Forces on the Rotor Bars of a Three Phase Induction Motor

A Research Essay by

Howard W Penrose, Ph.D., CMRP President, SUCCESS by DESIGN

http://www.motordoc.net howard@motordoc.net

The purpose of this paper and attachments is for research/education related to a number of documents discussing torsional stresses in Induction Motor Rotors.

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Forces on the Rotor Bars of a Three Phase Induction Motor

Howard W Penrose, Ph.D., CMRP President, SUCCESS by DESIGN

Introduction In this paper, we shall discuss the issue of torsional forces on three phase induction motor rotor bars and the implications. This particular issue has received some level of discussion, historically, in textbooks and manuals since the first induction machines in the late 1880’s. With earlier machines, these forces were viewed with limited importance because of the excess of materials used in construction. With changes to the construction of machines, and the reduction of materials used, this subject has increased in importance for both the manufacture and analysis of machines. At the present time, there are two specific areas of thought on the matter. The first is the traditional forces directly on the rotor bars and the second is a proposed theory of forces acting on a magnetic plane between the rotor bars and rotor bar slot walls. Both of the arguments result in similar findings, but only one can describe the effects observed in failed machines. This can be visualized using the following concept, in relation to the fields:

Another well-known application of magnetism and electromagnetism is the electric motor. Motor action means that movement is produced through the interaction of two magnetic fields. In generator action, you will recall, a current is produced when a loop of wire is rotated within a magnetic field. In motor action, motion is produced when current flows through a conductor in a magnetic field. Motor action is illustrated in [Figure 1]. In [Figure 1], a loop of wire is within a strong magnetic field. A voltage with polarity as shown is applied to the loop producing a current that flows into the loop on the left and out of the loop on the right. Thus, the left portion of the loop has a counterclockwise magnetic field, while the right portion has a clockwise magnetic field. The interaction of these magnetic fields produces torque, or a twisting motion, on the loop. This interaction of magnetic fields is more easily explained and understood using the two dimensional view of [Figure 1]. Notice the relationship of the flux lines around the left portion of the loop to the flux lines of the fixed field. They are of the same magnetic polarity below the loop but of the opposite magnetic polarity above the loop. The flux lines aid one another below the loop producing a strong field, but they oppose one another above the the loop producing a weak field. The

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loop is forced upward on this side toward the weak field. In the right portion of the loop, the opposite occurs.1

Figure 1: Motor Action on Conductors

With this in mind, where are the forces acting? Basic Electromagnetism In 1820, it was discovered that currents in wires generate magnetic fields and that wires carrying current in a magnetic field have forces on them. Between 1820 and 1840, the effect of coiling conductors in order to increase the strength of the magnetic fields was determined. The ability of such a coil in a magnetic field to develop torque was also discovered, as was the ability to focus the fields with iron.

With the realization that electric currents make magnetic fields, people immediately suggested that, somehow or other, magnets might also make electric fields. Various experiments were tried. For example, two wires were placed parallel to each other and a current was passed through one of them in the hope of finding a current in the other. The thought was that the magnetic field might in some way drag the electrons along the second wire, giving some such law as “likes prefer to move alike.” With the largest available current and the most sensitive galvanometer to detect any current, the result was negative. Large

1 Walls and Johnstone, DC/AC Principles: Analysis and Troubleshooting, West Publishing Co., New York, 1992

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magnets next to wires also produced no observed effects. Finally, Faraday discovered in 1840 the essential feature that had been missed – that electric effects exist only where there is something changing. If one of a pair of wires has a changing current, a current is induced in the other, or if a magnet is moved near an electric circuit, there is a current. This is the induction effect discovered by Faraday… We can easily understand one feature of magnetic induction from what we already know, although it was not known in Faraday’s time. It comes from the v x B force on a moving charge that is proportional to its velocity in a magnetic field. Suppose we have a wire which passes near a magnet… and that we connect the ends of the wire to a galvanometer. If we move the wire across the end of the magnet the galvanometer pointer moves. The magnet produces some vertical magnetic field, and when we push the wire across the field, the electrons in the wire feel a sideways force – at right angles to the field and to the motion. The force pushes the electrons along the wire. But why does this move the galvanometer, which is so far from the force? Because when the electrons which feel the magnetic force try to move, they push – by electric repulsion – the electrons a little farther down the wire; they, in turn, repel the electrons a little farther on, and so on for a long distance.2

The above forces are defined as Electromotive Force (EMF), which is mathematically defined as qv x B (Magnetic Force). The forces can be felt, directly. For instance, in a generator, a magnet generates a current that flows through the conductors. As the current demand of the load increases, the amount of force necessary to turn the shaft of the generator increases because a greater number of electrons pass a specific point per second (1 Amp = 6.28x1018 electrons/second, or 1 Coulomb/second)3 and there is more forces back on the magnetic field as it becomes harder to ‘push’ the electrons through the circuit. This is often demonstrated in physics classes when the leads of a small hand-cranked generator are shorted together, the student has a difficult time turning the shaft. This force is referred to as the Counter-Electromotive Force (CEMF) or back EMF. The forces on a conductor can be demonstrated, as well, using simple laboratory devices.

You have probably seen the dramatic demonstration of Lenz’s rule made with the gadget shown in [Figure 2]. It is an electromagnet… An aluminum ring is placed on the end of the magnet. When the coil is connected to an alternating-current generator by closing the switch, the ring flies into the air. The force comes, of course, from the induced currents in the ring. The fact that the ring flies away shows that the currents in it oppose the change of the field through it.

2 Feynman, et.al, The Feynman Lectures on Physics Volume 2: Mainly Electromagnetism and Motion, Addison-Wesley Publishing Company, Reading, MA, 1964 3 Penrose, Howard, Electrical Motor Diagnostics, SUCCESS by DESIGN, Connecticut, 2001.

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When the magnet is making a north pole at its top, the induced current in the ring is making a downward-point north pole. The ring and the coil are repelled just like two magnets with like poles opposite. If a thin radial cut is made in the ring, the force disappears, showing that it does indeed come from currents in the ring.4

Figure 2: Electromagnetic Forces on a Conductor

Another important factor is that electromagnetic fields will not penetrate a conductor, or conductive plate.

An interesting effect, similar in origin, occurs when a sheet of a perfect conductor. In a ‘perfect conductor’ there is no resistance whatsoever to the current. So if currents are generated in it, they can keep on going forever. In fact, the slightest EMF would generate an arbitrarily large current – which means that there can be no EMF’s at all. Any attempt to make a magnetic flux go through such a sheet generates currents that create opposite B fields – all with infinitesimal EMF’s, so with no flux entering.5

Now, if we set up an electromagnet with a flat plate of conductive material on a pendulum, as shown in Figure 3, and allow it to swing, it will continue until the electromagnet is energized. As soon as the pendulum comes to the center of the electromagnet, the eddy current forces (circulating currents) will cause it to come to an abrupt stop.

