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 𝑖 = 𝐼𝑚𝑎𝑥 cos 𝜔𝑡 Let 𝐹 (θ, t) = 𝑁 𝐼𝑚𝑎𝑥 cos 𝜃 cos 𝜔𝑡 𝑁 𝐼𝑚𝑎𝑥 = 2 cos 𝜔𝑡 − 𝜃 + 𝑁 𝐼𝑚𝑎𝑥 2 𝐹 (θ, t) = 𝐹𝑓 + 𝐹𝑏 cos 𝜔𝑡 + 𝜃 𝐹𝑓 represents a rotating mmf in the forward direction of " θ " 𝐹𝑏 represents a rotating mmf in the opposite direction. Equivalent Circuit The slip with respect to the forward field is: 𝑠𝑓 = 𝑛𝑠 −𝑛 𝑛𝑠 The slip with respect to the backward field is: 𝑠𝑏 = 𝑛𝑠 −(−𝑛) 𝑛𝑠 = 2−𝑠 𝑃𝑔𝑓 ′ 0.5 𝑅 2 2 = 𝐼𝑓 ( ) = 𝐼12 𝑅𝑓 𝑠 𝑃𝑔𝑏 ′ 0.5 𝑅 2 2 = 𝐼𝑏 ( ) = 𝐼12 𝑅𝑏 2−𝑠 𝑃𝑔𝑓 𝑇𝑓 = 𝜔𝑠 𝑃𝑔𝑏 𝑇𝑏 = 𝜔𝑠 𝑇 = 𝑇𝑓 − 𝑇𝑏 The mechanical power developed is : 𝑅𝐶𝐿)𝑓 = 𝑠 𝑃𝑔𝑓 𝑃𝑚 = 𝜔 𝑇 𝑅𝐶𝐿)𝑏 = (2 − 𝑠) 𝑃𝑔𝑏 𝑃𝑔𝑓 − 𝑃𝑔𝑏 = 1 − 𝑠 𝜔𝑠 𝜔𝑠 = 1−𝑠 𝑃𝑔𝑓 − 𝑃𝑔𝑏 The total rotor copper loss is: 𝑅𝐶𝐿 = 𝑅𝐶𝐿)𝑓 + 𝑅𝐶𝐿)𝑏 = 𝑠 𝑃𝑔𝑓 + (2 − 𝑠) 𝑃𝑔𝑏 The power output is : 𝑃𝑜 = 𝑃𝑚 − 𝑃𝑟𝑜𝑡 𝑃𝑟𝑜𝑡 = 𝑃𝑓+𝑤 + iron losses 𝑃𝑔 = (𝑃𝑔𝑓 + 𝑃𝑔𝑏 ) 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 𝑅1 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 Ω Ro = R1 + Rrot + (R2' / 4) 10.2 = 2.9 + Rrot + (2.7/4) Rrot = 6.625 Ω Zo = Vo / Io = 120 /3.5 = 34.3 Ω Xo = 32.75 Ω Xo = X1 + 0.5 ( Xm + X'2 ) 32.75 = 3.26 + 0.5 ( Xm + 3.26) Xm = 55.72 Ω 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 𝑛𝑠 − 𝑛 1800 − 1730 𝑠= = = 0.039 𝑛𝑠 1800 𝑍1 = 𝑅1 + 𝑗𝑋1 = 2.9 + j 3.26 The impedances of the rotor (referred to stator) with respect to the forward and backward rotating fluxes are : 𝑍′2𝑓 = 𝑍′2𝑏 = 0.5 𝑅′2 𝑠 0.5 𝑅′2 2−𝑠 + 0.5 𝑗𝑋′2 = 34.6 + j 1.63 + 0.5 𝑗𝑋′2 = 0.69 + j 1.63 The total forward and backward impedances are: 𝑍𝑓 = 𝑅𝑓 + 𝑗𝑋𝑓 = 𝑍′2𝑓 // 0.5 𝑗 𝑥𝑚 = 13 + j 16.8 𝑍𝑏 = 𝑅𝑏 + 𝑗𝑋𝑏 = 𝑍′2𝑏 // 0.5 𝑗 𝑥𝑚 = 0.614 + j 1.55 𝑍 = 16.5 + 𝑗 21. I1 = 120 / (16.5 + 𝑗 21.6) = 2.68 – j 3.5 =4.4∟-52.6º p.f. = cos (52.6) = 0.61 input power = Pi = 120 x 4.4 x 0.61 = 322 watt 𝑃𝑔𝑓 = 𝐼12 𝑅𝑓 = (4.4)2 𝑥 13 = 251.7 𝑤𝑎𝑡𝑡 𝑃𝑔𝑏 = 𝐼12 𝑅𝑏 = (4.4)2 𝑥 1.55 = 30 𝑤𝑎𝑡𝑡 The corresponding torques are: 𝑃𝑔𝑓 251.7 𝑇𝑓 = = = 1.335 𝑁. 