pp1

POWER POINT PRESENTATION ON

PROTECTION

BY K.P.KRISHNARAJENDRA

SUPERINTENDING ENGINEER,EL.,R.T.CIRCLE, KPTCL,BANGALORE

Mobile No. 9448350000

PROTECTION:

Accountability and Reliability:

• Accountability and Reliability of Power Generated, Transmitted and Distributed plays an important role in the field of power system while supplying power to all corner consumers.

Accountability:

• Measuring of Power at all levels to know the quantum of Generation, Energy transmitted and consumed.

• Magnitude of energy at Generation and Transmission level is high and cannot be measured directly.

• Hence instrument Transformers like CT’s and PT’s are used.

• The secondary of these will be 1/1Amps and 110V.

Current and Voltage Transformers:

• Different cores

• Different ratios.

Reliability:

• Some times the EHT/HT Lines / Transformers are disturbed due to faults.

• This will cause disturbance in power system and there is a chance of collapse of the system if the fault is not cleared.

• Hence proper protection system is a must.

Protection system:

• Protective relaying system detects such abnormal conditions in the system and gives trip command to the respective breakers to isolate faulty lines.

• The main inputs to the relays are Current, Current and Voltage, Breaker and Isolator status, input of all transformer mounted protection devices.

• These inputs are to be given as per the requirements.

Discrimination of protection:

• The protection scheme to be adopted is to be such that, it has to isolate only the faulty section.

Fault at F is far from the source and there are number of supply points in between, due to this fault relays provided at all the points will sense the fault causing tripping of healthy lines.

• Such tripping can be avoided by discriminating fault locations.

• By Time

• By Current

• By Time and Direction

• Directional Relay:

• In the relay there will be one Voltage coil and One current coil and a disc in between the two coil cores.

• Due to current in the coils, flux will be induced, and the flux induced in the voltage coil will lag the voltage by 90deg and the flux induced in current coil will be in phase with current

• Due to these two fluxes Torque (T) will be produced and is proportional to these two fluxes and sin of the angle between them

• Both these fluxes are proportional to voltage and current respectively. Hence the Torque (T) produced is proportional to VI and cos of angle between them.

• So the Directional relays are proportional to the power in the circuit.

Type of protections adopted:

• 400kV Lines – Main-1 and Main-II , both are distance relays with different characteristics.

• 220kV/110/66kV Lines – Main1 is distance and Main-II DOCR & DEFR.

• 33kV Lines- DOCR and DEFR

• 11kV Lines – OCR and EFR

• Power Transformers – Differential, REF, OCR, EFR, BZ,OSR,PRV, WTHT, OTHT.

• 400kV stations- Both side Bus Bar and LBB protections

• 220kV stations – Bus bar protection on 220kV side and LBB on 66kV side.

Backup protection:• 50 – Non directional Instantaneous

Over Current Relay• 50N - Non directional Instantaneous

Earth Fault Relay • 51 – Non directional Time element

Over Current Relay• 51N - Non directional Time element

Earth Fault Relay• 67R/Y/B– Directional Time element

Over Current Relay• 67N - Directional Time element EFR

• Local Breaker Backup(50Z)

• Differential (87)

• Restricted Earth fault(64)

• Bus Bar protection:

Other Trip and Auxiliary relays:

• Tripping relay - 86 ,96,186, 196,etc.

• TCSR - 195 , 295, 395,etc.

• DC supervision relays – 80A, 80B

• Over Voltage and Under Voltage relays

Backup OCR and EFR

• OCR setting range

50 to 200% in steps of 25%

• EFR setting range

10 to 40% in steps of 5% or

20 to 80% in steps of 10%

Distance Protection ( 21)

• Different Zones

• Carrier protection

• Broken conductor

• Fuse failure protection

• SOTF

• Line Differential Protection:

Differential protection:REF Protection:Checking Buchholtz relay Pressure relief deviceFire protection schemeBus Bar protection:LBB(Local breaker back up) protection

Parallel operation of Transformers.

