Category: electrical Machines

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

Power Developed by Synchronous Motor In this chapter, we will derive the expression for mechanical power developed (Pm) by a three-phase synchronous motor. Here, we will neglect the armature resistance Ra of the synchronous motor. Then, the armature copper loss will be zero and hence the mechanical power developed by the motor is equal to the input power (Pin) to the motor, i.e., $$mathrm{mathit{P_{m}}:=:mathit{P_{in}}}$$ Now, consider an under-excited (i.e.,Eb<V) three-phase synchronous motor having zero armature resistance (i.e.,Ra = 0), and is driving a mechanical load. The phasor diagram of one phase of this synchronous motor is shown in the figure. Because the motor is under-excited, thus it will operate at a lagging power factor, say (cos $phi$). From the phasor diagram, it is clear that $mathit{E_{r}}:=:I_{a}X_{s}$ and the armature current per phase $I_{a}$ lags the resultant EMF $mathit{E_{r}}$ by an angle of 90°. Therefore, the input power per phase to the motor is given by, $$mathrm{mathit{P_{in}}:=mathit{VI_{a}}:cos:phi :cdot cdot cdot (1)}$$ Since $mathit{P_{m}}$ is equal to $mathit{P_{in}}$, therefore, $$mathrm{mathit{P_{m}}:=mathit{VI_{a}}:cos:phi :cdot cdot cdot (2)}$$ From the phasor diagram, we have, $$mathrm{mathit{AB}:=mathit{I_{a}X_{s}}:cos:phi :=:mathit{E}_{mathit{b}}:sin:delta}$$ $$mathrm{therefore mathit{I_{a}:cosphi :=:frac{E_{b}:sindelta }{mathit{X_{s}}}}:cdot cdot cdot (3)}$$ Using equations (2) & (3), we obtain, $$mathit{P_{m}:=:frac{mathit{VE_{b}}sindelta }{X_{s}}}cdot cdot cdot (4)$$ This is the expression for mechanical power developed (Pm) per phase by the synchronous motor. For 3-phases of the motor, the developed mechanical power is given by, $$mathit{P_{m}:=:frac{3mathit{VE_{b}}sindelta }{X_{s}}}cdot cdot cdot (5)$$ Also, from the equations (4) & (5), it is clear that the mechanical power developed will be maximum when power angle ($delta$= 90°) electrical. Therefore, For per phase, $$mathit{P_{max}:=:frac{mathit{VE_{b}}}{X_{s}}}cdot cdot cdot (6)$$ For 3-phase, $$mathit{P_{max}:=:frac{mathit{3VE_{b}}}{X_{s}}}cdot cdot cdot (7)$$ Important Points The following important points may be noted about the mechanical power developed by a three-phase synchronous motor − The mechanical power developed by the synchronous motor increases with the increase in power angle ($delta$) and vice-versa. If power angle ($delta$) is zero, then the synchronous motor cannot develop mechanical power. When the field excitation of the synchronous motor is reduced to zero, i.e., ($E_{b}$ = 0), the mechanical power developed by the motor is also zero, i.e. the motor will come to a stop. Numerical Example A 3-phase, 4000 kW, 3.3 kV, 200 RPM, 50 Hz synchronous motor has per phase synchronous reactance of 1.5 $Omega$. At full-load, the power angle is 22° electrical. If the generated back EMF per phase is 1.7 kV, calculate the mechanical power developed. What will be the maximum mechanical power developed? Solution Given data, Voltage per phase, $mathit{V}:=:frac{3.3}{sqrt{3}}:=:1.9:kV$ Back EMF per phase,$mathit{E_{b}}:=:1.7:kV$ Synchronous reactance,$X_{s}:=:1.5Omega $ Power angle,$delta :=:22^{^{circ}}$ Therefore, the mechanical power developed by the motor is, $$mathrm{mathit{P_{m}}:=:frac{3:mathit{VE_{b}:sindelta} }{mathit{X_{s}}}:=:frac{3times 1.9times 1.7times mathrm{sin}:22^{circ}}{1.5}}$$ $$mathit{therefore P_{m}}:=:2.42times 10^{6}:W:=:2.42:MW$$ The mechanical power developed will be maximum when ($delta$ =90°), $$mathit{P_{max}}:=:frac{3mathit{VE_{b}}}{X_{s}}:=:frac{3times 1.9times 1.7}{1.5}$$ $$mathit{therefore P_{max}}:=:6.46:mathrm{MW}$$ Learning working make money

Methods of Starting 3-Phase Induction Motors The following four methods are extensively used for starting the three-phase induction motors − Direct-On-Line Starter Auto-Transformer Starter Star-Delta Starter Rotor Resistance Starter In this chapter, let”s discuss each of these starting methods in detail. Direct On Line Starter As the name implies, the direct-on line (D.O.L.) starter is one in which the three-phase induction motor is started by connecting it directly to a 3-phase balanced ac supply as shown in Figure-1. In this method, the induction motor draws a very high starting current about 4 to 10 times of the full-load current. It is because, the impedance of the motor at standstill is low. Therefore, the direct-on line (D.O.L.) method of starting is suitable for the motor of low power rating, usually up to 7.5 kW. Autotransformer Starter In this method of induction motor starting, a three-phase autotransformer is used to supply three-phase electricity to the motor. The autotransformer is mainly used to reduce the 35. Methods of Starting 3-Phase Induction Motors supply