Power Transformer Testing Procedures

POWER TRANSFORMER 1 Insulation Resistance Measurement Procedure a. 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.

1

2

Circuit

1W

2

1U - E 2U - E 1U - 2U

1

2

1

2

Megger Used MV & HV 5kV 5kV 5kV

2 Winding Resistance Measurement WINDING RESISTANCE METER

I I1

I P1

I P2

I I2

Circuit 1U-1N 1V-1N 1W-1N 2U-2N 2V-2N 2W-2N

1

2

1

2

1W

1

2

2

Tap No Resistance Ω All Taps

3 Magnetic Current Test

415 V AC

mA

1

2

1W

2

1

2

1

2

mA

mA

mA

Applied Voltage 1U-1V 1V-1W 1W-1U 1U-1N, 1V-1N,1W-1N

Measured Current 1U 1V 1W 1N

1

2

1W

2

1

2

1

2

Applied Voltage Measured Voltage 1U1N 1V1N 1W1N 1U1N 1V1N 1W1N 2U2N 415 415 415 Applied Voltage Measured Voltage 1U1V 1V1W 1W1U 1U1V 1V1W 1W1U 2U2V 415 415 415 -

2V2N

2W2N

2V2W

2W2U

Measured voltage

415 V AC

4 Magnetic Balance Test

5 Vector Group Confirmation. Dyn1

1U 2U

1U 2W

2n

2U

2V

2W 2V

1V 1

1V

1

Conditions: 1 -1U and 2u should be shorted. 2 -Apply 415V to 1U,1V&1W 3 -Satisfy the following conditions. 1U1W = 1U2n + 1W2n 1W2w = 1W2V 1V2v < 1V2w

Dyn11 1U

1U

2U 2n

2U

2V 2W 2V

2W

1

1V 3 -Satisfy the following conditions. 1U1V = 1U2n + 1V2n 1V2w = 1V2v 1W2w < 1W2v

1

1V

5 Vector Group Confirmation. 1U 2U

Ynd1

1U 2W 2V 2U 2W

1N 2V

1

1

1V

1V

3 -Satisfy the following conditions. 1U1N = 1U2v + 2v1N 1W2v = 1V2v 1V2v < 1V2w

YNyn0

1U

1U

2U 2n 2V

2W 2U 1N2n 2W

1

1N 2V

1V 3 -Satisfy the following conditions. 1W2w = 1V2v 1W2n = 1V2n 1U1N = 1U2n + 1N2n

6 7 8 9 10

Short Circuit Test and Differential Stability. REF Stability Test Cooler Circuits Temperature Indicators Calibration Auxilliary Protection Circuits

1

1V

V

6 Short Circuit Test and Differential Stability. Stable Condition.

Apply 3-Ph AC Voltage

N

R

Y

B

Diff Relay P1

S1

300/1 A S2

S1

S2

S1

S2

P2

B-Ph Id Y-Ph Id R-Ph Id P1

S1

500/1 A S1

S1

S2

S2

S2

P2

Temporary Short

Example: Short Circuit Current Calculation. Rating : 100MVA HV Voltage 220 kV Current 262.43 A % of Impedance (%Z) = 8.05%

LV 132kV 437.40 A

HV MVA = LV MVA = 100MVA √3 x 220000 x I1 = √3 x 123000 x I2 = 100,000,000 √3 x 220000 x I1 = 100,000,000 I1 = 100,000,000 / (√3 x 220000) I1 = 262.43A √3 x 123000 x I2 = 100,000,000 I2 = 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 side HV I primary = (415 x 262.43) / 17710 HV I primary = 6.15 A HV sec = 6.15 / 300 HV sec = 0.0205A LV I primary = (415 x 437.4) / 17710 LV I primary = 10.25 A LV sec = 10.25 / 500 LV sec = 0.0205 A

Unstable Condition Apply 3-Ph AC Voltage S1

S2

S1

S2

S1

S2

Id

Id

Id S1

S1

S1

S2

S2

S2

Temporary Short

7 REF Stability Test i)

Stable Condition.

Apply 3-Ph AC

Fault R

P1

Y

S1

B

S1

S1

HV REF

S2

P2

S2

S2

S2

S1

P2

P1

P2

P1

N

n S2

P1

S1

S1

S1

S1

LV REF

S2

P2

r

S2

y

SHORT

S2

b

ii)

Unstable Condition (Type 1)

Apply 3-Ph AC

S1

S1

S1

HV REF

S2

S1

S2

S1

S2

S2

S1

S2

S1

S1

LV REF

S2

S2

SHORT

S2

iii) Unstable Condition (Type 2)

S1

S1

S1

HV REF

S2

S2

S2

LOADING TRANSFORMER

S1

S1

S2

S1

S2

S1

LOADING TRANSFORMER

S1

LV REF

S2

S2

S2