Induction Motors Lecture Notes 2024 PDF

Technical collage –Ratmalana NDIT - 2nd Year year 2024 Electrical Machine – Induction Motor – Lecture note 01 Three-Phase Induction Machines Three phase induction motor:- As this type of motor does not require...

Technical collage –Ratmalana NDIT - 2nd Year year 2024 Electrical Machine – Induction Motor – Lecture note 01 Three-Phase Induction Machines Three phase induction motor:- As this type of motor does not require any starting device or we can say they are self starting induction motor This Motor consists of two major parts: Stator: As the name indicates stator is a stationary part of induction motor. Stator of three phase induction motor is made up of numbers of slots to construct a three phase winding circuit which is connected to three phase AC source. The three phase winding are arranged in such a manner in the slots that they produce a rotating magnetic field after three phase supply. Rotor: The rotor is a rotating part of induction motor. The rotor is connected to the mechanical load through the shaft. The rotor of the three phase induction motor are further classified as: 1. Squirrel cage rotor - Conductors are heavy copper or aluminum bars which fits in each slots & they are short circuited by the end rings 2. Slip ring rotor or wound rotor or phase wound rotor. Squirrel cage rotor Wound rotor Rotor of three phase induction motor consists of cylindrical laminated core with parallel slots that can carry conductors.. The slots are not exactly made parallel to the axis of the shaft but are slotted a little skewed because this arrangement reduces magnetic humming noise & can avoid stalling of motor. Operation principle of induction motor 1. If stator is energized by an A.C. current,rotating magnetic field is generated due to the applied current to the stator winding. 2. This flux produces magnetic field and the field revolves in the air gap between stator and rotor 1 3. So,the magnetic field induces a voltage in the short circuited bars of the rotor.This voltage drives current through the bars. 4. The interaction of the rotating flux and the rotor current generates the force that drives the motor and a torque is developed consequently. 5. The torque is proportional with the flux density and the rotor bar current(F=BIL) 6. The motor speed is less than the synchronous speed. 7. The direction of rotation of the rotor is the same as the direction of rotation of the revolving magnetic field in the air gap. Slip Slip Speed of the Induction Motor is defined as the difference between the synchronous speed (Ns)and the actual rotor speed(Nr). An induction motor cannot run at synchronous speed. Let us consider that the rotor of the induction motor is running at the synchronous speed. At this condition, the rotor conductors do not cut the flux and as a result, there is no generation of voltage, current and hence, no torque. Torque /slip and current /slip characteristics The slip speed of the induction motor is given as S =Ns -Nr……….…..(1) Ns = 120f /P The slip speed expressed as a fraction of the synchronous speed is called the Per Unit Slip or Fractional Slip. The per unit slip is generally called the Slip. It is denoted by s. per unit slip (s)= (Ns-Nr)/ Ns Persentage slip = (Ns-Nr) x 100/ Ns At standstill which rotor does not rotate s = 1 that is Nr = 0 At synchronous speed Nr = Ns , S= 0 2  Rotor frequency and slip f = NsP/120 When the rotor is stationary, the frequency of rotor current is same as the supply frequency. But when the rotor starts revolving, then the frequency of rotor current depends upon the relative speed or on slip speed. when the rotor rotates at speed Nr , the rotor conductors cut the rotating field at a speed N s -Nr , the frequency of the rotor of induced emf (fr) = P(Ns-Nr) /120 but s = (Ns-Nr )/ Ns s x Ns = Ns-Nr fr = s x NsP/120 fr = sf NCIT - 2015 December8.2 Actual speed of an a.c motor is 950RPM (a) Synchronous speed of motor (b) Number of poles (c) Percentage slip If synchronous speed is Ns ,supply frequency is f and number of poles are P f = NsP/120 assume supply frequency is 50Hz and nuber of poles are 4 synchronous speed = 120f /p =120 x 50 /4 =1500 assume supply frequency is 50Hz and number of poles are 6 synchronous speed = 120f /p =120 x 50 /6 =1000 assume supply frequency is 50Hz and number of poles are 8 synchronous speed = 120f /p =120 x 50 /8 =750 Actual rotor speed is 950. Therefore synchronous speed of the