Post on 22-Dec-2015
description
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
2
Protection Objectives
• Personnel Safety
Protection Objectives
• Equipment Protection
3
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
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
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
8
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
9
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
10
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.
11
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
12
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
13
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)
14
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)
15
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)
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
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)
18
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)
19
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
20
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)
21
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)
22
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)
23
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)
24
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)
25
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
26
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)
27
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)
28
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.
29
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
30
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
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
mechanical
withstand
through-fault. thermal
withstand
based on
transformer Z
25 x FLC
@ 2s
32
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)
33
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.
34
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)
35
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.
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
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)
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
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
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
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.
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)
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
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.
46
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
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.
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..
49
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.
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
51
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.
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.
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.
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.
55
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
56
Accurate One-Line Diagram
Equipment Ratings
• drive short-circuit
calculations
• affect asymmetrical
currents
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
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
59
Protective Devices
• Make & model
• CT Ratio
• Tap & time dial
settings (old days)
• setting files (today)
• Trip device type
• LT, ST, Inst settings
References
60
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