Calculation Example 2 - GE L30 Instruction Manual
Line current differential
system, ur series
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10.1 CT REQUIREMENTS
3.
This gives a total burden of:
This is less than the allowed 3 Ω, which is OK.
4.
The following procedure verifies the kneepoint voltage.
1.
The maximum voltage available from the
2.
The system X/R ratio
3.
The CT voltage for maximum phase fault is:
14000 A
V
---------------------------------- -
=
ratio of 300:1
4.
The CT voltage for maximum ground fault is:
12000 A
---------------------------------- -
V
=
ratio of 300:1
5.
The CT will provide acceptable performance in this application.
To check the performance of an IEC CT of class 5P20, 15 VA, ratio 1500:5 A, assume the following values:
•
maximum I
= 14 000 A
fp
•
maximum I
= 12 000 A
fg
•
impedance angle of source and line = 78°
•
CT secondary leads are 75 m of AWG 10.
The IEC rating requires the CT deliver up to 20 times the rated secondary current without exceeding a maximum ratio error
of 5%, to a burden of:
The total Burden = R
The following procedure verifies the kneepoint voltage.
1.
The maximum voltage available from the
2.
The system X/R ratio
3.
The CT voltage for maximum phase fault is:
14000 A
---------------------------------- -
V
=
ratio of 300:1
4.
The CT voltage for maximum ground fault is:
12000 A
V
---------------------------------- -
=
ratio of 300:1
10
5.
The CT will provide acceptable performance in this application.
10-2
0.75 Ω
R
=
CT
0.2 VA
0.008 Ω
R
=
----------------- -
=
r
(
)
2
5 A
3.75 Ω
×
×
------------------- -
R
=
2 75 m
L
1000 m
Total Burden
R
R
R
=
+
+
CT
r
CT
=
tan
78°
=
4.71
.
×
(
)
×
(
4.71
1
0.75
+
+
×
(
)
×
(
4.71
+
1
0.75
+
15 VA
Burden
=
----------------
(
)
2
5 A
= 0.008 + 0.52 = 0.528 Ω, which is less than the allowed 0.6 Ω, which is OK.
+ R
r
l
CT
=
tan
78°
=
4.71
.
×
(
)
×
(
4.71
+
1
0.75
+
×
(
)
×
(
4.71
1
0.75
+
+
L30 Line Current Differential System
×
0.26 Ω
0.528 Ω
=
2
=
0.75 Ω
0.008 Ω
0.52 Ω
=
+
+
L
(
⁄
) 400
×
1500 2000
300 V
=
=
0.008 Ω
)
0.26
271.26 V (< 300 V, which is OK)
+
=
0.008 Ω
)
0.52
+
=
291.89 V (< 300 V, which is OK)
0.6 Ω at the 5 A rated current
=
(
⁄
) 400
×
=
1500 2000
=
300 V
0.008 Ω
)
0.26
+
=
271.26 V (< 300 V, which is OK)
0.008 Ω
)
0.52
291.89 V (< 300 V, which is OK)
+
=
10 APPLICATION OF SETTINGS
1.28 Ω
=
.
.
10.1.3 CALCULATION EXAMPLE 2
.
(EQ 10.3)
(EQ 10.4)
(EQ 10.5)
(EQ 10.6)
(EQ 10.7)
(EQ 10.8)
(EQ 10.9)
GE Multilin
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