DHANALAKSHMI COLLEGE OF ENGINEERINGDEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING

EE2302 - ELECTRICAL MACHINES IIUNIT-I – SYNCHRONOUS GENERATOR

PART – B

Introduction

The induction machine was invented by NIKOLA TESLA in 1888. Right from its inception its ease of manufacture and its robustness have made it a very strong candidate for electromechanical energy conversion. It is available from fractional horsepower ratings to megawatt levels. It finds very wide usage in all various application areas. The induction machine is an AC electromechanical energy conversion device. The machine interfaces with the external world through two connections (ports) one mechanical and one electrical. The mechanical port is in the form of a rotating shaft and the electrical port is in the form of terminals where AC supply is connected. There are machines available to operate from three phase or single phase electrical input. Single phase machines are restricted to small power levels.

Alternating-current (ac) generators are commonly referred to as synchronous generators or alternators. A synchronous machine, whether it is a generator or a motor, operates at synchronous speed, that is, at the speed at which the magnetic field created by the field coils rotates. Expression for the synchronous speed N, in revolutions per minute (rpm) isNs = -12 0f PWhere f is the frequency in hertz (Hz) and P is the number of poles in the machine. Thus, for a 4-pole synchronous generator to generate power at 60 Hz, its speed of rotation must be 1800 rpm. On the other hand, a 4-pole synchronous motor operating from a 50-Hz source runs at only 1500 rpm. Any attempt to overload the synchronous motor may pull it out of synchronism and force it to stop. During our discussion of a direct-current (dc) generator we realized that the electromotive force (emf) induced in its armature coils is of the alternating type. Therefore, we can convert a dc generator to an ac generator by (a) replacing its commutator with a set of slip rings and (b) rotating the armature at a constant (synchronous) speed. The idea is novel but is not put into practice for the reasons we will mention shortly. We also recall that the relative motion of a conductor with respect to the Magnetic flux in a machine is responsible for the induced emf in that conductor. In other words, from the induced emf point of view it really does not matter whether the conductors (coils) rotate in a stationary magnetic field or a rotating magnetic field links a stationary conductor (coil). The former arrangement is preferred for dc generators, whereas the latter is more suitable for synchronous generators and is the topic of this chapter. Thus, the stationary member (stator) of a synchronous generator is commissioned as an armature, and the rotating member (rotor) carries the field winding to provide the required flux. There are numerous reasons for such an "inside-out'' construction of a synchronous generator, some of which are listed below.1. Most synchronous generators are built in much larger sizes than their dc counterparts. An increase in power capacity of a generator requires thicker conductors in its armature winding to carry high currents and to minimize copper losses. Deeper slots are therefore needed to house

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thicker conductors. Because the stator can be made large enough with fewer limitations, it inadvertently becomes the preferred member to house the armature conductors.2. Since the output of a synchronous generator is of the alternating type, the armature conductors in the stator can be directly connected to the transmission line. This eliminates the need for slip rings for ac power output.3. Since most of the heat is produced by the armature winding, an outer stationary member can be cooled more efficiently than an inner rotating member.4. Since the armature winding of a synchronous machine is more involved than the field winding, it is easier to construct it on the stationary member.5. Since the induced emf in the armature winding is quite high, it is easier to insulate it when it is wound inside the stationary member rather than the rotating member. A rigid frame also enables us to brace the armature winding more securely.6. The placement of a low-power field winding on the rotor presents no deterrent to the inside-out construction of a synchronous generator. The power to the field winding can be supplied via slip rings. If the field is supplied by permanent magnets, the slip rings can also be dispensed with.

Construction of a Synchronous Machine

The basic components of a synchronous machine are the stator, which houses the armature conductors, and a rotor, which provides the necessary field as outlined below.

Stator

The stator, also known as the armature, of a synchronous machine is made of thin laminations of highly permeable steel in order to reduce the core losses. The stator laminations are held together by a stator frame. The frame may be of cast iron or fabricated from mild steel plates. The frame is designed not to carry the flux but to provide mechanical support to the synchronous generator. The inside of the stator has a plurality of slots that are intended to accommodate thick armatureConductors (coils or windings). The armature conductors are symmetrically arranged to form a balanced poly phase winding. To this end, the number of slots per pole per phase must be an integer. The induced emf per phase in large synchronous generators is in kilovolts (kV) with a power handling capacity in megavolt amperes (MVA). The axial length of the stator core is comparatively short for slow-speed, large diameter generators. These generators have many poles and are left open on both ends for self-cooling. They are installed at locations where hydroelectric power generation is possible. The axial length of high-speed generators having 2 or 4 poles can be many times its diameter. These generators require forced air circulation for cooling and are totally enclosed. They are used when the rotors are driven by gas or steam turbines.

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Figure .1 A salient pole rotor.

Rotor

Two types of rotors are used in the design of synchronous generators, the cylindrical rotor and a salient-pole rotor. The rotor is rotated at the synchronous speed by a prime mover such as a steam turbine. The rotor has as many poles as the stator, and the rotor winding carries dc current so as to produce constant flux per pole. The field winding usually receives its power from a 115- or 230-V dc generator. The dc generator may be driven either by the same prime mover drivingthe synchronous generator or by a separate electric motor. The salient-pole rotor is used in low- and medium-speed generators because the windage loss is small at these speeds. It consists of an even set of outward projecting laminated poles. Each pole is dovetailed so that it fits into a wedge-shaped recess or is bolted onto a magnetic wheel called the spider. The field windingis placed around each pole, as indicated in Figure 7.1. The poles must alternate in polarity.

The cylindrical rotor is employed in a 2- or 4-pole, high-speed turbo-generator. It is made of a smooth solid forged steel cylinder with a number of slots on its outer periphery. These slots are designed to accommodate the field coils. The cylindrical construction offers the following benefits:1. It results in a quiet operation at high speed.2. It provides better balance than the salient-pole rotor.3. It reduces the windage loss.

Armature Windings

The stators (armatures) of most synchronous generators are wound with three distinct and independent windings to generate three-phase power. Each winding is said to represent one phase of a three-phase generator.

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The three windings are exactly alike in shape and form but are displaced from each other by exactly 120" electrical in order to ensure that the induced emfs in these windings are exactly 120" in time phase. The three-phase windings may be connected to form either a star (Y) or a delta (A) connection. If the windings are internally connected to form a Y connection, the neutral point is also brought out so that it can be properly grounded for safe operation.The double-layer winding is often used to wind the armature of a synchronous generator. As you may recall, a double-layer winding requires as many identical coils as there are slots in the stator. One side of each coil is placed at the bottom half of a slot, and the other side of the same coil fills the top half of another slot. In order to place the coils in this fashion, the coils must be pre-wound on the winding forms and then inserted into the slots. The number of coils per phase (or the number of slots per phase for a double layer winding) must be an integer. Since the coils must be distributed equally among the poles, the number of coils (slots) per pole per phase must also be an integer. In other words, if S is the number of slots in the armature, P is the number of poles, and q is the number of phases, then the number of coils per pole per phase is

n = s… pq

Where n must be an integer. The number of coils per pole per phase, n, is usually referred to as a phase group or phase belt. When the stator of a three-phase, 4-pole synchronous generator has 24 slots, the number of coils in each phase group is 2. There are 12 phase groups (poles x phases). All coils in a phase group (2 in this case) are connected in series.

Each coil in a phase group can be wound as a full-pitch coil. In other words, each coil in the armature can be made to span 180" electrical. Since the induced emfs in both sides of a full-pitch coil at any time are exactly in phase, theoretical yearning mandates the placement of full-pitch coils from the induced emf point of view. However, a full-pitch coil is rarely used. Instead, the generators are wound with fractional-pitch coils for the following reasons:

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1. A properly designed fractional-pitch coil reduces the distorting harmonics and produces a truer sinusoidal waveform.2. A fractional-pitch coil shortens the end connections of the windings and thereby not only saves copper but also reduces the copper loss in the coil.3. A shorter coil is easier to manage and reduces the end-turn build-up on both sides of the stator's stack. This slims down the overall length of the generator and minimizes the flux leakages.4. The elimination of high-frequency harmonics also cuts down the magnetic losses in the generator.The drawback of a fractional-pitch coil is that the induced emf in it is smaller than in a full-pitch coil. The reason is that the total flux linking the fractional-pitch coil is smaller than that of the full-pitch coil. The ratio of the flux linking the Fractional pitch coil to the flux that would link a full-pitch coil is called the pitch factor. Later, we will develop an equation to determine the pitch factor. To illustrate the placement of the phase windings in the slots of a synchronous generator, we make the following assumptions:

(a) All coils are identical.(b)Each coil is a fractional-pitch coil as long as a phase group contains more than one coil. All the coils in a phase group are connected in series.(c)Each phase group spans 180" electrical (one full pitch). Thus, the n coils in a phase group must be placed in such a way that the beginning end of the first coil is under the beginning of a pole and the finishing end of the nth coil is under the trailing end of the pole.

The electrical angle from the center of one slot to the center of an adjacent slot is known as the slot span or slot pitch. The coil span or coil pitch, the number of slots spanned by each coil, can be expressed in terms of either electrical degrees or the number of slots.

Pitch Factor

Owing to the spatial distribution of field windings on each pole of a cylindrical rotor, we can approximate the flux density emanating from the surface of a pole asB= Bm cosWhere B, is the maximum flux density per pole, The total flux linking a full-pitch coil is

p=∫B.ds

Whereds = Lrdm = 2Lrd p

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L is the axial length of the coil (rotor), r is radius of the rotor, and P is the number of poles. For a full-pitch coil, 8 varies from – π/2 to π/ 2. Evaluating the integral, we obtain 4LrBm

Qp = p

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Let us now assume that the coil span for a fractional-pitch coil is p. The flux linking the coil is maximum when the coil is symmetrically placed along the magnetic axis of the pole, as llustrated in Figure, Thus, the maximum flux linking the coil iscm=∫bm cos 2Lr d pp sin (ρ/2) = P kp Where kp = sin (ρ/2) is the pitch factor and kp< 1. If e is the induced emf in a full-pitch coil, then the induced emf in the fractional-pitch coil will be kpe,. We will use this fact later when we compute the induced emf in each phase group of a synchronous generator.

Distribution Factor

In order to make the induced emf approach a sinusoidal function, there is always more than one coil in a phase group. These coils are connected in series as depicted by the winding diagrams. Since the coils are displaced spatially from each other, the induced emfs in these coils are not in phase. If Ec is the induced emf in one coil and n is the number of coils in a phase group, the induced emf in the phase group Epg, is

Epg = kdnEc

Where kd is called the distribution factor and kd < 1.The distribution factor is unity only when all the coils are placed in the same slots. Since that defies the purpose of obtaining a sinusoidal waveform, the distribution factor is always less than unity. In order to verify Eq. (7.6), let us assume that there are n coils connected in series to form a phase group and the root-mean-square (rms) value of the induced emf in each coil is E,. Since the coils are distributed, the induced emf in each coil is out of phase with the next by the slot pitch y, as shown in Figure. Since the angle subtended by each phase-voltage phasor is y, the total angle for the phase group is ny. Therefore, we can write

1/2 Epg = L sin (nƳ)1/2 Ec = L sin (Ƴ/2)

From the above equations, we obtainEpg = sin (nƳ)/2 = n sin (nƳ)/2Ec sin (y/2) n sin (Ƴ/2)

Thus, the induced emf per phase group isEpg = n sin (nƳ)/2 Ec

n sin (Ƴ/2) =n KdEc

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Kd is the distribution factor

EMF equation of alternator.

Induced emf EquationLet us assume that the total flux emanating per pole of a round rotor revolving at an angular velocity of w, is QP. The maximum flux linking the fractional pitch coil is a, k, where kp = sin (ρ/2) is the pitch factor and ρ is the coil span in electrical degrees. As the flux revolves, the flux linking the coil at any time t can be expressed as

ϕc = ϕpkp cos ωtwhere ω = 2∏f is the angular frequency in rad/s.For a coil with Nc turns, the induced emf in the coil, from Faraday's law isEc = NcKpωϕp sinωtThe maximum value of the induced emf isEm = NcKpωϕp

and its rms value isEc =1/√2( Em)= 4.44f Nc ϕpkp

Since a phase group usually has more than one coil connected in series and each coil is displaced by a slot pitch, the induced emf in the phase groupEpg = nKdEc where n is the number of coils in a phase group and Kd is the distribution factor. For a given generator, the product KpKd is constant and is referred to as the winding factor. That is, the winding factor Kω isKω = KpKd

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The rms value of the induced emf in each phase group can be expressed in terms of the winding factor asEpg = 4.44 nNcKωfϕp

For a generator having P poles and a parallel paths, the induced emf per phase (phase voltage) is

Ea = P 4.44nNcKωfϕp

AThe factor PnNc/a in the above equation represents the actual number of turns per phase connected in series when there are a parallel paths. By taking into account the winding factor, k,, we can define the effective turns per phase as

Ne = PnNcKω A

Finally, we obtain an expression for the per-phase (no-load) voltage as

Ea = 4.44 fNeϕp

Note that above Equation is very similar to the one obtained for a transformer. In the case of a transformer, the effective number of turns is the same as the actual number of turns because each transformer winding consists of one coil that embraces the total flux in the magnetic core. The winding factor for a synchronous generator could also have been unity if (a) we used a full-pitch coil and (b) all the coils in a phase group were placed in the same slots.

The Equivalent Circuit

During our discussion of dc generators, we discerned that the terminal voltage of a dc generator is smaller than the generated voltage owing to (a) The voltage drop across its armature winding and (b) The decrease in the armature flux caused by the armature reaction. However, the terminal voltage of an ac generator depends upon the load and may be larger or smaller than the generated voltage. In fact, we aim to show that the terminal voltage may actually be higher than the generated voltage when the power factor (pf) is leading. For unity and lagging power factors, the terminal voltage is smaller than the generated voltage. Armature Resistance Voltage Drop

Let Ea be the per-phase generated voltage of a synchronous generator and Ia the per-phase current supplied by it to the load. If Ra is the per-phase resistance of the armature winding, then IaRa is the voltage drop across it. The IaRa voltage drop is in phase with the load current Ia. Since Ra also causes a power loss in the generator, it is kept as small as possible, especially for large machines.

Armature Leakage-Reactance Voltage Drop

The current Ia in the armature winding produces a flux. A part of the flux, the so-called leakage flux, links the armature winding only and gives rise to a leakage reactance Xa. The leakage

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reactance causes a voltage drop j I a X a , which leads Ia, by 90". The phasor diagrams depicting relationships between the per-phase generated voltage Ea the per-phase terminal voltage Va and the voltage drops Ia Ra, and jIa,Xa for three types of loads are shown in Figure.

Armature Reaction

The flux produced by the armature winding reacts with the flux set up by the poles on the rotor, causing the total flux to change. Such an interaction between the two fluxes is known as the armature reaction. To understand the effect of armature reaction on the terminal voltage of a synchronous generator, let us examine a sequence of events when the generator delivers a load at a unity power factor.(a) If p is the flux per pole in the generator under no load, then the generated voltage Ea must lag p by 90", as shown in Figure.(b) Since the power factor is unity, the phase current Ia, is in phase with the terminal phase voltage Va.

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(C) As the phase current fa passes through the armature winding, its magneto motive force (mmf) produces a flux ar which is in phase with Ia The effective flux per pole in the generator is the algebraic sum of the two fluxes; that is e = p + ar(d).The fluxar ,in turn, induces an emf Ear in the armature winding. Ear is called the armature reaction emf. The armature reaction emf Ear lags the

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Flux ar by 90 degree. Hence the effective generated voltage per-phase Ee is the algebraic sum of the no-load voltage Ear, and the armature reaction emf Ear. That is, Ee = Ea + Ear

The per-phase terminal voltage Va, is obtained by subtracting the voltage drops IaRa and jIaXa from Ee In other words,

Ee = Va +(IaRa + jIaXa)

From the phasor diagram, it should be obvious that the armature reaction has reduced the effective flux per pole when the power factor of the load is unity. Also, the terminal voltage is smaller than the generated voltage.

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By following the above sequence of events, we can obtain the phasor diagrams for the lagging and the leading power factors. From these figures it is evident that the resultant flux is (smaller/larger) with armature reaction for the (lagging/leading) power factor than without it. In addition, the terminal voltage Va, is (higher/lower) than the generated voltage Ea when the power factor is (leading/lagging). Since the flux per pole p is different for each of the three load conditions, the field current If must be adjusted each time the load is changed. Since the armature reaction emf Ear lags the current i, by 90°, we can also express it as

Ear = - jIaX m Where Xm a constant of proportionality. It is known as the magnetization reactance.

Both the magnetization reactance and the leakage reactance are present at the same time. It is rather difficult to separate one reactance from the other. For this reason, the two reactance are combined together and the sumXs = Xm + Xais called the synchronous reactance. The synchronous reactance is usually very large compared with the resistance of the armature winding. We can now define the synchronous impedance on a per-phase basis as

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Zs = Ra + jXs

The Equivalent Circuit and Phasor Diagrams The exact equivalent circuit of a synchronous generator on a per-phase basis embodying the synchronous reactance is given in Figure. The per-phase terminal voltage isVa = Ea - Ia(Ra + jXs) = E, - IaZsand the corresponding phasor diagrams for three types of loads are given in Figure.

Voltage Regulation:

When an alternator is subjected to a varying load, the voltage at the armature terminals varies to a certain extent, and the amount of this variation determines the regulation of the machine. When the alternator is loaded the terminal voltage decreases as the drops in the machine stars increasing and hence it will always be different than the induced emf. Voltage regulation of an alternator is defined as the change in terminal voltage from no load to full load expressed as a percentage of rated voltage when the load at a given power factor is removed without change in speed and excitation. Or the numerical value of the regulation is defined as the percentage rise in voltage when full load at the specified power-factor is switched off with speed and field current remaining unchanged expressed as a percentage of rated voltage.

