ELECTRICAL MACHINES III Fourth Year Prepared by : Prof. Ahmed Abdeltwab Types of small sized motors a. Single-phase motors. b. Special machines. a. Single-phase induction motors: (i) split-phase type (ii) capacitor start type (iii) capacitor start capacitor run type (iv) shaded-pole type AC series commutator motors: Universal motors. Repulsion motors. c. Single-phase synchronous motors 1) the reluctance type. 2) the hysteresis type. 2. Special Motors i) Stepper motors. ii) Permanent magnet motors. iii) Synchros System. iv) Servomechanism. Chapter 1 Single-Phase Induction Motors Construction of single phase induction motor Stator Rotor Principle of operation a) First consider that the rotor is stationary and the stator winding is connected to a single-phase supply. b) A pulsating flux is established in the machine along the axis of the stator winding. c) The pulsating stator flux induces current by transformer action in the rotor circuit. d) The rotor current produces a pulsating flux opposing the stator flux. e) As the angle between these fluxes is zero, no starting torque is developed. Revolving field theory a) For a sinusoidal distributed stator winding, the mmf along the position is: F(θ) = Ni cos θ π = πΌπππ₯ cos ππ‘ Let πΉ (θ, t) = π πΌπππ₯ cos π cos ππ‘ π πΌπππ₯ = 2 cos ππ‘ − π + π πΌπππ₯ 2 πΉ (θ, t) = πΉπ + πΉπ cos ππ‘ + π πΉπ represents a rotating mmf in the forward direction of " θ " πΉπ represents a rotating mmf in the opposite direction. Equivalent Circuit The slip with respect to the forward field is: π π = ππ −π ππ The slip with respect to the backward field is: π π = ππ −(−π) ππ = 2−π πππ ′ 0.5 π 2 2 = πΌπ ( ) = πΌ12 π π π πππ ′ 0.5 π 2 2 = πΌπ ( ) = πΌ12 π π 2−π πππ ππ = ππ πππ ππ = ππ π = ππ − ππ The mechanical power developed is : π πΆπΏ)π = π πππ ππ = π π π πΆπΏ)π = (2 − π ) πππ πππ − πππ = 1 − π ππ ππ = 1−π πππ − πππ The total rotor copper loss is: π πΆπΏ = π πΆπΏ)π + π πΆπΏ)π = π πππ + (2 − π ) πππ The power output is : ππ = ππ − ππππ‘ ππππ‘ = ππ+π€ + iron losses ππ = (πππ + πππ ) Torque Pulsations a) The pulsating torque results from the interactions of the forward flux with the backward rotor mmf and vice versa. b) The pulsating torque produces no average torque. c) It makes single-phase motors more noisier. d) The effects of the pulsating torque can be minimized by using rubber pads. Starting of single-phase induction motor a) Split-phase motors. b) Capacitor start motors. c) Capacitor start Capacitor run motors. (d) Shaded-pole motors. Split-phase induction motors a) The stator is provided with an auxiliary winding displaced in space 90° from the main winding. b) The two windings are designed to make phase shift between the two currents. Thus: starting winding : high R and low X . main winding : high X and low R . θ = 25° to 30° Capacitor start induction Motor Higher starting torque can be obtained if a capacitor is connected in series with the auxiliary winding θ = 80° Capacitor-Run Motors a) Permanent capacitor is connected in series with the auxiliary. b) This simplifies the construction and decreases the cost because the centrifugal switch is not needed. c) The power factor, torque pulsation, and efficiency are also improved because the motor runs as a two-phase motor. d) The motor will run more quietly. Capacitor start capacitor run motors Two capacitors, one for starting and one for running are used. The starting winding and the capacitor can be designed for perfect 2-phase operation at any load. The motor produces a constant torque and not a pulsating torque. The motor is vibration free and can be used in hospitals, and other places where silence is important. Shaded-pole induction motors a) It has salient pole construction. b) A shading ring is used on one portion of each pole. c) The flux in the