Category: electrical Machines

Energy Stored in a Magnetic Field In the previous chapter, we discussed that in an electromechanical energy conversion device, there is a medium of coupling between electrical and mechanical systems. In most of practical devices, magnetic field is used as the coupling medium. Therefore, an electromechanical energy conversion device comprises an electromagnetic system. Consequently, the energy stored in the coupling medium is in the form of the magnetic field. We can calculate the energy stored in the magnetic field of an electromechanical energy conversion system as described below. Consider a coil having N turns of conductor wire wound around a magnetic core as shown in Figure-1. This coil is energized from a voltage source of v volts. By applying KVL, the applied voltage to the coil to given by, $$mathrm{mathit{V:=:e:+:iR}cdot cdot cdot (1)}$$ Where, e is induced EMF in the coil due to electromagnetic induction. R is the resistance of the coil circuit. $mathit{i}$ is the current flowing the coil. The instantaneous power input to the electromagnetic system is given by, $$mathrm{mathit{p}:=:mathit{Vi:=:ileft ( e+iR right )}}$$ $$mathrm{Rightarrow mathit{p}:=:mathit{ie+ i^{mathrm{2}}}mathit{R}cdot cdot cdot (2)}$$ Now, let a direct voltage is applied to the circuit at time t = 0 and that at end of t = t1 seconds, and the current in the circuit has attained a value of I amperes. Then, during this time interval, the energy input the system is given by, $$mathrm{mathit{W}_{in}:=:int_{0}^{t_{mathrm{1}}}:mathit{p:dt}}$$ $$mathrm{Rightarrow mathit{W}_{in}:=:int_{0}^{t_{mathrm{1}}}:mathit{ie:dt}:+:int_{0}^{t_{mathrm{1}}}mathit{i^{mathrm{2}}R:dt}cdot cdot cdot (3)}$$ From Equation-3, it is clear that the total input energy consists of two parts − The first part is the energy stored in the magnetic field. The second part is the energy dissipated due to electrical resistance of the coil. Thus, the energy stored in the magnetic field of the system is, $$mathrm{mathit{W}_{mathit{f}}:=:int_{0}^{t_{mathrm{1}}}:mathit{ie:dt}:cdot cdot cdot (4)}$$ According to Faraday’s law of electromagnetic induction, we have, $$mathrm{mathit{e}:=:frac{mathit{dpsi }}{mathit{dt}}:=:frac{mathit{d}}{mathit{dt}}left ( mathit{Nphi } right ):=:mathit{N}frac{mathit{dphi }}{mathit{dt}}cdot cdot cdot (5)}$$ Where, $psi$ is the magnetic flux linkage and it is equal to $mathit{psi :=:Nphi }$. $$mathrm{therefore mathit{W_{f}}:=:int_{0}^{mathit{t_{mathrm{1}}}}frac{mathit{dpsi }}{mathit{dt}}mathit{i:dt}}$$ $$mathrm{Rightarrow mathit{W_{f}}:=:int_{0}^{psi_{mathrm{1}}}mathit{i:dpsi }cdot cdot cdot (6)}$$ Therefore, the equation (6) shows that the energy stored in the magnetic field is equal to the area between the ($psi -i$) curve (i.e., magnetization curve) for the electromagnetic system and the flux linkage ($psi$) axis as shown in Figure-2. For a linear electromagnetic system, the energy stored in the magnetic field is given by, $$mathrm{mathit{W_{f}}:=:int_{0}^{mathit{psi _{mathrm{1}}}}mathit{idpsi }:=:int_{0}^{psi_{mathrm{1}} }frac{psi }{mathit{L}}mathit{dpsi }}$$ Where, $psi:=:mathit{Nphi }:=:mathit{Li}$ and L is the self-inductance of the coil. $$mathrm{therefore mathit{W_{f}}:=:frac{psi ^{mathrm{2}}}{2mathit{L}}:=:frac{1}{2}mathit{Li^{mathrm{2}}}cdot cdot cdot (7)}$$ Concept of Coenergy Coenergy is an imaginary concept used to derive expressions for torque developed in an electromagnetic system. Thus, the coenergy has no physical significance in the system. Basically, the coenergy is the area between the $psi -i$ curve and the current axis and is denoted by $mathit{W_{f}^{”}}$ as shown above in Figure-2. Mathematically, the coenergy is given by, $$mathrm{mathit{W_{f}^{”}}:=:int_{0}^{i}psi mathit{di}:=:int_{0}^{i}mathit{Li:di}}$$ $$mathrm{Rightarrow mathit{W_{f}^{”}}:=:frac{1}{2}mathit{Li^{mathrm{2}}}cdot cdot cdot (8)}$$ From equations (7) and (8), it is clear that for a linear magnetic system, the energy stored in the magnetic field and the coenergy are equal. Learning working make money

