Filter
Top Products
⎯ 26 ⎯
6 F 2 S 0 8 5 7
Setting for phase angle matching
The phase angle difference between line currents on either side of the power transformer are
corrected by setting according to the hands of a clock and the transformer connections described in
IEC60076-1 as follows:
(When α-method is selected for [Phase matching])
If a winding is star-connected, set 1 (=star) for winding setting yd_p, yd_s, and yd_t. If
delta-connected, set 2 (=delta). Next, set the phase angle difference vec_s and vec_t from the
primary winding as a lagging angle winding expressed in hours. One hour corresponds to lagging
by thirty degrees.
Note: In the case of a zigzag connected winding, set 2 (=delta).
Example: Setting for star/star/delta transformer.
IEC60076-1
Setting
yd_p
yd_s
vec_s yd_t vec_t
Y y 0 d 11 1 1 0 2 11
yd_p: Because the primary winding is star-connected, set 1.
yd_s: Because the secondary winding is star-connected, set 1.
vec_s: Because the secondary winding is in phase with the primary winding, set 0.
yd_t: Because the tertiary winding is delta-connected, set 2.
vec_t: Because the tertiary winding lags the primary winding by 330°, set 11.
The settings for the transformer connections described in IEC60076-1 are listed in Table 2.2.5.2.
Note: The following calculation is performed in the relay for phase angle correction.
Table 2.2.5.1 Phase Angle Matching Calculation
O’clock Calculation Remarks
0
Ia’ = (2Ia − Ib − Ic)/ 3 Ib’ = (2Ib − Ic − Ia)/ 3
Ic’ = (2Ic − Ia − Ib)/ 3
1
Ia’ = (Ia – Ib)/
3 Ib’ = (Ib − Ic)/ 3 Ic’ = (Ic – Ia)/ 3
2
Ia’ = (Ia − 2Ib + Ic)/ 3 Ib’ = (Ia + Ib − 2Ic)/ 3
Ic’ = (Ib + Ic −2Ia)/ 3
3
Ia’ = (Ic − Ib)/
3 Ib’ = (Ia – Ic)/ 3 Ic’ = (Ib − Ia)/ 3
4
Ia’ = (2Ic − Ia − Ib)/ 3 Ib’ = (2Ia − Ib − Ic)/ 3
Ic’ = (2Ib − Ia − Ic)/ 3
5
Ia’ = (Ic – Ia)/
3 Ib’ = (Ia – Ib)/ 3 Ic’ = (Ib – Ic)/ 3
6
Ia’ = (Ib + Ic −2Ia)/ 3 Ib’ = (Ia − 2Ib + Ic)/ 3
Ic’ = (Ia + Ib − 2Ic)/ 3
7
Ia’ = (Ib − Ia)/
3 Ib’ = (Ic − Ib)/ 3 Ic’ = (Ia – Ic)/ 3
8
Ia’ = (2Ib − Ia − Ic)/ 3 Ib’ = (2Ic − Ia − Ib)/ 3
Ic’ = (2Ia − Ib − Ic)/ 3
9
Ia’ = (Ib – Ic)/
3 Ib’ = (Ic – Ia)/ 3 Ic’ = (Ia – Ib)/ 3
10
Ia’ = (Ia + Ib − 2Ic)/ 3 Ib’ = (Ib + Ic −2Ia)/ 3
Ic’ = (Ia − 2Ib + Ic)/ 3
11
Ia’ = (Ia – Ic)/
3 Ib’ = (Ib − Ia)/ 3 Ic’ = (Ic − Ib)/ 3
0
1
2
3
4
5
6
7
8
9
10
11
Setting value
Tertiary
Secondary
Primary