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Overcurrent Protection & Coordination for Industrial ApplicationsApplications

Doug Durand, P.E.Dominik Pieniazek, P.E.

2010 Industry Applications Society Annual Meeting - Houston, TXOctober 3-7, 2010

Slide 1

2

Agenda

▫ Introductionf O▫ Using Log-Log Paper & TCCs

▫ Types of Fault Current▫ Protective Devices & Characteristic

▫ Transformer Overcurrent Protection▫ Motor Overcurrent Protection▫ Conductor Overcurrent Protection▫ Generator Overcurrent ProtectionCurves

▫ Coordination Time Intervals (CTIs)▫ Effect of Fault Current Variations

▫ Generator Overcurrent Protection▫ Coordinating a System▫ Supplemental Material▫ Coordination Quizzes

▫ Multiple Source Buses▫ Partial Differential Relaying▫ Directional Overcurrent Coordination

▫ Hands-On Demonstration▫ References

Slide 2Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE

I t d tiIntroduction

Slide 3

4

Protection Objectives

• Personnel Safety


5


• Equipment Protection


6


• Service Continuity & Selective Fault Isolation13.8 kV

13.8 kV/480 V2.5 MVA

• Faults should be quickly detected and cleared with a minimum disruption of service.

480 V

5.75%

• Protective devices perform this function and must be adequately specified and coordinated.

• Errors in either specification or setting p gcan cause nuisance outages.


7

Types of Protection

Protective devices can provide the following assortment of protection, many of which can be coordinated. We’ll focus primarily on the last one, overcurrent

•DistanceHi h I d Diff ti l

overcurrent.

•High-Impedance Differential•Current Differential•Under/Overfrequency•Under/Overvoltage•Over Temperature•OverloadOverload•Overcurrent


8

Coordinating Overcurrent Devices

• Tools of the trade “in the good old days…”


9




10




11




12




13


• Tools of the trade today…


U i L L P & TCCUsing Log-Log Paper & TCCs

Slide 14

15

Ti C t Ch t i ti C (TCC)Log-Log Plots

1000

Time-Current Characteristic Curve (TCC)

Why log-log paper?effectively steady state

100

s

• Log-Log scale compresses values to a more manageable

1 minute I2t withstand curves plot as straight lines

1

10

In S

econ

ds values to a more manageable range.

• I2t withstand curves plot as

typical motor acceleration

0.1

1

Tim

e • I t withstand curves plot as straight lines.typical fault

clearing

5 cycles ( )

0.010.5 1 10 100` 1000 10000

(interrupting)

1 cycle (momentary)

FLC = 1 pu Fs = 13.9 pu Fp = 577 pu


Current in Amperes

16

000 CC

Plotting A Curve

1000

5000 hp Motor TCC

FLC = 598.9 A13.8 kV

100

s13.8/4.16 kV10 MVA

10

In S

econ

ds

Accel. Time = 2 s

4.16 kV

10 MVA6.5%

0.1

1

Tim

e I

M

0.01

0.1

0.5 1 10 100` 1000 10000

LRC = 3593.5 A

4 kV 5000 hp90% PF, 96% η, 598.9 A

3593.5 LRC, 2 s start


Current in Amperes

17

Plotting Fault Current & Scale Adjustment

10005000 hp Motor TCC with Fault on Motor Terminal

Adjustment

FLC = 598.9 A 13.8 kV

10013.8/4.16 kV10 MVA

10

Sec

onds

Accel. Time = 2 s 4.16 kV

6.5%

1

Tim

e In

15 kA

M

0 01

0.1

LRC = 3593.5 A

4 kV 5000 hp90% PF, 96% η, 598.9 A

3593.5 LRC, 2 s start15 kA


0.010.5 1 10 100` 1000 10000

Current in Amperes x 10 A

18

000 CC f

Voltage Scales

1000

5000 hp Motor TCC with Fault on Transformer Primary

45 kA @ 13.8 kV = ? @ 4.16 kV

13.8 kV

100

s

45 kA= (45 x 13.8/4.16)= 149.3 kA @ 4.16 kV 13.8/4.16 kV

10 MVA6.5%

10

In S

econ

ds

15 kA

4.16 kV

0.1

1

Tim

e I

4 kV 5000 hp90% PF 96% 598 9 A

15 kA

M

0.01

0.1

0.5 1 10 100` 1000 10000

90% PF, 96% η, 598.9 A 3593.5 LRC, 2 s start

149.3 kA15 kA


Current in Amperes x 10 Ax 100 A @ 4.16 kV

T f F lt C tTypes of Fault Currents

Slide 19

20

Fault Current Options

CurrentCrest/Peak

Momentary Interrupting/Breaking

Initial Symmetrical

ANSI• Momentary Symmetrical• Momentary Asymmetrical

IEC• Initial Symmetrical (Ik’’)• Peak (Ip)

• Momentary Crest• Interrupting Symmetrical• Adjusted Interrupting Symmetrical

• Breaking (Ib)• Asymmetrical Breaking (Ib,asym)


21

Fault Current Options

CurrentCrest/Peak

Momentary Interrupting/Breaking

Initial Symmetrical

• Symmetrical currents are most appropriate.• Momentary asymmetrical should be considered when setting

instantaneous functionsinstantaneous functions.• Use of duties not strictly appropriate, but okay.• Use of momentary/initial symmetrical currents lead to conservative CTIs.• Use of interrupting currents will lead to lower but still conservative CTIs


• Use of interrupting currents will lead to lower, but still conservative CTIs.

Protective Devices & CharacteristicProtective Devices & Characteristic Curves

Slide 22

23

Electromechanical Relays (EM)100

IFC 53 RELAY

Very Inverse TimeTime-Current

10

ND

S

Time-Current Curves

1

IME

IN S

EC

ON

Dia

l ng

s

10

0.1

T

Tim

e D

Set

tin

½1

23

1 10 100

0.01


MULTIPLES OF PICK-UP SETTING

24

Electromechanical Relays Pickup Calculation

4.16 kV

Pickup CalculationThe relay should pick-up for current values above the motor FLC ( ~ 600 A).

IFC53800/5 Set AT = 4For the IFC53 pictured, the available

ampere-tap (AT) settings are 0 5 0 6

M

4 kV

ampere-tap (AT) settings are 0.5, 0.6, 0.7, 0.8, 1, 1.2, 1.5, 2, 2.5, 3, & 4.

4 kV5000 hp

FLC = 598.9 ASF = 1.0

For this type of relay, the primary pickup current was calculated as:

PU = CT Ratio x AT

PU = (800/5) x 3 = 480 A (too low)= (800/5) x 4 = 640 A (107%, okay)


25

100

Electromechanical Relays 100

ON

DS

4.16 kV

IFC53800/5 Setting = 4 AT (640 A pickup)

TD = ??