Hence regulation can be expressed as% Regulation = (Eph – Vph / Vph ) x 100where Eph = induced emf /phase, Vph = rated terminal voltage/phase

Methods of finding Voltage Regulation: The voltage regulation of an alternator can be determined by different methods. In case of small generators it can be determined by direct loading whereas in case of large generators it cannot determined by direct loading but will be usually predetermined by different methods. Following are the different methods used for predetermination of regulation of alternators.1. Direct loading method2. EMF method or Synchronous impedance method3. MMF method or Ampere turns method4. ASA modified MMF method5. ZPF method or Potier triangle methodAll the above methods other than direct loading are valid for nonsalient pole machines only. As the alternators are manufactured in large capacity direct loading of alternators is not employed for determination of regulation. Other methods can be employed for predetermination of regulation.Hence the other methods of determination of regulations will be discussed in the following sections.

EMF method: This method is also known as synchronous impedance method. Here the magneticcircuit is assumed to be unsaturated. In this method the MMFs (fluxes) produced by rotor and stator are replaced by their equivalent emf, and hence called emf method. To predetermine the

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regulation by this method the following informations are to be determined. Armature resistance /phase of the alternator, open circuit and short circuit characteristics of the alternator.

OC & SC test on alternator:

Open Circuit Characteristic (O.C.C.)The open-circuit characteristic or magnetization curve is really the B-H curve of the completemagnetic circuit of the alternator. Indeed, in large turbo-alternators, where the air gap is relatively long, the curve shows a gradual bend. It is determined by inserting resistance in the field circuit and measuring corresponding value of terminal voltage and field current. Two voltmeters are connected across the armature terminals. The machine is run at rated speed and field current is increased gradually to If1 till armature voltage reaches rated value or even 25% more than the rated voltage. Figure illustrates a typical circuit for OC and SC test and figure illustrates OC and SC curve. The major portion of the exciting ampere-turns is required to force the flux across the air gap, the reluctance of which is assumed to be constant. A straight line called the air gap line can therefore be drawn as shown, dividing the excitation for any voltage into two portions, (a) that required to force the flux across the air gap, and (b) that required to force it through the remainder of the magnetic circuit. The shorter the air gap, the steeper is the air gap line.Procedure to conduct OC test:(i) Start the prime mover and adjust the speed to the synchronous speed of the alternator.(ii) Keep the field circuit rheostat in cut in position and switch on DC supply.(iii) Keep the TPST switch of the stator circuit in open position.(iv) Vary the field current from minimum in steps and take the readings of field current and stator terminal voltage, till the voltage read by the voltmeter reaches up to 110% of rated voltage. Reduce the field current and stop the machine.

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(v) Plot of terminal voltage/ phase vs field current gives the OC curve.

Short Circuit Characteristic (S.C.C.)The short-circuit characteristic, as its name implies, refers to the behaviour of the alternator when its armature is short-circuited. In a single-phase machine the armature terminals are short-circuited through an ammeter, but in a three-phase machine all three phases must be short-circuited. An ammeter is connected in series with each armature terminal, the three remaining ammeter terminals being short-circuited. The machine is run at rated speed and field current is increased gradually to If2 till armature current reaches rated value. The armature short-circuit current and the field current are found to be proportional to each other over a wide range, as shown in Figure, so that the shortcircuit characteristic is a straight line. Under short-circuit conditions the armature current is almost 90° out of phase with the voltage, and the armature mmf has a direct demagnetizing action on the field. The resultant ampere − turns inducing the armature emf are, therefore, very small and is equal to the difference between the field and the armature ampere − turns. This results in low mmf in the magnetic circuit, which remains in unsaturated condition and hence the small value of induced emf increases linearly with field current. This small induced armature emf is equal to the voltage drop in the winding itself, since the terminal voltage is zero by assumption. It is the voltage required to circulate the shortcircuitcurrent through the armature windings. The armature resistance is usually small compared withthe reactance.

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Fig. OCC and SCC of an alternator

Short-Circuit Ratio:The short-circuit ratio is defined as the ratio of the field current required to produce rated volts on open circuit to field current required to circulate full-load current with the armature short-circuited.Short-circuit ratio = If1/If2

Determination of synchronous impedance Zs:As the terminals of the stator are short circuited in SC test, the short circuit current is circulated against the impedance of the stator called the synchronous impedance. This impedance can be estimated form the oc and sc characteristics. The ratio of open circuit voltage to the short circuit current at a particular field current, or at a field current responsible for circulating the rated current is called the synchronous impedance. Synchronous impedance Zs = (open circuit voltage per phase)/(short circuit current per phase) for same IfHence Zs = (Voc) / (Isc)for same If

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From figure synchronous impedance Zs = V/IscArmature resistance Ra of the stator can be measured using Voltmeter – Ammeter method. UsingSynchronous impedance and armature resistance synchronous reactance and hence regulation can be calculated as follows using emf method.

Zs = √(Ra)2 + (XS)2 and Synchronous reactance Xs = √( Zs)2 - (Ra)2Hence induced emf per phase can be found as Eph = √[ (V cosϕ + IRa)2 + (V sin ± IXS)2]

Where V = phase voltage per phase = Vph , I = load current per phase in the above expression in second term + sign is for lagging pwer factor ans – sign is for leadingpower factor.

% Regulation = [(Eph – Vph / Vph )] x 100where Eph = induced emf /phase, Vph = rated terminal voltage/phaseSynchronous impedance method is easy but it gives approximate results. This method gives the value of regulation which is greater (poor) than the actual value and hence this method is called pessimistic method. The complete phasor diagram for the emf method is shown in figure

MMF method: This method is also known as amp - turns method. In this method the all the emfs produced by rotor and stator are replaced by their equivalent MMFs (fluxes), and hence called mmf method. In this method also it is assumed that the magnetic circuit is unsaturated. In this method boththe reactance drops are replaced by their equivalent mmfs. Figure 35 shows the complete phasordiagram for the mmf method. Similar to emf method OC and SC characteristics are used for the

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determination of regulation by mmf method. The details are shown in figure 36. Using the details it is possible determine the regulation at different power factors.

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From the phasor diagram it can be seen that the mmf required to produce the emf E1= ( V + IRa) is FR1.In large machines resistance drop may neglected.The mmf required to overcome the reactance drops is (A+Ax) as shown in phasor diagram. The mmf (A+Ax) can be found from SC characteristic as under SC condition both reactance drops will be present.

Following procedure can be used for determination of regulation by mmf method.(i) By conducting OC and SC test plot OCC and SCC as shown in figure 36.(ii) From the OCC find the field current If1 required to produce the voltage, E1= (V + IRa).(iii) From SCC find the magnitude of field current If2 ( A+Ax) to produce the required armature current. A+Ax can also found from ZPF characteristics.(iv) Draw If2 at angle (90+ ) from If1, where is the phase angle of current w. r. t voltage. Ifcurrent is leading, take the angle of If2 as (90- ) as shown in figure.(v) Determine the resultant field current, If and mark its magnitude on the field current axis.(vi) From OCC. find the voltage corresponding to If, which will be E0 and hence find theregulation. Because of the assumption of unsaturated magnetic circuit the regulation computed by this method will be less than the actual and hence this method of regulation is called optimistic method.

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ASA Modified MMF Method: Because of the unrealistic assumption of unsaturated magnetic circuit neither the emf method nor the mmf method are giving the realistic value of regulation. In spite of these short comings these methods are being used because of their simplicity. Hence ASA has modified mmf method for calculation of regulation. With reference to the phasor diagram of mmf method it can be seen that F = FR1 - ( A+Ax). In the mmf method the total mmf F computed is based on the assumption of unsaturated magnetic circuit which is unrealistic. In order to account for the partial saturation of the magnetic circuit it must be increased by a certain amount FF2 which can be computed from occ, scc and air gap lines as explained below referring to figure.

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If1 is the field current required to induce the rated voltage on open circuit. Draw If2 with length equal to field current required to circulate rated current during short circuit condition at an angle (90+ ) from If1. The resultant of If1 and If2 gives If (OF2 in figure). Extend OF2 upto F so that F2F accounts for the additional field current required for accounting the effect of partial saturation of magnetic circuit. F2F is found for voltage E (refer to phasor diagram of mmf method) as shown in figure. Project total field current OF to the field current axis and find corresponding voltage E0 using OCC. Hence regulation can found by ASA method which is more realistic.

Zero Power Factor ( ZPF) method: Potier Triangle Method:

During the operation of the alternator, resistance voltage drop IaRa and armature leakage reactance drop IaXL are actually emf quantities and the armature reaction reactance is a mmf quantity. To determine the regulation of the alternator by this method OCC, SCC and ZPF test details and characteristics are required. AS explained earlier oc and sc tests are conducted and OCC and SCC are drawn. ZPF test is conducted by connecting the alternator to ZPF load and exciting the alternator in such way that the alternator supplies the rated current at rated voltage running at rated speed. To plot ZPF characteristics only two points are required. One point is

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corresponding to the zero voltage and rated current that can be obtained from scc and the other at rated voltage and rated current under zpf load. This zero power factor curve appears like OCC but shifted by a factor IXL vertically and horizontally by armature reaction mmf as shown below in figure 42. Following are the steps to draw ZPF characteristics.By suitable tests plot OCC and SCC. Draw air gap line. Conduct ZPF test at full load for rated voltage and fix the point B. Draw the line BH with length equal to field current required to produce full load current on short circuit. Draw HD parallel to the air gap line so as to cut the OCC. Draw DE perpendicular to HB or parallel to voltage axis. Now, DE represents voltage drop IXL and BE represents the field current required to overcome the effect of armature reaction. Triangle BDE is called Potier triangle and XL is the Potier reactance. Find E from V, IRa, IXL and ϕUse the expression E = √(V cosϕ + IRa)2 + (V sinϕ) + IXL)2 to compute E. Find field currentcorresponding to E. Draw FG with magnitude equal to BE at angle (90+ϕ) from field current axis,where ϕ- is the phase angle of current from voltage vector E (internal phase angle).The resultant field current is given by OG. Mark this length on field current axis. From OCC find the corresponding E0. Find the regulation.

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Salient pole alternators and Blondel’s two reaction Theory:The details of synchronous generators developed so far is applicable to only round rotor or nonsalient pole alternators. In such machines the air gap is uniform throughout and hence the effect of mmf will be same whether it acts along the pole axis or the inter polar axis. Hence reactance of the sator is same throughout and hence it is called synchronous reactance. But in case salient pole machines the air gap is non-uniform and it is smaller along pole axis and is larger along the inter polar axis. These axes are called direct axis or d-axis and quadrature axis or q-axis. Hence the effect of mmf when acting along direct axis will be different than that when it is acting along quadrature axis. Hence the reactance of the stator cannot be same when the mmf is acting along d – axis and q- axis. As the length of the air gap is small along direct axis reluctance of the magnetic circuit is less and the air gap along the q – axis is larger and hence the along the quadrature axis will be comparatively higher. Hence along d-axis more flux is produced than q-axis. Therefore the reactance due to armature reaction will be differentalong d-axis and q-axis. These reactances areXad = direct axis reactance; Xaq = quadrature axis reactanceHence the effect of armature reaction in the case of a salient pole synchronous machine can be taken as two components - one acting along the direct axis (coinciding with the main field pole axis) and the other acting along the quadrature axis (inter-polar region or magnetic neutral axis) - and as such the mmf components of armature-reaction in a salient-pole machine cannot be considered as acting on the same magnetic circuit. Hence the effect of the armature reaction cannot be taken into account by considering only the synchronous reactance, in the case of a salient pole synchronous machine.In fact, the direct-axis component Fad acts over a magnetic circuit identical with that of the main field system and produces a comparable effect while the quadrature-axis component Faq acts along the interpolar axis, resulting in an altogether smaller effect and, in addition, a flux distribution totally different from that of Fad or the main field m.m.f. This explains why the application of cylindrical-rotor theory to salient-pole machines for predicting the performance gives results not conforming to the performance obtained from an actual test. Blondel’s two-reaction theory considers the effects of the quadrature and direct-axis components of the armature reaction separately. Neglecting saturation, their different effects are considered by assigning to each an appropriate value of armature-reaction “reactance,” respectively xad and xaq. The effects of armature resistance and true leakage reactance (XL) may be treated separately, or may be added to the armature reaction coefficients on the assumption that they are the same, for either the direct-axis or quadrature-axis components of the armature current (which is almost true). Thus the combined reactance values can be expressed as : Xsd = xad + xl and Xsq = xaq + xl for the direct- and cross-reaction axes respectively.In a salient-pole machine, xaq, the quadrature-axis reactance is smaller than xad, the direct-axisreactance, since the flux produced by a given current component in that axis is smaller as thereluctance of the magnetic path consists mostly of the interpolar spaces. It is essential to clearly note the difference between the quadrature and direct-axis components Iaq, and Iad of the armature current Ia, and the reactive and active components Iaa and Iar. Although both pairs are represented by phasors in phase quadrature, the former are related to the induced emf Et while the latter are referred to the terminal voltage V. These phasors are clearly indicated with reference to the phasor diagram of a (salient pole) synchronous generator supplying a lagging power factor (pf) load, shown in Fig.

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Iaq = Ia cos(Δ+ϕ); Iad = Ia sin(Δ+ϕ); and Ia = [(Iaq)2 + (Iad)2]Iaa = Ia cosϕ; Iar = Ia sinϕ; and Ia =√[(Iaa)2 + (Iar)2]Where " = torque or power angle and ! = the p.f. angle of the load.The phasor diagram Fig. shows the two reactance voltage components Iaq *Xsq and Iad * Xsd which are in quadrature with their respective components of the armature current. The resistance drop Ia x Ra is added in phase with Ia although we could take it as Iaq x Ra and Iad x Ra separately, which is unnecessary as Ia = Iad + jIaq.

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Power output of a Salient Pole Synchronous Machine:Neglecting the armature winding resistance, the power output of the generator is given by:P = V x Ia x cos ϕThis can be expressed in terms of ", by noting from Fig. 48 thatIa cosϕ = Iaq cosΔ + Iad sinΔV cosΔ = Eo − Iad * Xsd and V sin Δ = Iaq * XsqSubstituting the above expressions for power we get

P = V [(V sinΔ /Xsd) * cosΔ + (Eo − V cosΔ)/Xsd * sinΔ]On simplification we getP = (V * Eo/Xsd) sinΔ + V2 * (Xsd − Xsq)/(2 * Xsq * Xsq) * sin 2ΔThe above expression for power can also be written asP = (Eo * V * sin Δ /Xd) + V2 * (Xd − Xq) * sin 2Δ /(2 * Xq * Xq)The above expression for power consists of two terms first is called electromagnetic power and the second is called reluctance power.

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It is clear from the above expression that the power is a little more than that for a cylindrical rotor synchronous machine, as the first term alone represents the power for a cylindrical rotor synchronous machine. A term in (sin 2Δ) is added into the power – angle characteristic of a non-salient pole synchronous machine. This also shows that it is possible to generate an emf even if the excitation E0 is zero. However this magnitude is quite less compared with that obtained with a finite E0. Likewise it can be shown that the machine develops a torque - called the reluctance torque - as this torque is developed due to the variation of the reluctance in the magnetic circuit even if the excitation E0 is zero.

Determination of Xd and Xq by slip test:The direct and quadrature axis reactances Xd and Xq can be of a synchronous machine can beexperimentally determined by a simple test known as slip test. Basic circuit diagram for conducting this test is shown in figure. Here the armature terminals are supplied with a subnormal voltage of rated frequency with field circuit left open. The generator is driven by a prime mover at a slip speed which is slightly more or less than the synchronous speed. This is equivalent to the condition in which the armature mmf remains stationary and rotor rotates at a slip speed with respect to the armature mmf. As the rotor poles slip through the armature mmf the armature mmf will be in line with direct axis and quadrature axis alternately. When it is in line with the direct axis the armature mmf directly acts on the magnetic circuit and at this instant the voltage applied divided by armature current gives the direct axis synchronous reactance. When the armature mmf coincides with the quadrature axis then the voltage impressed divided by armature current gives the quadrature axis synchronous reactance. Since Xd > Xq the pointers of the ammeter reading the armature current will oscillate from a minimum to a maximum. Similarly the terminal voltage will also oscillate between the minimum and maximum.

Xd = Maximum voltage / minimum currentXq = Minimum voltage / maximum current.The figures below show the flux paths in direct and quadrature axis of a salient pole alternator.

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Synchronizing of alternators:

SynchronizingThe operation of connecting two alternators in parallel is known as synchronizing. Certain conditions must be fulfilled before this can be affected. The incoming machine must have its voltage and frequency equal to that of the bus bars and, should be in same phase with bus bar voltage. The instruments or apparatus for determining when these conditions are fulfilled are called synchroscopes. Synchronizing can be done with the help of (i) dark lamp method or (ii) by using synchroscope.

Reasons for operating in parallel:a) Handling larger loads.b) Maintenance can be done without power disruption.c) Increasing system reliability.d) Increased efficiency.

Conditions required for Paralleling:The figure below shows a synchronous generator G1 supplying power to a load, with another generator G2 about to be paralleled with G1 by closing switch S1. What conditions must be metbefore the switch can be closed and the 2 generators connected in parallel? Paralleling 2 or more generators must be done carefully as to avoid generator or other system component damage. Conditions to be satisfied are as follows:a) RMS line voltages must be equal.b) The generators to be paralleled must have the same phase sequence.c) The oncoming generator (the new generator) must have the same operating frequency as compared to the system frequency.

General Procedure for Paralleling Generators:

Consider the figure shown below. Suppose that generator G2 is to be connected to the running system as shown below:

1. Using Voltmeters, the field current of the oncoming generator should be adjusted until its terminal voltage is equal to the line voltage of the running system.

2. Check and verify phase sequence to be identical to the system phase sequence. There are 2 methods to do this:

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i. One way is using the 3 lamp method, where the lamps are stretched across the open terminals of the switch connecting the generator to the system (as shown in the figure below). As the phase changes between the 2 systems, the lamps first get bright (large phase difference) and then get dim (small phase difference). If all 3 lamps get bright and dark together, then the systems have the same phase sequence. If the lamps brighten in succession, then the systems have the opposite phase sequence, and one of the sequences must be reversed.

ii. Using a Synchroscope – a meter that measures the difference in phase angles (it does not check phase sequences only phase angles).

