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.

• 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 W_h α B_max^1.6f

Eddy current loss W_e α B_max² f²

For transformer, ( V / f ) α B_max

Therefore Wh = Af and We = Bf²

Total iron losses

          W_i = W_h + W_e

               = Af + Bf²

     W_i / 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 ( W_i / f ) for different frequencies, the constant part represents hysteresis loss and slope of graph represent eddy current loss at any frequency.

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 B_max Ai N

        ( V / f ) α B_max

Now

        W_h α B_max^1.6f

              = A B_max^1.6f

              = A ( V / f )^1.6f

              = A V^1.6 / f ^{– 0.6}

     Where A = Constant

        W_e α B_max² f²

             = B ( V / f )² f²

             = B ( V² )

     Where B = Constant

From given data

2000 = A V^1.6 / 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

W_h = A V^1.6 / f ^{– 0.6}

W_h = A V^1.6 / ( 1.1f ) ^{– 0.6}….. ( 3 )

From equation ( 1 ) and ( 3 )

( W_h ) / 2000 = ( 1.1f ) ^{– 0.6} / ( f ) ^{– 0.6}

           ( W_h ) = 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

W_h = A ( 1.1V )^1.6 / ( f ) ^{– 0.6}….( 4 )

W_e= B ( 1.1V )^1.6 ………..( 5 )

From equation ( 1 ) and ( 4 )

( W_h ) / 2000 = ( 1.1V ) ^1.6 / ( V ) ^1.6

                W_h = 2329.5 watt

From equation ( 2 ) and ( 5 )

( W_e ) / 1500 = ( 1.1V ) ² / ( V ) ²

               W_e = 1815 watt

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