GE STD15C User manual

GE
K-45307K
INSTRUCTIONS
TRANSFORMER
DIFFERENTIAL
RELAYS WITH
PERCENTAGE
AN!)
HARMONIC
RESTRAINT
TYPES
STD15C
and
STD16C
GE
Protection
and
Control
205
Great
Valley
Parkway
Malvern,
PA
19355-1337

GEK-45307
DESCRIPTION
3
APPLICATION
3
CALCULATION
OF
SETTINGS
5
Method
5
Current
Transformer Connections
5
Determination
of
CT
Turns
and
Type
STD
Relay
Tap
Setting
5
Current Transformer
Ratio
Error
6
Percent
Slope
Setting
8
I
Determination of
CT
Turns
and
STD
Relay
Tap
Setting
8
II Percent
Ratio
Error
9
IA
Repeat
-
CT
Turns
and
Relay
Tap
Setting
9
hA
Repeat
-
Percent
Ratio
Error
10
III
Percent
Slope
Setting
10
RATINGS
10
Models
12STD15B
and
12STD16B
10
Auxiliary
Relay
Control
Circuit
11
CHARACTERISTICS
ii
Pickup
and
Operating
Time
11
Overcurrent
Unit
Pickup
11
Percentage
Differential Characteristics
12
Harmonic
Restraint
Characteristics
12
BURDENS
13
CONSTRUCTION
14
Current
Transformers
14
Through—Current
Restraint Circuit
14
Differential-Current
Circuit
15
Overcurrent
Unit
15
Main
Operating
Unit
16
Target
and
Seal-in
Unit
16
Case
16
RECEIVING,
HANDLING
AND
STORAGE
17
ACCEPTANCE
TESTS
17
Visual
Inspection
17
Mechanical
Inspection
17
Electrical
Tests
17
Test
Equipment
18
INSTALLATION
PROCEDURE
18
Tests
18
Pickup
19
Harmonic
Current
Restraint
19
Through-Current
Restraint
20
Instantaneous
Overcurrent
Unit
21
Dropout
of
Main
Unit
21
Location
21
Mounting
21
Connections
21
ADJUSTMENTS
21
Tap
Plug Positioning
21
Percent
Slope
Setting
23
OPERATION
23
Targets
23
Disabling
of
Type
STD
Relay
23
MAINTENANCE
23
Contact Cleaning
23
PERIODIC
TESTS
24
Pickup
24
Harmonic
Current
Restraint
24
Through-Current
Restraint
25
SERVICING
25
RENEWAL
PARTS
26
2

GEK-45
307
TRANSFORMER
DIFFERENTIAL
RELAYS
WITH
PERCENTAGE
AND
HARMONIC RESTRAINT
STD15C
and
STD16C
INTRODUCTION
Relays
of
the
STD
type
are
transformer
differential
relays
provided
with
the
features
of percentage
and
harmonic restraint.
A
static decision
unit
controls
a
small
telephone-type
relay
that
provides
the
contact output.
Percentage
restraint
permits
accurate
discrimination
between
internal
and
external
faults
at
high
current,
while
harmonic
restraint
enables
the
relay
to
distinguish,
by
the
difference
in
waveform,
between
the
differential
current
caused
by
an
internal
fault,
and
that
of
transformer
magnetizing
inrush.
DES
CR1
PTI
ON
Each Type
STD
relay
is
a
single-phase
unit.
The
Type
STD15C
relay is designed
to
be
used
for
the
protection
of
two-winding
power
transformers
and
has
two
through-
current
restraint
circuits
and
one
differential—current
circuit.
The
Type
STD16C
relay
is designed
for
use
with
three—winding
power
transformers
and
has
three
through-current
restraint
circuits
and
one
differential—current
circuit.
It
may
also
be
used
for
four-circuit
transformer
protection
(see
Figure
1)
when
only
three
circuits
require
through-current
restraint,
while the
fourth circuit,
being
the
weakest,
needs
no
through-current
restraint.