voltage at starting and then connecting the motor to the full supply voltage as the motor attains a sufficient speed. The circuit arrangement of autotransformer starter is shown in Figure-2. The tapings on the autotransformer used for starting the induction motors are provided in such a way that when it is connected in the circuit, 60% to 80% of the supply voltage is applied to the motor. At the instant of starting, the autotransformer is put into the circuit and hence reduced voltage is applied to the motor. Consequently, the starting current is limited to a safe value. When the motor reaches about 80% of the rated speed, the autotransformer is removed from the circuit through a changeover switch, and the motor is then connected to the full supply voltage. The autotransformer starter has several advantages like low power loss, small starting current, etc. Therefore, this method is suitable for relatively larger induction motors of power ratings over 25 hp. Star-Delta Starter In this method, the three-phase induction motor is started as a star-connected motor and run as a delta-connected motor. The induction motors which are started by star-delta starter has the stator windings which are designed for delta operation and are connected in star during the starting period. When the motor attains a sufficient speed, the winding connections are changed from star to delta. Figure-3 shows the circuit arrangement of the star-delta starter. Here, the six terminals of the stator windings are connected to a changeover switch. At the instant of starting, the changeover switch connects the stator windings in star-configuration. As a result, each stator phase gets a voltage equal to V/√3, where V is the full line voltage. In this way, the stator windings get a reduced voltage during starting period. When the motor attains a specific speed, the changeover switch turns the connection of stator windings to delta. Each phase now gets the full line voltage V, and the motor runs at the normal speed. However, this method of starting of three-phase induction motor causes a large reduction in the starting torque of the motor. This method is best suited for medium sized induction motors, up to about 25 hp. Rotor Resistance Starter This method of starting is applicable to slip-ring induction motors only. In this method, a variable star-connected rheostat is inserted into the rotor circuit through slip rings, and full supply voltage is applied to the stator winding. The circuit arrangement of rotor resistance starter is shown in Figure-4. At the instant of starting, the handle of the star-connected rheostat is set in the ‘off’ position. Consequently, a maximum resistance is inserted in each phase of the rotor circuit, and reduces the starting current. At the same time, this resistance increases the starting torque. When the motor picks up speed, the external resistance is gradually removed from the rotor circuit by moving the rheostat handle. Once the motor attained about 80% of the normal speed, the handle is switched to the ‘on’ position, and thus the whole external resistance is removed from the rotor circuit. Learning working make money

Armature Reaction in Synchronous Machines Armature Reaction in an Alternator When a three-phase alternator is operating at no-load, there will be no current flowing through its armature winding. Hence, the magnetic flux produced in the air-gap will be due to rotor field poles only. But, when the alternator is loaded, the three-phase currents flowing through the armature winding will produce a rotating magnetic field in the air-gap. As a result, the resultant magnetic flux in the air-gap is changed. This effect is known as armature reaction, and may be defined as under − The current flowing through the armature winding of a three-phase alternator, the resulting magnetomotive force (MMF) produces a magnetic flux. This armature flux interacts with the main pole flux, and causing the resultant magnetic flux to become either less or more than the original main pole flux. This effect of armature flux on the main pole flux is called armature reaction. In a three-phase alternator, the effect of armature reaction depends upon the magnitude of the armature current and power factor of the load. Which means the power factor of the load determines whether the armature reaction flux distorts, opposes or assists the main field flux. The following discussion explains the nature of armature reaction in synchronous machines for different power factors − Unity Power Factor − When the alternator supplies a load at unity power factor, i.e. purely resistive load, the effect of armature reaction is to distort the main field flux. This is called cross-magnetizing effect of armature reaction. Lagging Power Factor − When the alternator supplies a load at