It
I
I
I
A
fs
CTs
CTp
tf
=
×
=
×
=
10459
1
9000
1 162
.
EQUATION14062 V1 EN-US
(Equation 36)
∠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.
12.4.11.3
Test of the boundary between zone 1 and zone 2, which is defined by the
parameter
ReachZ1
GUID-BE180E07-7D4B-4842-81FB-244DE8897430 v1
The trajectory of the impedance traverses the lens characteristic in zone 2
GUID-17F6AC8D-1F57-4245-A61D-776D469DD153 v1
Preliminary steady state test at 50 Hz
GUID-3D42E43D-873E-4271-9AAF-3B2789FF3B17 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
=
×
×
=
×
×
=
1 1
1 1 1435
0 1
13 8
11 44
1
.
.
.
.
.
,
,
,
EQUATION14065 V1 EN-US
(Equation 37)
∠
=
=
=
V
ForwardX
ForwardR
ts
arctan
arctan
.
.
59 33
8 19
82..14
°
EQUATION14058 V1 EN-US
(Equation 38)
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 39)
∠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 40)
∠I
tfs
= 0º
frequency of I
tfs
= 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:
•
VOLTAGE = 1.58 kV
•
CURRENT = 20918 A
•
R = 1.08%
•
X = 7.85%
•
ROTORANG = -3.04 rad
1MRK 502 067-UEN B
Section 12
Testing functionality by secondary injection
139
Commissioning manual