𝑚. 𝜔𝑠 188.5 𝑃𝑔𝑏 30 𝑇𝑏 = = = 0.159 𝑁. 𝑚. 𝜔𝑠 188.5 The resultant torque is : 𝑇 = 𝑇𝑓 − 𝑇𝑏 = 1.335 − 0.159 = 1.176 𝑁. 𝑚. The mechanical power developed is : 𝑃𝑚 = 1 − 𝑠 𝑃𝑔𝑓 − 𝑃𝑔𝑏 = 1 − 0.039 251.7 − 30 = 213 𝑤𝑎𝑡𝑡 Mechanical losses = 𝑃𝑟𝑜𝑡 = 𝐼𝑜2 𝑅𝑟𝑜𝑡 = 3.5 2 6.625 = 81 𝑤𝑎𝑡𝑡 The power output is : 𝑃𝑜 = 𝑃𝑚 − 𝑃𝑟𝑜𝑡 = 213 − 81 = 132 𝑤𝑎𝑡𝑡 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 𝑬𝟐𝒂 in the rotor: These voltages will produce currents 𝒊𝟐𝒎 and 𝒊𝟐𝒂 . There are two torques developed as a result of: a) the interaction of 𝝋𝒎 and I2a . b) the interaction of 𝝋𝒂 and 𝑰𝟐𝒎 . 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 𝑰𝒂 𝑽 𝑿𝒂 𝑽 = = 𝒔𝒊𝒏 𝜽𝒂 |𝒁𝒂 | 𝑿𝒂 𝑿𝒂 𝒘𝒉𝒆𝒓𝒆 𝜽𝒂 = −𝟏 𝑿𝒂 𝒕𝒂𝒏 𝑹𝒂 𝑽 If 𝑹𝒂 = 0 𝜽𝒂 = 90 𝑰𝒂 = If 𝑹𝒂 = ∞ 𝜽𝒂 = 0 𝑰𝒂 = 0 𝑿𝒂 The locus of 𝑰𝒂 is a semicircle having diameter 𝑽 𝑿𝒂 since Im is fixed, 𝑇𝑠𝑡 ∝ 𝐼𝑎 sin α ∝ 𝑙𝑒𝑛𝑔𝑡ℎ 𝐶𝐾 For maximum starting torque: 𝑇𝑠𝑡 ∝ 𝐼𝑎 sin α ∝ 𝑙𝑒𝑛𝑔𝑡ℎ 𝐷𝐾 ′ 𝜃𝑎 = 𝜃𝑚 2 𝑐𝑜𝑡 𝜃𝑎 = cot ( = 1+cos 𝜃𝑚 sin 𝜃𝑚 𝜃𝑚 2 )= 𝜃 cos 𝑚 2 𝜃 sin 2𝑚 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, 𝑻𝒔𝒕 ∝ 𝑰𝒂 𝒔𝒊𝒏 α ∝ 𝒍𝒆𝒏𝒈𝒕𝒉 𝑪𝑲 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 𝜑𝑓𝑚 and 𝜑𝑏𝑚 . b) The auxiliary winding flux is resolved into two revolving fluxes 𝜑𝑓𝑎 and 𝜑𝑏𝑎 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 𝜑𝑓𝑚 and 𝜑𝑏𝑚 . The voltages induced, -j Efa/a and j Eba/a in the main winding by 𝜑𝑓𝑎 and 𝜑𝑏𝑎 . Auxiliary equivalent circuit: Efa and Eba are the voltages induced by its own fluxes 𝜑𝑓𝑎 and 𝜑𝑏𝑎 . The voltages induced, j aEfm and -j aEbm in the main winding by 𝜑𝑓𝑚 and 𝜑𝑏𝑚 . 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.