Seminar Koordinasi Proteksi Arus Lebih

59
1 ion & Agenda 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 Day 2 Transformer Overcurrent Protection Motor Overcurrent Protection Conductor Overcurrent Protection Generator Overcurrent Protection Coordinating a System Introduction

description

overcurrent protection

Transcript of Seminar Koordinasi Proteksi Arus Lebih

Page 1: Seminar Koordinasi Proteksi Arus Lebih

1

ion &

Agenda

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

Day 2 ▫ Transformer Overcurrent Protection

▫ Motor Overcurrent Protection

▫ Conductor Overcurrent Protection

▫ Generator Overcurrent Protection

▫ Coordinating a System

Introduction

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Protection Objectives

• Personnel Safety

Protection Objectives

• Equipment Protection

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Protection Objectives

• Service Continuity & Selective Fault Isolation

13.8 kV

13.8 kV/480 V

2.5 MVA

5.75%

480 V

• Faults should be quickly

detected and cleared with a

minimum disruption of service.

• 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

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Coordinating Overcurrent Devices

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

Coordinating Overcurrent Devices

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

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Coordinating Overcurrent Devices

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

Coordinating Overcurrent Devices

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

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Coordinating Overcurrent Devices

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

Coordinating Overcurrent Devices

• Tools of the trade today…

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Tim

e In S

econds

Using Log-Log Paper & TCCs

Log-Log Plots Time-Current Characteristic Curve (TCC)

1000 effectively

steady state

Why log-log paper?

100

10

1

1 minute

typical motor

acceleration

typical fault

clearing

I2t withstand

curves plot as

straight lines

• Log-Log scale compresses

values to a more manageable

range.

• I2t withstand curves plot as

straight lines.

0.1

1 cycle

(momentary) 0.01

5 cycles

(interrupting)

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

0.5 1 10 100` 1000 10000

Current in Amperes

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4.1 5 kA

Tim

e In S

econds

Tim

e In S

econds

Plotting A Curve 5000 hp Motor TCC

1000

FLC = 598.9 A

13.8 kV

100

10

Accel. Time = 2 s

13.8/4.16 kV

10 MVA

6.5%

4.16 kV

1

0.1

0.01

LRA = 3593.5 A

M 4.16 kV 5000 hp

90% PF, 96% η, 598.9 A

3593.5 LRA, 2 s start

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

FLC = 598.9 A

13.8 kV

100

10

Accel. Time = 2 s

13.8/4.16 kV

10 MVA

6.5%

6 kV

1 1

0.1

M 4.16 kV 5000 hp

90% PF, 96% η, 598.9 A

3593.5 LRA, 2 s start

0.01

LRA = 3593.5 A 15 kA

0.5 1 10 100` 1000 10000

Current in Amperes x 10 A

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4.1 5 kA

Tim

e In S

econds

Voltage Scales

5000 hp Motor TCC with Fault on Transformer Primary

45 kA @ 13.8 kV

= ? @ 4.16 kV

= (45 x 13.8/4.16)

= 149.3 kA @ 4.16 kV

1000

100

10

45 kA

13.8 kV

13.8/4.16 kV

10 MVA

6.5%

6 kV

1 1

0.1

M 4.16 kV 5000 hp

90% PF, 96% η, 598.9 A

3593.5 LRA, 2 s start

0.01

15 kA 149.3 kA

0.5 1 10 100` 1000 10000

Current in Amperes x 100AA @ 4.16 kV

Types of Fault Currents

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Fault Current Options

Current

Crest/Peak

Momentary

Initial Symmetrical

Interrupting/Breaking

ANSI

• Momentary Symmetrical

• Momentary Asymmetrical

• Momentary Crest

• Interrupting Symmetrical

• Adjusted Interrupting Symmetrical

IEC

• Initial Symmetrical (Ik’’)

• Peak (Ip)

• Breaking (Ib)

• Asymmetrical Breaking (Ib,asym)

Fault Current Options

Current

Crest/Peak

Momentary

Initial Symmetrical

Interrupting/Breaking

• 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.

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TIM

E I

N S

EC

ON

DS

Tim

e D

ial

Se

ttin

gs

Protective Devices & Characteristic

Curves

Electromechanical Relays (EM)

100

10

IFC 53

RELAY Very

Inverse Time

Time-Current

Curves

1

10

3 2

0.1 1

½

0.01

1 10 100

MULTIPLES OF PICK-UP SETTING

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IFC 53 Relay Operating Times Fault Current 15 kA 10 kA

Multiple of Pick-up 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

TIM

E I

N S

EC

ON

DS

Tim

e D

ial S

ett

ing

s

Electromechanical Relays – Pickup Calculation

The relay should pick-up for current

values above the motor FLC ( ~ 600 A).

