Over Current Co Ord

Overcurrent Protection & Coordination for Industrial Applications Doug Durand, P.E. (PowerPoint assistance by Alok Gupta)

IEEE Continuing Education Seminar - Houston, TX February 16-17, 2010

Presenter Bio 1986 – Received BSEE degree from University of Texas at Austin and joined the Planning department of West Texas Utility 1987 – Joined Brown & Root (now KBR). Currently Senior Principal Engineer in charge of KBR’s Technical Services group. The group’s primary tasks involve power system studies including shortcircuit, load flow, motor starting, overcurrent coordination, underground cable ampacity, ground mat, arc-flash, harmonic load flow, transient stability, and relay configuration/setting. Professional Engineer (Texas) IEEE Member Assistant Technical Editor of the IEEE Std 399 – Brown Book (revision 1990)

1

Agenda Day 2

Day 1

▫ ▫ ▫ ▫ ▫ ▫ ▫ ▫

Introduction Information Required Using Log-Log Paper & TCCs Types of Fault Current Protective Devices & Characteristic Curves Coordination Time Intervals (CTIs) Effect of Fault Current Variations Multiple Source Buses

▫ ▫ ▫ ▫ ▫

Transformer Overcurrent Protection Motor Overcurrent Protection Conductor Overcurrent Protection Generator Overcurrent Protection Coordinating a System

Introduction

2

Protection Objectives • Personnel Safety

Protection Objectives • Equipment Protection

3

Protection Objectives • Service Continuity & Selective Fault Isolation 13.8 kV • Faults should be quickly detected and cleared with a minimum disruption of service. 13.8 kV/480 V 2.5 MVA 5.75%

480 V

• Protective devices perform this function and must be adequately specified and coordinated. • Errors in either specification or setting can cause nuisance outages.

Types of Protection • • • • • • • •

Distance High Impedance Differential Current Differential Under/Over Frequency Under/Over Voltage Over Temperature Overcurrent Overload

4

Coordinating Overcurrent Devices • Tools of the trade “in the good ole days…”

Coordinating Overcurrent Devices • Tools of the trade “in the good ole days…”

5

Coordinating Overcurrent Devices • Tools of the trade “in the good ole days…”

Coordinating Overcurrent Devices • Tools of the trade “in the good ole days…”

6

Coordinating Overcurrent Devices • Tools of the trade “in the good ole days…”

Coordinating Overcurrent Devices • Tools of the trade today…

7

Using Log-Log Paper & TCCs

Log-Log Plots Time-Current Characteristic Curve (TCC) 1000 effectively steady state

Time In Seconds

100

Why log-log paper? I2t withstand curves plot as straight lines

1 minute

• Log-Log scale compresses values to a more manageable range.

10 typical motor acceleration

• I2t withstand curves plot as straight lines.

1

0.1

typical fault clearing 5 cycles (interrupting)

1 cycle (momentary) 0.01 0.5

FLC = 1 pu

1

Fs = 13.9 pu 10

100`

Fp = 577 pu

1000

10000

Current in Amperes

8

Plotting A Curve 5000 hp Motor TCC 1000

FLC = 598.9 A

13.8 kV

Time In Seconds

100 13.8/4.16 kV 10 MVA 6.5%

10

Accel. Time = 2 s 4.16 kV 1

M 4.16 kV 5000 hp 90% PF, 96% η, 598.9 A 3593.5 LRA, 2 s start

0.1

LRA = 3593.5 A 0.01 0.5

1

10

100`

1000

10000

Current in Amperes

Plotting Fault Current & Scale Adjustment 5000 hp Motor TCC with Fault on Motor Terminal 1000 13.8 kV

FLC = 598.9 A

Time In Seconds

100 13.8/4.16 kV 10 MVA 6.5% 10

Accel. Time = 2 s

4.16 kV 15 kA

1

M 4.16 kV 5000 hp 90% PF, 96% η, 598.9 A 3593.5 LRA, 2 s start

0.1

15 kA

LRA = 3593.5 A 0.01 0.5

1

10

100`

1000

10000

Current in Amperes x 10 A

9

Voltage Scales 5000 hp Motor TCC with Fault on Transformer Primary

45 kA @ 13.8 kV = ? @ 4.16 kV

13.8 kV 45 kA

100

Time In Seconds

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

1000

13.8/4.16 kV 10 MVA 6.5% 10 4.16 kV 15 kA

1

M 4.16 kV 5000 hp 90% PF, 96% η, 598.9 A 3593.5 LRA, 2 s start

0.1

15 kA 0.01 0.5

1

10

100`

1000

149.3 kA

10000

100AA @ 4.16 kV Current in Amperes x 10

Types of Fault Currents

10

Fault Current Options Crest/Peak Current

Interrupting/Breaking

Momentary Initial Symmetrical

ANSI

IEC

• • • • •

• • • •

Momentary Symmetrical Momentary Asymmetrical Momentary Crest Interrupting Symmetrical Adjusted Interrupting Symmetrical