Conditions to be satisfied for paralleling of 2 or more transformers:

• Same steps of voltages

• Same impedance

• Same vector group

Parallel operation of 2 transformers

of differenct capacity:

• T-1 150MVA with 10% impedance

• T-2 100MVA with 8% impedance

• Taking 100MVA as base MVA

• % impedance of T1 = 10*100/150 = 6.7%

T1-6.7%

T2- 8%

• Total MVA = 250 MVA

• Load sharing by T-1 = 250*8/14.7 = 136 MVA

• Load sharing by T-2 = 250*6.7/14.7=114 MVA

• The load on T-2 is to be limited to 100MVA only

• To avoid overloading of T-2, the load that can be taken on T-1 has to be limited to 119MVA only.

• Thus with different impedances, the total load will be restricted to 88% only.

• As per the loading factor, the relay settings are to be made for over current relays.

• 2*100MVA paralleling:

• Paralleling of two 100MVA transformers of 12% and 9% impedance.

• Load sharing of T1=200*9/21 =86MVA

• Load sharing of T2=200*12/21=114MVA

• Load on T2 to be limited to 100MVA only.

• The total load that can be catered without overloading transformer T2 will be 75+100 = 175MVA, which will be 88% of total capacity.

• ON LOAD TAP CHANGER

• Tap changer winding will be provided on HV side of the transformer.

• There will be Main winding and the regulatory winding.

• Normally there are 17 taps in the OLTC with tap 5 as normal and 9 as mid tap.

• At minimum tap Main winding and the total regulatory winding will be in circuit with additive polarity.

• As the system voltage reduces, tap position is to be raised.

• As the tap is increased, some turns of regulatory winding will be excluded from the circuit. The polarity of the main winding and the regulatory winding will be of additive nature.

• At tap 9, only main winding will be in circuit.

• When the tap position is 10, some regulatory winding will be in the circuit and the polarity of the regulatory winding will be in opposite direction to that of the main winding.

• At maximum tap(17), polarity of entire regulatory winding will be opposite to the main winding.

• Vp/Vs = Np/Ns• Vs = Vp*Ns/Np. Since Ns is constant,

Vs is proportional to ratio of primary voltage to primary turns.

• Vs ά Vp/Np• Hence when primary voltage Vp

decreases or increases, Np is to be decreased or increased proportionately.

• Reactive Power Management:

• Assume 60MW load flow with 24MVAR

• MVA power 64.62 with current 565A

• If 25% of MVAR is compensated, MVA flow will be 62.6 with current 548A.

• I2R loss savings = 303.6-285.3 = 18.3 KW for a line of 5KM length with R=0.19/KM

• Savings in I2R Loss per year =18.3*24*30*12=158112Kwh

• When 25% of MVAR is compensated, additional load of 2MW can also be catered at 565Amps.

• Total units for 2MW additional load per year = 2000*24*30*12 = 17280000.

• Even at the rate of Rs.3/unit, Total additional revenue = 5.2 crore.

• For 50% reactive compensation of 12MVAR, additional load that can be catered will be 3.5MW and the annual revenue will be 9.1 crores.

• Hence at all the voltage levels action is to be taken to compensate reactive power.

• Consider 20MW load at 0.85pf• MVAR drawn at different pf

PF MVA MVAR MVAR improvement

Savings in MVA

0.85 23.53 12.40 - - -

0.87 22.29 11.33 1.10 8.7% 0.54

0.89 22.47 10.25 2.15 17% 1.06

0.90 22.22 9.65 2.71 22% 1.31

• If Savings in MVA is supplied to load of 0.9PF

MVA MW MVAR Savings in MU

Savings at Rs3/- in crores

0.54 0.486 12.40 4.20 1.26

1.06 0.954 11.33 8.25 2.48

1.31 1.179 10.25 10.20 3.06

• By improving the pf to 0.9 from 0.85, the energy savings for 20MW load will be 10.2MU in a year.

• If such savings is made in 12 stations, the total savings will be 122MU

• The average per day consumption of the state is 120MU.

