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5.6 S5 PROTECTION
To determine the appropriate value for the Stabilizing Resistor, use the following equation:
where: R
= resistance value of the stabilizing resistor, V
S
I
= current flowing through the stabilizing resistor and the 750/760,
S
I
= maximum secondary fault current magnitude
F
R
= internal resistance of the current transformer, and R
CT
A non-linear resistor is recommended if the peak fault voltage may be above the relays maximum of 2000 V. The following
calculation is done to determine if a non-linear resistor is required. When required, this should be provided by the end-user.
It is assumed that the ratio of the CT kneepoint (V
Next, the voltage that would result from a fault must be determined, neglecting saturation,
The peak value of this fault voltage would be:
If V
is greater than 2000 V, then a non-linear resistor must be used.
P
SAMPLE APPLICATION:
5
The CTs used in this example are 3000/1, 10P10, 15 VA, and the transformer used in the example is an 11 kV / 400 V,
2000 kVA. At 10P10 the voltage at which the CT will saturate will be 10 x 15 = 150 V. An equivalent IEEE description for
this CT would be 3000/1, C150.
We have:
R
= 3.7 Ω
CT
R
= 0.954 Ω assuming 600 feet of #12 wire
L
X(%) = impedance of transformer = 7% = 0.07
The rated transformer current through wye windings is given as:
5-56
Courtesy of NationalSwitchgear.com
V
s
R
----- -
=
=
s
I
s
) V
K
S
V
K
V
I
⋅
(
R
=
f
f
V
2 2
=
P
Figure 5–21: RESTRICTED EARTH FAULT SAMPLE APPLICATION
750/760 Feeder Management Relay
⋅
(
)
I
R
2R
+
F
CT
L
----------------------------------------- -
I
s
= voltage at which the 750/760 will operate
S
= resistance of attached wire leads
L
is to 2 for stability. Thus,
2V
=
S
2R
R
)
+
+
CT
L
S
⋅
V
⋅
(
V
V
)
–
k
f
K
2000 kVA
I
--------------------------- -
=
=
P
3 400 V
⋅
5 SETPOINTS
2887 A
GE Multilin
(EQ 5.5)
(EQ 5.6)
(EQ 5.7)
(EQ 5.8)
(EQ 5.9)
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