Consider the rotating magnetic field as equivalent to physical rotation of two stator poles N1 and S1. Consider an instant when two poles are at such a position where stator magnetic axis is vertical, along A-B as shown in the Fig. 1(a). At this instant, rotor poles are arbitrarily positioned as shown in the Fig. 1. At this instant, rotor is stationary and unlike poles will try to attract each other. Due to this rotor will be subjected to an instantaneous torque in anticlockwise direction as shown in the Fig. 1(a). Now stator poles are rotating very fast i.e. at a speed N s r.p.m. Due to inertia, before rotor hardly rotates in the direction of anticlockwise torque, to which it is subjected, the stator poles change their positions. Consider an instant half a period latter where stator poles are exactly reversed but due to inertia rotor is unable to rotate from its initial position. This is shown in the Fig. 1(b).

At this instant, due to the unlike poles trying to attract each other, the rotor will be subjected

to a torque in clockwise direction. This will tend to rotate rotor in the direction of rotating

magnetic field.

But before this happen, stator poles again change their position reversing the direction of

the torque exerted on the rotor.

Key Point : As a result, the average torque exerted on the rotor is zero. And hence the

synchronous motor is not self starting.

Note : The question is obvious that will happen if by chance the rotor position is in such a

way that the unlike rotor and stator poles are facing each other ? But owing to the large

inertia of the rotor, the rotor fails to rotate along with the stator poles. Hence again the

difference of position of magnetic axes gets created and rotor gets subjected to quickly

reversing torque. This is because the speed with which rotating magnetic field is rotating is so

high that it is unable to rotate the rotor from its initial position, due to the inertia of the rotor.

So under any case, whatever may be the starting position of the rotor, synchronous motor is

not self starting.

Behaviour of Synchronous Motor on Loading

When a d.c. motor or an induction motor is loaded, the speed of the motors drops. This is because the load torque demand increases then the torque produced by the motor. Hence motor draws more current to produce more torque to satisfy the load but its speed reduces. In case of synchronous motor speed always remains constant equal to the synchronous speed, irrespective of load condition. It is interesting to study how synchronous motor reacts to changes in the load condition. In a d.c. motor, armature develops an e.m.f.after motoring action starts, which opposes supply voltage, called back e.m.f. Eb. Hence if Ra the armature resistance and V is the supply voltage, we have established the relation for the armature current as, Ia = (V- Eb) / Ra ...... for a d.c. motorwhere Eb = ΦPNZ / 60A .........for a d.c. motor

In case of synchronous motor also, once rotor starts rotating at synchronous speed, the stationary stator (armature) conductors cut the flux produced by rotor. The only difference is

conductors are stationary and flux is rotating. Due to this there is an induced e.m.f. in the stator which according to Lenz's law opposes the supply voltage. This induced e.m.f. is called back

e.m.f. in case of synchronous motor. It is obtained as Ebph i.e. back e.m.f. per phase. This gets generated as the principle of alternator and hence alternating in nature and its magnitude can be

calculated by the equation,

or Ebph α Φ As speed is always synchronous, the frequency is constant and hence magnitude of such back e.m.f. can be controlled by changing the flux Φ produced by the rotor. Keypoint: So back e.m.f. in case of synchronous motor depends on the excitation given to the field winding and not on the speed, as speed is always constant.

As stator construction is similar to the armature of a three phase alternator, the impedance of the stator is called synchronous impedance of synchronous motor consisting of Ra as the stator winding resistance and Xs as the synchronous reactance. All the values are generally expressed on per phase basis. Zs = Ra + jXs Ω per phase So similar to the d.c. motor, we can write voltage equation for a synchronous motor as,

The difference is that this equation is vector equation as each quantity is alternating and has different phase. So addition is to be performed vectorially to obtain the result.

where Vph is the supply voltage per phase. The magnitude of Ebph is adjusted almost equal to Vph, on no load by controlling flux produced by rotor i.e. field winding.1.1 Ideal Condition on No Load The ideal condition on no load can be assumed by neglecting various losses in the motor. And Vph = Ebph

Under this condition, the magnetic locking between stator and rotor is in such a way that the magnetic axes of both, coincide with each other as shown in the Fig.1. As this is possible only under no losses condition, is said to be ideal in case of synchronous motor.