Initial Symmetrical (Ik’’) Peak (Ip) Breaking (Ib) Asymmetrical Breaking (Ib,asym)

Fault Current Options Crest/Peak Current

Momentary

Interrupting/Breaking

Initial Symmetrical

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

11

Protective Devices & Characteristic Curves

Electromechanical Relays (EM) 100

Very Inverse Time Time-Current Curves

1

10

3 2

0.1

Time Dial Settings

TIME IN SECONDS

10

IFC 53 RELAY

1 ½

0.01 1

10

100

MULTIPLES OF PICK-UP SETTING

12

Electromechanical Relays – Pickup Calculation The relay should pick-up for current values above the motor FLC ( ~ 600 A).

4.16 kV

IFC 53

For the IFC53 pictured, the available ampere-tap (AT) settings are 0.5, 0.6, 0.7, 0.8, 1, 1.2, 1.5, 2, 2.5, 3, & 4.

800/5

For this type of relay, the primary pickup current was calculated as: PU = CT Ratio x AT

4.16 kV 5000 hp FLC = 598.9 A SF = 1.0

Set AT = 4

M

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

Electromechanical Relays – Operating Time Calculation 4.16 kV

IFC 53 RELAY Very Inverse Time Time-Current Curves

10

IFC 53

800/5

Setting = 4 AT, ?? TD

10 kA 15 kA M

1.21 1 1.05

10

0.34 0.30

3 2

0.1 0.08 0.07

1 ½

0.01

15.6 1

Time Dial Settings

TIME IN SECONDS

100

23.4

4.16 kV 5000 hp 598.9 A, SF = 1

IFC 53 Relay Operating Times Fault Current

15 kA

10 kA

15000/640 = 23.4

10000/640 = 15.6

Time Dial ½

0.07 s

0.08 s

Time Dial 3

0.30 s

0.34 s

Time Dial 10

1.05 s

1.21 s

Multiple of Pick-up

10 100 MULTIPLES OF PICK-UP SETTING

13

Solid-State Relays (SS)

Microprocessor-Based Relays 52B

2000/5 01-52B

41-SWGR-01B 13.8 kV OCR F15B 400/5 01-F15B

Seconds

52B OC1 ANSI-Normal Inverse Pickup = 2.13 (0.05 – 20 xCT Sec) Time Dial = 0.96 Inst = 20 (0.05 – 20 xCT Sec) Time Delay = 0.01 s

F15B OC1 ANSI-Extremely Inverse Pickup = 8 (0.05 – 20 xCT Sec) Time Dial = 0.43 Inst = 20 (0.05 – 20 xCT Sec) Time Delay = 0.02 s

52B – 3P F15B – 3P 30 kA @ 13.8 kV

Amps X 100 (Plot Ref. kV=13.8)

14

Power CBs PWR MCB 3200 A

LT Pickup

16-SWGR-02A 0.48 kV PWR FCB 1600 A

LT Band Power MCB Seconds

Cutler-Hammer RMS 520 Series Sensor = 3200 LT Pickup = 1 (3200 Amps) LT Band = 4 ST Pickup = 2.5 (8000 Amps) ST Band = 0.3 (I^x)t = OUT

Power FCB Cutler-Hammer RMS 520 Series Sensor = 1200 LT Pickup = 1 (1200 Amps) LT Band = 2 ST Pickup = 4 (4500 Amps) ST Band = 0.1 (I^x)t = OUT

ST Pickup ST Band

Power MCB – 3P 47.4 kA @ 0.48 kV

Power FCB – 3P 90.2 kA @ 0.48 kV

Amps X 100 (Plot Ref. kV=0.48)

Insulated & Molded Case CB Insulated Case MCB 1200 A 16-SWGR-02A 0.48 kV Molded Case CB 250 A

