• VOLTAGE = 1.29 kV
• CURRENT = 20918 A
• R = 0.89%
• X=6.42%
• ROTORANG = -3.04 rad
Note that these values identify a point inside the lens characteristic in zone 1, that is close to the
boundary between zone 1 and zone 2. The START is issued, but no TRIP is performed.
Execution of the dynamic test
GUID-98707DB3-9A1A-4458-9704-25F8FCB94B79 v1
The test may be performed by using two states of a sequence tool that is a basic feature of test sets.
• State 1: pre-test condition.
Steady voltage and current are applied in order to get a steady high impedance, that is a point in the
plane R-X which is far away from the lens characteristic. Define the following three-phase
symmetrical quantities (the phase angle is related to phase L1):
V V
V
V
V
ts t RZ
VT s
VT p
= × × = × × =0 9 0 9 1435
0 1
13 8
9 36
1
. .
.
.
.
,
,
,
EQUATION14066 V1 EN-US (Equation 55)
∠ =
=
=V
ForwardX
ForwardR
ts
arctan arctan
.
.
59 33
8 19
82..14°
EQUATION14058 V1 EN-US (Equation 56)
frequency of V
ts
= 50 Hz
I
50s
= 0 A
I
tfs
= 0 A
• State 2: main test step.
Define the following three-phase symmetrical quantities (the phase angle is related to phase L1):
V V
V
V
V
ts t RZ
VT s
VT p
= × × = × × =0 9 0 9 1435
0 1
13 8
9 36
1
. .
.
.
.
,
,
,
EQUATION14066 V1 EN-US (Equation 57)
∠ =
=
=V
ForwardX
ForwardR
ts
arctan arctan
.
.
59 33
8 19
82..14°
EQUATION14058 V1 EN-US (Equation 58)
frequency of V
ts
= 50 Hz
I I
I
I
A
s
CTs
CTp
50 50
10459
1
9000
1 162= × = × = .
EQUATION14059 V1 EN-US (Equation 59)
∠I
50s
= 0º
frequency of I
50s
= 50 Hz
It I
I
I
A
fs
CTs
CTp
tf= × = × =10459
1
9000
1 162.
EQUATION14062 V1 EN-US (Equation 60)
∠I
tfs
= 180º
frequency of I
tfs
= 49.5 Hz
1MRK 504 165-UUS Rev. J Section 11
Testing functionality by secondary injection
Transformer protection RET670 155
Commissioning manual
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