Loss Of Excitation; Reverse Power - GE G60 Instruction Manual

9.1 SETTING EXAMPLE
For the example system, we have the following values:
The voltage supervision setting will be determined by a system study and may be disabled on either element if required. VT
fuse failure should supervise this element.
The choice of source is not critical for this application. The NEUTRL source is chosen for the following setting changes.
Make the following changes in EnerVista UR Setup or through the

1

LOSS OF EXCITATION

The reverse power element should be set at ½ the rated motoring power. The pickup is calculated as follows:
S
min
9
For the example system:
If the example system has delta-connected VTs instead of wye-connected VTs, and say the VT secondary setting is set to
115 V instead of 66 V, then the setting value of S
9-4
2
base kV CT ratio
-------------------------- - × -------------------- -
Z
=
b sec ( )
base MVA VT ratio
'(sec) ' Z ×
X X (sec)
=
d d b
X ( sec ) X × Z ( sec
=
d d b
( )
Z sec X
+
b d
------------------------------------------------ -
CENTER 1
=
2
(
Z sec
b
RADIUS 1 --------------------
=
2
PICKUP DELAY 1
X ( sec ) X '(sec)
+
d d
-------------------------------------------------
CENTER 2
=
2
(
X sec
d
-------------------- -
RADIUS 2
=
2
menu:
1 Rated Motoring Power (primary watts)
-- - × ------------------------------------------------------------------------------------------------------------------------------------------------------------------------- -
=
×
2 3 Phase CT Primary Phase VT Ratio Phase VT Secondary
×
1 22 10
-- - × -------------------------------------------------------------------- -
S
=
min
× ×
2 3 8000 A 157.5
min
1 22 10
S -- - × ----------------------------------------------------------------------- -
=
min
×
2 3 8000 A 157.5
G60 Generator Protection System
2
( )
18 kV 1600
----------------------------------- - × -------------- -
= =
211.765 MVA 157.5
0.216 15.54 Ω × 3.36 Ω
= =
) 1.967 × 15.54 Ω 30.57 Ω
= =
'(sec)
15.54Ω 3.36Ω
+
------------------------------------------ -
= =
2
)
15.54 Ω
--------------------- - 7.77 Ω
= =
2
0.06 seconds
=
30.57Ω 3.36Ω
+
------------------------------------------ - 16.97 Ω
= =
2
)
30.57 Ω
--------------------- - 15.28 Ω
= =
2

SETTINGS
× ×
6
W
0.044 pu
=
×
66 V
is calculated as follows:
6
× W
0.0253 pu
=
× ×
115 V
9 APPLICATION OF SETTINGS
9.1.7 LOSS OF EXCITATION
15.54 Ω
9.45 Ω

GROUPED ELEMENTS SETTING GROUP

9.1.8 REVERSE POWER

(EQ 9.7)
(EQ 9.8)
(EQ 9.9)
(EQ 9.10)
(EQ 9.11)
(EQ 9.12)
(EQ 9.13)
(EQ 9.14)
(EQ 9.15)
(EQ 9.16)
GE Multilin