Insulated Case MCB Seconds

Frame = 1250 Plug = 1200 A LT Pickup = Fixed (1200 A) LT Band = Fixed ST Pickup = 4 x (4000 A) ST Band = Fixed (I^2)t = IN Override = 14000 A

Molded Case CB HKD Size = 250 A Terminal Trip = Fixed Magnetic Trip = 10

Fault current < Inst. Override

Insulated Case MCB 11 kA @ 0.48 kV

Amps X 100 (Plot Ref. kV=0.48)

15

Insulated & Molded Case CB Insulated Case MCB 1200 A 16-SWGR-02A 0.48 kV Molded Case CB 250 A Insulated Case MCB Seconds

Frame = 1250 Plug = 1200 A LT Pickup = Fixed (1200 A) LT Band = Fixed ST Pickup = 4 x (4000 A) ST Band = Fixed (I^2)t = IN Override = 14000 A

Molded Case CB HKD Size = 250 A Terminal Trip = Fixed Magnetic Trip = 10

Fault current > Inst. Override

Insulated Case MCB 42 kA @ 0.48 kV

Amps X 100 (Plot Ref. kV=0.48)

Power Fuses MCC 1 4.16 kV Mtr Fuse

Seconds

Mtr Fuse JCL (2/03) Standard 5.08 kV 5R

Minimum Melting

Total Clearing

Mtr Fuse 15 kA @ 4.16 kV

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

16

Coordination Time Intervals (CTIs)

Coordination Time Intervals (CTIs) The CTI is the amount of time allowed between a primary device and its upstream backup.

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

Primary devices sense, operate & clear the fault first.

Main

Feeder

When two such devices are coordinated such that the primary device “should” operate first at all fault levels, they are “selectively” coordinated.

17

Coordination Time Intervals – EM In the good old days, Main

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

Feeder

It depends.

30 kA

Main

Feeder ?s

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

Coordination Time Intervals – EM On what did it depend? Remember the TD setting? It is continuously adjustable and not exact. So how do you really know where TD = 5? FIELD TESTING ! (not just hand set)

18

Coordination Time Intervals – EM Plotting the field test points.

“3x” means 3 times pickup 3 * 8 = 24 A (9.6 kA primary) 5 * 8 = 40 A (16 kA primary) 8 * 8 = 64 A (25.6 kA primary)

Seconds

Feeder

3x (9.6 kA), 3.3 s 5x (16 kA), 1.24 s 8x (25.6 kA), 0.63 s

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

Coordination Time Intervals – EM So now, if test points are not provided what should the CTI be?

Main

0.4 s

Feeder

But, if test points are provided what should the CTI be?

0.3 s

Seconds

30 kA Main w/ testing Main w/o testing Feeder

0.3 s 0.4 s

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

19

Coordination Time Intervals – EM Where does the 0.3 s or 0.4 s come from? 1. 2. 3.

breaker operating time (Feeder breaker) CT, relay errors (both) disk overtravel (Main relay only) Tested

Hand Set

breaker 5 cycle

0.08 s

0.08 s

Disk over travel

0.10 s

0.10 s

CT, relay errors

0.12 s

0.22 s

TOTAL

0.30 s

0.40 s

Main

Feeder

30 kA

Coordination Time Intervals – EM Red Book (per Section 5.7.2.1) Components

Obviously, CTIs can be a subjective issue. Buff Book (taken from Tables 15-1 & 15-2) Field Tested

Components 0.08 s 0.10 s 0.17 s 0.35 s

0.08 s 0.10 s 0.12 s 0.30 s

20

Coordination Time Intervals – EM & 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 = 0.2 s

CTI = 0.3 s

Main (SS)

(because disk OT is still in play) Feeder (SS)

Feeder (EM)

Coordination Time Intervals – EM & SS Main (EM)

Main (SS)

Feeder (SS)

Feeder (SS)

Main (SS)

Seconds

Main (EM)

Feeder (EM)

Feeder EM 0.3 s

0.2 s

disk OT still applicable Main (EM)

Main (SS)

Feeder (SS) 30 kA @ 13.8 kV

Feeder (EM) 30 kA @ 13.8 kV

Amps X 100 (Plot Ref. kV=13.8)

Amps X 100 (Plot Ref. kV=13.8)

21

Coordination Time Intervals – EM/SS with Banded Devices OC Relay combinations with banded devices EM Relay