IFC 53 RELAY

Very Inverse TimeTime-Current Curves

10

TIM

E IN

SE

CO

M

53

15 kA

TD = ??

10 kA

1

M

4 kV5000 hp

598.9 A, SF = 1

IFC 53 Relay Operating Times

1.051.21

0.34 ettin

gs

10

0.1 10000/640 = 15.6

15000/640 = 23.4Multiple of Pick-up

10 kA15 kAFault Current

IFC 53 Relay Operating Times0.300.34

Tim

e D

ial S

e

½1

23

0.080.07

0.01 1.21 s1.05 sTime Dial 10

0.34 s0.30 sTime Dial 3

0.08 s0.07 sTime Dial ½

23.415.6


1 10 100MULTIPLES OF PICK-UP SETTING

26

Solid-State Relays (SS)


27

Microprocessor-Based Relays

2000/5

41 SWGR 01B

01-52B

52B

41-SWGR-01B13.8 kV

OCR400/501-F15B

F15B

52B52BOC1ANSI-Normal InversePickup = 2.13 (0.05 – 20 xCT Sec)Time Dial = 0.96Inst = 20 (0.05 – 20 xCT Sec)Time Delay = 0.01 sF15B

OC1ANSI-Extremely Inverse

Seco

nds

ANSI-Extremely InversePickup = 8 (0.05 – 20 xCT Sec)Time Dial = 0.43Inst = 20 (0.05 – 20 xCT Sec)Time Delay = 0.02 s

F15B – 3P30 kA @ 13.8 kV

52B – 3P


Amps X 100 (Plot Ref. kV=13.8)

28

Power CBs

16-SWGR-02A0 48 kV

PWR MCB3200 A

LT Pickup

Power MCB

0.48 kV

PWR FCB1600 A

LT BandPower MCBCutler-Hammer RMS 520 SeriesSensor = 3200LT Pickup = 1 (3200 Amps)LT Band = 4ST Pickup = 2.5 (8000 Amps)ST Band = 0.3 (I^x)t = OUTPower FCB

Cutler-Hammer RMS 520 Series

Seco

nds

ST PickupSensor = 1200LT Pickup = 1 (1200 Amps)LT Band = 2ST Pickup = 4 (4500 Amps)ST Band = 0.1 (I^x)t = OUT

ST PickupST Band

Power MCB – 3P47.4 kA @ 0.48 kV

Power FCB – 3P90.2 kA @ 0.48 kV



29

Insulated & Molded Case CB

16-SWGR-02A

Insulated Case MCB1200 A

0.48 kV

Molded Case CB250 A

Insulated Case MCBFrame = 1250 Plug = 1200 ALT Pickup = Fixed (1200 A)LT Band = FixedST Pickup = 4 x (4000 A)ST Band = Fixed (I^2)t = INOverride = 14000 A

Molded Case CB

Seco

nds

Molded Case CBHKDSize = 250 ATerminal Trip = FixedMagnetic Trip = 10

Fault current < Inst. Override

Insulated Case MCB11 kA @ 0.48 kV

Inst. Override



30

Insulated & Molded Case CB

16-SWGR-02A

Insulated Case MCB1200 A

16 SWGR 02A0.48 kV

Molded Case CB250 A

Insulated Case MCBFrame = 1250 Plug = 1200 ALT Pickup = Fixed (1200 A)LT Band = FixedST Pickup = 4 x (4000 A)ST Band = Fixed (I^2)t = INOverride = 14000 AMolded Case CB

Seco

nds

Molded Case CBHKDSize = 250 ATerminal Trip = FixedMagnetic Trip = 10

Fault current > Inst. Override

Insulated Case MCB42 kA @ 0.48 kV



31

Power Fuses

MCC 14.16 kV

Mtr Fuse

Mtr FuseJCL (2/03)Standard 5.08 kV5R5R

Seco

nds

T t l

Minimum Melting

Total Clearing

Mtr Fuse 15 kA @ 4.16 kV


Amps X 10 (Plot Ref. kV=4.16 kV)

C di ti Ti I t l (CTI )Coordination Time Intervals (CTIs)

Slide 32

33

Coordination Time Intervals (CTIs)

The CTI is the amount of time allowed between a primary device and its upstream backup.

When two such de ices are

Backup devices wait for sufficient time to allow operation of primary devices. Main

devices are coordinated such that the primary device “should”

Primary devices sense, operate & clear the fault first

Feeder

operate first at all fault levels, they are “selectively” coordinated& clear the fault first. coordinated.


34

Coordination Time Intervals – EM

Main

In the good old (EM) days,

Wh i l CTI ld

Feeder

What typical CTI would we want between the feeder and the main breaker relays?

It depends

Main30 kA

Seco

nds

It depends. Feeder

? s

30 kA



35


On what did it depend?

Remember the TD setting?Remember the TD setting?

It is continuously adjustable and not exact.

So how do you really know where TD = 5?

FIELD TESTING !FIELD TESTING !(not just hand set)


36


Plotting the field test points.

Feeder

3x (9.6 kA), 3.3 s

5x (16 kA), 1.24 s

8x (25 6 kA) 0 63 sSe

cond

s

“3x” means 3 times pickup3 * 8 = 24 A (9.6 kA primary)5 * 8 40 A (16 kA i ) 8x (25.6 kA), 0.63 s5 * 8 = 40 A (16 kA primary)8 * 8 = 64 A (25.6 kA primary)

30 kA



37


So now, if test points are not provided what should the CTI b ?

Main

be?0.4 s Feeder

30 kA

Seco

nds

But, if test points are provided

30 kA

Main w/o testing

Main w/ testing

Feeder

0.3 swhat should the CTI be?

0.4 s0.3 s

30 kA



38


Where does the 0.3 s or 0.4 s come from?

1 b k ti ti (F d b k )

Main

1. breaker operating time2. CT, relay errors3 disk overtravel

(Feeder breaker) (both)

(Main relay only)

Tested Hand Set Feeder

Main3. disk overtravel (Main relay only)

breaker 5 cycle 0.08 s 0.08 s

Disk over travel 0.10 s 0.10 s

CT l 0 12 0 22

30 kA

CT, relay errors 0.12 s 0.22 s

TOTAL 0.30 s 0.40 s


39

Coordination Time Intervals – EMRed Book (per Section 5.7.2.1)

Components

Obviously, CTIs can be a subjective issue.

Buff Book (taken from Tables 15-1 & 15-2)

Components Field Tested

0.08 s0.10 s0.17 s0.35 s

0.08 s0.10 s0.12 s0.30 s


40

Coordination Time IntervalsEM & SSEM & SS

So, lets move forward a few years….

For a modern (static) relay what part of the margin can be dropped?