shaded part of the pole lags the flux in the unshaded part. d) The starting torque, efficiency and power factor are very low. Determination of Equivalent Circuit Parameters 1) Blocked rotor test: a) Measure π 1 by dc test. b) Rbl = Pbl / (Ibl)2 = R1 + R'2 c) Zbl = Vbl / Ibl d) Xbl can be calculated and take X1 = X'2 equal to 0.5 Xbl . No load test: i) The no load resistance Ro is calculated from: Ro = Po / ( Io) 2 Where Ro = R1 + Rrot + (R2' / 4) Knowing R2' from the blocked test, the value of Rrot can be obtained. ii) the no load impedance Zo is calculated from: Zo = Vo / Io iii) Xo can be calculated . Where Xo = X1 + 0.5 ( Xm + X'2 ) Then the value of Xm can be found. Example 1.1 The following test data are obtained from a 1/4 hp, 1-phase, 120 V, 60 Hz, 1730 rpm induction motor : Stator winding (main) resistance = 2.9 ohms Blocked rotor (standstill) test: V = 43 V, I = 5A, P = 140 W No-load test: I = 3.5 A, P = 125 W V = 120 V, (a) Obtain the double revolving field equivalent circuit for the motor. (b) Determine the rotational loss. Solution (a) R1 = 2.9 ohms PBL =5 2 (2.9 + R'2) = 140 w therefore R'2 = 2.7 ohms ZBL= 43/5 = 8.6 ohms 8.6 = (2.9 + + ( 2 2.7) Xl + X′2) 2 Xl + X′2 = 6.53 ohms Xl = X′2 = 6.53/2 = 3.26 ohms. Ro = Po / (Io)2 = 125 / (3.5)2 = 10.2 β¦ Ro = R1 + Rrot + (R2' / 4) 10.2 = 2.9 + Rrot + (2.7/4) Rrot = 6.625 β¦ Zo = Vo / Io = 120 /3.5 = 34.3 β¦ Xo = 32.75 β¦ Xo = X1 + 0.5 ( Xm + X'2 ) 32.75 = 3.26 + 0.5 ( Xm + 3.26) Xm = 55.72 β¦ Example 1.2 For the single-phase induction motor of Example 1.1, determine the input current, power, power factor, developed torque, output power, efficiency of the motor, air gap power, and rotor copper loss if the motor is running at the rated speed when connected to a 120 V supply. Solution ππ − π 1800 − 1730 π = = = 0.039 ππ 1800 π1 = π 1 + ππ1 = 2.9 + j 3.26 The impedances of the rotor (referred to stator) with respect to the forward and backward rotating fluxes are : π′2π = π′2π = 0.5 π ′2 π 0.5 π ′2 2−π + 0.5 ππ′2 = 34.6 + j 1.63 + 0.5 ππ′2 = 0.69 + j 1.63 The total forward and backward impedances are: ππ = π π + πππ = π′2π // 0.5 π π₯π = 13 + j 16.8 ππ = π π + πππ = π′2π // 0.5 π π₯π = 0.614 + j 1.55 π = 16.5 + π 21. I1 = 120 / (16.5 + π 21.6) = 2.68 – j 3.5 =4.4β-52.6º p.f. = cos (52.6) = 0.61 input power = Pi = 120 x 4.4 x 0.61 = 322 watt πππ = πΌ12 π π = (4.4)2 π₯ 13 = 251.7 π€ππ‘π‘ πππ = πΌ12 π π = (4.4)2 π₯ 1.55 = 30 π€ππ‘π‘ The corresponding torques are: πππ 251.7 ππ = = = 1.335 π. π. ππ 188.5 πππ 30 ππ = = = 0.159 π. π. ππ 188.5 The resultant torque is : π = ππ − ππ = 1.335 − 0.159 = 1.176 π. π. The mechanical power developed is : ππ = 1 − π πππ − πππ = 1 − 0.039 251.7 − 30 = 213 π€ππ‘π‘ Mechanical losses = ππππ‘ = πΌπ2 π πππ‘ = 3.5 2 6.625 = 81 π€ππ‘π‘ The power output is : ππ = ππ − ππππ‘ = 213 − 81 = 132 π€ππ‘π‘ Efficiency = 132 322 = 41% Example 1.3 The currents in the main and the auxiliary windings are as follows: The effective numbers of turns for the main and auxiliary windings are Nm and Na . The windings are placed in quadrature. a) Obtain expressions for the stator rotating mmf wave. b) Determine the magnitude and the phase angle of the auxiliary winding current to produce a balanced two-phase system. Solution (a) The stator mmf along a position defined by an angle θ (where θ = 00 defines the axis of the main winding) is contributed by both windings. Example : Prove that the starting torque of the split-phase induction motor is equal to: Where ( ( ) is the main current, ( ) is the auxiliary current, and ) is the angle between the two currents. Solution Assume that the cage rotor can be represented by an equivalent two-phase winding, having number of turns/phase N2 , resistance/phase R2 , and reactance/phase X2 (at the stator frequency f ). The current flowing through the main winding produces flux that induces voltage E2m in the rotor (by transformer action) given by: Similarly, flux in the auxiliary winding induces voltage π¬ππ in the rotor: These voltages will produce currents πππ and πππ . There are two torques developed as a result of: a) the interaction of ππ and I2a . b) the interaction of ππ and π°ππ . Since Then Therefore, Starting winding design: The starting winding can be designed to provide: 1) Maximum starting torque. 