Relay Co Ordination

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


• EXTREMELY INVERSE : n = 2 , K = 80


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


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


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


•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


•DESIRED RELAY D OPERATING TIME FOR FAULT AT F = 0.18 + 0.25 = 0.43SEC •DESIRED TMS FOR RELAY AT D =0.177


•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


•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


•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

Relay Co Ordination

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Relay Co Ordination