POWER TRANSFORMER

1 Insulation Resistance MeasurementProcedurea. For LT systems, use only 500V or 1000V Megger.b. For MV & HV systems, use 2500V or 5000V Megger.c. Remove Earth from Transformer neutral.d. Measure as per table below.

Circuit Megger UsedMV & HV

1U - E 5kV2U - E 5kV1U - 2U 5kV

2 Winding Resistance Measurement

Circuit Tap No Resistance Ω

1U-1N All Taps1V-1N1W-1N

2U-2N2V-2N2W-2N

1U

1V

1W

1N

2U

2V

2W

2N

1U 1V 1W 1N

2U 2V 2W 2N

WINDING RESISTANCE METERI I I

I I1 P1 P2 I2

3 Magnetic Current Test

Applied Voltage Measured Current1U-1V 1U1V-1W 1V1W-1U 1W1U-1N, 1V-1N,1W-1N 1N

4 Magnetic Balance Test

Applied Voltage Measured Voltage1U1N 1V1N 1W1N 1U1N 1V1N 1W1N 2U2N 2V2N 2W2N415 - - - - 415 - - - - 415 -

Applied Voltage Measured Voltage1U1V 1V1W 1W1U 1U1V 1V1W 1W1U 2U2V 2V2W 2W2U415 - - - - 415 - - - - 415 -

2U 2V 2W 2N

1U

1V

1W

1N

2U

2V

2W

2NmA

415 V

AC mA

mA

mA

1U

1V

1W

1N

2U

2V

2W

2N

415

V AC

Measured

volta

ge

5 Vector Group Confirmation.

Dyn1

Conditions:1 -1U and 2u should be shorted.2 -Apply 415V to 1U,1V&1W3 -Satisfy the following conditions.1U1W = 1U2n + 1W2n1W2w = 1W2V1V2v < 1V2w

Dyn11

3 -Satisfy the following conditions.1U1V = 1U2n + 1V2n1V2w = 1V2v1W2w < 1W2v

1U

1V1W

2W

2U

2V

1U 2U

1W

2V

2W 2n

1U

1V1W

2W

2U

2V

2U1U

1V1W

2W

2V

2n

5 Vector Group Confirmation.

Ynd1

3 -Satisfy the following conditions.1U1N = 1U2v + 2v1N1W2v = 1V2v1V2v < 1V2w

YNyn0

3 -Satisfy the following conditions.1W2w = 1V2v1W2n = 1V2n1U1N = 1U2n + 1N2n

6 Short Circuit Test and Differential Stability.7 REF Stability Test

1U

1V1W

2W

2V

2U

1N

1U

1V1W

2W

2V

2U

1U

1V1W

2W 2V

2U

1N2n

1U

1V1W

2W 2V

2U

2n

1N

8 Cooler Circuits9 Temperature Indicators Calibration

10 Auxilliary Protection Circuits

POWER TRANSFORMER

Megger Used

2U

2V

2W

2N

Measured

volta

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1V

6 Short Circuit Test and Differential Stability.

Stable Condition.

Diff RelayP1 S1

300/1 AS2 S1

S2 S1

P2 S2

B-Ph

Y-Ph

R-Ph

P1 S1

500/1 AS1 S2

S1 S2

S2

P2

Apply 3-Ph AC Voltage

N R Y B

Temporary Short

Id

Id

Id

Example:Short Circuit Current Calculation.

Rating : 100MVAHV LV

Voltage 220 kV 132kVCurrent 262.43 A 437.40 A% of Impedance (%Z) = 8.05%

HV MVA = LV MVA = 100MVA

√3 x 220000 x I1 = √3 x 123000 x I2 = 100,000,000√3 x 220000 x I1 = 100,000,000I1 = 100,000,000 / (√3 x 220000)I1 = 262.43A

√3 x 123000 x I2 = 100,000,000I2 = 100,000,000 / (√3 x 132000)I2 = 437.40A

%Zv = 8.05% = 220kV x % of impedance = 220000 x (8.05/100)

%Zv = 17,710 V

If We may apply 415 V to HV sideHV I primary = (415 x 262.43) / 17710HV I primary = 6.15 AHV sec = 6.15 / 300HV sec = 0.0205A

LV I primary = (415 x 437.4) / 17710LV I primary = 10.25 ALV sec = 10.25 / 500LV sec = 0.0205 A

Unstable Condition

S1

S2 S1

S2 S1

S2

S1

S1 S2

S1 S2

S2

Id

Apply 3-Ph AC Voltage

Temporary Short

Id

Id

7 REF Stability Test

i) Stable Condition.

R Y B

P1 S1 S1 S1

P2 S2 S2 S2

S2 S1 N

P2 P1P2 P1

nS2 S1

P1 S1 S1 S1

P2 S2 S2 S2

r y b

Apply 3-Ph AC

Fault

HV REF

LV REF

SHORT

ii)Unstable Condition (Type 1)

S1 S1 S1

S2 S2 S2

S2 S1

S2 S1

S1 S1 S1

S2 S2 S2

Apply 3-Ph AC

SHORT

LV REF

HV REF

iiiUnstable Condition (Type 2)

S1 S1 S1

S2 S2 S2

S2 S1

S2 S1

S1 S1 S1

S2 S2 S2

HV REF

LV REF

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