Disk overtravel

So if one of the two relays is static, we can use 0.2 s, right?

It depends

Main (EM) CTI = 0.3 s(b di k OT i

Main (SS)CTI = 0.2 s

Feeder (SS)

(because disk OT is still in play)

Feeder (EM)


41

Coordination Time Intervals

Main (EM) Main (SS)

Feeder (SS) Feeder (EM)

Main (EM) Main (SS)

Seco

nds

Feeder (SS)

0.3 s

disk OT still applicable

Feeder EM

0.2 s

disk OT still applicable

Feeder (SS)30 kA @ 13.8 kV

Main (EM)

Feeder (EM)30 kA @ 13.8 kV

Main (SS)


Amps X 100 (Plot Ref. kV=13.8)Amps X 100 (Plot Ref. kV=13.8)

42

Coordination Time Intervals EM/SS with Banded DevicesEM/SS with Banded Devices

OC Relay combinations with banded devices

disk over travelCT, relay errors

√ 0.1 s√ 0.12 s

EM Relay

Power

operating timeCTI

x -0.22 s

Static Trip or Molded Case Breaker

Power Fuse

disk over travelCT, relay errors

x -√ 0.12 s

Static Relay

operating timeCTI

x -0.12 s

Static Trip or Molded Case Breaker

Power Fuse


43

CTI – EM/SS with Banded Devices

EM-Banded SS-BandedEM Relay SS Relay

EM RelayPWR MCB SS RelayPWR MCB

PWR MCB PWR MCB

EM RelayPWR MCB SS RelayPWR MCB

Seco

nds

0.22 s 0.12 s

25 kA 25 kA


Amps X 100 (Plot Ref. kV=0.48) Amps X 100 (Plot Ref. kV=0.48)

44

CTI – Banded Devices• Banded characteristics include

tolerances & operating times.

• There is no intentional/ additional time delay needed between two banded devices.

• All that is required is clear space (CS)Se

cond

s

space (CS).



45

CTI – Banded Devices• Note that areas of mis-

coordination may exist even if the TCC looks good.

• Manufacturer of banded devices will typically not provide data below 0.01 sec.

Possible point of mis-coordination


46

Coordination Time Intervals SSummary

Buff Book (Table 15-3 – Minimum CTIsa)


Eff t f F lt C t V i tiEffect of Fault Current Variations

Slide 47

48

CTI & Fault Current Magnitude

Inverse relay characteristics imply

Main

Feeder

Relay Current

Operating Time

Main F2 = 20 kA

Seco

nds

F1 = 10 kA

For a fault current of 10 kA the CTI is 0 2 s

Feeder

0.06 s

0.2 sCTI is 0.2 s.

For a fault current of 20 kA the CTI is 0.06 s.

F2 = 20 kAF1 = 10 kAConsider a main-tie-main arrangement with a N.O. tie b k


Amps X 100 (Plot Ref. kV=13.8)breaker

49

Total Bus Fault versus Branch Currents

15 kA

10 kA

1 kA2 kA0.8 kA1.2 kA

M MM

• For a typical distribution bus all feeder relays will see a slightly different• For a typical distribution bus all feeder relays will see a slightly different maximum fault current.

• Years back, the simple approach was to use the total bus fault current as the basis of the CTI including main incomerthe basis of the CTI, including main incomer.

• Using the same current for the main led to a margin of conservatism.


50

Total Bus Fault versus Branch Currents

10 kAUsing Total Bus Using Actual

15 kAg

Fault Current of 15 kA

Maximum Relay Current of 10 kA

M

Main

FeederFeeder

Main

Seco

ndsMain

0.2 s0.8 s

15 kA10 kA15 kA



51

Curve Shaping

• Most modern relays include multiple OC El tOC Elements.

• Using a definite time characteristic

Seco

nds

(or delayed instantaneous) can eliminate the affect of varying fault current levels.

0.2 s

15 kA20 kA

10 kA



52

Curve Shaping – Danger of Independent OC UnitsIndependent OC Units

• Many software programs include the facility to plot integrated overcurrentfacility to plot integrated overcurrentunits, usually a 50/51.

• However, the OC units of many , ymodern relays are independent and remain active at all fault current levels.

Seco

nds

• Under certain setting conditions, such as with an extremely inverse characteristic, the intended definite time delay can be undercut and higher

0.2 s 0.1 stime delay can be undercut and higher fault levels.

15 kA20 kA

10 kA

40 kA


10 kA


M lti l S BMultiple Source Buses

Slide 53

54

Multiple Source Buses

• When a bus includes multiple sources, care must be taken to notcoordinate all source relays at the total fault current.

• Source relays should be plotted only to their respective fault y p y pcurrents or their “normalized” plots.

• Plotting the source curves to the total bus fault current will lead toPlotting the source curves to the total bus fault current will lead to much larger than actual CTIs.


55


Plot to Full Fault Plot to Actual2 3Plot to Full Fault Level

Plot to Actual Relay Current

2 2

12 kA 18 kA

2 3

1 130 kA 1

Seco

nds

0.2 s

1.1 s

30 kA12 kA 30 kA12 kA



56

Curve Shifting

• Many software packages include the facility to adjust/shift the characteristics of the source relays to line up at the bus maximum f lt tfault currents.

• Shifting allows relay operation to be considered on a common current basis (primarily the max).

• The shift factor (SF) is calculated using:( ) g

SF = Bus Fault / Relay CurrentSource Relay SF = 30/12 = 2.5 12 kA 18 kA

2 3

Feeder Relay SF = 30/30 = 1.030 kA 1


57

Curve Shifting2 5 x

Without shift factor With shift factor relay 2 2 3

2.5 x

both pickups = 3000 A. pickup shifts to 7500 A.

1

12 kA 18 kA

211

230 kA 1

Seco

nds

0.2 s 0.2 s

30 kA12 kA 30 kA



58


10 kA 5 kA10 kA 10 kA 5 kA10 kA

15 kA (Fa)

10 kA (Fb)

Bus A Bus B

Fa = 25 kA Fb = 25 kA

10 kA (Fb)

• Different fault locations cause different flows in tie.Different fault locations cause different flows in tie.SF(Fa) = 25 / (10 + 5) = 1.67SF(Fb) = 25 / 10 = 2.5

• Preparing a TCC for each unique location can confirm defining case• Preparing a TCC for each unique location can confirm defining case.

• Cases can be done for varying sources out of service & breaker logic used to enable different setting groups.


P ti l Diff ti l R l iPartial Differential Relaying

Slide 59

60

Partial Differential (Bus O.C.)R l iRelaying

• Commonly used on secondary selective systems with normally l d ti b kSource 1 Source 2 closed tie breakers.

