Universal Motors: Principle of Operation and Special Considerations

Lecture # 5 UniversalMotors

Principle of Operation A motor which can run on either alternating current (ac) or direct current (dc) is called universal motor. It is also known as single phase series motor. Now we need to answer the question of which of the dc motor configuration is used; is it a series dc motor or a shunt dc motor? In other word, the field winding of the universal motor should be connected in series or shunt with the armature winding? Shunt Field: In a dc shunt motor, Figure 2.1, the field winding has high resistance Rf and is connected in parallel with the armature winding Ra. Based on ohm s law, I = V/R, this high resistance Rf limits current flow in the field winding under dc operation. Figure 2-1 DC shuntmotor. 3

The field winding however, has large number of turns, which result in high inductive reactance Xf = .Lf when connected to an ac source, where the inductance is given by L = N2/R Both high resistance Rf and high inductive reactance Xf decrease the current flow in the field winding under ac operation. Therefore, the flux created is weak, since = N.I/R R . R , and R R is thereluctance. The highly inductive field winding tends to delay the reversal of the field winding current, which causes a considerable phase lagbetween the armature current and the flux produced by the field winding. Thus, a reduced average torque results, which may not be acceptable for the size of the motor. Therefore, A shunt dc motor is never designed for both dc and ac applications. 3

Series Field: In a dc series motor, Figure 2.2, the field winding and the armature winding are connected in series. As a result, the same currentflows through both windings. Therefore, the flux produced in field winding is in phase with the armature current even when it is connected to an ac source. The series winding has few number of turns; thus its inductive reactance is low under ac operation. A dc series motor specially designed for ac operation is called universal motor. A universal motor is wound and connected just like a dc series motor. That is, the field winding is connected in series with the armature winding. Figure 2-2 DC seriesmotor. 4

Some special consideration in building a universal motor Auniversal motor has a laminatedstator and field pole structure inorder to minimize the core-loss produced in them by the alternatingflux. Afewer numberof turns are used for ac applications to reducethe reactive voltage drop across the series fieldwinding. The effective conductors in the armature are increased in order tomake up for the reduction the flux in themotor. When a series motor is operated from a dcsource The current is unidirectional in both the field and armaturewindings. In other words, the flux produced by each pole and the direction of the current I in the armature conductors under that pole remain in the same direction at alltimes. Hence, the direction of the torque motor is constant. developed by the 5

When a series motor is operated from an acsource The current in the field and armature windings reverses itsdirection every half cycle as shown in Figure 2.3 for a two-pole seriesmotor. During the positive half cycle (Figure 2.3a), the flux produced by thefield winding is from right to left. For the marked direction of the current in the armature conductors, the motor develops a torque in the counterclockwise direction. During the negative half cycle (Figure 2.3b), the applied voltagehas reversed itspolarity. _ _ + Consequently, the currenthas reversed its direction. As a result, the fluxproduced by the poles is now directed from left toright. + (a) (b) Figure 2-3 Current and flux directions in a universal motor during (a) the positive and (b) The negative halfcycles.

Since the reversal in the current in the armature conductors is also accompanied by reversal in the direction of flux in the motor, the directionof the torque developed by the motor remainsunchanged. Hence, the motor continues its rotation in the counterclockwise direction. The instantaneous torque developed by the motor is givenby When the motor operates in the linear region (below the knee ofthe magnetization curve), the flux ?must be proportional to the field currentia When the permeability presence of the air-gap ensures that forall practical purposes the flux is inphase with the currentI of the magnetic core is relatively high,the Thus, the instantaneous torquedeveloped by the motor is (Figure2.4) Figure 2-4 The current and the torque developed by a universal motor. 8

Figure 2.5 shows the equivalent circuit, and the phasordiagram. The back emf Ea, the armature winding current Ia, and the flux per pole??p are in phase with each otheras shown. Note : Ea= ka ?p Rs and Xs are the resistance and reactance of the seriesfield winding. Ra and Xa are the resistance and reactance of thearmaturewinding. From Figure 2.5 we can calculate the induced emf Eain the armatureas Vs Ia Power factor: pf = cos(v-i) pf = cos( ) Figure 2-5 Equivalent circuit of a universal motor,and it sphasor diagram. 8

EXAMPLE 2-1 A120 Vrms, 60 Hz, universal motor operates at a speed of 8000 rpm on fullload and draws a current of 17.58Armsat a lagging power factor of0.912. The impedance of the series field winding is 0.65 + j1.2 the armature winding is 1.36 + j1.6 . Determine: (a) the induced emf in the armature Ea,(b) the power output Po,(c) the developed and shaft torque, and (d) the efficiency if the rotational loss is 80W. . The impedanceof SOLUTION The phase angle the motorcurrent: V a) s Ia From the equivalent circuit of themotor: = 120 0 17.58 24.22o [0.65 + 1.36 + j(1.2 + 1.6)]=74.1 24.22oV 9

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EXAMPLE 2-2 A 120 Vdc, universal motor, the same motor in Example 2-1, at the same load which draws the same current of 17.58 Adc. The resistance of the series field winding is 0.65 . The resistance of the armature winding is 1.36 . Determine: (a) the induced emf in the armature Ea,(b) the power outputPo, (c) the shaft torque Ts,and (d) the efficiency if the rotational loss is 80W. SOLUTION From the equivalent circuit of themotor: a) b) The power developed by themotor: P Pd d P Po o P Pin in ? ?? ? ? ?? ? ? ?? ? ? ?? ? P Pcu P Prot rot cu 11