lagging power factor, i.e. purely inductive load, the effect of armature reaction is partly demagnetizing and partly cross-magnetizing. This causes a reduction in generated voltage. Leading Power Factor − When the alternator supplies a load at leading power factor, i.e. purely capacitive load, the effect of the armature reaction is partly magnetizing and partly cross-magnetizing. This causes an increase in generated voltage. Armature Reaction in a Synchronous Motor When the synchronous machine is operated in motoring mode, the armature reaction flux is in phase opposition, which means the nature of armature reaction is reversed what is stated for the alternator. The following points explain the effects of armature reaction when the synchronous machine is operating in motoring mode − Lagging Power Factor − When the synchronous motor draws a current at a lagging power factor, the effect of armature reaction is partly magnetizing and partly cross-magnetizing. Leading Power Factor − When the synchronous motor draws a current at a leading power factor, the effect of armature reaction is partly demagnetizing and partly cross-magnetizing. Learning working make money

Fleming’s Left Hand and Right Hand Rules All electrical machines work on the principle of electromagnetic induction. According to this principle, if there is relative motion between a conductor and a magnetic field, then an EMF is induced in the conductor. On the other hand, if a current carrying conductor is placed in a magnetic field, the conductor experiences a force. For practical and analytical purposes, it is important to determine the direction of induced EMF and force acting on the conductor. Fleming’s hand rules are used for that. An English electrical engineer and physicist John Ambrose Fleming stated two rules in late 19th century to determine the direction of induced EMF and force acting on a current carrying conductor placed in a magnetic field. These rules popularly known as Fleming’s Left Hand Rule and Fleming’s Right Hand Rule. Basically, both left hand rule and right hand rule show a relationship between magnetic field, force and current. Fleming’s left hand rule is used to determine the direction of force acting on a current carrying conductor when it placed in a magnetic field, hence it is mainly applicable in electric motors. Whereas, Fleming’s right hand rule is used to determine the direction of induced EMF in a conductor moving relative to a magnetic field, thus it is mainly applicable in electric generators. Fleming’s Left Hand Rule Fleming’s left hand rule is particularly suitable to find the direction of force on a current carrying conductor in a magnetic field and it may be stated as under − Stretch out the forefinger, middle finger and thumb of your left hand so that they are at right angles (perpendicular) to one another as shown in figure 1. If the forefinger points in the direction of magnetic field, middle finger in the direction of current in the conductor, then the thumb will point in the direction of force on the conductor. In practice, Fleming’s left hand rule is applied to determine the direction of motion of conductor in electric motors. Fleming’s Right Hand Rule Fleming’s right hand rule is particularly suitable to determine the direction of induced EMF and hence electric current in a conductor when there is a relative motion between the conductor and magnetic field. Fleming’s left hand rule may be stated as under − Stretch out the forefinger, middle finger and thumb of your right hand so that they are at right angles (perpendicular) to one another as shown in figure 2. If the forefinger points in the direction of magnetic field, thumb in the direction of motion of the conductor, then the middle finger will point in the direction of induced EMF or current. In practice, Fleming’s right hand rule is used to determine the direction of induced EMF and current in the electric generators. Comparison of Fleming’s Left Hand Rule and Right Hand Rule The following table gives a brief comparison of Fleming’s left hand and right hand rules − Parameters Fleming’s Left Hand Rule Fleming’s Right Hand Rule Purpose Fleming’s LHR is used to determine the direction of force acting on a current carrying conductor in a magnetic field. Fleming’s RHR is used to find the direction of induced EMF or current in a conductor. Use Fleming’s left hand rule is mainly applicable in electric motors. Fleming’s right hand rule is applicable in electric generators. Learning working make money

