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 69)
∠I
tfs
= 180º
frequency of I
tfs
= 50.5 Hz
Expected result: start of the protection function and trip in zone 1 when trip conditions are
fulfilled.
The trajectory of the impedance does not enter the lens characteristic
GUID-189FECED-12A7-4A41-BCCF-D93C0FF9FA41 v1
Preliminary steady state test at 50 Hz
GUID-448D5357-6CC2-438E-83CA-25FD381B7DFF 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 RvsZ
VT s
VT p
= × × = × × =1 1 1 1 5899
0 1
13 8
47 02. .
.
.
.
,
,
,
EQUATION14069 V1 EN-US (Equation 70)
29.60
arctan 180 arctan 180 90.56
0.29
ts
ReverseX
V
ReverseR
Ð = - ° = - ° = - °
æ ö æ ö
ç ÷ ç ÷
è ø è ø
EQUATION14068 V1 EN-US (Equation 71)
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 72)
∠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 73)
∠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 0 rad. For this particular injection the service values
are:
• VOLTAGE= 6.49 kV
• CURRENT= 20918 A
• R= –0.32%
• X=–32.57%
• ROTORANG= 0.08 rad
Note that these values identify a point outside the lens characteristic, even if it is close to the
point SE. Neither START nor TRIP is issued.
Execution of the dynamic test
GUID-31EF4F76-8FC8-4952-896C-B3A8A5FD635A v1
The test may be performed by using two states of a sequence tool that is a basic feature of test sets.
1MRK 505 378-UEN Rev. K Section 11
Testing functionality by secondary injection
Line differential protection RED670 147
Commissioning manual
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