3. Check and verify generator frequency is same as that of the system frequency. This is done bywatching a frequency of brightening and dimming of the lamps until the frequencies are close by making them to change very slowly.

4. Once the frequencies are nearly equal, the voltages in the 2 systems will change phase with respect to each other very slowly. The phase changes are observed, and when the phase angles are equal, the switch connecting the 2 systems is closed.

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Synchronizing Current:If two alternators generating exactly the same emf are perfectly synchronized, there is no resultant emf acting on the local circuit consisting of their two armatures connected in parallel. No current circulates between the two and no power is transferred from one to the other. Under this condition emf of alternator 1, i.e. E1 is equal to and in phase opposition to emf of alternator 2, i.e. E2 as shown in the Figure .There is, apparently, no force tending to keep them in synchronism, but as soon as the conditions are disturbed a synchronizing force is developed, tending to keep the whole system stable. Suppose one alternator falls behind a little in phase by an angle. The two alternator emfs now produce a resultant voltage and this acts on the local circuit consisting of the two armature windings and the joining connections. In alternators, the synchronous reactance is large compared with the resistance, so that the resultant circulating current Is is very nearly in quadrature with the resultant emf Er acting on the circuit. Figure represents a single phase case, where E1 and E2 represent the two induced emfs, the latter having fallen back slightly in phase. The resultant emf, Er, is almost in quadrature with both the emfs, and gives rise to a current, Is, lagging behind Er by an angle approximating to a right angle. It is, thus, seen that E1 and Is are almost in phase. The first alternator is generating a power E1 Is cos 1, which is positive, while the second one is generating a power E2 Is cos 2, which is negative, since cos 2 is negative. In other words, the first alternator is supplying the second with power, the difference between the two amounts of power represents the copper losses occasioned by the current Is flowing through the circuit which possesses resistance. This power output of the first alternator tends to retard it, while the power input to the second one tends to accelerate it till such a time that E1 and E2 are again in phase opposition and the machines once again work in perfect synchronism. So, the action helps to keep both machines in stable synchronism. The current, Is, is called the synchronizing current.

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Synchronizing Power:

Suppose that one alternator has fallen behind its ideal position by an electrical angle q, measured in radians. Since E1 and E2 are assumed equal and q is very small Er is very nearly equal to qE1.Moreover, since Er is practically in quadrature with E1 and Is may be assumed to be in phase with E1 as a first approximation. The synchronizing power may, therefore, be taken as,Ps = E1 Is and Is = Er / 2Zs and Er = E1Ps = E1

2/ 2Zs or Ps = E12/ 2Xs

Where Zs is the synchronous impedance, Zs = Xs when the resistance is neglected.When one alternator is considered as running on a set of bus bars the power capacity of which is very large compared with its own, the combined reactance of the others sets connected to the bus bars is negligible, so that , in this case Zs = Xs is the synchronous reactance of the one alternator under consideration.Total synchronizing power Psy = 3E1

2/ 2Zs orPsy = 3qE12/ 2XsWhen the machine is connected to an infinite bus bar the synchronizing power is given byPsy = 3qE12/ Zs orPsy = 3qE12/ XsAnd synchronizing torque Tsy = Psy x 60 / 2∏ Ns

Alternators with a large ratio of reactance to resistance are superior from a synchronizing point of view to those which have a smaller ratio, as then the synchronizing current Is cannot be considered as being in phase with E1. Thus, while reactance is bad from a regulation point of view, it is good for synchronizing purposes. It is also good from the point of view of self-protection in the even of a fault.

List of question

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1. Find the no load phase and line voltages of a star connected 3 phase, 6 pole alternator which runs at 1200 rpm, having flux per pole of 0.1wb sinusoidally distributed. It’s stator has 54 slots having double layer winding. Each coil has 8 turns and the coil is chorded by 1 slot.

2. A 3300V, 3phase star connected alternator has a full load current of 100A. On short circuit a field current of 5A was necessary to produce full load current. The emf on open circuit for the same excitation was 900V. The armature resistance was 0.8Ω/phase. Determine the full load voltage regulation for (1)0.8pf lagging (2)0.8pf leading.

3. A 3 phase, 50Hz star connected 2000kA for a certain field excitation. With the same excitation, the open circuit voltage was 900V. The resistance between a pair of terminals was 0.12 Ω. Find the full load regulation at UPF and 0.8pf lagging. Draw the phasor diagrams.

4. A 3 phase 16 pole alternator has a star connected winding with 144 slots and 10 conductors per slot. The flux per pole 0.04wb and is sinusoidally distributed. The speed is 375 rpm. Find the frequency, phase emf and line emf. The coil span is 1600 electrical.

5. (a) Describe the principle and construction of slow speed operation generator with neat diagram. (b) Derive the emf equation of alternator.

6. What are the methods of determining regulation of alternator? Discuss each briefly.

7. Explain the procedure for POTIER method to calculate voltage regulation of alternator.

8. For a salient pole synchronous machine, prove the d-axis synchronous reactance Xd, can be obtained from its OCC and SCC. Neglect armature resistance.

9. Explain the condition for parallel operation of 3 phase alternator with neat diagram.

10. From the following test results, determine the regulation of a 2kV single phase alternator, delivering a current of 100A at 0.8pf leading. Test results: full load current of 100A is produced on short circuit by a field excitation of 2.5A. An emf of 500V is produced on open circuit by the same field current. The armature resistance is 0.8Ω.

11. Two alternators working in parallel supply the following loads (i) lighting load of 500kW, (ii) 1000kW at 0.9pf, (iii) 500kW at 0.9pf lead, (iv) 800kW at 0.8 lag. One alternator is supplying 1500kW at 0.95pf lagging. Calculate the load on the other machines.

UNIT II – SYNCHRONOUS MOTOR

PART – B

Introduction

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A synchronous motor, as the name suggests, runs under steady-state conditions at a fixed speed called the synchronous speed. The synchronous speed, as discussed in the preceding chapter, depends only upon (a) the frequency of the applied voltage and (b) the number of poles in the machine. In other words, the speed of a synchronous motor is independent of the load as long as the load is within the capability of the motor. If the load torque exceeds the maximum torque that can be developed by the motor, the motor simply comes to rest and the average torque developed by it is zero. For this reason, a synchronous motor is not inherently self-starting. Therefore, it must be brought up almost to its synchronous speed by some auxiliary means before it can be synchronized to the Because of its constant speed-torque characteristic, a small synchronous motor is used as a timing device. A large synchronous motor may be used not only to drive a certain load but also to improve the overall power factor (pf) of an industrial plant because it can be operated at a leading power factor. However, when a synchronous motor is operated at no load just to improve the power factor, it is usually referred to as a synchronous condenser. A synchronous motor can be either a single-phase or a polyphase motor. Only three-phase synchronous motors are discussed in this chapter, but the development is valid for any polyphase synchronous motor.

characteristic features of synchronous motor.

A synchronous motor is electrically the same as a synchronous generator. However, it is not self-starting. When the motor is equipped with a damper winding, it develops a starting torque owing to the induced current in its damper winding. The damper winding may be either a three-phase winding or a squirrel-cage winding. When the motor attains a speed nearly equal to its synchronous speed, the field winding is excited and the motor pulls into synchronism. Once the motor attains its synchronous speed, the damper winding becomes ineffective. The motor keeps on rotating at its synchronous speed as long as the load torque is less than the pull-out torque. The pull-out torque is the maximum torque the motor can handle without pulling out of synchronism. The equations for the power (or torque) developed by a synchronous motor

are the same as those for a synchronous generator. The only difference is that the motor rotates in the same direction as the torque it develops. On the other hand, the generator is forced to rotate by the prime mover in a direction opposite to the torque developed. For any given load, a synchronous motor can be made to operate at either a

lagging power factor, unity power factor, or a leading power factor. The field current (and thereby the excitation voltage) that results in the armature current in phase with the terminal voltage (unity power factor) at any load is known as the normal excitation. A decrease in the field current gives rise to under excitation and a lagging power factor. Likewise, an increase in the field current from its normal value results in over excitation and a leading power factor.

An overexcited synchronous motor with a leading power factor is not only used as a motor to supply the needed mechanical power but also acts as a power factor correction device. In fact,

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there are some synchronous motors called synchronous condensers that are designed just to control the power factor.

Construction and Operation of a Synchronous Motor

The armature of a synchronous motor is exactly the same as that of a synchronous generator. It has a large number of slots that are designed to house the three identical, double-layer phase windings. The phase windings are spatially displaced by 120" electrical from one another and are excited by a balanced three phase source. The phase windings upon excitation produce a uniform magnetic field that rotates along the periphery of the air-gap at the synchronous speed. If m is the maximum value of the flux produced by the maximum current I, in each phase, the strength of the uniform revolving magnetic field is

r =1.5 m

The synchronous speed in revolutions per minute (rpm) at which the flux revolves around the periphery of the air-gap is

120f Ns = 120f

P

where f is the frequency of the three-phase power source and P is the number of poles in the motor.

The rotor of the synchronous motor has a field winding that produces the constant flux in the motor in exactly the same fashion as it does in a synchronous generator. Once the field winding is excited by a direct-current (dc) source, it produces alternate poles on the surface of the rotor. Thus far, it must be obvious that there is no difference between a synchronous motor and a synchronous generator.

Let us now assume that the rotor is at rest (standstill condition) and the field winding is excited to produce alternate poles on its periphery. The revolving field created by the armature can be visualized as if two magnets, a north pole and a south pole, are rotating at a constant (synchronous) speed just above the poles of the rotor. When the south pole of the revolving field is just above the north pole of the rotor, the force of attraction between them tends to move the rotor in the direction of the revolving field. Owing to the heavy mass of the rotor, it takes time before it can start moving, but by then the revolving field has reversed its polarity. Now the force of repulsion between the two like polarity poles tends to move the rotor in the opposite direction. As the rotor tries to rotate in the opposite direction, the revolving field has reversed its polarity once again. Thus, each pole on the rotor is acted upon by a rapidly reversing force of equal magnitude in both directions. The average torque thus developed by the rotor is zero. Therefore, a synchronous motor is not self-starting. Hence, to start a synchronous motor we must either

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provide some means for it to develop starting torque by itself or drive the rotor at nearly its synchronous speed by another prime mover. Then synchronize it by exciting the field winding.

To make a synchronous motor self-starting, an additional winding known as the damper winding (the induction winding or the amortisseur winding) is provided in the pole faces of the motor. The damper winding, also called a squirrel cage winding, is a short-circuit winding. For small machines, a squirrel-cage winding requires the placement of the rotor laminations in a mold and then forcing of the molten conducting material (often aluminum) into the slots. The mold has cavities on either side of the rotor, which are filled by the molten conducting material at the same time. The conducting material from one end of the slot to the other forms a conducting bar. The conducting bars are shorted by the end rings as shown in Figure 8.1. The entire one-piece construction looks like a squirrel cage and hence the name. For large machines, the squirrel-cage winding may be formed by driving metal bars into the slots one at a time and then shorting them with annular conducting strips on both ends.

The damper winding in a synchronous motor may also be of the wound-rotor type. The wound-rotor winding is used when we want to (a) control the speed of the motor and (b) develop high starting torque. A three-phase winding with as many poles as there are in the armature is placed in the rotor slots. One end of the three-phase windings is internally connected to form a common node while the other ends are connected to the slip rings. External resistance can then be

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included in series with each phase winding, as shown in Figure in order to increase the starting torque.

In any case, the damper winding forms a closed loop. The uniform revolving field induces an electromotive force (emf) in the damper winding which, in turn, results in an induced current in it. As explained in Chapter 3, the induced current exerts a torque on the damper-winding conductors and forces them to rotate in the direction of the revolving field. Under no load, the rotor speed is almost (but not exactly) equal to the synchronous speed of the motor. This, in fact, is the principle of operation of an induction motor, as explained in Chapter 3. A synchronous motor is, therefore, brought to its no-load speed as an induction motor. During the rotor acceleration period, the field winding must be short through an appropriate bank of resistors. The field winding should never be left

open because it can develop high voltage, just like the secondary winding of a step-up transformer. Once the rotor attains a steady speed, the short circuit is removed and the field winding is energized by connecting it to a dc source. The field poles thus formed on the rotor's periphery pull the rotor in step with the revolving field. In other words, the strong field poles are locked in step with the revolving poles of opposite polarity created by the armature winding

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(armature poles). The motor is then said to be synchronized. The damper winding becomes ineffective when the rotor rotates at synchronous speed.

At no load, the magnetic axes of the armature poles and the rotor poles are nearly aligned, as shown in Figure 8.3a. The magnetic lines of force are perpendicular to the rotor surface. Thus, they exert no torque on the rotor. The rotor poles start slipping behind the armature poles as the load on the motor is increased, as depicted in Figure 8.3b. The magnetic lines of force then have a component parallel to the armature surface and exert a force on the rotor. The angle, more appropriately the power angle, between the two magnetic axes keeps increasing with increasing load on the motor. The motor reaches an unstable region at about 60" electrical of angular displacement between the magnetic axes. Any further increase in the load may pull the motor out of synchronism.

The damper winding not only enables a synchronous motor to develop the starting torque but also serves another useful purpose. It tends to minimize motor hunting. Hunting comprises the successive overshoots and undershoots in the motor speed due to sudden changes in the load. When the load on a synchronous motor is changed suddenly, it takes time for the motor to adjust its power angle

owing to its inertia. During these power-angle adjustments, the speed of the motor fluctuates above and below its synchronous speed. These changes in speed induce current in the damper winding, thereby creating a torque that opposes the change. For example, when the load on the motor is suddenly increased, the rotor tends to slow down owing to an increase in the applied torque. As soon as it slows down, the current induced in the damper-winding conductors exerts an accelerating force on the rotor in the direction of its rotation. On the other hand, if the load is suddenly reduced, the inertia of the motor tends to increase the rotor speed. Again, a current is induced in the damper-winding conductors. However, the induced current is in the opposite direction now. Thus, it creates a torque in the opposite direction and forces the rotor to slow down. It is possible for hunting to become intolerably severe if the motor is not equipped with a damper winding.

The methods of starting synchronous motor.

If the rotating magnetic field of the stator in a synchronous motor rotates at a low enough speed, there will be no problem for the rotor to accelerate and to lock in with the stator’s magnetic field. The speed of the stator magnetic field can then be increased to its rated op-erating speed by gradually increasing the supply frequency f up to its normal 50- or 60-Hz value.

This approach to starting of synchronous motors makes a lot of sense, but there is a big problem: Where from can we get the variable frequency supply? The usual power supply system’s enerally regulate the frequency to be 50 or 60 Hz as the case may be. However, variable-frequency voltage source can be obtained from a dedicated generator only in the

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Olden days and such a situation were obviously impractical except for very unusual or special drive applications.

But the present day solid state power converters offer an easy solution to this. We now have the rectifier- inverter and cycloconverters, which can be used to convert a constant frequency AC supply to a variable frequency AC supply. With the development of such modern solid-state variable-frequency drive packages, it is thus possible to continuously control the frequency of the supply connected to the synchronous motor all the way from a fraction of a hertz up to and even above the normal rated frequency. If such a variable-frequency drive unit is included in a motor-control circuit to achieve speed control, then starting the synchronous motor is very easy-simply adjust the frequency to a very low value for starting, and then raise it up to the desired operating frequency for normal running.

When a synchronous motor is operated at a speed lower than the rated speed, its internal generated voltage (usually called the counter EMF) EA = Kφω will be smaller than normal. As such the terminal voltage applied to the motor must be reduced proportionally with the frequency in order to keep the stator current within the rated value. Generally, the voltage in any variable-frequency power supply varies roughly linearly with the output frequency.

Basically there are three methods that are used to start a synchronous motor:

• To reduce the speed of the rotating magnetic field of the stator to a low enough value that the rotor can easily accelerate and lock in with it during one half-cycle of the rotating magnetic field’s rotation. This is done by reducing the frequency of the applied electric power. This method is usually followed in the case of inverter-fed synchronous motor operating under variable speed drive applications.

• To use an external prime mover to accelerate the rotor of synchronous motor near to its synchronous speed and then supply the rotor as well as stator. Ofcourse care should be taken to ensure that the direction of rotation of the rotor as well as that of the rotating magnetic field of the stator are the same. This method is usually followed in the laboratory- the synchronous machine is started as a generator and is then connected to the supply mains by following the synchronization or paralleling procedure. Then the power supply to the prime mover is disconnected so that the synchronous machine will continue to operate as a motor.

• To use damper windings or amortisseur windings if these are provided in the ma-chine. The damper windings or amortisseur windings are provided in most of the large synchronous motors in order to nullify the oscillations of the rotor whenever the synchronous machine is subjected to a periodically varying load.

Each of these methods of starting a synchronous motor are described below in detail.

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1. Motor Starting by reducing the supply Frequency

If the rotating magnetic field of the stator in a synchronous motor rotates at a low enough speed, there will be no problem for the rotor to accelerate and to lock in with the stator’s magnetic field. The speed of the stator magnetic field can then be increased to its rated op-erating speed by gradually increasing the supply frequency f up to its normal 50- or 60-Hz value.

This approach to starting of synchronous motors makes a lot of sense, but there is a big problem: Where from can we get the variable frequency supply? The usual power supply systems generally regulate the frequency to be 50 or 60 Hz as the case may be. However, variable-frequency voltage source can be obtained from a dedicated generator only in theolden days and such a situation was obviously impractical except for very unusual or special drive applications.

But the present day solid state power converters offer an easy solution to this. We now have the rectifier- inverter and cycloconverters, which can be used to convert a constant frequency AC supply to a variable frequency AC supply. With the development of such modern solid-state variable-frequency drive packages, it is thus possible to continuously control the frequency of the supply connected to the synchronous motor all the way from a fraction of a hertz up to and even above the normal rated frequency. If such a variable-frequency drive unit is included in a motor-control circuit to achieve speed control, then starting the synchronous motor is very easy-simply adjust the frequency to a very low value for starting, and then raise it up to the desired operating frequency for normal running.