APPLICATION
The
current
transformer
ratios
and
relay
taps
should be
selected
to
obtain
the
maximum
sensitivity
without
risking
thermal
overload
of
the
relay
or
current
transformer
(CT),
or
the
possibility
of
misoperation.
Therefore,
current
transformer
ratios
in
the
various windings
of
the
power
transformer
should
be
selected
with the
following
points
in
mind:
1.
The
lower
the
relay
tap
and
the
lower
the
CT
ratio
selected, the
higher
will
be
the
sensitivity.
However, the
lowest
CT
ratio
and
the
lowest
relay
tap
may
not
be
compatible
with
some
of
the
following
restrictions.
Where
a
choice
is
available
of
increasing
either
the
CT
ratio
or
the
relay
tap,
it
is
desirable
to
increase
the
CT
ratio
in
preference
to
the
relay
tap.
Since
the
relay
burden is
likely
to
be small
compared
to
the lead
burden,
increasing
the
CT
ratio
tends
to
improve
the
relative
performance
of
the
CT5,
as
a
result
of
reducing
the
maximum
secondary
fault
current and
increasing
the
accuracy
of
the
CTs.
These
1nstrucron5
do
not
purport
to cover
i.11
details
or
variations
in
equipaent
nez
e provide
for
every
possible
ctingency
to
be
met in
connection
with
installation,
operation
or
IlntnCe. Should
further
nforma
ton
be
desi red
or
should
parti
cular
probleme arise
which
are
not
covered
ziWficiently
for
the
purchaser’s
purposes,
the
matter
should
be
referred
to the
General
Electric
Company.
To
the
extent
required
the
products
described herein
meet
applicable
ANSI,
IEt
andN
standards,
but
no
such
assurance
is
given
with
respect
to
local
codes
and
ordinances because
they
varç greatly.
3

GEK—45307
2.
The
CT
secondary current
should not
exceed
the continuous
thermal
rating
of
the
CT secondary
winding.
3.
The
relay
current
corresponding
to
maximum kVA (on
a
forced-cooled
basis)
should not
exceed twice
tap
value, which
is
the
thermal rating of
the relay.
4.
The CT
ratios
should
be
high
enough that the
secondary
currents will
not damage
the relay
under maximum
internal
fault conditions
(refer to RATINGS).
5.
The
relay
current corresponding
to rated
kVA
of
the
power
transformer
(on
self-
cooled
basis) should
not
exceed the
relay tap value
selected
(magnetizing
inrush
night
operate
the
instantaneous overcurrent
unit). If the
transformer
under
consideration
does
not
have
a
self—cooled
rating,
the
transformer
manufacturer should
be
consulted for
the
“equivalent
self—cooled
rating”;
that
is the rating
of
a
self-cooled
transformer
that would
have the
same
magnetizing
inrush
characteristics
as the transformer
being considered.
6. The
current
transformer
tap chosen must
be
able
to
supply
the
relay
with
8
times rated
relay
tap
current,
with
an
error
of less than 20%
of
the total
current.
If
the
current
transformers produce
an error
of
greater
than
20%
at
less
than
8
times
tap
value,
the
harmonic
content of
the
secondary
current
may
be
sufficient
to cause
false
restraint on
internal faults.
7. The CT ratios
should
be
selected
to
provide balanced secondary
current
on
external
faults. Since
it
is rarely
possible to
match the secondary
currents
exactly
by
selection of
current
transformer ratios,
ratio—matching
taps
are
provided
on
the
relay by
means
of which the
currents may usually
be
matched
within
5%.
When
the
protected
transformer
is
equipped
with
load-ratio
control
it
is
obvious
that a close match
cannot be obtained
at
all points
of the ratio-
changing
range. In this case,
the secondary
currents are
matched
at
the middle
of the
range and the
percentage_differential
characteristic
of
the relay
is
relied
upon to prevent
relay operation
on the unbalanced
current
which
flows
when the
load-ratio
control
is
at the ends
of the range.
