Hysteresis Loss and Eddy Current Loss

Electronics & ElectricalElectronDigital Electronics

Hysteresis Loss

When a magnetic material is subjected to cycle of magnetisation (i.e. it is magnetised first in one direction and then in the other), a power loss occurs due to molecular friction in the material i.e. the magnetic domains of the material resist being turned first in one direction and then in the other. Therefore, energy is required in the material to overcome this opposition. This loss being in the form of heat and is termed as hysteresis loss. The effect of hysteresis loss is the rise of temperature of the machine.

The formula for the calculation of hysteresis loss is devised by Steinmetz, known as Steinmetz hysteresis law. He found that the area of hysteresis loop of a magnetic material is directly proportional to $1.6^{th}$ power of the maximum flux density.

$$\mathrm{Area \:of \:hysteresis \:loop\:\propto \mathit{B^{1.6}_{max}}}$$

$$\mathrm{\Longrightarrow Hysteresis\: energy \:loss =\mathit{\eta B^{1.6}_{max}}}$$

Where, η is a proportionality constant called as hysteresis coefficient. Its value depends upon the nature of magnetic material i.e. the smaller the value of hysteresis coefficient of a material, the lesser is the hysteresis loss.

If f is the frequency of magnetisation and V is the volume of the magnetic material in m3, then,

$$\mathrm{Hysteresis\: power\: loss,\mathit{P_{h}=\eta B^{1.6}_{max}fV}\: Watts}$$

The hysteresis loss can be reduced by using silicon steel to make the core of electric machines.

Numerical Example(1)

A power transformer has a core made up of a magnetic material for which hysteresis coefficient is 120 $J/m^{3}$.Its volume is 9000 $cm^{3}$ and the maximum flux density is 1.45 $Wb/m^{3}$.What is the hysteresis loss in watts if the frequency of magnetisation is 50 Hz?

Solution

$$\mathrm{Hysteresis\: power\: loss,\mathit{P_{h}=\eta B^{1.6}_{max}fV}}$$

$$\mathrm{\Longrightarrow \mathit{P_{h}}=120\times 1.45^{1.6}\times 50\times (9000\times10^{-6})=97.855 W}$$

Eddy Current Loss

When a magnetic material is subjected to a changing magnetic field, a voltage is induced in the material according to Faraday’s law of electromagnetic induction. Since the material is conducting, the induced voltage circulates currents within the body of the magnetic material. These circulating currents are known as eddy currents. These eddy currents causes $\mathit{I^{2}R}$ loss in the material, known as eddy current loss. The eddy current loss also results in the increase in temperature of the material.

$$\mathrm{Eddy\: current\:power \:loss,\mathit{P_{e}=K_{e}B^{2}_{max} f^{2}t^{2}V} \:Watts}$$

Where,

  • $K_{e}$ = Eddy current coefficient,
  • $B_{max}$ = Maximum flux density,
  • f = frequency of magnetisation or flux,
  • t = thickness of lamination, and
  • V = Volume of magnetic material.

The eddy current loss can be reduced as follows −

  • By using thin sheets, called laminations which are insulated from each other by a thin coating of varnish, instead of using a solid block of magnetic material.
  • Using a magnetic material of high resistivity (e.g. silicon steel).

Numerical Example 2

The flux in a magnetic core is alternating sinusoidally at 50 Hz. The maximum flux density is 1.8 $Wb/m^{2}$. The eddy current loss then amounts to 180 W. Determine the eddy current loss in the core when the frequency is 60 Hz and the flux density is 1.3 Wb/m2.

Solution

$$\mathrm{\because \:Eddy\: current \:loss,\mathit {P_{e}\propto B^{2}_{max}f^{2}}}$$

  • Case 1 − When $$\mathrm{\mathit{B_{max1}} = 1.8 Wb/m^{2} and f1 = 50 Hz, then \mathit{P_{e1}}\propto(1.8)^{2}\times(50)^{2}}$$

  • Case 2 − When $$\mathrm{\mathit{B_{max2}} = 1.3 Wb/m^{2} and f2 = 60 Hz, then \mathit{P_{e2}}\propto(1.3)^{2}\times(60)^{2}}$$

Therefore,

$$\mathrm{\frac{\mathit{P_{e2}}{P_{e1}}}=\frac{(1.3)^{2}\times(60)^{2}}{(1.8)^{2}\times(50)^{2}}=0.751}$$

$$\mathrm{\Longrightarrow \mathit{P_{e2}}=0.751\times \mathit{P_{e1}}=0.751\times 180=135.18 W}$$

raja
Published on 23-Jul-2021 12:55:18
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