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I.L. 40-201.9
Please note that the information applies to LP and LN as well.
f : = 60
------ -
asin
1
t r ( ):
=
----------
----------------------
2 f ⋅
f π ⋅
i:
=
0..48
π
⋅
X
:
=
sin
i
----- -
i
24
π
π
i .
y
:
=
sin
----- -
+
-- -
i
24
6
K-4
⋅
r:
=
0 0.05..1
,
r
r
2
0
0.05
>>
0.1
0.15
0.2
0.25
0.3
0.35
.04
0.45
0.5
0.55
0.6
0.65
0.7
0.75
0.8
0.85
0.9
0.95
1
1.05
1.1
1.15
1.2
1.25
1.3
1.35
1.4
Figure 3. Square Wave Duration - 60 Hz
Power Automation and Protection Division
2
t(r)•1000
Where r = IKEY/IT [rms]
8.33
i.e. IKEY = 0.5,
8.15
<<[msec]
7.96
7.77
IKEY time = t (r) *1000
7.58
IT computation
7.39
7.2
IT = C0xI0+C1xI1+C2xI2
7.01
Example:
6.81
6.62
Applying balanced 3-phase fault of
6.42
5A rms requires C1=1 to obtain IT
6.21
of 5A rms
60.1
5.8
5.59
5.37
5.15
4.90
4.67
4.43
4.17
3.89
IKEY
3.61
3.3
2.96
2.58
2.15
1.6
0.75
60 Hz
IT = 5 A
I< t(r) >I
i
i
REL 352 Version 1.00
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