Phase Distance Through Power Transformers; Example - GE D60 Instruction Manual

10.5 PHASE DISTANCE THROUGH POWER TRANSFORMERS

c) ZONE 4 SETTING IN APPLICATION (A)
As the transformer is not located between the potential source and the reach point for Zone 4, the reach impedance must
not include the positive-sequence impedance of the transformer. Because both VTs and CTs are located on the same side
as the intended reach point, no correction for the transformer ratio is required. The primary impedance must be only re-cal-
culated to secondary quantities:
d) ZONE 1 SETTING IN APPLICATION (B)
As the transformer is not located between the potential source and the reach point for Z1, the reach impedance must not
include the positive-sequence impedance of the transformer. The CTs are located on the other side of the transformer, thus
transformer ratio must be included:
e) ZONE 4 SETTING IN APPLICATION (B)
As the transformer is located between the potential source and the reach point for Zone 4, the reach impedance must
include the positive-sequence impedance of the transformer. The VTs are located on the other side of the transformer, thus
transformer ratio must be included:
Given the following for the system shown in the previous section:
Z = 30 Ω ∠85° (intended reach of Zone 1)
X
Z = 0.06 Ω ∠88° (intended reach of Zone 4)
H
n = 8000:5 = 1600 (located at H)
CT
n = 315000:120 = 2625 (located at X)
VT
Transformer: 13.8/315 kV, 150 MVA, 10%, delta/wye, 315 kV side lagging 30°
Transformer impedance:
The Zone 1 settings are:
PHS DIST Z1 REACH:
PHS DIST Z1 RCA:
PHS DIST Z1 XMFR VOL CONNECTION:
PHS DIST Z1 XMFR CUR CONNECTION:
The Zone 4 settings are:
PHS DIST Z4 REACH:
PHS DIST Z4 RCA:
10
PHS DIST Z4 XMFR VOL CONNECTION:
PHS DIST Z4 XMFR CUR CONNECTION:
10-14
Z
4
Z
=
1
( (
Z Z at H
=
4 T
10
--------- -
Z ( at H )
=
T
100
13.8
× ---------- -
Z 30
=
1
315
"0.80"
"85"
"None"
"Dy1"
Z ( 0.127 90° ∠ 0.006 88° ∠
= +
4
"2.60"
"89"
"Yd11"
"None"
D60 Line Distance Protection System
n
CT
× ---------
Z
=
H
n
VT
V n
 
H CT
------ - ---------
Z × ×
 
X
V n
X VT
V n
 
X CT
) ) × ------ - × ---------
Z
+
 
H
V n
H VT
2
( 13.8 )
------------------ -
× 0.127Ω 90° ∠
=
150
1600
× ------------ - ∠
0.8011Ω 85°
=
2625
315 1600
---------- - ------------ -
) × × 2.601Ω 89.4°
=
13.8 2625
10 APPLICATION OF SETTINGS

10.5.2 EXAMPLE

∠
(EQ 10.5)
(EQ 10.6)
(EQ 10.7)
(EQ 10.8)
(EQ 10.9)
(EQ 10.10)
GE Multilin