• So by improving the pf from 0.85 to 0.90 in 12 stations, energy for an additional one day can be catered.

• The savings in terms of rupee at Rs3/Kwh will be 36.72 crores per year.

• Battery set charging:• Filling of electrolyte to all the cells as

per supplier recommendations.• Allow the cells to stand for 8-12Hrs.• Record SG and Voltage of each cell.• Start initial charging at the rate of 6%

AH of current.• Hourly readings of SG and voltage are

to be recorded while charging .• Charge the cells for a minimum time of

50Hrs,till the SG & Voltage are constant for 3 consecutive hrs.

• Allow the cells to stand for 8-12 Hrs.• Start discharging of Battery set at the

rate of 10%AH for 10Hrs. During discharge the voltage of each cell should not drop below 1.85V.

• Record SG and Voltage before starting 2nd cycle of charging.

• Start second cycle of charging similar to 1st cycle for a minimum period of 30Hrs.

• Hourly readings are to be recorded.• Allow the cells to stand for 8-12Hrs.

• Start discharging of Battery set at the rate of 10%AH (30Amps for 300AH)for 10Hrs. During discharge the voltage of each cell should not drop below 1.85V.

• Record SG and Voltage after discharge.• Again charge the Battery set in Boost

mode.• During charging, if the temperature of

the cells exceeds 50 deg, current is to be reduced and such reduction in current is to be compensated by extending the time.

• Checking of Healthiness of Battery Set:• Select some of the cells as pilot cells.• Record SG and voltage of pilot cells and the l

DC load voltage.• Switch off AC supply to Battery charger.• Record the pilot cells readings and DC load

voltage at 30 minutes interval.• If there is no change in the readings for 8 to

10 hours, the Battery set is in good condition.

• This test is to be carried out once in 3 or 6 months.

• Recording of instantaneous parameters of all IF points and cross checking.

I1 I2 I3 V1 V2 V3 Load PF

Cross check Load = 3VI*PF

• Monthly Energy consumption at all voltage levels is to be done and the error , if any, is to be recorded only in percentage.

• When any breaker is taken for maintenance, trip the breaker through relays and also check for remote/local operations before charging.

• All maintenance works carried out are to be entered in a register.

• Reading of Drawings

• Wiring Schedule • A- Current circuit for primary protection

• B- Current circuit for Bus bar protection

• C- Current circuit for Backup protection

• D- Current circuit for metering circuit.

• E- Voltage circuit.

• J- Main DC

• K- Control DC

• L- DC supply for indication and annunciation circuit.

• P- DC supply for Bus bar and LBB protection.

• U- Spare contact wiring.

• H- Main Ac supply and control AC supply for lighting and heating.

• M- AC control supply for motor circuit.

• Recording of Interruption and its advantages.

• Checking of Battery set & Battery Charger.

• Checking of DC Ground

CASE STUDIES:

Failure of 11kV PT at Adugodi. Damage to control cables during

11kV feeder faults. Tripping of 20MVA Transformer on

Differential at HAL factory. Non-tripping of Breaker for line fault Tripping of Transformers on BZ

during winter.

Tripping of Healthy 11kV lines on HS when test charged.

Differential trip relay operated indication and annunciation frequently at HSR.

Tripping of Transformers on Differential/REF for external faults.

Tripping of 220kV breaker at HSR without any relay indication.

Relay Co-ordination:Following points are to be considered.

CT ratio used Fault MVA at each voltage level Current setting adopted Time delay to be adopted. Curve selected Relay operating time

Calculation of fault MVA of a station: Assume fault MVA of 66kV bus at

sending end station as 1000MVA Line length as 8KMs with impedance of

line as 0.2ohms/KM 20MVA Transformer with 10%

impedance.

• Source impedance Zs = 662/1000=4.36

• Line Impedance ZL = 0.2*8 = 1.6

• Total impedance Z = 4.36+1.6= 5.96

• Fault MVA f 66kVBus = 662/5.96

= 731MVA

• 66kV Fault current = 731*1000/(1.72*66)

= 6400Amps.

