD ALLOY OR SILVER ) WIRE
ENCLOSED IN A CONTAINER WITH METAL
CONTACTS AND QUENCHING MEDIUM
FUSES
• IT NORMALLY FOLLOWS THE CHARACTERISTICS
OF I^2 * t = K , HOWEVER ENCLOSED
FUSES HAVE A CHARACTERISTIC OF I^3.5
* t = K
• FUSES HAVE A CHARACTERISTIC OF
LIMITING THE PROSPECTIVE FAULT
CURRENT TO A LOWER VALUE
OVERCURRENT RELAY
A RELAY WHICH OPERATES WHEN THE CURRENT THROUGH IT EXCEEDS A PRESET
VALUE
THE VALUE OF THE PRESET CURRENT ABOVE WHICH THE RELAY OPERATES IS CALLED THE PICK UP VALUE OF THE RELAY
AN OVERCURRENT RELAY IS NORMALLY USED FOR PROTECTION OF DISTRIBUTION FEEDERS , MOTORS AND TRANSFORMERS
TYPES OF OVERCURRENT RELAYS
• CONSTRUCTION WISE:
• ELECTROMAGNETIC:
a)ATTRACTED ARMATURE YPE
b)INDUCTION DISC TYPE
• STATIC
• NUMERIC
• APLICATION WISE:
• INVERSE DEFINITE MINIMUM TIME RELAY
(IDMT) RELAY
• DEFINITE TIME RELAY
• INSTANTANEOUS OVERCURRENT RELAY
IDMT RELAY CHARACTERISTICS
• GENERAL RELAY EQUATION :
t = K/((I/In)^n – 1)
t = RELAY OPERATING TIME
I = FAULT CURRENT
In = RELAY PLUG SETTING ( PICK UP )
n = CONSTANT
K = CONSTANT
IDMT RELAY CHARACTERISTICS
• AS PER BRITISH STANDARD DIFFERENT
CHARACTERISTICS ARE OBTAINED BY
SUBSTITUTING VALUES OF n AND K
• NORMAL INVERSE : n = 0.02 , K = 0.14
OPERATING TIME AT TEN TIMES PS = 3.0 SECS
• VERY INVERSE : n = 1 , K = 13.5
OPERATING TIME AT TEN TIMES PS = 1.5 SECS
• EXTREMELY INVERSE : n = 2 , K = 80
OPERATING TIME AT TEN TIMES PS = 0.808 SECS
RELAY SETTINGS
• IDMT RELAYS :
PS = PLUG SETTING (FOR SETTING PU)
TMS = TIME MULTIPLIER SETTING( OPTG TIME)
RELAY FOLLOWS THE CHARACTERISTICS FOR
VARIOUS VALUES ( x PS) OF OPERATING
CURRENT
RELAY SETTINGS
• DEFINITE TIME RELAY :
PU= PICK UP CURRENT
t = OPERATING TIME
RELAY OPERATES ABOVE PU CURRENT AFTER
SET TIME DELAY t
• INSTANTANEOUS RELAY :
PU = PICK UP CURRENT
RELAY OPERATES ABOVE PU CURRENT
INSTANTLY
METHODS OF CO-ORDINATION
• TIME GRADED RELAY CO-ORDINATION
• CURRENT GRADED RELAY CO-ORDINATION
• CURRENT AND TIME GRADED RELAY CO-
ORDINATION
• USE OF DIRECTIONAL RELAYS
TIME GRADED RELAY
COORDINATION
SOURCE A B C FAULT
F
FOR FAULT AT LOCATION F OPERATING TIME OF RELAY AT C < B < A DISADVANTAGE: OPERATING TIME MAXIMUM FOR FAULTS NEAREST TO THE SOURCE APPLICATION:SHORT DISTANCE RADIAL FEEDERS WHERE FAULT CURRENT IS ALMOST SAME FOR SECTIONS SA, AB ,BC AND CD
S
D
CURRENT GRADED RELAY
COORDINATION
SOURCE A B C D
S F3 F1 F2
FAULT LEVEL F1 > F2 > F3 RELAYS AT A , B AND C SET TO PICK UP AS PER FAULT CURRENT DUE TO FAULTS AT F1 , F2 AND F3 RESPECTIVELY DEFINITE TIME RELAYS USED ( TIME SETTING OF ALL RELAYS IS SAME
80%
CURRENT GRADED RELAY
COORDINATION
ADVANTAGE :
