**2. Eddy Current Losses:**

- The term “eddy currents” is applied to those electric currents which circulate within a
*mass of’ conducting material when the latter**is situated in a varying magnetic field.*The conducting material may be considered as consisting of large number of closed conducting paths, each of which behaves like a short circuited*winding.*The varying magnetic field induces eddy e.m.fs. in these closed elemental paths giving rise to eddy currents. These eddy currents produce loss in power resulting in heating of materials. This loss is of considerable importance as it effects the efficiency and heating of electrical machines. *The eddy currents produce*a*magnetic field of their own which opposes the main magnetic field. As the effect of eddy currents is not uniform over the cross-section of the material, this results in*a*flux distribution which is not uniform, the flux density in the outer portions being greater than that at the centre. Thus there is*a*reduction in effective cross-section.**The**effect of eddy currents upon flux distribution is chiefly of importance in transformers where the material otherwise would be worked at*a*uniform flux density.*- The magnetic materials used for varying magnetic fields are laminated (made up of thin sheets insulated from each other) so as to reduce eddy currents and associated losses, as by laminating, the
*area of path of eddy currents is reduced giving rise to*a*large value of resistance.*

*Eddy **Current **Loss in Thin Sheets: *

Fig. 28 shows a thin plate of thickness *t *and width *b, *thickness being *considerably smaller than the width. *Let us suppose that this sheet carries a flux, *B *= *B _{max} *sinωt,

the field running parallel to the axis of the sheet. Eddy currents would flow in the sheet in the elemental paths a shown in Fig. 28.

Let us consider 1 metre axial length and also 1 metre width of sheet. Take an elemental path of thickness *dx *at a distance *x *from the axis. Flux enclosed by 1 metre length of this path,

ɸ= B (flux density) × 2x × 1 (area) = 2xB

= 2xB sin ωt

– The thickness t of laminated material is of considerable importance as regards to eddy current loss. Eqn. (16) shows that eddy current losses are reduced by using thinner plates and a material of higher resistivity.

– Eqn. (16) is derived by *assuming that magnetic field is uniform throughout and parallel to the axis and also the sheets are very thin. *Therefore, as such, *it should not be applied to cases where frequency is abnormal, the sheets thick or the variation of flux density is not sinusoidal with time. *

The *eddy current loss can be *reduced in the following ways :

- It has been found that this loss can be minimised by building up the required cross-section for the flux path by stacking thin pieces known as
*“laminations”*as shown in Fig. 29. Since the e.m.fs. set up in the material by the varying flux are usually of small magnitude, the natural oxide on the surface of the sheet iron or steel from which the laminations are punched will effectively insulate the laminations from one another, and thus limit each eddy current path to a single lamination.

2. This loss may also be reduced by grinding the ferromagnetic material to a powder and mixing it with a binder that effectively insulates the particles from one another. This mixture is then formed under pressure into the desired shape and heat treated. Magnetic cores for use in communication equipment are frequently made by this process.

- It may be noted that
*iron losses*(hysteresis and eddy current losses)*if allowed to take place unchecked, not only reduce the efficiency of electrical equipment but also raise the temperature of the core.*Hence these losses should be kept*as small*as*is economically possible.*

*Other Factors Affecting Core Loss*

*Rivets and bolts.*Laminations must be held together. A common method is to use rivets bolts, but they*tend to short circuit lamination insulation.*Rivets and bolts very often form part of a closed magnetic loop which, under A.C. operation, may*cause a serious loss of electrical energy.**Burrs.*As dies become worn, the burrs that they form on punched edges of lamination stock become larger and larger. Thus burrs are not always removed by deburring operation, and they cut*through sheet insulation forming contacts between laminations and allow excessive eddy currents to flow.*These are commonly referred to as*intersheet eddy currents.*Machining operations that are performed on stacked laminations, such as the grinding of induction motor stators or broaching of solenoid faces, often smear the lamination together, forming good connections between sheets. Many schemes, such as treating the faces with acid or alkali, have been tried in an attempt to eat away the metal particles after laminations have been assembled into apparatus, but this attempted cure often causes other troubles, such as rusting. Burning of the burrs with a flame has also been attempted with varying success*Pressure.*Mechanical strains set up in magnetic materials either from*cold working**during**rolling or punching*or*distortion during final assembly operations,*affect magnetic properties inseveral ways. The hysteresis and eddy current losses as well as the magnetising current are increased.*T**he increased hysteresis loss due to punching depends upon the material, its**thickness, and the configuration of the grains.*The loss may be as high as 300 per cent of the loss in the unpunched sheet. The strains produced by punching operations increases the loss primarily in the narrow section next to the punched edge. The per cent increase in hysteresis loss is much lower for high flux densities than for low ones.*The**effect of tension on silicon steel is to increase the permeability until high flux densities are**reached.**The effect of compression, however, is much more marked than that of tension and reduces**the permeability**considerably.*

*Total iron losses*

- Total iron loss is the sum of hysteresis and eddy current loss and is given by the relation:

The total iron loss can be calculated with the aid of above expression but it gives good results only in the case of static *machinery and not in the case of dynamo electric machinery. The reason **being that this **expression is applicable when the field is simply alternating, but when the reversal of **flux are due **to rotational magnetic field, the losses do not follow the same laws. Therefore in practice, **iron loss curves **are used which represent the loss in watt per kg as a function of flux density. *

**Example ****17**. *Calculate the energy lost per hour in a specimen of iron subjected to magnetisation at 50 c/s. The specimen weighs 50 kg and the hysteresis loop is equivalent in area to 250 joules per **cubic metre. Density of iron is **7500kg/m ^{3}. *

**Solution**. Loss = 250 joules/m^{3}/cycle.

**Example 18**. *The hysteresis loop of a specimen of iron having 10 kg mass is equivalent in area **to 250 joules/m ^{3} of i*

*ron. Find the loss of energy per hour at the rate of 50*c/

*s. Assume the density of*

*iron as 7500 kg/m*

^{3}.**Solution. **Energy loss/m^{3}/cycle = 250 joules

Density of iron = 7500 kg/m^{3}

**Example 19**. *The hysteresis loop for a certain magnetic material is drawn to the following scale: *1 cm = 4 *AT */ cm *and *1 *cm *= *0.05 Wb */*m ^{2}. The loop area is 90 cm^{2} and density of given material 7800 kg */

*m*/

^{3}. Calculate the hysteresis loss in watts*kg at 50*c / s.

**Solution**. Hysteresis loss = *xy *× (area of *B-H *loop) joules/m^{3}/cycle

Also 1 cm = 4 AT/cm = 400 AT/m

and 1 cm = 0.05 Wb/m^{2}

*i.e., x= **400 *

*y **=0.05 *

loss = 400 0.05 90 = 1800 joules/m^{3}/cycle

**Example 20**. *In a certain transformer, the hysteresis loss is 300 *W *when the maximum flux density is 0.9 Wb**/*m^{2} *and the frequency 50 *c/s. *What would be the hysteresis loss if the maximum flux density were increased to *1.1 *Wb*/m^{2} *and the frequency reduced to 40 *c/s. *Assume the hysteresis loss over this range to be proportional to *

**Solution.** We know that,

**Example 21**. *Calculate the energy loss per hour by hysteresis in 50 kg iron core which **is **subjected to a sinusoidal flux alternating at 50 *c / s. *The hysteresis loop for the core has an area of 150 cm ^{2}, when plotted to a scale of *1 cm =

*0.008 Wb*/

*m*1 cm =

^{2},*20 AT*/

*cm.*

**Solution**. Here 1 cm = 20 AT/cm = 2000 AT/m

1 cm = 0.008 Wb/m^{2}

Area of hysteresis loop = 150 cm^{2}

Loss = *xy *× (area of *B*/*H loop) *joules/m^{3}/cycle

= 2000 × 0.008 × 150 = 2400 joules/m^{3}/cycle

Taking density of iron 7500 kg/m^{3},

**Example 22**. *Calculate the loss per kg in a specimen of alloy steel for a maximum flux
*density of

*1.*

*1 Wb*/m

^{2}

*and a frequency of 50 Hz, using 0.5*mm

*thick sheets. The resistivity of alloy resistivity of alloy steel is 30×10*m.

^{-8 Ω}*The density is 7800 kg*

*/*

*m*/

^{3}. Hysteresis loss in each cycle is 380 W-s*m*

^{3}.**Solution**. Maximum flux density, *B*_{max}* *= 1.1 Wb/m^{2}

Frequency = 50 Hz

Thickness of sheet = 0.5 mm

Resistivity of alloy steel, ρ = 30 × 10^{-8 } Ωm

Density = 7800 kg/m^{3}

Hysteresis loss/cycle = 380 W-s/m^{3}

**Total iron loss:**

Eddy current loss is given by,

**Example ****23***.A specimen of cold rolled grain oriented 0.3 mm thick stampings has a resistivity of 4.8 × 10 ^{-7 }*

*Ωm. The published hysteresis loop is essentially rectangular in form, with a coercive*

*force, of 12 AT/m for all values of peak flux densities upto 1.6 Wb/m*

^{2}. The manufacturer quotes the*iron loss in material*

*as*

*1.2 W /*

*kg, with a sinusoidal flux density 1.0 Wb/*

*m*

^{2}*(peak) at 100 Hz. Calculate*

*the loss in the*

*material from its properties and compare it with the quoted value. The density is 7650 kg/m*

^{3}.**Solution.** Thickness of stamping = 0.3 mm

Resistivity, ρ = 4.8 × 10^{-7} Ωm

Density = 7650 kg/m^{3}

The material is subjected to a peak flux density of 1 Wb/m^{2} and therefore the value of coercive force is 12 AT/m for all densities. The hysteresis loop is thus a rectangle with sides 2 × 1 = 2 Wb/m^{2}.

Loss per cycle = area of hysteresis loop = 24 × 2 = 48 W-s/m^{3}

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