4 Feynman, et.al 5 Feynman, et.al

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If you now cut several slots in the conductive material, it will severely reduce the amount of eddy currents, allowing the conductor to swing through the field with reduced forces on it.

Figure 3: Pendulum

Elementary Three Phase Induction Machine The three phase induction motor is the most common types of electric machine in use today. We are going to start by covering basic principle, then general assembly of the machine.

A field just like that of a rotating magnet can be made with an arrangement of coils… We take a torus of iron (that is, a ring of iron like a donut) and wind six coils on it… Continuing the process [of energizing pole pairs], we get a sequence of fields shown in the rest of the figure. If the process is done smoothly, we have a ‘rotating’ magnetic field. We can easily get the required sequence of currents by connecting the coils to a three-phase power line, which provides just such a sequence of currents. “Three-phase power” is made in a generator using the principle…, except that there are three loops fastened together on the same shaft in a symmetrical way – that is, with an angle of 120o from one loop to the next. When the coils are rotated as a unit, the emf is a maximum in one, then in the next, and so on in a regular sequence. There are many practical advantages of three-phase power. One of them is the possibility of making a rotating magnetic field. The torque produced on a conductor by such a rotating field is easily shown by standing a metal ring on an insulating table just above the torus, as shown in [Figure 4]. The rotating field causes the ring to spin about a vertical

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axis. The basic elements seen here are quite the same as those at play in a large commercial three-phase induction motor.6

Figure 4: Rotating Field Torque on a Conducting Ring

In a proper electric machine, instead of the coils mounted in line with the toroid, the coils are mounted such that the coils face in towards the bore of the stator core where the rotor will be mounted. “An induction motor operates on the basis of interaction of induced rotor currents and the air-gap field. If the rotor is allowed to run under the torque developed by this interaction, the machine will operate as a motor.”7

Furthermore, because the current in each phase of the three-phase supply attains its respective maximum value at a different instant of time, the centerline of magnetic flux shifts from C to B to A, assuming this to be the phase sequence of the applied voltage. The shifting magnetic field has the same effect on the squirrel cage rotor as that produced by a magnet sweeping around the rotor, as shown in [Figure 5]. The rotating magnetic field set up by the three-phase current in the stator passes through the many windows formed by the bars in the squirrel cage rotor. This behavior is simulated in [Figure 5], where the moving magnets represent the ‘rotating poles’ set up in the stator and representative window of the stationary squirrel cage rotor is in the process of being swept by the clockwise rotating of these ‘rotating poles’ (stator flux). At the instant shown, there are three magnetic

6 Feynman, et.al. 7 Nasar, Syed, Theory and Problems of Electric Machines and Electromechanics, McGraw-Hill, New York, 1981

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lines directed downward through the window, and one line directed upward, for a net flux of two lines in the downward direction. As the flux rotation proceeds, the net downward flux through the window is reduced to zero and then increases in the upward direction. For the instant shown, the net flux is in the downward direction but becoming less and less. The decreasing flux through the window induces a current in the associated squirrel cage bars in a direction to delay the change in flux (Lenz’s law). The result, as shown in [Figure 5], is a counter-clockwise induced current that produces an additional downward contribution of flux.

Figure 5: Rotating Fields and Rotor

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The interaction of the magnetic field caused by the induced current in the squirrel-cage bars with the magnetic field of the stator produces a mechanical force on the bars. This force, for the instant shown in [Figure 5], causes the rotor to experience a torque in the direction of the stator flux.8

Methods of Evaluating Torque on Rotor Bars For the purposes of understanding the differences between the two concepts of forces on the rotor bars versus on the plane between the rotor bar and slot wall, a chapter from Syed Nasar’s Theory and Problems of Electric Machines and Electromechanics (Electric Machines) and from Syed Nasar and Ion Bodea’s book, The Induction Machine Handbook (Induction Machines), are included as appendixes. In the Electric Machines book, Dr. Nasar discusses the traditional method of evaluating the operation of electric machines through the interaction of the rotating magnetic field and rotor bar currents. As described on page 87, “Equation (5.4) describes the rotating magnetic field produced by the stator of the induction motor. This field cuts the rotor conductors, and thereby voltages are induced in these conductors. The induced voltage give rise to rotor currents, which interact with the air-gap field to produce a torque, which is maintained as long as the rotating field and induced rotor currents exist. Consequently, the motor starts rotating at a speed n<ns in the direction of the rotating field. (See Problem 5.26).”9 The chapter then covers the equivalent circuit circuit of induction machines, rotor currents and the interaction of the rotating fields with those currents. The Induction Machine book provides the only book-published theory on moving the forces on the rotor bar to forces on a plane between the rotor bar and slot walls. The chapter starts on P. 15, “In both cases the winding arrangement on the part of the machine – the primary – connected to the grid (the stator in general) should produce a traveling field in the machine airgap. This traveling field will induce voltages in conductors on the part of the machine not connected to the grid (the rotor, or the mover in general), - the secondary. If the windings on the secondary (rotor) are closed, a.c. currents occur in the rotor. The interaction between the primary field and secondary currents produces torques from zero to rotor speed onward. The rotor speed at which the rotor currents are zero is called the ideal no-load (or synchronous) speed. The rotor windings may be multiphase (wound rotors) or made of bars short-circuited by end rings (cage windings).”10 In the same chapter, the Induction Machine book then states that “The electromagnetic traveling field produced by the stator currents exists in the airgap and crosses the rotor teeth to embrace the rotor winding (rotor cage). Only a small fraction of it radially traverses the top of the rotor slot which contains conductor material. It is thus evident that, with rotor and stator conductors in slots, there are no main forces experienced by the conductors themselves. Therefore, the method of forces experienced by conductors in fields does not apply directly to rotary Ims with conductors in slots. The current occurs

8 Hubert, Charles, Preventive Maintenance of Electrical Equipment, McGraw-Hill, New York, 1955 9 Nasar, Syed 10 Boldea, Ion and Nasar, Syed, The Induction Machine Handbook, CRC Press, Boca Raton, FL, 2002

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in the rotor cage (in slots) because the magnetic traveling flux produced by the stator in any rotor cage loop varies in time even at zero speed. If the cage rotor rotates at speed n (in rps), the stator-producing traveling flux density in the airgap moves with respect to the rotor with the relative speed.”11 “According to the Maxwell stress tensor theory, at the surface border between mediums with different magnetic fields and permeabilities, the magnetic field produces forces. The interaction force component perpendicular to the rotor slot wall is…”12 The chapter then goes on to discuss how the forces are of a positive value against one wall and a negative value on the other (Figure 6) and that the total value of the two forces acts on the walls of the rotor slots.

