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:
• 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
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 (Equation 51)
∠ =
=
=V
ForwardX
ForwardR
ts
arctan arctan
.
.
59 33
8 19
82..14°
EQUATION14058 V1 EN (Equation 52)
frequency of V
ts
= 50 Hz
I
50s
= 0 A
Section 12 1MRK502053-UEN B
Testing functionality by secondary injection
134 Generator protection REG670 2.0 IEC
Commissioning manual