Unbalance Example - GE MULTILIN 269 MOTOR MANAGEMENT RELAY Series Instruction Manual

269 UNBALANCE EXAMPLE

The unbalance algorithm of the 269 makes 2 assump-
tions:
1) The three phase supply is a true three phase supply.
Example. Phase magnitudes 3.9, 5, 5
From fig. 1: a=3.9 ∠ 0
Symmetrical component analysis of unbalance (the ratio of negative sequence current to positive sequence current)
in this example yields:
+
2
1
(I x I
I I
3
n 2 a
= =
+
1
I I (I xI
P 1 3 a
∠
3.9 0 + (1 120)
=
∠
3. 9 0 + 1 120 (5 -112.95) + (1 120)
∠
3.9 0 + 5 127. 05 + 5 232.95
=
∠
3. 9 0 + 5 7.05 + 5 352.95
3. 9 - 3. 01+ j 3.99 - 3. 01- j 3
=
3.9 + 4. 96 + j 0. 61+ 4. 96 - j 0.61
-2.12
=
13.82
= − 0 1534
.
Therefore, unbalance is | -0.1534 | × 100% = 15.34 %
When a motor is lightly loaded however, the ratio of
negative sequence to positive sequence current will
increase as the positive sequence current becomes a
relatively small value. This may result in nuisance trips
even though a lightly loaded motor can withstand rela-
Figure 1
b=5 ∠ -112.95 c=5 ∠ 112.95
+
xI )
∠
b c
where x = 1 120 = -0. 5 + j 0.866
+
2
x I )
b c
∠ ∠ − + ∠
2
( 5 112 95 . )
∠ ∠
∠ ∠
∠ ∠
.99
2) There is no zero sequence current flowing (no
ground fault).
For simplicity, the 3φ may be drawn in the shape of a
triangle (three vectors must cancel each other). This
also makes it plain to see that no phasor could change
in magnitude without corresponding magnitudes and/or
phase angles changing. From magnitudes, phase an-
gles can always be derived using simple trigonometry.
∠
1 120 5 112 95 ( . )
∠ ∠
2
( 5 112 95 . )
tively large amounts of unbalance. The 269 derates
unbalance below Full Load by multiplying the unbal-
ance by Iavg/IFLC.
APPENDIX A
Figure 2
A-1