1) TIME SETTING OF RELAY NEAR THE
SOURCE IS LESS
2) CHEAP , AS ONLY DEFINITE TIME RELAYS
CAN BE USED
DISADVANTAGE:
1) RELAYS TEND TO OVERREACH
2) HENCE ONLY 80% OF THE FEEDER IS
NORMALLY COVERED
3) TRANSIENT STABILITY IS NOT GOOD
TIME AND CURRENT GRADED RELAY
CO-ORDINATION
COMBINATION OF TIME AND CURRENT GRADING
USED
RETAINS ADVANTAGES OF BOTH SYSTEMS WHILE
ELIMINATING THE DISADVANTAGES OF BOTH
IDMT RELAYS NORMALLY USED
ALSO COMBINATION OF BOTH DEFINITE TIME
AND IDMT RELAYS CAN BE USED
TIME AND CURRENT GRADED RELAY CO-
ORDINATION
SINCE RELAYS HAVE INVERSE TIME CURRENT
CHARACTERISTIUCS , FAULTS NEAREST TO
THE RELAY ARE CLEARED FASTER
RELAYS DO NOT OVERREACH AS OPERATION
CAN BE DELAYED FOR SAME PICK UP VALUES
TIME AND CURRENT GRADED RELAY CO-
ORDINATION
IDMT RELAY CHARACTERISTIC IS INVERSE
FOR CURRENTS UP TO 10 TIMES THE PS
BETWEEN 10 AND 20 TIMES THE PS THE
RELAY HAS AN ALMOST DEFINITE TIME
CHARACTERISTICS
RELAY PLUG SETTING IS DONE AS PER THE
FULL LOAD CURRENT RATING OF THE
FEEDER
TIME AND CURRENT GRADED RELAY CO-
ORDINATION
RELAYS ARE SET TO PICK UP AT HIGHER
CURRENTS TOWARDS THE SOURCE
OPERATING TIME IS ALSO SET
PROGRESSIVELY HIGH TOWARDS THE SOURCE
DATA FOR RELAY CO-ORDINATION
ONE LINE DIAGRAM OF THE POWER SYSTEM
NETWORK
THE MAXIMUM AND MINIMUM FAULT LEVELS
AT VARIOUS BUSES
THE RATIO OF ALL CT AND PT USED FOR
PROTECTIVE RELAYS IN THE SYSTEM
DATA FOR RELAY CO-ORDINATION
THE NORMAL AND MAXIMUM LOAD RATING
OF ALL FEEDERS
RATING AND STARTING CURRENT OF
LARGEST MOTOR AT EACH BUS
SHORT TIME RATING OF ALL POWER
EQUIPMENT , CABLES/ OH LINES AND BUS
BARS
TYPICAL EXAMPLE OF O/C RELAY CO-
ORDINATION
ONE LINE DIAGRAM :
25 MW UNIT1 31.25MVA
31.5 MVA 0.14 P.U
A
B C
D E
F G
FUSE
LT MOTOR
HT MOTOR H
11/0.433 KV 2MVA 0.06 PU
GEN
11/6.9 KV 12 MVA 0.08 PU
B1 B2
B3 B4
SIMILAR TO FEEDER BC SIMILAR TO UNIT 1
FAULT LEVEL DATA
SR NO BUS NUMBER
(VOLTAGE)
MAX(MVA)/KA MIN(MVA)
1 B1 ( 11KV) 450/23.63 225
2 B2 ( 11KV) 400/20.99 200
3 B3 ( 433V) 33/44 30
4 B4 (6.9KV) 150/12.55 140
5 HT MOTOR 130/10.87 120
6 LT MOTOR 22/29.33 20 F
D
D
G
G
LT M
A
E
E
CT RATIO DATA
SR NO LOCATION NUMBER CT RATIO
1 A 2000/1
2 B 1000/1
3 C 1000/1
4 D 125/1
5 E 1000/1
6 F 3000/1
7 G 1200/1
8 H 50/1
9 LT MOTOR NA(FUSE 160 A)
SHORT TIME RATING DATA
SR NO COMPONENT DESIGNATION RATING(MVA/SEC)
1 B1 40KA /3SEC
2 B2 40KA/3SEC
3 B3 50KA/1SEC
4 B4 21KA/1SEC
5 CABLE B1-B2 20KA/1SEC
6 CABLE LT MOTOR 20KA/1SEC
7 CABLE HT MOTOR 20KA/1SEC
MOTOR RELATED DATA
HIGHEST RATING HT MOTOR
RATING : 450 KW
FULL LOAD CURRENT : 45A
STARTING CURRENT : 248A
STARTING TIME : 5SEC
HIGHEST RATING LT MOTOR
RATING : 37KW
FULL LOAD CURRENT : 66A