motor will be 1000 Therefore number of poles = 6 Percentage slip = (Ns-Nr) x 100/ Ns =(1000 – 950) x 100 / 1000 = 5% NCIT - 2013 December 6.2 Actual speed of the motor is 1450RPM.Calculate (a) Synchronous speed of motor -1500 (b) Number of poles -4 3 (c) Percentage slip =3.33% NCIT - 2012 december6.3 Six pole three phase induction motor is connected to 400 volt, 50Hz supply.When the percentage slip is 4%,calculate the synchronous speed and speed of the motor. No of poles of the motor =6 Supply frequency = 50Hz If synchronous speed is Ns, Ns = 120f /P = 120 x 50 /6 =100RPM Percentage slip =4% =(Ns-Nr) x 100/ Ns = (1000 –Nr) x 100/1000 (1000 - Nr)10=4 Nr = 999.6RPM NCIT – 2016 DECEMBE 6.D 400V, 50Hz, 8pole ,3phase induction motor runs at 720 RPM.calculate i) Synchronous speed ii) Slip speed iii) Percentage slip iv) Rotor Frequency 1) To find synchronous speed applying f= NSP /120 50 = Ns x 8 /120 Ns =50 x 120 /8 = 750RPM 2)To find slip speed applying Ns –Nr 750 – 720 = 30RPM 3)To find percentage slip applying s = (Ns –Nr)100 /Ns s = (750 – 720)100 / 750 =4% 4) To find rotor frequency applying fr =sf Fr = [(750 -720)/750]50 = 0.04 x 50 =2Hz Power flow diagram of induction motor At rated speed iron loss in the rotor core is negligible because iron loss is a function of rotor frequency and the frequency of rotor current is very small.( fr = sf ) 4 Let T be the torque in newton meters exerted on the rotor by the rotating magnetic field rotating at synchronous speed Ns. Power transferred from stator to rotor(i.e. rotor input) = 2𝜋NsT/60 This input power to the rotor is termed torque in synchronous watts. When the rotor is rotating at a speed of Nr RPM, total mechanical power developed by rotor is 2𝜋NrT/60 watts. Difference between power transferred to rotor via magnetic field of air gap and mechanical power developed by the rotor is equal to I2R loss in the rotor winding 2𝜋NsT/60 -2𝜋NrT/60 = rotor I2R loss rotor I2R loss = 2𝜋T(Ns -Nr) /60 rotor I2R loss / rotor input =2𝜋T(Ns -Nr) /60 /2𝜋NsT/60 =(Ns -Nr) /Ns = s rotor I2R loss = s x rotor input NCIT - December 2014 8.3 Input power of 3 phase motor is 50 kw. If the stator loss is 1.2 kw, when the slip is 0.03,calculate the generated mechanical power of the motor Input power to rotor = Input power to stator – stator losses = 50 kw -1.2kw =48.8kw Rotor copper loss =slip x rotor input 5  = 0.03 x 48.8 = 1.464watt When neglecting rotor iron losses mechanical power developed by the rotor = Rotor input –rotor copper loss = 48.8 – 1.464 =47.336watt NCIT - 2012 august6.2 A 20 kw,3 phase ,415v,50Hz,6pole squirrel cage type induction motor operates at a speed of 960RPM. At 0.75 power factor lagging. The stator power loss and the rotational power loss at the full load operation are 600watts and 800watts respectively. At this full load operation calculate, 1.The synchronous speed of the motor 2. Per unit slip 3. The frequency of the rotor induced EMF 4.Rotor copper loss 5.Gross mechanical power developed 6. The efficiency of the motor Input power to Power transferred Total mechanical Rotor out put stator winding to Rotor power developed or motor out I2R loss in stator I2R loss in rotor Friction and winding winding windage loss Rotor speed 960RPm Frequency = 50Hz This will be a 6 pole motor 1.Synchronous speed of 6 pole motor = Ns = 120f/p =120 x 50 /6 = 1000Rpm 2. Per unit slip = (Ns - Nr)/ Ns =(1000 – 960)/1000 = 0.04 3. The frequency of the rotor induced EMF (fr) = sf = 0.04 x 50 =2Hz 4.to find rotor cu loss Motor output = 20kw Rotational power loss -800watt =0.8kw 6  5. Mechanical power developed(gross) = 20 + 0.8 =20.8kw Power transferred to Rotor - I2R loss in rotor winding = Total mechanical power developed But I2R loss in rotor winding = s x Power transferred to Rotor Power transferred to Rotor - s x Power transferred to Rotor = Total mechanical power developed Power transferred to Rotor (1-s) = Total mechanical power developed Power transferred to Rotor = 20.8 /(1`-0.04) =21.66kw rotor copper loss = 21.66 -20.8 = 0.86kw Stator power loss = 600watts = 0.6kw Input power to stator = 21.66 +0.6 =22.26kw 6.Efficiency% = output x 100/input = 20 x 100 / 22.26 =89.84% NCIT -2011 Aug 6(ii) A 415v,50Hz,3phase,4pole squirrel cage induction motor develops a full load torque of 200Nm.The frequency of rotor induced EMF being 2Hz.The stator power losses are 1 kW and the rotational losses are equivalent to torque of 15Nm.Calculate i) The per unit slip ii) The output power