GE G30 Instruction Manual page 249

5 SETTINGS
It may be suggested that phase relationship for the ACB sequence can be returned the transformer nameplate values by
connecting source phases A, B and C to transformer terminals A, C, and B respectively. Although this restores the name-
plate phase shifts, it causes incorrect identification of phases B and C within the relay, and is therefore not recommended.
All information presented in this manual is based on connecting the relay phase A, B and C terminals to the power system
phases A, B, and C respectively. The transformer types and phase relationships presented are for a system phase
sequence of ABC, in accordance with the standards for power transformers. Users with a system phase sequence of ACB
must determine the transformer type for this sequence.
If a power system with ACB rotation is connected to the Wye winding terminals 1, 2, and 3, respectively, from a Y/d30 trans-
former, select a Power Rotation setting of ACB into the relay and enter data for the Y/d330 transformer type.
e) MAGNITUDE COMPENSATION
Transformer protection presents problems in the application of current transformers. CTs should be matched to the current
rating of each transformer winding, so that normal current through the power transformer is equal on the secondary side of
the CT on different windings. However, because only standard CT ratios are available, this matching may not be exact.
In our example, the transformer has a voltage ratio of 220 kV / 69 kV (i.e. about 3.188 to 1) and a compensating CT ratio is
500 A to 1500 A (i.e. 1 to 3). Historically, this would have resulted in a steady state current at the differential relay. Interpos-
ing CTs or tapped relay windings were used to minimize this error.
The G30 automatically corrects for CT mismatch errors. All currents are magnitude compensated to be in units of the CTs
of one winding before the calculation of differential and restraint quantities.
The reference winding (w ) is the winding to which all currents are referred. This means that the differential and restraint
ref
currents will be in per unit of nominal of the CTs on the reference winding. This is important to know, because the settings of
the operate characteristic of the percent differential element (pickup, breakpoints 1 and 2) are entered in terms of the same
per unit of nominal.
The reference winding is chosen by the relay to be the winding which has the smallest margin of CT primary current with
respect to winding rated current, meaning that the CTs on the reference winding will most likely begin to saturate before
those on other windings with heavy through currents. The characteristics of the reference winding CTs determine how the
percent differential element operate characteristic should be set.
The G30 determines the reference winding as follows:
1. Calculate the rated current (I
Note: enter the self-cooled MVA rating for the P
2. Calculate the CT margin (I
3. Choose the winding with the lowest CT margin:
In our example, the reference winding is chosen as follows.
1. Calculate the rated current for windings 1 and 2:
P 1 [ ]
rated
I 1 [ ] ----------------------------------- -
=
rated
× 1 [ ]
3 V
nom
2. With these rated currents, calculate the CT margin for windings 1 and 2:
CT primary 1 [ ]
1 [ ] ------------------------------------- -
I
=
margin
I
rated
2 [ ] I < 1 [ ]
3. Since I
margin margin
The reference winding is shown in
.
REFERENCE WINDING
GE Multilin
) for each winding:
rated
P [ ] w
rated
[ ] ------------------------------------ -
I w
=
rated
3 × V [ ]
nom
rated
) for each winding:
margin
[ ]
CT primary w
-------------------------------------- - , where w
I
=
margin
[ ]
I w
rated
100 MVA
-------------------------------- 262.4 A
,
= =
×
3 220 kV
500 A
-------------------- -
1.91 I
,
= =
1 [ ]
262.4 A
, the reference winding w
ref

ACTUAL VALUES METERING
G30 Generator Protection System
,
, where w 1 2
=
w
setting.
1 2 ,
=
P 2 [ ]
rated
I 2 [ ] ----------------------------------- -
= =
rated
× 2 [ ]
3 V
nom
CT primary 2 [ ]
2 [ ] ------------------------------------- -
= =
margin
2 [ ]
I
rated
is winding 2.
 
TRANSFORMER
5.4 SYSTEM SETUP
(EQ 5.7)
(EQ 5.8)
100 MVA
---------------------------- - 836.7 A
(EQ 5.9)
=
×
3 69 kV
1500 A
-------------------- -
1.79
(EQ 5.10)
=
836.7 A
DIFFERENTIAL AND RESTRAINT
5-109
5