• CT wiring automatically discriminates between faults on Bus A and Bus B.

51ASource 1 Source 251B

Ip2Is2Is1 0Is1+Is2Ip1

between faults on Bus A and Bus B.

• CT wiring ensures that main breaker relay sees the same current as the f lt d f dIp1+Ip2

Is2 Is2

Bus A Bus BIp2

faulted feeder.

• 51A trips Main A & tie; 51B trips Main B & tie.

Ip1+Ip2

Feeder A Feeder BMain B & tie.

• Eliminates need for relay on tie breaker & saves coordination step.


61

Partial Differential Relaying

Source 1 Source 2

• Scheme works with a source or tie breaker open.

51ASource 1 Source 251B

00Is1 Is10Ip1• The relay in the open source must

remain in operation.

Is1 Is1

Bus A Bus BIp1

Open

Ip1

• Relay metering functions can be misleading due to CT summation wiring.

Feeder A Feeder B

Ip1

• Separate metering must be provided on dedicated CTs or before the currents are summed.


62

Partial Differential Relaying

Source 1 Source 2 Source 3

• Scheme will work for any number of sources or bus ties.

51ASource 1 Source 2 51B

Ip2Is2

I 2+

Is1

Is2+

0Is1+Is2+Is3Ip1

Source 3

Ip3Is3

• A dedicated relay is needed for each bus section.

Ip1+Ip2+Ip3

Is2+Is3

Is2+ Is3

Bus A Bus BIp2+Ip3

• Partial differential schemes simplify the coordination of multiple source buses by Ip1+Ip2+Ip3

Feeder A Feeder B

p yensuring the main relay for each bus always see the same current as the faulted feederfeeder.


Di ti l O t R l iDirectional Overcurrent Relaying

Slide 63

64

Directional Current Relaying

67 67

Bus A Bus B

• Directional overcurrent (67) relays should be used on double ended line ups• Directional overcurrent (67) relays should be used on double-ended line-ups with normally closed ties and buses with multiple sources.

• Protection is intended to provide more sensitive and faster detection of faults in the upstream supply systemin the upstream supply system.

• Directional device provides backup protection to the transformer differential protection.


T f O t P t tiTransformer Overcurrent Protection

Slide 65

66

Transformer Overcurrent ProtectionNEC Table 450.3(A) defines overcurrent setting requirements for primary & secondary protection pickup settings.


67

Transformer Overcurrent Protection

• C37.91 defines the ANSI withstand protection limits.

• Withstand curve defines thermal & mechanical limits of a transformer experiencing a through-fault.

mechanical withstand

• Requirement to protect for mechanical damage is based on frequency of through faults & t f i

thermal withstand

transformer size.

• Right-hand side (thermal) used for setting primary protection.

25 x FLC @ 2s

based on transformer Z

• Left-hand side (mechanical) used for setting secondary protection.


68


PrimaryFLC = 2.4 MVA/(√3 x 13.8) = 100.4 ARelay PU must be ≤ 600% FLC = 602 4 A

PWR-MCB

Relay pickup

Relay PU must be ≤ 600% FLC = 602.4 AUsing a relay setting of 2.0 x CT, the relay PU = 2 x 200 = 400 A400 / 100.4 = 398% so okay 2.4 MVA, 5.75% Z

∆-Y Resistor Ground

SecondaryFLC = 2.4 MVA / (√3 x 0.48) = 2887 AMCB Trip must be ≤ 250% FLC = 7217 A

Seco

nds

∆ Y Resistor Ground

R-Primary ti l ti

13.8 kVMCB Trip must be ≤ 250% FLC 7217 ABreaker Trip = 3200 A per bus rating3200 / 2887 = 111% (okay)

Ti d l d d l l f t ti

optional time delay settings

13.8/0.48 kV2.4 MVA5 75%

R-Primary

Time delay depends on level of protection desired.

480 V

5.75%

PWR-MCB3200 A



69

Transformer Overcurrent Protection∆-Y Connections – Phase-To-Phase Faults∆-Y Connections Phase-To-Phase Faults

0 0.866

0.5

1.0

A

B

a

b

• A phase-phase fault on the secondary

0.50.866C

c

b

appears more severe in one phase on the primary.

• Setting the CTI based on a three

Seco

nds

• Setting the CTI based on a three-phase fault is not as conservative as for a phase-phase fault.

0.3 s 0.25 s

• The secondary curve could be shifted or a slightly larger CTI used, but can be ignored if primary/ secondary selectivity is not critical.

30 kA30 x 0.867 = 26 kA


yAmps X 100 (Plot Ref. kV=13.8)

70

Transformer Overcurrent Protection∆ Y C ti Ph T G d F lt∆-Y Connections – Phase-To-Ground Faults

1.00.577

0.5770

Aa

0 0

2.4 MVA, 5.75% Z∆-Y Resistor Ground

0.577

0

C

B

c

b0 0

2.4 MVA, 5.75% Z∆-Y Solid Ground

R-Primary

58%

• A one per unit phase-ground fault on the secondary appears as a 58% (1/√3) phase fault on the primary.

PWR-MCB

Seco

nds

y

13 8 kV( √ ) p p y

• The transformer damage curve is shifted 58% to the left to ensure protection.

13.8 kV

13.8/0.48 kV2.4 MVA

R-Primary

protection.

45 kA @ 0.48 kV480 V

2.4 MVA5.75%

PWR-MCB3200 A



71


Inrush Current

• Use of 8-12 times FLC @ 0 1 s is anPWR-MCB

• Use of 8-12 times FLC @ 0.1 s is an empirical approach based on EM relays.


• The instantaneous peak value of the inrush current can actually be much higher than 12 times FLC. Se

cond

s


13.8 kV

• The inrush is not over at 0.1 s, the dot just represents a typical rms equivalent of the inrush from 8-12 x FLC

13.8/0.48 kV2.4 MVA5 75%

R-Primary

equivalent of the inrush from energization to this point in time. (typical)

480 V

5.75%

PWR-MCB3200 A

45 kA @ 0.48 kV



72


Setting the primary inst. protection

• The primary relay instantaneousPWR-MCB

• The primary relay instantaneous (50) setting should clear both the inrush & the secondary fault current.


• It was common to use the asymmetrical rms value of secondary fault current (1.6 x sym) t t bli h th i t t

Seco

nds


13.8 kVto establish the instantaneous pickup, but most modern relays filter out the DC component.

8-12 x FLC 13.8/0.48 kV2.4 MVA5 75%

R-Primary

(typical)

480 V

5.75%

PWR-MCB3200 A FpFs



73


∆-Y Connection & Ground Faults1.0

0.577 a0.577

0.577

0.577

0

A

B

a

c

b0 0

H XZ0

C

Phase Currents Zero Sequence Network

• A secondary L-G fault is not sensed by the ground (zero sequence) devices on the primary (∆) side.

