Interposing CT

Interposing Current TransformersFor use in conjunction with

Type MBCH Transformer Differential Protection

2

Description Reference No.

Single-phase transformer 0.577 – 1.732/1A GJ0104 010



Three-phase transformer group 0.577 – 1.732/1A GJ0104 050



Table 1

Interposing Current Transformers

General

Transformer differential relayscompare the phase and magnitude ofthe current entering one winding ofthe transformer with that leaving viathe other winding(s). Any difference inphase or magnitude between themeasured quantities will cause currentto flow through the operate windingof the relay. If this current exceeds therelay setting, tripping of thetransformer circuit breakers will beinitiated.

To enable a comparison to be made,the differential scheme should bearranged so that the relay will seerated current when the full loadcurrent flows in the protected circuit.In order to achieve this, the linecurrent transformers must be matchedto the normal full load current of thetransformer. Where this is not thecase it is necessary to use an auxiliaryinterposing current transformer toprovide amplitude correction.

The connection of the line CTs shouldcompensate for any phase shiftarising across the transformer.Alternatively the necessary phasecorrection may also be provided bythe use of an interposing CT.

The interposing current transformerslisted in Table 1 are available forcurrent ratio and phase anglematching of the line CTs used withtransformer differential relay typeMBCH (Publication Nos. R6070 andR6017).

The general winding arrangement isas shown in Figure 1, whilst Table 2gives details of the number of turns oneach of the transformers whosecurrent ranges are noted in Table 1.

Figure 1: Disposition of windings on interposing transformer

0.577 – 1.732/1A 2.886 – 8.66/1A 2.886 – 8.66/5A1 – 2 5 1 12 – 3 5 1 13 – 4 5 1 14 – 5 5 1 15 – 6 125 25 25X – 7 25 5 57 – 8 25 5 58 – 9 25 5 5

S1 – S2 125 125 25S3 – S4 90 90 18

Table 2

Number of TurnsTransformer Rating

PrimaryWinding

Taps

1 2 3 4 56

7 8 9

S1 S2 S3 S4 P1 P2

X

Output to relay

Removable jumper

Input fromline CTs

3

Main and interposing currenttransformers

WARNING: do not open thesecondary circuit of a live CT since thehigh voltage produced may be lethalto personnel and could damageinsulation.

Before the power circuit is energised,two precautions are essentialregarding the interposing currenttransformer:

– the primary and the secondarywindings of the CT must haveconnections made to at least onesection of each winding

– neither set of connections mustinclude any break.

The above are very important, asdangerously high voltages could bepresent on leads and winding taps ifthe warning is not observed.

When the chosen connections havebeen made to the various taps on thewindings of the interposing CT to givethe desired ratio, (see Figure 1) it isimportant to realise that any part ofany winding section that was not usedin making these connections, shouldnot be linked or connected toanything else, otherwise the accuracyand effectiveness of the CT will begreatly impaired.

Main transformertap change range

The required interposing CT ratiosshould be calculated using the powertransformer transformation ratio whichcorresponds to the mid-tap point onthe tap change range. The examplesused in this publication assume thatthe mid-tap point coincides with thenominal transformation ratio.

Design

The interposing transformers may besupplied as single or triple unitassemblies. The design is of the air-insulated type, each phase unitconsisting of a pair of high-gradesilicon steel ‘C’ cores, thus minimisingthe magnetising current and therebylimiting the current error. Figure 16shows the physical arrangement anddimensions of these transformers.

The following examples illustrate theprocedure.

EXAMPLE 1

Single phase transformers

In a single phase transformer therated primary and secondary currentsare 195A and 780A respectively asshown in Figure 2. The outputs fromthe respective current transformers arethus 4.875A and 3.9A.

In order to match these two currentvalues, so that no relay differentialcurrent flows under healthyconditions, an interposing CT will berequired to adjust the main line CToutput of 3.9A to a value of 4.875A.This is illustrated in Figure 2.