Power Fuse

Static Trip or Molded Case Breaker

Static Relay

Power Fuse

Static Trip or Molded Case Breaker

disk over travel √ CT, relay errors √ operating time x CTI

0.1 s 0.12 s 0.22 s

disk over travel x CT, relay errors √ operating time x CTI

0.12 s 0.12 s

Coordination Time Intervals – EM/SS with Banded Devices EM-Banded

SS-Banded

EM Relay

SS Relay

PWR MCB

PWR MCB

PWR MCB

EM Relay

SS Relay

Seconds

PWR MCB

0.22 s

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

0.12 s

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

22

Coordination Time Intervals – Banded Devices • Banded characteristics include tolerances & operating times.

Seconds

• There is no intentional/additional time delay needed between two banded devices. • All that is required is clear space (CS).

Amps X 100 (Plot Ref. kV=0.48)

Coordination Time Intervals – Summary

Buff Book (Table 15-3 – Minimum CTIsa)

23

Effect of Fault Current Variations

CTI & Fault Current Magnitude Inverse relay characteristics imply Operating Time

For a fault current of 10 kA the CTI is 0.2 s. For a fault current of 20 kA the CTI is 0.06 s. If CTI is set based on 10 kA, it reduces to less than the desired value at 20 kA.

Feeder

F1 = 10 kA Seconds

Relay Current

Main

Main

F2 = 20 kA

Feeder

0.2 s 0.06 s

F1 = 10 kA

F2 = 20 kA

Amps X 100 (Plot Ref. kV=13.8)

24

Total Bus Fault versus Branch Currents 10 kA 15 kA

1.2 kA

0.8 kA

M

1 kA

2 kA

M

M

•

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 incomer.

•

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

Total Bus Fault versus Branch Currents

Using Total Bus Fault Current of 15 kA

10 kA

Using Actual Maximum Relay Current of 10 kA

15 kA

Feeder

Feeder

M

0.2 s

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

Seconds

Main Main

0.8 s

10 kA

15 kA

Amps X 100 (Plot Ref. kV=13.8)

25

Curve Shaping

Using a definite time characteristic (or delayed instantaneous) can eliminate the affect of varying fault current levels.

Seconds

Most modern relays include multiple OC Elements.

0.2 s 20 kA 15 kA 10 kA Amps X 100 (Plot Ref. kV=13.8)

Multiple Source Buses

26

Multiple Source Buses

•

When a bus includes multiple sources, care must be taken to not coordinate all source relays at the total fault current.

•

Source relays should be plotted only to their respective fault currents or their “normalized” plots.

•

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

Multiple Source Buses

Plot to Full Fault Level

2

Plot to Actual Relay Current

3

18 kA

12 kA

2

2

30 kA

1

1

Seconds

1

1.1 s

0.2 s

12 kA

30 kA

Amps X 100 (Plot Ref. kV=13.8)

12 kA

30 kA

Amps X 100 (Plot Ref. kV=13.8)

27

Adjusted Pickup Method

•

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

•

The shift factor (SF) is calculated using: SF = Bus Fault / Relay Current

Adjusted Pickup Method

Without shift factor both pickups = 3000 A.

2

12 kA

1

30 kA

1

With shift factor relay 2 pickup shifts to 7500 A. SF = (30/12) * 3000 A

1

2

Seconds

2

3

18 kA

12 kA

0.2 s

0.2 s

30 kA

30 kA

Amps X 100 (Plot Ref. kV=13.8)

Amps X 100 (Plot Ref. kV=13.8)

28

Multiple Source Buses

10 kA Bus A

10 kA

5 kA

Bus B

15 kA (Fa) 10 kA (Fb) Fb = 25 kA

Fa = 25 kA

• Different fault locations cause different flows in tie. SF(Fa) = 25 / (10 + 5) = 1.67 SF(Fb) = 25 / 10 =2 • 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.

Partial Differential Relaying

Source 1 Ip1

Is1

51A

51B

Is1+Is2

0

Is2 Bus A

Source 2 Is2

Ip2

Is2 Bus B

Ip2

Ip1+Ip2 Feeder A

Feeder B

• Typically used on double-ended systems with normally closed ties. • Automatically discriminates between Bus A and Bus B faults. • Eliminates need for relay on tie breaker. • Saves coordination step.