4.16 kV

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.

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)

800/5

M

4.16 kV

5000 hp FLC =

598.9 A SF =

1.0

IFC 53

Set AT = 4

Electromechanical Relays – Operating Time Calculation 100

IFC 53

RELAY Very

Inverse Time Time-

Current Curves

10

800/5

10 kA

15 kA

4.16 kV

IFC

Setting = 4 AT, ?? TD 53

1.21 1

1.05

M

4.16 kV

10 5000 hp 598.9 A, SF = 1

0.34 0.30

3

2

0.1 1

0.087 ½

0.01 15.6 23.4

1 10 100

MULTIPLES OF PICK-UP SETTING

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Seconds

Solid-State Relays (SS)

Microprocessor-Based Relays

2000/5

01-52B

52B

41-SWGR-01B 13.8 kV

400/5 OCR F15B

01-F15B

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

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

52B – 3P F15B – 3P

3.8 kV 30 kA @ 1

Amps X 100 (Plot Ref. kV=13.8)

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PWR MCB

3200 A

16-SWGR-02A

0.48 kV

PWR FCB

1600 A

Seconds

Seconds

Power CBs

LT Pickup

LT Band

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

Power MCB 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

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

Molded Case CB HKD

Size = 250 A

Terminal Trip = Fixed

Magnetic Trip = 10

Insulated Case MCB 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

Fault current

< Inst. Override

Insulated Case MCB

11 kA @ 0.48 kV

Amps X 100 (Plot Ref. kV=0.48)

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Seconds

Seconds

Insulated & Molded Case CB

Insulated Case MCB

1200 A

16-SWGR-02A

0.48 kV

Molded Case CB

250 A

Molded Case CB HKD

Size = 250 A

Terminal Trip = Fixed

Magnetic Trip = 10

Insulated Case MCB 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

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 JCL

(2/03) Standard

5.08 kV

5R

Mtr Fuse

Minimum

Melting

Total

Clearing

Mtr Fuse

15 kA @

4.16 kV

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

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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.

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Seconds

Coordination Time Intervals – EM

In the good old days,

What typical CTI would we

want between the feeder and

the main breaker relays?

Main

Main

Feeder

30 kA

It depends.

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)

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Seconds

Seconds

Coordination Time Intervals – EM

Plotting the field test points.

Feeder

“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)

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?

Main w/ testing

Feeder

30 kA

Main w/o testing

0.3 s

0.3 s 0.4 s

30 kA

Amps X 100 (Plot Ref. kV=13.8)

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Components

Coordination Time Intervals – EM

Where does the 0.3 s or 0.4 s come from?

1. breaker operating time

(Feeder breaker)

2. CT, relay errors

3. disk overtravel

(both)

(Main relay only)

Main

breaker 5 cycle

Disk over travel

CT, relay errors

TOTAL

Tested

0.08 s

0.10 s

0.12 s

0.30 s

Hand Set

0.08 s

0.10 s

0.22 s

0.40 s

Feeder

30 kA

Coordination Time Intervals – EM

Red Book (per Section 5.7.2.1)

Obviously, CTIs can be a

subjective issue.

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

Components Field Tested

0.08 s

0.10 s

0.17 s

0.08 s

0.10 s

0.12 s

0.35 s 0.30 s

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Seconds

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?

So if one of the two relays is static, we can

use 0.2 s, right?

Disk overtravel

It depends

Main (EM)

CTI = 0.3 s (because disk OT is

still in play)

CTI = 0.2 s Main (SS)

Feeder (SS) Feeder (EM)

Coordination Time Intervals – EM & SS

Main (EM) Main (SS)

Feeder (SS) Feeder (EM)

Main (EM) Main (SS)

Feeder (SS) 0.3 s

Feeder EM

0.2 s

disk OT still applicable

Main (EM)

Feeder (SS)

30 kA @ 13.8 kV

Main (SS)

Feeder (EM)

30 kA @ 13.8 kV

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

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Seconds

Coordination Time Intervals – EM/SS with Banded Devices

OC Relay combinations with banded devices

EM Relay disk over travel √ 0.1 s

Power

Fuse

Static Trip or

Molded Case

Breaker

Static Relay

operating time

CTI disk over travel

x -

0.22 s

x - CT, relay errors √ 0.12 s

Power

Fuse

Static Trip or

Molded Case

Breaker

operating time

CTI

x -

0.12 s

Coordination Time Intervals – EM/SS with Banded Devices

EM-Banded EM Relay SS-Banded

SS Relay

PWR MCB PWR MCB

PWR MCB EM Relay PWR MCB SS Relay

0.22 s 0.12 s

25 kA 25 kA

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

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Seconds

Coordination Time Intervals – 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).