Using taps S1 – S2 on the secondarywinding gives 25 turns. Thus thenumber of turns (Tp) required on theprimary input winding will be givenby:

Tp = 25 x 4.8753.9

= 31.25

The current imbalance due to the useof 31 turns is limited to:

Error = 31.25 – 3131.25

x 100% = 0.8%

The tap positions required to give 31turns on the primary winding areshown in Table No. 2 to be nos. 4–7.

It may be noted that an interposingtransformer could also have beenused in conjunction with the 200/5ACTs. This would have increased theoutput current from 4.875A to 5.0,and the interposing CT alreadyconsidered above would haverequired a ratio of 3.9/5A instead of3.9A/4.875A.

Figure 2

Physical location

The interposing transformers, whereused, should be located in the samepanel as the associated differentialrelay. The electrical proximitymaximises the transformer outputvoltage applied to the relay underinternal fault conditions and therebyensures high speed relay operationwith minimum transformer dimensions.

Transformer output

The maximum output voltagesavailable from the interposing CTwindings S1–S2, and S1–S4 (seeFigure 1) are indicated by themagnetisation characteristics shown inFigure 15.

It should be noted that the S3–S4winding should not be used on itsown as the output winding to therelay. The reason is that this windingdoes not produce sufficient outputvoltage by itself to maintain thequoted relay operating time. For star-connected output windings, sectionS1–S2 may be used either alone orwith section S3–S4 connected inseries.

For delta-connected output windings,where any one phase winding wouldbe required to energise two poles ofthe relay in series, windings (S1–S2)and (S3–S4) must be connected inseries.

With the link between terminals (6)and (X) connected as necessary, theincoming input connections to P1 andP2 may be linked across to theappropriate input winding terminalsaccording to the number of turnsrequired as indicated in Table 2.

X 6

4S2

S1

P1

P2

3.9A

Interposing currenttransformer

1000/5A780A195A

Single phasetransformer

4.875A

200/5A

7

Differentialrelay

4

In omitting to provide a secondinterposing CT the effective settingwould be increased by the ratio:

54.875

x 100% = 1.025%

This would give a setting of:

1.025 x 20% = 20.5%on the 20% tap

This slight increase in effectiveprotection setting would normally beacceptable and thus a secondinterposing CT may be consideredunnecessary.

EXAMPLE 2

Three phase transformer(see Figures 3 and 11)

30MVA 11/66kV, Delta-star

11kV Winding:

Normal current at 11kV

30 MVA3 x 11kV

= 1575A

As the 11kV winding is delta-connected the associated currenttransformers should be star-connected,and during rated load conditions willgive the following current per pilotphase:

15751600

x 1.0A = 0.984A

This current is sufficiently close to therelay rated value (1A) that nocorrection is required.

66kV Winding:Rated current at 66kV

= 30 MVA3 x 66kV

=262.5A

Normally, delta connected currenttransformers would be associated withthe star-winding of the maintransformer. However, since the latterin this case are star connected, thenecessary phase and amplitudecorrection may be carried out using astar-delta connected interposingtransformer, as shown in Figures 3and 11.

The output current of the 300/1A HVcurrent transformers will be:

262.5300

x 1.0A = 0.875A

Thus the required ratio of theinterposing CT will be 0.875/

Figure 3: Interposing transformer connections for example No. 2

a

S1 S2 S3

S4

b

c

Todifferential

relay

P2

P1226X A

226X B

226X C

N

To lineCT’s

S1 S2 S3

S4

S1 S2 S3 S4

P2

P1

P2

P1

0.984A. Using output windings (S1 –S2) and (S3 –S4) connected in seriesgives 125 + 90 = 215T.

Hence input winding turns should be:

215 x 0.9843 x 0.875 = 139.6 say 140 turns

The input taps giving 140 turns areNos. 2 – 6 as shown in Figure 3.

