ABB CIRCUIT SHIELD 87T - Page 12

Transformer
Differential
Relay
IB
7.6
.
1.7
-
4
Page
12
12
.
CT
performance
check
(
continued
)
:
Assuming
only
the
relay
is
connected
to
the
ct
’
s
:
High
side
ct
’
s
in
wye
2.82
ohms
Low
side
ct
’
s
in
delta
5.64
ohms
2.3
RL
H
+
RB
3.4
RL
L
+
3
RB
Relay
Burdens
from
Table
3
:
RBH
0.04
ohms
(
4.5
A
tap
)
0.05
ohms
(
3.2
A
tap
)
Calculate
maximum
allowed
one
-
way
lead
resistance
:
RL
H
RB
L
(
2.82
(
5.64
-
3
*
0.04
)
/
3.4
1.62
ohms
max
.
0.05
)
/
2 . 3
1.20
ohms
max
.
R L
L
13
.
Select
minimum
operating
current
setting
:
For
a
minimum
operating
current
of
25
%
relative
to
full
load
current
:
IH
S
High
side
tap
=
(
25
%
x
5.02
J
/
3.2
5.02
A
(
from
step
4
)
3.2
A
(
from
step
7
)
=
39
%
of
tap
.
(
D
/
(
100
+
D
)
)
x
R
X
T
From
page
4
:
Im
W T
.
39
.
39
*
(
100
+
D
)
/
D
.
39
*
(
125
/
25
)
=
D
=
25
%
from
step
10
1.95
(
Minimum
Restraint
setting
)
R
R
Ca
1
culations
for
Three
-
Winding
Transformers
:
Tap
selections
are
calculated
two
at
a
time
,
based
on
the
primary
rating
of
the
transformer
-
and
not
on
the
ratings
of
the
individual
secondary
windings
.
For
the
following
example
assume
transformer
ratings
:
Tertiary
12.4
kv
,
7.5
/
10
Mva
de
1
ta
Secondary
69
kv
,
30
/
40
Mva
wye
Primary
161
kv
,
30
/
40
Mva
OA
/
FA
wye
1200
:
5
C
200
ct
’
s
:
(
all
are
multiratio
)
ct
connections
:
600
:
5
C
200
1200
:
5
C
800
de
1
ta
de
1
ta
wye
primary
to
tertiary
-
16
%
primary
to
secondary
-
9.5
%
impedances
:
(
based
on
OA
rating
)
Tertiary
Primary
Secondary
S
amp
1
e
.
c
a
.
1
c
u
J
.
a
t
i
o
.
n
:
465
A
(
@
1
OMva
)
(
1862
A
@
40
Mva
)
2
.
Max
through
fault
currents
(
assuming
only
source
is
infinite
bus
on
primary
side
)
11640
A
335
A
(
@
40
Mva
)
143
A
(
@
40
Mva
)
1
.
Max
load
currents
3523
A
1
509
A
1000
:
5
600
:
5
3
.
Choose
ct
ratios
400
:
5
2.32
A
(
9.31
A
@
40
Mva
)
2.79
A
1
.
79
A
4
.
CT
secondary
currents
at
maximum
load
2.32
A
(
9.31
A
@
40
Mva
)
4.83
A
3.10
A
5
.
Relay
currents
at
maximum
load
6
.
Ratio
of
relay
currents
(
Note
:
for
this
calculation
,
the
same
transformer
rating
must
-
be
used
for
all
windings
;
in
this
case
the
40
Mva
rating
.
)
1
arger
/
smal
1
er
=
secondary
/
primary
=
4.83
/
3.10
1.558
.
Ai
2
9.31
/
3.10
tertiary
/
primary
3.003
.
1
arger
/
smal
1
er
A i
3
tertiary
/
secondary
=
9.31
/
4.83
1.928
.
1
arger
/
smaller
A
2
3
(
Example
continued
on
page
23
)
Courtesy of NationalSwitchgear.com