Figure 6: Induction Machine Torque Theory

The book states: “Despite this reality, the principle of IM is traditionally explained by forces on currents in a magnetic field.”13 However, the book immediately states that it would be correct to evaluate the forces on the rotor by moving the position of the rotor bars, mathematically: “It may be demonstrated that, mathematically, it is correct to ‘move’ the rotor currents from rotor slots, eliminate the slots and place them in an infinitely thin conductor sheet on the rotor surface to replace the actual slot-cage rotor configuration. This way the tangential force will be exerted directly on the ‘rotor’ conductors. Let us use this concept to explain further the operation modes of IM.”14 The remainder of the chapter goes on to discuss the operation of an induction machine in terms of torque on the rotor. 11 Boldea, Ion and Nasar, Syed 12 Boldea, Ion and Nasar, Syed 13 Boldea, Ion and Nasar, Syed 14 Boldea, Ion and Nasar, Syed

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In the motor design book, Electric Machines: Steady-State Theory and Dynamic Performance, Dr. Mulukutla Sarma states:

In our discussion that followed Equation 5.2.33, the third possible method of producing constant torque was to cause the mmf axis of stator and rotor to rotate at such speeds relative to their windings that they remain stationary with respect to each other. If the stator and rotor windings are polyphase and carry polyphase ac, then both the stator-mmf and rotor-mmf axes may be caused to rotate relative to their windings. Such a machine will have polyphase stator ac excitation at ws, polyphase rotor ac excitation at wr, and the rotor speed wm satisfy Equation 5.3.17. Let us consider the rotor speed wm = ws – wr and the same phase sequence of sources. A rotating magnetic field of constant amplitude, rotating at ws rad/s relative to the stator, is produced because of polyphase stator excitation. A rotating magnetic field of constant amplitude, rotating at wr rad/s relative to the rotor, is also produced because of polyphase rotor excitation. The speed of rotation of the rotor magnetic field relative to the stator is (wm + wr) or ws if the rotor is rotating with a positive speed of rotation wm in the direction of rotating fields. If so, the condition for energy conversion at constant torque is satisfied. Such a situation is diagrammatically illustrated in Figure 5.3.15 [Figure 7]. The machine under these conditions is operating as a double-fed polyphase machine. Normally, in an induction machine with polyphase stator and rotor windings, only a source to excite the stator is employed, and the rotor excitation at the appropriate frequency is induced from the stator winding. The device is thus known as an induction machine.”15

Figure 7: Figure 5.3.15 from Dr. Sarma’s Book

15 Sarma, Mulukutla, Electric Machines: Steady-State Theory and Dynamic Performance, PWS Publishing Co., Boston, MA, 1996

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The traditional method of determining electromagnetic induced torque on the rotor bars is also important from the aspect of rotor heating, per Richard Nailen’s book, Managing Motors:

In most motors, rotor bar temperature is increased by another factor: the deep bar effect. The rotor bar currents during acceleration are at a fairly high frequency, because the rotor ‘sees’ the relative speed of the stator magnetic field compared to physical rpm of the rotor. When that actual rpm is low, rotor frequency is high and at locked rotor will be the full line frequency. Under this condition, the effective current-carrying depth of the rotor bar becomes only about 3/8-inch for copper and half-an-inch for aluminum or brass, no matter how deep the bars actually are. Thus, the apparent ac resistance of the bars becomes much higher at low speed than at full speed, raising the I2R loss and the bar temperature just that much more. If hot enough, the bars lose strength, so that they cannot resist even slight bending force from the shorting rings for more than a fraction of normal motor life. If heated still further, the bars may actually melt.16

Figure 8: Laminated Cores

The Important Difference Between Methods (Conclusion) The traditional method of calculating torque based upon stator magnetic fields and rotor current provides a more accurate method of calculating the operation of the electric 16 Nailen, Richard, Managing Motors, Barks Publications, Inc., Chicago, IL, 1996.

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machine and allows for a more accurate understanding of machine failure analysis. For instance, in motor heating, it assists in the explanation of the failure of electric motor rotor bars due to too many starts of the machine. Additionally, the necessity for the theory to still have to use the traditional method of evaluating forces on the rotor bar brings the theory, itself into question. In other cases where proposed concepts of calculating torque directly on the teeth, or core, requires that the basic principles and laws of electromagnetism, discussed on page 4 and 5 of this document be ignored. As shown in Figure 8, the stator has laminations, the rotor core is constructed in the same way. In order for the forces to work directly in the rotor teeth, the rotor core teeth would have to be solid, not laminated. This is one of the reasons why the theory in Boldea and Nasar’s book has to discuss these forces in terms of a perpendicular magnetic plane and not the rotor teeth. In this way, the ‘magnetic plane’ acts in a similar way as shown on p. 1 and 2 of this document. However, the description still shows that there must be torque produced as a direct result of the rotor bars and, therefore, by the most basic of Faraday’s laws, torsional forces acting on the conductors. Therefore, it is the conclusion of this paper that it is still necessary to accept traditional methods of calculating torque on the rotor bars of the electric machine in order to describe real-world applications as the rotor laminations are seperated from each other to reduce eddy currents. Bibliography Sarma, Mulukutla, Electric Machines: Steady-State Theory and Dynamic Performance, PWS Publishing Company, Boston, MA, 1996 Feynman, et.al., The Feynman Lectures on Physics: Mainly Electromagnetism and Motion, Addison-Wesley Publishing Co, Reading, MA, 1964 Nasar, Syed A., Theory and Problem of Electric Machines and Electromechanics, McGraw-Hill, Inc., New York, 1981 Nailen, Richard, Managing Motors, Barks Publications, Inc., Chicago, IL, 1996 Walls, Ron and Johnstone, Wes, DC/AC Principles: Analysis and Troubleshooting, West Publishing Company, New York, 1992 Penrose, Charles, The D’Este Steam Engineers’ Manual, Julian D’Este Company, Boston, MA, 1913 Hubert, Charles, Preventive Maintenance of Electrical Equipment, McGraw-Hill, New York, 1955 Boldea and Nasar, The Induction Machine Handbook, CRC Press, Boca Raton, FL, 2002