• Source impedance Zs = 662/731= 5.96

• Tfr.Impedance ZT = 0.1*66/20=21.78

• Total impedance ZHV = 5.96+21.78= 27.74

• Impedance referred to 11kV side

ZLV = ZHV *KVLV2/ KVHV2

= 27.74*112/662 = 0.771

• 11kV fault MVA = 112/0.771 = 157MVA

• 11kV Fault current = 157*1000/(1.72*11)

= 8240Amps.

Relay co-ordination:

• 11kV fault MVA = 157

• 11kV fault current = 8.24 KA

• If referred to 66kV fault = 1.37 KASEL 400/1 100% 0.2 3.43 5.5 1100ms

I Line 400/1 100% 0.15 3.43 5.5 825ms

Tfr. 200/1 100% 0.125 6.87 3.6 450ms

Bank 1200/1 100% 0.1 6.87 3.6 360ms

Feeder 400/1 100% 0.05 20.6 2.2 110ms

For Incoming line and 20MVA transformer 75% of the setting is sufficient.

SEL 400/1 100% 0.1 3.43 5.5 550ms

I Line 400/1 75% 0.1 4.60 4.3 430ms

Tfr. 200/1 75% 0.1 9.16 3.0 300ms

Bank 1200/1 100% 0.075 6.87 3.6 270ms

Feeder 400/1 100% 0.05 20.60 2.2 110ms

• For 66kV fault If = 6400 Amps

SEL 400/1 100%

0.1 16 2.4 240ms

I Line 400/1 75% 0.1 21.3 2.2 220ms

Tfr 200/1 75% 0.1 42.7 <2.0 <200ms

Power Transformers Testing:

– Ratio test

– IR test

– Short Circuit Test

– Winding Resistance test

– Magnetic Balance test

– Magnetizing current test

– Checking vector group

– No load test

– Load loss test

– Separate source test

– Induced over voltage test

– HV excitation test.

Transformers protections:

• All transformer mounted devices.

• Fire protection scheme.

Station Equipments & P.C.tests

CT’s and PT’s:

• IR Test

• Polarity test

• Ratio Test

Breakers:

• IR Test

• Contact Resistance

• Operating Timings

• Breaker Operations.

Control & Relay Panel

• Checking of all Relays

• Checking of Protection Scheme

• Checking of Indication Scheme

• Checking of Annunciation Scheme

Isolators:

• Contact Resistance

• Control

• Indication

• Wiring schedule of 20MVA, 66/11kV Power transformer

TFR. BANK

DCDB

ACDB

PTMB CB

CRPCTMB

Thank youThank you

• Calculation of Fault MVA of a station:

• Assume Fault MVA on 220kV Bus of 400kV station as 6000MVA

• Source Impedance at 400kV station Zs=kV*kV/MVA

= 220*220/6000

= 8.07 ohms

220kV RS

66kV Bus

400kV RS

220kV Bus 220kV Bus 66kV

Lin

es

Tfr-1

Tfr-2

• Let the Impedance of the line is

2.5 ohms/KMs.• Impedance of the line of 20KMs is 50 ohms• Total Impedance up to 220kV bus at 220kV

station = 8.07 +50.0 = 58.0 ohms• Fault MVA at 220kV bus of 220kV station

= 220*220/58.0 = 835 MVA

Assume 2 nos. of 20MVA Transformers and

% Impedance of each transformer = 10%

Impedance of the Transformer

= (10/100)*220*220/20 = 48.4ohms

• Total impedance up to Transformers• = (58.0 + 48.4/2) = 82.2 ohms• Impedance referred on to 66kV side• = ZHV (kVLV*KVLV/KVHV*KVHV)• = 82.2 (66*66/220*220)• = 7.4 ohms• Fault MVA on 66kV side• = 66*66/7.4=592MVA• Fault current = 592 *1000/1.732x66• = 5178 Amps.• Taking fault MVA as reference, relay settings of

the station are calculated

• For 3 seconds curve, the approximate time multiplier details are shown below.