2) Maximum starting torque per ampere. Maximum starting torque of split-phase motor Example: For the split phase motor shown in figure, derive an expression for the auxiliary winding resistance in order to obtain maximum torque at starting, assuming specified number of turns of the starting winding “Na”. Solution π°π π½ πΏπ π½ = = πππ π½π |ππ | πΏπ πΏπ πππππ π½π = −π πΏπ πππ πΉπ π½ If πΉπ = 0 π½π = 90 π°π = If πΉπ = ∞ π½π = 0 π°π = 0 πΏπ The locus of π°π is a semicircle having diameter π½ πΏπ since Im is fixed, ππ π‘ ∝ πΌπ sin α ∝ πππππ‘β πΆπΎ For maximum starting torque: ππ π‘ ∝ πΌπ sin α ∝ πππππ‘β π·πΎ ′ ππ = ππ 2 πππ‘ ππ = cot ( = 1+cos ππ sin ππ ππ 2 )= π cos π 2 π sin 2π In capacitor start motors, find the value of capacitor to obtain maximum starting torque? Solution a) If Xc is infinitely large, Ia is zero. b) If Xc = Xa, Ia is a maximum, is equal to |V|/Ra, and is in phase with the supply voltage V . c) The locus of Ia is a semicircle having diameter= |V| /Ra. since Im is fixed, π»ππ ∝ π°π πππ α ∝ ππππππ πͺπ² Knowing that: Then: Since Then: Maximum starting torque/amp of capacitor start motor a) the starting current is represented by OC b) the starting torque is represented by CK c) the starting torque/amp. = CK / OC this ratio is maximum when OC is tangent to the circle. Example 1.4 A 4-pole, single-phase, 120 V, 60 Hz induction motor gave the following standstill impedances when tested at rated frequency: Main winding: Zm =1.5 +j4.o ohms Auxiliary winding:Za = 3 + j 6.0 ohms a)Determine the value of external resistance to be inserted in series with the auxiliary winding to obtain maximum starting torque as a resistor split-phase motor. b) Determine the value of the capacitor to be inserted in series with the auxiliary winding to obtain maximum starting torque as a capacitor start motor. c) Determine the value of the capacitor to be inserted in series with the auxiliary winding to obtain maximum starting torque per ampere of the starting current as a capacitor-start motor. d) Compare the starting torques and starting currents in parts (a), (b), and (c) expressed as per unit of the starting torque without any external element in the auxiliary circuit, when operated at 120 V, 60 Hz. Solution Resistor split-phase motor: Equivalent circuit for a capacitor run motor a) The main winding flux can be resolved into two revolving fluxes πππ and πππ . b) The auxiliary winding flux is resolved into two revolving fluxes πππ and πππ c) The four revolving fluxes induce voltages in the two windings. Main equivalent circuit: Efm and Ebm are the voltages induced by its own fluxes πππ and πππ . The voltages induced, -j Efa/a and j Eba/a in the main winding by πππ and πππ . Auxiliary equivalent circuit: Efa and Eba are the voltages induced by its own fluxes πππ and πππ . The voltages induced, j aEfm and -j aEbm in the main winding by πππ and πππ . a Va = la (Zc + a2Zf + a2Zb) + ja Efm — ja Ebm Efa = la . a2 Zf Eba = la a2 Zb Efm = Im Zf Ebm = Im Zb For starting, slip s =1 and Rf = Rb. The starting torque is: Example 1.5 A single-phase 120 V, 60 Hz, four-pole capacitor-run motor has the following equivalent circuit parameters: a) Draw the equivalent circuit at slip 0.05. b) Determine the total starting current and the starting torque of the motor at rated voltage.