Equivalent Circuit and Power Factor of Synchronous Motor Equivalent Circuit of Synchronous Motor A synchronous motor is a doubly excited system which means it is connected to two electrical systems, where a three-phase ac supply is connected to the armature winding and a DC supply to the rotor winding. Figure-1 shows the per phase equivalent circuit of a three-phase synchronous motor. Here,V is the voltage per phase applied to the motor, Rais the armature resistance per phase, and $X_{s}$ is the synchronous reactance per phase. These two parameters (i.e. $R_{a}$ and $X_{s}$) give the synchronous impedance per phase ($Z_{s}$) of the motor. From the equivalent circuit of the synchronous motor as shown in Figure-1, we can write its voltage equation as, $$mathrm{mathit{V}:=:mathit{E_{b}+I_{a}Z_{s}}}$$ $$mathrm{Rightarrow mathit{V}:=:mathit{E_{b}+I_{a}left ( R_{a} +jX_{s}right )}}$$ Where, the bold letters represent the phasor quantities. Therefore, the armature current per phase is given by $$mathrm{I_{a}:=:frac{mathit{V-E_{b}}}{mathit{Z_{s}}}:=:frac{mathit{V-E_{b}}}{left ( mathit{R_{a}+jX_{s}}right )}:=:frac{mathit{E_{r}}}{mathit{R_{a}+jX_{s}}}}$$ Where,$mathit{E_{r}}$ is the resultant voltage in the armature circuit. The armature current and synchronous impedance of a synchronous motor are the phasor quantities having magnitude and phase angle. Thus, the magnitude of the armature current is given by, $$mathrm{left|I_{a} right|:=:frac{mathit{V-E_{b}}}{mathit{Z_{s}}}:=:frac{mathit{E_{r}}}{mathit{Z_{s}}}}$$ And, the magnitude of the synchronous impedance is given by, $$mathrm{left|Z_{s} right|:=:sqrt{mathit{R_{a}^{mathrm{2}}+X_{s}^{mathrm{2}}}}}$$ The equivalent circuit and the above equations helps a lot in understanding the operation of a synchronous motor as, When the field excitation is such that $mathit{E_{b}=V}$, then the synchronous motor is said to be normally excited. When the field excitation is such that $mathit{E_{b}<V}$, then the synchronous motor is said to be under excited. When the field excitation is such that $mathit{E_{b}> V}$, then the synchronous motor is said to be over excited. As we shall see in the next section that the excitation of the synchronous motor affects its power factor. Power Factor of Synchronous Motor One of the most important characteristics of a synchronous motor is that it can be made to operate at leading, lagging or unity power factor by changing the field excitation. The following discussion explains how the change in field excitation affects the power factor of a synchronous motor − When the rotor excitation current is such that it produces all the required magnetic flux, then there is no need of extra reactive power in the machine. Consequently, the motor will operate at unity power factor. When the rotor excitation current is less than that required, i.e., the motor is under-excited. In this case, the motor will draw reactive power from the supply to provide the remaining flux. Thus, the motor will operate at a lagging power factor. When the rotor excitation current is more than that required, i.e. the motor is over-excited. In this case, the motor will supply reactive power to the 3-phase line, and behaves as a source of reactive power. Consequently, the motor will operate at a leading power factor. Hence, we may conclude that a synchronous motor absorbs reactive power when it is under-excited and delivers reactive power when it is over-excited. Learning working make money

Output Power of 3-Phase Alternator Consider a three-phase alternator with a cylindrical rotor and is operating at a lagging power factor. Let, E = per phase induced EMF V = per phase terminal voltage Ia = per phase armature current cos$phi$ = power factor (lagging) of load $delta$ = power angle (angle between E and V) Therefore, the output power of a three-phase alternator is given by, $$mathrm{mathit{P_{0}}:=:3mathit{VI_{a}cosphi }cdot cdot cdot (1)}$$ Approximate Output Power of 3-Phase Alternator In a three-phase alternator, the resistance $R_{a}$ of the armature circuit is very small as compared to the synchronous reactance $X_{s}$ of the machine. Thus, we can neglect the armature resistance ($R_{a}$), after that we get approximate equivalent circuit of the alternator as shown in Figure-1. The phasor diagram of the circuit is also shown in Figure-1. From the phasor diagram, we get, $$mathrm{mathit{AB}:=:mathit{I_{a}X_{s}cosphi }:=:mathit{E}:mathrm{sindelta }}$$ $$mathrm{Rightarrow