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Penrose, Howard, Motor Circuit Analysis, SUCCESS by DESIGN, 2001 About the Author Howard W Penrose, Ph.D., CMRP, is the President of SUCCESS by DESIGN, a reliability and maintenance services consultant and publisher, and the Founding Executive Director of the Institute of Electrical Motor Diagnostics, Inc. He has over 20 years in the energy, reliability and maintenance industries with experience from the shop floor to academia and manufacturing to military. Dr. Penrose’s personal clients include the UAW/GM Worldwide Facilities Group for all GM facilities energy and maintenance best practices, US Steel motor system maintenance and management programs, Amtrak electrical reliability and Reliability Center, Inc. as their electrical scene investigator and trainer. In the past, Dr. Penrose worked as an electric motor repair journeyman, field service manager, Adjunct Professor of Industrial Engineering at the University of Illinois at Chicago, Senior Research Engineer at the UIC Energy Resources Center, General Manager of an electrical motor diagnostics instrument vendor, Vice President of a US Military R&M consulting firm and President of an industrial R&M consulting firm. He has repaired, troubleshot, designed, installed or researched a great many technologies that have been, or will be, introduced into industry, including being involved in the reliability investigation of Left Ventricle Heart Device motors and design improvements in association with the University of Virginia. He has coordinated US Department of Energy and Utility projects including the industry-funded modifications of the US DOE’s MotorMaster Plus software and the development of the Pacific Gas & Electric Motor System Performance Analysis Tool project. Dr. Penrose’s UAW/GM WFG team won the 2006 Quality Network Planned Maintenance ‘People Make Quality Happen’ award for their work in improving the steam systems within General Motors facilities by adopting the US Department of Energy tools and best practices. Dr. Penrose is a past Chair of the Chicago Section of the Institute of Electrical and Electronics Engineers (IEEE), and presently leads several IEEE Standards Authority efforts and is a USA representative to ISO for industrial reliability standards. He is a member of the Vibration Institute, Electrical Manufacturing and Coil Winding Association, International Maintenance Institute, Society of Maintenance and Reliability Professionals, NETA and a Life Member of MENSA. He has numerous articles, books and professional papers published in a number of industrial topics and is a US Department of Energy MotorMaster Certified Professional as well as a trained vibration analyst, infrared analyst and motor diagnostics professional.

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Chapter

Polyphase lnduction Motors

5.T GENERAL REMARXS

The induction motor is probably thc most common of all nrotors. Like rhc dc machinc. aninduction motor €onsists of a stator and a rotor. the lalter mounled on bearings and separated fromth€ slator by an air gap. The srator core. made up of punchings (or tarninarions), ca.ries sloGembedded conductors. These conductors are in(erconnected in a rlredetermin€d fasbidl andconslitule the armature windings

All€rnating current is supplied to rhe slator windings. and the currenrs in rhe roror windings arcindrced by thc magnetic field of lhe srar()r currents. The rotor of rhe induction machine iscylindrical and carries eithcr (1) conducting bars short-circuited at both ends by conducting rings, asin ̂ cage-.ype machine (Fig. 5-l)i or (2) a polyphasc \rinding with rerminats broughr out lo stip ringsfor €xrernal conncctions, as in a vround-rotot mochine (Ftg.5-2). A wound-rotor winding is simitar rothat of lhe stator. Sometimcs thc cage-lype machinc is called a brushless nachine and rhewound-ro(or machine tetmed a slip-ing machine.

Flr .5- l Flt.5-2

An induction motor operales on the basis of intcraction of irduced rotor currents and the air-gapfreld. If the rotor is allowed to run under the lorque devetoped by this interaction, the machine willop€rale as a mok)r. On the othcr hand, thc robr may bc driv€n by an exremat agency bcyond asp€ed such rhat thc machine begins ro detivcr electric powerl it thcn operarcs as an rnctuctiongeneralor. Alnr){r ,nvaridbly. inducrion 'nachine\ a 'c u.ed.rs mrrr,rrs.

5.2 MMFS OF ARMATURE VYINDINGS

As in a dc machine, rhere arc ofren s€vcral independcnr sers of windings on rhe stakx ot aninduclion motor. For instan.e, ̂ thrce-phose winding is shown in Fig. 5-3. where each sror con.arnsiwo coil sides. Sucb a wioding is a doubte-tayer wtnding. Noticc atso frorn Fig. 5_3 that h js afour'pofe winding, bur ihat rhe pok pitch (of 9 teeth) js stightly grealer lhan the coit pitch (of 8lecth). Thus, it is a tdctional-pitch (or chorded, winding. Finally, observc that in rhis case thereare threc slols p€r pole per phase_ If the number of slors pcr pole p€r phase werc nonintegral, rhewinding woufd be koown as a lructional-slot windinp.

85

86 POLYPHASE INDUCTION MOTORS lcHAP. 5

""ii(

,(.r: O. phas Ar r, pha$ ar a, pn s .'rt. s3

(6 )

\ -\Y-L/Z- /'. - x-T--<_ -/

l t . '

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Fl8. $4

CHAP. 5l BOLYPHASF INDUCTION MMORS

Because the armature winding consists of interconnected coils, ir is advantageous to consider themmf of a single full-pitch €oil having N tums. From Fig. -5-4(d) ir is evident thar rhe machine haslvro poles; and from Afipere's circuilal law it follows that the mmf has the uniform value M (At)between rhe coil sides, as depicted in Fig- 5-4(b). Thus, the mmf per pote is Ni/2, as indicated in Figt4(c)- This also represents, to a difierent scale, rhe flux-density distriburion. The mmf distributionof Fig. 5-4(c) can be Fourier analyzed, and the fundamental comF)n€nt is given by

(r l)

87

*,r., ,y: 41141 ... "'

where r measures circumferenlial distance around the stator and where n rhe potc pilch (or coilpilch), is lhe circumfer€ntial dislance betweeo adjac€nt poles. lf t=I\tsin@r (A), (-tl)

g lx, t) = 0.s NI st)n .'t c " 4 1e9 (s.2')

t bcing the rms value of LTo eliminat€ the harmonics from the mmfs, thc armature winding is appropriatety distribured

over the stator periphery, as in Fig. 5-3. Thus, we may assome rhar rhc mmf produccd by eachphase (of a three-phasc winding, say) is sinusoidal in space. Io a rhrec-phase induction nrlchine lhcmmfs are displaced from each other by I20' (elecrrical) in space:

'*) (.r.3)

gc- g. sin l@t + t:o"1.,r" if Il* 12x.)\ 7 |

where g, is the amplitude ol each mmf. For lhc N,turn coil. considering only rhe fundamcnlal,

5.3 PRODUCTION OT ROTATING MAGNETTC F'IELDS

Adding the three mmfs of (-i3), we obrain rhe resuttanr mmf as

9(t . r)= ( iJ)

ll is seen that the mmf is a wave, of amplitudc L59",. rhat rravets circumferenrially at spced

q = y @ t , \

rela(ive ro the staror. We call r,, thc ryn.&rrnous Delocity. Notc thal lhe wavetengrh is

^ - 2 * ' ' - ' - '- ; = " t m t

If the machine has p poles, (5.5) rnay be rewriften in the form

,, =synchronous speed =ryf, w^t

ir^ = ir- stn dr cos

9" = 9- sin tut - l2U'\ cos I --2 -\ 7

L5,9. s inior 3 l\ 7 J

(s.6\

(s.7)

where t = @lzt is the staror cunent (and mmf-roralional) frequency.E{uation (tJ) describes rhe rotaling magncric fietd produced by the slaror of rhc induction

motor. This field cuts the rolor conduclors, and thereby votlages arc induced in these conduc-tols. The induced voltages give rise ro rotor currents. which interact with the air_saD ficld to

