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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
Slide 4Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
5
Protection Objectives
• Equipment Protection
Slide 5Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
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Protection Objectives
• 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.
Slide 6Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
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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
Slide 7Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
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Coordinating Overcurrent Devices
• Tools of the trade “in the good old days…”
Slide 8Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
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Coordinating Overcurrent Devices
• Tools of the trade “in the good old days…”
Slide 9Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
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Coordinating Overcurrent Devices
• Tools of the trade “in the good old days…”
Slide 10Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
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Coordinating Overcurrent Devices
• Tools of the trade “in the good old days…”
Slide 11Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
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Coordinating Overcurrent Devices
• Tools of the trade “in the good old days…”
Slide 12Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
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Coordinating Overcurrent Devices
• Tools of the trade today…
Slide 13Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
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
Slide 15Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
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
Slide 16Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
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
Slide 17Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
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
Slide 18Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
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)
Slide 20Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
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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
Slide 21Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
• 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
Slide 23Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
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)
Slide 24Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
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
Slide 25Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
1 10 100MULTIPLES OF PICK-UP SETTING
26
Solid-State Relays (SS)
Slide 26Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
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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
Slide 27Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
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
Slide 28Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
Amps X 100 (Plot Ref. kV=0.48)
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
Slide 29Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
Amps X 100 (Plot Ref. kV=0.48)
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
Slide 30Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
Amps X 100 (Plot Ref. kV=0.48)
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
Slide 31Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
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.
Slide 33Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
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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
Slide 34Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
Amps X 100 (Plot Ref. kV=13.8)
35
Coordination Time Intervals – EM
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)
Slide 35Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
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Coordination Time Intervals – EM
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
Slide 36Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
Amps X 100 (Plot Ref. kV=13.8)
37
Coordination Time Intervals – EM
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
Slide 37Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
Amps X 100 (Plot Ref. kV=13.8)
38
Coordination Time Intervals – EM
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
Slide 38Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
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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
Slide 39Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
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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)
Slide 40Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
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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)
Slide 41Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
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
Slide 42Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
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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
Slide 43Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
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).
Slide 44Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
Amps X 100 (Plot Ref. kV=0.48)
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
Slide 45Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
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Coordination Time Intervals SSummary
Buff Book (Table 15-3 – Minimum CTIsa)
Slide 46Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
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
Slide 48Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
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.
Slide 49Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
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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
Slide 50Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
Amps X 100 (Plot Ref. kV=13.8)Amps X 100 (Plot Ref. kV=13.8)
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
Slide 51Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
Amps X 100 (Plot Ref. kV=13.8)
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
Slide 52Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
10 kA
Amps X 100 (Plot Ref. kV=13.8)
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.
Slide 54Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
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Multiple Source Buses
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
Slide 55Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
Amps X 100 (Plot Ref. kV=13.8)Amps X 100 (Plot Ref. kV=13.8)
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
Slide 56Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
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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
Slide 57Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
Amps X 100 (Plot Ref. kV=13.8)Amps X 100 (Plot Ref. kV=13.8)
58
Multiple Source Buses
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.
Slide 58Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
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.
Slide 60Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
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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.
Slide 61Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
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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.
Slide 62Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
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.
Slide 64Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
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.
Slide 66Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
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.
Slide 67Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
68
Transformer Overcurrent Protection
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
Slide 68Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
Amps X 10 (Plot Ref. kV=13.8)
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
Slide 69Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
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
Slide 70Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
Amps X 10 (Plot Ref. kV=13.8)
71
Transformer Overcurrent Protection
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.
2.4 MVA, 5.75% Z∆-Y Resistor Ground
• The instantaneous peak value of the inrush current can actually be much higher than 12 times FLC. Se
cond
s
∆ Y Resistor Ground
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
Slide 71Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
Amps X 10 (Plot Ref. kV=13.8)
72
Transformer Overcurrent Protection
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.
2.4 MVA, 5.75% Z∆-Y Resistor Ground
• 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
∆ Y Resistor Ground
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
Slide 72Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
Amps X 10 (Plot Ref. kV=13.8)
73
Transformer Overcurrent Protection
∆-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.
Slide 73Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
74
Transformer Overcurrent Protection
∆-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.