mathit{I_{a}cosphi }:=:frac{mathit{E:mathrm{sindelta }}}{mathit{X_{s}}}cdot cdot cdot (2)}$$ Now, from equation (1) & (2), we get, $$mathrm{mathit{P_{0}}:=:frac{3mathit{EV:mathrm{sindelta }}}{mathit{X_{s}}}cdot cdot cdot (3)}$$ The expression in Equation-3 gives the approximate output power of a three-phase alternator. When the alternator is operating at a constant speed with a constant field current, then Xs and E both are constant and therefore the terminal voltage V is also constant. Thus, from Equation-3, we can observe, $$mathrm{mathit{P_{0}}propto :mathrm{sindelta }}$$ We know, when $delta$ = 90°, then $$mathrm{mathrm{sin:90^{circ}}:=:1}$$ Thus, the alternator supplies maximum power at $$ =90°, and it is given by, $$mathrm{mathit{P_{max}}:=:frac{3mathit{EV}}{mathit{X_{s}}}}cdot cdot cdot (4)$$ The maximum power given by Equation-4 is called the static stability limit of the alternator. Numerical Example A 3-phase, 11 kV, 3 MVA star-connected alternator has a per phase synchronous reactance of 10 $Omega$. Its excitation is such that the generated line EMF is 15 kV. When the alternator is connected to infinite busbars. Calculate the maximum output power of the alternator at the given excitation when armature resistance is neglected. Solution Given data, $$mathrm{mathrm{Line:voltage,}mathit{V_{L}}:=:11:kV:=:11000:V}$$ $$mathrm{therefore:mathrm{Terminal:voltage:per:phase,}mathit{V}:=:frac{11000}{sqrt{3}}:=:6350.85:V}$$ $$mathrm{mathrm{Generated:line:EMF}:=:15:kV:=:15000:V}$$ $$mathrm{therefore:mathrm{Generated:EMF:per:phase,}mathit{E}:=:frac{15000}{sqrt{3}}:=:8660.25:V}$$ $$mathrm{mathrm{Synchronous: reactance: per: phase,}:mathit{X_{s}}:=:10:Omega }$$ Therefore, the maximum power output of the alternator will be, $$mathrm{mathit{P_{max}}:=:frac{3mathit{EV}}{mathit{X_{s}}}:=:frac{3times 8660.25times 6350.85}{10}}$$ $$mathrm{therefore mathit{P_{max}}:=:16499times 10^{3}W:=:16499:mathrm{kW}}$$ Learning working make money

Back EMF in DC Motor In a DC motor, when the armature rotates under the influence of the driving torque, the armature conductors move through the magnetic field, and therefore an EMF is induced in them by the generator action. This induced EMF in the armature conductors acts in opposite direction to the applied voltage $mathit{V_{s}}$ and is known as the back EMF or counter EMF. The magnitude of the back EMF is given by, $$mathrm{mathit{E_{b}}:=:frac{mathit{NPphi Z}}{mathrm{60}mathit{A}}:cdot cdot cdot (1)}$$ The back EMF $mathit{E_{b}}$ is always less than the applied voltage $mathit{V_{s}}$. However, this difference is small when the DC motor is running under normal conditions. In DC motors, the back EMF $mathit{E_{b}}$ induced in the armature opposes the applied voltage, thus the applied voltage has to overcome this EMF $mathit{E_{b}}$ to force a current $mathit{I_{a}}$ in the armature circuit for motor action. The required power to overcome this opposition is given by, $$mathrm{mathit{P_{m}}:=:mathit{E_{b}I_{a}}:cdot cdot cdot (2)}$$ The power $mathit{P_{m}}$ is one which actually converted into mechanical power. For this reason, the power $mathit{P_{m}}$ is also called as electrical equivalent of mechanical power developed. Consider a shunt DC motor whose electrical equivalent circuit is shown in Figure-1. When a DC voltage $mathit{V_{s}}$ is applied across terminals of the motor, the field electromagnets are excited and armature conductors are supplied with current. Hence, a driving torque acts on the armature which begins to rotate. When the armature rotates, a back EMF is induced in the armature conductors that opposes the applied voltage $mathit{V_{s}}$. This applied voltage has to force a current through the armature conductors against the back EMF. The voltage equation of the dc motor can be expressed as, $$mathrm{mathit{V_{s}:=:E_{b}+I_{a}R_{a}}:cdot cdot cdot (3)}$$ Where,$mathit{R_{a}}$ is the resistance of the armature circuit. Then, the armature current of the DC motor is given by, $$mathrm{mathit{I_{a}}:=:frac{mathit{V_{s}-E_{b}}}{mathit{R_{a}}}:cdot cdot cdot (4)}$$ Since the applied voltage $mathit{V_{s}}$ and armature resistance $mathit{R_{a}}$ are usually fixed for a given motor, then the value of $mathit{E_{b}}$ will determine the current