CT Requirements En

7/27/2019 CT Requirements En

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Schuster, Werben, 2.5.2000

CT-requirements for Pilot-Wire Differential Relays

Rel. 1.1

Introduction..............................................................................................................................................................2

1 Test summary / conclusion ........................................................ ............................................................ ..........3

2 Test concept ................................................................ ....................................................... ..............................3

3 Simulation / Test................................................................ .......................................................... ....................4

3.1 Primary current ........................................................... ......................................................... ....................4

3.2 NETOMAC Data for 10P10 CT with 1A sum. CT..................................................................................53.3 NETOMAC Data for Class X CT with 5A sum. CT................................................................................6

3.4 Combination of both simulations .............................................................................................................73.5 7SD502, 7SD600 Algorithm with real CTs on both sides .......................................................................8

3.6 7SD502, 7SD600 Algorithm with real CTs on one and ideal on the other side....................................10

4 Summary.................... ........................................................... ..................................................... ....................12


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Introduction

The CT requirements given in the SIEMENS line differential protection manuals for 7SD60, 7SD502

are:

1)Npr

kd

IIn max with: Ikd max = max. steady-state through fault current

INpr = primary rated current of CT

If this requirement is fulfilled no saturation will occur under steady state through fault conditions on

both line sides.

2) 4/5

5/4

2

1

n

n7SD502 with: n operational ALF of the CTs

3) 3/4

4/3

2

1

n

n7SD60 with: n operational ALF of the CTs

If this requirement is fulfilled saturation of both CTs will not deviate substantially.

These requirements were developed for the classical differential protection relays (7SD24,

7SD20), which were designed 20 years ago.

These relays worked with instantaneous samples. They tripped as soon as the instantaneous samples of

the differential current exceeded the tripping characteristic.

Therefore it was important to avoid too much differential current caused by saturation.

Diagram 1 gives the following information:

- one non-saturated current in combination with an ideal saturated current that occurs after 5 ms and

lasts for 5ms

- differential- and stabilising current caused by this saturation, for instantaneous and filtered values

- instantaneous and filtered differential- and stabilising currents in the tripping characteristic as a

function of the time

The i_diff / i_rest-diagram (left-side bottom in Diagram 1) shows how instantaneous differential-

and restrain current is changing during of a period (10ms). After of a period saturation starts and

the operating point of the instantaneous values moves into the tripping area (after of a period the

operation point is at zero and the cycle starts again).

fundamentalfilter

filter

1

-1 T 2T

t

1

T 2Tt

idiff

irest

t2TT

2

idiff

1

1 2 irest

trip

restrain

trip

restrain

Irest

Idiff

1

1 2

1,5

Irest

0,3

Idiff

T

T


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Diagram 1

Numerical relays like the 7SD502, 7SD600 perform a filtering of the fundamental of the differential

current (sin-, cos-filter with 1 period filter length). The influence of saturation on the fundamental ofthe differential current is much smaller than the influence on instantaneous differential samples (see

diagram 1). The operation point remains steady at one location within the restraint area.

For numerical line differential relays the requirements in the manuals are very conservative. In fact,

for numerical relays it can be allowed to not fulfil these requirements and to allowed steady state

saturation. Of course if the requirements are fulfilled no further considerations need to be done. If the

requirements are not fulfilled, quite complex considerations need to be performed to ensure that the

protection remains stable for external fault, since it is very difficult to develop less conservative

general valid requirements.

1 Test summary / conclusion

For 3-phase faults all CTs used by CLP fulfil the manual requirements. For A-phase faults the first

manual requirement is not fulfilled. Steady state saturation can occur.

The second requirement that allows deviation of the CT-ALFs only within a specified range is

fulfilled.

Since this second requirement is fulfilled steady state saturation on both ends is almost the same.

Therefore and due to the filtering the differential current caused by steady state saturation is very small

and cannot cause mal-function.

This document proves, that 7SD502, 7SD600 combined with the CTs used by CLP remains stable

under worst case condition, which is an external A-phase fault with max. fault current and max. DC-component. No matter that requirement 1) is not fulfilled.

2 Test concept

It shall be proven that no mal-function occurs under worst case through fault condition.

The concept is to simulate the worst case saturation numerically. The simulation provides data files

which include the saturated and non-saturated currents with a specified sample frequency of 2 kHz

(Data is provided in the comtrade-format).