8.
In
some applications,
the
power
transformer will
be
connected
to the
high
voltage
or low voltage
system through four
breakers
(as shown in
Figure
1)
as
for example
in
a
ring—bus
arrangement.
In
this case,
the
CT
ratios
must
be
selected
so
that
the secondary
windings
will not
be
thermally
overloaded
on
load current
flowing around
the ring in
addition to
the
transformer
load
current.
It
is recommended that
CTs
on
each of the
two low
voltage
(or
high
voltage)
breakers
be
connected
to a
separate restraining
winding
to assure
restraint
on heavy
through-fault
current
flowing
around the
ring bus.
It is not
desirable
to
protect
two
parallel transformer
banks with
one
set
of
differential
protection,
since
the
sensitivity of the
protection
would
be
reduced.
In addition,
if
the
banks can be
switched separately,
there is
a
possibility
of
false
operation on magnetizing
inrush
to
one transformer
bank, causing
a
“sympathetic
inrush”
into the bank
already energized.
In
this
case,
the
harmonics
tend to
flow
between
the banks,
with the
possibility
that there will
be
insufficient
harmonics
in
the
relay
current
to
restrain the
relay.
Typical
elementary
diagrams
for the
STD15C and
STD16C are
illustrated
in Figures
2
and 3.
4

GEK—45307
CALCULATION
OF SETTINGS
METHOD
The
calculations
required for determining
the proper relay
and
CT
taps are
outlined
below.
A
sample calculation, for the
transformer shown in
Figure
4,
is
then
given.
CURRENT
TRANSFORMER CONNECTIONS
Power
Transformer Connections
Current
Transformer
Connections
Delta-Wye
Wye-Delta
Wye-Del ta
Del ta-Wye
Delta-Delta
Wye—Wye
Wye-Wye
Delta-Delta
Delta-Zigzag
with
O
phase
Delta—Delta
shift
between primary
and
secondary
DETERMINATION
OF CT
TURNS
AND
TYPE STD RELAY TAP SETTING
1.
Determine
the maximum line currents
(Max.
I)
on the basis
that each power
transformer
winding may carry the maximum
forced—cooled rated kVA
of
the
transformer.
— Maximum Transformer kVA
Max. -
(Line kV)
2. Determine
the full—load rated
line
currents (100%
I)
on
the basis that
each
power transformer winding may carry the
full self-cooled rated kVA of
the
transformer,
or
the “equivalent”
self-cooled ratings.
— 100%
Transformer kVA
100%
‘P -
(Line
kV)
Actually,
this calculation
does not mean that all
windings
will necessarily
carry
these maximum
load currents continuously. This is
only a convenient way
of
calculating
the
currents in the other
windings in
proportion to
their
voltage ratings.
This
is the requirement for selecting the
relay tap
setting
so that the relay will
not
operate
for
any external
fault.
3.
Select
CT
ratios so that the secondary current
corresponding
to maximum
Ip does
not exceed the CT
secondary
thermal
rating
(5
amperes).
In the case
where a
transformer is connected to
a
ring
bus, for
example,
the CT ratio
should be
selected so that the CT
thermal rating will not
be
exceeded by the maximum
load
current
in
either breaker. Also, select CT ratios so
that the
relay currents
can
be properly matched by means
of the relay taps. (Highest
current not more
than 3
times
lowest
current).
For
Wye-connected
CTs
Tap Current
= 100%
I
5

GEK-45307
For Delta-connected
CTs
Tap
Current
=
LnL
N
where
N is
the
number
of CT secondary
turns.
4. Check
the
matching
of
relay
currents
to
relay
taps,
to keep the
mismatch
error
as low as possible.