Working of 3-Phase Alternator A 3-phase alternator is a synchronous machine that converts mechanical energy into 3-phase electrical energy through the process of electromagnetic induction. As we discussed in previous chapters, a 3-phase alternator, also called a 3-phase synchronous generator, has a stationary armature and a rotating magnetic field. In the three-phase alternator, the rotor winding (serves as field winding) is energized from a DC supply and alternate north and south poles are developed on the rotor. Operation of Three-Phase Alternator When the rotor is rotated (say in anticlockwise direction) by a prime mover (engine, turbine, etc.), the stator winding (serves as armature winding) is cut by the magnetic flux of the rotor poles. Due to electromagnetic induction, an EMF is induced in the armature winding. This induced EMF is alternating one because the north and south poles of the rotor alternately pass the armature winding conductors. We can determine the direction of the induced EMF by Fleming’s right hand rule. The electrical equivalent circuit of a star-connected armature winding and dc field winding three-phase alternator is shown in Figure-1. When the rotor is rotated a three-phase voltage is generated in the armature winding. The magnitude of generated voltage depends upon the speed of the rotation of rotor and the DC excitation current. However, the magnitude of generated voltage in each phase of the armature is the same, but displaced by 120° electrical from each other in space as shown in the phasor diagram. Frequency of Generated Voltage In a three-phase alternator, the frequency of generated voltage depends upon the speed of rotation and the number of field poles in machine. Let N = speed of rotation in RPM P = number of field poles Then, the frequency of generated voltage is given by, $$mathrm{mathit{f}:=:frac{mathit{NP}}{120}:mathrm{Hz}:cdot cdot cdot (1)}$$ It should be noted that N is the synchronous speed because the alternator is a synchronous machine whose rotor always rotates at the synchronous speed. EMF Equation of Three-Phase Alternator The mathematical relation which gives the value of EMF induced in the armature winding of a three-phase alternator is termed as its EMF equation. Let N = speed of rotation in RPM P = number of field poles on rotor $phi$ = flux per pole in weber Z = number of armature conductors per phase Then, in one revolution, each stator conductor is cut by a flux of $mathit{Pphi }$ Weber, i.e., $$mathrm{mathit{dphi }:=:mathit{Pphi }}$$ Also, time taken to complete one revolution is, $$mathrm{mathit{dt }:=:frac{60}{mathit{N}}}$$ Therefore, the average EMF induced in each armature conductor is, $$mathrm{mathrm{EMF :per:conductor}:=:mathit{frac{dphi }{dt}}:=:frac{mathit{Pphi }}{(60/mathit{N})}:=:frac{mathit{Pphi N}}{mathrm{60}}}$$ Since Z is the total number of conductors in the armature winding per phase, then $$mathrm{mathrm{Avg.:EMF:per:phase, }mathit{E_{av}/mathrm{phase}}:=:mathit{Ztimes }frac{mathit{Pphi N}}{mathrm{60}}}$$ $$mathrm{because mathit{N}:=:frac{120mathit{f}}{mathit{P}}}$$ Then, $$mathrm{mathit{E_{av}/}mathrm{phase}:=:frac{mathit{Pphi Z}}{60}times frac{120mathit{f}}{mathit{P}}:=:2mathit{fphi Z}:mathrm{Volts}}$$ Now, the RMS value of generated EMF per phase is given by, $$mathrm{mathit{E_{mathrm{RMS}}/}mathrm{phase}:=:left ( mathit{E_{av}/mathrm{phase}} right )times mathrm{form:factor}}$$ In practice, we consider that a three-phase alternator generates a sinusoidal voltage, whose form factor is 1.11. $$mathrm{mathit{E_{mathrm{RMS}}/}mathrm{phase}:=:2mathit{fphi Z}times 1.11}$$ $$mathrm{therefore mathit{E_{mathrm{RMS}}/}mathrm{phase}:=:2.22mathit{fphi Z}:mathrm{volts}:cdot cdot cdot (2)}$$ Sometimes, number of turns (T) per phase rather than number of conductors per phase are specified. In that case, we have, $$mathrm{mathit{Z}:=:2mathit{T}}$$ $$mathrm{therefore mathit{E_{mathrm{RMS}}/}mathrm{phase}:=:mathit{E_{ph}}:=:4.44mathit{fphi Z}:mathrm{volts}:cdot cdot cdot (3)}$$ The expressions in equations (2) & (3) are known as EMF equation of three-phase alternator. Numerical Example (1) What is the frequency of the voltage generated by a three-phase alternator having 6 poles and rotating at 1200 RPM? Solution Given data, P = 6; N = 1200 RPM $$mathrm{mathrm{Frequency,}mathit{f}:=:frac{mathit{NP}}{120}:=:frac{1200times 6}{120}}$$ $$mathrm{thereforemathit{f} :=:60:Hz}$$ Numerical Example (2) The armature of a 4-pole, 3-phase, 50 Hz alternator has 24 slots and 10 conductors per slot. A flux of 0.03 Wb is entering the armature from one pole. Calculate the induced EMF per phase. Solution $$mathrm{mathrm{Total:number:of:conductors}:=:24times 10:=:240}$$ $$mathrm{mathrm{Number:of:conductors:per:phase,}mathit{Z}:=:frac{240}{3}:=:80}$$ $$mathrm{therefore mathit{E_{ph}}:=:2.22mathit{fphi Z}:=:2.22times 50times 0.03times 80}$$ $$mathrm{mathit{E_{ph}}:=:266.4:V}$$ Learning working make money

