power transformer testing procedures
Post on 26-Oct-2014
622 Views
Preview:
TRANSCRIPT
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 2W2N
415 - - - - 415 - - - - 415 -
Applied Voltage Measured Voltage1U1V 1V1W 1W1U 1U1V 1V1W 1W1U 2U2V 2V2W 2W2U
415 - - - - 415 - - - - 415 -
2U
2V
2W
2N
1U
1V
1W
1N
2U
2V
2W
2N
mA
415
V A
C mA
mA
mA
1U
1V
1W
1N
2U
2V
2W
2N
415
V A
C
Mea
sure
d v
olta
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
1V1W
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
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 P1
P2 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
iii) Unstable Condition (Type 2)
S1 S1 S1
S2 S2 S2
S2 S1
S2 S1
S1 S1 S1
S2 S2 S2
HV REF
LV REF
LOADING TRANSFORMER
LOADING TRANSFORMER
top related