29

Partial Differential Relaying Embellish notes Source 1 Ip1

Is1

51A

51B

0

Is1

Is1 Bus A

Source 2 0

Scheme works even with one source out of service.

0 Open

Is1

Bus B

Ip1

However, the relay in the open source must remain in operation since the relay in the remaining source sees no operating current.

Ip1 Feeder A

Feeder B

Partial Differential Relaying Show same 3 source 1 tie example

30

Directional Current Relaying

67

Bus A

67

Bus B

• 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 system.

Transformer Overcurrent Protection

31

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

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 thermal withstand based on transformer Z

25 x FLC @ 2s

32

Transformer Overcurrent Protection Requirement to protect for mechanical damage is based on frequency of through faults & transformer size. Right-hand side (thermal) used for setting primary protection. Left-hand side (mechanical) used for setting secondary protection.

mechanical withstand thermal withstand 25 x FLC @2s

based on transformer Z

Transformer Overcurrent Protection FLC = 2.4 MVA/(√3 x 13.8) = 100.4 A Relay PU must be ≤ 600% FLC = 602.4 A

PWR-MCB

400 / 100.4 = 398% so okay Time Dial depends on level of protection desired.

Seconds

Using a relay setting of 2.0 x CT, the relay PU = 2 x 200 = 400 A

2.4 MVA, 5.75% Z ∆-Y Resistor Ground

13.8 kV

13.8/0.48 kV 2.4 MVA 5.75% PWR-MCB 480 V

Amps X 10 (Plot Ref. kV=13.8)

33

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

0.5

a

0 .5

0

1.0 B

0.866 b

0 .5

0.5

c

C

0.866

• Assume 1:1 turns ratio X1 = X2 = X0 3-Phase fault current = base current. • Phase-phase fault current is typically 0.866 per unit. • This current transforms to 0.5 (0.866/√3) per unit on the delta phase winding and adds up to 1.0 per unit on the line-side.

Add TCC here to show primary relay clipped at 100% secondary fault current coordinated with a secondary relay clipped at 86% of the secondary fault current. Trim down notes on left and summarize that we normally don’t mess with this issue…

• The CTIs need to be set assuming 1.15 (1/0.866) per unit fault current.

Transformer Overcurrent Protection ∆-Y Connections – Phase-To-Ground Faults 1.0

0.577 A

B

a 0

0

0.577

0 0

0.577

0 b

Add TCC with a shifted and non-shifted transformer damage curve….

c

C

• Assume 1:1 turns ratio X1 = X2 = X0 3-Phase fault current = base current. • The 1 per unit phase-ground fault current transforms to 0.577 (1/√3) per unit on the delta phase. • The transformer damage curve is shifted to the left to ensure protection.

34

Will be deleted once the previous page is fixed. For a solidly-grounded secondary

For resistancegrounded secondary PWR-MCB

PWR-MCB

2.4 MVA, 5.75% Z ∆-Y Solid Ground

Seconds

2.4 MVA, 5.75% Z ∆-Y Resistor Ground

13.8 kV

13.8/0.48 kV 2.4 MVA 5.75%

• Withstand curve shifted left due to ∆-Y xfmr. • More difficult to provide protection.

PWR-MCB 480 V

Amps X 10 (Plot Ref. kV=13.8)

Amps X 10 (Plot Ref. kV=13.8)

Transformer Overcurrent Protection Inrush Current PWR-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.

Seconds

2.4 MVA, 5.75% Z ∆-Y Resistor Ground

• The inrush is not over at 0.1 s, the dot just represents a typical rms equivalent of the inrush from energization to this point in time.

13.8 kV

13.8/0.48 kV 2.4 MVA 5.75%

8-12 x FLC (typical)

PWR-MCB 480 V

Amps X 10 (Plot Ref. kV=13.8)

35

Transformer Overcurrent Protection Setting the primary inst. protection PWR-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) to establish the instantaneous pickup, but most modern relays filter out the DC component.

Seconds

2.4 MVA, 5.75% Z ∆-Y Resistor Ground

13.8 kV

13.8/0.48 kV 2.4 MVA 5.75%

8-12 x FLC (typical)

PWR-MCB 480 V

Amps X 10 (Plot Ref. kV=13.8)

Transformer Overcurrent Protection ∆-Y Connection & Ground Faults

0

0

0

• A secondary L-G fault is not sensed by the ground devices on the primary (∆) side. • Low-resistance and solidly grounded systems are coordinated as separately derived systems.