88 POI.YPHASE INDUCTION MOTORS ICHAP. 5

field and induced rotorio the direction of the

S.4 SLIP! MACHINE EQUIVALDNT CIRCUITS

The acrualspeed, r, of rhe rotor is often r€larcd to rhe synchronous spced, r,, via the sli:

produce a torque, which is maintained as long as the rotaling magneticcurrents €xist. Consequently, the motor starts rolaling at a speed n <n,rotatins field. (Se€ Problem 5.26.)

or thc percent slitr. lfils.Ai standstill (r = l), thc roraring magneric Retd prodrccd by rhe stator has the samc sp€ed wirh

rcspecl tolhc rotor windings as wilh respect to the sralor windings. Thus. rhe frequency of lhe rotorcrrrrcnrs, r. is the same as the frequency of the stator cunents, /r. Al synchronous speed (s = 0).there is no relative molion between the rotating field and the rolor, and the frequcncy of rororcurrent is zero. (lndeed, rhc rotor curent is zcro.) Ar intermediate specds the roror cuE-cnifrequency is proporrional to the sliD,

(58)

(5.e)wherefore t is known as the slip frcquency. Noling rhat thc rotor currents arc of slip frcquency. wchave the rotor equivalenl circui! (on a per-phasc basis) of Fig. S-s(a). which gives rh€ roror currcnr.

, - --"E::\,/Fl + (sxJ

Hcrc, E: is thc ioduced rolor emf al standstilli X, is thc rotor lcakage reactance per pnas€ atstandstill; and R? is the rotor resistancc per phase. This may also be written as

r ' = -�-�E'' f7p=- -, i l a l + x i

For (-trr) we redraw rhe circuit of Fig. 5-5(d) as Fig. 5-5(r).

(s.to)

llili'u

tIl ift$

Fl*.5-5

In order to include lhe siaror circuir, the induction molor may be viewed as a rransformer wIlh anair gap, having a va'iable resisrance in th€ sccondary 1*" 1.11011. Thus rhe prirnary of thclransformer corrcsponds to the slator of the inducrion moror, whcreas-thu secondary' corresponos rorlre rotor on a per-phase basis. Because of rhe arr gap, however, th! vatue of itre magnerangreactance. x-. tenG ro_b€ row as compared io that of a truc transformer. As in a rransfomer, wenave a muruat nux trnkrng borh rhc srator and roror. reprcsented by the magnerizing reacraoce andvarious leakage Ruxes. For insrancc. rhe rotat roror teakage itux is d-cnoted iy x2 tn r-ig.

5-5rt) Considerinq rhe rotor as being coupled to th€ stator as the secondary of a transformer is

L"]rtl'a . ri" ""-""", we mav draw tie ciriuit shown in Fig' 5-6 To develop this circuit further'

*. i."a i" *J** tft" .otor q;antities as referred to the slator' For lhis Purpose we must know the

transformation ralio, as in a transformer.

t*ng. t5

Thevoltagelransformal ionfat iointheinduct ionmoto'must inc|udetheef iectofthestatorandrotor windintdistributions. It can be sho$/n that' for a cage-lype rotor' lhe rolor resrstance per

-, -^' |L!jJ\'

cHA". t POLYPI]ASE INDUCTION MOTORS

phase, Rr, referred to the stator, is

Ri= a"h where

Here [.r =winding factor (see Problem 5.3) of the stator having N! series_connectedphas€

t,r=winding {actor of the rolor having Nl= p/4 series'connected tums pe' phase, for a

cage rotor, where P is the number of poles

''tt = number of Phases on the slalo(

m, =number of bars per pole Pair

R? = resistance of one bar

Sirnilarly,xi= a'x2

where Xi is (he rotot leakage reactance P€r phase' referred to the stalor'

(')

nt. S7

(s.t t)

turns per

(s.12)

90 POLYPHASE INDUCTION MOTORS

P"- Pt r iR,

This power is dissipated in lhe net resistance R:/s, whence

If wc subtra€t the rolor (standstill) resistive bss from P', we obtain

p t = p s , I i R i = { t s ) p .

ICHAP 5

Bearing in mind both the similarities and the difierences between an induction motor and a

trun"formci, we now refer the rotor quantities to the stator to obtain from Fig 5_6 lhe exacl

equivalent circuit (pet phase) shown in Fig 5-7(a) For reasons that will become immediatelv clear'

we split Rils as

&=n l+&n " t

to obtain the circuil shown in Fig. 5-7(t). Here, -R, is simply lhe per-phase s(andstill roror resistance

referred to lhe stalor and Ri(l - s)/s is a per-phase dvnamic .esistance that depends on thc rotor

specd and corresponds to the k)ad on the motor. Nolice that all the parameters shown in Fig 5-7

are standstill values.

5.5 CALCULATIONS FROM EQUIVALENT CIRCUTS

The major usefulness of an equivalcnl circuit of 6n induction motor is in the calcnlalion of itsperformance. All calculalions are made on a peFphase basis assuming a balanced oP€ralion ot'themachinet the tolal quantities arc then oblaincd by using lhe appropriate multip'ying factor.

Ft 5.t. PoPGr fow In .!| bducfon nolor.

Figurc 5-{t(a) is Fig.5-7(6) with R," omitted. (Corc losses, most of which ar€ in the slator. willbc included only in efrciency calculations.) In Fie. t8(b) we show apProximately the powcr flowand various power losses in one phasc of the machine. The Power crossing thc air 8ap' P'. is lhedifierence beN€en lhe input power' P = Vrlrcos ,r, and the slator resistive loss: that is

& = 1 : +

(-t 13)

(s.H)

thc dcveloped el€ctromagnctic

(5. 15)

cnAP. 5l POLYPHASE INDUCTTON MOTORS

This is the power that appears across the resistance Ri(1 - r)/s, which corresponds io the load. Therotational (mechanical) loss, P,, may be sublracted from Pd to oblain the shaft outpDt powcr,P". Thus

9l

(s. th)

and rhe efficiency, '1, is the ratio P/4.