EMF Equation of Transformer For electrical transformer, the EMF equation is a mathematical expression used to find the magnitude of induced EMF in the windings of the transformer. Consider a transformer as shown in the figure. If N1 and N2 are the number of turns in primary and secondary windings. When we apply an alternating voltage V1 of frequency f to the primary winding, an alternating magnetic flux $phi$ is produced by the primary winding in the core. If we assume sinusoidal AC voltage, then the magnetic flux can be given by, $$mathrm{mathit{phi }:=:phi _{m}:mathrm{sin}:mathit{omega t}:cdot cdot cdot (1)}$$ Now, according to principle of electromagnetic induction, the instantaneous value of EMF e1 induced in the primary winding is given by, $$mathrm{mathit{e_{mathrm{1}}}:=:mathit{-N_{mathrm{1}}}frac{mathit{dphi }}{mathit{dt}}}$$ $$mathrm{Rightarrow mathit{e_{mathrm{1}}}:=:mathit{-N_{mathrm{1}}}frac{mathit{d}}{mathit{dt}}left ( phi _{m}: mathrm{sin}:mathit{omega t}right )}$$ $$mathrm{Rightarrow mathit{e_{mathrm{1}}}:=:mathit{-N_{mathrm{1}}}:mathit{omega phi :cos:omega t}}$$ $$mathrm{Rightarrow mathit{e_{mathrm{1}}}:=:-mathrm{2}mathit{pi fN_{mathrm{1}}}:mathit{phi_{m} :cos:omega t}}$$ Where, $$mathrm{mathit{omega :=:mathrm{2}pi f}}$$ $$mathrm{because -mathit{cos:omega t}:=:mathrm{sin}left ( mathit{omega t-mathrm{90^{circ}}} right )}$$ Therefore, $$mathrm{mathit{e_{mathrm{1}}}:=:mathrm{2}mathit{phi fN_{mathrm{1}}}:mathit{phi_{m}:mathrm{sin}left ( mathit{omega t-mathrm{90^{circ}}} right )}}:cdot cdot cdot (2)$$ Equation (2) may be written as, $$mathrm{mathit{e_{mathrm{1}}}:=:mathit{E_{m_{mathrm{1}}}}mathrm{sin}left ( mathit{omega t-mathrm{90^{circ}}} right ):cdot cdot cdot (3)}$$ Where,$mathit{E_{m_{mathrm{1}}}}$ is the maximum value of induced EMF $mathit{e_{mathrm{1}}}$. $$mathrm{mathit{E_{mathrm{m1}}}:=:mathrm{2}mathit{pi fN_{mathrm{1}}}:mathit{phi_{m}}}$$ Now, for sinusoidal supply, the RMS value $mathit{E_{mathrm{1}}}$ of the primary winding EMF is given by, $$mathrm{mathit{E_{mathrm{1}}}:=:frac{mathit{E_{mmathrm{1}}}}{sqrt{2}}:=:frac{2mathit{pi fN_{mathrm{1}}}phi_{m}}{sqrt{2}}}$$ $$mathrm{thereforemathit{E_{mathrm{1}}}:=:4.44:mathit{fphi _{m}N_{mathrm{1}}}:cdot cdot cdot (4)}$$ Similarly, the RMS value E2 of the secondary winding EMF is, $$mathrm{mathit{E_{mathrm{2}}}:=:4.44:mathit{fphi _{m}N_{mathrm{2}}}:cdot cdot cdot (5)}$$ In general, $$mathrm{mathit{E}:=:4.44:mathit{fphi _{m}N}:cdot cdot cdot (6)}$$ Equation (6) is known as EMF equation of a transformer. For a given transformer, if we divide the EMF equation by the supply frequency, we get, $$mathrm{frac{mathit{E}}{mathit{f}}:=:4.44:phi _{m}mathit{N}:=:mathrm{Constant}}$$ Which means the induced EMF per unit frequency is constant but it is not same on both primary and secondary side of the given transformer. Also, from equations (4) and (5), we have, $$mathrm{frac{mathit{E_{mathrm{1}}}}{mathit{E_{mathrm{2}}}}:=:frac{mathit{N_{mathrm{1}}}}{mathit{N_{mathrm{2}}}}:or:frac{mathit{E_{mathrm{1}}}}{mathit{N_{mathrm{1}}}}:=:frac{mathit{E_{mathrm{2}}}}{mathit{N_{mathrm{2}}}}}$$ Hence, in a transformer, the induced EMF per turn in the primary winding is equal to the induced EMF per turn in the secondary winding. Numerical Example A single phase 3300/240 V, 50 Hz transformer has a maximum magnetic flux of 0.0315 Wb in the core. Calculate the number of turns in primary and secondary windings. Solution Given data, $$mathrm{mathit{E_{mathrm{1}}:=:mathrm{3300}:mathrm{V}:mathrm{and}:mathit{E_{mathrm{2}}:=:mathrm{240}:V}}}$$ $$mathrm{mathit{f}:=:50:Hz;:phi _{m}:=:0.0315:Wb}$$ The EMF equation of the transformer is, $$mathrm{mathit{E}:=:4.44:mathit{fphi _{m}N}}$$ Therefore, for primary winding, $$mathrm{mathit{N_{mathrm{1}}}:=:frac{mathit{E_{mathrm{1}}}}{4.44:mathit{fphi _{m}}}:=:frac{3300}{4.44times 50times 0.0315}}$$ $$mathrm{mathit{N_{mathrm{1}}}:=:471.9:=:472}$$ Also, for secondary winding, $$mathrm{mathit{N_{mathrm{2}}}:=:frac{mathit{E_{mathrm{2}}}}{4.44:mathit{fphi _{m}}}:=:frac{240}{4.44times 50times 0.0315}}$$ $$mathrm{mathit{N_{mathrm{2}}}:=:34.32:=:35}$$ It is not possible for a winding to have part of a turn. Thus, the number of turns should be a whole number. Learning working make money

Single-Phase Induction Motor As the name implies, these induction motors are operated on a single-phase AC supply. Single-phase induction motors are the most familiar electric motors because these are commonly used in domestic and commercial appliances like fans, pumps, washing machines, air conditioners, refrigerators, etc. Although single-phase induction motors are relatively less efficient than three-phase induction motors, but they are extensively used as the substitute for three-phase induction motors in low power applications. A typical single-phase induction motor consists of two main parts − a stator and a rotor. The stator of a single-phase induction motor carries a single-phase winding, whereas the rotor has the squirrel-cage construction. How to Make an Induction Motor Self