It I
I
I
A
fs
CTs
CTp
tf= × = × =10459
1
9000
1 162.
EQUATION14062 V1 EN (Equation 46)
∠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
Preliminary steady state test at 50 Hz
• 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 (Equation 47)
∠ =
=
=V
ForwardX
ForwardR
ts
arctan arctan
.
.
59 33
8 19
82..14°
EQUATION14058 V1 EN (Equation 48)
frequency of V
ts
= 50 Hz
I I
I
I
A
s
CTs
CTp
50 50
10459
1
9000
1 162= × = × = .
EQUATION14059 V1 EN (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 (Equation 50)
∠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:
1MRK 504 165-UUS - Section 11
Testing functionality by secondary injection
Transformer protection RET670 2.2 ANSI 163
Commissioning manual