When a synchronous motor is operated at a speed lower than the rated speed, its internal generated voltage (usually called the counter EMF) EA = Kφω will be smaller than normal. As such the terminal voltage applied to the motor must be reduced proportionally with the frequency in order to keep the stator current within the rated value. Generally, the voltage in any variable-frequency power supply varies roughly linearly with the output frequency

2. Motor Starting with an External Motor

The second method of starting a synchronous motor is to attach an external starting motor (pony motor) to it and bring the synchronous machine to near about its rated speed (but not exactly equal to it, as the synchronization process may fail to indicate the point of closure of the main switch connecting the synchronous machine to the supply system) with the pony motor. Then the output of the synchronous machine can be synchronized or paralleled with its power supply system as a generator, and the pony motor can be detached from the shaft of the machine or the supply to the pony motor can be disconnected. Once the pony motor is turned OFF, the shaft of the machine slows down, the speed of the rotor magnetic field BR falls behind Bnet, momentarily and the synchronous machine continues to operate as a motor. As soon as it begins to operates as a motor the synchronous motor can be loaded in the usual manner just like any motor.

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This whole procedure is not as cumbersome as it sounds, since many synchronous mo-tors are parts of motor-generator sets, and the synchronous machine in the motor-generator set may be started with the other machine serving as the starting motor. Moreover, the starting motor is required to overcome only the mechanical inertia of the synchronous ma-chine without any mechanical load (load is attached only after the synchronous machine is paralleled to the power supply system). Since only the motor’s inertia must be overcome, the starting motor can have a much smaller rating than the synchronous motor it is going to start. Generally most of the large synchronous motors have brushless excitation systems mounted on their shafts. It is then possible to use these exciters as the starting motors. For many medium-size to large synchronous motors, an external starting motor or starting by using the exciter may be the only possible solution, because the power systems they are tied to may not be able to handle the starting currents needed to use the damper (amortisseur) winding approach described next.

3. Motor Starting by using damper (Amortisseur) Winding

As already mentioned earlier most of the large synchronous motors are provided with damper windings, in order to nullify the oscillations of the rotor whenever the synchronous machine is subjected to a periodically varying load. Damper windings are special bars laid into slots cut in the pole face of a synchronous machine and then shorted out on each end by a large shorting ring, similar to the squirrel cage rotor bars. A pole face with a set of damper wind-ings is shown in Figure..

When the stator of such a synchronous machine is connected to the 3-Phase AC sup-ply, the machine starts as a 3-Phase induction machine due to the presence of the damper bars, just like a squirrel cage induction motor. Just as in the case of a 3-Phase squirrel cage induction motor, the applied voltage must be suitably reduced so as to limit the starting cur-rent to the safe rated value. Once the motor picks up to a speed near about its synchronous speed, the DC supply to its field winding is connected and the synchronous motor pulls into step i.e. it continues to operate as a Synchronous motor running at its synchronous speed.

Equivalent circuit model and phasor diagram of a synchronous motor

The behavior of a synchronous motor can be predicted by considering its equivalent circuit on similar lines to that of a synchronous generator as described below. Equivalent circuit model and phasor diagram of a synchronous motor The equivalent-circuit model for one armature phase of a cylindrical rotor three phase synchronous motor is shown in Figure below exactly similar to that of a synchronous generator except that the current flows in to the armature from the supply. Applying Kirchhoff’s voltage law to Figure below

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VT = IaRa + jIaXl + jIaXas + EfCombining reactances, Xs = Xl + XasVT = Ef + Ia(Ra + jXs)or VT = Ef + IaZswhere:Ra = armature resistance (/phase)Xl = armature leakage reactance (/phase)Xs = synchronous reactance (/phase)Zs = synchronous impedance (/phase)VT = applied voltage/phase (V)Ia = armature current/phase(A)

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A phasor diagram shown in Figure above, illustrates the method of determining the counter EMFwhich is obtained from the phasor equation;Ef = VT − IaZsThe phase angle " between the terminal voltage VT and the excitation voltage Ef in Figure above is usually termed the torque angle. The torque angle is also called the load angle or power angle.

Effect of changes in load on, Ia, , and p. f. of synchronous motor: The effects of changes in mechanical or shaft load on armature current, power angle, and power factor can be seen from the phasor diagram shown in Figure below; As already stated, the applied stator voltage, frequency, and field excitation are assumed, constant. The initial load conditions, are represented by the thick lines. The effect of increasing the shaft load to twice its initial value is represented by the light lines indicating the new steady state conditions. While drawing the phasor diagrams to show new steady-state conditions, the line of action of the new jIaXs phasor must be perpendicular to the new Ia phasor. Furthermore, as shown in figure if the excitation is not changed, increasing the shaft load causes the locus of the Ef phasor to follow a circular arc, thereby increasing its phase angle with increasing shaft load. Note also that an increase in shaft load is also accompanied by a decrease in i; resulting in an increase in power factor.As additional load is placed on the machine, the rotor continues to increase its angle of lag relative to the rotating magnetic field, thereby increasing both the angle of lag of the counter EMF phasor and the magnitude of the stator current. It is interesting to note that during all this load variation, however, except for the duration of transient conditions whereby the rotor assumes a new position in relation to the rotating magnetic field, the average speed of the machine does not change. As the load is being increased, a final point is reached at which a further increase in " fails to cause a corresponding increase in motor torque, and the rotor pulls out of synchronism. In fact as stated earlier, the rotor poles at this point, will fall behind the

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stator poles such that they now come under the influence of like poles and the force of attraction no longer exists. Thus, the point of maximum torque occurs at a power angle of approximately 90# for a cylindrical-rotor machine. This maximum value of torque that causes a synchronous motor to pull out of synchronism is called the pull-out torque. In actual practice, the motor will never be operated at power angles close to 900 as armature current will be many timesits rated value at this load

Effect of changes in excitation on the performance synchronous motorIncreasing the strength of the magnets will increase the magnetic attraction, and thereby cause the rotor magnets to have a closer alignment with the corresponding opposite poles of the rotating magnetic poles of the stator. This will obviously result in a smaller power angle. This fact can also be seen from power angle equation. When the shaft load is assumed to be constant, the steady-state value of Ef sinΔ must also be constant. An increase in Ef will cause a transient increase in Ef sinΔ, and the rotor will accelerate. As the rotor changes its angular position, Δ decreases until Ef sin" has the same steady-state value as before, at which time the rotor is again operating at synchronous speed, as it should run only at the synchronous speed. This change in angular position of the rotor magnets relative to the poles of rotating magnetic field of the stator occurs in a fraction of a second. The effect of changes in field excitation on armature current, power angle, and power factor of a synchronous motor operating with a constant shaft load, from a constant voltage, constant frequency supply, is illustrated in figure below.

Ef1 sinΔ1 = Ef2 sinΔ 2 = Ef3 sin Δ 3 = Ef sin ΔThis is shown in Figure below, where the locus of the tip of the Ef phasor is a straight line parallel to the VT phasor. Similarly,Ia1 cos ϕ i1 = Ia2 cos ϕ i2 = Ia3 cos ϕ i3 = Ia cos ϕ iThis is also shown in Figure below, where the locus of the tip of the Ia phasor is a line perpendicular to the phasor VT.

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Note that increasing the excitation from Ef1 to Ef3 caused the phase angle of the current phasor with respect to the terminal voltage VT (and hence the power factor) to go from lagging to leading. The value of field excitation that results in unity power factor is called normal excitation. Excitation greater than normal is called over excitation, and excitation less than normal is called under excitation. Further, as indicated in Figure, when operating in the overexcited mode, |Ef | > |VT |. A synchronous motor operating under over excited condition is called a synchronous condenser.

Phasor diagram showing effect of changes in field excitation on armature current, power angle and power factor of a synchronous motor

V and inverted V curve of synchronous motor:

Graphs of armature current vs. field current of synchronous motors are called V curves and are shown in Figure below for typical values of synchronous motor loads. The curves are related to the phasor diagram shown in figure below, and illustrate the effect of the variation of field excitation on armature current and power factor. It can be easily noted from these curves that an increase in shaft loads require an increase in field excitation in order to maintain the power factor at unity.

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The points marked a, b, and c on the upper curve corresponds to the operating conditions of the phasor diagrams shown. Note that for P = 0, the lagging power factor operation is electrically equivalent to an inductor and the leading power factor operation is electrically equivalent to a capacitor. Leading power factor operation with P = 0 is sometimes referred to as synchronous condenser or synchronous capacitor operation. Typically, the synchronous machine V-curves are provided by the manufacturer so that the user can determine the resulting operation under a given set of conditions.

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Hunting.

To make a synchronous motor self-starting, an additional winding known as the damper winding (the induction winding or the amortisseur winding) is provided in the pole faces of the motor. The damper winding, also called a squirrelcage winding, is a short-circuit winding. For small machines, a squirrel-cage winding requires the placement of the rotor laminations in a mold and then forcing of the molten conducting material (often aluminum) into the slots. The mold hascavities on either side of the rotor, which are filled by the molten conducting material at the same time. The conducting material from one end of the slot to the other forms a conducting bar. The conducting bars are shorted by the end rings as shown in Figure The entire one-piece construction looks like a squirrel cage and hence the name. For large machines, the squirrel-cage winding may be formed by driving metal bars into the slots one at a time and then shorting them with annular conducting strips on both ends.

The damper winding in a synchronous motor may also be of the wound-rotor type. The wound-rotor winding is used when we want to (a) control the speed of the motor and (b) develop high starting torque. A three-phase winding with as many poles as there are in the armature is placed in the rotor slots. One end of the three-phase windings is internally connected to form a common node while the other ends are connected to the slip rings. External resistance can then be

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included in series with each phase winding, as shown in Figure in order to increase the starting torque. In any case, the damper winding forms a closed loop. The uniform revolving field induces an electromotive force (emf) in the damper winding which, in turn,results in an induced current in it.The induced current exerts a torque on the damper-winding conductors and forces them to rotate in the direction of the revolving field. Under no load, the rotor speed is almost (but not exactly) equal to the synchronous speed of the motor. This, in fact, is the principle of operation of an induction motor. A synchronous motor is, therefore, brought to its no-load speed as an induction motor. During the rotor acceleration period, the field winding must be shorted through an appropriate bank of resistors. The field winding should never be left open because it can develop high voltage, just like the secondary winding of a step-up transformer. Once the rotor attains a steady speed, the short circuit is removed and the field winding is energized by connecting it to a dc source. The field poles thus formed on the rotor's periphery pull the rotor in step with the revolving field. In other words, the strong field poles are locked in step with the revolving poles of opposite polarity created by the armature winding (armature poles). The motor is then said to be synchronized. The damper winding becomes ineffective when the rotor rotates at synchronous speed. At no load, the magnetic axes of the armature poles and the rotor poles are nearly aligned, as shown in Figure. The magnetic lines of force are perpendicular to the rotor surface. Thus, they exert no torque on the rotor. The rotor poles start slipping behind the armature poles as the load on the motor is increased,as depicted in Figure. The magnetic lines of force then have a componentparallel to the armature surface and exert a force on the rotor. The angle, more appropriately the power angle, between the two magnetic axes keeps increasing with increasing load on the motor. The motor reaches an unstable region at about 60" electrical of angular displacement between the magnetic axes. Any further increase in the load may pull the motor out of synchronism.

The damper winding not only enables a synchronous motor to develop the starting torque but also serves another useful purpose. It tends to minimize motor hunting. Hunting comprises the successive overshoots and undershoots in the motor speed due to sudden changes in the load. When the load on a synchronous motor is changed suddenly, it takes time for the motor to adjust its power angle owing to its inertia. During these power-angle adjustments, the speed of the motor fluctuates above and below its synchronous speed. These changes in speed induce current in the damper winding, thereby creating a torque that opposes the change. For example, when the load on the motor is suddenly increased, the rotor tends to slow down owing to an increase in the applied torque. As soon as it slows down, the current induced in the damper-winding conductors exerts an accelerating force on the rotor in the direction of its rotation. On the other hand, if theload is suddenly reduced, the inertia of the motor tends to increase the rotor speed. Again, a current is induced in the damper-winding conductors. However, the induced current is in the opposite direction now. Thus, it creates a torque in the opposite direction and forces the rotor to slow down. It is possible for hunting to become intolerably severe if the motor is not equipped with a damper winding.

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List of questions

1. (a) Explain the methods of starting synchronous motor against high-torque loads. (b) Explain various torques associated with synchronous motor.

2. (a) Draw the equivalent circuit and phasor diagram of a synchronous motor. (b) Explain the significance of V and inverted V curves.

3. (a). Explain the working of synchronous motor with different excitations (b) List out the main characteristic features of synchronous motor.

4. Discuss the following (i) Constant excitation circle. (ii) Constant power circle.

5. Derive the mechanical power developed per phase of a synchronous motor.

6. A 3300V, 3 phase synchronous motor running at 1500 rpm has its excitation kept constant corresponding to no-load terminal voltage of 3000V. Determine the power input, power factor and torque developed for an armature current of 250A if the synchronous reactance is 5 Ω per phase and armature resistance is neglected.

7. A synchronous motor having 40% reactance and negligible resistance is to be operated at rated voltage at UPF,0.8pf lag,0.6pf lag, 0.8pf lead and 0.6pf lead. What are the values of induced emf.8. A 75 kW, 400V, 4 pole, 3 phase, star connected synchronous motor has a resistance and synchronous reactance per phase of 0.04Ω and 0.4 Ω respectively. Compute for full load 0.8pf lead the open circuit emf per phase and gross mechanicalpower developed. Assume an efficiency of 92.5%.

9. A 6600V, 3 phase, star connected synchronous motor draws a full load current of 80A at 0.8pf leading. The armature resistance is 2.2Ω and reactance of 22Ω per phase. If the stray loss of the machine is 3200W. Find (i) Emf induced(ii)Output power (iii) Efficiency of the machine.

10. A 2000V, 3 phase, 4 pole, star connected synchronous motor runs at 1500rpm. The excitation is constant and corresponding to an open circuit voltage of 2000V. The resistance is negligible in comparison with synchronous reactance of 3.5Ω /ph. For an armature current of 200A.Determine (i) power factor (ii) Power input (iii) Torque developed.

11. Explain V-curves and inverted V-curves.12. Explain the various starting methods of a synchronous motor.

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Unit III – THREE PHASE INDUCTION MOTOR

PART – B

Introduction

The direct-current (dc) and synchronous motors we have discussed thus far have one thing in common: both are the doubly-fed type. These motors have direct current in their field windings and alternating current (ac) in their armature windings. Since the electrical power is delivered directly to the armature of a dc motor via a commutator, it can also be referred to as a conduction motor. We now consider a motor in which the rotor receives its power not by conduction but by induction and is therefore called an induction motor. A winding that receives its power exclusively by induction constitutes a transformer. Therefore, an induction motor is a ransformer with a rotating secondary winding. From the above discussion, the following must be evident:

1. An induction motor is a singly-fed motor. Therefore, it does not require a commutator, slip-rings, or brushes. In fact, there are no moving contacts between the stator and the rotor. This results in a motor that is rugged, reliable, and almost maintenance free. 2. The absence of brushes eliminates the electrical loss due to the brush voltage drop and the mechanical loss due to friction between the brushes and commutator or the slip-rings. Thus, an induction motor has a relatively high efficiency.3. An induction motor carries alternating current in both the stator and the rotor windings.4. An induction motor is a rotating transformer in which the secondary winding receives energy by induction while it rotates.

There are two basic types of induction motors: single-phase induction motors and polyphase induction motors. Single-phase induction motors are favored for domestic applications. A large number of these motors are built in the fractional horsepower range. On the other hand, polyphase induction motors cover the entire spectrum of horsepower ratings and are preferably installed at locations where a polyphase power source is easily accessible. Owing to the widespread generation and transmission of three-phase power most polyphase induction motors are of the three-phase type. In this chapter, we confine our discussion exclusively to three-phase induction motors. The theoretical development, however, can be easily extended to an n-phase induction motor where n ≥ 2.

ConstructionIn actual practice, the three coils form three windings distributed over several slots. Thesewindings may be connected in star or delta and three terminations are brought out. Theseare conventional three phase windings which are discussed in greater detail in the chapterson alternators. Such windings are present n the stator as well as rotor. A photograph of

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the stator of an induction machine is shown in fig. 9. A close up of the windings is shown in fig. 10.the several turns that makeup a coil are seen in this picture. The three terminations are connected to rings on which three brushes make a sliding contact. As the rotor rotates the brushes slip over the rings and provide means of connecting stationary external circuit elements to the rotating windings. A schematic of these arrangements is shown in fig. 13. A photograph of a wound rotor for an induction machine is shown in fig. fig shows a close up of the slip ring portion. Brushes are not shown in this picture. Induction machines, which have these kinds of windings and terminals that are brought out, are called slip ring machines. The reader may note that in order that torque is produced current must flow in the rotor. To achieve that, the stationary brush terminals must either be shorted, or connected to a circuit allowing current flow. Sometimes a star connected resistor bank is connected so that the developed starting torque is higher. There are also other forms of power electronic circuitry that may be connected to the rotor terminals to achieve various functions. The popularity of the induction machine however, stems from another variety of rotor

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that is used. This rotor has slots into which copper or aluminum bars are inserted. Thesebars are then shorted by rings that are brazed on to each of the rotor ends. Figure showsa simple schematic.