drawn by the DC motor. If the speed of the DC motor is high, then the value of back EMF is large and hence the motor will draw less armature current and vice-versa. Significance of Back EMF in DC Motor The back EMF in a DC motor makes it a self-regulating machine, which means it makes the motor to draw a sufficient amount of armature current to develop the torque required by the mechanical load. Now, from Equation-4, we may explain the importance of back EMF in the DC motor as − Case 1 − Motor running on no load In this case, the dc motor requires a small torque to overcome the frictional and windage losses. Thus, the armature current $mathit{I_{a}}$ drawn by the motor is small and the back EMF is nearly equal to the supply voltage. Case 2 − Motor load is changed suddenly In this case, when load is suddenly attached to the motor shaft, the armature slows down. Consequently, the speed at which the armature conductors move through the magnetic field is decreased and hence the back EMF decreases. This decreased back EMF allows a larger current to pass through the armature conductors, and larger armature current means high driving torque. Hence, it is clear that the driving torque increases as the motor speed decreases. The reduction in motor speed stops when the armature current is sufficient to produce the increased torque required by the mechanical load. Consider another situation, where the load on the motor is decreased. In this case, the driving torque is momentarily more than the requirement so that the armature is accelerated. The increase in the armature speed increases the back EMF, and causes the armature current to decrease. Once the armature current is just sufficient to produce the reduced driving torque required by the load, the motor will stop accelerating. This discussion clears that the back EMF in a dc motor automatically regulates the flow of armature current to meet the load requirements. Learning working make money

Losses in DC Machines In DC machines (generator or motor), the losses may be classified into three categories namely, Copper losses Iron or core losses Mechanical losses All these losses appear as heat and hence raise the temperature of the machine. They also reduce the efficiency of the machine. Copper Losses In dc machines, the losses that occur due to resistance of the various windings of the machine are called copper losses. The copper losses are also known as I2R losses because these losses occur due to current flowing through the resistance of the windings. The major copper losses that occur in dc machines are as, $$mathrm{mathrm{Armature:copper:loss}:=:mathit{I_{a}^{mathrm{2}}R_{a}}}$$ $$mathrm{mathrm{Series:field:copper:loss}:=:mathit{I_{se}^{mathrm{2}}R_{se}}}$$ $$mathrm{mathrm{Shunt:field:copper:loss}:=:mathit{I_{sh}^{mathrm{2}}R_{sh}}}$$ In dc machines, there is also a brush contact loss due to brush contact resistance. In practical calculation, this loss is generally included in armature copper loss. Iron Losses The iron losses occur in core of the armature of a DC machine due to rotation of the armature in the magnetic field. Because these losses occur in core of the armature, these are also called core losses. There are two types iron or core losses namely hysteresis loss and eddy current loss. Hysteresis Loss The core loss that occurs in core of the armature of a dc machine due to magnetic field reversal in the armature core when it passes under the successive magnetic poles of different polarity is called hysteresis loss. The hysteresis loss is given by the following empirical formula, $$mathrm{mathrm{Hysteresis:loss,}mathit{P_{h}}:=:mathit{k_{h}B_{max}^{mathrm{1.6}}fV}}$$ Where, $mathit{k_{h}}$ is the Steinmetz’s hysteresis coefficient, $mathit{B_{max}}$ the maximum flux density,f is the frequency of magnetic reversal, and V is the volume of armature core. The hysteresis loss in dc machines can be reduced by making the armature core of such materials that have a low value of Steinmetz’s hysteresis coefficient like silicon steel. Eddy Current Loss When the armature of a DC machine rotates in the magnetic field of the poles, an EMF is induced in core of the armature which circulates eddy currents in it. The power loss due to these eddy