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Electric Machine Losses and Efficiency with Examples
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,
f 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,
t 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,
$$\eta=\frac{P_{0}}{P_{i}}\times\:100=\frac{20}{20.65}\times\:100=96.85\%$$