17/01/2021

To perform short circuit test on single phase transformer

 

Significance

The short circuit on the single phase transformer is performed in order to find out equivalent circuit parameters of the transformer. This test is performed to find out

  • Copper losses at full load as well as any load
  • Equivalent impedance as referred to primary (Z01) or secondary side (Z02)
  • Equivalent leakage reactance as referred to primary (X01)  or secondary side (X02)
  • Equivalent resistance as referred to primary (R01)  or secondary side (R02)
  • Voltage drop as referred to primary or secondary side
  • Voltage regulation at any load and power factor
  • Voltage transformation ratio

Instruments required

Instrument

Specification

AC voltmeter

 

AC ammeter

 

Wattmeter

 

Multi meter

 

Single phase auto transformer

 

 Theory

  • In this test, the low voltage winding is usually short circuited by thick wire or ammeter and supply is given to the high voltage winding. 
  • The supply to the high voltage winding is given by using auto transformer. 
  • The applied voltage Vsc is gradually increases till the ammeter shows the full load current. 
  • As the applied voltage Vsc is very low ( only 5 – 10% of the normal voltage ), core flux is very small, iron losses are small so it is neglected.
  •  The wattmeter Wsc shows full load copper losses of the transformer.
  • Full load copper losses of both winding Wsc = Isc2 R01

       ⸫  R01 = Wsc / Isc2

  • Equivalent impedance as referred to primary side

        Zsc = Z01 = Vsc / Isc

       ⸫  X01 = √ ( Z012 – R012 )

  • The voltage transformation ratio ( k ) can easily find out by taking reading of HV winding and LV winding. 
  • The transformer equivalent circuit parameter can easily determined as referred to secondary side after calculating voltage transformation ratio ( k ).

       R02 = R01 / k2

       X02 = X01 / k2

      Z02 = Z01 / k2

Circuit diagram

short-circuit-test-on-the-transformer


Procedure

  • Make connection as per circuit diagram
  • Keep the voltage of the auto transformer with zero voltage position
  • Switch on the supply and slowly increase the auto transformer voltage until the ammeter shows the full load current
  • Note down the reading of ammeter, voltmeter and wattmeter.

Precaution

  • It should be kept in mind that setting of the auto transformer must be in minimum or zero voltage position before switch on the supply.

Observation table

Short circuit voltage

Short circuit current

Short circuit power

Vsc

Isc

Wsc

 

 

 

Calculation 

Short circuit input Power

Wsc = Vsc Isc Cos Φsc 

 

Short circuit power factor

& power factor angle

         

Cos Φsc = Wsc / Vsc Isc   

Φsc = Cos – 1( Wsc / Vsc Isc )

 

Transformer winding resistance as referred to primary side

R01 = Wsc / Isc2

Transformer winding impedance as referred to primary side

 

Vsc / Isc

 

Transformer winding reactance as referred to primary side

 

X01 = √ ( Z012 – R012 )

 

Voltage transformation ratio ( k )  

k = I1 / I2 = Isc / I2

 

Transformer winding resistance as referred to secondary side

R02 = R01 / k2

 

Transformer winding impedance as referred to secondary side

 

Z02 = Z01 / k2

 

Transformer winding reactance as referred to secondary side

 

X02 = X01 / k2

 

Full load copper losses

Isc2 R01

Voltage drop as referred to primary side

Isc ( R01 + j X01 )

Voltage drop as referred to secondary side

I2 ( R02 + j X02 )

Voltage regulation at any load and any power factor

+ sign for lagging power factor

– sign for leading power factor

0V2 = No load voltage

I1 R01 Cos Ф2 ± I1 X01 Sin Ф2 / ( 0V2 )

                   OR

I2 R02 Cos Ф2 ± I2 X02 Sin Ф2 / ( 0V2 )

 

 Conclusion

  • The wattmeter shows full load copper losses and iron losses of the transformer. 
  • As the supply voltage is only small amount of rated voltage, iron losses is neglected therefore the wattmeter shows full load copper losses.

Which winding of the transformer is short circuited? LV winding or HV winding. Give reason

  • The LV winding of the transformer is short circuited. 
  • Let us try to understand reason for that, consider a 10 kVA, 1100 V / 220 V transformer, if supply is given to LV side, voltage required for full load current flows through winding during short circuit of HV winding lies between 5% to 10% of rated value i.e ( 220 × 5 ) / 100 = 11 V to ( 220 × 10 ) / 100 = 22 V which is very small. 
  • At low voltage ( 11 V to 22 V ), high precision reading would not be obtained by ordinary meter.

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