Fig. 1 Magnetic locking under no load condition

As magnitude of Ebph and Vph is same and opposes the phasor diagram for this condition can be shown as in the Fig. 2.

In practice this is impossible. Motor has to supply mechanical losses and iron losses alongwith small copper losses. Let us see how it can be explained in case of synchronous motor.1.2 Synchronous Motor on No Load (With Losses) We have seen that Ebph and Vph are magnitudewise same, which is adjusted by controlling field current, in turn controlling the flux. Now due to the various losses practically present on no load, the magnetic locking exists between stator and rotor but in such a way that there exists a small angle difference between the axes of two magnetic fields as shown in the Fig.3.

Fig. 3 Magnetic locking under practical condition So the rotor axis falls back with respect to stator axis by angle 'δ' as shown in the Fig.3 This angle decides the amount of current required to produce the torque to supply various losses. Hence this angle is called load angle, power angle, coupling angle, torque angle or angle of retardation and denoted as δ as mentioned earlier. The magnetic locking still exists between the two and rotor rotates at synchronous speed alongwith rotating magnetic field maintaining angle difference between the axes of two fields, as shown in the Fig. 3(b). The flux lines between the two get stretched due to such retardation of rotor axis with respect to stator. Now though Ebph = Vph , Ebph will not be located in exact opposition with Vph , but will get displaced from its initial position by angle'δ' as shown in the Fig. 4(a).

Fig. 4(a) Phasor diagram for no load condition with losses Hence the vector difference between the two, Ebph and Vph is not zero but give rise to a phasor 'OB' as shown.

This resultant decides the amount of current Iaph to be drawn to produce the torque which meets the various losses present in the synchronous motor. Under no load condition, δ is very small and hence ERph is also very small.

So current drawn by the motor is also very small on no load which is the case in all the various type of motors.1.3 Synchronous Motor on Load As the load on the synchronous motor increases, there is no change in its speed. But what gets affected is the load angle 'δ' i.e. the angle by which rotor axis retards with respect to stator axis. Hence as load increases, δ increases but speed remains synchronous. As δ increases, though Ebph and Vph magnitudes are same, displacement of Ebph from its ideal position increases.

As synchronous impedance is constant, the magnitude of Iaph drawn by the motor increases as load increases. This current produces the necessary torque which satisfied the increased load demand. The magnetic locking still exists between the rotor and stator. The phasor diagrams showing ERph increases as load increases are shown in the Fig. 4(b) and (c).

Fig. 4 So from the above discussion it is clear that on no load, current drawn by the motor is very small. This is because the stator and the rotor magnetic axes are almost matching transformer each other i.e. load angle δ is very small. As load increases, rotor magnetic axis starts retarding with respect to stator axis i.e. load angle δ increases maintaining the magnetic locking condition. And hence in case of the synchronous motor load affects the angle δ without affecting the speed. As δ increases, the magnitude of ERph increases which shows that motor draws more current from the supply. This satisfies the increased load torque demand. Key point: So torque produced in the synchronous motor depends on the load angle 'δ' for small values of and to be precise depends on 'sinδ'. The load angle 'δ' is measured in degrees electrical. As angle δ increases, the magnetic flux lines producing the force of attraction between the two get more and more stretched. This weakens the force maintaining the magnetic locking, though torque produced by the motor increases. As δ reaches upto 90o electrical i.e. half a pole pitch, the stretched flux lines get broken and hence magnetic locking between the stator and rotor no longer exists. The motor comes out of synchronism. So torque produced at δ equal to 90oelectrical is the maximum torque, a synchronous motor can produce, maintaining magnetic locking i.e. synchronism. Such s torque is called pull out torque. The relationship between torque produced and load angle is shown in the Fig 5.