With this data the following test is performed:

The data is fed into a 7SD502, 7SD600 algorithm simulation, which will provide the Idiff/Istab

operation points in the stabilising characteristic as a function of the time.

This methodology is necessary, because we have to consider not only the primary CT but also the

summation CT. Furthermore the saturation of the summation CT will influence the saturation of the

primary CT since a saturated summation CT causes a different burden for the primary CT than a non-

saturated summation CT.

This kind of simulation can not be performed with a common CT-simulation program. A line-

simulation program is necessary because of the mutual influence of primary and summation CT.

The business of SIEMENS EV NP department is the line design for all voltage levels. They are using

the line simulation program NETOMAC. By the use of this program EV NP calculated the saturated

currents on the secondary sides of the summation CTs.

CLP uses three different primary current CT:


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- old class X CT (400/5A)

- new class X CT (400/5A)

- new 10P10 CT (400/1A)

The NETOMAC simulations were performed for the new 10P10 CT and the new class X CT.

Saturation of old and new class X CT will almost not deviate due to same CT data and almost same

ALFs. Therefor the simulation of the new class X CT should be sufficient.

3 Simulation / Test

The test is performed for two cases. First, for the real case with the two CT combinations CLP is using

on both ends.

And second, a worst, theoretical case with one real CT combination on one end (10P10 with 1A sum.

CT) and ideal CTs on the other end. This case causes the maximum theoretic differential current.

Please notice, that this case is just for demonstration. This case cannot occur since both CT

combinations on both sides will saturate under the considered condition.

The required CT data (most important: magnetization curves) for the NETOMAC-simulation were

given by the primary CT manufacturer RITZ and the summation-CT manufacturer SIEMENS

A&D CD.

3.1 Primary current

The primary current is calculated with the following information:

Effective short circuit current 20 kA

System time constant 30ms

Complete asymmetry

(max. DC component)

0 degree

The current curve is shown in diagram 2. This current is the input signal of the NETOMAC simulation

(interval equals 0,5ms due to 2 kHz sampling frequency of NETOMAC simulation) .

Diagram 2: primary current in [kA]

0 50 100 150 20040

20

0

20

40

60

time in [ms]

Primarycurrentin[kA]

48.626

28.185

pi

2000 i interval.


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3.2 NETOMAC Data for 10P10 CT with 1A sum. CT

The data at the secondary side of the summation CT, calculated by the line-simulation program

NETOMAC is given in the following diagram.

The red curve is the saturated current on the secondary side of the summation CT. The blue curve is

the ideal non-saturated current. Especially in the beginning during strong asymmetry (because of DC-

component) heavy saturation occurs. After approx. 75ms steady state can be seen, still with some

saturation.

(interval equals 0,5ms due to 2 kHz sampling frequency of NETOMAC simulation)

Diagram 3

0 50 100 150 2005

0

5

10

time in [ms]

secondarysumm.

CTcurren

tin[A]

7.026

4.073

si

pi

2000 i interval.


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3.3 NETOMAC Data for Class X CT with 5A sum. CT

The data at the secondary side of the summation CT, calculated by the line-simulation program

NETOMAC is given in the following diagram.

The red curve is the saturated current on the secondary side of the summation CT. The blue curve is

the ideal non-saturated current. Especially in the beginning during strong asymmetry (because of DC-

component) heavy saturation occurs. After approx. 100ms steady state can be seen, still with some

saturation.


Diagram 4

0 50 100 150 20010

5

0

5

time in [ms]

secondarysumm.

CTcurren

tin[A]

4.082

7.01

si

pi

2000 i interval.


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3.4 Combination of both simulations

The saturation of the Class X CT in combination with the 5A summation CT is stronger than the

saturation of the 10P10 CT in combination with the 1A summation CT. Under steady state thesaturation for both combinations is almost the same. The reason for this is that the second CT

requirement (limited ALF deviation) is fulfilled.

In case of exactly the same steady state saturation on both sides, no differential current would occur.

Therefore the second requirement (limited ALF deviation) is important if saturation occurs.


Diagram 5

0 50 100 150 2006

4

2

0

2

4

6

time in [ms]

secondarysumm.

CTcurrentin[A]

4.473

4.06

psi

xs( )i

2000 i interval.


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3.5 7SD502, 7SD600 Algorithm with real CTs on both sides

The instantaneous values for the differential (red curve) and the stabilising current (blue curve) are

given in the following diagram. It can be seen that the largest differential current occurs in thebeginning when the primary current is asymmetric (DC-component). After 300ms both CT-

combinations are in a steady state and the differential current is at its minimum.

