**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}^{1.6}f

Eddy current loss W_{e} α B_{max}^{2
}f^{2}

For transformer, ( V / f ) α B_{max}

Therefore Wh = Af and We = Bf^{2}

Total^{ }iron losses

W_{i} = W_{h} + W_{e}

= Af + Bf^{2}

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}^{1.6}f

= A B_{max}^{1.6}f

= A ( V / f )^{1.6}f

= A V^{1.6} / f ^{– 0.6}

Where A = Constant

W_{e} α B_{max}^{2 }f^{2}

= B ( V / f )^{2 }f^{2}

= B ( V^{2} )

Where B = Constant

From given data

2000 = A V^{1.6} / f ^{– 0.6} …. ( 1
) and

1500 = B ( V^{2} )………. ( 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

*You may also like :*

What do you mean by voltage harmonics and current
harmonics?

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Describe the effect of harmonics on Power factor.

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