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Such a rotor is called squirrel cage rotor. This rotor behaves like a short-circuited windingand hence the machine is able to perform electromechanical energy conversion. This typeof rotor is easy to manufacture, has no sliding contacts and is very robust. It is this feature that makes induction machine suitable for use even in hazardous environments and reliable operation is achieved. The disadvantage of this type of rotor is that the motor behavior cannot be altered by connecting anything to the rotor — there are no rotor terminals. Fig. 15 shows a photograph of a squirrel cage rotor. The rotor also has a fan attached to it. This is for cooling purposes. The bars ( white lines on the surface) are embedded in the rotor iron which forms the magnetic circuit. The white lines correspond to the visible portion of the rotor bar. Sometimes two rotor bars are used per slot to achieve some degree of variability in the starting and running performances. It is to make use of the fact that while high rotor

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resistance is desirable from the point of view of starting torque, low rotor resistance is desirable from efficiency considerations while the machine is running. Such rotors are called double cage rotors or deep-bar rotors.To summarize the salient features construction1. The stator of the 3 - phase induction machine consists of normal distributed AC windings.2. Balanced three phase voltages impressed on the stator, cause balanced three phase currents to flow in the stator.3. These stator currents cause a rotating flux pattern (the pattern is a flux distribution which is sinusoidal with respect to the space angle) in the air gap.4. The rotating flux pattern causes three phase induced e.m.f.s in rotor windings (again normal ac windings). These windings, if shorted, carry three phase-balanced currents. Torque is produced as a result of interaction of the currents and the air gap flux.5. The rotor may also take the form of a squirrel cage arrangement, which behaves in a manner similar to the short-circuited three phase windings.

Equivalent CircuitIt is often required to make quantitative predictions about the behavior of the induction machine, under various operating conditions. For this purpose, it is convenient to represent the machine as an equivalent circuit under sinusoidal steady state operating conditions. Since the operation is balanced, a single-phase equivalent circuit is sufficient for most purposes. In order to derive the equivalent circuit, let us consider a machine with an open circuited rotor. Since no current can flow and as a consequence no torque can be produced, the situation is like a transformer open-circuited on the secondary (rotor). The equivalent circuit under this condition can be drawn as shown in fig.

The Rotating Magnetic Field

The principle of operation of the induction machine is based on the generation of a rotating magnetic field. Let us understand this idea better. Click on the following steps in sequence to get a graphical picture. It is suggested that the reader read the text before clicking the link.

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• Consider a cosine wave from 0 to 360. This sine wave is plotted with unit amplitude.• Now allow the amplitude of the sine wave to vary with respect to time in a simisoidalfashion with a frequency of 50Hz.Let the maximum value of the amplitude is, say, 10 units. This waveform is a pulsating sine wave.

iapk = Im cos 2_.50.t

• Now consider a second sine wave, which is displaced by 120 from the first (lagging)• and allow its amplitude to vary in a similar manner, but with a 120time lag.

ibpk = Im cos(2_.50.t − 120)Similarly consider a third sine wave, which is at 240 lag. . .• and allow its amplitude to change as well with a 240 time lag. Now we have threepulsating sine waves.

icpk = Im cos(2_.50.t − 240)

Let us see what happens if we sum up the values of these three sine waves at every angle. The result really speaks about Tesla’s genius. What we get is a constant amplitude travelling sine wave! In a three phase induction machine, there are three sets of windings — phase A winding,phase B and phase C windings. These are excited by a balanced three-phase voltage supply.This would result in a balanced three phase current. Equations 1— 3 represent the currentsthat flow in the three phase windings. Note that they have a 120 time lag between them.Further, in an induction machine, the windings are not all located in the same place. They are distributed in the machine 120 away from each other (more about this in the section on alternators). The correct terminology would be to say that the windings have their axes separated in space by 120. This is the reason for using the phase A, B and C since waves separated in space as well by 120.

When currents flow through the coils, they generate mmfs. Since mmf is proportional tocurrent, these waveforms also represent the mmf generated by the coils and the total mmf. Further, due to magnetic material in the machine (iron), these mmfs generate magnetic flux,which is proportional to the mmf (we may assume that iron is infinitely permeable and non-linear effects such as hysterisis are neglected). Thus the waveforms seen above would also represent the flux generated within the machine. The net result as we have seen is a travelling flux wave. The x-axis would represent the space angle in the machine as one travels around the air gap. The first pulsating waveform seen earlier would then represent the a-phase flux, the second represents the b-phase flux and the third represents the c-phase.

This may be better visualized in a polar plot. The angles of the polar plot represent the space angle in the machine, i.e., angle as one travels around the stator bore of the machine. Click on the links below to see the development on a polar axes.

• This plot shows the pulsating wave at the zero degree axes. The amplitude is maximum at zero degree axes and is zero at 90 axis. Positive parts of the waveform are shown in red while negative in blue. Note that the waveform is pulsating at the 0−180 axis and red and blue alternate in any given side. This corresponds to the sinewave current changing polarity. Note that

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the maximum amplitude of the sine wave is reached only along the 0−180 axis. At all other angles, the amplitude does not reach a maximum of this value. It however reaches a maximum value which is less than that of the peak occurring at the 0 − 180 axis. More exactly, the maximum reached at any space angle would be equal to times the peak at the 0 − 180 axis. Further, at any space angle the time variation is sinusoidal with the frequency and phase lag being that of the excitation, and amplitude being that corresponding to the space angle.

• This plot shows the pulsating waveforms of all three cosines. Note that the first is pulsating about the 0 − 180 axis, the second about the120 − 300axis and the third at 240 − 360axis.• This plot shows the travelling wave in a circular trajectory. Note that while individual pulsating waves have maximum amplitude of 10, the resultant has amplitude of 15. If f1 is the amplitude of the flux waveform in each phase, the travelling wave can then be represented as

Principles of Torque ProductionIn the earlier section, we saw how a rotating flux is produced. Now let us consider a rotor, which is placed in this field. Let the rotor have a coil such that the coil sides are placed diametrically opposite each other. This is shown in the fig. 1. Since the flux generated by the stator rotates flux linked by this rotor coil also changes.

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Since the flux pattern is varying sinusoidal in space, as the flux waveform rotates, the flux linkage varies sinusoidal. The rate of variation of this flux linkage will then be equal to the speed of rotation of the air gap flux produced. This sinusoidal variation of the flux linkage produces a sinusoidal induced emf in the rotor coil. If the coil is short circuited, this induced emf will cause a current flow in the coil as per Lenz’s law. Now imagine a second coil on the rotor whose axis is 120 away from the first. This is shown in fig. 2. The flux linkage in this coil will also vary sinusoidal with respect to time and therefore cause an induced voltage varying sinusoidal with time. However the flux linkages in these two coils will have a phase difference of 120 (the rotating flux wave will have to travel 120 in order to cause a similar flux linkage variation as in the first coil), and hence the time varying voltages induced in the coils will also have a 120 phase difference. A third coil placed a further 120 away is shown in fig. 3. This will have a time varying induced emf lagging 240 in time with respect to the first. When these three coils are shorted upon themselves currents flow in them as per Lenz’s law. The mechanism by which torque is produced may now be understood as follows. The diagram in fig. 4 shows a view of the rotor seen from one end. Positive current is said to

flow in these coils when current flows out of the page in a, b, c conductors and into a′ , b′ and c′ respectively. If we look at the voltage induced in these coils as phasors, the diagram looks as shown in fig. The main flux is taken as the reference phasor. Considering that the induced emfis −d /dt where is the flux linkage, the diagram is drawn as shown.

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As usual, the horizontal component of these phasors gives the instantaneous values of the induced emf in these coils. Let these coils be purely resistive. Then these emf phasors also represent the currents flowing in these coils. If we consider the instant t = 0, it can be seen that1. The field flux is along 0 axis.2. The current in a phase coil is zero.3. The current in b phase coil is −√3/ 2 units.4. The current in c phase coil is +√3/2 units.These currents act to produce mmf and flux along the axes of the respective coils. Let us consider the space around b′ and c coil sides. The situation is shown in fig. The resulting flux pattern causes a tendency to move in the anticlockwise direction. This is easy to see through the so called whiplash rule. Alternatively, since the force on a current

carrying conductor is F = q(v X B), it can be seen that the torque produced tends to rotate the rotor counter-clockwise. The magnitude of the torque would increase with the current magnitude in the coils. This current is in turn dependent on the magnitude of the main field flux and its speed of rotation. Therefore one may say that motion of the main field tends to drag the rotor along with it. When the rotor is free to move and begins moving, the motion reduces the relative speed between the main field and the rotor coils. Less emf would therefore be induced and thetorque would come down. Depending on the torque requirement for the load, the difference in speed between the rotor and the main field settles down at some particular value. From the

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foregoing, the following may be noted. 1. The torque produced depends on a non-zero relative speed between the field and the rotor.

2. It is therefore not possible for the rotor to run continuously at the same speed of the field. This is so because in such a condition, no emf would be induced in the rotor and hence no rotor current, no torque. 3. The frequency of currents induced in the rotor coils and their magnitude depends onthis difference in speed. These are important conclusions. The speed of the main field is known as the synchronous speed, ns. If the actual speed of the rotor is nr then the ratio

s = N s − N r Ns

is known as slip and is frequently expressed as a percentage. Typically induction machines are designed to operate at about less than 4 percent slip at full load. It is instructive to see the situation if the rotor resistance is neglected and is considered to be purely inductive. The phasor diagram of voltages and the currents would then look as shown in fig.

At t = 0, one can see that current in a phase coil is at negative maximum, while b and c phases have positive current of 0.5 units. Now if we consider the current flux profiles at coil sides a, b′ , c, the picture that emerges is shown in fig. Since main flux at the a coil side is close to zero, there is very little torque produced from there. There is a tendency to move due to the b′ and c coil sides, but they are in opposite directions however. Hence there is no net torque on the rotor. This brings up another important conclusion — the resistance of the rotor is an important part of torque production in the induction machine. While a high resistance rotor is better suited for torque production, it would also be lossy.

Power flow diagram.

An induction motor can be basically described as a rotating transformer. Its input is a 3 phase system of voltages and currents. For an ordinary transformer, the output is electric power from the secondary windings. The secondary windings in an induction motor (the rotor) are shorted out, so no electrical output exists from normal induction motors. Instead, the output is mechanical. The relationship between the input electric power and the output mechanical power of this motor is shown below:

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The input power to an induction motor Pin is in the form of 3-phase electric voltages and currents. The first losses encountered in the machine are I2R losses in the stator windings (the stator copper loss PSCL). Then, some amount of power is lost as hysteresis and eddy currents in the stator (Pcore). The power remaining at this point is transferred to the rotor of the machine across the air gap between the stator and rotor. This power is called the air gap power PAG of the machine. After the power is transferred to the rotor, some of it is lost as I2R losses (the rotor copper loss PRCL), and the rest is converted from electrical to mechanical form (Pconv). Finally, friction and windage losses PF&W and stray losses Pmisc are subtracted. The remaining power is the output of the motor Pout.

The core losses do not always appear in the power-flow diagram at the point shown in the figure above. Because of the nature of the core losses, where they are accounted for in the machine is somewhat arbitrary. The core losses of an induction motor come partially from the stator circuit and partially from the rotor circuit. Since an induction motor normally operates at a speed near synchronous speed, the relative motion of the magnetic fields over the rotor surface is quite slow, and the rotor core losses are very tiny compared to the stator core losses. Since the largest fraction of the core losses comes from the stator circuit, all the core losses are lumped together at that point on the diagram. These losses are represented in the induction motor equivalent circuit by the resistor RC (or the conductance GC). If core losses are just given by a number (X watts) instead of as a circuit element, they are often lumped together with the mechanical losses and subtracted at the point on the diagram where the mechanical losses are located.

The higher the speed of an induction motor, the higher the friction, windage, and stray losses. On the other hand, the higher the speed of the motor (up to nsync), the lower its core losses. Therefore, these three categories of losses are sometimes lumped together and called rotational losses. The total rotational losses of a motor are often considered to be constant with changing speed, since the component losses change in opposite directions with a change in speed.

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Induction Motor Torque-Speed Characteristics

The torque-speed relationship will be examined first from the physical viewpoint of the motor’s magnetic filed behaviour and then, a general equation for torque as a function of slip will be derived from the induction motor equivalent circuit.

No-load Condition

Assume that the induction rotor is already rotating at no load conditions, hence its rotating speed is near to synchronous speed. The net magnetic field Bnet is produced by the magnetization current IM . The magnitude of IM and Bnet is directly proportional to voltage E1 . If E1 is constant, then Bnet is constant. In an actual machine, E1 varies as the load changes due to the stator impedances R1 and X1 which cause varying volt drops with varying loads. However, the volt drop at R1 and X1 is so small, that E1 is assumed to remain constant throughout.

At no-load, the rotor slip is very small, and so the relative motion between rotor and magnetic field is very small, and the rotor frequency is also very small. Since the relative motion is small, the voltage ER induced in the bars of the rotor is very small, and the resulting current flow IR is also very small. Since the rotor frequency is small, the reactance of the rotor is nearly zero, and the max rotor current IR is almost in phase with the rotor voltage ER . The rotor current produces a small magnetic field BR at an angle slightly greater than 90 degrees behind Bnet. The stator current must be quite large even at no-load since it must supply most of Bnet .

The induced torque which is keeping the rotor running, is given by:

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In terms of magnitude, the induced torque will be small due to small rotor magnetic field.On-load Conditions

As the motor’s load increases, its slip increases, and the rotor speed falls. Since the rotor speed is slower, there is now more relative motion between rotor and stator magnetic fields. Greater relative motion means a stronger rotor voltage ER which in turn produces a larger rotor current IR . With large rotor current, the rotor magnetic field BR also increases. However, the angle between rotor current and BR changes as well.

Since the rotor slip is larger, the rotor frequency rises (fr =sfe) and the rotor reactance increases (ωLR). Therefore, the rotor current now lags further behind the rotor voltage, and the rotor magnetic field shifts with the current. The rotor current now has increased compared to no-load and the angle δ has increased. The increase in BR tends to increase the torque, while the increase in angle δ tends to decrease the torque (τind is proportional to sin δ, and δ>90º). Since the first effect is larger than the second one, the overall induced torque increases to supply the motor’s increased load.

As the load on the shaft is increased, the sin δ term decreases more than the BR term increases (the value is going towards the 0 cross over point for a sine wave). At that point, a further increase in load decreases τind and the motor stops. This effect is known as pullout torque.

Modelling the torque-speed characteristics of an induction motor

Looking at the induction motor characteristics, a summary on the behaviour of torque:

Note:

a) Rotor magnetic field will increase as the rotor current will increase (provided that the rotor core is not saturated). Current flow will increase as slip increase (reduction in velocity) b) The net magnetic field density will remain constant since it is proportional to E1 (refer to equivalent induction motor equivalent circuit). Since E1 is assumed to be constant, hence Bnet will assume to be constant.c) The angle will increase as slip increases. Hence the sin value will reduce until as such that the reduction of sin d will be greater than the increase of BR (pullout torque). Since is greater than 90 degrees, as such that:

The torque speed curve may be divided into 3 regions of operations:

a) Linear region or low slip regionb) Moderate slip region located until the pullout torque level.c) High slip region

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Typical values of pullout torque would be at about 200% to 250% of the rated full load torque of the induction machine. The starting torque would be about 150% than the rated full load torque; hence induction motor may be started at full load.

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Comments on the Induction Motor Torque Speed Curve

a) Induced Torque is zero at synchronous speed.b) The graph is nearly linear between no load and full load (at near synchronous speeds).c) Max torque is known as pull out torque or breakdown torqued) Starting torque is very large.e) Torque for a given slip value would change to the square of the applied voltage.f) If the rotor were driven faster than synchronous speed, the motor would then become a generator.g) If we reverse the direction of the stator magnetic field, it would act as a braking action to the rotor – plugging.

Since Pconv may be derived as follows:

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Hence we may plot a similar characteristic to show the amount of power converted throughout the variation of load.

Torque Equation

The torque produced by three phase induction motor depends upon the following three factors:Firstly the magnitude of rotor current, secondly the flux which interact with the rotor of three phase induction motor and is responsible for producing emf in the rotor part of induction motor, lastly the power factor of rotor of the three phase induction motor.Combining all these factors together we get the equation of torque as-

Where, T is the torque produced by induction motor,φ is flux responsible of producing induced emf,I2 is rotor current,cosθ2 is the power factor of rotor circuit.The flux φ produced by the stator is proportional to stator emf E1.i.e φ ∝ E1

We know that transformation ratio K is defined as the ratio of secondary voltage (rotor voltage) to that of primary voltage (stator voltage).

Rotor current I2 is defined as the ratio of rotor induced emf under running condition , sE2 to total impedance, Z2 of rotor side,

and total impedance Z2 on rotor side is given by ,

Putting this value in above equation we get,

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We know that power factor is defined as ratio of resistance to that of impedance. The power factor of the rotor circuit is

Putting the value of flux φ, rotor current I2, power factor cosθ2 in the equation of torque we get,

Combining similar term we get,

Removing proportionality constant we get,

Where ns is synchronous speed in r. p. s, ns = Ns / 60. So, finally the equation of torque becomes,

Derivation of K in torque equation.In case of three phase induction motor, there occur copper losses in rotor. These rotor copper losses are expressed asPc = 3I2

2R2

We know that rotor current,

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Substitute this value of I2 in the equation of rotor copper losses, Pc. So, we get

The ratio of P2 : Pc : Pm = 1 : s : (1 - s)Where P2 is the rotor input,Pc is the rotor copper losses,Pm is the mechanical power developed.

Substitute the value of Pc in above equation we get,

On simplifying we get,

The mechanical power developed Pm = Tω,

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Substituting the value of Pm

We know that the rotor speed N = Ns(1 - s)Substituting this value of rotor speed in above equation we get,

Ns is speed in revolution per minute (rpm) and ns is speed in revolution per sec (rps) and the relation between the two is

Substitute this value of Ns in above equation and simplifying it we get

Comparing both the equations, we get, constant K = 3 / 2πns

Equation of Starting Torque of Three Phase Induction Motor

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Starting torque is the torque produced by induction motor when it is started. We know that at start the rotor speed, N is zero.

