1 November 2017

Losses in the Transformer

The following losses are occurs in the transformer.

  • This loss is due to rapid reversal of magnetism due to alternating flux.
  • It depends on the loop area of the material. The hysteresis loss is given by

        Wh = hBmax1.6fV Joule / second or watt
Where h = Steinmetz constant, its value depends on the core material
     Bmax = Maximum flux density in weber / meter2
            f = Supply frequency in Hz
          V = Volume of the core in meter3
  • An emf is induced in the core due to alternating flux in the core according to the laws of electromagnetic Induction. 
  • Although this emf is small, it sets up large current in the core due to small core resistance.
  • The loss due to this current is known as eddy current loss. The eddy current loss is given by
         We = K Bmax1.6f2t2  
Where K = Constant
     Bmax = Maximum flux density in weber / meter2
            f = Supply frequency in Hz
            t = Thickness of lamination in mm
  • The eddy current loss varies as ( I ) Square of maximum flux density ( II ) Square of supply frequency ( III ) Square of thickness of lamination.
  • If the supply voltage keeps constant, the core loss remains constant. The core loss does not depend on the load current.

  • The value of core loss can find out by open circuit test.

These losses are due to ohmic resistances of the transformer winding.
  • Copper losses in the primary winding = I12 R1
  • Copper losses in the secondary winding = I22 R2
  • Total copper losses = I12 R1 + I22 R2
  • Total copper losses at full load ( Wc ) = I12 R01

                                                                    = I22 R02
Where R01 = Transformer winding resistance as referred to primary side
            R02 = Transformer winding resistance as referred to secondary  side
Copper Losses Wc α ( I )2
  • It means that copper loss varies as square of the current.

Total copper losses at half load
Wc  α ( I / 2 )2 R
        =I2R / 4
  • Therefore total copper loss at half load is reduced four time that of full load copper losses.

Total copper losses at twice full load
       Wc α ( 2I )2 R
             = 4I2 R
  • Therefore total copper loss at twice full load is four times that of full load copper losses. 
  • The copper loss depends on the value of square load current and its value can find out by short circuit test.

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