Diagram 6 instantaneous values of differential and stabilising current [pu]

The next diagram shows the filtered values of differential- and stabilising current. The positive

influence of the filter algorithm can be seen (as explained in the introduction).

Diagram 7 filtered values of differential and stabilising current [pu]

0 50 100 150 200 250 300 350 400200

0

200

400

600

time in [ms]

current[pu]

408.063

163.896

iDiffi

iStabi

40020 i

0 50 100 150 200 250 300 350 4000

100

200

300

time in [ms]

filterdcurrent[pu]

300

4.862 103.

Diffeffi

Stabgli

40020 i


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The next diagram shows the operation points of differential- and stabilising current compared to the

tripping characteristic as a function of the time.

The progress of the operation point starts at zero. Within the first 5-6 ms no differential current occurs,

since there is no saturation. The operation points are close to the x-axle. The operation point moves far

into the region of additional stabilisation, since this region starts where the stabilising current exceeds

4*IN. After approx. 5-6ms saturation starts and causes a differential current. The operation point

moves away from the x-axle, closer to the tripping characteristic. Even if the operation point would

move into the tripping area for a limited time, the differential protection would not trip due to

blocking. This blocking caused by a strong external fault lasts for 1 second (additional stabilisation,

saturation detector).

With increasing time the operation point moves away from tripping characteristic, since the

differential current is decreasing. Under steady state, the operating point is far away from the tripping

area and quite close to the x-axle. The requirement of no saturation under steady-state would move the

operating point on the x-axle. Anyhow, it can be seen that the steady-state saturation can not cause

mal-function, since the operation point is far away from the tripping zone.

Diagram 8 filtered values of differential and stabilising [pu] given in the trippingcharacteristic as a function of the time

0 50 100 150 2000

20

40

60

80

100

Stabilizing Current [pu]

DifferentialCurrent[pu]

100

0

Diffeffl

IDIFFgr

ST1 b.

yast2c

2200 Stabgll

a, b, c,


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3.6 7SD502, 7SD600 Algorithmwith real CTs on one and ideal on the other side

Just for demonstration on how good the filter algorithm works in case of saturation, these diagrams arepresenting the not realistic, but worst case. One side considers the real CT combination of 10P10 and

1A sum. CT. The other side considers ideal CTs that do not saturate.

The signals fed into the 7SD502/60 algorithm are given in Diagram 3.

Diagram 9 Instantaneous values of differential and stabilising current [pu]

Diagram 10 Filtered values of differential and stabilising current [pu]

0 50 100 150 2000

100

200

300

time in [ms]

filterdcurrent[pu]

300

1.662 103.

Diffeffi

Stabgli

20020 i

20 40 60 80 100 120 140 160 180 200200

0

200

400

600

time in [ms]

current[pu

]

449.792

192.528

iDiffi

iStabi

20020 i


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The next diagram shows the operation points of differential- and stabilising current compared to the

tripping characteristic as a function of the time.

After approx. 10ms the operation point moves into the tripping area. But even this would cause no

tripping due to differential-protection-blocking because of the saturation detector (additional

stabilising region).

With increasing time the operation point moves out off the tripping area. since the differential current

is decreasing. Under steady-state, the operating point is not close to the tripping area, even in this non

realistic worst case. After 1 second, when the blocking expires no tripping would occur.

Diagram 11 Filtered values of differential and stabilising [pu] given in the tripping

characteristic as a function of the time

0 50 100 150 200 250 3000

50

100

150

Stabilizing Current [pu]

DifferentialCurrent[pu

]

170

0

Diffeffl

IDIFFgr

ST1 b.

yast2c

3000 Stabgll

a, b, c,


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4 Summary

The CT requirements in future given in the SIEMENS line differential protection manuals for 7SD60,

7SD502, the application guide and the CT-dimensioning program, based on these results, are:

1)Npr

kd

I

In max with: Ikd max = max. steady-state through fault current

INpr = primary rated current of CT

If this requirement is fulfilled no saturation will occur under steady state through fault conditions on

both line sides.

2) 2/3

3/2

2

1

n

n7SD502, 7SD60 with: n operational ALF of the CTs

Additionally it must be checked, if the accuracy limiting factors are between the limit 2).

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