Amps X 100 (Plot Ref. kV=0.48)

Coordination Time Intervals – Summary

Buff Book (Table 15-3 – Minimum CTIsa)

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Seconds

Effect of Fault Current Variations

CTI & Fault Current Magnitude

Inverse relay characteristics

imply

Main

Relay

Current

Operating

Time

Feeder

F1 = 10 kA

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.

Main

Feeder

0.2 s

F2 = 20 kA

0.06 s

If CTI is set based on 10 kA, it

reduces to less than the

desired value at 20 kA.

F1 = 10 kA

F2 = 20 kA

Amps X 100 (Plot Ref. kV=13.8)

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Seconds

Total Bus Fault versus Branch Currents

10 kA

15 kA

1.2 kA 0.8 kA 2 kA 1 kA

M 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 15 kA

Using Actual

Maximum Relay

Current of 10 kA

Feeder Feeder

M

Main

0.2 s

Main

0.8 s

15 kA 10 kA 15 kA

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

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Seconds

Curve Shaping

Most modern relays

include multiple OC

Elements.

Using a definite time

characteristic (or delayed

instantaneous) can

eliminate the affect of

varying fault current levels.

0.2 s

20 kA

15 kA 10 kA

Amps X 100 (Plot Ref. kV=13.8)

Multiple Source Buses

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Seconds

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 3

12 kA 18 kA

Plot to Actual

Relay Current

2 2

1 30 kA 1 1

0.2 s

1.1 s

12 kA 30 kA 12 kA 30 kA

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

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Seconds

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 3

12 kA 18 kA

With shift factor relay 2

pickup shifts to 7500 A.

SF = (30/12) * 3000 A

1

30 kA 1 1 2 2

0.2 s 0.2 s

12 kA 30 kA 30 kA

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

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Multiple Source Buses

10 kA 10 kA 5 kA Bus A Bus B

15 kA (Fa)

10 kA (Fb)

Fa = 25 kA Fb = 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

51A

Is1+Is2

51B

Source 2

• Typically used on double-ended

systems with normally closed

Ip1 Is1

Is2

0 Is2

Is2

Ip2 ties.

• Automatically discriminates

Bus A Bus B Ip2

between Bus A and Bus B faults.

Ip1+Ip2

Feeder A Feeder B

• Eliminates need for relay on tie

breaker.

• Saves coordination step.

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Partial Differential Relaying

Embellish notes

Source 1

51A

51B

Source 2

Scheme works even with one

Ip1 Is1 0

Is1

Is1 0

Is1

0

Open

source out of service.

However, the relay in the open

source must remain in operation

Bus A Bus B Ip1

Ip1

since the relay in the remaining

source sees no operating current.

Feeder A Feeder B

Partial Differential Relaying

Show same 3 source 1 tie example

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Directional Current Relaying

67 67

Bus A 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

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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

mechanical

withstand

through-fault. thermal

withstand

based on

transformer Z

25 x FLC

@ 2s

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Seconds

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.

mechanical

withstand

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

based on

transformer Z

thermal

withstand

25 x FLC

@ 2 s

Transformer Overcurrent Protection

FLC = 2.4 MVA/(√3 x 13.8) = 100.4 A

Relay PU must be ≤ 600% FLC = 602.4 A

Using a relay setting of 2.0 x CT, the relay

PU = 2 x 200 = 400 A

400 / 100.4 = 398% so okay

Time Dial depends on level of protection

desired.

PWR-MCB

13.8 kV

2.4 MVA, 5.75% Z

∆-Y Resistor Ground

13.8/0.48 kV

2.4 MVA

5.75%

PWR-MCB

480 V

Amps X 10 (Plot Ref. kV=13.8)

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0.577

0

0.577

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

A

primary relay clipped

1.0 0 B

0.5

C

0.866

b

c

0.866

at 100% secondary

fault current

coordinated with a

• 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.

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

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…

Transformer Overcurrent Protection

∆-Y Connections – Phase-To-Ground Faults 1.0

0.577 a A

0 0

B b

c

C

Add TCC with a

shifted and non-shifted

transformer damage

curve….

• 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.

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Seconds

Seconds

Will be deleted once the previous page is fixed.

