Data Structure
Networking
RDBMS
Operating System
Java
MS Excel
iOS
HTML
CSS
Android
Python
C Programming
C++
C#
MongoDB
MySQL
Javascript
PHP
Physics
Chemistry
Biology
Mathematics
English
Economics
Psychology
Social Studies
Fashion Studies
Legal Studies
- Selected Reading
- UPSC IAS Exams Notes
- Developer's Best Practices
- Questions and Answers
- Effective Resume Writing
- HR Interview Questions
- Computer Glossary
- Who is Who
Losses and Efficiency of an Alternator
Losses in an Alternator
The losses that occur in an alternator can be divided into the following categories −
Copper Losses or I2R Losses
The copper losses or I2R losses occur in the armature winding and rotor winding of the alternator. These losses occur due to the resistance of the windings.
$$\mathrm{Armature\:winding\:cu\:loss = {πΌ^{2}_{π}} π _{π}}$$
$$\mathrm{Rotor\:winding\:cu\:loss={πΌ^{2}_{π}} π _{π}}$$
Core Losses
The core losses or iron losses occur in the pole faces, teeth and stator core of the alternator. The core losses in the alternator occur because the various iron parts of the machine are subjected to the varying magnetic field. The core losses consist of eddy current loss and hysteresis loss.
$$\mathrm{Core\:Losses,\:π_{π} = Hysteresis\:Loss (π_{β}) + Eddy\:Current\:Loss(π_{π} )}$$
Hysteresis Loss – The hysteresis loss occurs in the cores of the alternator since any given iron part is subjected to magnetic reversal as the magnetic field varies. When the magnetic reversal occurs, some amount of power has to be spent to overcome the magnetic friction, which is called as hysteresis loss. The hysteresis loss is given by,
$$\mathrm{Hysteresis\:Loss,\:π_{β} = πΎ_{β}\:{π΅^{1.6}_{max}}π\:π\:Watts}$$
Eddy Current Loss – When the armature and rotor cores of the alternator are subjected to the changing magnetic field, an EMF is induced in the cores which circulates eddy currents in the cores. The power loss due to these eddy currents is known as eddy current loss and is given by,
$$\mathrm{Eddy\:Current\:Loss, π_{π} = πΎ_{π}\:{π΅^{2}_{max}}\:π^{2}\:π‘^{2}\:π\:Watt}$$
Mechanical Losses
In the alternators, there are two types of mechanical losses viz. friction losses and windage losses. The friction losses occur due to the friction in the moving parts such as bearings etc. while the windage losses occur due to the friction between the moving parts of the machine and the air inside the alternator’s casing.
$$\mathrm{Mechancial\:losses = Friction\:losses + Windage\:losses}$$
Miscellaneous Losses
All the losses in the alternator which cannot be easily accounted for are known as miscellaneous losses. The miscellaneous losses in the alternator may be result of the following −
Distorted flux due to the effect of armature reaction.
Non-uniform distribution of the current over the cross-section of the armature conductors, etc.
In practice, the miscellaneous losses are taken to be 1% of the full-load losses.
Note −
The core losses and mechanical losses together are known as rotational losses, i.e.,
$$\mathrm{Rotational\:losses,\:π_{π} = Core\:losses + Mechanical\:losses}$$
All the losses occur in the alternator are converted in to heat and result in the increase of temperature and decrease in the efficiency of the alternator.
Efficiency of Alternator
The efficiency of the alternator is defined as the ratio of output power to input power. Therefore, the per unit efficiency of the alternator is given by,
$$\mathrm{Efficiency,\:π =\frac{Output\:power (π_{out})}{Input\:Power\:(π_{in})}}$$
Also, the percentage efficiency is,
$$\mathrm{Efficiency,\:π =\frac{Output\:power (π_{out})}{Input\:Power\:(π_{in})} \times 100 =\frac{π_{out}}{π_{out} + Losses} \times 100}$$
Expression for Efficiency of Alternator
Consider a 3-phase alternator which is supplying a load at lagging power factor.
Let,
$$\mathrm{π = Terminal\:voltage\:per\:phase}$$
$$\mathrm{πΌ_{π} = Armature\:current\:per\:phase}$$
$$\mathrm{cos\:φ = Power\:factor\:of\:the \:load}$$
Therefore, the power output of the alternator is given by,
$$\mathrm{π_{ππ’π‘} = 3ππΌ_{π}\:cos\:φ}$$
The armature copper losses of the alternator are,
$$\mathrm{Armature\:cu\:losses = 3{πΌ^{2}_{π}} π _{π}}$$
And the field winding losses are,
$$\mathrm{Field\:winding\:losses = π_{π}\:πΌ_{π}}$$
Where,
ππ is the DC voltage across the rotor field winding, and
πΌπ is the DC current through the rotor field winding.
Also, the rotational losses are given by,
$$\mathrm{Rotational\:losses,\:π_{π} = Core\:losses + Mechanical\:losses}$$
Now, if ππππ π. represents the miscellaneous losses in the alternator, then
$$\mathrm{Total\:Losses = 3{πΌ^{2}_{π}} π _{π} + π_{π}\:πΌ_{π} + π_{π} + π_{misc}.}$$
As the rotor of the alternator rotates at a constant speed so that the rotational losses are constant. Also, the field winding losses are constant. If the miscellaneous losses are also assumed to be constant, then
$$\mathrm{Total\:constant\:losses,\:π_{πΆ} = π_{π} + π_{misc}. + π_{π}\:πΌ_{π}}$$
The copper losses in the armature winding vary as the square of armature current in the armature winding. As the armature current varies and hence the armature copper losses. For this reason, the armature copper losses are also known as variable losses.
$$\mathrm{∴ \:Variable\:losses = 3{πΌ^{2}_{π}}π _{π}}$$
Therefore, the efficiency of the alternator can be written as,
$$\mathrm{π =\frac{π_{out}}{π_{out} + Losses}=\frac{3ππΌ_{π}\:cos\:φ}{3ππΌ_{π}\:cos\:φ + 3{πΌ^{2}_{π}}π _{π} + π_{πΆ}}}$$
Condition for Maximum Efficiency of Alternator
The efficiency of the alternator will be maximum when the variable losses are equal to the constant losses of the machine, i.e.,
$$\mathrm{Variable\:losses = Cosntant\:Losses}$$
$$\mathrm{\Rightarrow \:3{πΌ^{2}_{π}}π _{π}= π_{πΆ}}$$
In practice, the maximum efficiency of an alternator occurs at about 85% of full load.
3K+ Views
