It is seen that the core losses of transformer includes,

1. Hysteresis loss

2. Eddy current loss

For a given volume and thickness of laminations, these losses depend on the operating frequency, maximum flux density in the core and the voltage.

The hysteresis loss is given by Steinmet'z relation,

P

_{h }= K_{h }B_{m}^{1.67}_{ }f v watts
i.e. P

_{h }= A B_{m}^{1.67}_{ }f watts ..............(1)
where A = constant assuming constant voltage

The eddy current loss is given by,

P

_{e }= K_{e }B_{m}^{2}_{ }f^{2}_{ }_{ }t^{2}_{ }watts
i.e. P

_{e }= B B_{m}^{2}_{ }_{ }f^{2}_{ }_{ }watts .........(2)
where B = constant for given thickness t of core

Thus the total core loss becomes,

P

_{i }= P_{h }+ P_{e }= A B_{m}^{1.67}_{ }_{ }f +B B_{m}^{2}_{ }_{ }_{ }_{ }f^{2}.........(3)_{ }_{ }
Practically conduct two tests on transformers at two different frequencies f

_{1 }and f_{2}, keeping maximum flux density in the core same. The results are to be used in the equations (1), (2) and (3) to obtain the constants A and B. Thus the core losses i.e. iron losses can be separated into hysteresis and eddy current losses.**Example**1 : A single phase transformer shows 63 W core losses at 40 Hz while 110 W at 60 Hz. Both the tests are performed at same value of maximum flux density in the core. Find hysteresis and eddy current losses at 50 Hz frequency.

**Solution**: P

_{i1 }= 63 W, f

_{1 }= 40 Hz, P

_{i2 }= 110 W, f

_{2 }= 60 Hz, B

_{m}is same.

As B

_{m}is same, it can be absorbed in the constants A and B. Thus we can write,
P

_{h }= Af while P_{e }= B_{ }_{ }f^{2}_{ }_{ }**.**P

^{.}._{i }= P

_{h }+ P

_{e }= Af + B

_{ }

_{ }f

^{2}

_{ }

**.**63 = A x 40 + B x (40)

^{.}._{}

^{2}

_{ }...........(1)

and 110 = A x 60 + B x (60)

_{}^{2}_{ }............(2)
Solving (1) and (2) we get,

A = 1.0584, B = 0.0129

Thus two losses at 50 Hz are,

P

_{h }= Af = 1.0584 x 50 = 52.92 W
P

_{e }= B_{ }_{ }f^{2}= 0.0129 x 50_{}^{2}= 32.25 W
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