Lecture 12 Overcurrent Protection.pdf

1
Lecture 12: Overcurrent
Protection
Example of time-overcurrent
protection
ELCT 751 Overcurrent Protection 1
p
Directional overcurrent protection
Distribution system protection
Fuses
Example of overcurrent
protection
1 2
52-1 52-2
Source
51-1 51-2
3
System is simplified for
illustration of principles Load
Example
Bu
s
Max
Fault
Min Fault
CT
Ratio
Load
1 1395 A 1208 A 200:5 60 A
1 1395 A 1208 A 200:5 60 A
2 695 A 602 A 100:5 50 A
3 521 A 451 A - 100 A
10
100
100
Tvi I 1

( )
Tvi I 2

( )
Tvi I 5

( )
Time
[sec]
10
M is the
Multiplier
Setting
Very Inverse Time Overcurrent Relay Curves
0.1 1 10 100
0.1
1
0.1
Tvi I 10

( )
20
0.5 I
Current [per unit of pickup setting]
1 10
1
M=1
M=2
M=5
M=10
–51-2 (Downstream) Setting
• sees load current of 100 A
• set pickup at 2 x load = 200 A
• on sec of CT: 200 A x (5/100) = 10
A (10 A tap or 200% of CT sec)
• time multiple M = 1 (fastest since no
downstream coordination)
• check pickup on min fault: 451/200
= 2.25 x pickup (2.5 to 3 is better,
but this is OK)
–51-1 (Upstream) Setting
• sees load current 150 A
• pickup: 2 x 150 A = 300 A
• on CT sec: 300 A x 5/200 = 7.5 A
(7.5/5 = 1.5) Set pickup at 150% of
CT
• set to backup 51-2 for min fault at 3 if
possible: 451/300 = 1.5 (eventually
trips which is good)
• check coordination with 51-2

2
–coordination between 51-1 and 51-2
• coordinate at max fault seen by both
relays (max fault at 2 is 695A)
• 51-1 sees 695 x 5/200 = 17.4 A sec
or 17.4/7.5 = 2.3 x pickup
• 51-2 sees 695 x 5/100 = 34.8 A sec or
34.8/10 = 3.5 x pickup
• 51-2 on M = 1 curve gives Top = 0.4
sec at 3.5 times pickup, from time-
current curves
10
100
100
Tvi I 1

( )
Tvi I 2

( )
Tvi I 5

( )
Time
[sec]
10
M is the
Multiplier
Setting
0.1 1 10 100
0.1
1
0.1
Tvi I 10

( )
20
0.5 I
1 10
1
M=1
M=2
M=5
M=10
3.5
0.4
2.3
–Coordination (cont’d)
• allow 0.3 sec coordination margin
between adjacent relays
• 51-1 needs to operate after 0.4+0.3
= 0.7 sec
• at 2.3 times pickup, M=1 gives
Top = 0.8 sec
10
100
100
Tvi I 1

( )
Tvi I 2

( )
Tvi I 5

( )
Time
[sec]
10
M is the
Multiplier
Setting
0.1 1 10 100
0.1
1
0.1
Tvi I 10

( )
20
0.5 I
1 10
1
M=1
M=2
M=5
M=10
3.5
0.4
0.8
2.3
Summary of settings for example
Bus CT Ratio
Pickup [% of
CT sec 5A]
M = Time
Multiplier
1 200:5 150 % 1
1 200:5 150 % 1
2 100:5 200 % 1
Coordination Margin = 0.4 seconds
for max fault on bus 2
Comments on example
–There are often downstream devices
that we need to coordinate with
–Settings of relay depend on
technology
gy
• electromechanical relay settings
entered via taps and dials
• electronic relay settings entered into
software via keyboard or via dials +
keyboard

3
Comments
• Simple overcurrent relays are not
sufficient for network systems:
–Loop type subtransmission
High voltage transmission
–High-voltage transmission
networks
• These use directional
overcurrent or directional
distance relays
Distribution Systems
• Power transmission lines
interconnect generating plants
and main substations
• Power subtransmission lines
connect main substations with
other substations and large
customers
Distribution Systems
• Power distribution lines deliver
power to all other customers,
and to loads within the
premises of large customers
• Power distribution lines may be
medium-voltage or low-voltage:
Distribution systems
–Medium-voltage systems use
nominal voltages above 1 kV and
below 60 kV
–Low-voltage systems use nominal
g y
voltages under 1 kV
• Medium-voltage lines may be
utility lines or contained within
a large industrial system
Distribution systems
• Medium-voltage utility systems
will be considered first
–Overhead lines
–Underground lines
Underground lines
• Medium-voltage industrial
systems and low-voltage
industrial systems will be
considered later
Overview
• Most power distribution systems use
overcurrent protection
– Protects system against short circuits
– Also helps protect apparatus from
Also helps protect apparatus from
damage (secondary)
• Hardware used (switchgear, relays)
varies by system type, but function
is ordinarily overcurrent protection