36

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.

Motor Overcurrent Protection

37

Motor Overcurrent Protection • Fuse provides short-circuit protection.

GE Multilin 469 Standard O/L Curve Pickup = 1.01 X FLC Curve Multiplier = 3

• 49 or 51 device provide motor overload protection.

Bussmann JCL Size 9R Hot 1000 hp 4.16 kV 650% LRC Seconds

• Overload pickup depends on motor FLC and service factor. • The time delay for the 49/51 protection is based on motor stall time.

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

Motor Overcurrent Protection GE Multilin 469 Standard O/L Curve Pickup = 1.01 X FLC Curve Multiplier = 3

• In the past, instantaneous OC protection was avoided on contactorfed motors since the contactors could not clear high short-circuits. Hot 1000 hp 4.16 kV 650% LRC Seconds

• With modern relays, a delayed instantaneous can be used if its setting is coordinated with the contactor interrupting rating.

Bussmann JCL Size 9R

Contactor 6 kA Int. M Amps X 10 (Plot Ref. kV=4.16)

38

Motor Overcurrent Protection • The instantaneous setting for a breaker-fed motor must be set to pass the motor asymmetrical inrush. • Can be done with a pickup over the asymmetrical current.

GE Multilin 469 Standard O/L Curve Pickup = 1.01 X FLC Curve Multiplier = 3

1000 hp 4.16 kV 650% LRC

Hot

Seconds

• Can be done using a lower pickup with time delay to allow DC component to decay out.

3 kA @ 4.16 kV

M

Amps X 10 (Plot Ref. kV=4.16)

Conductor Overcurrent Protection

39

Conductor Overcurrent Protection LV Cables NEC 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 Ratings (A) Fuses & Fixed-Trip Circuit Breakers – cites all standard ratings (B) Adjustable Trip Circuit Breakers – Rating shall be equal to maximum setting (C) Restricted Access Adjustable-Trip Circuit Breakers – Rating can be equal to setting if access is restricted

Conductor Overcurrent Protection MV Cables NEC 240.101 (A) Rating or Setting of Overcurrent Protective Devices Fuse rating ≤ 3 times conductor ampacity Relay setting ≤ 6 times conductor ampacity

40

Conductor Overcurrent Protection • The insulation temperature rating is typically used as the operating temperature (To).

Seconds

• The final temperature (Tf) depends on the insulation type (typically 150 deg. C or 250 deg. C).

1 – 3/C 350 kcmil Copper Rubber Tc = 90 deg. C

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

Amps X 100 (Plot Ref. V=600)

Generator Overcurrent Protection

41

Generator Overcurrent Protection •

A generators fault current contribution decays over time.

•

Overcurrent protection must allow both for moderate overloads & be sensitive enough to detect the steady state contribution to a system fault.

Generator Overcurrent Protection FLC/Xd

•

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

•

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

•

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

42

Generator 51V Pickup Setting Example

Fg

19500 kVA 903 A Xd = 280%

1200/5

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

Generator 51V Pickup Setting Example 51V Setting > 1354/1200 Using 1.15, 51V pickup With old EM relays, 51V pickup (no restraint)

= 1.13 = 1.15 x 1200 A = 1380 A

= 25% of 1380 A = 345 A (> 322.5 A, not good)

With new relays a lower MF can be set, such that 51V pickup (no restraint) = 15% of 1380 A = 207 A (< 322.5, so okay)

43

Generator 51V Settings on TCC

Seconds

15% x Pickup

Pickup = 1.15 x CT-Sec

GTG-101A No Load Constant Excitation Total Fault Current

• Limited guidance on overcurrent protection (C37.102 Section 4.1.1) with respect to time delay. • Want to avoid nuisance tripping, especially on islanded systems, so higher TDs are better.

30 kA

Amps X 10 (Plot Ref. kV=12.47)

Coordinating a System

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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.

Coordinating a System Showing a vertical slice of the system can reduce crowding, but still be hard to read. Upstream equipment is shown on multiple and redundant TCCs.

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TCC Zone Map

TCC-6 TCC-3 TCC-2

TCC-5 TCC-1

TCC-Comp

TCC-4

TCC-307J TCC-101J

TCC-212J

Coordinating a System: TCC-1 Zone Map

•Motor starting & protection is adequate. •Cable withstand protection is adequate. •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 required

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Coordinating a System: TCC-2 Zone 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 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.