A universal motor can bemanufactured in two differentways: 1. Non-Compensated type with concentratedpoles. 2. Compensated type with distributedfield. The compensated type is preferred for high power rating appliancesand the non-compensated for low power ratedappliances. Both the compensated and non-compensated have construction similarto that of a dc seriesmotor. 1. Non-Compensated Type Motor: The non-compensated motor has 2 salientpoles and it is laminated, Figure2.6. The armature is of wound type and thelaminated core is either straight or skewedslots. The leads of the armature winding areconnected to thecommutator. Figure 2-6 Two salientpoles. 12

The number of commutator segments is increased, andhigh-resistance brushes are used to reduce the harmful effects of sparking at thebrushes. The skew in the armature slots Figure 2.7 serves for twopurposes: It reduces the magnetichum. It aids in reducing the locking tendency of rotor, which is called magnetic locking. Magneticlocking is a condition during which the rotor teeth remains locked under the stator teeth due to magnetic attraction between the stator androtor. Figure 2-7 Skew in thearmature. 2. Compensated TypeMotor: The compensated motor consists of distributed field winding and thestator core is similar to that of split-phase motor, Figure2.8. We already know that split phase motors consist of an auxiliary windingin 13

Similar to the split phase motors, thecompensated type also consists of an additionalwinding. The compensating winding helps in reducing the reactance voltage which is caused due to alternating flux, when the motor runs with the aid of anACsupply. A universal motor with a conductively coupled compensating winding is shown in Figure2.9a. The corresponding phasordiagram (Figure 2.9b) shows the improvement in the powerfactor. Figure 2-8Compensating winding . Figure 2-9 (a) Equivalent circuit and (b) phasor diagram for a conductively compensated universal motor. (a) (b) 16

Characteristics of Universal Motor It can be shown, that there is 29.3% less torque on a machine operating on ac as opposed to dc, for a motor designed to operate at the samepeak flux level and RMS value ofcurrent. Speed control: Speed control of universal motor is very important, and thefollowing methods are employed for the speed control of universalmotors. 1. Resistance Method: In this method of speed control a variableresistance is connected in series with the motor, Figure2.11. The amount of resistance in the circuit can be changed. Afoot pedal is used for thispurpose. Usually this method is employed formotors used in sewing machines. Figure 2-11Resistance speed control. 18

Characteristics of Universal Motor 2. Centrifugal MechanismMethod: This method is involved whenever the application involves a numberof speeds. Best example is home food and fruitmixers. Here a centrifugal device is attached to the motor,Figure 2.12. If themotor rises above the specified speed set by the knob, the centrifugal device opens the contact and the resistance R comes in contact with the circuit. This causes the motor speed to decrease below the setspeed. When the motor runs slower than the speed set by the knob, the contact is established and resistance is short circuited. This causes the speed toincrease. The variations in speed are noticeable asthe process is repeated in a rapidmanner. The resistance R is connected across thegovernor points shown in Figure 2.12. A capacitor C is used across the contacts points to reducesparking produced due to the opening and closing of thepoints. Figure 2-12Centrifugal mechanism speedcontrol. 19

3.Tapping-Field Method: In this method, the flux is reduced, and hencespeed is increased, by decreasing the number of turns of the series field winding as shown in Figure2-13. In this method a number of tapping from fieldwinding are broughtoutside. Now how this tapping arrangement isestablished? The tapping arrangement is established by twoways: Figure 2-13Centrifugal mechanism speedcontrol. The field poles are wound in varioussections with different series of wire and taps are brought out from eachsection. The second method, Nichromeresistance wire, Figure 2-14, is wound over one of thefield poles and the taps are brought out from thiswire. Figure 2-14Nichrome resistance wire. 20

Can the direction of rotation be reversed for these types ofmotors? Yes, the direction of rotation canbe reversed by changing the direction of flow of current through the armature or fieldwinding. This can also be done by interchanging the leadson the brush holders, Figure2-15. Figure 2-15 Changeling the direction of a universal motor. Applications Of The UniversalMotor Auniversal motor is used quite extensively in the fractionalhorsepower range. Some applications that require variation in speed with load are saws and routers, sewing machines, portable machine tools, vacuum cleaners, and kitchenequipment s. 19

One may ask a very logicalquestion: With all these drawbacks, why do we use a universalmotor? Some of the reasons are givenbelow: 1. Auniversal motor is needed when it is required to operatewith complete satisfaction on dc and acsupply. 2. The universal motor satisfies the requirements when we need amotor to operate on ac supply at a speed in excess of 3000 rpm (2-pole induction motor operating at 50 Hz). Since the power developed is proportional to the motor speed, a high-speed motor develops more power for the same size than a lows-peedmotor. 3. When we need a motor that automatically adjusts its speed underload, the universal motor is suitable for that purpose. Its speed is high when the load is light and low when the load isheavy. 20

1. A 240-V, 50-Hz, 2-pole universal motor operates at a speed of12000rpm The motoron full load and draws a current of 6.5Aat 0.94pf lagging. parameters are Ra= 6.15 , Xa= 9.4 , Rs= 4.55 , and Xs= 3.2 . Calculate (a) the induced emf of the motor, (b) the shaft torque,and (c) the efficiency if the rotational loss is 65W. 2. If the motor of Problem 1 draws a current of 12.81 A at 0.74 pf lagging when its load is increased, determine the operating speed of themotor. Assume that the motor is operating in the linearregion. 3. Resolve Problem 1 if the universal motor supplied by 240-V dc andruns at the same load which draws the same current of6.5A. 4. Resolve Problem 2 if the universal motor supplied by 240-V dc andruns at the same load which draws the same current of12.81A. 21