L i t d lidl d d t th d f Y• Low-resistance and solidly-grounded systems on the secondary of a ∆-Y transformer are therefore coordinated separately from the upstream systems.


74


∆-Y Connection & Ground Faults

• The ground resistor size is selected to limit the fault current while still providing sufficient current for coordination.

• The resistor ratings include a maximum continuous current that must be considered.


M t O t P t tiMotor Overcurrent Protection

Slide 75

76

Motor Overcurrent Protection

• Fuse provides short-circuit protection.

GE Multilin 469Standard O/L CurvePickup = 1.01 X FLCCurve Multiplier = 3

• 49 or 51 device provide motor overload protection.

1000 hp

BussmannJCL Size 9R

Hot

• Overload pickup depends on motor FLC and service factor.

Seco

nds

1000 hp4 kV

650% LRC

• The time delay for the 49/51 protection is based on motor stall time.

M3 kA @ 4.16 kV



77


• In the past, instantaneous OC protection was avoided on contactor-fed motors since the contactors could


not clear high short-circuits.

• With modern relays, a definite time 1000 hp

BussmannJCL Size 9R

Hot

unit can be used if its setting is coordinated with the contactor interrupting rating.

Seco

nds

1000 hp4 kV

650% LRC

Contactor6 kA Int.

M3 kA @ 4.16 kV



78


• The instantaneous or definite time setting for a breaker-fed motor must be set to pass the motor asymmetrical i h


inrush.

• Can be done with a pickup over the asymmetrical current

5000 hp4 kV

Hot

asymmetrical current.

• Can be done using a lower pickup and time delay to allow the DC component

Seco

nds

650% LRC

time delay to allow the DC component to decay out.

M3 kA @ 4.16 kV



C d t O t P t tiConductor Overcurrent Protection

Slide 79

80

Conductor Overcurrent Protection

LV CablesNEC 240.4 Protection of Conductors – conductors shall be protected against overcurrent in accordance with their ampacities

(B) Devices Rated 800 A or Less – the next higher standard device rating shall be permitted

(C) Devices Rated over 800 A – the ampacity of the conductors shall be ≥ the device rating

NEC 240 6 Standard Ampere RatingsNEC 240.6 Standard Ampere Ratings(A) Fuses & Fixed-Trip Circuit Breakers – cites all standard ratings

(B) Adjustable Trip Circuit Breakers – Rating shall be equal to maximum ttisetting

(C) Restricted Access Adjustable-Trip Circuit Breakers – Rating can be equal to setting if access is restricted


81


MV Feeders & Branch CircuitsNEC 240.101 (A) Rating or Setting of Overcurrent Protective Devices

Fuse rating ≤ 3 times conductor ampacityRelay setting ≤ 6 times conductor ampacity

MV M t C d tMV Motor ConductorsNEC 430.224 Size of Conductors

Conductors ampacity shall be greater than the overload setting.p y g g


82


• The insulation temperature rating is typically used as the operating temperature (To).p ( o)

• The final temperature (Tf) depends on the insulation type (typically

1 – 3/C 350 kcmilCopper RubberTo = 90 deg. C

Seco

nds

on the insulation type (typically 150 deg. C or 250 deg. C).

• When calculated by hand you• When calculated by hand, you only need one point and then draw in at a -2 slope.


Amps X 100 (Plot Ref. V=600)

G t O t P t tiGenerator Overcurrent Protection

Slide 83

84

Generator Overcurrent Protection

• A generator’s fault current contribution decays over time.

FLC/Xd FLC

• Overcurrent protection must allow both for moderate overloads & be sensitive enough to detect the steady t t t ib ti t t f lt

GTG-101ANo LoadConstant Excitation state contribution to a system fault.

• Voltage controlled/ restrained relays (51V) are commonly used

Seco

nds

Constant ExcitationAC Fault Current

(51V) are commonly used.

• The pickup at full restraint is typically ≥ 150% of Full Load Current (FLC)

Interrupting contribution (FLC/X’d)

≥ 150% of Full Load Current (FLC).

• The pickup at no restraint must be < FLC/Xd

.

Momentary contribution (FLC/X”d)


FLC/XdAmps X 10 (Plot Ref. kV=12.47)

85

Generator 51V Pickup S tti E lSetting Example

Fg

19500 kVA903 AXd = 280%

1200/5

Fg

51V

12.47 kV

Fg = FLC/Xd = 903 / 2.8 = 322.5 A

51V pickup (full restraint) > 150% FLC = 1354 A

51V pickup (no restraint) < 322.5 A


86

Generator 51V Pickup S tti E lSetting Example

51V Setting > 1354/1200 = 1.13Using 1.15, 51V pickup = 1.15 x 1200 A = 1380 A

With old EM relays, 51V pickup (no restraint) = 25% of 1380 A

= 345 A (> 322 5 A not good)= 345 A (> 322.5 A, not good)

With new relays a lower MF can be set, such that 51V i k ( t i t) 15% f 1380 Apickup (no restraint) = 15% of 1380 A

= 207 A (< 322.5, so okay)


87

Generator 51V Settings on TCC

15% x PickupPickup = 1.15 x CT-Sec • Limited guidance on overcurrent

protection (C37.102 Section 4.1.1) with respect to time delayfull

decreasing voltage

GTG-101ANo LoadConstant Excitation

with respect to time delay.

• Want to avoid nuisance tripping, especially on islanded systems, so

no restraint

restrainto tage

Total Fault Current

Seco

nds

especially on islanded systems, so higher TDs are better.

30 kA



C di ti S tCoordinating a System

Slide 88

89

Coordinating a System

• TCCs show both protection & coordination.

• Most OC settings should be shown/confirmed on TCCs.

• Showing too much on a single TCC can make it impossible to read.


90


• Showing a vertical slice of the system ycan reduce crowding, but still be hard to read.

• Upstream equipment is shown on multipleshown on multiple and redundant TCCs.


91


• A set of overlapping TCCs can be used to limit the amount of information on each curve and demonstrate coordination of the system from the bottom up.

• Protection settings should be based on equipment ratings and available spare capacity – not simply on the present operating load

d i t ll d i tand installed equipment.

• Typical TCCs can be used to establish settings for similar installationsinstallations.

• Device settings defined on a given TCC are used as the starting point in the next upstream TCCpoint in the next upstream TCC.

• The curves can be shown on an overall one-line of the system to illustrate the TCC coverage (Zone Map)


illustrate the TCC coverage (Zone Map).