Start? The major drawback of a single-phase induction motor is that it is not self-starting, but requires some starting mechanism. In single-phase induction motors, the stator winding produces a pulsating magnetic field which varies in sinusoidal manner. Hence, this magnetic field reverses its polarity after each half-cycle of AC, but does not rotate in space. As a result, this alternating magnetic field does not produce rotation in a stationary rotor. Although, if the rotor is rotated in one direction by some external mean, it will continue to run in the direction of rotation. However, this method of starting a single-phase induction motor is not convenient in practice. Therefore, in order to make a single-phase induction motor self-starting, we are somehow required to produce a revolving magnetic field inside the motor. This can be achieved by converting a single-phase AC supply into two-phase AC supply by providing an additional winding. Thus, a single-phase induction motor consists of two windings on its stator, namely main winding and starting winding. These windings are 90° out of phase with each other. Types of Single-Phase Induction Motors Depending upon the method employed to make the motor self-starting, single-phase induction motors may be classified into the following three types − Split-phase induction motor Capacitor-start induction motor Capacitor-start capacitor-run induction motor Let us now discuss each of these Induction Motors in greater detail. Split-Phase Induction Motor A split-phase induction motor is a type of single-phase induction motor in which the stator consists of two windings namely, a starting winding and a main winding, where the starting winding is displaced by 90° electrical from the main winding. The starting winding operates only during the starting period of the motor. The starting and main windings are so designed that the starting winding has high resistance and relatively low reactance, while the main winding has relatively low resistance and high reactance so that currents flowing in the two windings have a reasonable phase difference ($alpha$) of about 25° to 30°. Now, when the starting winding of the motor is connected to a source of single-phase ac supply, the starting winding carries a current $mathit{I_{s}}$, while the main winding carries a current $mathit{I_{m}}$ as shown in Figure-1. As the starting winding is made highly resistive whereas the main winding highly inductive. Consequently, the currents $mathit{I_{s}}$ and $mathit{I_{m}}$ in the two windings have a reasonable phase difference of about 25° to 30°. As a result, a weak revolving magnetic field is produced inside the motor which starts it. Capacitor Start Induction Motor The type of single-phase induction motor in which a capacitor C is connected in series with the starting winding as shown in Figure-2. This capacitor is called starting capacitor. The value of starting capacitor is so chosen that the starting current $mathit{I_{s}}$ leads the current $mathit{I_{m}}$ through the main winding by about 80°. Once the motor attains about 75% of the rated speed, a centrifugal switch isolates the starting winding from the circuit. The motor then operates as a single-phase induction motor and continues to accelerate till it reaches the normal speed. Therefore, in this type of single-phase induction motor, the capacitor in series with the starting winding introduces a phase shift between the two windings so that the motor can start itself. Capacitor-Start Capacitor Run Induction Motor This motor is almost identical to a capacitor-start induction motor except that the starting winding is not disconnected from the motor circuit. Therefore, in case of capacitor-start capacitor-run induction motor, both the windings (starting winding & main winding) remain connected to the supply during starting as well as running period. In this motor, two capacitors C1 and C2 are used in the starting winding as show in Figure-3. The capacitor C1 having small capacitance value is used for optimum running of the motor and hence permanently connected in series with the starting winding, whereas the larger capacitor C2 is connected in parallel with C1 and it remains in the circuit during starting period only. The starting capacitor C2 is isolated from the circuit by a centrifugal switch when the motor reaches about 75% of rated speed. The motor is then runs as a single-phase induction motor. Applications of Single-Phase Induction Motors Single-phase induction motors are mainly used in domestic and commercial appliances like fans, ACs, refrigerators, air-cooler, washing machines, etc. Given below are some of the applications of different kinds of single-phase induction motors − Split-Phase Induction Motors − These motors are best suited for moderate starting-torque applications like fans, washing machines, oil burners, small machine tools, etc. Capacitor-Start Induction Motors − These motors are suitable for applications that require relatively high-starting torque like compressors, large fans, pumps and high-inertia loads, etc. Capacitor-Start Capacitor-Run Induction Motors − These motors are suitable for constant torque, and vibration free applications like in hospital appliances, studio equipment and in many other appliances that are used in such areas where silence is important. Learning working make money

