The following losses are occurs in the transformer.
- The core loss depends upon the supply voltage.
- The coreloss practically remains constant for all loads because the core flux remains constant at any load conditions.
- This loss includes hysteresis loss and eddy current loss.
- 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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