Rockwell Automation 865 Differential Protection Relay User Manual
Page 151
Applications
8-9
865-UM001A-EN-P – July 2009
Formula to solve needed CT burden
By replacing the complex power terms with corresponding
resistances in equation 8.7, we get
Equation
8.9
L
W
CT
N
CT
ALF
A
R
R
R
R
R
k
k
+
+
+
=
where the nominal burden resistance is
2
sec
NCT
N
N
I
S
R =
R
CT
= Winding resistance (See Figure 8.3)
R
W
= Wiring resistance (from CT to the relay and back)
R
L
= Resistance of the protection relay input
S
N
= Nominal burden of the CT
I
NCTsec
= Nominal secondary current of the CT
By solving S
N
and substituting k
A
according Equation 8.9 , we get
Equation
8.10
(
)
2
sec
NCT
CT
L
W
CT
NCT
ALF
NTra
SET
N
I
R
R
R
R
I
k
I
cI
S
⎥
⎦
⎤
⎢
⎣
⎡
−
+
+
>
Example
1
Transformer:
16 MVA YNd11 Z
k
= 10%
110 kV / 21 kV (84 A / 440 A)
CTs on HW side:
100/5 5P10
Winding resistance R
CT
= 0.07 Ω
(RCT depends on the CT type, INCT and power rating. Let's say
that the selected CT type, 100 A and an initial guess of 15 VA yields
to 0.07 Ω.)
Safety factor c = 4.
(Transformer differential, earthed Y.)