currents is known as eddy current loss. The eddy current loss is given by, $$mathrm{mathrm{Eddy:current:loss,}mathit{P_{e}}:=:mathit{k_{e}B_{max}^{mathrm{2}}f^{mathrm{2}}t^{mathrm{2}}V}}$$ Where,$mathit{K_{e}}$ is a constant of proportionality, and tis the thickness of lamination. From the expression for eddy current loss it is clear that the eddy current loss depends upon the square of thickness of lamination. Therefore, to reduce this loss, the armature core is built up of thin laminations that are insulated from each other by a thin layer of varnish. Mechanical Losses The power losses due to friction and windage in a dc machine are known as mechanical losses. In a dc machine, the friction loss occurs in form of bearing friction, brush friction, etc. while the windage loss occurs due to air friction of rotating armature. The mechanical losses depend upon the speed of the machine. But these losses are practically constant for a given speed. Note− Iron or core losses and mechanical losses together are known as stray losses. Constant and Variable Losses In DC machines, we may group the above discussed losses in the following two categories − Constant Losses Variable Losses Those losses in a DC machine that remain constant at all loads are called constant losses. These losses include − iron losses, shunt field copper loss, and mechanical losses. Those losses in a DC machine that vary with load are known as variable losses. The variable losses in a DC machine are − armature copper loss and series field copper loss. Total losses in a DC machine = Constant losses + Variable losses Learning working make money

Discuss Electrical Machines Electrical Machines is a core subject within electrical engineering discipline that deals with the design, operation and applications of energy conversion devices. A system that converts electrical energy into other forms of energy is known as an Electrical Machine. The purpose of the tutorial is to introduce and explain the fundamental concepts in Electrical Machines, which include Basic Concepts of Electromechanical Energy Conversion Devices, Transformers, DC Machines (Motor and Generator), Induction Motors, and Synchronous Machines (Alternator and Motor). Learning working make money

Electrical Machines – Useful Resources The following resources contain additional information on Electrical Machines. Please use them to get more in-depth knowledge on this. Useful Video Courses 169 Lectures 13.5 hours 36 Lectures 4 hours 28 Lectures 3 hours 64 Lectures 7.5 hours 41 Lectures 2 hours 16 Lectures 1.5 hours Learning working make money

Types of DC Machines A DC machine is a device which converts electrical energy in the form of direct current into mechanical energy or mechanical energy into electrical energy in the form of direct current. Therefore, a DC machine is basically an electromechanical energy conversion device. Based on the energy conversion, DC machines can be classified into the following two types − DC Motor DC Generator The basic construction of both DC motor and DC generator is almost similar. However, the fundamental principles involved in the operation of DC motor and DC generator are different. Every DC machine essentially consists of a system of conductors and system of magnets or electromagnets. The system of conductor is called armature, and in case of DC machines it is mounted on a movable shaft. The system of magnets or electromagnets is known as the field system, and it produces the required working magnetic flux. DC Motor The block diagram of a DC motor is shown in Figure-1. When the DC machine is designed to convert DC electrical energy into rotational mechanical energy, it is called a DC motor. Therefore, in case of a DC motor, the electrical energy is supplied to the machine through input terminals and mechanical energy output is taken from the shaft in the form of rotation of the shaft. DC Generator The block diagram of a DC generator is shown in Figure-2. A DC machine which can convert mechanical energy input into electrical energy output is known as a DC generator. Thus, in a DC generator, the mechanical energy from a source like engine, turbine, etc. is supplied to the DC machine in the form of rotational energy of the shaft and the DC electrical energy is received as output from the armature terminals. Learning working make money

