GE 750 Instruction Manual page 201
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CHAPTER 5: SETPOINTS
750 FEEDER MANAGEMENT RELAY – INSTRUCTION MANUAL
FIGURE 5–21: Restricted Earth Fault Sample Application
We have: R
= 3.7 Ω, R
= 0.954 Ω (assuming 600 feet of #12 wire), and X(%) = impedance
CT
L
of transformer = 7% = 0.07
The rated transformer current through wye windings is given as:
and the maximum fault current is:
I
MAXf
Therefore, the secondary full load current is:
and the maximum secondary fault current is:
I
Smax
A V
/ V
ratio of 2 is assumed to ensure operation. As such,
K
S
V
= I
(R
+ 2R
) = 77.05 V and
S
f
CT
L
V
= 2V
= 154.1 V
K
S
To calculate the size of the stabilizing resistor, assume I
current, that is:
I
PICKUP
This means also (assuming 1% for CT magnetizing current):
I
RELAY PICKUP
and therefore:
2000 kVA
I
2887 A
------------------------- -
=
=
P
3 400 V
⋅
I
2887 A
P
41243 A
------------ -
-----------------
=
=
=
( )
X %
0.07
2887 A
I
0.962 A
=
-----------------
=
SFLC
3000
0.962 A
13.74 A
------------------ -
=
=
=
0.07
×
0.3
2887 A
866 A (Primary)
=
=
866 A
(
4
×
0.01
)
0.248 A
------------- -
=
–
=
3000
S5 PROTECTION
(EQ 5.10)
(EQ 5.11)
I
(EQ 5.12)
f
to be 30% rated transformer
PICKUP
(EQ 5.13)
I
(EQ 5.14)
=
S
(EQ 5.9)
5 - 75
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