Calculate
the
percentage
of
mismatch
as
follows: on
two—winding
transformers,
determine
the
ratio of
the two relay currents
and the
tap values
selected.
The
differences
between
these ratios, divided
by
the
smaller
ratio, is
the percent
of mismatch.
The
mismatch normally
should not
exceed
5%.
For
three-winding
transformers,
the percent of
mismatch error
should be
checked
for
all combinations
of currents or
taps.
If taps cannot
be selected
to keep this
percentage
error within
allowable
limits,
it will
be
necessary to
choose a
different
CT ratio on
one or
more
lines,
to obtain a better
match between
relay currents
and
relay taps.
5.
Check to see
that the sum
of
the
relay currents
that
will
be
applied to
the
relay for
a
fault
at
the
terminals of the power
transformer
is
less than 220
amperes
RMS for
1
second.
If the
period
during
which a fault current
flows
in
the
relay can
be
definitely limited
to a shorter
time,
a
higher
current can be
accommodated
in accordance with
the
relation:
(Amperes)?
x
seconds
48,400
Also
check that the
suni
of
the
multiples of
tap current
on an
internal
or
external
fault
do
not
exceed
150.
CURRENT
TRANSFORMER
RATIO
ERROR
The
CT ratio
error must be
less
than 20% at
8
times relay rated
tap
current.
This
is based on
the
instantaneous
unit
being
set at its
normal setting,
which
is
8
times
tap
rating.
If
the
instantaneous
unit
pickup
is
raised
above
this value,
the
20%
figure
must
be
reduced,
as
described
in the CHARACTERISTICS
section.
The
calculations
listed below
are for the
worst—fault condition,
as
far as CT
performance
is concerned,
which is an internal
ground
fault
between
the
CT
and
the
transformer
winding, with none
of
the
fault
current
supplied through
the neutral of
the
protected
transformer.
1.
Determine
the
burden
on each CT, using
the
following
expressions:
a. For
Wye-connected
CTs
=
+ Ne+2f
+ 2R
Ohms
(Equation
1)
b. For Delta-connected
CTs
Z
= 2B
+ Ne
+
2f
+ 2R Ohms
(Equation
2)
1000
6

GEK-45307
where
B
STD
relay
total
burden
(see
Table
I)
N
= number
of
turns
in
bushing
CT
e
bushing
CT
resistance
per turn,
milliohms
(at
maximum
expected
tempera
ture)
f
busing
CT
resistance
per
lead,
milliohms
(at
maximum
expected
temperature)
R
one-way
control
cable
lead
resistance
(at
maximum
expected
temperature)
TABLE
I
Total
Burden
for
60 Cycle
Relays
STD
TAPS
AMPS
8
X
TAP
AMPS
BURDEN
OHMS
(B)
MIN
P.U.
AMPS
2.9
23.2
0.180 0.87
3.2
25.6
0.156
0.96
3.5
28.0 0.140
1.05
3.8
30.4
0.120
1.14
4.2
33.6
0.112
1.26
4.6
36.8
0.096 1.38
5.0
40.0
0.088
1.50
8.7
69.6
0.048
2.61
2.
Determine
CT
secondary
current
for
8
times
tap
setting.
=
8
x
STD relay
tap
rating
(Note:
For
the
location
of
fault
assumed,
all
the fault
current
is
supplied
by
one CT,
so
that
CT
current
and
relay
current
are the
same,
regardless of
whether
the
CTs
are
connected
in
wye
or
delta.)
3.
Determine
secondary
CT
voltage
required
at
8
times
tap
setting.
Esec
=
IZ
4.
From excitation
curve
of
particular
tap
of
current
transformer
being
used,
determine
excitation
current
IE.
corresponding to
this
secondary
voltage,
Esec.
5.
Determine
the
percent
error
in
each
CT
by the
expression:
%
error
=
1 E
X
100
Is
This
should
not
exceed
20%
of any
set
of CTs.