Fig. 5 Torque angle characteristic

Operation of S.M. at constant Load Variable Excitation

We have seen previously that when load changes, for constant excitation, current drawn by the motor increases. But if excitation i.e. field current is changed keeping load constant, the synchronous motor reacts by by changing its power factor of operation. This is most interesting feature of synchronous motor. Let us see the details of such operation. Consider a synchronous motor operating at a certain load. The corresponding load angle is δ. At start, consider normal behaviour of the synchronous motor, where excitation is adjusted to get Eb = V i.e. induced e.m.f. is equal to applied voltage. Such an excitation is called Normal Excitation of the motor. Motor is drawing certain current from the supply and power input to the motor is say Pin. The power factor of the motor is lagging in nature as shown in the Fig. 1(a).Now when excitation is changed, changes but there is hardly any change in the losses of the motor. So the power input also remains same for constant load demanding same power output. Now Pin = √3 VL IL cos Φ = 3 (Vph Iph cos Φ) Most of the times, the voltage applied to the motor is constant. Hence for constant power input as Vph is constant, 'Iph cos Φ' remains constant. Note : So far this entire operation of variable excitation it is necessary to remember that the cosine component of armature current, Ia cosΦ remains constant. So motor adjusts its cos Φ i.e. p.f. nature and value so that I a cos Φ remains constant when excitation of the motor is changed keeping load constant. This is the reason why synchronous motor reacts by changing its power factor to variable excitation conditions.1.1 Under Excitation When the excitation is adjusted in such a way that the magnitude of induced e.m.f. is less than the applied voltage (Eb < V) the excitation is called Under Excitation. Due to this, ER increases in magnitude. This means for constant Zs, current drawn by the motor increases. But ER phase shifts in such a way that, phasor Ia also shifts (as ER ^ Ia = θ) to keep Ia cos Φ component constant. This is shown in the Fig. 1(b). So in under excited condition, current drawn by the motor increases. The p.f. cos Φ decreases and becomes more and more lagging in nature.1.2 Over Excitation

The excitation to the field winding for which the induced e.m.f. becomes greater than applied voltage (Eb < V), is called over excitation. Due to increased magnitude of Eb, ER also increases in magnitude. But the phase of ER also changes. Now = ER ^ Ia = θ is constant, hence Ia also changes its phase. So Φ changes. The Ia increases to keep Ia cos Φ constant as shown in Fig.1(c). The phase of ER changes so that Ia becomes leading with respect to Vph in over excited condition. So power factor of the motor becomes leading in nature. So overexcited synchronous motor works on leading power factor. So power factor decreases as over excitation increases but it becomes more and more leading in nature.1.3 Critical Excitation When the excitation is changed, the power factor changes. The excitation for which the power factor of the motor is unity (cos Φ = 1) is called critical excitation. Then Iaph is in phase with Vph. Now Ia cos Φ must be constant, cos Φ = 1 is at its maximum hence motor has to draw minimum current from supply for unity power factor condition. So for critical excitation, cos Φ = 1 and current drawn by the motor is minimum compared to current drawn by the motor for various excitation conditions. This is shown in the Fig. 1(d).

Fig. 1 Constant load variable excitation operation

V-Curves and Inverted V-Curves

From the previous article, it is clear that if excitation is varied from very low (under excitation) to very high (over excitation) value, then current Ia decreases, becomes minimum at unity p.f. and then again increases. But initial lagging current becomes unity and then becomes leading in nature. This can be shown as in the Fig. 1.

Fig. 1

Excitation can be increased by increasing the field current passing through the field winding of synchronous motor. If graph of armature current drawn by the motor (Ia) against field current (If) is plotted, then its shape looks like an english alphabet V. If such graphs are obtained at various load conditions we get family of curves, all looking like V. Such curves are called V-curves of synchronous motor. These are shown in the Fig. 2a). As against this, if the power factor (cos Φ) is plotted against field current (I f), then the shape of the graph looks like an inverted V. Such curves obtained by plotting p.f. against I f, at various load conditions are called Inverted V-curves of synchronous motor. These curves are shown in the Fig. 2(b).

Fig. 2 V-curves and Inverted V-curves

1.1 Experimental Setup to Obtain V-Curves Fig. 3 shows an experimental setup to obtain V-curves and Inverted V-curves of synchronous motor. Stator is connected top three phase supply through wattmeters and ammeter. The two wattmeter method is used to measure input power of motor. The ammeter is reading line current which is same as armature (stator) current. Voltmeter is reading line voltage.

