P.3.325
Date Code 20151029 Protection Manual SEL-411L Relay
Protection Functions
Protection Application Examples
For a ground fault, the total burden (Z
burden
) is 2 • 1.25 = 2.5 ,i.e., Z
burden
(2.5 < 2.7
We obtain similar results for a fault at S. Therefore, we need not worry about
CT saturation for this application.
Minimum Fault Levels. In calculating the minimum fault current, we
assume a resistive fault with load.
Equation 3.116 provides the total fault current for a resistive ground fault
during load conditions:
Equation 3.116
where d is the fault location in per unit, and R
f
is the fault resistance.
Equation 3.116 is applicable to any two-terminal line. For highly resistive
faults (R
f
>> system impedances), this equation reduces to the following:
Equation 3.117
The following equation provides the ratio of the remote and local currents
(I
AR
/I
AL
) for this fault.
Equation 3.118
Where C
1
and C
0
are the positive-sequence and zero-sequence current
distribution factors provided by the following:
Equation 3.119
For a fault in the middle of the line with R
f
= 100 , we can apply the values
from Table 3.166 to Equation 3.116–Equation 3.118.
I
1F
= 2631 A
We then divide by the base current, as in the following:
Equation 3.120
Z
burden
800 V
7.6 A 28 1+•
----------------------------------------
I
1F
3V
LN
I
LOAD
– Z1S d+ ZL1• • •
2 Z1S d Z1L• +Z1R 1 d–Z1L• +• •
Z1S Z1L Z1R++
-------------------------------------------------------------------------------------------------------------------
Z0S d Z0L• +Z0R 1 d–Z0L• +•
Z0S Z0L Z0R++
---------------------------------------------------------------------------------------------------------- 3R
f
• ++
-----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------=
K
P
21C
1
–1C
0
–3
I
LOAD
I
1F
----------------•
–+•
2C
1
C
0
3
I
Load
I
1F
------------• ++•
----------------------------------------------------------------------------------------------=
C
1
Z1R 1 d–+ Z1L•
Z1S Z1L Z1R++
----------------------------------------------------= C
0
Z0R 1 d–+ Z0L•
Z0S Z0L Z0R++
----------------------------------------------------=
2631 A
2000 A
------------------ 1.3=
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