5.6 APPNOXIMA'IT EQUVALENT C'TRCUIT PARAMETERS FROM TFST DATA

Sometimes the equivalent circuit ot the induction motor is app(oximaled by thc onc shown ioFig- 5-9. The parameters of the appro,(imate circuil can be obtained from the following two tests.

nr.5,

Ncl,ord T€61

In this test, rated voltage is applicd to rh€ mschine and it is allowed to run on no-load. Inputpowcr (corrcclcd for friction and windage loss), voltage, and currenl are measuredi thcse, reduced toper-phasc values, arc dcnoted by Ar y,r and ,o. re$pectively. When the machine runs on noload,lhe slip is close Io zero and the circuil in Fig. 5-9 to the righl of Ihe shunt branch is takcn to bc anopcn circuit. Thus. thc parametcrs R- and X- are found from

R. =ff (s.17\

(5.18\

(-t/e)

(5.20)

n.=n,*. '4.=ft

x.= x,* .'4=!EEErn (5.19\ and (5.20), the constanr a'� is rhe same as in (51l). The sraror resisrance per phase,

R,, can be directly measured, aod, knowing & from (-tt9), we €an determine R!=azR:, the rolorresistance refened to the stator. There is no simple method of deremining Xt ^nd Xt-o'�X2 *pararely. The total value given by (tU) is sometimes cqually divided between Xr and Xj.

x. f o- 'r

'"={#-frBLock.d-Rotor Tst

In this test, the rotor of the rnachinc is blocked (s = 1). and a reduced vollagc is applied to rhemachine so that the raled currenl flows through lhe stator windings. The input power, voltage. andcuneni are recorded and reduced to per-phase values; these are denoted. respcctively, by P", %, and1,. ln this test, the iron losses are assumed lo be negligible and the shunt branch of lhe circuit shownin Fig. 5-9 is consid€red to be absent. The paramelers are thus found from

92 POLYPHASE INDUCTION MOTORS lcHAP. 5

Solved pmblems

5.1. An N-turn winding is made up ot coits dishibuted in stots. as fte winding shown in Fig.5-3. The volrages induced in these coils are displaced from one anolher in piase try ttre stotangle a. The resultant voltage at the terminats of rhe N_turn winding js then the phasorsum of rhe coil votrages. Find an expression fot the distibution factot, kd. wherc

td _ _ maeniluds of -resulrant vottagesum or magn uctes ot Individuat coi l v, ! t lagcs

If,t p be Ih€ nunber of p(tesi e, rhc number of stors: and a, thc nudber of phases. Then (,, _q! , !*. s is rhe number or slors per pole per phas. The stor ansre " i, gi,* tt" "i*_*,

" = ' ! Q z = r mQ - q

The phasor addilion of volrages (for 4 = 3) is shown itr Fig. 5_t0. from rhe geonerry of which wc sc_l"=#=m=ffi (5.2 t)

which is the desired resutr.

rt. $ro n8.s-ll

fie voftage induccd in a fracrional-pirch coit is .educed by a fador knowtl as thc pikh lacto..j;;iiji,Tl"."o ro rhe vorrase induced in a rur-pirch c;ir. oa"" "" *p."__i""'r1,.,r,.

In a sinusoidalty disrributed nux densirv we show a.tu[_pirch and a tractional_pitch coit in Fig.5-ll The cojl sp€n of lhe fult_pirch coit is equat lo the pote pirch, r lf,r the coil sDan or rhchactionat-pirch coil be I < a as shown. n" n," ri,u,e ,r,. i. "ir.,r_0i",,'li,i*i,,"[i' o.r-,rionar ro ihe sh'ded a.ca in Fis. 5{r. whereas rhe fl* r;"rii-g ,r,"i"i;iJ i.i;";;;;i,;;"i1" ,""enrre arq undq rh€ cure. rhe pirch racro. is rhc-r"- ,r," -,i. Jr "'"i;;;;i;; ]il" ,",^,

u- f '","2to" I 1'. i"y* ="n# ls 22)Notice that in (t22), p and ? may be measured in any convenienr udir.

5.3. Calculate the distrib(ion factor (problem 5.1), rhe pircb factor (probtem 5.2), and thewinding factot, ft, - tdk , Ior the stator winding of Fig. 5_3.

\ , +p \

CHAP, 5] IOLYPTIASE INDUCTION MOTORS

From Fig.5,3, n=3, p=4, and O=j6. ,nrus,

J6 - r,{Pq ut,j,- t "-oru)-2tr

SubsiitutinS rhese in (t2I) yictds

o, #fi. 'u*Aho. Fig. 5-3 shows lhat ,:9 stots anrl B = S stors. Hence, from (-r22).

r" sn li - sin x." = {,.e8s

,(" = kd*p = (0.96X1.91i5)= t.94s

A 4-pole, 3-phase induction motor is en€rcondition for which rhe srip is 0.03. o",'-'-':o-l':'1."*H'

"npplv and is running at a load

j::1"-""i* '" i;;i.t;;; ;i# ;;:Hffi ,'il:::'""fr'.f Jil'.T#i l:,["":ff::rrame, in rpm; (d) speed of the rotor I.ot"ring .nu!""iii i"ii.;;;."-

'"-' iotarrng magneric field wi(h respeq b the stator

n' -t4[ ' tzora'tp

- --i- - lrun rpm

\a) n = (r _ r)r, = (1_0.03X1s00)= t746 rpmlb) /,= yr= (0.03)(({)) = t.ri H!(c) The p potcs on lbc srator irduce an equar number of potes on rhc .oior. Now, thu sanearsumcnt rhar tcd ro (r4) can bc aoDried_ro thc roror. Thus, ,,,"-.ri.,l-a""* j,-",rg

masnedc nerd wnosc speea, cbuoc tiihe ntor. s

t2|l1, t20st,* = p - - ; = * '

But rh. speed of rh€ rotor retarive ro therolor lietd with respcd ro ,h" ",uu,. i,

' "utot i' ' = (! -r)t' Thercfore lhc sPced of lhe

s " = n ' + n = n 'i.e.. in this cssc, t80O rpm.

(d) z*ro.

A 60-Hz induction molor has 2 potes and runs ar 3Sl0 rpm. Calcutate (a) rhe synchronousspeed and (r) rh€ percenr stip.

n, = I?14 = l2q6o) = 3600 .P-

, = &:r= i@9_{]! = sn25 : r r*

Using rhe rotor equivatent circuit of Fig. 5-5(r), show that anTIIT-,T'l_""'"c

ro,rque when its roror resisrance Gcgardedreaxage reaclance. Alt quantiti€s are on a per-phase basis.From Fig. 5-5(,), rbe devetoped power, pd, is given by

93

s.4.

5.5.

(a )

(b)

induclion motor will have aas variabl€) is equal to its

1

5.6.

p": f i& Bn,: r-- (1)

POLYPHASE INDUCTION MOTORS

and the rotor.urent L is such ihat

"= Ej' " lRJsf + X1

Also, lhe mechanical af,gular velocjty is

d- = (l s),,!

where o, h the synchronous angular velocity. Thesc ibree equalions gi!e:

- Eis R'�d R; + s:x :

For a maximun 'L we musl havc dT./rR? = 0. which leads Io

R i + s ' x 1 2 R 1 = o o r R r = r x :

Al slafling. s = l, this becomes Rr = &.

ICHAP 5

12)

(J)

14)

Using only the rolor cir€uil (as in Problem 5.6), calculat€ th€ k)rqoc dcvcbped per phdsc bva 6-pole, 60-Hz, 3-phase induction motof at a slip of 5"/., if the nl(nor dcvelops a maximump€r-phase torquc T.1 = 300 N .m while running at 7tn rpm. Thc rotor leakaSe rcaclanccis 3.0 O per phasc.