Losses and Efficiency of 3-Phase Alternator Losses in Three-Phase Alternator The losses that occur in a three-phase alternator may be divided into the following four categories − Copper losses Iron or core losses Mechanical losses Stray load losses Read through this section to find out more about the types of losses that occur in a three-phase alternator. Copper Losses The copper losses occur in the armature winding and rotor winding of the alternator due to their resistance when currents flowing through them. Thus, these losses are also called I2R losses. Iron or Core Losses The iron or core losses occur in iron parts like stator core and rotor core of the alternator. These losses consist of hysteresis loss and eddy current loss. The core losses occur because various iron parts of the alternator are subjected to the changing magnetic field. Mechanical Losses The mechanical losses occur in the moving or rotating parts like rotor, shaft, bearings, etc. of the alternator. There are two main types of mechanical losses namely, frictional losses and windage losses. The frictional losses are due to friction of bearings in the alternator, while the windage losses are due to friction between the rotating parts of the alternator and the air inside the casing of the alternator. Stray Load Losses This category includes those losses in the alternator which cannot be easily determined. These losses are also called miscellaneous losses. The stray load losses may be because of the following reasons − Distortion of main field flux due to armature reaction. Non-uniform distribution of current over the area of cross-section of armature conductors. In practical calculations, we take the stray load losses 1% of the full-load losses. Note The iron losses and mechanical losses together are called rotational losses because these losses occur in the alternator due to rotation of the rotor. All these losses that occur in an alternator are converted into heat and result in the increase of the temperature and decrease in the efficiency of the alternator. Efficiency of Three-Phase Alternator The ratio of output power to input power of an alternator is called efficiency of the alternator. The efficiency is usually expressed in percentage. $mathrm{mathrm{Efficiency,} : eta :=:frac{Output:Power}{Input:Power}times 100%:=:frac{Output:Power}{Output:Power+Losses}times 100%}$ Now, we will derive the expression for efficiency of a three-phase alternator. For that consider a three-phase alternator operating at a lagging power factor. Let, V= terminal voltage per phase Ia = armature current per phase cos $phi$ = load power factor (lagging) Thus, the output power of a three-phase alternator is given by, $$mathrm{mathit{P_{0}}:=:3:mathit{VI_{a}cos:phi }}$$ The losses in the alternator are, $$mathrm{mathrm{Armature: copper: loss,}mathit{P_{cu}}:=:3:mathit{I_{mathit{a}}^{mathrm{2}}R_{a}}}$$ $$mathrm{mathrm{Field :winding: copper: loss}:=:mathit{V_{f}I_{f}}}$$ Where,Vf is the DC voltage across field winding and If is the DC field current. $$mathrm{mathrm{Rotational: losses,}mathit{P_{r}}:=:mathrm{Core:losses:+:Mechanical:losses}}$$ $$mathrm{mathrm{Stray:load:losses}mathit{P_{s}}}$$ $$mathrm{therefore mathrm{Total:losses:in:alternator,}mathit{P_{loss}}:=:3:mathit{I_{a}^{2}R_{a}:+:P_{r}:+:P_{s}:+:V_{f}I_{f}}}$$ Since, the speed of rotation of the rotor is constant so the rotational losses are constant. The field winding copper losses are also constant. If we assume stray-load losses to be constant. Then, we have, $$mathrm{mathrm{Total:constant:losses,}mathit{P_{c}}:=:mathit{P_{r}:+:P_{s}:+:V_{f}I_{f}}}$$ $$mathrm{therefore:mathrm{Variable:losses} :=:mathrm{3}mathit{I_{a}^{mathrm{2}}R_{a}}}$$ Hence, the efficiency of the alternator is given by, $$mathrm{eta :=:frac{mathit{P_{0}}}{mathit{P_{0}+mathrm{Losses}}}:=:frac{3mathit{VI_{a}cosphi }}{3mathit{VI_{a}cosphi :+:mathrm{3}mathit{I_{a}^{mathrm{2}}R_{a}}:+P_{c}}}cdot cdot cdot (1)}$$ Equation-1 can be used to determine the efficiency of a three-phase alternator. Condition for Maximum Efficiency The efficiency of an alternator will be maximum when the variable losses are equal to the