3.4
SEL-587Z Instruction Manual Date Code 20020903
Relay Elements
High-Impedance Protection
Balance two considerations when calculating the differential setting, for
example, a low-value setting for maximum sensitivity for in-zone faults
against a high-value setting to ensure stability for a saturated CT for through
faults. First, calculate the minimum setting for the latter condition. Figure 3.5
shows the equivalent circuit for a through fault with one CT saturating.
Figure 3.5 Equivalent Circuit for Through Fault.
Because the CT has saturated, the inductance (X5) falls to zero, effectively
reducing the CT impedance to R
CT
, the CT internal resistance. The following
equation expresses the voltage developed across the high-impedance element:
Equation 3.1
where
V
r
= voltage across the high-impedance element
I
F
= maximum external fault current
N = CT ratio
R
CT
= the CT secondary winding and lead resistance up to
the CT terminals
R
LEAD
= the one-way resistance of lead from junction points
to the most distant CT
P = 1 for three-phase faults and 2 for single-phase-to-
ground faults
Multiplier P is 2 for single-phase-to-ground faults because the fault current
flows through both of the faulted-phase CT cables. During a balanced three-
phase fault, the fault current in the return cable is 0; only the one-way cable
resistance is involved so the P multiplier is 1.
The voltage is directly proportional to the CT secondary resistance and the CT
secondary leads. Keep this voltage as low as possible by parallelling all the CT
secondaries in the switchyard as closely as possible to the current
transformers. Use the highest possible CT ratio, particularly for very high
fault currents.
Calculate the voltage developed across the high-impedance element:
Equation 3.2
87
R
2000
R
CT
R
LEAD
Saturated CT
X 5X 4X 3X 2
X 1
MOV
V
r
R
CT
PR
LEAD
• +()
I
F
N
----
• =
V
r
I
F
N
----
R
CT
R
LEAD
P• ()+()• =
50000
400
---------------
0.82 0.6 1•
()+()• =
178 V (nearest volt)=
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