of the motor in Kw iii) The gross mechanical power developed iv) The rotor i2R power loss (rotor copper loss) v) The input power to the stator vi) The efficiency of the motor If Synchronous speed of 4 pole motor is Ns, applying f =Ns P /120 Ns = 120 f /p =120 x 50 /4 =1500 rpm Rotor frequency(fr) =2hz To find slip applying fr =Sf 2 = S x 50 1) The per unit Slip(S) =2 /50 =0.04 To find full load rotor speed(Nr) applying S =(Ns –Nr) /Ns 0.04 = (1500 –Nr ) /1500 Nr = 1500 -60 =1440rpm 7  Stator power losses =1Kw Rotational loss absorb torque =15Nm To find rotational loss applying p =2𝜋NrT /60 Rotational loss = p = 2x𝜋x1440x15 /60 =2.26Kw Torque developed in the rotor =200Nm To find gross mechanical power developed in the motor applying p =2𝜋NrT /60 2) gross mechanical power developed in the motor =2x𝜋x1440x200 /60 =30.159Kw 3) output power of the motor = mechanical power developed in the motor – rotational losses = 30.159 – 2.26 =27.899Kw Input power to the rotor (1-S) = mechanical power developed in the rotor Input power to the rotor = 30.159 /(1 – 0.04)= 31.416 Kw 4) Rotor cu loss = Input power to the rotor - mechanical power developed in the motor Rotor cu loss =(31.416 – 30.159) = 1.257Kw Stator power loss =1Kw 5) Input power to the stator = stator power loss + input power to the rotor = 1Kw +31.416Kw=32.416Kw 6) Efficiency = out put power of the motor /in put power of the motor = 27.899 /32.416=0.86 0.86 x100 =86% Wound rotor induction motor To start the motor, all of the resistance of the wye-connected speed controller is inserted in the rotor circuit. The stator circuit's energized from the three-phase line. The voltage induced in the rotor develops currents in the rotor circuit. The rotor currents, how ever, are limited in value by the resistance of the speed controller. As a result, the stator current also is limited in value. In other words, to minimize the starting surge of current to a wound-rotor induction motor, insert the full resistance of the speed controller in the rotor circuit. The starting torque is affected by the resistance inserted in the rotor 8  secondary. With resistance in the secondary, the power factor of the rotor is high or close to unity. This means that the rotor current is nearly in phase with the rotor-induced voltage. If the rotor current is in phase with the rotor-induced voltage, then the rotor magnetic poles are being produced at the same time as the stator poles. This creates a strong magnetic effect, which creates a strong starting torque. As the motor accelerates, steps of resistance in the wye-connected speed controller can be cut out of the rotor circuit until the motor accelerates to its rated speed. Squirrel cage induction motors draw 500% to over 1000% of full load current (FLC) during starting. While this is not a severe problem for small motors, it is for large (10’s of kW) motors. Placing resistance in series with the rotor windings not only decreases start current, locked rotor current (LRC) but also increases the starting torque, locked rotor torque (LRT). The figure below shows that by increasing the rotor resistance from R0 to R1 to R2, the breakdown torque peak is shifted left to zero speed. Note that this torque peak is much higher than the starting torque available with no rotor resistance (R 0). Slip is proportional to rotor resistance, and pullout torque is proportional to slip. Thus, high torque is produced while starting. The resistance decreases the torque available at full running speed. But that resistance is shorted out by the time the rotor is started. A shorted rotor operates like a squirrel cage rotor. The heat generated during starting is mostly dissipated external to the motor in the starting resistance. The complication and maintenance associated with brushes and slip rings is a disadvantage of the wound rotor as compared to the simple squirrel cage rotor. This motor is suited for starting highly inertial loads. A high starting resistance makes the high pull out torque available at zero speed. Speed Control Motor speed may be varied by putting variable resistance back into the rotor circuit. This reduces the rotor current and speed. A higher resistance at R2 reduces the speed further. This speed control technique is only useful over a range of 50% to 100% of full speed. 