92

Phase TCC Zone Map

TCC-6

TCC-2

TCC-3

TCC-5TCC-Comp

TCC-1TCC-4

TCC-307J


TCC-212JTCC-101J

93

Coordinating a System: TCC-1Zone Map

Seco

nds

• Motor starting & protection is adequate.• Cable withstand protection is adequate.Cab e s a d p o ec o s adequa e• The MCC main breaker may trip for faults

above 11 kA, but this cannot be helped.• The switchgear feeder breaker is selective

with the MCC main breaker, although not


necessarily requiredAmps X 100 (Plot Ref. kV=0.48)

94


• The switchgear feeder breaker settings established on TCC-1 set the basis for this curve.

• The main breaker is set to be selective with the

Seco

nds

feeder at all fault levels.• A CTI marker is not required since the

characteristic curves include all margins and breaker operating times.

• The main breaker curve is clipped at its through-fault current instead of the total bus fault current to allow tighter coordination of the upstream relay. (See TCC-3)



95


• The LV switchgear main breaker settings

Seco

nds

g gestablished on TCC-2 set the basis for this curve.

• The transformer damage curve is based on frequent faults and is not shifted since the transformer is resistance grounded.

• The primary side OC relay is selective with the secondary main and provides adequate transformer and feeder cable protection.

• The OC relay instantaneous high enough to pass the secondary fault current and transformer inrush


the secondary fault current and transformer inrush current. Amps X 100 (Plot Ref. kV=0.48)

96

Coordinating a System: TCC-307JZone Map

Thi t th b i f th t d i

Seco

nds

• This curve sets the basis for the upstream devices since its motor is the largest on the MCC.

• Motor starting and overload protection is acceptable.

• Motor feeder cable protection is acceptable• Motor feeder cable protection is acceptable• The motor relay includes a definite time unit to

provide enhanced protection.• The definite time function is delay to allow the

asymmetrical inrush current to pass


asymmetrical inrush current to pass.Amps X 10 (Plot Ref. kV=4.16)

97


• The 307J motor relay settings established on TCC-307J set the basis for this curve.

• The tie breaker relay curve is plotted to the total b f lt t t b ti

Seco

nds

bus fault current to be conservative.• The main breaker relay curve is plotted to its

let-through current.• A coordination step is provided between the tie

and main relay although this decision isand main relay although this decision is discretionary.

• All devices are selectively coordinated at all fault current levels.

• The definite time functions insulate the CTIs


The definite time functions insulate the CTIs from minor fault current variations.


98


Seco

nds

• The MV MCC main breaker settings established on TCC-4 set the basis for this curve.

• The transformer damage curve is based on frequent faults and is not shifted since the transformer is resistance groundedresistance grounded.

• The primary side OC relay is selective with the secondary main and provides adequate transformer and feeder cable protection.

• The OC relay instantaneous high enough to pass the f f


secondary fault current and transformer inrush current.Amps X 10 (Plot Ref. kV=13.8)

99

Coordinating a System: TCC-CompZone Map

Seco

nds

• Due to the compressor size, this curve may set the basis for the MV switchgear main breaker.

• Motor starting and overload protection is acceptable.p

• Short-circuit protection is provided by the relay/breaker instead of a fuse as with the 1000 hp motor.

• The short-circuit protection is delayed 50 ms to


avoid nuisance tripping.Amps X 10 (Plot Ref. kV=13.8)

100


• The feeder breaker settings established on TCC-3, TCC-4, and TCC-Comp are shown as the basis for this curve.Th tti f f d 52A1 (t th 2 4 MVA)

Seco

nds

• The settings for feeder 52A1 (to the 2.4 MVA) could be omitted since it does not define any requirements.

• A coordination step is provided between the tie and main relay although this decision isand main relay although this decision is discretionary.

• All devices are selectively coordinated at all fault current levels.

• The definite time functions insulate the CTIs from


minor fault current variations.Amps X 10 (Plot Ref. kV=13.8)

101

Ground TCC Zone Map

TCC-G1

TCC-G2


S l t l M t i lSupplemental Material

Slide 102

103

Current Transformer Basics103

5150

5150

5150

5150

Don’t let polarity marks fool you!


104


ia

51

51N

5150

iaibic ia+ib+ic

51Ia

ia+ib+ic

Residual CT connection Protected Bus

51G

IbIc

Zero sequence CT

51

q

Bus NOT Protected


105


Understand How CTs work!

IEEE Guide for the Application of CurrentTransformers Used for ProtectiveRelaying Purposes - IEEE Std C37.110


106

Basic Guides for Protective R l S tti

106

Relay Settings

Suggested “Rules of Thumb” for MV EquipmentSuggested Rules of Thumb for MV Equipment

• Transformers• Bus• Bus• Feeders• MotorsMotors• Capacitors


107

Basic Guides for P t ti R l S tti

107

Protective Relay Settings

Suggested “Rules of Thumb” for MV Equipment gg•The intent of this section is to provide a range of “typical” settings. It is the engineer’s responsibility to verify the application on an individual basis.

•This section does NOT apply to equipment 600 V and below.

•Care must be taken when coordinating a microprocessor TOC element with an electromechanical relay downstream The electromechanical relay mayan electromechanical relay downstream. The electromechanical relay may respond to a fundamental phasor magnitude, true RMS, or rectified magnitude .


108

Rules of Thumb…(above 600 V)P T fPower TransformersPhase Relays (delta – wye)

Primary – Phase Settings•CT Ratio: 200% FLA•Set pickup to comply with NEC 450-3, but as a rule of thumb setting should be less than 300% of transformer self cooled rating or 150%should be less than 300% of transformer self cooled rating or 150% of transformer maximum rating.•Try to set the time dial such that pickup time for maximum through fault is in the neighborhood of 1.0 seconds or less. If higher, ensure that ANSI d i t t d ddamage points are not exceeded.•Set instantaneous at between 160% and 200% of maximum through fault (assume infinite bus). Ensure that available system short circuit allows this. •Time Dial set at 1.0 to 1.5 seconds at maximum fault. Do not exceedTime Dial set at 1.0 to 1.5 seconds at maximum fault. Do not exceed 2.0 seconds which is the mechanical damage point.


109

Rules of Thumb…(above 600 V)P T fPower TransformersPrimary Ground Relay Settings

Primary – Ground Settings•Set 50G if primary winding is delta connected•Set 50G if primary winding is delta connected.•Provide time delay (approx 20 msec) when setting digital relays with zero sequence CTs. No time delay when using elecro-mechanical relays with zero sequence CTs.•CT considerations:

•Residually connected neutral. CT mismatch and residual magnetizationwill not allow the most sensitive setting. Recommend to delay above inrush.•Zero sequence CT Care must be taken to ensure that cables areZero sequence CT. Care must be taken to ensure that cables areproperly placed and cable shields are properly terminated.

iaibic i ib i Ia

ia+ib+icIa

50N

5150

ic ia+ib+ic 50G

IaIbIc

IaIbIc


Residual CT connection Zero sequence CT

110

Rules of Thumb…( above 600 V)

Power TransformersPrimary Fuse Phase Protection

Primary Fuse Rating of power transformer: 135% FLA < Fuse < 250% FLA Try to stay in the range of 150%135% FLA < Fuse < 250% FLA. Try to stay in the range of 150%.