If
it
does,
it
will be
necessary
to
choose
a
higher
tap
on that set
of CTs,
and repeat
the
calculations
on selection
of
relay
taps,
mismatch
error,
and percent
ratio
error.
7

GE:K-45307
PERCENT
SLOPE
SETTING
The proper
percent
slope
required
is
determined
by the
sum
of:
a.
The
maximum
range
of manual
taps
and
the
load—ratio-control,
or automatic
tap
changing
means,
in percent.
b.
The
maximum
percent
of
mismatch
of
the
relay
taps.
Set
the desired
percent
slope
by
means
of
R3
(See
Figure
6A).
The
percentage
slope
setting
selected
should
be
greater
than
the
ratio
of
maximum
total
error
current
to
the
smaller
of the
through
currents.
In general,
if the
total
error
current
does not
exceed
20%,
the
25%
setting
is
used.
If
it
exceeds
20%,
but
not
35%,
the
40%
setting
is
used.
If the
movable
lead
is
used
(as
in
Figure
1,
for
example)
the
percent
slope
setting
should
be
chosen
about
twice
as high,
since
the movable
lead
provides
no restraint.
EXAMPLE
(REFER
TO FIGURE
4)
I.
Determination
of
CT
Turns
and
STD Relay
Tap
Settings
1.
Transformer
and
Line
A
BC
2.
Maximum
I
= 3750/
J3
(Line
kV)
19.7
49.5
157
3.
100%
Ip
= 3000/
J
(Line
kV)
15.7
39.6
125
4.
Assume
CT
turns
(N)
20 20
60
5.
Maximum
I
secondary
(less
than
5a)
0.98
2.47
2.62
6.
100%
I
secondary
0.79
1.98
2.08
7.
CT connections
Delta
Wye
Delta
8.
Relay
Current
for
100%
I
Sec.
1.37
1.98
3.60
Select
a relay
tap for
one
of the
line
currents
and
calculate
what
the
currents
in
other
lines
would
be if
they
were
increased
in
the
same
ratio.
If
any
current
is
greater
than
V3
times
any other,
the
8.7
tap
should
be
chosen
for it,
and
new
ideal
relay
taps
calculated
for
the
other
lines.
9.
Ideal
Relay
Taps
(Set
C
8.7)
3.31
4.78 8.7
10.
Try
Relay
Taps
3.2
4.6 8.7
11.
Check
Mismatch
Error
Ratio
of
Taps
on
Lines
B-A
=
1.43
Ratio
of
Sec. Lines
Currents
1.98
-
1.37
—
1.44
Mismatch
1.44
- 43
=
0.7%
Ratio
of Taps
on
LinesC-B
-!.j-
=
1.89
Ratio
of
Sec.
Line
Currents
=
1.82
1.89
-
1.82
Mismatch
1
8
=
3.8%
8

GE
K—45
307
Ratio
of
Taps
on Lines
C-A
2.72
Ratio of
Sec.Line Currents
= 2.63
Mismatch
2.72 -2.63
=
(All
are
less than
5%; therefore 1
mismatch error
is
not
excessive)
12. Check
that
the sum
of the
maximum relay currents
is
less than 220 amps
for
1
second, and therefore,
short-time rating
of relay
is
not exceeded.
II. Percent Ratio
Error
ASSUME (all
measured at
their maximum
One-way
CONTROL
CABLE RESISTANCE
Bushing
A CT
resistance
per turn
B
II
C
II II
Bushing A CT
resistance
per lead
B
II II
II II
Ii
II II
expected
temperatures)
(R)
= 0.284
ohms
Ce) =
4 rnilliohms
(e)
2.5
Ce)
= 2.3
(f)
= 75
niilliohms
(f)=525
(f)
= 18.6
1.
Burdens
on
CTs,
using
Equation
1
or Equation
2 from page
6.
a.