So, the equation of starting torque is easily obtained by simply putting the value of s = 1 in the equation of torque of the three phase induction motor,

Maximum Torque Condition for Three Phase Induction MotorIn the equation of torque,

The rotor resistance, rotor inductive reactance and synchronous speed of induction motor remains constant. The supply voltage to the three phase induction motor is usually rated and remains constant so the stator emf also remains the constant. The transformation ratio is defined as the ratio of rotor emf to that of stator emf. So if stator emf remains constant then rotor emf also remains constant.If we want to find the maximum value of some quantity then we have to differentiate that quantity with respect to some variable parameter and then put it equal to zero. In this case we have to find the condition for maximum torque so we have to differentiate torque with respect to some variable quantity which is slip, s in this case as all other parameters in the equation of torque remains constant.So, for torque to be maximum

Now differentiate the above equation by using division rule of differentiation. On differentiating and after putting the terms equal to zero we get,

Neglecting the negative value of slip we get

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So, when slip s = R2 / X2, the torque will be maximum and this slip is called maximum slip Sm and it is defined as the ratio of rotor resistance to that of rotor reactance.NOTE : At starting S = 1, so the maximum starting torque occur when rotor resistance is equal to rotor reactance.Equation of Maximum TorqueThe equation of torque is

The torque will be maximum when slip s = R2 / X2

Substituting the value of this slip in above equation we get the maximum value of torque as,

CONCLUSION : From the above equation it is concluded thatThe maximum torque is directly proportional to square of rotor induced emf at the standstill. The maximum torque is inversely proportional to rotor reactance. The maximum torque is independent of rotor resistance. The slip at which maximum torque occur depends upon rotor resistance, R2. So, by varying the rotor resistance, maximum torque can be obtained at any required slip.

From here we can say:

a) Torque is related to the square of the applied voltageb) Torque is also inversely proportional to the machine impedancesc) Slip during maximum torque is dependent upon rotor resistanced) Torque is also independent to rotor resistance as shown in the maximum torque equation.

By adding more resistance to the machine impedances, we can vary:

a) Starting torqueb) Max pull out speed

Equivalent Circuit of 3-Phase Induction Motor

In a 3-phase induction motor, the stator winding is connected to 3-phase supply and the rotor winding is short-circuited. The energy is transferred magnetically from the stator winding to the short-circuited, rotor winding. Therefore, an induction motor may be considered to be a

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transformer with a rotating secondary (short-circuited). The stator winding corresponds to transformer primary and the rotor finding corresponds to transformer secondary. In view of the similarity of the flux and voltage conditions to those in a transformer, one can expect that the equivalent circuit of an induction motor will be similar to that of a transformer. Figure shows the equivalent circuit (though not the only one) per phase for an induction motor. Let us discuss the stator and rotor circuits separately.

Stator circuit. In the stator, the events are very similar to those in the transformer primal y. The applied voltage per phase to the stator is V1 and R1 and X1 are the stator resistance and leakage reactance per phase respectively. The applied voltage V1 produces a magnetic flux which links the stator winding (i.e., primary) as well as the rotor winding (i.e., secondary). As a result, self-induced e.m.f. E1 is induced in the stator winding and mutually induced e.m.f. E'2(= s E2 = s K E1 where K is transformation ratio) is induced in the rotor winding. The flow of stator current I1 causes voltage drops in R1 and X1. V1 = -E1 + I1(R1 + j X1) ...phasor sumWhen the motor is at no-load, the stator winding draws a current I0. It has two components viz., (i) which supplies the no-load motor losses and (ii) magnetizing component Im which sets up magnetic flux in the core and the airgap. The parallel combination of Rc and Xm, therefore, represents the no-load motor losses and the production of magnetic flux respectively. I0 = Iw + Im Rotor circuit. Here R2 and X2 represent the rotor resistance and standstill rotor reactance per phase respectively. At any slip s, the rotor reactance will be s X2 The induced voltage/phase in the rotor is E'2 = s E2 = s K E1. Since the rotor winding is short-circuited, the whole of e.m.f. E'2 is used up in circulating the rotor current I'2. E'2 = I'2 (R2 + j s X2 ) The rotor current I'2 is reflected as I"2 (= K I'2) in the stator. The phasor sum of I"2 and I0 gives the stator current I1. It is important to note that input to the primary and output from the secondary of a transformer are electrical. However, in an induction motor, the inputs to the stator and rotor are electrical but the output from the rotor is mechanical. To facilitate calculations, it is desirable and necessary to replace the mechanical load by an equivalent electrical load. We then have the transformer equivalent circuit of the induction motor. It may be noted that even though the frequencies of stator and rotor currents are different, yet the magnetic fields due to them rotate at synchronous speed Ns. The stator currents produce a

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magnetic flux which rotates at a speed Ns. At slip s, the speed of rotation of the rotor field relative to the rotor surface in the direction of rotation of the rotor is= 120 f ' = 120sf =sNs P P

But the rotor is revolving at a speed of N relative to the stator core. Therefore, the speed of rotor field relative to stator core= sNs + N = (Ns - N) + N = NsThus no matter what the value of slip s, the stator and rotor magnetic fields are synchronous with each other when seen by an observer stationed in space. Consequently, the 3-phase induction motor can be regarded as being equivalent to a transformer having an air-gap separating the iron portions of the magnetic circuit carrying the primary and secondary windings. Figure shows the phasor diagram of induction motor.

Equivalent Circuit of the RotorWe shall now see how mechanical load of the motor is replaced by the equivalent electrical load. Figure shows the equivalent circuit per phase of the rotor at slip s. The rotor phase current is given by;I2’ =sE2/√(R2

2 +(sX2)2 Mathematically, this value is unaltered by writing it as:I2’ =E2/√(R2 /s) 2 +(X2)2

As shown in Figure we now have a rotor circuit that has a fixed reactance X2 connected in series with a variable resistance R2/s and supplied with constant voltage E2. Note that Figure transfers the variable to the resistance without altering power or power factor conditions.

The quantity R2/s is greater than R2 since s is a fraction. Therefore, R2/s can be divided into a fixed part R2 and a variable part (R2/s - R2) i.e.,

(R2/S) = R2 + R2(1/s-1)

Figure shows the equivalent circuit per phase of a 3-phase induction motor. Note that mechanical load on the motor has been replaced by an equivalent electrical resistance RL given by;RL = R2(1/s-1)

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Note that circuit shown in Figure is similar to the equivalent circuit of a transformer with secondary load equal to R2 given by eq. (i). The rotor e.m.f. in the equivalent circuit now depends only on the transformation ratio K (= E2/E1). Therefore; induction motor can be represented as an equivalent transformer connected to a variable-resistance load RL given by eq. (i). The power delivered to RL represents the total mechanical power developed in the rotor. Since the equivalent circuit of Figure is that of a transformer, the secondary (i.e.,rotor) values can be transferred to primary (i.e., stator) through the appropriate use of transformation ratio K. Recall that when shifting resistance/reactance from secondary to primary, it should be divided by K2 whereas current should be multiplied by K. The equivalent circuit of an induction motor referred to primary is shown in Figure.

Note that the element (i.e., R'L) enclosed in the dotted box is the equivalentelectrical resistance related to the mechanical load on the motor. The followingpoints may be noted from the equivalent circuit of the induction motor:(i) At no-load, the slip is practically zero and the load R'L is infinite. Thiscondition resembles that in a transformer whose secondary winding isopen-circuited.(ii) At standstill, the slip is unity and the load R'L is zero. This conditionresembles that in a transformer whose secondary winding is short-circuited.(iii) When the motor is running under load, the value of R'L will depend uponthe value of the slip s. This condition resembles that in a transformer whose secondary is supplying variable and purely resistive load. (iv) The equivalent electrical resistance R'L related to mechanical load is slip or speed dependent. If the slip s increases, the load R'L decreases and

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the rotor current increases and motor will develop more mechanical power. This is expected because the slip of the motor increases with the increase of load on the motor shaft.

The no load and blocked rotor tests of a three phase induction motor.

The No-Load Test

In this case the rated voltage is impressed upon the stator windings and the motor operates freely without any load. This test, therefore, is similar to the open-circuit test on the transformer except that friction and windage loss is associated with an induction motor. Since the slip is nearly zero, the impedance of the rotor circuit is almost infinite. The per-phase approximate equivalent circuit of the motor with the rotor circuit open is shown in Figure. Let Woc, Ioc, and Voc, be the power input, the input current, and the rated applied voltage on a per-phase basis under no-load condition. In order to represent the core loss by an equivalent resistance Rc we must subtract the friction and windage loss from the power input. The friction and windage loss can be measured by coupling the motor under test to another motor with a calibrated output and running it at the no-load speed of the induction motor. Let P be the friction and windage loss on a per-phasebasis. Then, the power loss in Rc is

The Blocked-Rotor Test

This test, also called the locked-rotor test, is very similar to the short-circuit test of a transformer. In this case, the rotor is held stationary by applying external torque to the shaft. The stator field winding is connected to a variable three-phase supply. The voltage is carefully increased from zero to a level at which the motor draws the rated current. At this time, the readings of the line current, the applied line voltage, and the power input are taken by using the two-wattmeter method, as illustrated in Figure. Since the rotor-circuit impedance is relatively small under blocked-rotor condition (s = l), the applied voltage is considerably lower than the rated voltage ofthe motor. Thus, the excitation current is quite small and can be neglected. Under this assumption, the approximate equivalent circuit of the motor is given in Figure 9.10 on a per-phase basis. The total series impedance isZe = R1 + R2 + j(Xl + X2) = Re + jXe,

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List of questions

1. Explain in detail, the principle of working of a three phase induction motor. (8) (N/D-11)

2. Explain in detail, the constructional details of a squirrel cage 3 phase induction motor with neat diagram. (16)

3. The power input to the rotor of 400 V, 50 Hz, 6 pole, 3 phase induction motor is 75 kW. The rotor electromotive force is observed to make 100 complete alternations per minute. Calculate slip, rotor speed, rotor copper loss per phase and mechanical power developed. (16)

4. A 6 pole induction motor is fed from 50 Hz supply. If the frequency of rotor emf at full load is 2 Hz, find full load speed and slip. (6)

5. The power input to the rotor of a 3 phase, 50 Hz, 6 pole induction motor is 80 kW. The rotor emf makes 100 complete alternations per minute. Determine the following:a)Slipb) Motor speedc) Mechanical power developedd) Rotor copper loss per phasee)Rotor resistance per phase if rotor current is 65 Af) Torque developed

6. Develop the equivalent circuit for a three phase slip ring induction motor. (16) (N/D-11)

7. In a 3 phase induction motor, the maximum torque is 2 times the full load torque and starting torque is equal to the full load torque. Calculate the full load speed and speed at which maximum torque occurs. (16) (A/M-05)

8. Explain in detail, the slip-torque characteristic of three phase induction motor. (8) (N/D-11)

9. Explain the effect of varying rotor resistance of an induction motor on its speed-torque characteristics. (8)

10. Derive the condition for maximum starting torque of a three phase induction motor. (8)

11. Derive the various torque equations of a three phase induction motor. (16)

12.A 3300 V, 10 pole, 50 Hz three phase star connected induction motor has slip ring rotor resistance per phase = 0.015 Ω and standstill reactance per phase = 0.25 Ω. If the motor runs at 2.5 percent slip on full load, determine the following:a) Speed of the motorb) Speed at which the torque will be maximumc) The ratio of maximum torque to full load torque (16)

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13. Explain the power stages on a 3 phase induction motor and also their relationships with gross torque Tg and overall efficiency. (8) (A/M-05)

14. A 220 V, 3 phase, 4 pole, 50 Hz, star connected induction motor is rated at 3.73 kW. The equivalent circuit parameters are, R1 = 0.45 Ω, X1 = 0.8 Ω, R2’ = 0.4 Ω, X2’ = 0.8 Ω and B0 = – 1/30 mho. The stator core loss is 50 W and rotational losses are 150 W. For a slip of 4 %, find the input current, power factor, air gap power, mechanical power, electromagnetic torque, output power and efficiency. Also draw the equivalent circuit and mark the given parameters.

(16) (N/D-06)

15. A 3 phase, 400 V induction motor gave the following test readings:No load test : 400 V; 1250 W; 9 ABlocked rotor test : 150 V; 4 kW; 38 ADraw the circle diagram. If the normal rating is 14.91 kW, calculate the full load current and slip from the circle diagram. (16) (A/M-05)

16.In a 6 pole, 3 phase 50 Hz motor with star connection, the rotor resistance per phase is 0.03 Ω, reactance at standstill is 1.5 Ω per phase and emf between the slip rings on O.C. is 175 V. Calculate the following:a)Slip at speed of 950 rpm b)Rotor emf per phasec)Rotor frequency d)Reactance at a speed of 950 rpm (16)

17. Draw the circle diagram for a 20 HP, 50 Hz, 3 phase star connected induction motor with the following data:No load test : 400 V; 9 A and pf 0.2 laggingBlock rotor test : 200 V; 50 A and pf 0.4 laggingDetermine the line current and efficiency at full load. Assume stator and rotor copper losses on blocked rotor to be equal. (16)

18. Explain in detail, the construction and working principle of a double cage induction motor.

UNIT IV – STARTING AND SPEED CONTROL OF

THREE PHASE INDUCTION MOTOR

PART-B

Starting Of Induction Motor.

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A 3-phase induction motor is theoretically self-starting. The stator of an induction motor consists of 3-phase windings, which when connected to a 3-phase supply creates a rotating magnetic field. This will link and cut the rotor conductors which in turn will induce a current in the rotor conductors and create a rotor magnetic field. The magnetic field created by the rotor will interact with the rotating magnetic field in the stator and produce rotation.

Therefore, 3-phase induction motors employ a starting method not to provide a starting torque at the rotor, but because of the following reasons;

1) Reduce heavy starting currents and prevent motor from overheating.

2) Provide overload and no-voltage protection.

There are many methods in use to start 3-phase induction motors. Some of the common methods are;

Direct On-Line Starter (DOL)

Star-Delta Starter

Auto Transformer Starter

Rotor Impedance Starter

Power Electronics Starter

At the time of starting, the rotor speed is zero and the per-unit slip is unity. Therefore, the starting current, from the approximate equivalent circuit

I2s = V1/(Re = j Xe)

where R, = R, + R, and X, = X, + X,. The corresponding value of the starting torque is

T = 3V12 / (ω(R2+X2)

Since the effective rotor resistance, R,, is very small at the time of starting compared with its value at rated slip, R,/s, the starting current may be as much as 400% to 800% of the full-load urrent. On the other hand, the starting torque may only be 200% to 350% of the full-load torque. Such a high starting current is usually unacceptable because it results in an excessive line voltage drop which, in turn, may affect the operation of other machines operating on the same powersource. Since the starting current is directly proportional to the applied voltage, Equation suggests that the starting current can be reduced by impressing a low voltage across motor terminals at the time of starting. However, it is evident from Equation that a decrease in the applied voltage results in a decline in the starting torque.

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Therefore, we can employ the low-voltage starting only for those applications that do not require high starting torques. For instance, a fan load requires almost no starting torque except for the loss due to friction. The induction motor driving a fan load can be started using low-voltage starting. The starting current can also be decreased by increasing the rotor resistance. As mentioned earlier, an increase in the rotor resistance also results in an increase in the starting torque which, of course, is desired for those loads requiring high starting torques. However, a high rotor resistance (a) reduces the torque developed at full load, (b) produces high rotor copper loss, and (c) causes a reduction in motor efficiency at full load. These drawbacks, however, do not represent a problem for wound-rotor motors. For these motors, we can easily incorporate high external resistance in series with the rotor windings at the time of starting and remove it when the motor operates at full load.

For rotors using squirrelcage winding (die-cast rotors), the change in resistance from a high value at starting to a low value at full load is accomplished by using quite a few different designs, as shown in Figure 9.12. In each design, the underlying principle is to achieve a high rotor resistance at starting and a low rotor resistance at the rated speed. At starting, the frequency of the rotor is the same as the frequency of the applied source. At full load, however, the rotorfrequency is very low (usually less than 10 Hz). Thus, the skin effect is more pronounced at starting than at full load. Hence, the rotor resistance is higher at

starting than at full load owing to the skin effect alone. Also, as the currents are induced in the rotor bars, they produce a secondary magnetic field. Part of the secondary magnetic field links only the rotor conductor and manifests itself as the leakage flux. The leakage flux increases as we move radially away from the air-gap toward the shaft, and is quite significant at starting. Thus, in a multicage rotor at starting, the inner cage presents a high leakage reactance compared with the outer cage. Owing to the high leakage reactance of the inner cage, the rotor current tends to confine itself in the outer cage. If the cross-sectional area of the outer cage is smaller than that of the inner cage, it presents a comparatively high resistance at starting. When the

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motor operates at full load, the rotor frequency is low. Thereby, the leakage flux is low. In this case, the current tends to distribute equally among all the cages. As a result, the rotor resistance is low.

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Another technique that is commonly used to increase the rotor resistance and lessen the effects of harmonics in an induction motor is called skewing. In this case, the rotor bars are skewed with respect to the rotor shaft, as illustrated in Figure 9.15. Skew is usually given in terms of bars. The minimum skew must be one bar to avoid cogging. Skews of more than one bar are commonly used.

Rotor Impedance Starter

This method allows external resistance to be connected to the rotor through slip rings and rushes. Initially, the rotor resistance is set to maximum and is then gradually decreased as the motor peed increases, until it becomes zero. The rotor impedance starting mechanism is usually very bulky and expensive when compared with other methods. It also has very high maintenance costs. Also, a considerable amount of heat is generated through the resistors when current runs through them. The starting frequency is also limited in this method. However, the rotor impedance method allows the motor to be started while on load. Figure shows the connection of a 3phase induction motor with rotor resistance starter.

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Direct On-Line Starter (DOL)

The Direct On-Line (DOL) starter is the simplest and the most inexpensive of all starting methods and is usually used for squirrel cage induction motors. It directly connects the contacts of the motor to the full supply voltage. The starting current is very large, normally 6 to 8 times the rated current. The starting torque is likely to be 0.75 to 2 times the full load torque. 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

There are safety mechanisms inside the DOL starter which provides protection to the motor as well as the operator of the motor.The power and control circuits of induction motor with DOL starter are shown in figure.