EXAMPLE 3(see Figures 4, 5, 6, 7 and 12)

Three winding transformer

Taking the case of a three-windingtransformer where load current flowsin all three windings (as illustrated inFigure 4) to apply differentialprotection to such a transformer, thechoice of line current transformerratios ought to be based on the MVArating of the primary input winding,rather than on the MVA ratings of theindividual windings.

Since in many cases the line CT ratioswill have already been chosen interms of the MVA ratings of thewindings, interposing CTs will benecessary in order to match thesecondary currents.

It will be found that more accuratematching taps may be chosen on theinterposing CTs if the most lightlyrated main transformer winding beconsidered first. In the case being

considered this will be the 13.45kVwinding supplying only 100MVA.

13.45kV side:Rated current =

100 MVA3 x 13.45kV

= 4,293A

Line current transformer ratio:5000/5A

∴ Line CT output current

= 42935000

x 5.0A = 4.293A

During external fault conditions,however, the line CT output currentmust be related to the primary inputMVA, and thus becomes:

Rated Current(for external fault consideration)

= 400 MVA3 x 13.45kV

= 17,170AThe interposing transformerassociated with the 13.45kV CTs willbe star-star connected. From an outputpoint of view the S1–S2 windingwould suffice, however it is preferableto include winding S3–S4 giving atotal of 43 turns per secondary phase.This is because the higher the numberof output winding turns, the more turnswill be required on the input side, orprimary, for a given current ratio, thus

Figure 4

500/138kV

100MVA

13.45kV

300MVA400MVA

5

the smaller will be the tap percentageerror:

Input turns required to produce5A in output winding:

5 x 4317.17

= 12.52 say 12 turns

In order to obtain 12 turns on theinterposing transformer input winding,use terminals 3 and 8 with link 6–Xremoved and terminal nos 5 and Xconnected. See Figure 5.

Output winding current will now be:

Is =12 x 17.17

43= 4.792A

Regard 4.8A as rated current towhich must be matched the currentsfrom the 500kV and 138kVinterposing CTs respectively.

500kV side:

Rated current = 400 MVA3 x 500kV

= 462A

Secondary current of line CTs500/5A:

462500

x 5.0A = 4.62A

∴ Required ratio of interposing CT:

4.624.8/ 3

ie. 4.62/2.77A

Interposer input winding turns

2.77 x 434.62


Error in interposing CT secondarywinding current

26 – 25.7825.9

x 100% = 0.853%

Interposing CT input taps togive 26 turns

nos. 4 and 6 (see Figure 6)

138kV side:

Rated current base 400MVA:

400 MVA3 x 138kV

= 1673A

Secondary current of line CTs1250/5A:

16731250

x 5.0A = 6.69A

The ratio of the interposing CTsassociated with the 138kV line CTswill therefore be:

Figure 5: Interposing transformer associated with 13.45kV CTs

3

3

S1 S2 S3 S4

Torelay

X

X

3X

To13.45kVline CTs

8

8

8

5

5

5

S1 S2 S3 S4

S1 S2 S3 S4

S1 S2 S3 S4

Torelay

P2 P1

46X

To 500kVline CTs

9

46X9

46X9

S1 S2 S3

S4

S1 S2 S3

S4

P2 P1

P2 P1

Figure 6: Interposing transformer associated with 500kV CTs

Figure 7: Interposing transformer associated with 138kV CTs

25

25Torelay

X

To 138kVline CTs

9

X9

25X9

S1 S2 S3 S4

S1 S2 S3

S4

S1 S2 S3

S4

6.694.8/ 3

ie. 6.69/2.77A

Interposing transformer input turns:

2.77 x 436.69

= 17.8 say 18T

Error in interposing secondarywinding current:

18 x 17.817.8

x 100% = 1.12%

Interposing CT input taps to give18 turns

2 and 9 (See Figure 7) with link(5 – X) reconnected

EXAMPLE 4

Transformer fed via mesh busbararrangement (see Figures 8 and 13)

Where a transformer is fed from amesh busbar arrangement andcontrolled by two circuit breakers in

6

Figure 8: Interposing CTs associated “mesh busbar” transformer

the mesh, the rating of the transformerwill normally be somewhat lower thanthat of other feeders emanating fromthe mesh.