constant losses, i.e., $$mathrm{mathit{P_{c}}:=:3:mathit{I_{a}^{mathrm{2}}R_{a}}cdot cdot cdot (2)}$$ In practice, the maximum efficiency of an alternator usually occurs at about 85% of the full rated load. Numerical Example A three-phase alternator has a per phase terminal voltage of 230 V, and the per phase armature current of 14.4 A. The resistance of the armature circuit of the alternator is 0.5Ω, and the constant losses are 200 watts. Calculate the efficiency and maximum efficiency of the alternator if it is suppling a load at 0.8 lagging power factor. Solution $$mathrm{mathrm{Efficiency,}eta :=:frac{3mathit{VI_{a}cosphi }}{3mathit{VI_{a}cosphi :+:mathrm{3}mathit{I_{a}^{mathrm{2}}R_{a}}:+P_{c}}}}$$ $$mathrm{Rightarrow:eta :=:frac{3times 230times 14.4times 0.8}{left ( 3times 230times 14.4times 0.8 right ):+:left ( 3times 14.4^{2}times 0.5 right ):+:200}}$$ $$mathrm{therefore eta :=:0.9395:=:93.95%}$$ For maximum efficiency of the alternator, $$mathrm{mathit{P_{c}}:=:3:mathit{I_{a}^{mathrm{2}}R_{a}}}$$ $$mathrm{therefore eta_{max} :=:frac{3mathit{VI_{a}cosphi }}{3mathit{VI_{a}cosphi :+:mathrm{2}mathit{P_{c}}}}:=:frac{3times 230times 14.4times 0.8}{left ( 3times 230times 14.4times 0.8 right ):+:left (2times 200 right )}}$$ $$mathrm{therefore eta_{max} :=:0.9521:=:95.21%}$$ Learning working make money

Ideal and Practical Transformers Ideal Transformer An ideal transformer is an imaginary model of the transformer which possesses the following characteristics − The primary and secondary windings have negligible (or zero) resistance. It has no leakage flux, i.e., whole of the flux flows through the magnetic core of the transformer. The magnetic core has infinite permeability, which means it requires negligible MMF to establish flux in the core. There are no losses due winding resistances, hysteresis and eddy currents. Hence, its efficiency is 100 %. Working of an Ideal Transformer We may analyze the operation of an ideal transformer either on no-load or on-load, which is discussed in the following sections. Ideal Transformer on No-Load Consider an ideal transformer on no-load, i.e., its secondary winding is open circuited, as shown in Figure-1. And, the primary winding is a coil of pure inductance. When an alternating voltage $mathit{V_{mathrm{1}}}$ is applied to the primary winding, it draws a very small magnetizing current $mathit{I_{mathit{m}}}$ to establish flux in the core, which lags behind the applied voltage by 90°. The magnetizing current Im produces an alternating flux $mathit{phi_{m}}$ in the core which is proportional to and in phase with it. This alternating flux ($mathit{phi_{m}}$) links the primary and secondary windings magnetically and induces an EMF $mathit{E_{mathrm{1}}}$ in the primary winding and an EMF $mathit{E_{mathrm{2}}}$ in the secondary winding. The EMF induced in the primary winding $mathit{E_{mathrm{1}}}$ is equal to and opposite of the applied voltage $mathit{V_{mathrm{1}}}$ (according to Lenz’s law). The EMFs $mathit{E_{mathrm{1}}}$ and $mathit{E_{mathrm{2}}}$ lag behind the flux ($mathit{phi_{m}}$) by 90°, however their magnitudes depend upon the number of turns in the primary and secondary windings. Also, the EMFs $mathit{E_{mathrm{1}}}$ and $mathit{E_{mathrm{2}}}$ are in phase with each other, while $mathit{E_{mathrm{1}}}$ is equal to $mathit{V_{mathrm{1}}}$ and 180° out of phase with it. Ideal Transformer on On-Load When a load is connected across terminals of the secondary winding of the ideal transformer, the transformer is said to be loaded and a load current flows through the secondary winding and load. Consider an inductive load of impedance connected across the secondary winding of the ideal transformer as shown in Figure-2. Then, the secondary winding EMF $mathit{E_{mathrm{2}}}$ will cause a current $mathit{I_{mathrm{2}}}$ to flow through the secondary winding and load, which is given by, $$mathrm{mathit{I_{mathrm{2}}}:=:frac{mathit{E_{mathrm{2}}}}{mathit{Z_{mathit{L}}}}:=:frac{mathit{V_{mathrm{2}}}}{mathit{Z_{mathit{L}}}}}$$ Where, for an ideal transformer, the secondary winding EMF $mathit{E_{mathrm{2}}}$ is equal to the secondary winding terminal voltage $mathit{V_{mathrm{2}}}$. Since we considered an inductive load, therefore, the current $mathit{I_{mathrm{2}}}$ will lag behind $mathit{E_{mathrm{2}}}$ or $mathit{V_{mathrm{2}}}$ by an angle of $mathit{phi_{mathrm{2}}}$. Also, the no-load current $mathit{I_{mathrm{0}}}$ being neglected because the transformer is ideal one. The current flowing in the secondary winding ($mathit{I_{mathrm{2}}}$) sets up an MMF ($mathit{I_{mathrm{2}}}mathit{N_{mathrm{2}}}$) which produces a flux $mathit{phi_{mathrm{2}}}$ in opposite direction to the main flux ($mathit{phi_{mathit{m}}}$). As a result, the total flux in the core changes from its original value, however, the flux in the core should not change from its original value. Therefore, to maintain the flux in the core at its original value, the primary current must develop an MMF which can counter-balance the demagnetizing effect of the secondary MMF $mathit{I_{mathrm{2}}}mathit{N_{mathrm{2}}}$. Hence, the primary current $mathit{I_{mathrm{1}}}$ must flow so that $$mathrm{mathit{I_{mathrm{1}}}mathit{N_{mathrm{1}}}:=:mathit{I_{mathrm{2}}}mathit{N_{mathrm{2}}}}$$ Therefore, the primary winding must draw enough current to neutralize the demagnetizing effect of the secondary current so that the main flux in the core remains constant. Hence, when the secondary current ($mathit{I_{mathrm{2}}}$) increases, the primary current ($mathit{I_{mathrm{1}}}$) also increases in the same manner and keeps the mutual flux ($mathit{phi_{mathit{m}}}$) constant. In an ideal transformer on-load, the secondary current $mathit{I_{mathrm{2}}}$ lags behind the secondary terminal voltage $mathit{V_{mathrm{2}}}$ by an angle of $mathit{phi _{mathrm{2}}}$. Practical Transformer A practical transformer is one which possesses the following characteristics − The primary and secondary windings have finite resistance. There is a leakage flux, i.e., whole of the flux is not confined to the magnetic core. The magnetic core has finite permeability, hence a considerable amount of MMF is require to establish flux in the core. There are losses in the transformer due to winding resistances, hysteresis and eddy currents. Therefore, the efficiency of a practical transformer is always less than 100 %. The analytical model of a typical practical transformer is shown in Figure-3. Characteristics of a Practical Transformer Following are the important characteristics of a Practical Transformer − Winding Resistances The windings of a transformer are usually made up of copper conductors. Therefore, both the primary and secondary windings will have winding resistances, which produce the copper loss or $mathit{i^{mathrm{2}} mathit{R}}$ loss in the transformer. The primary winding resistance $mathit{R_{mathrm{1}}}$ and the secondary winding resistance $mathit{R_{mathrm{2}}}$ act in series with the respective windings as shown in Figure-3. Iron Losses or Core Losses The core of the transformer is subjected to the alternating magnetic flux, hence the eddy current loss and hysteresis loss occur in the core. The hysteresis loss and eddy current loss together are known as iron loss or core loss. The iron loss of the transformer depends upon the supply frequency, maximum flux density in the core, volume of the core and thickness of the laminations etc. In a practical transformer, the magnitude of iron loss is practically constant and very small. Leakage Flux The current through the primary winding produces a magnetic flux. The flux $mathit{phi _{mathit{m}}}$ which links both primary and secondary windings is the useful flux and is known as mutual flux. However, a fraction of the flux ($mathit{phi _{mathrm{1}}}$) produced by the primary current does not link with the secondary winding. When a load is connected across the secondary winding, a current flows through it and produces a flux ($mathit{phi _{mathrm{2}}}$), which links only with the secondary winding. Thus, the part of $mathit{phi _{mathrm{1}}}$, and the flux $mathit{phi _{mathrm{2}}}$ that link only their respective winding are known as leakage flux. The leakage flux has its path through the air which has very high reluctance. Therefore, the effect of primary leakage flux ($mathit{phi _{mathrm{1}}}$) is to introduce an inductive reactance ($ mathit{X_{mathrm{1}}}$) in series with the primary winding. Similarly, the secondary leakage flux