9  Rotor EMF and reactance under running condition When a balanced three-phase induction motor is excited by a balanced three-phase source, the currents in the phase windings must be equal in magnitude and 120 -degree electrical apart in phase. Thus, a three-phase induction motor can be represented on a per-phase basis by an equivalent circuit at any slip s. When an induction motor is stationary, the stator and rotor windings form the equivalent of a transformer as shown in Figure. The rotor e.m.f. at standstill is given by E2 =(N2/N1)E1 where E1 is the supply voltage per phase to the stator. E2 is the induce voltage per phase of the rotor. When rotor is stationary(at standstill condition S=1), then the frequency of the rotor EMF is the same as that of the stator supply frequency. The value of EMF induced in the rotor at standstill(E2) is maximum because the relative speed between the rotor and the revolving stator flux is maximum(E2α Ns). In fact , the motor is equivalent to a three phase transformer with a short circuited rotor secondary. When rotor starts running , the relative speed between rotor and the rotating stator flux is decreased. Hence the rotor induced EMF which is directly proportional to this relative speed, is also decreased(and may disappear altogether if rotor speed were to become equal to the speed of stator flux) E2r α Ns -Nr.( E2r = rotor EMF under running condition) E2r /E2 =(Ns –Nr) /Ns =S Therefore under running condition E2r = SE2 Rotor impedance and current Rotor resistance :-The rotor resistance R2 is unaffected by frequency or slip, and hence remains constant. Rotor reactance:- Rotor reactance varies with the frequency of the rotor current. At standstill, reactance per phase, X2 = 2𝜋fL When running, reactance per phase, Xr = 2𝜋frL but fr =Sf Xr = 2𝜋𝑆fL =S x 2𝜋fL Xr = sX2 10  Above figure represents the rotor circuit when running NCIT 2007 Dec 05:-A six pole three phase ,50Hz,wound rotor induction motor(slip ring motor) is running at 960 rpm at the full load condition. The rotor winding has connected as star connection and has a rotor resistance/phase 0.2Ω and rotor reactance per phase is0.02H.The EMF generated between slip rings at standstill is 220v. for this full load condition I. Total impedance per phase II. The EMF induced in the rotor per phase III. The rotor current per phase when slip rings are short circuited. Synchronous speed of 6 pole motor at 50Hz frequency =Ns = 120f /p =120 x 50 /6 =1000rpm Rotor speed =960rpm Slip =(1000 -960)/1000 = 0.04 When the motor is at stationary, the frequency of the rotor current is same as supply frequency Rotor frequency at running condition =fr =sf =0.04 x 50 =2Hz When the brushes short circuited Rotor reactance per phase at stand still (X2)=2 x 𝜋 x fr x L =2 x 𝜋 x 50 x 0.02 =6.28Ω Rotor reactance per phase at 0.04 slip =2 x 𝜋 x fr x L =2 x 𝜋 x 2 x 0.02 =0.251Ω or( s x2 = 0.04 x 6.28 =0.251Ω) Rotor resistance per phase = 0.2Ω 1) Rotor impedance per phase = √(0.22 +0.2512) =0.321Ω EMF generated between slip rings (line voltage) at standstill=220 2) The EMF induced in the rotor per phase (phase voltage) at standstill=220 /√3 =127V The EMF induced in the rotor per phase at 0.04 slip = SE0 = 0.04 x 127 =5.08V When the slip rings are short circuited to find rotor current per phase applying v= iz 3) Rotor current per phase at 0.04 slip = Vph /Z = 5.08 /0.321 =15.82A NCIT – 2009 JUNE 07 :- A three phase 4 pole 50Hz wound rotor induction motor is operating at 1440RPM under its full load condition.The star connected rotor has the rotor resistance per phase 0.4Ω 11 and standstill rotor reactance per phase 2Ω.The induced EMF between slip rings at standstill is 180V.At the starting of motor , a star connected external resistor bank having rotor resistance per phase 5Ω is connected with slip rings. i) At the starting calculate a) Rotor resistance per phase b) Rotor current per phase ii) At running under full load with short circuited slip rings a) Rotor impedance per phase b) Rotor current per phase Synchronous speed of 4 pole motor at 50Hz frequency =Ns = 120f /p =120 x 50 /4 =1500rpm Per unit slip at 1440RPM =(1500 -1440) /1500 =0.04 At standstill condition:- Resistance of the rotor circuit per phase =0.4 +5 = 5.4Ω Reactance of the rotor circuit =2Ω Impedance of the rotor circuit per phase =√(5.42 +22) =5.76Ω EMF between slip rings at standstill(line voltage) =180V The EMF induced in the rotor per phase (phase voltage) at standstill =180 /√3 =103.92V Rotor current per phase =V/Z =103.92 /5.76 =18.04A At running under full load with short circuited slip rings:- Reactance of the