Because the current transformerswithin the mesh will have a highercurrent rating than those associatedwith the transformer, their use asdifferential protection CTs for thetransformer will result in a highereffective fault setting of the differentialprotection system.

A better location for the differentialprotection CTs is in the direct feed tothe transformer, at point ‘P’ inFigure 13. Where suitable CTs areavailable at point ‘P’, the applicationand selection of suitable interposingCTs is exactly as described inExample 2.

In those cases where the only CTsavailable for the transformerdifferential protection are located atthe circuit breakers within the ‘mesh’,then the protection arrangement willbe as shown in Figure 13.

EXAMPLE

40MVA 132/33kV Star-DeltaTransformer

132kV (Mesh side):

Line current transformer ratio :500/1A

Transformer rated current (at 132kV):

40 MVA3 x 132kV

= 175A

∴ Current in current transformersecondary winding:

175A500

x 1.0 = 0.35A

Preferred current ratio ofinterposing CT:

1.03

0.35= 0.35/0.577A

NB. ∆ output windingHowever, the maximum inputwinding turns are:220 turns (see Figure 8)

∴ Increase in current throughthe transformer:

0.35 x 220(125 + 90)

= 0.358A

Note: It has already been stated(see Page 3) that for delta connectedoutput windings both windings(S1 – S2) and (S3 – S4) must beconnected in series.

Pilot current to the relay:

3 x 0.358 = 0.62A

33kV side:

33kV Interposing CT:

star/star connection

Output winding turns:

125 turns winding (S1 – S2)

33kV Line current:

40 MVA3 x 33kV

= 700A

33kV CT Secondary current:700/800 x 1 = 0.875A

Ratio of 33kV Interposing CT:

0.875/0.62A

Input winding turns (see Figure 8):

125 x 0.620.875


Referring to Figure 13 it will be notedthat when the rated current of thetransformer flows, the correspondingcurrent flowing in the relay circuit willbe 0.62A. Thus the effectiveprotection setting will be:

1.00.62

= 1.6 x selected relay setting

The above increase in setting is due tothe nominal primary rating of the linecurrent transformer being appreciablygreater than the rating of thetransformer being protected.This condition is unavoidable.

EXAMPLE 5

Star delta transformer with earthingtransformer on delta side(see Figures 9, 10 and 14).

75MVA – 132/33kV

Where a three phase earthingtransformer is associated with a delta-connected LV winding as in Figure14, a zero sequence path is providedwithin the protected zone. This permitsthe passage of zero sequence currentsthrough the differential relay, underexternal earth fault conditions.

It is necessary to block such currentsby inserting a star-delta-starinterposing current transformer asshown in Figure 14. This same

1 6 X 9

1 6 X 9

1 6 X 9

To132kVline CTs

5

2X9

5

2X9

5

2X9

To33kV

line CTs

Two transformers thus

From 2nd set of 132kV line CTs

MBCHrelay

S1S2S3S4

P2P1

S1S2S3S4

P2P1

S1S2S3S4

P2P1

P2 P1

S1 S2

P2 P1

S1 S2

P2 P1

S1 S2

7

interposing CT may be used for ratiocorrection required due to anymismatch between the HV and LV linecurrent transformers.

132kV side:

Normal current at 132kV:

75 MVA3 x 132kV

= 328A

Current transformer ratio: 400/5A

Secondary current:

328400

x 5.0A = 4.1A

Interposing transformer required:star/delta ratio 4.1/5/ 3

A(per phase)Figure 9 shows details of thetransformer.