Fig. 3 Experimental setup for V-curves

A rheostat in a potential divider arrangement is used in the field circuit. By controlling the voltage by rheostat, the field current can be changed. Hence motor can be subjected to variable excitation condition to note down the readings.Observation Table :

Now IL = Ia, per phase value can be determined, from the stator winding connections. IL = Iaph for stator connection IL/√3 = Iaph for delta connection The power factor can be obtained as

The result table can be prepared as :

The graph can be plotted from this result table.1) Ia Vs If → V-curve2) cosΦ Vs If → Inverted V-curve The entire procedure can be repeated for various load conditions to obtain family of V-curves and Inverted V-curves.

Expression for Back E.M.F or Induced E.M.F. per Phase in S.M.

Case i) Under excitation, Ebph < Vph . Zs = Ra + j Xs = | Zs | ∟θ Ω θ = tan-1(Xs/Ra) ERph ^ Iaph = θ, Ia lags always by angle θ. Vph = Phase voltage applied ERph = Back e.m.f. induced per phase ERph = Ia x Zs V ... per phase Let p.f. be cosΦ, lagging as under excited, Vph ^ Iaph = Φ Phasor diagram is shown in the Fig. 1.

Fig. 1 Phasor diagram for under excited condition

Applying cosine rule to Δ OAB, (Ebph)2 = (Vph)2 + (ERph)2 - 2Vph ERph x (Vph ^ ERph) but Vph ^ ERph = x = θ - Φ (Ebph)2 = (Vph)2 + (ERph)2 - 2Vph ERph x (θ - Φ) ......(1) where ERph = Iaph x Zs

Applying sine rule to Δ OAB, Ebph/sinx = ERph/sinδ

So once Ebph is calculated, load angle δ can be determined by using sine rule.Case ii) Over excitation, Ebph > Vph

p.f. is leading in nature. ERph ^ Iaph = θ Vph ^ Iaph = Φ The phasor diagram is shown in the Fig. 2.

Fig.2 Phasor diagram for overexcited condition

Applying cosine rule to Δ OAB, (Ebph)2 = (Vph)2 + (ERph)2 - 2Vph ERph x cos(Vph ^ ERph) Vph ^ ERph = θ + Φ... (Ebph)2 = (Vph)2 + (ERph)2 - 2 Vph ERph cos(θ + Φ) .......(3)

But θ + Φ is generally greater than 90o

... cos (θ + Φ) becomes negative, hence for leading p.f., Ebph > Vph . Applying sine rule to Δ OAB, Ebph/sin( ERph ^ Vph) = ERph/sinδ

Hence load angle δ can be calculated once Ebph is known.Case iii) Critical excitation In this case Ebph ≈ Vph, but p.f. of synchronous motor is unity.... cos = 1 ... Φ = 0o

i.e. Vph and Iaph are in phase and ERph ^ Iaph = θ Phasor diagram is shown in the Fig. 3.

Fig. 3 Phasor diagram for unity p.f. condition

Applying cosine rule to OAB, (Ebph)2 = (Vph)2 + (ERph)2 - 2Vph ERph cos θ ............(5) Applying sine rule to OAB, Ebph/sinθ = ERph/sinδ

where ERph = Iaph x Zs V

Power Flow in Synchronous MotorNet input to the synchronous motor is the three phase input to the stator.... Pin = √3 VL IL cosΦ W where VL = Applied Line Voltage IL = Line current drawn by the motor cosΦ = operating p.f. of synchronous motor or Pin = 3 ([er phase power) = 3 x Vph Iaph cosΦ W Now in stator, due to its resistance Ra per phase there are stator copper losses. Total stator copper losses = 3 x (Iaph)2 x Ra W... The remaining power is converted to the mechanical power, called gross mechanical power developed by the motor denoted as Pm.... Pm = Pin - Stator copper losses Now P = T x ω... Pm = Tg x (2πNs/60) as speed is always Ns

