3 Winding Transformer Commissioning Test

8/20/2019 3 Winding Transformer Commissioning Test

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J.M. PANG & SEAH PTE LTD 1

A Case Study of the

Commissioning Test of aDifferential Protection Scheme for

a Three Winding Transformer

he transformer is one of the most important hardware in the electrical power system. An

important consideration in transformer protection is the high cost of the transformer and the

relative long outage time that occurs when a transformer fails. The most common form of

transformer protection is differential protection, whose zone of protection is bounded by the location of the

protection current transformers (CTs) between the transformer’s high voltage and low voltage terminals.

The differential protection should operate for a fault within the zone of the CT and should not operate for a

fault outside the zone of the CT. The major advantage of the differential protection is the fast speed of

operation to detect a fault as compared to transformer

protection based on overcurrent relays. The use of

overcurrent relays for transformer protection will result in

a comparatively longer operation time for a fault because

of the need to co-ordinate with upstream overcurrent

relays. A differential protection does not require co-

ordination with other protection scheme, and whose

speed of operation is essentially instantaneous for all in-

zone faults. The CT ratio, polarity and connection must

all be correct in order for the correct operation of the

differential protection. Hence it is imperative that the

differential protection be commissioned before first timeenergization of the transformer. The article will share my

experience in the commissioning of a differential

protection for a three winding transformer at a power

station as shown in Figure 10.1.

Three Winding Transformers

The equivalent circuit for a three winding transformer can be represented by a star of impedance as shown

in Figure 10.2. The impedance of any of these branches can be determined by considering the short circuit

T

PowerGrid

300MVA Yd 11, 230/19kV

230kV

19kV

58/23/35MVA 19/6.6/22kV

G

CT2

22kV

CT1

6.6kV

19kV250MW

CT3

19kV primary CT1 - located at transformer6.6kV secondary CT2 - located at 6.6kV switchgear22kV tertiary CT3 - located at 22kV switchgear

PowerGrid

300MVA Yd 11, 230/19kV

230kV

19kV

58/23/35MVA 19/6.6/22kV

G

CT2

22kV

CT1

6.6kV

19kV250MW

CT3

19kV primary CT1 - located at transformer6.6kV secondary CT2 - located at 6.6kV switchgear22kV tertiary CT3 - located at 22kV switchgear

FIGURE 10.1 : Single Line Diagram

Chapter

10




impedance between any pairs of windings with the third winding in open circuit. Therefore we will have:

Zps - impedance at primary with secondary short-circuit and Tertiary open circuit.

Zpt - impedance at primary with Tertiary short-circuit and secondary open circuit.

Zst - impedance at secondary with Tertiary short-circuit and primary open circuit.

The impedance values of Zps,

Zpt and Zst have physical meaning and

can be measured. However the

equivalent circuit values of Zp, Zs and Zt

are fictitious and cannot be measured.

It is possible to have negative values for

one of the impedance. The values of

Zps, Zpt, Zst may be based on different

base values and hence these impedance

values must be converted to the same

base values before the start of any

calculations. The capacity of the

primary winding is often chosen as the

common base values. Table 10.1 shows the nameplate data and the normalized values of impedance for

Zps, Zpt and Zst.

The normalized values of the impedance to 58MVA base are ;

Zps = Zp + Zs = 31%

Zpt = Zp + Zt = 19%

Zst = Zs + Zt = 62%

Solving the three equations,

Zp = - 6% ; Zs = 37% ; Zt = 25%

It = 60% x Ip

Zt= 25%Zs= 37%

Zp = -6%

Is = 40% x Ip

Ip

Is

Primary19kV

Tertiary22kV

Secondary6.6kV

Ip

It

Primary19kV

Secondary6.6kV

Tertiary22kV

It = 60% x Ip

Zt= 25%Zs= 37%

Zp = -6%

Is = 40% x Ip

Ip

Is

Primary19kV

Tertiary22kV

Secondary6.6kV

Ip

It

Primary19kV

Secondary6.6kV

Tertiary22kV

FIGURE 10.2 : Equivalent Circuit of 3 Winding Transformer

Capacity VoltageNameplateImpedance

Impedance at58MVA Base

Phase angleDifference

Primary 58MVA Primary = 19kV Zps = 12.5%, 23MVA Zps = 31%, Primary toSecondary = 30

o

Secondary 23MVA Secondary = 6.6kV Zpt = 12%, 35MVA Zpt = 19%, Primary toTertiary = 30

o

Tertiary 35MVA Tertiary = 22kV Zst = 62%, 58MVA Zst = 62%, Secondary toTertiary = 0

o

TABLE 10.1 : Impedance and Vector Group Data




Commissioning

An external 3 phase, 400 volts diesel generator was connected at the 19kV side with three different

combinations of open/short circuit at the 6.6kV and 22kV.

