20
The output current, per phase pilot, of the 300/1A current transformers is given by:
262.43 x 1
300
This should be adjusted by the interposing transformer so that 0.984A flows into the
relay.
The windings S1–S2, and S3–S4 must be used in series as output windings giving
(125 + 90) = 215 turns.
Primary turns (Tp) required, therefore are given by:
Is/√3
Ip
0.984 x 215
√3 x 0.875
say Tp = 140 turns
ie. connect each phase pilot from the 300/1A current transformers to primary
terminal nos. 2 and 6 (see Figure 11 and Table 1). Complete connections to the
interposing transformer as given below:
4.4.3 Three winding transformer
An example of the three winding transformer is shown in Figure 13. The voltage and
power rating of each winding is indicated. The current transformer ratios are chosen
as a function of the winding voltage and power rating of the particular windings with
which they are associated.
For transformer differential protection matching of current transformers is correct
when the CT ratios are determined on a basis of associated winding voltage only.
500kV Winding:
Based on 400 MVA the rated current is given:
400 x 10
6
√3 x 500 x 10
3
Secondary current from 500/5 current transformers:
462 x 5
500
The 500/5 star connected CTs are associated with the 500kV star winding, and thus
the transition to delta connected secondaries must be made by means of an
In = = 462A
Is = = 4.62A
Is = = 0.875A
∴ Primary turns (Tp) = = 139.6
Tp = x Ts
A
B
C
To
Relay
To 66kV
line
current
transformer
P1
P1
P1
A
B
C
N
S1
S1
S1
S2
S2
S2
S3
S3
S3
S4
S4
S4
P2
P2
P2
6
6
6
2
2
2
To Next Page
Loading...
