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Figure F.5 – Secondary current in case of maximum DC offset

# Figure F.5 – Secondary current in case of maximum DC offset

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– 124 –

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;

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.

– 126 –

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

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

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

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Figure F.5 – Secondary current in case of maximum DC offset

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