rotor circuit per phase= SX2 =0.04 x2 =0.08Ω Resistance of the rotor circuit per phase =0.04Ω Impedance per phase = √(0.082 +0.042) =0.089Ω The EMF induced in the rotor per phase (phase voltage) at 0.04 slip = SE2 = 0.04 x 103.92 =4.1568V Rotor current per phase = V/Z = 4.1568 /0.089 =46.7A NCIT 2012 august7.3:-A three phase wound rotor induction motor produces an EMF 60v between slip rings at standstill. The resistance and the reactance per phase of the motor at standstill are 0.4Ω and 12 3Ω respectively.The rotor circuit has connected with a star connected resistor bank having the resistance of 4Ω each. Calculate the rotor current per phase and the power factor, i) When the motor is starting with resistor bank ii) When the motor is operating at the slip of 4%with the short circuited slip rings Rotor voltage between slip rings (line voltage) at standstill =60V Rotor voltage per phase = 60 /√3 =34.64V When the rotor is starting with resistor bank Total rotor resistance =4+0.4 =4.4Ω Rotor reactance at standstill =3Ω Rotor impedance at standstill = √(4.42 +32) =√28.36 =5.325Ω Rotor current at standstill = 34.64 /5.325 =6.5A Rotor power factor at standstill =R /Z=4.4/5.325 =0.826 When the motor is operating at the slip of 4%with the short circuited slip rings Rotor voltage per phase = SE0 = (4/100) 34.64 =1.3856V Rotor reactance per phase =SX0 = (4/100) 3 =0.12V Rotor impedance without resistor bank = √ (0.42 +0.122) =0.4176Ω Rotor current at 4% slip with short circuited slip rings = 1.3856 /0.4176 =3.318A Rotor power factor at 4% slip with short circuited slip rings =R/Z = 0.4 /0.4176 =0.957 Relation between torque and rotor power factor In d.c motor the torque (Ta) is proportional to product of armature current and flux per pole(Ta =IaФ). Similarly in the case of an induction motor torque also proportional to product of flux per stator pole and rotor current. However there is another factor to be taken in to account i.e the power factor of the rotor(cosϕ2) T 𝛼 Ф I2cos ϕ2 T = KФ I2cos ϕ2 I2 = rotor current at standstiil ϕ2 = Angle between rotor EMF and rotor current Ф =flux per stator pole K = constant 13  Denoting rotor emf at standstill by E2.we have that E2 𝛼 Ф T = K E2 I2cos ϕ2 Standstill or starting torque(Tst) = K E2 I2cos ϕ2 Starting torque E2 =rotor EMF per phase at standstill R2 = rotor resistance /phase X2 = rotor reactance / phase at standstill Z2 = √R22 +X22 = rotor impedance/phase at standstill Torque at standstill or starting torque = T = KФ I2cos ϕ2 = Tst I2 = E2 / Z2 = E2 /√(R22 +X22) For series LR circuit cosѲ2 = R2 /Z2= R2 / √R22 +X22 Tst = K ФE2 /√(R22 +X22) x R2 / √(R22 +X22) Tst = K ФE2 R2 /( R22 +X22 ) If supply voltage(V) is constant then flux (Ф) is constant and hence E2 both are constant. Tst = K2 R2 /( R22 +X22 ) At full load At running condition Induced voltage of the rotor =SE2 and reactance of the rotor=SX2 TFL = K ФSE2 R2 /( R22 +(SX2)2 ) TFL 𝛼 S R2 /( R22 +(SX2)2 ) At maximum torque Requirement of maximum torque is R2 = SX2 Tmax 𝛼 S R2 /( R22 +(SX2)2 Tmax 𝛼 S. SX2 /( 2(SX2)2 ) = 1/ 2X2 Full load and maximum torque ratio TFL / Tmax = SR2 /( R22 +(SX2)2 )/ 1/ 2X2 = S R2 2X2/ R22 +(SX2)2 Dividing both the numerator and the denominator by X22 TFL / Tmax = 2SR2/X2 / (R22/X22 + S2) assuming R/X2 = a and s = full load slip 14  TFL / Tmax = 2aS / (a2 + S2) Starting to maximum torque ratio Tst / Tmax = R2 /( R22 +X22 ) /1/ 2X2 = R22X2 / ( R22 +X22 ) Dividing both the numerator and the denominator by X22 Tst / Tmax =(R22 /X2) / (R22/X22 +1) assuming R/X2 = a Tst / Tmax = 2a / (a2 + 1) TFL / Tmax = 2as / (a2 + s2) 2007december 06 A 50Hz,8pole induction motor has a full load slip of 0.04 per unit. The rotor resistance is 0.001Ω per phase and the stand still reactance is 0.005Ω/phase. Find the ratio of the maximum to full load torque and the speed at which maximum torque occurs. assuming R/X2 = a and s = full load slip R/X2 = 0.001 / 0.005 =0.2 TFL / Tmax = 2 x 0.2 x 0.04 /0.22 + 0.042 = 0.016 / 0.04 + 0.0016 = 0.3846 Synchronous speed of 8 pole motor = 50 x 120 /8 =750RPM At maximum torque R 2 = S X2 S = 0.001/ 0.005 = 0.2 S = Ns –NR/ NS = 750 – NR / 750 = 0.2 Nr = 750-150 = 600 Direct On Line Starter Direct On Line Starter :-Method is a common method of starting of Cage Induction Motor. The motor is connected by means of a starter across the full supply voltage. This starter has two basic protections as Over-load protection and no volt protection. 