Construction of Synchronous Machine Like any other rotating electrical machine, a synchronous machine (generator or motor) has two essential parts namely, Stator − It is a stationary part of the machine and carries the armature winding. Rotor − It is the rotating part of the machine. The rotor of a synchronous machine produces the main field flux. In this chapter, let”s discuss how the Stator and Rotor of a Synchronous Machine is constructed. Stator Construction The stator of a synchronous machine includes various parts like frame, stator core, stator windings and cooling mechanisms, etc. The frame is the outer part of the machine, and made up of cast iron for small-sized machines, and of welded steel for large-sized machines. The frame encloses the whole machine assembly and protects it from mechanical and environmental impacts. The stator core is a hollow cylinder which is made up of high-grade silicon steel laminations. The silicon steel laminations reduce the hysteresis and eddy-current losses in the machine. A number of evenly spaced slots are provided on the inner periphery of the stator core. A three-phase winding is put in these slots. When current flows through the stator winding, it produces a sinusoidal magnetic field and hence the EMF. Rotor Construction In synchronous machines, there are two types of rotor constructions used namely, salient-pole rotor and cylindrical rotor. Salient-Pole Rotor The term salient means projecting. Hence, a salient-pole rotor is one that consists of field poles projecting out from the surface of the rotor core as shown in Figure-2. Since the rotor is subjected to a changing magnetic field, therefore it is made up of thin steel lamination to reduce the eddy current loss. The field poles of same dimensions are constructed by stacking laminations of the required length and then riveted together. After wounding the field coils around each pole core, poles are fitted to a steel spider keyed to the rotor shaft. At the outer end of each pole, damper bars are provided to dame out the oscillations of rotor during sudden changes in the load. Though, the synchronous machines using the salient-pole rotor have a non-uniform air gap, where the air gap is minimum under the pole centers and it is maximum in between the field poles. The pole face (outer end of the pole) is so shaped that length of the radial air-gap increases from the pole center to the pole tip so it can result a sinusoidal distribution of flux in the air-gap. This will ensure the smooth operation of synchronous machine. In the salient-pole rotor, the individual field coils are connected in series so that they can give alternate north and south poles. The ends of the field coils are connected to a source of DC supply though brushes and slip-rings. The synchronous machines having salient-pole rotor usually have a large number of field poles, and operate at lower speeds. These machines have a larger diameter and a shorter axial-length. Cylindrical Rotor This type of synchronous machine rotor construction has a smooth cylindrical structure. In case of cylindrical rotor, there are no physical poles projecting outward. The cylindrical rotor is made from slid forgings of high-grade Ni-Cr-Mo steel. On the outer periphery of the rotor, evenly spaced slots are cut in about two-third part of the rotor and theses slots are parallel to the rotor shaft. The field windings are placed in these slots. The field windings are excited from a source of DC supply. The un-slotted part of the rotor forms pole-faces. The synchronous machines that use cylindrical rotor have a smaller diameter and longer axial length. The cylindrical rotor construction limits the effect of centrifugal forces. Consequently, the cylindrical rotor construction is mainly used in high-speed synchronous machines. Also, this rotor construction provides a greater mechanical strength and permits relatively more accurate dynamic balancing to the machine. The major advantage of having a synchronous machine using cylindrical rotor is that it makes less mechanical losses. Since the cylindrical rotor provides a uniform air-gap in the machine, hence their operation is less noisy. Learning working make money

