If there is a mismatch between CTs at line ends, then the lowest primary CT rated current should be used as a
reference current for p.u. calculations (assuming that the load current cannot continuously exceed this value). This
means that the recommended settings Phase Is1 = 0.2 pu is equal to 0.2*(the lowest primary CT rated value). The
same consideration applies for other current settings such as Phase Is2.
13.2 SENSITIVITY UNDER HEAVY LOADS
The sensitivity of the phase current differential protection is governed by its settings and also the magnitude of
load current in the system. For a three-ended system, with devices LOCAL, REMOTE1, and REMOTE2, the following
applies:
|I
di
| = |(Ī
LOCAL
+ Ī
REMOTE1
+ Ī
REMOTE2
)|
|I
bias
|= 0.5 (|I
LOCAL
| + |I
REMOTE1
| + |I
REMOTE2
|)
Assume a load current of I
L
flowing from end LOCAL to REMOTE1 and REMOTE2. Assume also a high resistance
single-end fault of current I
F
, being fed from end LOCAL. In the worst case, we can also assume I
F
to be in phase
with I
L
:
I
LOCAL
= I
L
+ I
F
I
REMOTE1
= -y*I
L
where 0<y<1
I
REMOTE2
= - (1-y) I
L
|I
d
i
|= |I
F
|
|I
bias
| = |I
L
| + 0.5 |I
F
|
Phase current differential protection sensitivity when |I
bias
| < Phase Is2:
The phase current differential protection would operate if |I
diff
| > Phase k1 |I
bias
| + Phase Is1
therefore:
|I
F
| > (Phase k1 |I
L
| + Phase Is1) / (1 - 0.5 Phase k1)
For Phase Is1 = 0.2 pu, Phase k1 = 30% and Phase Is2 =2.0 pu, then
● for |I
L
| = 1.0 pu, the phase current differential protection would operate if |I
F
| > 0.59 pu
● for |I
L
| = 1.59 pu, the phase current differential protection would operate if |I
F
| > 0.80 pu
If |I
F
| = 0.80 pu and |I| = 1.59 pu, then |I
bias
| = 1.99 pu, which reaches the limit of the low percentage bias curve.
Phase current differential protection sensitivity when |
bias
| > Phase Is2:
The phase current differential protection would operate if |I
diff
| > Phase k2 |I
bias
| - (Phase k2 - Phase k1) Phase Is2 +
Phase Is1
therefore:
|I
F
| > (Phase k2 |I
L
| - (Phase k2 - Phase k1) Phase Is2 + Phase Is1) / (1 - 0.5 Phase k2)
For Phase Is1= 0.2 pu, Phase k1 = 30%, Phase Is2 = 2.0 pu and Phase k2 = 100%, then,
● for |I
L
| = 2.0 pu, the phase current differential protection would operate if |I
F
| > 1.6 pu
● for |I
L
| = 2.5 pu, the phase current differential protection would operate if |I
F
| > 2.6 pu
Fault resistance coverage
Assuming the fault resistance R
F
is much higher than the line impedance and source impedance, then for a 33 kV
system and 400/1 CT:
Chapter 6 - Current Differential Protection P543i/P545i
126 P54x1i-TM-EN-1
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