Current Reversal Guard; Instantaneous Overcurrent Protection; Teed Feeder Protection; Apparent Impedance - GE MiCOM P40 Agile Technical Manual

Application Notes
MiCOM P40 Agile P442, P444
Where:
∆R
width of the power swing detection band
∆f
power swing frequency (f
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 resistance
LOAD

3.4.1.13 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 time-delays are:
tREVERSAL GUARD
'Aid Dist Delay' =

3.4.1.14 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
Line Impedance
Fault current seen by relay
Case 2:
Source Impedance
Line Impedance
Fault current seen by relay
The overcurrent setting must be in excess of 2251A. To provide an adequate safety margin a
setting ≥120% the minimum calculated should be chosen, say 2800A.
3.4.2

Teed feeder protection

The application of distance relays to three terminal lines is fairly common. However, several
problems arise when applying distance protection to three terminal lines.
3.4.2.1

Apparent Impedance

Figure 50 shows a typical three terminal line arrangement. For a fault at the busbars of
terminal B the impedance seen by a relay at terminal A will be equal to:
Z =
a
– f
A
=
98 ms (typical).
=
=
=
=
=
=
=
=
Z + Z + [ Z .(I /I ) ]
at bt bt c a
)
B
0
2
230 / 5000 =
48.4Ω
(230000 / √3) / (10.58 + 48.4)
2251A
2
230 / 3000 =
48.4 Ω
(230000 / √3) / (17.63 + 48.4)
2011A
P44x/EN AP/Hb6
(AP) 5-75
10.58 Ω
17.63 Ω