Manual

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
LWCT
NCT
ALFA
RRR
RR
kk
++
+
=
where the nominal burden resistance is
2
secNCT
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
secNCTCTLWCT
NCTALF
NTraSET
N
IRRRR
Ik
IcI
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.)