Unbalance Test - GE 469 Manual

7 TESTING
The 469 measures the ratio of negative sequence current ( I
percent is used as the unbalance level when motor load exceeds FLA. When the average phase current is
below FLA, the unbalance value is derated to prevent nuisance tripping as positive sequence current is much
smaller and negative sequence current remains relatively constant. A sample calculation is given below.
I
2
--- -
The derating formula is:
I
1
Ic=1000A
@ 247°
Figure 7–2: THREE PHASE EXAMPLE FOR UNBALANCE CALCULATION
Symmetrical component analysis of vectors using the mathematic vector convention yields a ratio of negative
sequence current to positive sequence current as shown:
1
2
( )
-- - I a I aI
+ +
I
a b c
3
2
=
--- - ------------------------------------------- -
I 1
2
( )
1 -- - I aI a I
+ +
a b c
3
∠ ( ∠
I
780 0° + 1 120°
2
--- - = ------------------------------------------------------------------------------------------------------------------------------------------------------------------- -
I
∠ ( ∠
780 0° + 1 120°
1
∠
I
780 0° + 1000 127°
2
--- - = ----------------------------------------------------------------------------------------------- -
∠
I
780 0° + 1000 7°
1
If FLA = 1000, then:
780 1000 1000
+ +
I ------------------------------------------------ - A
=
avg
3
I
and since = 926.7 A
avg
469 Unbalance = – 0.1532
GE Power Management
I
× avg ×
---------- - 100%
FLA
POWER SYSTEM
VECTOR
CONVENTION
Ia=780A
@ 0°
Ib=1000A
@ 113°
∠
where a = 1 120° =
2
) ( ∠ ) ( ∠
1000 – 113° + 1 120°
) 1000 ( ∠ ) ( ∠
– 113° + 1 120°
∠ ∠
+ 1000 233° 780 601.8
= -------------------------------------------------------------------------------------------------- -
∠ ∠
+ 1000 353° 780
926.7 A
=
<
1000 = FLA the 469 unbalance is:
926.7
× ×
-------------- - 100% = 14.2%
1000
469 Motor Management Relay
7.3 ADDITIONAL FUNCTIONAL TESTING
) to positive sequence current ( I
2
MATHEMATICAL
VECTOR
CONVENTION
Ic=1000A
@ 113°
Ib=1000A
@ -113°
– 0.5 + j 0.886
) 1000 133° ( ∠ )
2
) ( ∠ )
1000 133°
j 798.6 601.8
– + –
j 121.9 992.5 j 121.9
+ 992.5 + +

7.3.3 UNBALANCE TEST

). This value as a
1
Ia=780A
@ 0°
j 798.6
– – 423.6
= ----------------- - = – 0.1532
+ 2765
7-11
7