4
HV
SYSTEM
FEEDER #1
FEEDER #2
FEEDER #3
115:12.47 kV
THREE-PHASE DISTRIBUTION TRANSFORMER BANK
FUSE
THREE-PHASE LINE
SINGLE-PHASE LINE
CAPACITOR BANK
RECLOSER CIRCUIT BREAKER
Overview
• MV utility system uses power
circuit breakers or automatic
circuit reclosers at substation
i ll hi h
• To save time, call this the
station breaker in either case
• Taps may be protected by fuses,
reclosers, or sectionalizers
Line Sectionalizing
• Line sectionalizing may be done with
reclosers, sectionalizers, fuses, or
switches
– Typical system showed a single feeder
sectionalizing recloser (radial feeder)
sectionalizing recloser (radial feeder)
– A switch (either load-break or
disconnect) could be used
– Load-break switches may be equipped
with remote controls (considered later)
Line Sectionalizing
• In the simple example system, only
one recloser was used
– Placed near the end of the reach of
the station breaker
Set for automatic operation with a
– Set for automatic operation with a
preset number of trip and reclose
operations
– Covers the entire feeder with rapid
fault clearing
– Restores service for temporary faults
Line Sectionalizing
FEEDER #1
FEEDER #2
Breaker Zone Recloser Zone
Overlapping zones of protection:
Station breaker overlaps recloser zone.
Recloser sectionalizes line, increases
reliability.
FEEDER #3
Automatic Circuit Recloser
Station Circuit Breaker
Line Sectionalizing
• Fault at X is cleared by recloser
FEEDER #1
FEEDER #2
FEEDER #2
FEEDER #3
Breaker Zone Recloser Zone

5
Line Sectionalizing
• Fault at X is cleared by recloser
• No permanent outage for
temporary fault
• Outage only downstream of
recloser for permanent fault
Fuses
• The simplest form of overcurrent
protection is the fuse
• Types of fuses common on MV
yp
distribution:
–Expulsion fuse
–Current-limiting fuse
Expulsion Fuse
• Expulsion fuse in open-type
cutout shown below
–Fuse link (inside tube) produces
an arc on melting
an arc on melting
–Expulsion tube emits deionizing
gases
–Gases are expelled from tube and
fault is cleared at current zero
Expulsion Fuse Cutout
LINE TERMINAL
BRACKET
EXPULSION TUBE
LINE TERMINAL
PORCELAIN
SUPPORT
OPEN TYPE FUSE CUTOUT
(NOT TO SCALE)
Fusible Link
TIN FUSIBLE ELEMENT
TIN-PLATED BUTTON HEAD
CUTAWAY VIEW OF TYPICAL FUSE LINK
TUBE
LEADER
CUTAWAY VIEW OF TYPICAL FUSE LINK
(NOT TO SCALE)
Tin fusible element shown
Other types are also available
Link fits inside expulsion tube

6
Expulsion fuse clearing fault
AVAILABLE CURRENT
X/R INFINITE
FAULT CURRENT
MELT TIME
CIRCUIT VOLTAGE
TRANSIENT RECOVERY VOLTAGE
FUSE
VOLTAGE
TIME
Current-Limiting Fuse
• Current-limiting fuse clears the fault
quickly by forcing the current to zero
– Long fusible element packed in silica
sand
sand
– Sand confines the arc to small area
– Produces high pressures and very high
resistance
– Resistance forces current to zero
Current-Limiting Fuse Clearing
AVAILABLE CURRENT
/
FAULT CURRENT
CIRCUIT
MELT TIME
X/R INFINITE
CIRCUIT
VOLTAGE
FUSE VOLTAGE
TIME
Recovery Voltage
• Compare the expulsion fuse
voltage to the current-limiting
fuse voltage
fuse voltage
–Expulsion fuse clears near peak
voltage. Ringing transient due to
interrupting RLC circuit
Recovery Voltage
• Compare the expulsion fuse
voltage to the current-limiting
fuse voltage.
C t li iti f l
–Current-limiting fuse clears near
zero system voltage
–Fuse voltage controlled by careful
design of link
–Little ringing due to system
Distribution Transformer Fusing
• Fuse protecting transformer must:
– clear short-circuit in transformer
– not damage fuse link on inrush current,
load pickup, short-time overloads
– coordinate with upstream line
sectionalizing devices (fuse or recloser
upstream)
– provide a degree of protection for
severe overloads