STARTING CURRENT :363A
STARTING TIME : 3SEC
FEEDER RELATED DATA
FEEDER F
NORMAL LOADING : 2000A
FEEDER G
NORMAL LOADING : 700A
CO-ORDINATION INTERVALS
TYPE OF RELAYS ASSUMED : INDUCTION DISC
TYPE OVERCURRENT RELAYS
BREAKER TO FUSE INTERVAL: 0.15 SEC
BREAKER TO BREAKER INTERVBAL : 0.25 SEC
INSTANTANEOUS ELEMENT OPERATING TIME:60mSEC
LT MOTOR CALCULATIONS
•STARTING CURRENT OF LT MOTOR = 363A •FUSE RATING = 160A (SLOW BLOWING) •CURRENT THROUGH FUSE DURING MOTOR STARTING = 363/160 = 2.26 TIMES RATED A •BLOWING TIME FROM CHARACTERISTIC = 14SEC •FAULT CURRENT FOR FAULT AT LT MOTOR TERMINALS = 29330A •= 29330/160 = 183 TIMES FUSE RATING •FUSE BLOWING TIME FROM CURVE = 0.03 SEC
MCC INCOMER (F) CALCULATIONS
•NORMAL LOAD AT MCC INCOMER (F) = 2000A •STARTING CURRENT OF LARGEST LT MOTOR =363 A •MAXIMUM LOAD ON INCOMER (F) = 2000+363 = 2363 A •CT RATIO OF INCOMER (F) = 3000/1 •CURRENT(Ip)AT F = 2363/3000 =0.788A
•OPERATING TIME DESIRED FOR FAULT AT LT MOTOR TERMINALS = 0.03+ 0.15 = 0.18SEC
MCC INCOMER (F) CALCULATIONS
• FAULT CURRENT THROUGH FOR LT MOTOR TERMINAL FAULT = 29335 A • CURRENT THROUGH CT SEC = 29335/3000= 9.77A • CURRENT THROUGH RELAY AT F = 9.77/0.788=12.4 • DESIRED TMS =0.066
TRF HT (D) CALCULATIONS
•RELAY USED AT D IS NORM INV + HIGH SET
•CURRENT THROUGH D FOR FAULT AT B3(MAX) = 44000/25.4( TRF RATIO) = 1732A •CURRENT THROUGH RELAY AT D = 1732/125(CTR) =13.85A
•HIGH SET ELEMENT SETTING = 1.3 * 13.85 = 18.0A (SET 30 % HIGHER THAN MAX LT FAULT LEVEL) •RELAY D HIGH SET ELEMENT SETTING = 18A
TRF HT (D) CALCULATIONS
•CURRENT THROUGH D FOR MAX LOADING AT F =105A •PICK UP OF RELAY AT D = 105/125(CTR) = 0.84A •FAULT CURRENT THROUGH D FOR FAULT AT F = 44000/25.4( TRF RATIO) = 1732 A •FAULT CURRENT THROUGH RELAY AT D 1732/125 = 13.85 A = 13.85 / 0.84 = 16.5 TIMES THE PICK UP
TRF HT (D) CALCULATIONS
•DESIRED RELAY D OPERATING TIME FOR FAULT AT F = 0.18 + 0.25 = 0.43SEC •DESIRED TMS FOR RELAY AT D =0.177
TRF HT (D) CALCULATIONS
•RELAY D HIGH SET P U CHECK
•FAULT CURRENT FOR MIN FAULT AT D = 10500A
•CURRENT THROUGH RELAY AT D = 10500/125 = 84A
•HIGH SET RELAY SET TO PU AT 18A • RELAY OPERATING TIME 0.06 SECS
11 KV I/C (C) CALCULATIONS
•RELAY USED AT C IS NORM INV
•MAX LOAD THROUGH C = 630 + 105 ( FL AMP OF BOTH TRANSFORMERS) = 735A
•PICK UP= 735/1000 = 0. 735A •FAULT CURRENT THROUGH C FOR FAULT AT D = 20995 A
•CURRENT THROUGH RELAY AT C = 20995/1000 = 21A
11 KV I/C (C) CALCULATIONS
• CURRENT THROUGH RELAY AT C = 21/0.75= 28 TIMES • DESIRED OPERATING TIME FOR FAULT AT D = 0.06 + 0.25 = 0.31 SECS • TMS OF RELAY AT C =0.152
11 KV O/G (B) CALCULATIONS
