26/11/2020

To Separate out Iron losses in the Transformer

Introduction

  • There are no moving parts in the transformer therefore the mechanical losses are zero. 
  • The open circuit test and short circuit test on transformer is performed in order to find out core losses and copper losses.  
  • We will study how to separate hysteresis loss and eddy current loss.

separation-of-losses-in-the-transformer


  • Figure shows the circuit diagram which is similar to open circuit test except that the variable frequency, variable voltage supply is required at the input side. 
  • The variable frequency, variable voltage supply is obtained either by inverter / cycloconverter / adjusting excitation and speed of alternator. 
  • The applied voltage and frequency is adjusted such that ratio of voltage to frequency remains constant. 
  • Note down the reading of wattmeter which indicates total iron losses.

Now

Hysteresis loss Wh α Bmax1.6f

Eddy current loss We α Bmax2 f2

For transformer, ( V / f ) α Bmax

Therefore Wh = Af and We = Bf2

Total iron losses

          Wi = Wh + We

               = Af + Bf2

     Wi / f = A + Bf

  • The hysteresis loss and eddy current loss can find out at any frequency by knowing constant value of A and B.
  • It means that if we draw a graph of ( Wi / f ) for different frequencies, the constant part represents hysteresis loss and slope of graph represent eddy current loss at any frequency.
separation-of-losses-in-the-transformer


Example

A transformer with normal voltage and supply frequency has eddy current loss and hysteresis loss 1500 W and 2000 W respectively. Find out both the losses for the following conditions

( a ) frequency increases by 10% with normal supply

( b ) Supply voltage increases by 10% with normal frequency

Solution

Primary / secondary voltage of transformer

        V = 4.44 f Bmax Ai N

        ( V / f ) α Bmax

Now

        Wh α Bmax1.6f

              = A Bmax1.6f

              = A ( V / f )1.6f

              = A V1.6 / f – 0.6

     Where A = Constant

        We α Bmax2 f2

             = B ( V / f )2 f2

             = B ( V2 )

     Where B = Constant

From given data

2000 = A V1.6 / f – 0.6 …. ( 1 )  and

1500 = B ( V2 )………. ( 2 )

( a ) Frequency increases by 10% keeping supply voltage constant

New frequency = f + ( 10/100 ) f = 1.1f

Wh = A V1.6 / f – 0.6

Wh = A V1.6 / ( 1.1f ) – 0.6….. ( 3 )

From equation ( 1 ) and ( 3 )

( Wh ) / 2000 = ( 1.1f ) – 0.6 / ( f ) – 0.6

           ( Wh ) = 1889 watt

The eddy current loss does not affected by changing only frequency.

( b ) Supply voltage increases by 10%

New supply voltage = V + ( 10/100 ) V

                                 = 1.1 V

Wh = A ( 1.1V )1.6 / ( f ) – 0.6….( 4 )

We= B ( 1.1V )1.6 ………..( 5 )

From equation ( 1 ) and ( 4 )

( Wh ) / 2000 = ( 1.1V ) 1.6 / ( V ) 1.6

                Wh = 2329.5 watt

From equation ( 2 ) and ( 5 )

( We ) / 1500 = ( 1.1V ) 2 / ( V ) 2

               We = 1815 watt

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