I I
I
I
A
s
CTs
CTp
50 50
10459
1
9000
1 162= × = × = .
EQUATION14059 V1 EN-US (Equation 49)
∠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 50)
∠I
tfs
= 180º
frequency of I
tfs
= 49.5 Hz
Expected result: start of the protection function and trip in zone 2 when trip conditions are fulfilled.
The trajectory of the impedance traverses the lens characteristic in zone 1
GUID-12F82DD8-A4CE-48C9-944B-0CF3B3F6C9F9 v1
Preliminary steady state test at 50 Hz
GUID-AA953F5F-2389-4F05-8BBD-D8DC9AF3B0E4 v1
• Go to Main menu /Test /Function status /Impedance protection /OutOfStep(78,Ucos) /
OOSPPAM(78,Ucos):1 /Outputs to check the available service values of the function block
OOSPPAM.
• Apply 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 51)
∠ =
=
=V
ForwardX
ForwardR
ts
arctan arctan
.
.
59 33
8 19
82..14°
EQUATION14058 V1 EN-US (Equation 52)
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 53)
∠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 54)
∠I
tfs
= 0º
frequency of I
tf
= 50 Hz
• Check that the service values (VOLTAGE, CURRENT, R(%), X(%)) are according to the injected
quantities and that ROTORANG is close to 3.14 rad. For this particular injection the service values
are:
Section 11 1MRK 504 165-UUS Rev. J
Testing functionality by secondary injection
154 Transformer protection RET670
Commissioning manual
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