Concept of Induced EMF According to principle of electromagnetic induction, when the magnetic flux linking to a conductor or coil changes, an EMF is induced in the conductor or coil. In practice, the following two ways are used to bring the change in the magnetic flux linkage. Method 1 − Conductor is moving in a stationary magnetic field We can move a conductor or coil in a stationary magnetic field in such a way that the magnetic flux linking to the conductor or coil changes in magnitude. Consequently, an EMF is induced in the conductor. This induced EMF is known as dynamically induced EMF. It is so called because the EMF induced in a conductor which is in motion. Example of dynamically induced EMF is the EMF generated in the AC and DC generators. Method 2 − A stationary conductor is placed in a changing magnetic field When a stationary conductor or coil is placed in a moving or changing magnetic field, an EMF is induced in the conductor or coil. The EMF induced in this way is known as statically induced EMF. It is so called because the EMF is induced in a conductor which is stationary. The EMF induced in a transformer is an example of statically induced EMF. Therefore, from the discussion, it is clear that the induced EMF can be classified into two major types namely, Dynamically Induced EMF Statically Induced EMF Dynamically Induced EMF As discussed in the above section that the dynamically induced EMF is one which induced in a moving conductor or coil placed in a stationary magnetic field. The expression for the dynamically induced EMF can be derived as follows − Consider a single conductor of length l meters located in a uniform magnetic field of magnetic flux density B Wb/m2 as shown in Figure-1. This conductor is moving at right angles relative to the magnetic field with a velocity of v m/s. Now, if the conductor moves through a small distance dx in time dt seconds, then the area swept by the conductor is given by, $$mathrm{mathit{A:=:ltimes dx:}mathrm{m^{mathrm{2}}}}$$ Therefore, the magnetic flux cut by the conductor is given by, $$mathrm{mathit{dphi }:=:mathrm{Flux:densitytimes Area: swept}}$$ $$mathrm{Rightarrow mathit{dphi }:=:mathit{Btimes ltimes dx}:mathrm{Wb}}$$ According to Faraday’s law of electromagnetic induction, the EMF induced in the conductor is given by, $$mathrm{mathit{e}:=:mathit{N}frac{mathit{dphi }}{mathit{dt}}:=:mathit{N}frac{mathit{Bldx}}{mathit{dt}}}$$ Since, we have taken only a single conductor, thus N = 1. $$mathrm{mathit{e}:=:mathit{Blv}:mathrm{volts}cdot cdot cdot (1)}$$ Where, v = dx/dt, velocity of the conductor in the magnetic field. If there is angular motion of the conductor in the magnetic field and the conductor moves at an angle θ relative to the magnetic field as shown in Figure-2. Then, the velocity at which the conductor moves across the magnetic field is equal to “vsinθ”. Thus, the induced EMF is given by, $$mathrm{mathit{e}:=:mathit{B:l:v}:mathrm{sinmathit{theta }}:mathrm{volts}cdot cdot cdot (2)}$$ Statically Induced EMF When a stationary conductor is placed in a changing magnetic field, the induced EMF in the conductor is known as statically induced EMF. The statically induced EMF is further classified into following two types − Self-Induced EMF Mutually Induced EMF Self Induced EMF When EMF is induced in a conductor or coil due to change of its own magnetic flux linkage, it is known as self-induced EMF. Consider a coil of N turn as shown in Figure-3. The current flowing through the coil establishes a magnetic field in the coil. If the current in the coil changes, then the magnetic flux linking the coil also changes. This changing magnetic field induces an EMF in the coil according to the Faraday’s law of electromagnetic induction. This EMF is known as self-induced EMF and the magnitude of the self-induced EMF is given by, $$mathrm{mathit{e}:=:mathit{N}frac{mathit{dphi }}{mathit{dt}}}$$ Mutually Induced EMF The EMF induced in a coil due to the changing magnetic field of a neighboring coil is known as mutually induced EMF. Consider two coils X and Y placed adjacent to each other as shown in Figure-4. Here, a fraction of the magnetic flux produced by the coil X links with the coil Y. This magnetic flux of coil X which is common to both coils X and Y is known as mutual flux ($mathit{phi _{m}}$). If the current in coil X is changed, then the mutual flux also changes and hence EMF is induced in both the coils. Where, the EMF induced in coil X is known as self-induced EMF and the EMF induced in coil Y is called mutually induced EMF. According to Faraday’s law, the magnitude of the mutually induced EMF is given by, $$mathrm{mathit{e_{m}}:=:mathit{N_{Y}}frac{mathit{dphi _{m}}}{mathit{dt}}}$$ Where,$mathit{N_{Y}}$ is the number of turns in coil Y and $frac{mathit{dphi _{m}}}{mathit{dt}}$ is rate of change of mutual flux. Learning working make money