120(i5{D .^_...

.. = E!!L- 7{,=,, ..' l2r

Fron (4) of Problem 5.6,

r . r x i . r " x l 2 (s ' t s )r: - i ' RT-;FI t; {, ̂ T

where ee have us€d R, - r 'x C, 'n\(qtrcnr l) .

- 2(7), . , - ; j ( : l (xr) ,?4 N m

The rolor of a 3-phase,6)-Hz. -polc induclion mot()r takes 120 kW at 3 Hz. Dclcrminc (d)the rotor speed and (b) the rotor coppcr bsscs.

r - ( l rh, . . ( l 0.05xl l {n) = l7l0 rpm

roror coppcr ross = r x (rotor inpur) = (ll.{)5)(120)= 6 kw

The molor of Problem 5.ll has a stator coppcr loss of 3 kW, a mechanical loss of 2 kW. and astator core ioss of l.?kW. Calculate (d) thc molor output at the shaft and (r) theelficiency. Neglect rotor core loss.

Frcm Problem 5.8, the rotor input is 120 kW and the rotor copper bss is 6 kW.

m o r o r o u t p u t = l 2 ( , - 6 - 2 = l 1 2 k W

'not( ) . inpul = l2 l ,+: l+ 1.7 = 124.7 kW

outout l l2 ^^ -^ .

", - tY' - tt\*t = 'r, '.n

(r) By (t t5).

(a )

(b )

OrAP. 5l POLYPTIASE INDUCTION MOTORS

5.10- A Gpole, 3-phase, 60-Hz induction motor takes 48 kW in power at 1140 rPm- The statorcopper loss is l.4kW, stalor core loss is 1.6kW, and rotor mechanical losses areI kW. Find the motor efrciency.

95

e

* = Y ' - t r \ f ; i = u n , n * - n D ( n l t 4 4 ^ ^ -1200

roror i rput = slaro! oorpdt = (stator input) (stator loss)=48-(1.4+1.6)=45kW

.otor output = (l r)x (rctor input)= (1 0.05X45)= 42.75 twmoior output = (rotor ourpul) _ (rolational losses) - 42.75 | = 4l .75 kv'l

motorcff ic icncy=ff=8?%

A slip-ring induction motor, having a synchronous specd of 1800 rpm, runs al n =

1710 rpm when the rotor resistance per phase is 0.2 O. The molo. is required to develop aconstan( torque down to a speed of n* = 1440 tpm. Using the rotor circuil of Fig. 5-5(b),explain how this goal may be accomplished. The rotor leakage reactance al standslill is 2 o

Fom (4) of Probl€m 5.6, we may wrire

t =, *i*?*vherc k = Eilu, h a positilc consta l. ll h €asy to verify that AT.las ̂ nd AT,lARa arc always olopposire si8ns. Thus, if ]i is to stay fixed as r increases (i.e. th€ spccd decreases), R, musl also@nrinuously incrcasc, attaining its m (imum value al r*. We then hav€ lhe quadralic equalion

s R : r ' R ,R|.r-;ry1= R;:- r': i l

for rhis maximum vatuc, Ri. Substituting lhc numerical daia

I

, I

tgxj- t7l0 - . --' t*00R 2 = o . z A

. . , l l l 0 0 1 , 1 4 0 _ 0 ,"

I II{J{J

X t = 2 { l

and solving, wc obtain Ri=0.tt(). Thus, a conlinuously variable exlernal resislor, of maximumresistance 0.8-0.2=0.6O, lnusl be in*rted in thc roktr c ircui t .

5.12. The synchronous speed of an induction motor is 90{} rpm. Uodcr a blocked-rotor condition,the input powcr to (he motor is 45 kW at 193.6 A. The stalor resistancc per phasc is 0.2 Oand the iransformation ratio is a =2. Calculate (d) th€ ohmic value of the rotor resisianceper phase and (6) the tnotor slarting lorque- The stator and rotor are wye-connected.

(a) From ('le),

R ' + a ' R , . ,

whence Rr = 0.05 O.

(r) Referred Io the stator, rhe.olor resistanc€ pcr phase is Ri-a'R,=0.2O. fhen

"r*ring r..qu" = lI4= 111:l!I(q4)= 218.6 N. m

5.13. A lphase induction molor has th€ per-phase circuit parameleB shown in Fig. 5-12. Atwhat slip will the developed power be maximum?

u.r*0..=$#ff

I

98 POLYPHASE INDUCTTON MOTORS

xD x^;------4Al-+ - - _ / Y Y \ _ _

<8i

)

Fk. rtf,

5.t7. Compute ihe srarting current and starring lorque of the moior of p.obtem 5.t6.Us rhe comprere cncuit or Fi{. 5-r-3-t wirl !*e

probtem 5.16(b) aDd probr€m 5.1412 2 s . 3 v . R ^ = o . t e a . x h = 0 . 4 s t r . x i = o z a . n : 1 , = n : = 0 . i i r . - - 1 , ; ; , ' . ' . .

a = roi, - n-r#?*o; iify" - 2et) 2 A

The startinS torque is givcfl by the erprcssion dcvetopcd in pmblcm 5.12(r):

r=,=ffi= rz.s r.,s.18.

5.I9.

1:-r--,,1" 1,"" of probtem i.t4, using the comptete circuir of Fig. 5-t3, catcutarc (d) r^,wcrcrossing lhc air gap, (D) dcveloped power.

moror. compare wirh the corresponding re(scJr,l",jiii,tril,i:;1,11 (d) ourput k)rquc or rhe

From Fig. 5,13. wnh r = r/40,

2. = 2,, r ! * 6 5 = o.tv + iu.4e + 4 + jo.2

wh.nce Z, = 4.U6tl. Moreovcr, as computed in probtcm 5.16(r). %l=225.3V. Then.

n-!=f f i=sz.ue(a\ 4 = 3rj,& = 3(sj.o6)?)= 31.?g4 kw

P, = O - s)4 = (0.975X33.7s,4) = 32.93q kwP" = A _ (81n W) = 32. tlr) kw

* n 32 110r"=- = r;ffiaj_ 174.e N.^The results are in cxc€lenr agreemenr with rh.rs€ obrained in probtcm 5.14.