Relay output winding turns:

(S1 – S4) 25 + 18 = 43 turns

Input winding turns:

43 x 54.1 x 3


33kV Side:

Normal current:

75 MVA3 x 33kV

= 1312A

Current transformer ratio: 1500/5A

Secondary current:

1312 x 51500

= 4.37A

Interposing transformer required:

– star/delta/star

Secondary winding S3 – S4 is usedfor the delta connection. The relay isconnected to windings S1 – S2 = 25turns.

Interposing CT primary turns:

54.37

x 25 = 28.6 say 29 turns

Interposing CT connections are shownin Figure 10.

7X

Torelay

6

From132kVline CTs

5

7X65

7X65

S1S2S3S4P2P1

S1S2S3S4P2P1

S1S2S3S4P2P1

Figure 9: Interposing transformer connections for example No. 5

9X

Torelay

6

From33kV

line CTs

1

9X61

9X61

S1S2

S3

S4P2P1

S1S2

S3

S4P2P1

S1S2

S3

S4P2P1

Figure 10: Interposing CTs for example No.5

8

Figure 11: Interposing CTs

Figure 12: Three winding transformer showing interposing CTs

2526

28

2324

27

MBCH

2526

28

2324

27

MBCH

2526

28

2324

27

MBCH

0.568/0.875A

30MVA11/66kV1600/1A 300/1A

2224

28

2123

27

MBCH

500kV500/5A

2625

MBCH

MBCH

1250/5A

5000/5A

17.17/4.8A

17.17/4.8A

2.77/7.0A

400MVA / 100MVA / 300MVA

13.45kV 138kV

4.62/2.77A

2224

28

2123

272625

2224

28

2123

272625

9

Figure 13: Mesh-bus arrangement showing interposing CTs

Figure 14: Star/Delta transformer with earthing-transformer on delta-side

2224

28

2123

27

MBCH

500/1A

2625

MBCH

MBCH

800/1A132/33kV [Y[D1]

2224

28

2123

272625

2224

28

2123

272625

0.35/0.358A

Point ‘P’Mesh busbar arrangement

400MVA

0.62/0.875A

0.35/0.358A

2526

28

2324

27

MBCH

75MVA132kV/33kV

400/5A 1500/5A

2624

2526

28

2324 MBCH

26

2526

28

2324 MBCH

26

Earthingtransformer

NER

5.0 4.37A4.1/2.89A

10

Figure 15: Magnetisation characteristic for 0.58 1.73/1A2.89 8.66/1A

0

20

40

60

80

100

120

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1 Amperes

Volta

ge

NB. For 5A output windings (S1 – S2, S1 – S41)

Multiply Voltage scale by 0.2Current scale by 5.0

Knee point voltage S1 – S4

Knee point voltage S1 – S2

11

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12

34

56

78

91

23

45

67

89

12

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56

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9

S1 S

2 S

3 S

4 P

1 P

2S1

S2

S3

S4

P1

P2

S1 S

2 S

3 S

4 P

1 P

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83.5

(1Ø

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116

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343

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16

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117 5.5

145

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taps

St Leonards Works, Stafford, ST17 4LX EnglandTel: 44 (0) 1785 223251 Fax: 44 (0) 1785 212232 Email: [email protected] Internet: www.alstom.com

©2000 ALSTOM T&D Protection & Control Ltd

Our policy is one of continuous development. Accordingly the design of our products may change at any time. Whilst every effort is made to produce up to date literature, this brochure shouldonly be regarded as a guide and is intended for information purposes only. Its contents do not constitute an offer for sale or advice on the application of any product referred to in it.

ALSTOM T&D Protection & Control Ltd cannot be held responsible for any reliance on any decisions taken on its contents without specific advice.

Publication R6077C Printed in England.

ALSTOM T&D Protection & Control Ltd

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Interposing CT