Figure F.5 – Secondary current in case of maximum DC offset
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Lm =
IEC 60255-121:2014 © IEC 2014
0,5 ⋅ U sat
I 70 – I 20 ⋅ 2πf
(
)
(F.2)
where U sat is the saturation voltage where the curve is practically horizontal and I 70 and I 20
are the exciting current at 70 % and 20 % of U sat respectively.
If the magnetization curve is based on tests at rated frequency the losses have some
influence. If we assume that the phase angle of Z m is 45° the magnetizing reactance will be
X m = 2 ⋅ Zm and the magnetizing inductance can be estimated as follows:
Lm =
(I
2 ⋅ 0,5 ⋅ U sat
70
)
– I 20 ⋅ 2πf
(F.3)
The secondary time constant for a high remanence type CT is generally a few seconds. It is
important to verify that the CT model does not have a much smaller secondary time constant.
If it should have a value in the same order of magnitude as the primary time constant it may
have a major influence on the tests. In such cases the model should be more close to a CT of
class TPZ. The transient behaviour of the CT will be changed and the DC component in the
current will be damped very fast.
If the magnetization curve of the basic CT is recorded at rated frequency the magnetizing
inductance can be estimated to 56,3 H according to Equation (F.3) and the secondary time
constant can be calculated as follows:
Ts =
Lm
56,3
=
= 3,8 s
Rct + Rb 5 + 10
If the burden is decreased below the rated burden then the secondary time constant will
increase and vice versa. Therefore increasing the burden should not be a method to force the
CT into saturation. In such cases there is a risk that the transient performance of the CT
changes and can influence the tests.
When the size of the CT is changed, the magnetizing curve has to be scaled accordingly. For
example if the CT size is increased to a K td factor equal to 2, the voltage and current values
of the magnetization curve should be doubled. In this way, the magnetizing inductance will
remain the same which means that the secondary time constant also will be unchanged.
IEC 60255-121:2014 © IEC 2014
– 125 –
Annex G
(informative)
Informative guide for dimensioning of CTs for distance protection
G.1
General
This annex describes the practical procedure when dimensioning CTs for distance protection.
Two different cases are presented here. The first case describes a method to verify if a given
CT fulfils the requirements of a specific application. The other case describes a method to
provide the CT manufacturer with necessary CT data for the application. We will show one
example for each case.
In both cases we consider the same distance relay. In these examples it is assumed that the
relay manufacturer has combined the requirements for different fault positions and specified
the CT requirements with Equations (G.1) and (G.2). The CTs shall have a rated equivalent
limiting secondary e.m.f. E al that is larger than or equal to the required rated equivalent
limiting secondary e.m.f. E alreqC for close-in fault and E alreqZone1 for zone 1 fault as shown
below:
Eal ≥ Ealreqc =
Eal ≥ EalreqZone1 =
(
I fc
⋅ K totC ⋅ I sr Rct + Rw + Raddbu
Ipr
)
I fZone1
⋅ K totZone1 ⋅ I sr (R ct +R w + Raddbu )
Ipr
(G.1)
(G.2)
where
I fc
is the maximum primary fault current through the CT in case of close-in forward
and reverse faults; both three-phase faults and phase to earth faults shall be
considered;
I fZone1
is the maximum primary fault current through the CT in case of a fault at the end
of zone 1; both three-phase faults and phase to earth faults shall be considered;
I pr
is the CT rated primary current;
I sr
is the CT rated secondary current;
R ct
is the CT secondary winding resistance;
Rw
is the resistance of the secondary wire; for phase to earth faults the loop
resistance containing the phase and neutral wires (double length) shall be used
and for three-phase faults the phase wire (single length) can be used;
R addbu
is the total additional burden from the distance relay and any other relays
connected to the same CT core;
K totC
is the necessary total over-dimensioning factor for close-in forward and reverse
faults;
K totC
is 2 for the primary time constant T p ≤ 50 ms;
is 3 for the primary time constant T p > 50 ms;
is the necessary total over-dimensioning factor for zone 1 faults;
K totC
K totZone1
K totZone1
K totZone1
is 4 for the primary time constant T p ≤ 30 ms;
is 7 for the primary time constant T p > 30 ms.
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IEC 60255-121:2014 © IEC 2014
The values of the over-dimensioning factors K totC and K totZone1 are product specific and
provided as an example. The manufacturer of the distance protection will supply these
factors.
We also assume that the secondary wire and additional burden are the same for the two
examples. The resistance of the secondary wires can be calculated with the following
expression:
Rw = ρ ⋅
I
Ω
A
In our examples the single length of the secondary wire is 200 m and the cross-section area is
2
2,5 mm 2 . The resistivity for copper at 75 °C is 0,021 Ω mm /m. With this value the R w = 1,7 Ω.
The total additional burden in our example is 0,3 Ω.