15  Main or Power Circuit Control Circuit Direct Online starter power circuit has fuse unit (F1), Main power contactor (K1), thermal overload releaser (F2). control circuit consist components like Main power contactor coil and contacts, start button, stop push button and overload relay. Working Principle 1) To start the motor, start push button(S1) is pressed. After that, the main power contactor coil(K1) energized due to electromechanical action and close the main contacts(K1) in power circuit and close the auxiliary contacts(K1) in control circuit. It applies full line voltage to the motor terminals. and the motor starts running. The motor will draw a very high inrush current for a short time. 2) After that start push button get released and supply to contactor coil continued through latching contact(K1) provided to the main contractor of D O L starter. 3) At time of stopping the motor we need to pressed stop push button(S0) on DOL starter circuit, this disconnect supply to the coil(K1) and contactor coil get de-energized, and it releases self holding contact(K1) and motor supply get disconnected. In order to avoid excessive voltage drops in the supply line due to high starting currents, the DOL starter is used only for motors with a rating of less than 5KW Reasons for reduced voltage starting At the time of starting the motors will draw 5 to 7 times of its full load current from its supply. This may cause some voltage drop in the supply lines and voltage across other loads gets reduced. Hence to limit this starting current, the reduced voltage is applied to the motor at the time of starting through star-delta starter, auto transformer starter etc.. But the disadvantage of this method is that the starting torque gets reduced. 16 Star delta starter (manual) This starter uses a TPDT (triple pole double throw) switch and it connects the stator winding in star during the starting condition and delta in running condition. Due to this star connection, the applied voltage to the motor is reduced by the factor 1/√3. The current in each phase is also reduced by the same factor so that line current at starting becomes 1/√3 x 1/√3 = 1/3 of the current motors. Since T 𝛼 V2, the stating torque reduced (1/√3)2 =33.3% of the normal value. When the motor picks up the speed, the TPDT switch is thrown manually on the other side such that the winding is now connected in delta across the supply. So the normal voltage is applied to the motor (because in delta connection voltage is same, V L =VP) and hence the motor runs at normal speed. This stating system has interlocks to prevent motor starting with the switch in run position. This method is cheap and maintenance free as compared to other methods. Auto Transformer Starter In this method, a three-phase auto transformer is connected in series with the motor. Auto transformer reduces the voltage applied to the motor and this reduced voltage results the less current through the motor. When the 50% tap is used, the starting torque developed by the motor would be 0.52 or 25% full load torque As in the figure TPDT switch is used for connecting the auto transformer in the circuit for starting purposes. When motor gathers 80% of the normal speed, the change over switch is thrown into run position which connects the motor directly to the supply. Starting of Single Phase Induction Motors The single phases induction motors are classified based on the method of starting method and in fact are known by the same name descriptive of the method. 17 1. Split – phase Induction Motor The main winding which is highly inductive and lesser resistive is connected across the line in the usual manner. The auxiliary or starting winding has a greater resistance and lesser inductive reactance as compared to main winding. Current in the starting winding lags the supply voltage by lesser angle. The current in the main winding lags the supply voltage with greater angle due to its high inductance. Now there are two currents which are not in phase with each other. As shown in the figure. IM is the current in the main winding and IA is the current in the starting winding. The angle between the two currents is ϕ as shown in the figure and the starting torque is proportional to sin of an angle‘ϕ‘. These two currents are out of phase about 250 -300 and produces weak magnetic field which starts the motor. Since the currents in two windings are not equal, the rotating magnetic field is not uniform. Hence the starting