* K1M Main contactorThe DOL starter consists of a coil operated contactor

K1M controlled by start and stop push buttons. On pressing the start push button S1, the contactor coil K1M is energized from line L1. The three mains contacts (1-2), (3-4), and (5-6) in fig. (1) are closed. The motor is thus connected to the supply. When the stop push

button S2 is pressed, the supply through the contactor K1M is disconnected. Since the K1M is de-energized, the main contacts (1-2), (3-4), and (5-6) are opened. The supply to motor is disconnected and the motor stops

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Star-Delta Starter

The star delta starting is a very common type of starter and extensively used, compared to the other types of the starters. This method used reduced supply voltage in starting. Figure(2) shows the connection of a 3phase induction motor with a star – delta starter.

The method achieved low starting current by first connecting the stator winding in star configuration, and then after the motor reaches a certain speed, throw switch changes the winding arrangements from star to delta configuration.

By connecting the stator windings, first in star and then in delta, the line current drawn by the motor at starting is reduced to one-third as compared to starting current with the windings connected in delta. At the time of starting when the stator

where V is the line voltage. Since the torque developed windings are start connected, each stator phase gets voltage

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by an induction motor is proportional to the square of the applied voltage, star- delta starting reduced the starting torque to one – third that obtainable by direct delta starting.

K2M Main Contactor K3M Delta Contactor K1M Star Contactor

F1 Thermal Overload Relay

Auto Transformer Starter

The operation principle of auto transformer method is similar to the star delta starter method. The starting current is limited by (using a three phase auto transformer) reduce the initial stator applied voltage.

The auto transformer starter is more expensive, more complicated in operation and bulkier in construction when compared with the star – delta starter method. But an auto transformer starter is suitable for both star and delta connected motors, and the starting current and torque can be adjusted to a desired value by taking the correct tapping from the auto transformer. When the star delta method is considered, voltage can be adjusted only by factor of.

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Figure (3) shows the connection of a 3phase induction motor with auto transformer starter.

Speed control of induction motors

The speed of an induction motor is given as

N = 120f/p (1-S).

So obviously the speed of an induction motor can be controlled by varying any of three factors namely supply frequency f, number of pole P or slip S.

1. Pole changing

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2. Stator voltage control

3. Supply frequency control

4. Rotor resistance control

5. Slip energy recovery.

The basic principles of these methods are described below

Pole changing

The number of stator poles can be change by

Multiple stator windings

Method of consequent poles

Pole amplitude modulation (PWM)

The methods of speed control by pole changing are suitable for cage motors only because the cage rotor automatically develops number of poles equal to the poles of stator winding.

Multiple stator windings

In this method the stator is provided with two separate windings which are wound for two different pole numbers. One winding is energized at a time. Suppose that a motor has two windings for 6 and 4 poles. For 50 Hz supply the synchronous speed will be 1000 and 1500 rpm respectively. If the full load slip is 5% in each case, the operating speeds will be 950 rpm and 1425 rpm respectively. This method is less efficient and more costly, and therefore, used only when absolutely necessary.

Method of consequent poles

In this method a single stator winding is divided into few coil groups. The terminals of all these groups are brought out. The number of poles can be changed with only simple changes in coil connections. In practice, the stator winding is divided only in two coil groups. The number of poles can be changed in the ratio of 2:1.

Fig.(1) shows one phase of a stator winding consisting of 4 coils divided into two groups a – b and c – d. Group a – b consists of odd numbered coils(1,3) and connected in series. Group c – d has even numbered coils (2, 4) connected in series. The terminals a,b,c,d are taken out as shown.

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The coils can be made to carry current in given directions by connecting coil groups either in series or parallel shown in fig. (1-b) and fig.(1-c) respectively.With this connection, there will be a total of 4 poles giving a synchronous speed of 1500 rpm for 50 Hz system. If the current through the coils of group a – b is reversed (fig.2), then all coils will produce north (N) poles.

In order to complete the magnetic path, the flux of the pole groups must pass through the spaces between the groups, thus inducing magnetic poles of opposite polarity (S poles) in the inter – pole spaces.

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Stator Voltage Control

The torque developed by an induction motor is proportional to the square of the applied voltage. The variation of speed torque curves with respect to the applied voltage is shown in fig.(3). These curves show that the slip at maximum torque sm remains same, while the value of stall torque comes down with decrease in applied voltage.

Further, we also note that the starting torque is also lower at lower voltages. Thus, even if a given voltage level is sufficient for achieving the running torque, the machine may not start. This method of trying to control the speed is best suited for loads that require very little starting torque,ଶ but their torque requirement may increase with speed T.

Supply Frequency Control

The synchronous speed of an induction motor is given by

N = 120f / p

The synchronous speed and, therefore, the speed of motor can be controlled by varying the supply frequency.

The emf induced in the stator of an induction motor is given byTherefore, if the supply frequency is change, E1 will also change to maintain the same air gap flux. If the stator voltage drop is neglected the terminal voltage V1 is equal to E1 . in order to avoid saturation and to minimize losses, motor is operated at rated air gap flux by varying terminal voltage with frequency so as to maintain ( v/f ) ratio constant at rated value.

This type of control is known as constant volt in per hertz. Thus, the speed control of an induction motor using variable frequency supply requires a variable voltage power source.

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Injecting an EMF in the Rotor CircuitThe speed of a wound-rotor induction motor can also be changed by injecting an emf in the rotor circuit, as shown in Figure 9.19. For proper operation, the frequency of the injected emf must be equal to the rotor frequency. However, there is no restriction on the phase of the injected emf. If the injected emf is in phase with the induced emf in the rotor, the rotor current increases. In this case, the rotor circuit manifests itself as if it has a low resistance. On the other hand, if the injected emf is in phase opposition to the induced emf in the rotor circuit, the rotor current decreases. The decrease in the rotor current is analogous to the increase in the rotor resistance. Thus, changing the phase of the injected voltage is equivalent to changing the rotor resistance. The change in the rotor resistance is accompanied by the change in the operating speed of the motor. Further control in the speed can also be achieved by varying the magnitude of the injected emf.

Rotor Resistance Control

We have already discussed the effect of changes in the rotor resistance on the Speed-torque characteristic of an induction motor. This method of speed control is suitable only for wound-rotor induction motors. The operating speed of the motor can be decreased by adding external resistance in the rotor circuit. However, an increase in the rotor resistance causes (a) an increase in the rotor copper loss, (b) an increase in the operating temperature of the motor, and (c) a reduction in the motor efficiency. Because of these drawbacks, this method of speed controlcan be used only for short periods.

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This method allows external resistance to be connected to the rotor through slip rings and brushes. Initially, the rotor resistance is set to maximum and is then gradually decreased as the motor speed increases, until it becomes zero.

The rotor impedance starting mechanism is usually very bulky and expensive when compared with other methods. It also has very high maintenance costs. Also, a considerable amount of heat is generated through the resistors when current runs through them. The starting frequency is also limited in this method. However, the rotor impedance method allows the motor to be started while on load. Figure (4) shows the connection of a 3phase induction motor with rotor resistance starter.

In wound rotor induction motor, it is possible to change the shape of the torque – speed curve by inserting extra resistance into rotor circuit of the machine. The resulting torque – speed characteristic curves are shown in fig.(4).

This method of speed control is very simple. It is possible to have a large starting torque and low starting current at small value of slip.

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The major disadvantage of this method is that the efficiency is low due to additional losses in resistors connected in the rotor circuit. Because of convenience and simplicity, it is often employed when speed is to be reduced for a short period only (cranes).

Cascade operation of induction motors to obtain variable speed.

The power drawn from the rotor terminals could be spent more usefully. Apart from using the heat generated in meaning full ways, the slip ring output could be connected to another induction machine. The stator of the second machine would carry slip frequency currents of the first machine which would generate some useful mechanical power. A still better option would be to mechanically couple the shafts of the two machines together. This sort of a connection is called cascade connection and it gives some measure of speed control as shown below.

Let the frequency of supply given to the first machine be f1 , its number poles be p1, and its slip of operation be s1 . Let f2, p2 and s2 be the corresponding quantities for the second machine. The frequency of currents flowing in the rotor of the first machine and hence in the stator of the second machine is s1f1. Therefore f2 = s1f1. Since the machines are coupled at the shaft, the speed of the rotor is common for both. Hence, if n is the speed of the rotor in radians,

n =f1

(1 − s1) = ±s1f1

(1 − s2).p1 p2

Note that while giving the rotor output of the first machine to the stator of the second, the resultant stator mmf of the second machine may set up an air-gap flux which rotates in the same direction as that of the rotor, or opposes it. this results in values for speed as

n =

f1

or n =

f1

(s2 negligible)p1 + p2 p1 − p2

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The latter expression is for the case where the second machine is connected in opposite phase sequence to the first. The cascade connected system can therefore run at two possiblespeeds.

Speed control through rotor terminals can be considered in a much more general way. Consider the induction machine equivalent circuit of fig. , where the rotor circuit has been terminated with a voltage source Er .

If the rotor terminals are shorted, it behaves like a normal induction machine. This is equivalent to saying that across the rotor terminals a voltage source of zero magnitude is connected. Different situations could then be considered if this voltage source Er had a non-zero magnitude. Let the power consumed by that source be Pr . Then considering the rotor side circuit power dissipation per phase

sE1I2′ cos φ2 = I2

′ R2′ + Pr .

Clearly now, the value of s can be changed by the value of Pr . For Pr = 0, the machine is like a normal machine with a short circuited rotor. As Pr becomes positive, for all other circuit conditions remaining constant, s increases or in the other words, speed reduces. As Pr becomes negative,the right hand side of the equation and hence the slip decreases. The physical interpretation is that we now have an active source connected on the rotor side which is able to supply part of the rotor copper losses. When Pr = −I2′2R2 the entire copper loss is supplied by the external source. The RHS and hence the slip is zero. This corresponds to operation at synchronous speed. In general the circuitry connected to the rotor may not be a simple resistor or a machine but a power electronic circuit which can process this power requirement. This circuit may drive a machine or recover power back to the mains. Such circuits are called static kramer drives.

List of question

1. With neat diagrams explains the working of any two types of starters used for squirrel cage type 3 phase induction motor. 2. Discuss the various starting methods of induction motors.3. Explain the different speed control methods of phase wound induction motor.4. Explain the various schemes of starting squirrel cage induction motor. 5. Explain the speed control of 3 phase squirrel cage induction motor by pole changing. 6. Discuss the theory of star – delta starter. 7. Explain briefly the various speed control schemes of induction motors. 8. Explain in detail the slip power recovery scheme. 9. Explain the various techniques of sped control of induction motor from rotor side control. 10. Explain the cascade operation of induction motors to obtain variable speed.

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UNIT V – SINGLE PHASE INDUCTION MOTORS AND SPECIAL MACHINES

PART-B

Single phase motor.

Introduction

As the name suggests, these motors are used on single-phase supply. Single-phase motors are the most familiar of all electric motors because they are extensively used in home appliances, shops, offices etc. It is true that single-phase motors are less efficient substitute for 3-phase motors but 3-phase power is normally not available except in large commercial and industrial establishments. Since electric power was originally generated and distributed for lighting only, millions of homes were given single-phase supply. This led to the development of single-phase motors. Even where 3-phase mains are present, the single-phase supply may be obtained by using one of the three lines and the neutral. In this chapter, we shall focus our attention on the construction, working and characteristics of commonly used single-phase motors.

A single phase induction motor is very similar to a 3-phase squirrel cage induction motor. It has (i) a squirrel-cage rotor identical to a 3-phase motor and (ii) a single-phase winding on the stator.

Unlike a 3-phase induction motor, a single-phase induction motor is not self-starting but requires some starting means. The single-phase stator winding produces a magnetic field that pulsates in strength in a sinusoidal manner. The field polarity reverses after each half cycle but the field does not rotate. Consequently, the alternating flux cannot produce rotation in a stationary squirrel-cage rotor. However, if the rotor of a single-phase motor is rotated in one direction by some mechanical means, it will continue to run in the direction of rotation. As a matter of fact, the rotor quickly accelerates until it reaches a speed slightly below the synchronous speed. Once the motor is running at this speed, it will continue to rotate even though single-phase current is flowing through the stator winding. This method of starting is generally not convenient for large motors. Nor can it be employed fur a motor located at some inaccessible spot.

Fig.shows single-phase induction motor having a squirrel cage rotor and a single phase distributed stator winding. Such a motor inherently docs not develop any starting torque and, therefore, will not start torotate if the stator winding is connected to single-phase a.c. supply. However, if the Fig. rotor is started by auxiliary means, the motor will quickly attain me final speed. This strange behaviour of single-phase induction motor can be explained on the basis of double-field revolving theory.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.

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Double-Field Revolving Theory

The double-field revolving theory is proposed to explain this dilemma of no torque at start and yet torque once rotated. This theory is based on the fact that an alternating sinusoidal flux can be represented by two revolving fluxes, each equal to one-half of the maximum value of alternating flux and each rotating at synchronous speed in opposite directions.

The above statement will now be proved. The instantaneous value of flux due to the stator current of a single-phase induction motor is given by;

=mcosωt

Consider two rotating magnetic fluxes 1 and 2 each of magnitude m/2 and rotating in opposite directions with angular velocity ω. Let the two fluxes start rotating from OX axis at

t = 0. After time t seconds, the angle through which the flux vectors have rotated is at. Resolving the flux vectors along-X-axis and Y-axis, we have

Thus the resultant flux vector is = mcosωt along X-axis. Therefore, an alternating field can be replaced by two relating fields of half its amplitude rotating in opposite directions at synchronous speed. Note that the resultant vector of two revolving flux vectors is a stationary vector that oscillates in length with time along X-axis. When the rotating flux vectors are in phase. The resultant vector is = m; when out of phase by 180° the resultant vector = 0.

Let us explain the operation of single-phase induction motor by double-field revolving theory.

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PRINCIPLE OF OPERATION Suppose the rotor is at rest and 1-phase supply is given to stator winding. The current flowing in the stator winding gives rise to an mmf whose axis is along the winding and it is a pulsating mmf, stationary in space and varying in magnitude, as a function of time, varying from positive maximum to zero to negative maximum and this pulsating mmf induces currents in the short-circuited rotor of the motor which gives rise to an mmf. The currents in the rotor are induced due to transformer action and the direction of the currents is such that the mmf so developed opposes the stator mmf. The axis of the rotor mmf is same as that of the stator mmf. Since the torque developed is proportional to sine of the angle between the two mmf and since the angle is zero, the net torque acting on the rotor is zero and hence the rotor remains stationary.

For analytical purposes a pulsating field can be resolved into two revolving fields of constant magnitude and rotating in opposite directions as shown in Fig. and each field has a magnitude equal to half the maximum length of the original pulsating phasor. These component waves rotate in opposite direction at synchronous speed. The forward (anticlockwise) and backward-rotating (clockwise) mmf waves f and b are shown in Fig. In case of 3-phase induction motor there is only one forward rotating magnetic field and hence torque is developed and the motor is self-starting. However, in single phase induction motor each of these component mmf waves produces induction motor action but the corresponding torques are in opposite direction. With the rotor at rest the forward and backward field produce equal torques but opposite in direction and hence no net torque is developed on the motor and the motor remains stationary. If the forward and backward air gap fields remained equal when the rotor is revolving, each of the component fields would produce a torque-speed characteristic

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similar to that of a polyphase induction motor with negligible leakage impedance as shown by the dashed curves f and b in Fig.

The resultant torque-speed characteristic which is the algebraic sum of the two component curves shows that if the motor were started by auxiliary means it would produce torque in what-ever direction it was started.

Assume that the rotor is started by spinning the rotor or by using auxiliary circuit, in say clockwise direction. The flux rotating in the clockwise direction is the forward rotating flux and that in the other direction is the backward rotating flux. The slip w.r.t. the forward flux will be

The rotor rotates opposite to the rotation of the backward flux. Therefore, the slip w.r.t the backward flux will be

Thus for forward rotating flux, slip is s (less than unity) and for backward rotating flux, the slip is 2-s (greater than unity) since for usual rotor resistance/reactance ratios, the torque at slips of less than unity are greater than those at slips of more than unity, the resultant torque will be in the direction of the rotation of the forward flux. Thus if the motor is once started, it will develop net torque in the direction in which it has been started and will function as a motor.

Making Single-Phase Induction Motor Self-Starting

The single-phase induction motor is not self-starting and it is undesirable to resort tomechanical spinning of the shaft or pulling a belt to start it. To make a single-phaseinduction motor self-starting, we should somehow produce a revolving stator magneticfield. This may be achieved by converting a single-phase supply into two-phase supplythrough the use of an additional winding. When the motor attains sufficient speed, thestarting means (i.e., additional winding) maybe removed depending upon the type of themotor. As a matter of fact, single-phase induction motors are classified and namedaccording to the method employed to make them self-starting.

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(i) Split-phase motors-started by two phase motor action through the use of an auxiliary or starting winding.

(ii) Capacitor motors-started by two-phase motor action through the use of an auxiliary winding and a capacitor.

(iii) Shaded-pole motors-started by the motion of the magnetic field produced by means of a shading coil around a portion of the pole structure.

1. Split – phase Induction Motor

The stator of a split – phase induction motor has two windings, the main winding and the auxiliary winding. These windings are displaced in space by 90 electric degrees as shown in figure (4-a).

The auxiliary winding is made of thin wire so that it has a high R/X ratio as compared to the main winding which has thick super enamel copper wire. When the two stator windings areenergized from a single – phase supply, the current Im and Ia in the main winding and auxiliary winding lag behind the supply voltage V, and Ia leading the current Im as shown in figure (4-b).

This means the current through auxiliary winding reaches maximum value first and the mmf or flux due to Ia lies along the axis of the auxiliary winding and after some time the current Im reaches maximum value and the mmf due to Im lies along the main winding axis. Thus the motor becomes a 2 – phase unbalanced motor. Because of these two fields a starting torque is

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developed and the motor becomes a self starting motor. After the motor starts, the auxiliary winding is disconnected usually by means of centrifugal switch that operates at about 75% of synchronous speed. Finally the motor runs because the main winding. Since this being single phase some level of humming noise is always associated with the motor during running. The power rating of such motors generally lies between 60- 250W.Characteristics

Due to their low cost, split – phase induction motors are most popular single – phase motors in the market

Since the starting winding is made of thin wire, the current density is high and the winding heats up quickly. If the starting period exceeds 5 seconds, the winding may burn out unless the motor is protected by built – in thermal relay. This motor is, therefore, suitable where starting periods are not frequent.

