APPENDIX A
A-1
269 UNBALANCE EXAMPLE
The unbalance algorithm of the 269 makes 2 assump-
tions:
1) The three phase supply is a true three phase supply.
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.
Example. Phase magnitudes 3.9, 5, 5
Figure 1 Figure 2
From fig. 1: a=3.9 ∠ 0 b=5 ∠ -112.95 c=5 ∠ 112.95
Symmetrical component analysis of unbalance (the ratio of negative sequence current to positive sequence current)
in this example yields:
I
I
=
I
I
=
(I x I xI )
(I xI x I )
where x =1 120 = -0.5+ j 0.866
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
n
P
2
1
1
3
a
b c
1
3
a b
2
c
2
2
+ +
∠
=
∠ ∠ ∠ − + ∠ ∠
∠ ∠ ∠ ∠ ∠
=
∠ ∠ ∠
∠ ∠ ∠
=
( . ) ( . )
( . )
5 112 95 1 120 5 112 95
5 112 95
.99
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-
tively large amounts of unbalance. The 269 derates
unbalance below Full Load by multiplying the unbal-
ance by Iavg/IFLC.
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