Areva MiCOM P441, Technical Manual

Page: 276 / 636

P44x/EN AP/E33 Application Notes
Page 102/220 MiCOM P441/P442 & P444
3.1.12 Power Swing Band
Typically, the
∆
R and
∆
X band settings are both set between 10 - 30% of R3Ph. This gives
a secondary impedance between 0.6 and 1.8
Ω
. For convenience, 1.0
Ω
could be set.
The width of the power swing band is calculated as follows:
∆
R = 1.3
×
tan(
π × ∆
f
× ∆
t)
×
R
LOAD
Assuming that the load corresponds to 60° angles between sources and if the resistive reach
is set so that Rlim = R
LOAD
/2, the following is obtained:
∆
R = 0.032
× ∆
f
×
R
LOAD
To ensure that a power swing frequency of 5 Hz is detected, the following is obtained:
∆
R = 0.16
×
R
LOAD
Where:
∆
R width of the power swing detection band
∆
f power swing frequency (f
A
– f
B
)
Rlim resistive reach of the starting characteristic (=R3ph-R4ph)
Z network impedance corresponding to the sum of the reverse (Z4) and
forward (Z3) impedances
R
LOAD
load resistance
3.1.13 Current Reversal Guard
The current reversal guard timer available with POP schemes needs a non-zero setting
when the reach of the zone 2 elements is greater than 1.5 times the impedance of the
protected line. In this example, their reach is only 1.3 times the protected line impedance.
Therefore, current reversal guard logic does not need to be used and the recommended
settings for scheme timers are:
tREVERSAL GUARD = 0
Tp = 98ms (typical).
3.1.14 Instantaneous Overcurrent Protection
To provide parallel high-speed fault clearance to the distance protection, it is possible to use
the I>3 element as an instantaneous highset. It must be ensured that the element will only
respond to faults on the protected line. The worst case scenario for this is when only one of
the parallel lines is in service.
Two cases must be considered. The first case is a fault at Blue River substation with the
relay seeing fault current contribution via Green Valley. The second case is a fault at Green
Valley with the relay seeing fault current contribution via Blue River.
Case 1:
Source Impedance = 230
2
/ 5000 = 10.58
Ω
Line Impedance = 48.4
Ω
Fault current seen by relay = (230000 /
√
3) / (10.58 + 48.4)
= 2251A