Electric Machine Losses and Efficiency with Examples

ElectronElectronics & ElectricalDigital Electronics

There are two types of electric machines viz. static electric machines (e.g. transformer) and rotating electric machines (e.g. motors and generators). All electric machines are not ideal, therefore they have some losses due to which efficiency being less than 100%. In general, we come across three popular electric machines viz. transformer, DC machines (motor & generator) and AC machines (motor & generator), therefore we will discuss the loss and efficiency in these machines one by one.

Losses in a DC machines

The losses in DC machines may be divided into three categories as −

Copper Losses

Copper losses occur due to currents in windings of the machines. Therefore, in DC machines, the copper losses being,

$$Armature\:Copper\:Loss=I_{a}^{2}R_{a}\:\:Watts$$

$$Shunt\:Field\:Copper\:Loss=I_{sh}^{2}R_{sh}\:\:Watts$$

$$Series\:Field\:Copper\:Loss=I_{se}^{2}R_{se}\:\:Watts$$

The copper losses are variable losses because they are the function of winding current.

Iron or Core Losses

The iron losses occur in the armature of a DC machine and due to rotation of the armature in the magnetic field. The iron losses are of two types −

Hysteresis Loss

The hysteresis loss occurs in the armature of DC machine because any part of the armature is subjected to magnetic field reversal as it passes under successive poles. Experimentally, it is found that

$$Hysteresis\:Loss,P_{h}=K_{h}B_{max}^{1.6}fV\:\:Watts$$

Where,

  • Kh is the hysteresis coefficient,

  • Bmax is maximum flux density,

  • is the magnetic reversal frequency,

  • V is the volume of armature in m3

To reduce the hysteresis loss, the armature core is mode of such materials which have low value of hysteresis coefficient like silicon steel.

Eddy Current Loss

When the armature rotates in the magnetic field, an EMF is induced in it, due to which eddy currents circulate in the armature core. The power loss due to these eddy currents is termed as eddy current loss. In order to reduce the eddy current loss, the armature core is built up of thin laminations which are insulated from each other by a thin layer of varnish. Therefore,

$$Eddy\:Current\:Loss,P_{e}=K_{e}B_{max}^{2}f^{2}t^{2}V\:\:Watts$$

Where,

  • Ke is a proportionality constant,

  • Bmax is maximum flux density,

  • f is the frequency of magnetic flux,

  • is the thickness of each lamination,

  • V is the volume of armature core.

Since, the parameters of eddy current and hysteresis losses are constant, thus the iron losses of a DC machine are constant losses.

Mechanical Losses

The losses due to friction and windage are called as mechanical losses. The friction losses are like bearing friction, brush friction etc. and the windage loss is like air friction of rotating armature. The mechanical losses occur in the moving parts of the machines and depend upon speed of the machine.

Note – Iron losses and mechanical losses together are known as stray losses, i.e.

Strey Losses = Iron losses + Mechanical losses

Losses in a Transformer

The power losses in a transformer are of two types −

  • Iron or Core Losses

  • Copper Losses

Iron or Core Losses

The irons losses consist of hysteresis and eddy current losses and occur in the core of the transformer due to alternating flux. The iron losses of the transformer can be determined by the open-circuit test.

$$Hysteresis\:Loss,P_{h}=K_{h}B_{max}^{1.6}fV\:\:Watts$$

$$Eddy\:Current\:Loss,P_{e}=K_{e}B_{max}^{2}f^{2}t^{2}V\:\:$$

Also,

$$Iron\:or\:core\:Losses,P_{i}=P_{h}+P_{e}=Constant\:Losses$$

The hysteresis losses can be minimised using silicon steel whereas the eddy current losses can be reduced using core made up of thin laminations.

Copper Losses

Copper losses occur in the primary and secondary windings of the transformer due to their resistance. These can be determined by short circuit test.

$$Copper\:Losses,P_{cu}=I_{1}^{2}R_{1}+I_{2}^{2}R_{2}$$

Losses in Rotating AC Machines

The losses occur in rotating AC machines are also same as those are in DC machines. These losses can be classified into two categories as −

Fixed or Constant Losses

  • Stator iron loss

  • Friction and windage loss

Variable Losses

  • Stator Copper Loss

  • Rotor Copper Loss

Electric Machine Efficiency

The efficiency of an electric machine is defined as the ratio of the output power to the input power, i.e.

$$Efficiency,\eta=\frac{Output\:Power(P_{0})}{Input\:Power(P_{i})}$$

$$\because\:Input\:Power=Output\:Power+Losses$$

$$\therefore\:Efficiency,\eta=\frac{Output\:Power(P_{0})}{Output\:Power(P_{0})+Losses}=(1+\frac{Output\:Power(p_{0})}{Losses})$$

Numerical Example #1

The armature resistance of a compound long shunt DC motor is 0.0858 Ω. It has shunt and series field resistances of 60 Ω and 0.06 Ω respectively. The motor draws a total current of 100 A. If the shunt field current and series field current are 2 A, determine the total cu loss of the motor.

Solution

Armature Current,

$$I_{a}=I_{r}+I_{sh}=100+2=102$$

Therefore, Armature Cu Loss,

$$=I_{a}^{2}R_{a}=102^{2}\times\:0.0858=892.66\:W$$

Series Field Cu Loss,

$$=I_{se}^{2}R_{se}=I_{a}^{2}R_{se}=102^{2}\times\:0.06=624.24\:W$$

Shunt Field Cu Loss,

$$=I_{sh}^{2}R_{sh}=2^{2}\times\:60=240\:W$$

∴ Total Cu Losses,

$$P_{cu}=I_{a}^{2}R_{a}+I_{se}^{2}R_{se}+I_{sh}^{2}R_{sh}$$

⇒ Total Cu Losses,

$$P_{cu}=892.66+624.24+240=1756.9\:W$$

Numerical Example #2

A power transformer has a core material for which hysteresis coefficient is 120 J/m3and eddy current-loss coefficient is 250. Its volume is 10000 cm3 and the maximum flux density is 1.18 Wb/m2 . The core is built up of thin laminations of thickness 8 mm. What is the total iron/core loss in watts, if the frequency of the alternating current is 50 Hz?

Solution

The hysteresis power loss is given by,

$$P_{h}=K_{h}B_{max}^{1.6}fV$$

$$=120\times\:1.18^{1.6}\times\:50\times\:10000\times\:10^{-6}$$

$$=78.19\:W$$

And, the eddy current loss is given by,

$$P_{e}=K_{e}B_{max}^{2}f^{2}t^{2}V$$

$$=250\times\:1.18^{2}\times\:50^{2}\times\:(8\times\:10^{-3})^{2}\times\:10000\times\:10^{-6}$$

$$=0.557\:W$$

Therefore,

Total core losses = $78.19 + 0.557 = 78.747\:W$

Numerical Example #3

In a 25 kVA transformer, the iron loss is 250 W and full load copper loss is 400 W. Find the efficiency at full load at 0.8 power factor lagging.

Solution

Full Load Output,

$$P_{0}=25\times\:0.8=20\:kW$$

Total Full Load Losses,

$$= 250 + 400 = 650 W = 0.65 kW$$

Full Load Input Power,

$$P_{i}= 20 + 0.65 = 20.65\:kW$$

Therefore, Full Load Efficiency,

$$\eta=\frac{P_{0}}{P_{i}}\times\:100=\frac{20}{20.65}\times\:100=96.85\%$$

raja
Published on 02-Jul-2021 09:16:42
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