Line
A, Z
= 2
(0.156)
+ (20x4)
+
(2.0 x 75)
1000
= 0.312
+ 0.205
+ 0.568
=
1.085
+ 2 (.284)
b. Line
3,
Z = 0.096
+ (20
x
2.5)
+ (2.0 x 52.5)
1000
A
T.o85
25.6
27.8
1
.00
3.4%
+ 0.568
B
0.8
36.8
29.4
50
136%
C
0.833
69.6
58.0
0.5
0.8%
Exciting
current
on line
B is too high;
should
improve CT
performance.
IA - Repeat
— CT
Turns
and
Relay
Tap
Setting
try
higher tap
on CT
to
1. 100%
2.
Try
CT
turns
(necessary to change
C
also
for proper matching)
3. 100% 1 secondary
4.
Relay Current
5. Ideal
Relay
Taps (Set
C
8.7)
6.
Use Relay
Taps
7.
Mismatch
Error is
less
than
5%
= 0.096 + 0.138
+ 0.568
0.80
c.
Line C,
Z 2 (0.048)
+ (60
x
2.3)+(2
x 18.6)
+ 0.568
= 0.096
+ 0.180
+
0.568 0.833
2. Impedance,
ohms
3.
8
times
tap,
amperes
4. E 5
CT voltage require
(IZ)
5.
JE
required,
from excitation
curve
6. % Ratio Error
15.7 39.6
125
20
40 80
0.79 0.99 1.56
1.37
0.99
2.70
4.40
3.19
8.7
4.6
3.2 8.7
9

GEK-45307
hA
Repeat - Percent Ratio
Error
1.
Burden
on
CTs
Line
A,
Z
=
0.192
+
0.205
+ 0.568
= 0.965
Line
B,
Z
0.156
+
0.188
+
0.568
=
0.912
Line
C,
7
= 0.096
+
0.226
+
0.568
= 0.890
2.
Impedance
Ohms
0.965
0.912
0.890
3.
8
times
Tap,
Amperes
36.8 25.6 69.6
4.
CT voltage
required
(17)
35.6 23.4
61.9
5.
‘C
required, from
excitation
curve
1.1 0.25
0.17
5.
%
of
Ratio
Error
3.1%
1.0%
0.3%
Percent
error
is
less
than
20%,
so
CT
taps
and
relay
taps
are
satisfactory.
III
Percent
Slope
Setting
1.
Assume
load
ratio
control
maximum
range
10.0%
2.
Relay
tap
mismatch,
from
IA above (Lines
A-B)
4.6%
Use
25%
setting
14.6%
RATI
NGS
MODELS
12STD15C AND
12STD16C
Continuous
Rating
The
through-current
transformer
and
differential-current
transformer
will
stand
twice
tap value
for
any
combination
of
taps,
or they
will
stand
twice
tap
value
if
all
but
one
of
the
restraint
windings
carry 0 current,
and
the
full restraint
current
(equal
to
twice
tap value)
flows
through
the
differential—current
transformer.
Short
Time
Rating
(Thermal)
220
amperes
for
1
second,
measured
in
the primary
of
any
transformer
of
the
type
STD
relay.
Higher
currents
may
be
applied
for
shorter
lengths
of
time
in
accordance
with
the
following
equation:
= 48,400
where
=
current
amperes
t
=
time
in
seconds.
Short
Time
(Electrical
For
both
the
STD15C
and
STD16C the
sum
of
the
multiples
of
tap
current fed
to
the
relay
from
the
several
sets
of
current
transformers
should not
exceed
150.
These
multiples
should
be
calculated
on the
basis of
RMS
symmetrical
fault
current.
This
limitation
is
a
result
of
the
voltage
rating
of
the
rectifiers
in
the
through—
current
restraint
circuit.
Note
that
in
Figure
1
external
fault
current
can
flow
through
circuit
breakers
52—1
and
52—2
without
being
limited
by
the
transformer
i
in
peda
n
ce.