torque is small. A centrifugal switch which is normally closed is connected in series with the starting winding. When the motor comes up to speed about 75% of the synchronous speed it opens automatically due to centrifugal force and puts the starting winding out of the circuit. It is important that the centrifugal switch should open otherwise the auxiliary winding being made of thin wire will be overheated and thus burns. The starting torque is low, typically 100% to 200% of the rated torque. The motor draws high starting current, approximately 700% to 1,000% of the rated current. The maximum generated torque ranges from 250% to 350% of the rated torque It is used in washing machines, fans, blowers, centrifugal pumps, wood working tools, grinders and various other low starting torque applications. 18 2. Capacitor – Start MotorCapacitors are used to improve the starting and running performance of the single phase inductions motors. The construction of this type of motors is similar to the split phase motor. The difference is that in series with the auxiliary winding the capacitor is connected. The capacitor – start motor is identical to a split – phase motor except that the starting winding has as many turns as the main winding. The capacitive circuit draws a leading current, this feature used in this type to increase the split phase angle α between the two currents Im and Ia. The starting winding is opened by the centrifugal switch when the motor attains about 75% of synchronous speed. Capacitor employed is short time duty rating and electrolytic type.The value of capacitor is so chosen that Ia leads Im by about 90 o so that the starting torque is maximum for certain values of Ia and Im. Since the capacitor is in series with the start circuit, it creates (5-b) more starting torque, typically 200% to 400% of the rated torque. And the starting current, usually 450% to 575% of the rated current, is much lower than the split-phase due to the larger wire in the start circuit. Again, the starting winding is opened by the centrifugal switch when the motor attains about 75% of synchronous speed. The motor then operates as a single – phase induction motor and continues to accelerate till it reaches the normal speed. Capacitor – start motors are used where high starting torque is required and where high starting period may be long e.g. to drive: a) Compressors b) large fans c) pumps Permanent – Split Capacitor Motor In this motor, as shown in fig.(6-a), the capacitor that is connected in series with the auxiliary winding is not cut out after starting and is left in the circuit all the time. This simplifies the construction and decreases the cost because the centrifugal switch is not needed. The capacitor is of paper type. The 19 capacitor is permanently connected to the starting winding. In case of the paper capacitor, the value of the capacitance is small since it is difficult and becomes uneconomical to manufacture paper capacitor of higher value. The power factor, torque pulsation, and efficiency are also improved because the motor runs as a two – phase motor. The motor will run more quietly. The typical starting torque of the PSC motor is low, from 30% to 150% of the rated torque. PSC motors have low starting current, usually less than 200% of the rated current, making them excellent for applications with high on/off cycle rates. The motor design can easily be altered for use with speed controllers. They can also be designed for optimum efficiency and High-Power Factor (PF) at the rated load. They re considered to be the most reliable of the single-phase motors, mainly because no centrifugal starting switch is required. An electrolytic capacitor cannot be used for continuous operation, hence, oil-filled paper capacitor is used which is larger in size and more expensive. These motor are used where the required starting torque is low such as air – moving equipment i.e. fans, blowers etc. Shaded Pole Induction Motor These motors have a salient pole construction. A shaded band consisting of a short – circuited copper turn, known as a shading coil, is used on one portion of each pole, as shown in Each shaded pole has its own exciting coil. About 1/3 rd portion of each pole core is surrounded, by a strap of copper forming a shading coil. When the flux is increasing in the pole, a portion of the flux attempts to pass through the shaded tip or shaded portion of the pole. This flux induces voltage and hence