7
Inrush/Load Pickup Curve
10.
100
T [SEC]
3 x 3.62 = 10.9 A
2 x 3.62 = 7.24 A
COLD-LOAD
PICKUP CURVE
FUSE MAXIMUM CLEARING
TIME CURRENT CURVE
FUSE MINIMUM MELT
TIME CURRENT CURVE
For illustration.
1 10
0.1
1.0
100 1000
0.01 I [A]
6 x 3.62 = 21.7 A
12 x 3.62 = 43.4 A
25 x 3.62 = 90.5 A
INRUSH CURVE
Use actual inrush
pickup and fuse
curves.
1 0
10.
100
T [SEC]
3 x 3.62 = 10.9 A
2 x 3.62 = 7.24 A
COLD-LOAD
PICKUP CURVE
FUSE MAXIMUM CLEARING
TIME CURRENT CURVE
FUSE MINIMUM MELT
TIME CURRENT CURVE
1 10
0.1
1.0
100 1000
0.01 I [A]
6 x 3.62 = 21.7 A
12 x 3.62 = 43.4 A
25 x 3.62 = 90.5 A
INRUSH CURVE
Fuse Ratio
• Ratio of minimum melt current to
transformer full-load current
– High fuse ratio allows more
transformer heating on overload but
i f i h t
is more secure from inrush current
– Low fuse ratio allows less transformer
heating on overload but is less secure
from inrush current
– Typical fuse ratios are around 2 to 4
Transformer Damage Curves
• ANSI C-57 standards give damage
curves, which should lie above and
to the right of fuse maximum
clearing curve
clearing curve
– Compromise with damage curve is
preferable to miscoordination with
upstream device
– Miscoordination is preferable to fuse
damage on load pickup
Shunt Capacitor Fusing
• Shunt capacitors for power-factor
correction are usually fused
–Will not usually protect against
device failure
device failure
–Can prevent tank rupture
–Capacitor manufacturer can supply
rupture curves for coordination
with fuses
1 0 .
1 0 0
F U S E C U R V E S N O T O K
T A N K M A Y R U P T U R E
T [S E C ]
Typical capacitor rupture curve
1 0
0 .1
1 .0
1 0 0 1 0 0 0
0 .0 1
1 0 0 0 0
F U S E C U R V E S O K
T A N K W IL L N O T R U P T U R E
I [A ]

8
Capacitor Fusing
• Curve shown for illustration only
–Use actual curves from
manufacturer
F se sho ld ithstand 135% of
• Fuse should withstand 135% of
nominal capacitor current
–Allows 10% overvoltage + 15%
overcapacitance +10% harmonic
current
Capacitor Fusing
• Grounded-wye capacitor with one
phase completely shorted:
–fault current = available single
line-ground short-circuit current
line ground short circuit current
• Ungrounded-wye capacitor with
one phase completely shorted:
–maximum fault current = 3 X
normal capacitor current
Ungrounded wye capacitor with
phase a shorted to neutral
Vab
Vca
Vbc Van
Vbn
Vcn
Ia
Ib
Ic
Normal
Vcn
Vbn
Ia
Ib
Ic
Phase a shorted
to neutral
Capacitor Inrush
• High-frequency damped
sinusoid: estimate inrush I2 t
d t it
and compare to capacitor
– Rule of thumb:
I2 t = K Isc[kA] Ic[A]
Capacitor Inrush
I2 t = K Isc[kA] Ic[A]
–if X/R=0.5 then K = 3
–if X/R=1 0 then K = 4
if X/R=1.0 then K = 4
–if X/R=2.0 then K = 6
–if X/R=5.0 then K = 14
–if X/R=10 then K = 27
Capacitor Inrush
• Isolated capacitors rarely have
inrush problems
• If capacitors are connected back
to back, see ANSI C37 for more
,
information about capacitor
switching
–Inrush for back-to-back capacitor
banks is a severe problem that
should be carefully considered

Lecture 12 Overcurrent Protection.pdf

Recommended

Recommended

More Related Content

Similar to Lecture 12 Overcurrent Protection.pdf

Similar to Lecture 12 Overcurrent Protection.pdf (20)

More from ChadWood16

More from ChadWood16 (9)

Recently uploaded

Recently uploaded (20)

Lecture 12 Overcurrent Protection.pdf