(b)

(c)

(d)

:^,""r9"- 119-r-.1t"" motoris usualty staned by apptying a reduccd votrase across rhe nurorisucb a voltage may be obtained from an autor.ansformer. rr a motoi is a U" "i"*" *5070 of full-load rorque and if the fulr-vokage srarnng currenr is 5 times the fu -road currcnr.d€tcrmine the percent reduclion in rhe apptied voltage (hat i", ,r," p"."*i ,p'., ".autorransformer). Thc fu -toad slip is 4%.

r = I:4j

(Ii)"* = roror cunenr at srarr if tul vokasc is apprred(Ij)s = roror curenr ai srari if rcduced volrage is apptied

k =.ario of reduced vottage to fufl voliage

lcHAP. 5

( r )

cnAP. 5l 99FOLYPTIASE INDUCTION MOTORS

At a given dip-in particular,ar s: I rotor current nay be considered proportional

(r')s ,.0t;-^

(li)s =,c(ri)* - &5lin

Applyine (/) at reduced-vdtage stai and al full-load, and substiluting (2), ee obtain

t2)

r, - (ks!b\' LlT- \ rt"" | |

fron *bi.h k -t).T17 = 70.716.

j - rxr'xo.oo

5.2), A 3-phase, 400-V, wye-connected inducljoo .nolor takes lh€ fulfload current at 45 V wilhthe rotor blocked. Th€ full-load slip is 4ol. Calculale the lappings r( on a 3'phas€autotransformer to limil lhe starting current to 4 tim€s the full-load current For such alimitarion, determine the ralio of thc starting torque to full_bad torque.

By th€ (apprcximale) currcnl'vol(aSe proPorlionalily,

,r ?l{loIF 45

bbclcd{oror current, IE is lhc lull-load curent. and ts is the slaninS

o='#ft=#n=o'No*, from (l) ol Prchlem 5.1'r,

.t

a n d f = k*here 1, is lhc full-voltagecuref,i. But il is Siven thal

,tt

hI

i

I-! = (r.A)'- = t4)2tr.t4) = u M

5.2I. Thc motor of Problem 5.an employs rhe wye"delta rtalet shown in Fig. 5-l4i thal is, lhephases are €onnected in wye at the time of starting and are swilched to delia when the motoris running. The full-load slip is 4% and the motor draws approximately 9 times the full-loadcurrent if started dirccdy from the mains. Determine thc ralio of starling lorque to full-load

It. tla. Sda.n r o.r W cl'|GDod .o wt t'd s*hch..m D cortt.gond to alE d.lL c..tFfoc

100 POLYPHASE INDUCTION MOTORS lcHAP. _5

When de ph.ss are swirched to defta_ rh

;ffi "':;:ji,,"",'t#if*HT"""":f t1;ff "ifi :".i'.:,;;x..:H"::il:'ii:".ii:t:J:

f :(;.; ',0*,=,*

&

rt,$r5Fk. r-16

t" l:-r-ol']1

a hiSh €rarring rorque in .a

cage-rvpe moror, a doubre-cage roror rs used. Thelorms ot a slot and of the bars of thrtigrt". ."ri.r"n"" th"n il-"-i_n

"- !'e-two^ca8es are shown in Fig S_15 The outer cage has a

11y, 1,1,;1 ;{ei ri,'"1'";,'{' ;1,: J,*1,:.l:!::,.ffi""#jil';.;ill;*;li T;[XH:;equlvatent circuit for such a rotor i*" t "u" rt " p"r-pr,""

-"liu- '-"" 's SNen in Fig 5'16 Suppose thit' for a cerlain motor,

& = 0 . t O R " = 1 . 2 f l X t = 2 A X , = 1 r lDet€rmine (he.ratio of the torques provided by the two cages at (a) star tii| and (b)2y" slip.(4) From Fis. 5-t6. ar r = I,

zi = (0 ]Y + Q)' = 4.ot a1z i= (1 .2 ) ,+ ( t f=244a"

power i.pur ro ih€ inn€r cagea pu = r?R = 0.tfpoee. i.pur lo lh€ oure. cage- pb = IZRo= 1.211"

tHF*i*#:"**. = + = * = H (+l = "i er='#(#)=,*(r) Simitarty, ar s = 0.02.

zt=( f f i | *py=ua,

z1=ff i '+ey,=No1s,

c= n: (_i.f = rn.}

5,23. At standstil, the impedanc€s of the rnner and outer cages of an induction mok,r areZ , = 0 . O 2 + j 2 A Z , = O . 2 + j I d l

(per-phase vatues). Ar what stiD"q*ir u* t " "i*,i,;*;: r:li.

will the torques contribured bv tbe two caees

{,

CI{AP. 5] POLYPI{ASE INDUCTION MOTORS

l.€t r be the r€quired slip. Then,

10 t

_ - ,R, - - , R"

T /L\ ) Rt tZ- \1 R,T" \t"J R., \Z) R.,

s-24- At a slip of 3olo, for the Sphase motor of Problem 5.23, the rotor input phas€ voltage is45V. Calculate (a) lhe motor line curent and (b) the torques contribuled by the twocages. The motor has 4 poles and operates at 60 Hz.

b = 0 . 6 7 + j 2 A

which are in parallel to form Z. Hence,

Z - = 6 . 6 7 + j l A

' = # f f i l k = ( ) e s + i r ' s = I q l q o

(a) n=l=f i=zse(r) n, = 1?!({0 = 1666,0-

rnrur ro,cu. = f - i66ffffi = 3r5 N..

From Problem 5.23 we conclude that al s = 0.(R eith€r ca8e conlributes 315/2 = 157.5 N m.

s.E.

s.Tt, A 3-phase, dislributed armaiurc winding has 12 poles and 180 slols. Thc coil pitch i$ 14 slots Calcutate (a) the distribution facto., (r) tbe pitch factor, (.) thc winding fador.Ans. (a) 0.95t; (r) o-95: (.) o.9l

A l-Dhase. 60-Hz inducrion moro. has 8 poler and ope.ate *ith a slip of 0.05 fo. a cenainload. Compute (in rpm) tne (r) sFed of the rotor with resp€ct lo the slalo., (r) sPed of the rotorwith respect to the slato' hagrelic field, (.) speed of the rotor daAnelic 6eld wirh respect to the .otor,(d) sp€ed of ihe rotor magnelic 6eld with respect lo the stabr, (a) spoed of the rolo. field wiih respectto the staror feld. A" (a) 8s5 .pn; (r) a5 rpmi (.) 45 rpm: (l) q)0 rpni (e)

z:=(!) '+er=a",t{ '*e

zz= (w) '+6Y=!] ! ' ,1

to.ozff4J* r) = to.zr(Lf,-'l -.) r = 0.03

Supplementary Problems

verify rhe vafidily ot (a) (5.4r, (bl (5.7').

Explain why an inductiotr molor will not run (a) at ihe synchronous speed, (r) in a d;cction oppositero rhe rorati.g magnetic feld.

5.4.

( hrPttr 2

( o }s l l t ( l ( I l o \ \ sPu( l s \ND ( )P l iRA I l ( ) \P I { lN ( l P l , l l s

l h r L r h L ( l ! i n r r i r L l r r r \ ' r ' ) . r r ! r l l \ r i r i r l ) ( n l | l i n { r r L I r r r r ' ! ' f r r r ' L r ' L i l I

, i r r l | ' r ' l " r r " t i "

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