P54x/EN AP/La4 Application Notes
(AP) 6-52
MiCOM P543, P544, P545 & P546
AP
Considering the charging current on the circuit shown in Figure 16, the following calculation
is done:
Ich = 0.58 A ( 45 + 30 + 10 ) = 49.3 A
Susceptance = C = Ich/V
B = 49.3 A/( 275/ 3) kV primary
B = 0.31 x 10-3 S primary.
As the CT ratio on the three ends are different, it is necessary to apply a correction factor in
order to ensure secondary currents balance for all conditions:
To calculate the correction factor (CF), the same primary current must be used even this
current is not the expected load transfer for every branch. This will ensure secondary current
balance for all conditions.
A good approximation to calculate the correction factor, would be to use the primary rated
current of the smallest CT ratio as a base current. In this case we will use the primary rated
CT current at End C, in order to correct the secondary currents to the relay rated current:
For End A 1200 A
Secondary current =1200x 5/4000 = 1.5 A
CF = 5/1.5 = 3.33
For End B 1200 A
Secondary current =1200x 5/4000 = 1.5 A
CF = 5/1.5 = 3.33
For End C 1200 A prim = 5 A sec
Secondary current =1200x 5/1200 = 5 A
CF = 5/5 = 1
As mentioned on example 3.1.1, the following settings are recommended:
Is1 = 0.2 In
Is2 = 2 In
K1 = 30%
K2 = 100%
Therefore,
settings in secondary values for each end are:
Is1 = 0.2In =1 A
Is2 = 2In =10 A
Note: Settings shown in primary values at ends A and B appear different
compared with end C. This is not a problem as the currents at ends A
and B will be multiplied by the Correction Factor, when the differential
calculation is done. There would not be a requirement to alter settings
by CF as the relay works in secondary values.
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