• ASSUMED THAT FAULT LEVEL AT B1 AND B IS SAME • ASSUMED THAT FAULT LEVEL AT D AND B2 IS SAME • FOR FAULT AT B2 OPERATING TIME OF RELAY AT C IS O.31 SECS • OPENING OF EITHER C OR B CAUSES POWER INTERRUPTION TO B2 • HENCE PICK UP AND TMS OF RELAY AT B IS SAME AS THAT OF RELAY AT C ( NO TIME GRADING)
11 KV I/C TO B1 (A) CALCULATIONS
•FULL LOAD CURRENT OF GENERATOR = 1640A
•CT RATIO AT A = 2000/1
•PICK UP OF RELAY AT A = 1640/2000=0.82A •ASSUMED THAT FAULT LEVEL IS SAME AT B AND B1 •FAULT CURRENT THROUGH B FOR FAULT AT B = 23600 A
•CURRENT THROUGH RELAY AT B = 23600/1000 = 23.6A
11 KV I/C TO B1 (A)
CALCULATIONS
•PICK UP OF RELAY AT B = 0.75A
•CURRENT THROUGH RELAY AT B = 23.6/0.75 = 31.5 TIMES PICK UP •OPERATING TIME OF RELAY AT B FOR FAULT AT B =0.31 •TMS OF RELAY AT B = 0.158
11 KV I/C TO B1 (A)
CALCULATIONS
•FAULT CURRENT THROUGH A FOR FAULT AT B = 23600 A
•CURRENT THROUGH RELAY AT A = 23600/2000 =11.80 •CURRENT THROUGH RELAY AT A = 11.80/0.82=14.4 TIMES
•DESIRED OPTG TIME IS 0.31 + 0.25 =0.56SEC • TMS OF RELAY AT A =0.22
6.9 KV I/C (G) CALCULATIONS
•RELAY AT G IS NORM INV
•MAXIMUM LOAD ON G IS 700 + 248 = 948A
•CT RATIO AT G = 1200/1
•PICK UP OF RELAY AT G = 948/1200 = 0.79A •CURRENT THROUGH G FOR FAULT AT HT MOTOR TERMINAL = 10877 A •CURRENT THROUGH RELAY AT G = 10877/1200 = 9.06A
6.9 KV I/C (G) CALCULATIONS
•CURRENT THROUGH RELAY AT G = 9.06/0.79 = 11.5 TIMES PICK UP •RELAY OPERATING TIME REQ = 0.06 + 0.25 = 0.31 SEC
•RELAY TMS =0.11
•CURRENT THROUGH RELAY AT G FOR FAULT AT B4= 12551/1200 = 10.45A = 10.45/0.79=13.22 TIMES PICK UP
•FOR TMS = 0.11 OP TIME = 0.3 SEC
TRF HT (E) CALCULATIONS
•RELAY AT E NI + HIGH SET •MAX LOAD ON E = 630A
•CT RATIO AT E = 1000/1
•PLUG SETTING OF RELAY AT E = 630/1000 = 0.635A •FAULT CURRENT THROUGH E FOR FAULT AT G = 12551/ 1.6(TRF RATIO) = 7844A •CURRENT THROUGH RELAY AT E = 7844/1000 = 7.84 A
TRF HT (E) CALCULATIONS
•CURRENT THROUGH RELAY AT E IS 7.84/0.635 =12.35 TIMES THE PS •DESIRED OP TIME FOR FAULT AT B4 = 0.31 + 0.25 = 0.56 SEC •TMS OF RELAY AT E =0.206
TRF HT (E) CALCULATIONS
•HIGH SET ELEMET
•CURRENT THROUGH E FOR MAX FAULT AT G = 12551/ 1.6(TRF RATIO) = 7844A
•CURRENT THROUGH RELAY AT E = 7844/1000= 7.84A •HIGH SET ELEMENT SETTING = 1.3 * 7.84 =10.2 •CURRENT THROUGH E FOR MIN FAULT AT E 10500/1000 = 10.5A
•HENCE HIGH SET WILL OPERATE FOR MIN FAULT AT E
FINAL RELAY SETTINGS
SR
NO
LOCATION RELAY
CURVE
PICK
UP
TMS HS
1 A NI 0.82A 0.22 -
2 B NI 0.75A 0.158 -
3 C NI 0.75A 0.152 -
4 D NI +HS 0.84A 0.177 20A
5 E NI +HS 0.635A 0.206 10A
6 F NI 0.788A 0.066 -
7 G NI
0.79A 0.11 -
TYPICAL FUSE CHARACTERISTIC
OPERATING
CURRENT(%)
THANK YOU
top related