Slide 74Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
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
Slide 76Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
Amps X 10 (Plot Ref. kV=4.16)
77
Motor Overcurrent Protection
• In the past, instantaneous OC protection was avoided on contactor-fed motors since the contactors could
GE Multilin 469Standard O/L CurvePickup = 1.01 X FLCCurve Multiplier = 3
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
Slide 77Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
Amps X 10 (Plot Ref. kV=4.16)
78
Motor Overcurrent Protection
• The instantaneous or definite time setting for a breaker-fed motor must be set to pass the motor asymmetrical i h
GE Multilin 469Standard O/L CurvePickup = 1.01 X FLCCurve Multiplier = 3
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
Slide 78Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
Amps X 10 (Plot Ref. kV=4.16)
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
Slide 80Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
81
Conductor Overcurrent Protection
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
Slide 81Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
82
Conductor Overcurrent Protection
• 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.
Slide 82Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
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)
Slide 84Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
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
Slide 85Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
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)
Slide 86Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
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
Slide 87Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
Amps X 10 (Plot Ref. kV=12.47)
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.
Slide 89Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
90
Coordinating a System
• 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.
Slide 90Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
91
Coordinating a System
• 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)
Slide 91Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
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
Slide 92Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
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
Slide 93Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
necessarily requiredAmps X 100 (Plot Ref. kV=0.48)
94
Coordinating a System: TCC-2Zone Map
• 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)
Slide 94Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
Amps X 100 (Plot Ref. kV=0.48)
95
Coordinating a System: TCC-3Zone Map
• 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
Slide 95Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
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
Slide 96Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
asymmetrical inrush current to pass.Amps X 10 (Plot Ref. kV=4.16)
97
Coordinating a System: TCC-4Zone Map
• 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
Slide 97Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
The definite time functions insulate the CTIs from minor fault current variations.
Amps X 10 (Plot Ref. kV=4.16)
98
Coordinating a System: TCC-5Zone Map
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
Slide 98Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
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
Slide 99Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
avoid nuisance tripping.Amps X 10 (Plot Ref. kV=13.8)
100
Coordinating a System: TCC-6Zone Map
• 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
Slide 100Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
minor fault current variations.Amps X 10 (Plot Ref. kV=13.8)
101
Ground TCC Zone Map
TCC-G1
TCC-G2
Slide 101Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
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!
Slide 103Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
104
Current Transformer Basics104
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
Slide 104Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
105
Current Transformer Basics105
Understand How CTs work!
IEEE Guide for the Application of CurrentTransformers Used for ProtectiveRelaying Purposes - IEEE Std C37.110
Slide 105Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
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
Slide 106Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
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 .
Slide 107Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
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.
Slide 108Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
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
Slide 109Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
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.
Slide 110Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
111
Rules of Thumb…( above 600 V)
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
Slide 111Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
Protect resistor using I t curve. Typical resistor is rated for 10 seconds at nominal current (to be specified at time of order).
112
Rules of Thumb…( above 600 V)
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).
Slide 112Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
113
Rules of Thumb…( above 600 V)
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.
Slide 113Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
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.
Slide 114Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
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
Slide 115Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
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
Slide 116Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
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
Slide 117Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
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,
Slide 118Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
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.
Slide 119Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
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
Slide 122Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
15 kA @ 13.8 kV
Amps X 100 (Plot Ref. kV=13.8)
123
Coordination Quiz #2
Does this TCC look okay??
• 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
Slide 123Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
30 kA @ 13.8 kV
Amps X 100 (Plot Ref. kV=13.8)
124
Coordination Quiz #3
Does this TCC look okay??
• 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
Slide 124Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
50 kA @ 13.8 kV
Amps X 100 (Plot Ref. kV=13.8)
125
Coordination Quiz #4
Does this TCC look okay??
• 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
Slide 125Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
the primary relay.Amps X 10 (Plot Ref. kV=13.8)
126
Coordination Quiz #5
Does this TCC look okay??
• 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.
Slide 126Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
Amps X 10 (Plot Ref. kV=13.8)
127
Coordination Quiz #6
Does this TCC look okay??
• 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
Slide 127Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
Amps X 100 (Plot Ref. kV=0.48)
128
Coordination Quiz #7
Does this TCC look okay??
• 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
Slide 128Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
SF.@
Amps X 100 (Plot Ref. kV=13.8)
129
Coordination Quiz #8
Does this TCC look okay??
• 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
Slide 129Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE
@
Amps X 100 (Plot Ref. kV=12.5)
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
Slide 131Overcurrent Coordination for Industrial Applications IEEE IAS 2010 - Copyright: IEEE