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Power Automation and Protection Division
f : = 50
r
------ -
asin
2
1
t r ( ):
=
----------
----------------------
2 f ⋅
f π ⋅
i:
=
0..48
π
⋅
X
:
=
sin
i
----- -
i
24
π
π
i .
y
:
=
sin
----- -
+
-- -
i
24
6
.
REL 352 Version 1.00
⋅
r:
=
0 0.05..1
,
2
r
t(r)•1000
0
10
0.05
9.77
>>
0.1
9.55
0.15
9.32
0.2
9.1
0.25
8.87
0.3
8.64
0.35
8.41
0.4
8.17
0.45
7.94
0.5
7.7
0.55
7.46
0.6
7.21
0.65
6.96
0.7
6.7
0.75
6.44
0.8
6.17
0.85
5.9
0.9
5.61
0.95
5.31
1
5
1.05
4.67
1.1
4.33
1.15
3.95
1.2
3.55
1.25
3.1
1.3
2.58
1.35
1.93
1.4
0.9
Figure 4. Square Wave Duration - 50 Hz
Where r = IKEY/IT [rms]
i.e. IKEY = 0.1,
<<[msec]
IT = 1 A
IKEY time = t (r) *1000
IT computation
IT = C0xI0+C1xI1+C2xI2
Example:
Applying balanced 3-phase fault of
1A rms requires C1=1 to obtain IT
of 1A rms
IKEY
i
I.L. 40-201.9
50 Hz
I< t(r) >I
i
K-5
A
P
P
E
N
D
I
X
K
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