|I
F
| = (Vph-n /R
F
)(1/CT ratio) pu = 47.63/R
F
pu
Based on the analysis above, the phase current differential protection detects a fault current in excess of 0.59 pu
with a load current of 1 pu flowing. The fault resistance is less than 47.63/0.59 = 81 ohms in this case.
With a short time overload current of 2.0 pu, the phase current differential protection can detect a fault resistance
of 47.63/1.6 = 30 ohms or lower.
13.3 PERMISSIVE INTERTRIPPING
The permissive intertrip (PIT) timer can be set between 0 and 200 ms. This time should be set to provide
discrimination with other protection devices. For example, if there is a genuine busbar fault, the time delay should
be set to allow busbar protection to clear the fault. A typical setting may be 100 to 150 ms.
13.4
CT RATIO CORRECTION SETTING GUIDELINES
If there is no mismatch between current transformers at line ends, Ph CT Corr’tion should be set to 1 in all devices
Note:
The following consideration includes references to currents used in Neutral Current Differential protection and protection of
feeders with In-zone transformers. If not applicable they may be ignored.
If the CTs at line ends have different primary ratings, then one of them is considered as a reference (the one with
the lowest primary rating). Ph CT Corr’tion should be set to 1 in the device connected to the reference CT. For all
other devices Ph CT Corr’tion is calculated as follows:
Ph CT Corr'tion
IEDx
= (Phase CT Primary
IEDx
) / (Phase CT Primary
REFERENCE
)
Settings Phase Is1, Phase Is2, Id High Set, Diff Is1, Diff Is2 must be calculated as pu of the reference primary
rating, then converted into local primary (secondary) values. For setting Phase Is1 in IEDX, the equations would be:
Phase Is1 [p.u.] = (Phase Is1
IABSOLUTE PRI
[A]) / (Phase CT Primary
REFERENCE
[A])
Phase Is1
IEDx PRI.
[A] = (Phase Is1 [p.u.]) (Phase CT Primary
IEDx
)
Phase Is1
IEDx SEC.
[A] = (Phase Is1 [p.u.]) (Phase CT Sec'y
IEDx
)
The same considerations apply to settings Phase Is2, Id High Set, In Diff Is1, Diff Is2.
Example:
Assume that we have a three-ended application with line ends X, Y, Z (500/5, 800/5, 200/1) and Current Differential
settings in absolute primary values:
Phase Is1
ABSOLUTE PRI.
= 100A
Phase Is2
ABSOLUTE PRI.
= 1000A
Id High Set
ABSOLUTE PRI.
= 3000A
Di Is1
ABSOLUTE PRI.
= 50A
Di Is2
ABSOLUTE PRI.
= 1000A
Setting
End X End Y End Z
Phase CT Primary 500A 800A 200A
Phase CT Sec’y 5A 5A 1A
Ph CT Corr’tion 500/200 = 2.5 800/200 = 4 1 (reference)
Phase Is1 [p.u.] 100/200 = 0.5 100/200 = 0.5 100/200 = 0.5
P543i/P545i Chapter 6 - Current Differential Protection
P54x1i-TM-EN-1 127
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