I x Times Time Multiplier I x Times Time Multiplier

2 Times 10.0 8 Times 3.4



6 Times 3.8 20 Tomes 2.2

7 Times 3.6

• Pickup current (Ip) = Rated Secondary current of CT x current setting

• If CT secondary is 1 Amps • For 100% setting Ip = 1 x 1 = 1Amps• Relay operates at current = or < 1 Amps• Plug setting multiplier (PSM) and is ratio of fault

current If in the relay to Ipickup• PSM = If/Ip• Assume If = 3200 A and CTR 400/1A• If = 3200/400 = 8 Amps• PSM = 8/1Amps• For time set of 0.1, the operating time

3.4x0.1=0.34

• Relay Co-ordination:

• Assume fault level of 66kV = 600MVA• Source Impedance Zs = 66*66/600 = 7.26 ohms• Transformer impedance Zt=(10/100)(66*66/20)

= 21.78 ohms

• Total impedance Z = 7.26 + 21.78 = 29.04 ohms• Impedance referred to 11kV = 29.04(11*11/66*66)

= 0.807 ohms11

kV F

eed

ers

CB Bank20MVA-10%

66kV 11kV

CB

400/1 1200/1200/1

400/1

• Fault MVA on 11kV side = 11*11/0.807=150MVA• Fault current If = 150x1000/1.732x11 = 7880Amps• Fault current referred to 66kV = 7880/6= 1312Amps• Secondary fault current (PSM)• 11kV Feeder = 7880/400 = 19.7Amps t – 0.05S• Bank = 7880/1200 = 6.57Amps t – 0.10S• Transformer = 1312/200 = 6.57Amps t – 0.15S• I/C Line = 1312/400 = 3.3 Amps t – 0.20S• As per the curve operating time required by the

relays – 0.11S – 0.35S – 0.53S – 1.6S• Here we can reduce the time delay of incoming

line either to 0.15 or even 0.1S.

• If CTs of transformers are of 300/1A,for the same settings the secondary fault current will be (1312/300 ) 4.4 Amps and the operating time required for the relay is 0.6S and the same can be accepted.

• If the current setting is kept at 75%, Ip will be 0.75A and the PSM will be 26.3,8.8, 8.8 and 4.4 respectively.

• The operating time will be <0.1S, 0.32S, 0.48 S and 1.0 S

• Other wise the current setting is to be changed to 0.75 for which the relay operating time will be 0.8S.

• With the above details, blind time delay co-ordination can not be adopted

• If CTs of line are changed to 300/1A from 400/1A, with 100% setting, the fault current will be (1312/300) 4.4A and the relay operating time will be about 0.8S and the time delay of 0.2S is OK.

Breakers:

• Interlocks used in breakers

• Antipumping

• Pole discrepancy trip

• CTD

AUTO RECLOSERECLOSE

LOCKOUT CONT01

TBI-35

63AR

TBI-3602

CCBLOCKOUT

INDICATION

TBI-23

TBI-24L10

L9

63AGX

(N) H2TB1-41

240V, AC50HZ

TB1-40(P)

