Category: electrical Machines

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

Electrical Machines – Quick Guide Electromechanical Energy Conversion Today, electrical energy is the most widely used form of energy for performing several industrial, commercial and domestic functions such as pumping water, fans, coolers, air conditioning, refrigeration, etc. Since, most of processes require the conversion of electrical energy into mechanical energy. Also, the mechanical energy is converted into electrical energy. Hence, this clears that we need a mechanism to convert the electrical energy into mechanical energy and mechanical energy into electrical energy and such a mechanism is known as electromechanical energy conversion device. Electromechanical Energy Conversion Device Thus, a device which can convert electrical energy into mechanical energy or mechanical energy into electrical energy is known as electromechanical energy conversion device. The electric generators and electric motors are the examples of electromechanical energy conversion device. In any electromechanical energy conversion device, the conversion of electrical energy into mechanical energy and vice-versa takes place through the medium of an electric field or a magnetic field. Though, in most of the practical electromechanical energy conversion devices, magnetic field is used as the coupling medium between electrical and mechanical systems. The electromechanical energy conversion devices can be classified into two types − Gross motion devices (like motors and generators) Incremental motion devices (such as electromagnetic relays, measuring instruments, loudspeakers, etc.) The device which converts electrical energy into mechanical energy is known as electric motor. The device which converts mechanical energy into electrical energy is known as electric generator. In an electric motor, when a current carrying conductor is placed in a changing (or rotating) magnetic field, the conductor experiences a mechanical force. In case of a generator, when a conductor moves in a magnetic field, an EMF is induced in the conductor. Although, these two electromagnetic effects occur simultaneously, when the energy conversion takes place from electrical to mechanical and vice-versa in all the electromechanical energy conversion devices. Energy Balance Equation The energy balance equation is an expression which shows the complete process of energy conversion. In an electromechanical energy conversion device, the total input energy is equal to the sum of three components − Energy dissipated or lost Energy stored Useful output energy Therefore, for an electric motor, the energy balance equation can be written as, Electrical energy input = Energy dissipated + Energy stored + Mechanical energy output Where, The electrical energy input is the electricity supplied from the main supply. Energy stored is equal to sum of the energy stored in the magnetic field and in the mechanical system in the form of potential and kinetic energies. The energy dissipated is equal to sum of energy loss in electric resistance, energy loss in magnetic core (hysteresis loss + eddy current loss) and mechanical losses (windage and friction losses). For an electric generator, the energy balance equation can be written as, Mechanical energy input = Electrical energy output + Energy stored + Energy dissipated Where, the mechanical energy input is the mechanical energy obtained from a turbine, engine, etc. to turn the shaft of the generator. 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

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

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

Speed Regulation and Speed Control Speed Regulation of Induction Motors The speed regulation of induction motor is defined as the change in the motor speed with change in load. It is expressed as a fraction or percentage of full-load speed, i.e., $$mathrm{mathrm{Speed:regulation}:=:mathit{frac{N_{nl}-N_{fl}}{N_{fl}}}times 100%}$$ Where,$mathit{N_{nl}}$ is the no-load speed of the motor and $mathit{N_{fl}}$ is the full-load speed of the motor. The speed regulation of an induction motor is about 3% to 5%. Due to this small speed regulation, the induction motors are classified as the constant speed motors. Speed Control of Three-Phase Induction Motors The speed of a three-phase induction motor is given by, $$mathrm{mathit{N_{r}}:=:left ( 1-mathit{s} right )mathit{N_{s}}:cdot cdot cdot (1)}$$ Where,s is the slip and $mathit{N_{s}}$ is the synchronous speed in RPM. $$mathrm{mathit{N_{s}}:=:frac{120mathit{f}}{mathit{P}}:cdot cdot cdot (2)}$$ From equations (1) & (2), it is clear that the speed of a three-phase induction motor can be varied by changing the following − Frequency (f) of AC supply, Number of stator poles (P), and Slip (s). In practice, the change in supply frequency is generally not possible because the commercial electric supplies have a constant frequency. Hence, the speed of a three-phase induction motor can be changed either by changing the number of stator poles (P) or the slip (s). We shall now discuss the speed control of squirrel-cage and slip-ring induction motors. Speed Control of Squirrel-Cage Induction Motors The speed control of squirrel-cage induction motors is changed by changing the number of stator poles. By the pole changing method, there are only two or four speeds are possible. In a two-speed induction motor, one stator winding is provided, which may be switched through a suitable control equipment to provide the two speeds. Where, one speed is half of the other. For example, the stator winding may be connected for either 4 or 8 stator poles, giving synchronous speeds of 1500 RPM and 750 RPM, when the motor is supplied from a source of 50 Hz AC supply. In a four speed induction motor, two separate stator windings are provided each of which provides two speeds. Following are the major disadvantages of pole changing method of speed control − This method cannot be used to obtain gradual continuous speed control. It makes the motor design and switching of the interconnection of stator windings more complicated. This method can provide a maximum of four different speeds for any one motor due to design and interconnection complications. Speed Control of Slip-Ring Induction Motors The speed of a slip-ring induction motor can be varied by changing the motor slip. The following methods are employed for changing the slip and hence the speed − By the changing the stator line voltage. By changing the resistance of the rotor circuit. By adding and changing a foreign voltage in the rotor circuit. Numerical Example For a three-phase induction motor, the no-load speed of the motor is 900 RPM and its full-load speed is 880 RPM. Find the speed regulation of the motor. Solution Given data, $mathit{N_{nl}}$ = 900 RPM $mathit{N_{fl}}$ = 880 RPM $$mathrm{thereforemathrm{Speed:regulation}:=:mathit{frac{N_{nl}-N_{fl}}{N_{fl}}}times 100%}$$ $$mathrm{Rightarrow mathrm{Speed:regulation}:=:frac{900-880}{880}times 100%:=:2.273%}$$ Learning working make money