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Example of the Logarithmic Interpolation
Alarm 2 current is set to 6 kA. What is the maximum number of
operations according Table 4.13.
The current 6 kA lies between points 2 and 3 in the table. That
gives value for the index k. Using
k
=
2
C
=
10000
k
C
=
80
k+1
I
=
31 kA
k+1
I
=
1.25 kA
k
and the Equation 4.2 and Equation 4.3, the device calculates
10000
ln
80
=
=
n
. 1
5038
31000
ln
1250
=
⋅
=
. 1
5038
a
10000
1250
Using
Equation the device gets the number of permitted
operations for current 6 kA.
⋅
6
454
10
=
=
C
945
. 1
5038
6000
Thus the maximum number of current breaking at 6 kA is 945.
This can be verified with the original breaker curve in Figure
4.4. Indeed, the figure shows that at 6 kA the operation count is
between 900 and 1000. A useful alarm level for operation-left,
could be in this case for example 50 being about five per cent of
the maximum.
Example of Operation Counter Decrementing when the CB is
breaking a current
Alarm2 is set to 6 kA. CBFP is supervising trip relay T1 and trip
signal of an overcurrent stage detecting a two phase fault is
connected to this trip relay T1. The interrupted phase currents
are 12.5 kA, 12.5 kA and 1.5 kA. How much are Alarm2
counters decremented?
Using Equation 4.1 and values n and a from the previous example,
the device gets the number of permitted operation at 10 kA.
⋅
6
454
10
=
=
C
313
10
kA
. 1
5038
12500
At alarm level 2, 6 kA, the corresponding number of operations
is calculated according
Supporting Functions
⋅
6
454
10
4-15
857-UM001A-EN-P – July 2009
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