2. Capacitor – Start Motor

Capacitors are used to improve the starting and

running performance of the single phase inductions motors.

The capacitor – start motor is identical to a split – phase motor except that the starting winding has as many turns as the main winding. Moreover, a capacitor C is connected in series with the starting winding as shown in figure (5-a).

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The value of capacitor is so chosen that Ia leads Im by about 90o (Fig.5-b) so that the starting torque is maximum for certain values of Ia and Im. 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.

The typical torque – speed characteristic is shown in fig (5-c).

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Characteristics

Although starting characteristics of a capacitor – start motor are better than those of a split – phase motor, both machines possess the same running characteristics because the main windings are identical.

The phase angle between the two currents is about 90

compared to about 25o in a split – phase motor. Consequently, for the same starting torque, the current in the starting winding is only about half that in a split – phase motor. Therefore, the starting winding of a capacitor start motor heats up less quickly and is well suited to applications involving either frequent or prolonged starting periods.

Capacitor – start motors are used where high starting torque is required and where high starting period may be long e.g. to drive:

a)Compressorsb) large fans c) pumps d)high inertia loads

The power rating of such motors lies between 120W and 0.75 kW.

Capacitor-Start Capacitor-Run Motor

This motor is identical to a capacitor-start motor except that starting winding is not opened after starting so that both the windings remain connected to the supply when running as well as at starting. Two designs are generally used.

(i) In one design, a single capacitor C is used for both starting and running as shown in Fig. This design eliminates the need of a centrifugal switch and at the same time improves the power factor and efficiency of the motor.

(ii) In the other design, two capacitors C1 and C2 are used in the starting winding as shown in Fig. The smaller capacitor C1 required for optimum running conditions is permanently connected in series with the starting winding. The much larger capacitor C2 is connected in parallel with C1 for optimum starting and remains in the circuit during starting. The starting capacitor C1 is disconnected when the motor approaches about 75% of synchronous speed. The motor then runs as a single-phase induction motor.

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Characteristics

(i) The starting winding and the capacitor can be designed for perfect 2-phase operation at any load. The motor then produces a constant torque and not a pulsating torque as in other single-phase motors.

(ii) Because of constant torque, the motor is vibration free and can be used in: (a) hospitals (6) studios and (c) other places where silence is important.

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 fig

When alternating current flow in the field winding, an alternating flux is produced in the field core. A portion of this flux links with the shading coil, which behaves as short – circuited secondary of a transformer. A voltage is induced in the shading coil, and this voltage circulates a current in it. The induced current produces a flux called the induced flux which opposes the main core flux. The shading coil, thus, causes the flux in the shaded portion to lag behind the flux in the unshaded portion of the pole. At the same time, the main flux and the shaded pole flux are

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displaced in space. This displacement is less than 90o. Since there is time and space displacement between the two fluxes, the conditions for setting up a rotating magnetic field are produced. Under the action of the rotating flux a starting torque is developed on the cage rotor. The direction of this rotating field (flux) is from the unshaded to the shaded portion of the pole.

The typical torque-speed characteristic is shown in fig.

Characteristic

The salient features of this motor are extremely simple construction and absence of centrifugal switch

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Since starting torque, efficiency and power factor are very low, these motors are only suitable for low power applications e.g. to drive: Small fans b) toys c) hair driers. The power rating of such motors is up to about 30 W.

Equivalent Circuit of Single – Phase Induction Motor

When the stator of single phase induction motor is connected to single – phase supply, the stator current produces a pulsating flux. According to the double – revolving field theory, the pulsating air – gap flux in the motor at standstill can be resolved into two equal and opposite fluxes with the motor. Since the magnitude of each rotating flux is one – half of the alternating flux, it is convenient to assume that the two rotating fluxes are acting on two separate rotors. Thus, a single – phase induction motor may be considered as consisting of two motors having a common stator winding and two imaginary rotors, which rotate in opposite directions. The standstill impedance of each rotor referred to the main stator winding isZ = r2ˊ/2 +jX2ˊ/2

The equivalent circuit of single – phase induction motor at standstill is shown in fig.(9).

R1m = resistance of stator winding

X 1m = leakage reactance of stator winding

XM = total magnetizing reactance

r2ˊ = resistance of rotor referred to the stator

X2ˊ = leakage reactance of rotor referred to the stator

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In this diagram, the portion of the equivalent circuit representing the effects of air gap flux is split into two portions. The first portion shows the effect of forward rotating flux, and the second portion shows the effect of backward rotating flux.

The forward flux induces a voltage Emf in the main stator winding. The backward rotating flux induces a voltage Emb in the main stator winding. The resultant induced voltage in the main stator winding is Em , where

Em = Emf + Emb

At standstill, Emf = Emb

Now suppose that the motor is started with the help of an auxiliary winding. The auxiliary winding is switched out after the motor gains it normal speed.

The effective rotor resistance of an induction motor depends on the slip of the rotor. The slip of the rotor with respect to the forward rotating flux is S. The slip the rotor with respect to the backward rotating flux is (2-S).

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When the forward and backward slips are taken in account, the result is the equivalent circuit shown in fig which represents the motor running on the main winding alone.

The simplified equivalent circuit of single – phase induction motor with only main winding energized is shown in fig.

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Hysteresis motor.A hysteresis motor is a synchronous motor without salient (or projected) poles and without dc excitation which starts by virtue of the hysteresis losses induced in its hardened steel secondary member by the revolving filed of the primary and operates normally at synchronous speed and runs on hysteresis torque because of the retentivity of the secondary core. It is a single-phase motor whose operation depends upon the hysteresis effect i.e., magnetization produced in a ferromagnetic material lags behind the magnetizing force.

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Working Principle

When stator is energized, it produces rotating magnetic field. The main and auxiliary, both the windings must be supplied continuously at start as well as in running conditions so as to maintain the rotating magnetic field. The rotor, initially, starts to rotate due to eddy-current torque and hysteresis torque developed on the rotor. Once the speed is near about the synchronous, the stator pulls rotor into synchronism. In such case, as relative motion between stator field and rotor field vanishes, so the torque due to eddy-currents vanisehes. When the rotor is rotating in the synchronous speed, the stator revolving filed flux produces poles on the rotor as shown in Fig. Due to the hysteresis effect, rotor pole axis lags behind the axis of rotating magnetic field. Due to this, rotor poles get attracted towards the moving stator poles. Thus rotor gets subjected to torque called hysteresis torque. This torque is constant at all speeds.

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When the stator field moved forward, due to high residual magnetism (i.e. retentivity) the rotor pole strength remains maintained. So higher the retentivity, higher is the hysteresis torque. The hyteresis torque is independent of the rotor speed.

The high retentivity ensures the continuous magnetic locking between stator and rotor. Due to principle of magnetic locking, the motor either rotates at synchronous speed or not at all. Only hysteresis torque is present which keeps rotor running at synchronous speed.

stepper motor.

A stepper motor is an electro mechanical device, which converts electrical pulses into discrete mechanical movements. The shaft or spindle of a stepper motor rotates in discrete step increments when electrical command pulses are applied to it in the proper sequence. The sequence of the applied pulses is directly related to the direction of motor shafts rotation. The speed of the motor shafts rotation is directly related to the frequency of the input pulses and the length of rotation of input pulses applied.

3. Stepper Motor Types

There are three basic stepper motor types. They are:

Variable-reluctance Permanent-magnet Hybrid Variable-reluctance (VR)

This type of stepper motor has been around for a long time. It is probably the easiest to understand from a structural point of view. Figure 1 shows a cross section of a typical V.R. stepper motor. This type of motor consists of a soft iron multi-toothed rotor and wound stator. When the stator windings are energized with DC current the poles become magnetized. Rotation occurs when the rotor teeth are attracted to energized stator poles.

Permanent Magnet (PM)

Often referred as a “tin can” or “canstack” motor the permanent magnet step motor is a low cost and low-resolution type motor with typical step angles of 7.50 to 150. PM motors as the name implies have permanent magnets added to the motor structure. The rotor no longer has teeth as with the VR motor. Instead the rotor is magnetized with alternating north and south poles situated in a straight line parallel to the rotor shaft. These magnetized motor poles provide increased magnetic flux intensity and because of this the PM motor exhibits improved torque characteristics when compared with the VR type.

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Figure 1. variable reluctance (VR) motor Figure 2. Principle of a PM or tincan motor

Torque Generation

The torque produced by a stepper motor depends on several factors.

The step rate

The drive current in the windings The drive design or type

In a stepper motor a torque is developed when the magnetic fluxes of the rotor and stator are displaced from each other. The stator is made up of high permeability magnetic material. The presence of this high permeability material causes the magnetic flux to be confined for the most part to the paths defined by the stator structure in the same fashion that currents are confined to the conductors of an electronic circuit. This serves to concentrate the flux at the stator poles. The torque output produced by the motor is proportional to the intensity of the magnetic flux generated when the winding is energized.

The basic relationship which defines the intensity of the magnetic flux is defined by:

H = (N ×i) ÷1 Where

N = The number of winding turns i = current

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H = Magnetic field intensity l = Magnetic flux path length

This relationship shows that the magnetic flux intensity and consequently the torque is proportional to the number of winding turns and the current and inversely proportional the lengths of the magnetic flux path. From this basic relationship one can see that the same frame size stepper motor could have very different torque output capabilities simply by changing the winding parameters. More detailed information on how the capability of the motor can be found in the application note entitled “Drive Circuit Basics”.

8. Phase, Poles and Stepping Angles

Usually stepper motors have two phases, but three – and five-phase motors also exist.

A bipolar motor with two phases has one winding/phase and a unipolar motor has one winding, with a center tap per phase. Sometimes the unipolar stepper motor is referred to as a “four-phase motor”, even though it only has two phases.

Motors that have two separate windings per phase also exist- these can be driven in either bipolar or unipolar mode.

A pole can be defined as one of the regions in a magnetized body where the magnetic flux density is concentrated. Both the rotor and the stator of a step motor have poles. Figure 5 contains a simplified picture of a two-phase stepper motor having 2 poles for each phase on the stator and 2 poles on the rotor. In reality several more poles are added to both the rotor and stator structure in order to increase the number of steps per revolution of the motor, or in other words to provide a smaller basic stepping angle. The permanent magnet stepper motor contains an equal number of rotor and stator pole pairs. Typically the PM motor has 12 poles pairs. The stator has 12 pole pairs per phase. The hybrid type stepper motor has rotor with teeth. The rotor is split into two parts, separated by a permanent magnet - making half of the teeth south poles and half north poles. The number of pole pairs is equal to the number of teeth on one of the rotor halves. The stator of a hybrid motor also has teeth to build up a higher number of equivalent poles = 360/teeth pitch) compared to the main poles, on which the winding coils are wound. Usually 4 main poles are used for 3.6 hybrids and 8 for 1.8 and 0.9 degree types.

It is the relationship between the number of rotor poles and the equivalent stator poles, and the number the number of phases that determines the full-step angle of a stepper motor.

Step Angle = 360 ÷(N ph × Ph) = 360 / N

Nph = Number of equivalent poles per phase = number of rotor poles Ph = Number of phases

N = Total number of poles for all phases together

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If the rotor and stator tooth pitch is unequal, a more-complicated relationship exists.

Stepping mode

The following are the most common drive modes.

Wave Drive (1 phase on)

Full Step Drive (2 phase on) Half Step Drive (1 + 2 phases on) Micro stepping (Continuously varying motor currents) For the following discussions please refer to the figure 1.

In Wave Drive only one winding is energized at any given time. The stator is energized

according to sequence A → B ← A → B and the rotor steps from position

8 → 2 → 4 → 6 . For unipolar and bipolar wound motors with the same winding parameters this excitation mode would result in the same mechanical position. The disadvantage of this drive mode is that in the unipolar wound motor y9ou are only using 25% and in the bipolar motor only 50% of the total motor winding at any given time. This means that you are not getting the maximum torque output from the motor.

In Full Step Drive you are energizing two phases at any given time. The stator is

energized according to the sequence AB → A B → A B → A B and the rotor steps from position 1 → 3 → 5 → 7 . Full step mode results in the same angular movement as 1 phase on drive but the mechanical position is offset by one half of a full step. The torque output of the unipolar wound motor is lower than the bipolar motor since the unipolar motor uses only 50% of the available winding while the bipolar motor uses the entire winding.

Half Step Drive combines both wave and full step (1 & 2 phases on) drive modes. Every second step only one phase is energized and during the other steps one phase on each stator. The stator is energized according to the sequence AC → B → A B → A → A B → B → A B → A and the rotor steps from position

1 → 2 → 3 → 4 → 5 → 6 → 7 → 8 . This results in angular movements that are half of those in 1 or 2 phases on drive modes. Half stepping can reduce a phenomena referred to as resonance which can be experienced in 1 or 2 phases on drive modes.

The excitation sequences for the above drive modes are summarized in Table 1.

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In Micro stepping Drive the currents in the winding are continuously varying to be able to break up one full step into many smaller discrete steps. More information on micro stepping can be found in the micro stepping chapter.

Reluctance Motor

It is a single-phase synchronous motor which does not require d.c. excitation to the rotor. Its operation is based upon the following principle:

Whenever a piece of ferromagnetic material is located in a magnetic field; a force is exerted on the material, tending to align the material so that reluctance of the magnetic path that passes through the material is minimum.

Construction

A reluctance motor (also called synchronous reluctance motor) consists of: (i)A stator carrying a single-phase winding along with an auxiliary winding to produce a synchronous-revolving magnetic field. (ii)A squirrel-cage rotor having unsymmetrical magnetic construction. This is achieved by symmetrically removing some of the teeth from the squirrel-cage rotor to produce salient poles on the rotor. As shown in Fig. (9.24 (i)), 4 sailent poles have been produced on me rotor. The salient poles created on the rotor must be equal to the poles on the stator.

Note that rotor salient poles offer low reductance to the stator flux and, therefore, become strongly magnetized.

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Operation

(i)When single-phase stator having an auxiliary winding is energized, a synchronously-revolving field is produced. The motor starts as a standard squirrel-cage induction motor and will accelerate to near its synchronous speed.

(ii)As the rotor approaches synchronous speed, the rotating stator flux will exert reluctance torque on the rotor poles tending to align the salient-pole axis with the axis of the rotating field. The rotor assumes a position where its salient poles lock with the poles of the revolving field [See Fig. (9.24 (ii))]. Consequently, the motor will continue to run at the speed of revolving flux i.e., at the synchronous speed.

(iii)When we apply a mechanical load, the rotor poles fall slightly behind the stator poles, while continuing to turn at synchronous speed. As the load on the motor is increased, the mechanical angle between the poles increases progressively. Nevertheless, magnetic attraction keeps the rotor locked to the rotating flux. If the load is increased beyond the amount under which the reluctance torque can maintain synchronous speed, the rotor drops out of step with the revolving field. The speed, then, drops to some value at which the slip is sufficient to develop the necessary torque to drive the load by induction-motor action.

Characteristics

(i) These motors have poor torque, power factor and efficiency.

(ii) These motors cannot accelerate high-inertia loads to synchronous speed. (iii) The pull-in and pull-out torques of such motors are weak.

Despite the above drawbacks, the reluctance motor is cheaper than any other type of synchronous motor. They are widely used for constant-speed applications such as timing devices, signalling devices etc.

Repulsion-Induction Motor

The repulsion-induction motor produces a high starting torque entirely due to repulsion motor action. When running, it functions through a combination of induction-motor and repulsion motor action.

Construction

Fig. shows the connections of a 4-pole repulsion-induction motor for 230 V operation. It consists of a stator and a rotor (or armature).

(i) The stator carries a single distributed winding fed from single-phase supply.

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(ii)The rotor is provided with two independent windings placed one inside the other. The inner winding is a squirrel-cage winding with rotor bars permanently short-circuited. Placed over the squirrel cage winding is a repulsion commutator armature winding. The repulsion winding is connected to a commutator on which ride short-circuited brushes. There is no centrifugal device and the repulsion winding functions at all times.

Operation

(i)When single-phase supply is given to the stator winding, the repulsion winding (i.e., outer winding) is active. Consequently, the motor starts as a repulsion motor with a corresponding high starting torque.

(ii)As the motor speed increases, the current shifts from the outer to inner winding due to the decreasing impedance of the inner winding with increasing speed. Consequently, at running speed, the squirrel cage winding carries the greater part of rotor current. This shifting of repulsion-motor action to induction-motor action is thus achieved without any switching arrangement.

(iii) It may be seen that the motor starts as a repulsion motor. When running, it functions through a combination of principle of induction and repulsion; the former being predominant.

Characteristics

(i) The no-load speed of a repulsion-induction motor is somewhat above the synchronous speed because of the effect of repulsion winding. However, the speed at full-load is slightly less than the synchronous speed as in an induction motor.

(ii) The speed regulation of the motor is about 6%.

(iii) The starting torque is 2.25 to 3 times the full-load torque; the lower value being for large motors. The starting current is 3 to 4 times the full-load current.

This type of motor is used for applications requiring a high starting torque with essentially a constant running speed. The common sizes are 0.25 to 5 H.P.

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List of questions

1. Give the classification of single phase motors .Explain any two types of single phase induction motors. 2. Explain the double field revolving theory for operation of single phase induction motor. 3. Explain the operation of shaded pole induction motor with diagram.4. Develop equivalent circuit of a single phase induction motor ignoring core losses.5. Explain the working principle of single phase induction motor .Mention its four applications. 6. What is the principle and working of hysteresis motor? Explain briefly.7. Explain the construction and working of stepper motor. 8. Explain the principle of operation and applications of reluctance motor.9. Explain the principle of operation and applications of repulsion motor and hysteresis motor.10. Explain about no load and blocked rotor test of single phase induction motor.

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