Calculating Settings For Four Step Residual Overcurrent Protection Zero Or Negative Sequence Direction Ef4Ptoc(51N_67N) - ABB REL650 Applications Manual

Section 3
REL650 setting examples
3.1.10
70
source impedance at line end A should be maximized (minimum short circuit power)
for this calculation. We get the phase current If
To assure that step 3 detects all short circuits on the adjacent lines out from the local
busbar a phase-phase short circuit is applied at fault points 9 and 10. The source
impedance at line end B should be maximized (minimum short circuit power) at this
calculation. Resulting phase current is I
Delay of I> should be set larger than the delay of distance protection zone 2
(normally 0.4 s), this means normally at least 0.8 s.
Step 3 should be set so that the maximum load current out on the line does not give
unwanted trip.
The maximum load current can be estimated as:
S
= =
I max
Load ,max
×
3 V
min
2142-ANSI V1 EN
The current setting of step 2 is, if possible, chosen as:
I
× Load , max £
1 2 . I
h step
GUID-C3C741A6-8E7B-4E1E-803A-8BC5CD82831C V1 EN
840
× £ £
1 2 . I
step 3
. 0 95
GUID-871F169F-1B5E-43C5-8628-1E010031999B V1 EN
where ƞ is the reset ratio of the overcurrent function. For the overcurrent function in
the IED, ƞ = 0.95.
In this case it is not possible to fulfil the requirement above. Priority is given to
selectivity, choosing setting 1100 A and accepting that the phase overcurrent
protection cannot act as remote back-up protection for the remote lines.
2. Set t3 to 0.8 s
Calculating settings for four step residual overcurrent
protection, zero or negative sequence direction
EF4PTOC(51N_67N)
The residual overcurrent protection is more difficult to set as the ground-fault current is
highly dependent of the network configuration in the power system. In order to achieve
setting that assure selective fault clearance a large number of calculations have to be made
with different fault locations, different switching states in the system and different
ground-fault types. Below one example of setting of four step residual overcurrent
protection for a line in a meshed solidly grounded system is given.
fault9,10,min
180
=
0.84 kA
× ×
3 0.9 138
£ ×
0 7 . min( I
3 fault , 7 , 8 , 9 10
× £
0 7 . 460 or 1061
1MRK 506 334-UUS A
= 0.46 kA
ault7,8,min
= 0.46 kA
)
min
£
I 322
step 3
Application manual
(Equation 21)
(Equation 22)
(Equation 23)