This is the gross mechanical torque developed. In d.c. motor, electrical equivalent of gross mechanical power developed is Eb x Ia, similar in synchronous motor the electrical equivalent of gross mechanical power developed is given by, Pm = 3 Ebph x Iaph x cos (Ebph ^ Iaph) i) For lagging p.f., Ebph ^ Iaph = Φ - δii) For leading p.f., Ebph ^ Iaph = Φ + δiii) For unity p.f., Ebph ^ Iaph = δNote : While calculating angle between Ebph and Iaph from phasor diagram, it is necessary to reverse Ebph phasor. After reversing Ebph, as it is in opposition to Vph, angle between Ebph and Iaph must be determined. In general,

Positive sign for leading p.f. Neglecting sign for lagging p.f. Net output of the motor then can be obtained by subtracting friction and windage i.e. mechanical losses from gross mechanical power developed.... Pout = Pm - mechanical losses.

where Tshaft = Shaft torque available to load. Pout = Power available to load... Overall efficiency = Pout/Pin

Alternative Expression for Power Developed by a Synchronous Motor Consider the phasor diagram of a synchronous motor running on leading power factor cosΦ as shown in the Fig. 1.

Fig. 1 The line CD is drawn at an angle θ to AB. The lines AC and DE are perpendicular to CD and AE.

and angle between AB = Ebph and Iaph is also ψ. The mechanical per phase power developed is given by,

In triangle OBD, BD = OB cosψ = Ia Zs cosψ OD = OB sin ψ = Ia Zs sin Now BD = CD - BC = AE - BC

Substituting in (2), Ia Zs cosψ = Vph cos (θ-δ) - Eb cosθ

All values are per phase values Substituting (3) in (1),

This is the expression for the mechanical power developed interms of the load angle δ and the internal machine angle θ, for constant voltage Vph and constant Eph i.e. excitation.

Condition for Maximum Power Developed

The value of δ for which the mechanical power developed is maximum can be obtained as,

Note : Thus when Ra is negligible, θ = 90o for maximum power developed. The corresponding torque is called pull out torque.1.1 The Value of Maximum Power Developed The value of maximum power developed can be obtained by substituting θ =δ in the equation of Pm.

When Ra is negligible, θ = 90o and cos (θ) = 0 hence,

... Ra = Zs cosθ and Xs = Zs sinθ Substituting cosθ = Ra/Zs in equation (6b) we get,

Solving the above quadratic in Eb we get,

As Eb is completely dependent on excitation, the equation (8) gives the excitation limits for any load for a synchronous motor. If the excitation exceeds this limit, the motor falls out of step.1.2 Condition for Excitation When Motor Develops (Pm ) R max

Let us find excitation condition for maximum power developed. The excitation controls E b. Hence the condition of excitation can be obtained as,

Assume load constant hence δ constant.

but θ = δ for Pm = (Pm)max

Substitute cosθ = Ra/Zs

This is the required condition of excitation.Note : Note that this is not maximum value of but this is the value of foe which power developed is maximum. The corresponding value of maximum power is,

Hunting in Synchronous Motor

It is seen that, when synchronous motor is on no load, the stator and rotor pole axes almost coincide with each other.When motor is loaded, the rotor axis falls back with respect to stator. The angle by which rotor retards is called load angle or angle of retardation δ. If the load connected to the motor is suddenly changed by a large amount, then rotor tries to retard to take its new equilibrium position. But due to inertia of the rotor, it can not achieve its final position instantaneously. While achieving its new position due to inertia it passes beyond its final position corresponding to new load. This will produce more torque than what is demanded. This will try reduce the load angle and rotor swings in other direction. So there is periodic swinging of the rotor on both sides of the new equilibrium position, corresponding to the load. Such a swing is shown in the Fig. 1.

Fig. 1 Hunting in synchronous motor Such oscillations of the rotor about its new equilibrium position, due to sudden application or removal of load is called swinging or hunting in synchronous motor. Due to such hunting, the load angle changes its value about its final value δ. As changes, for same excitation i.e. Ebph the current drawn by the motor also changes. Hence during hunting there are changes in the current drawn by the motor which may cause problem to the other appliances connected to the same line. The changes in armature current due to hunting is shown in the Fig. 2.