PWR-MCB

For resistance-

grounded secondary

PWR-MCB

For a solidly-grounded

secondary

2.4 MVA, 5.75% Z ∆-Y Resistor Ground

2.4 MVA, 5.75% Z

∆-Y Solid Ground

13.8 kV

13.8/0.48 kV

2.4 MVA

5.75%

PWR-MCB

480 V

• Withstand

curve shifted

left due to

∆-Y xfmr.

• More difficult

to provide

protection.

Amps X 10 (Plot Ref. kV=13.8)

Amps X 10 (Plot Ref. kV=13.8)

Transformer Overcurrent Protection

Inrush Current

• 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.

• 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.

PWR-MCB

13.8 kV

13.8/0.48 kV

2.4 MVA

5.75%

PWR-MCB

480 V

2.4 MVA, 5.75% Z

∆-Y Resistor Ground

8-12 x FLC

(typical)

Amps X 10 (Plot Ref. kV=13.8)

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Seconds

Transformer Overcurrent Protection

Setting the primary inst. protection

• 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.

PWR-MCB

13.8 kV

13.8/0.48 kV

2.4 MVA

5.75%

PWR-MCB

480 V

2.4 MVA, 5.75% Z

∆-Y Resistor Ground

8-12 x FLC

(typical)

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.

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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

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Motor Overcurrent Protection

Seconds

Seconds

• Fuse provides short-circuit

protection.

• 49 or 51 device provide motor

overload protection.

GE Multilin 469

Standard O/L Curve

Pickup = 1.01 X FLC

Curve Multiplier = 3

Bussmann

JCL Size 9R

• Overload pickup depends on

motor FLC and service factor.

• The time delay for the 49/51

protection is based on motor

stall time.

1000 hp

4.16 kV

650% LRC

Hot

M

Amps X 10 (Plot Ref. kV=4.16)

Motor Overcurrent Protection

• In the past, instantaneous OC

protection was avoided on contactor-

fed motors since the contactors could

not clear high short-circuits.

• With modern relays, a delayed

instantaneous can be used if its

setting is coordinated with the

contactor interrupting rating.

1000 hp 4.16 kV

650% LRC

Hot

GE Multilin 469

Standard O/L Curve

Pickup = 1.01 X FLC

Curve Multiplier = 3

Bussmann

JCL Size 9R

Contactor

6 kA Int.

M

Amps X 10 (Plot Ref. kV=4.16)

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38

Motor Overcurrent Protection

th

Seconds

• The instantaneous setting for a

breaker-fed motor must be set to pass

the motor asymmetrical inrush.

GE Multilin 469

Standard O/L Curve

Pickup = 1.01 X FLC

Curve Multiplier = 3

• Can be done with a pickup over the

asymmetrical current.

• Can be done using a lower pickup wi

time delay to allow DC component to

decay out.

1000 hp

4.16 kV

650% LRC

Hot

3 kA @ 4.16 kV

M

Amps X 10 (Plot Ref. kV=4.16)

Conductor Overcurrent Protection

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40

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

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41

Seconds

Conductor Overcurrent Protection

• The insulation temperature rating

is typically used as the operating

temperature (To).

• The final temperature (Tf) depends

on the insulation type (typically

150 deg. C or 250 deg. C).

• When calculated by hand, you

only need one point and then draw

in at a -2 slope.

1 – 3/C 350 kcmil

Copper Rubber

Tc = 90 deg. C

Amps X 100 (Plot Ref. V=600)

Generator Overcurrent Protection

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42

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.

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43

Generator 51V Pickup Setting Example

Fg

1200/5

19500 kVA

903 A

Xd = 280%

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 = 1.13

Using 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)

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)

Page 43: Seminar Koordinasi Proteksi Arus Lebih

44

Seconds

Generator 51V Settings on TCC

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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45

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-Comp

TCC-1

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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47

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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48

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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50

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

SWGR-1

OCR 600/5

SWGR-3

OCR 600/5

2000/5 OCR

Main

Does this TCC look good??

400/5 TR-FDR1

0.301 s

OCR

TR-FDR2

• 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.

0.3 s

Coordination Pop Quiz #2

400/5

2000/5

FDR-1

OCR

Main-1

OCR

FDR-2

Does this TCC look good??

• 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.

0.615 s

0.304 s

• The main relay time delay is

actually too fast since the CTI at

30 kA is less than 0.2s.

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52

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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53

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??

• 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.

0.03 s

• 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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54

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.

0.302 s

• The 51V curve will shift left and

lose selectivity with the feeder if

a close-in fault occurs and the

voltage drops.

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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56

Accurate One-Line Diagram

Equipment Ratings

• drive short-circuit

calculations

• affect asymmetrical

currents

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57

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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58

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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59

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

120