GE 750 User Manual

12-
38
750/760 Feeder Management Relay GE Power Management
12.5 SENSITIVE GROUND CURRENT 12 S5 PROTECTION
12
c) 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 / 400V, 2000kVA
. 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.
Figure 12–17: SAMPLE APPLICATION
R
CT
= 3.7
Ω
R
L
= 0.954
Ω
assuming 600 feet of #12 wire
X ( % ) = impedance of transformer = 7% = 0.07
Rated transformer current through wye windings:
Maximum fault current is:
Therefore, the secondary full load current is:
and the maximum secondary fault current is:
A
V
K
/ V
S
ratio of 2 is assumed to ensure operation.
V
S
= I
f
( R
CT
+ 2 R
L
) = 77.05 V
V
K
= 2 V
S
= 154.1 V
I
P
2000 kVA
3400 V
⋅
----------------------------
2887 A
==
I
MAXf
I
P
X %
()
--------------
2887 A
0.07
------------------- 41243 A
== =
I
SFLC
2887 A
3000
------------------- 0.962 A
==
I
Smax
0.962 A
0.07
--------------------- 13.74 A
I
f
===

GE 750 Specifications

General

Brand	GE
Model	750
Category	Relays
Language	English