Primary fuse rating of power transformer should be approximately 200% FLA if transformer has a secondary main.

Generally use E-rated fuses. Note that TOC characteristics of fuses are not allthe same.


111


Power TransformersSecondary Resistance Grounded

Secondary Low-Resistance Grounded•Set pickup for 20% to 50% of maximum ground fault. Note that ground resistors typically have a continuous rating of 25-50% of nominal. This valuecan be specified when purchasing the equipmentcan be specified when purchasing the equipment.

Example: 2000 A main breaker (2000:5 CTs), it may make sense to specify an 400 A ground resistor with a continuous rating of 50% (200 A) such that a 2000:5 residually connected CT input can be used with a minimum pickup (0.1 x CT = 0.5 A secondary, 200 A primary).

•Set the time dial such that at the time to trip is 2 0 seconds at maximumSet the time dial such that at the time to trip is 2.0 seconds at maximum ground fault

•Protect resistor using I²t curve Typical resistor is rated for 10 seconds at


Protect resistor using I t curve. Typical resistor is rated for 10 seconds at nominal current (to be specified at time of order).

112


Power TransformersSecondary Solidly Grounded

Secondary Solidly Grounded (for balanced three phase industrial loads)•If secondary is solidly grounded and neutral relay is available (using CT on X0 bushing) set pickup at approximately 50% of phase element and ensureX0 bushing), set pickup at approximately 50% of phase element and ensuretransformer 2 second damage point is protected. Coordinate TOC with main breaker (or partial differential) ground relay.

•Decrease the primary phase element by 58% (to account for transformer damage curve shift). This is the equivalent current seen on the primary (delta) for a secondary ground fault (refer to the Symmetrical Components

t ti O t 5th 2010 b D K t Ed h ff)presentation on Oct 5th, 2010 by Dr. Kurt Ederhoff).


113


Protection for TransformerSecondary Faults on Solidly Grounded Systems

It is certainly preferable to rely on the50T

87T

It is certainly preferable to rely on the transformer primary phase overcurrent relayfor backup of transformer secondary ground faults. However, downstream coordination does not always afford us that

50T51T

51 coordination does not always afford us that luxury (shifting the transformer damage curve and associated transformer primary relay 58%).

NTPhase-Gnd

For solidly grounded transformer secondary installations, an argument can be made that the 87T is the primary protection and 51NT is the backup protection for a transformer

51Main

51N

the backup protection for a transformer secondary ground fault. This will allow you to set the 50T/51T relay without consideration of the 58% shift.


114

Rules of Thumb…( above 600 V)Power TransformersPrimary Neutral (wye – delta)

P i Sid W G d d T fPrimary Side Wye-Grounded Transformer If primary is solidly grounded and neutral relay is available, set pickup at approximately 50% of phase element. This must coordinate with upstream line protection devices (i.e. 21P, 21G, 67, 67G …). If it’s at the utility level, p ( , , , ) y ,they will review and provide settings.

For generator step-up transformers (GSU), the HV 51NT should typically be the g p p ( ), yp ylast device to trip for upstream ground faults. Ensure that the GSU damage curveand the H0 grounding conductor is protected.


115

Rules of Thumb…( above 600 V)O

F f lt b t th

50T51T

50T51T

Directional OvercurrentConsideration for transformer secondary fault

For a fault between the transformer and main breaker, the partialdifferential bus relays will

51B

differential bus relays willnot detect current (otherthan motor contribution).

Both transformer primarctio

n

ectio

n

Both transformer primaryovercurrent relays will detect see the same current. A directional

N.C. 6767

Trip

Dire

c

Trip

Dire

overcurrent relay is requiredto prevent tripping of both transformers via 50T/51T.

51B

Set 67 pickup at 40% of transformer FLA.Coordinate with time curve with 50T/51T


Coordinate with time curve with 50T/51T

116

Rules of Thumb…( above 600 V)O

50T51T

50T51T

Directional OvercurrentConsideration for transformer secondary fault

For a fault between the transformer and main breaker, the main and tiebreaker relays will all seebreaker relays will all seethe same current (otherthan motor contribution).

Th ti b k ill t i f ll d

51Tie

51M 51M

n on The tie breaker will trip followed by the respective transformerprimary overcurrent. A directional overcurrent relay is required

N.C. 6767

Trip

Dire

ctio

n

Trip

Dire

ctio

y qto prevent loss of one bus.

Set 67 pickup at 40% of transformer FLA.Coordinate with time curve with 51Tie


117

Rules of Thumb…( above 600 V)B d F dBus and Feeders

Bus Relays (Main Breaker or Partial Differential):Pickup set between 100% and 125% FLA (150% FLA maximum)p ( )Set to coordinate with transformer primary protective relaying

Do not enable the instantaneous overcurrent element on main breaker relays!

Feeder Relays:Set pickup to comply with NEC 240-100 (limited to 600% of rated ampacity of conductor). Actually, pickup permitted by NEC is slightly higher.Keep it down in the neighborhood of 200%. The intent is NOT to provide overload protection. The intent is to provide short-circuit protection.

Set time dial as required to coordinate with downstream devices while protectingSet time dial as required to coordinate with downstream devices while protecting conductor against damage.

Enable instantaneous element only if the load has a notable impedance (i.e. f ) f f


transformer, motor, capacitor, etc) or if the load is the end of a radial circuit.

118

Rules of Thumb…( above 600 V)Induction MotorsInduction Motors

Pickup set at 101% - 120% Nameplate Rating depending on Service Factor and normal load.

Motor < 1,500 hp Set at 1.15 x FLAMotor > 1,500 hp Set just above FLA x S.F.

Instantaneous Trip set at 200% LRC. A higher pickup may be used depending on system p g p p y p g yavailable short circuit, however, do not lower below 160% LRC unlessyou know that the relay filters/removes the DC component. Ensure that the instantaneous trip setting will not cause a motor starter to attempt interrupting a fault beyond its rating.p g y g

Ground Overcurrent. For Zero Sequence CT (BYZ) set ground Trip at 10A primary and Alarm at 5A primary. Set for instantaneous if using electromechanical and set at 20 msec delay (minimum) if using digital relays.y ( ) g g yFor solidly grounded systems, ensure that the ground trip setting will not cause a motor starter to attempt interrupting a fault beyond its rating.