current in 20 the copper ring, and by Lenz's law the direction of current is such that it opposes the flux entering the coil. Hence in the beginning, the greater portion of the flux passes through un shaded side of each pole. When the flux reaches its maximum value, its rate of change is zero thereby the e.m.f. and hence current in the shading coil becomes zero. Large amount of main flux then links with the coil. After the main flux tends to decrease, the current induced in the shading coil now tends to prevent the flux linking with shading coil from decreasing. The shading coils thus causes the flux in shaded portion as shown to lag the flux in shaded portion of the pole. Because the shaded-pole motor lacks a start winding, starting switch or capacitor, it is electrically simple and inexpensive. Also, the speed can be controlled merely by varying voltage, or through a multi-tap winding. Its low starting torque is typically 25% to 75% of the rated torque. It is a high slip motor with a running speed 7% to 10% below the synchronous speed. Generally, efficiency of this motor type is very low (below 20%). Since starting torque, efficiency and power factor are very low, these motors are only suitable for low power applications as small fans, toys, hair driers etc. The power rating of such motors is up to about 30 W Capacitor - Start Capacitor – Run The capacitor-start capacitor-run induction motor is also known as two value capacitor motor. The schematic diagram of a capacitor-start capacitorrun induction motor is shown below. The capacitor-start capacitor-run induction motor consists of a squirrel cage rotor and its stator has two windings, named starting or auxiliary winding and the main or running winding. The two windings are displaced by an angle of 90° in the space. This motor uses two capacitors − the starting capacitor (CS) and the running capacitor (CR). The two capacitors are connected in parallel at the instant of starting.The starting capacitor is larger in value and is of the ac electrolytic type. The running capacitor permanently connected in series with the starting winding, is of smaller value and is of the paper oil type. Typical values of these capacitors for a 0.5 hp are about Cs =300 F, Cr =40 F 21  In order to obtain a high starting torque, a large starting current is required. For this, the capacitive reactance in the starting winding should be low. Hence, for capacitive reactance to be small, the value of starting capacitor (CS) should be large. During the normal operation of the motor, the rated line current should be smaller than the starting current. Therefore, the capacitive reactance of the running capacitor should be high and the value of the running capacitor (CR) should be small. The phasor diagram of the capacitor-start capacitor-run motor is shown below. At starting both the capacitors are in the circuit, therefore, the phase angle φ is greater than 90°. When the starting capacitor (CS) is disconnected from the circuit, then the phase angle becomes 90° electrical. Following are the primary characteristics of a capacitor-start capacitor-run induction motor − These motors have quiet and smooth running operation. They have high efficiency and power factor These motors produce constant torque and not a pulsating torque. Because of the constant torque, the motor is vibration free. Ability to start heavy loads Figure shows the torque-speed curves of various kinds of single-phase AC induction motors. 22 Universal Motor: It is also known as Single Phase Series Motor. They can be operated either on ac or dc supply.At approximately the same speed and output. It is also known as Single Phase Series Motor. The armature is of the same type as ordinary series motor. When the motor runs on AC voltage, the alternating flux causes a reactance voltage, which limits the current to a much lower level than would be produced with DC voltage. AC supply also induces more significant eddy currents than are produced when the motor operates on DC supply. To curtail eddy current losses, universal motors use laminated cores (rather than solid iron), which increases their resistance and reduces eddy currents. These motors are used in sewing machines, vacuum cleaners, kitchen appliances, drills, hair dryers etc. The direction of rotation can be reversed by reversing the current in the field circuit or armature. The speed torque characteristics for dc and ac input supply is shown above. : 23