The three combinations were ;

Connection 19kV 6.6kV 22kV

Figure 10.3A Generator connection Short circuit Open circuit

Figure 10.3B Generator connection Open circuit Short circuit

Figure 10.3C Generator connection Short circuit Short circuit

The KVA rating of the external generator was calculated using the impedance data of the three

winding transformer of Table 10.1. The required KVA rating of the external generator for the three

different connections were ;

Connection KVA rating of GeneratorPercentage of

Full Load Current

Figure 10.3A 3 x 400 x 119 = 82 KVA 7%

Figure 10.3B 3 x 400 x 195 = 135 KVA 11%

Figure 10.3C 3 x 400 x 416 = 288 KVA 23%

A 3 phase, 400 volts, 300KVA generator was selected for the commissioning test. The calculated

value of 416A at the 19kV winding is about 23% of the full load current and so considered high enough to

commission the differential protection relay. Table 10.2 is the summary of the measured values for Figure

10.3A, Figure 10.3B and Figure 10.3C under external fault conditions. Incorrect CT polarity at the 22kV CT

was suspected because ;

High value of differential current for the case of Figures 10.3B and 10.3C

2 IBias = differential current for the case of Figure 10.3B

The star point of the 22kV CT was reversed to rectify the incorrect CT polarity. Table 10.3 is the

summary of the measured values with the correct CT polarity. The differential protection was considered

stable for the out of zone fault because the differential current was zero. The short circuit at the 22kV side

was relocated to create an in-zone fault and the differential current was non-zero and of high value. The

differential relay was considered commissioned.




Conclusion

In this case study, the polarity of the 22kV CT was incorrect and fortunately discovered during the

commissioning test. If such a commissioning test was not done, the differential relay will operate under

normal loading of the transformer, or during an external fault condition.

In-Zone Fault In-Zone Fault In-Zone Fault Out Zone Fault Out Zone Fault Out Zone Fault

External GeneratorConnected to 19kV

Side

Short Circuitat 22kV

Short Circuitat 6.6kV

Short Circuit at22kV and 6.6kV


Open Circuit

(Fig. 10.3B)

Short Circuit at6.6kV and 22kVOpen Circuit

(Fig. 10.3A)


(Fig. 10.3C)

I red 344A 344 A 344A 162A 105A 344A

I yellow 344A 344 A 344A 162A 105A 344A19kV

I blue 344A 344 A 344A 162A 105A 344A

I red 421A 0A 0A 0A 289A 421A

I yellow 421A 0A 0A 0A 289A 421A6.6kV

Iblue 421A 0A 0A 0A 289A 421A

I red 0A 169A 0A 137A 0A 169A

I yellow 0A 169A 0A 137A 0A 169A22kV

I blue 0A 169A 0A 137A 0A 169A

I red 140m A 154m A 97m A 91m A 59m A 197m A

I yellow 140m A 154m A 97m A 91m A 59m A 197m ABias

I blue 140m A 154m A 97m A 91m A 59m A 197m A

I red 113m A 81m A 194m A 0A 0A 0A

I yellow 113m A 81m A 194m A 0A 0A 0ADiff.

I blue 113m A 81m A 194m A 0A 0A 0A

TABLE 10.3 : Measured Value with Correct CT Polarity

Out of Zone Fault Out of Zone Fault Out of Zone FaultExternal Generator

Connected to19 kV Side

22kV ShortCircuit and 6.6kV

Open Circuit(Fig. 10.3B)

6.6kV ShortCircuit and 22kV

Open Circuit(Fig. 10.3A)

22kV and 6.6kVShort Circuit(Fig. 10.3C)

I red 162 A 105 A 344 A

I yellow 162 A 105 A 344 A19kV

I blue 162 A 105 A 344 A

I red 0 A 289 A 421 A

I yellow 0 A 289 A 421 A6.6kV

I blue 0 A 289 A 421 A

I red 137 A 0 A 169 A

I yellow 137 A 0 A 169 A22kV

I blue 137 A 0 A 169 A

I red 91 m A 59 m A 197 m A

I yellow 91 m A 59 m A 197 m ABias

I blue 91 m A 59 m A 197 m A

I red 183 m A 0 A 226 m A

I yellow 183 m A 0 A 226 m ADiff.

I blue 183 m A 0 A 226 m A

TABLE 10.2 : Measured Value with Incorrect CT Polarity




-- END --

~

400V3Ø Generator

Is

Zt= 25%

Z = 31%

Zs= 37%

~Ip

Zp = -6%

400V3Ø Generator

i = 0.4 x Full load

0.31 x 19 Current

= 6.79% x 1762

= 119A

i

~

400V3Ø Generator

Is

Zt= 25%

Z = 31%

Zs= 37%

~Ip

Zp = -6%

400V3Ø Generator

i = 0.4 x Full load

0.31 x 19 Current

= 6.79% x 1762

= 119A

i

FIGURE 10.3A : Single Line Diagram

~

400V3Ø Generator

It

Zt= 25%

Z = 19%

Zs= 37%

~Ip

Zp = -6%

400V3Ø Generator

i

i = 0.4 x Full load

0.19 x 19 Current

= 11.08% x 1762

= 195A

~

400V3Ø Generator

It

Zt= 25%

Z = 19%

Zs= 37%

~Ip

Zp = -6%

400V3Ø Generator

i

i = 0.4 x Full load

0.19 x 19 Current

= 11.08% x 1762

= 195A

FIGURE 10.3B : Single Line Diagram

i = 0.4 x Full load

0.089 x 19 Current

= 23.65% x 1762

= 416A

~

400V3Ø Generator

Is

Zt= 25%

Z = 8.9%

Zs= 37%

~Ip

Zp = -6%

400V3Ø Generator

i

It

i = 0.4 x Full load

0.089 x 19 Current

= 23.65% x 1762

= 416A

~

400V3Ø Generator

Is

Zt= 25%

Z = 8.9%

Zs= 37%

~Ip

Zp = -6%

400V3Ø Generator

i

It

FIGURE 10.3C : Single Line Diagram

3 Winding Transformer Commissioning Test

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