GE 750 Instruction Manual page 153

S5 Protection
GE Multilin
Although the 750/760 is designed for feeder protection, it can provide Restricted
Earth Fault protection on transformers that do not have dedicated protection. To use
the 750/760 for this type of protection, a stabilizing resistor and possibly a non-
linear resistor will be required.
The inclusion of a stabilizing resistor encourages the circulating fault current to flow
via the magnetizing impedance of the saturated current transformer thus
minimizing spill current in the REF relay. A non-linear resistor will be required where
the voltage across the inputs would be greater than 2000 V. Refer to Restricted
Earth Fault Inputs on page 3–9 for the connections required to use the 750/760 to
perform Restricted Earth Fault protection.
To determine the appropriate value for the Stabilizing Resistor, use the following
equation:
R
s
where: R = resistance value of the stabilizing resistor,
S
V = voltage at which the 750/760 will operate
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
attached wire leads
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,
V
f
The peak value of this fault voltage would be:
V
P
If V is greater than 2000 V, then a non-linear resistor must be used.
P
Sample Application:
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.
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V I ⋅ ( R 2R )
+
s F CT L
= ----- - = ------------------------------------------
I I
s s
) V
K S
V 2V
=
K S
I ⋅ ( R 2R R )
= + +
f CT L S
2 2 ⋅ V ⋅ ( V V )
= –
k f K
750/760
Feeder Management Relay
(EQ 5.5)
= resistance of
L
is to 2 for stability. Thus,
(EQ 5.6)
(EQ 5.7)
(EQ 5.8)
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