Fig. 2 Current variations during hunting

If such oscillations continue for longer period, there are large fluctuations in the current. If such variations synchronous with the natural period of oscillation of the rotor, the amplitude of the

swing may become so great that motor may come out of synchronism. At this instant mechanical stresses on the rotor are sever and current drawn by the motor is also very large. So motor gets subjected to large mechanical and electrical stresses.Note : Hence hunting is not desirable phenomenon from motor point of view and must be prevented.1.1 Use of Damper Winding to Prevent Hunting It is mentioned earlier that in the slots provided in the pole faces, a short circuited winding is placed. This is called damper winding. When rotor starts oscillating i.e. when hunting starts a relative motion between damper winding and the rotating magnetic field is created. Due to this relative motion, e.m.f. gets induced in the damper winding. According to Lenz's law, the direction of induced e.m.f. is always so as to oppose the cause producing it. The cause is the hunting. So such induced e.m.f. oppose the hunting. The induced e.m.f. tries to damp the oscillations as quickly as possible. Thus hunting is minimised due to damper winding. The time required by the rotor to take its final equilibrium position after hunting is called as setting time of the rotor. If the load angle is plotted against time, the schematic representation of hunting can be obtained as shown in the Fig. 3. It is shown in the diagram that due to damper winding the setting time of the rotor reduces considreably.

Fig. Effect of damper winding on hunting

Synchronous Condensers

When synchronous motor is over excited it takes leading p.f. current. If synchronous motor is on no load, where load angle δ is very small and it is over excited (Eb > V) then power factor angle increases almost upto 90o. And motor runs with almost zero leading power factor condition. This is shown in the phasor diagram Fig. 1.

Fig. 1 Synchronous condenser This characteristics is similar to a normal capacitor which takes leading power factor current. Hence over excited synchronous motor operating on no load condition is called as synchronous condenser or synchronous capacitor. This is the property due to which synchronous motor is used as a phase advancer or as power improvement device.1.1 Disadvantage of Low Power Factor In various industries, many machines are of induction motor type. The lighting and heating loads are supplied through transformers. The induction motors and transformers draw lagging current from the supply. Hence the overall power factor is very low and lagging in nature. The power is given by, P = VI cosΦ .............. single phase... I = P/(VcosΦ) The supply voltage is constant and hence for supplying a fixed power P, the current is inversely proportional to the p.f. cosΦ. Let P = KW is to be supplied with a voltage of 230 V then,Case i) cosΦ = 0.8, I = (5 x103)/(230 x 0.8) = 27.17 ACase ii) cos = 0.6,

I = (5 x103)/(230 x 0.6) = 36.23 A Thus as p.f. decreases, becomes low, the current drawn from the supply increases to supply same power to the load. But if p.f. maintained high, the current drawn from supply is less. The high current due to low p.f. has following disadvantages :1. For higher current, conductor size required is more which increases the cost.2. The p.f. is given by cosΦ = Active power/ Apparent = (P in KW)/ (S i.e. KVA rating) Thus for fixed active power P, low p.f. demands large KVA rating alternators and transformers. This increases the cost.3. Large current means more copper losses and poor efficiency.4. Large current causes large voltage drops in transmission lines, alternators and other equipments. This results into poor regulation. To compensate such drop extra equipments is necessary which further increases the cost.Note : Hence power factor improvement is must practice. Hence the supply authorities encourage consumers to improve the p.f.1.1 Use of Synchronous Condenser in Power Factor Improvement The low power factor increases the cost of generation, distribution and transmission of the electrical energy. Hence such low power factor needs to be corrected. Such power factor correction is possible by connecting synchronous motor across the supply and operating it on no load with over excitation.

Now let Vph is the voltage applied and I1ph is the current lagging Vph by angle Φ1. This power factor Φ1 is very low, lagging. The synchronous motor acting as a synchronous condenser is now connected across the same supply. This draws a leading current of I2ph. The total current drawn from the supply is now phasor of Iph and I2ph. This total current IT now lags Vph by smaller angle Φ due to which effective power factor gets improved. This is shown in the Fig. 2.

Fig. 2 Power factor correction by synchronous condenser

This is how the synchronous motor as a synchronous condenser is used to improve power factor of the combined load.