Mechanical Jam set 150% FLA at 2 sec, unless application does not allow this (i.e. grinder,


pp ( gcrusher, etc).

119

Rules of Thumb…( above 600 V)Capacitors

Capacitor Bank:For individual protection, the Fuse protecting the capacitor is chosen such that itsFor individual protection, the Fuse protecting the capacitor is chosen such that its continuous current capability is greater than or equal to 135% of rated capacitor current. The feeder cable should be sized as such for continuous operation. This over rating is due to 10% for allowable overvoltage conditions, 15% for

it kVAR ti t l (thi l t t 15% t d i ti fcapacitor kVAR rating tolerance (this correlates to 15% percent deviation from nominal capacitance) and 10% for overcurrent due to harmonics.

For unbalance, set Alarm for loss of one capacitor, set Trip for overvoltage of p p g110% rated (nameplate).

For feeder protection, set Pickup at 135% of FLA, set Time Dial at 1.0, set 50P element above maximum inrush and include a slight time delay toset 50P element above maximum inrush and include a slight time delay to coordinate with individual fuse clear time. Plot TOC to protect the capacitor case rupture curve.


Note: Systems with high harmonic content require special attention.

H d O D t tiHands-On Demonstration

Slide 120

C di ti Q iCoordination Quizzes

Slide 121

122

Coordination Quiz #1

Does this TCC look okay??

• There is no need to maintain a

Main2000/5 OCR

/

SWGR-1

• There is no need to maintain a coordination interval between feeder breakers.

TR-FDR2600/5 OCR

TR-FDR1400/5 OCR

• The CTI between the main and feeder 2 is appropriate unless all relays are electromechanical and h d t

Main-POC1TR-FDR2-P

OC1

TR-FDR1-POC1

Seco

nds

hand set.

• Fix – base the setting of the feeder 2 relay on its downstream

0.3 s

0.3 s

OC1

feeder 2 relay on its downstream equipment and lower the time delay if possible.

Main-3PTR-FDR2-3PTR-FDR1-3P15 kA @ 13 8 kV


15 kA @ 13.8 kV


123



• The CTIs shown between mainMain-1

2000/5 OCR

SWGR-3 • The CTIs shown between main and both feeders are sufficient.

• Assuming testing EM relays, the

FDR-2600/5 OCR

FDR-1400/5 OCR

SWGR 3

Main P Assuming testing EM relays, the 0.62 s CTI cannot be reduced since the 0.30 s CTI is at the limit.

Main-POC1

FDR-1-P

Seco

nds

• The main relay time delay is actually too fast since the CTI at 30 kA is less than 0.2 s.

0.62 s

0.30 sFDR-2-POC1

FDR 1 POC1

• Fix – raise the time delay setting of the main relay.

Main-3PFDR-2-3PFDR-1-3P30 kA @ 13 8 kV


30 kA @ 13.8 kV


124



• The marked CTIs are okay but

Main-32000/5 OCR

SWGR-4

• The marked CTIs are okay, but….

• A main should never include an instantaneous setting.

FDR--2600/5 OCR

FDR--1400/5 OCR

instantaneous setting.

• Fix – delete the instantaneous on the main relay and raise the time

0 47 s

Main-3-POC1

FDR--1-POC1

Seco

nds

ydelay to maintain a 0.2s CTI at 50 kA.

0.47 s

0.33 sFDR--2-POC1

OC1

Main-3-3PFDR--2-3PFDR--1-3P50 kA @ 13 8 kV


50 kA @ 13.8 kV


125



• Primary relay pickup is 525% of• Primary relay pickup is 525% of transformer FLC, thus okay.

• Transformer frequent faultTransformer frequent fault protection is not provided by the primary relay, but this is okay –adequate protection is provided b th d i

Seco

nds

by the secondary main.

• Cable withstand protection is inadequateinadequate.

• Fix – Add instantaneous setting to the primary relay


the primary relay.Amps X 10 (Plot Ref. kV=13.8)

126



• Selectivity between Relay14 on• Selectivity between Relay14 on the transformer primary and CB44 on the secondary is not provided, but this can be acceptable.

Seco

nds

• Relay 14 is not, however, selectively coordinated with f d b k CB46feeder breaker CB46.

• Fix – raise Relay14 time delay setting and add CTI marker

0.08 ssetting and add CTI marker.



127



• Crossing of feeder characteristics• Crossing of feeder characteristics is no problem.

• There is no need to maintain anLVMain There is no need to maintain an intentional time margin between two LV static trip units – clear space is sufficient.

a

LVFDR2

Seco

nds

• Fix – lower the main breaker short-time delay band.

0.21 s

LVFDR2

LVFDR1

LVMain – 3P30 kA @ 0.48 kV

LVFDR2 – 3PLVFDR1 – 3P45 kA @ 0.48 kV



128



• The source relays should not be 10 kA 5 kA

plotted to the full bus fault level unless their plots are shifted based on: SF = Total fault current / relay

Source1 - POC1 SF = Total fault current / relay

current.

• Assuming each relay actually

OC1

Feeder - POC1

Source2 - POC1

Seco

nds

Assuming each relay actually sees only half of the total fault current, the CTI is actually much higher than 0.3 s.

0.3 s

OC1

• Fix – plot the source relays to their actual fault current or apply SF

Source1 - 3PSource2 – 3PFeeder – 3P15 kA @ 13.8 kV


SF.@


129



• There are two curves to be• There are two curves to be concerned with for a 51V – full restraint and zero restraint.

• Assuming the full restraint curve• Assuming the full restraint curve is shown, it is coordinated too tightly with the feeder.

Th 51V ill hift l ft d

51V - POC1

FDR-5 - POC1

Seco

nds

• The 51V curve will shift left and lose selectivity with the feeder if a close-in fault occurs and the voltage drops.

0.30 sOC1

g p

• Fix – show both 51V curves and raise time delay.

51V – 3PFDR-5 – 3P15 kA @ 13.8 kV


@


R fReferences

Slide 130

131

Selected References

• IEEE Std 242 – Buff Book• IEEE Std 141 – Red Book• IEEE Std 399 – Brown Book• IEEE C37.90 – Relays• IEEE C37.91 – Transformer Protection

IEEE C37 102 G id f AC G t P t ti• IEEE C37.102 – Guide for AC Generator Protection• NFPA 70 – National Electrical Code• Applied Protective Relaying – Westinghouse• Protective Relaying – BlackburnProtective Relaying Blackburn• Protective Relaying Theory and Applications – ABB Power T&D Company• Protective Relaying for Power Systems – IEEE Press• Protective Relaying for Power Systems II – IEEE Press• AC Motor Protection – Stanley E. Zocholl• Industrial and Commercial Power System Applications Series – ABB • Analyzing and Applying Current Transformers - Stanley E. Zocholl


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