SH

8SHH7H5

H13

50

H15

H4

EARTHING

SW-1 SW-2

63CA

L18L15

TBI-17L19

TBI-17L17

TBI-17L16

ACFYBBD

AC

FA

IL I

ND

ICA

TIO

N

DC

FIA

LU

RE A

NN

UC

IATIO

N TBI-32TBI-29

TBI-16L13L11L5L3L1

AC

SA

IL I

ND

ICA

TIO

N

RY

DC

FA

IL I

ND

ICA

TIO

N

LO

W O

IL L

EV

EL

LO

W C

AS

PR

ES

SU

RE

LO

W A

IR P

RES

SU

RE

ACFDCC630A63AA

TBI-28TBI-26TBI-22TBI-21TBI-19TBI-17L14L12L7L6L4L2

TBI-21TBI-25TBI-20TBI-18

IL

223K

SPACE HEATER/RECEPTACLE/ILLUMINATION LAMP CONTROL CKT

COMPRESSOR MOTORCONTROL CIRCUIT

63AG

M5ACF - YB

M3

ACF - RY

M1

49M-1

MI3A

88ACM

88ACM

88ACM

88ACM

M5A

M3A

M1A 49M

49M

49M

M11

M9

M7

ACM

TT

TB1-41

TB1-40

415VAC50HZ

TB1-33

TB1-35

H2

H5

H3

H18A

M13

TB2

TB2

TB2

TB2 111

131

155

177

199

2111

2313

2515

2717

2919

12 14 16 18 20 22 24 26 28 112 4 6 8 10 12 14 16 18 20

403622

3824

4026

4228

4430

4632

4834

5036

5238

54

35 37 39 41 43 45 47 49 51 5321 23 25 27 29 31 33 35 37 39

SPARE AUX S W 52a 52a-10 NOS 52b-10 NOS

ALARM CONTACTS

J7

J4

63AGXGLRLDCC52T252T1

L52C

52Y

52b

52a

52Y

8DJ1(+)

TB1-1

J2(-)

TB1-2

K3

IC

K7TB1-6

K9

K11

K13

52b

71CJ29

J27DCC

E2

J3

J25R1

J23

J2136D1

PB

PB

DCC

K29

K27

63AGX

K23

K57

K55

52b52a

63AGX

52B52b

J11J9

52A52a

J3 DCC

DCC

63AL

63CL

J31

J5

TNC

K25

-52TRIP

43LRREMOTE

PO

ST

CLO

SE

PR

E C

LO

SE

43LRREMOTE

PO

ST

CLO

SE

43LRLOCAL

PR

E C

LO

SE

TB1-13

PROTE-CTIONTRIP

REMOTETRIP

K51 K55 K61K41

TRIPCKTSUP

K27 TB1-15TB1

-14

K53TB1-12

TB1-11

TB1-10K21 K23

TB1-8

REMOTETRIP

PROTE-CTIONTRIP

TB1-9

11-52CLOSE

TNC

K5

43LRLOCAL

43LRREMOTE

TB1-5 TB1-3K1

TBN1-4

REMOTECLOSE

AUTORECLOSE

K3

K3

43LRLOCAL

TO REMOTE CONTROL PANEL FOR CIRCUIT BREAKER CONTRON

110VDC

TO ENERGISE 86 RELAYK103

TO ANNUN. CKT.

4

20

14

19

13

3

L103 L135

U6

U4

U2

22

15U5

U3

U1

21

16

21

TBB.1

TBB.6

TBB.3

TBB.5

TBB.4

TBB.2

SPARE

K101

2423

65

1817

110V DC SUPPLY FROM SHT. 9

9

TRAFO. WDG. TEMP HIGH TRIP

K301

B33 X 3 WD: H408W1352

10

7

8

30A10

K302

10

30B

30C TRAFO. OIL TEMP HIGH TRIP

7B7.14

TO ZONE E1

TRAFO. BUCH. TRIP

490T

49WT

63T

TRAFO. MK

K355

K353TO ZONE E4

K351

TB7.17

TB7.16

TB7.15

TO ENERGISE 86 RELAYK103

TO ANNUN. CKT.

4

20

14

19

13

3

L103 L135

U12

U10

U8

22

15U11

U9

U7

21

16

21

TBB.7

TBB.12

TBB.9

TBB.11

TBB.10

TBB.8

SPARE

K101

2423

65

1817

110V DC SUPPLY

9

TRAFO. PRV TRIP

K301

B33 X 3 WD: H408W1352

10

7

8

30D10

K302

10

30E

30F SPARE

7B7.18

OIL SURGE (OLTC)

IC

PRV-T

63T-OLTC

K361

K359

K357

TB7.21

TB7.20

TB7.19

FROM ZONE D8 FROM ZONE D5

pp1

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