G.2
Example 1
Verify that the CT fulfils the requirements for the distance protection in the following
application shown in Figure G.1. Zone 1 is 80 % of the line length.
Nominal system voltage 110 kV, 50 Hz
A IIff
ZA1s == 0,318
0,318 ++ j8,0
j8,0 Ω
Ω
ZA1s
TpA
80ms
ms
T
pA ==80
ZA0s == 0,5
0,5 ZA1s
ZA1s
ZA0s
Z
Z<
B
Line
length
= 10
kmkm
Line
length
= 10
Z1line
= 0,35
+ j4,0
ΩΩ
Z1line
= 0,35
+ j4,0
Z0line
4 Z1
line
Z0line
= 4= Z1
line
ZB1s
ZB1s= =0,095
0,095+ +j3,0
j3,0ΩΩ
TpB
100ms
ms
TpB==100
ZB0s
ZB0s= =ZB1s
ZB1s
IEC
0196/14
Figure G.1 – Distance relay example 1
The existing CTs have the following data: 1 000/1 A, TPX 30 VA, the rated symmetrical shortcircuit current factor K ssc = 10, the rated transient dimensioning factor K td = 2, the secondary
winding resistance R ct = 15 Ω and the rated burden R b = 30 Ω. (This CT is approximately the
same as a 5P20, 30 VA and R ct = 15 Ω.)
From the data the E al can be calculated:
Eal = K ssc ⋅ K td ⋅ Isn ⋅ (RCT + Rb ) = 10 ⋅ 2 ⋅1 ⋅ (15 + 30 ) = 900 V
We shall know the currents flowing through the CT for faults at the different fault positions.
Calculations of the fault currents give the following results shown in Table G.1. The nominal
system voltage is 110 kV. The equivalent voltage source 121 kV has been used in the fault
current calculations.
IEC 60255-121:2014 © IEC 2014
– 127 –
Table G.1 – Fault currents
Fault position
Fault current through the CT [kA]
Three-phase fault
Phase to earth fault
Close-in forward fault, I fCfw
8,7
11,4
Close-in reverse fault, I fCrev
10,0
8,0
Zone 1 fault, I fZone1
6,2
5,3
The primary time constant is required in order to choose the total over-dimensioning factor to
be used in calculating the required rated equivalent limiting secondary e.m.f. We can see that
close-in forward phase to earth fault will be the dimensioning case for the close-in faults. The
primary time constant for close-in forward fault is 80 ms and we shall use the total overdimensioning factor K totC = 3 in Equation (G.1).
For the zone 1 fault we need to calculate the primary time constant for both three-phase fault
and phase to earth fault to be able to know what over-dimensioning factor K totZone1 to be
used. The positive sequence impedance is:
Z1Zone1 = ZA1s + 0,8 ⋅ Z1line = 0,318 + j8,0 + 0,8 ⋅ (0,35 + j4,0 ) = 0,598 + j11,2
The primary time constant for three-phase fault is:
TpZone1pp =
11,2
L1
X1
=
=
= 0,060 s
R1 ω ⋅ R1 100 ⋅ π ⋅ 0,598
For the phase to earth fault we shall consider the primary time constant for:
Zpe = 2 ⋅ Z1zone1 + Z 0 zone1 = 2(ZA1s + 0,8 ⋅ Z1line ) + (ZA0s + 0,8 ⋅ Z 0line ) =
2(0,598 + j11,2 ) + 0,5(0,318 + j8,0 ) + 0,8 ⋅ 4 ⋅ (0,35 + j4,0 ) = 2,475 + j39,2
The primary time constant for phase to earth fault is:
TpZone1pe =
Lpe
Xpe
39,2
=
=
= 0,050 s
Rpe ω ⋅ Rpe 100 ⋅ π ⋅ 2,475
Both primary time constants are >30 ms. Therefore we shall use the total over-dimensioning
factor K totZone1 = 7 in Equation (G.2).
We can now calculate the required rated equivalent limiting secondary e.m.f. according to
Equations (G.1) and (G.2). In this case we only need to consider forward phase to earth fault
for close-in faults. The R w in this case is the loop resistance with double length of the
secondary wire.
I
11400
EalreqC = fCfw ⋅ K totC ⋅ I sr (Rct + R w + Raddbu ) =
⋅ 3 ⋅ 1 ⋅ (15 + 2 ⋅ 1,7 + 0,3) = 640 V
Ipr
1000
For the zone 1 case we need to check both three-phase fault and phase to earth fault. The
fault current is higher for the three-phase fault but the burden is smaller as we only need to
consider single length of the secondary wire.
EalreqZone1 =
I fZone1pp
Ipr
⋅ K totZone1 ⋅ Isr (Rct + R w + Raddbu ) =
6200
⋅ 7 ⋅ 1 ⋅ (15 + 1,7 + 0,3) = 738 V
1000