Coordinating a System: TCC-3 Zone Map

• •

•

•

The LV switchgear main breaker settings established 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 current.

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Coordinating a System: TCC-307J Zone Map

•

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 The motor relay includes a second definite time unit to provide enhanced protection. The definite time function is delay to allow the asymmetrical inrush current to pass.

• • • •

Coordinating a System: TCC-4 Zone 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 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 is discretionary. All devices are selectively coordinated at all fault current levels. The definite time functions insulation the CTIs from minor fault current variations..

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Coordinating a System: TCC-5 Zone Map

•

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 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 current.

•

•

•

Coordinating a System: TCC-Comp Zone Map

•

• •

•

Due to the compressor size, this curve may set the basis for the MV switchgear main breaker. Motor starting and overload protection is acceptable. Short-circuit protection is provided by the relay/breaker instead of a fuse as with the 1000 hp motor. The short-circuit protection is delay 50 ms to avoid nuisance tripping.

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Coordinating a System: TCC-6 Zone Map

•

•

•

• •

The feeder breaker settings established on TCC-3, TCC-4, and TCC-Comp are shown as the basis for this curve. 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 is discretionary. All devices are selectively coordinated at all fault current levels. The definite time functions insulation the CTIs from minor fault current variations.

TCC Zone Map

TCC-G1

TCC-G2

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Coordination Pop Quiz #1 2000/5

OCR

Does this TCC look good??

Main SWGR-1 400/5

OCR 600/5 TR-FDR1

OCR

•

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

•

The CTI between the main and feeder 2 is appropriate unless all relays are electromechanical and hand set.

TR-FDR2

0.301 s 0.3 s

Coordination Pop Quiz #2 2000/5

Does this TCC look good??

OCR

Main-1 SWGR-3 400/5

OCR

600/5 FDR-1

•

The CTIs shown between main and both feeders are sufficient.

•

Assuming testing EM relays, the 0.615s CTI cannot be reduced since the 0.304s CTI is at the limit.

•

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

OCR

FDR-2

0.615 s 0.304 s

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Coordination Pop Quiz #3 Does this TCC look good?? •

The marked CTIs are okay, but….

•

A main should never include an instantaneous setting.

0.472 s 0.325 s

Coordination Pop Quiz #4 Does this TCC look good?? •

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

•

Transformer frequent fault protection is not provided by the primary relay, but this is okay – adequate protection is provided by the secondary main.

•

Cable withstand protection is inadequate. Adding a 50 to the primary relay would be appropriate and fix this.

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Coordination Pop Quiz #5 Does this TCC look good?? •

Crossing of feeder characteristics is no problem.

•

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

0.021 s

Coordination Pop Quiz #6 Does this TCC look good??

0.03 s

•

The source relays should not be plotted to the full bus fault level unless their plots are shifted based on: SF = Total fault current / relay current.

•

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

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Coordination Pop Quiz #7 Does this TCC look good?? •

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

•

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

•

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

0.302 s

Coordination Software • Computer-aided coordination software programs originated in the late 1980s. • The accuracy of the device characteristic curves was often highly questionable. • There are numerous, much more powerful programs available today, many of which are very mature. • Even still, the accuracy of the protective devices functions and characteristics is still extremely critical.

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Coordination Software • For many years clients maintained separate impedance models for power studies and protective device models for coordination studies. • Integrated models are now the norm and are required to support arc-flash studies.

Information Required

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Accurate One-Line Diagram

Equipment Ratings

• drive short-circuit calculations • affect asymmetrical currents

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Equipment Ratings

• drives short-circuit calculations • affects transformer through-fault protection • inrush affects settings

Equipment Ratings

• • • •

size and construction ampacity insulation type/rating affects I2t damage curve

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Equipment Ratings

• bus continuous rating affects relay pickup (different for LV and MV) • short-circuit rating must be adequate but does not affect overcurrent coordination • differing fault current affect coordination

Equipment Ratings

• adds to short-circuit • LRC, acceleration time, & hot/ cold stall times affect time curve selection

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Protective Devices • Make & model • CT Ratio • Tap & time dial settings (old days) • setting files (today) • Trip device type • LT, ST, Inst settings

References

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Selected References • • • • • • • • •

Applied Protective Relaying – Westinghouse Protective Relaying – Blackburn 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 – Guide for AC Generator Protection NFPA 70 – National Electrical Code

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