Comparison of Synchronous and Induction Motor

Applications of Three Phase Synchronous Motor

The important characteristics of the synchronous motor is its constant speed irrespective of the load conditions, and variable power factor operation. As seen earlier its power factor can be controlled by controlling its excitation. For overexcitation its power factor is leading in nature, which is very important from the power factor correction point of view.

       Due to constant speed characteristics, it is used in machine tools, motor generator sets, synchronous clocks, stroboscopic devices, timing devices, belt driven reciprocating compressors, fans and blowers, centrifugal pumps, vacuum pumps, pulp grinders, textile mills, paper mills line shafts, rolling mills, cement mills etc.

       The synchronous motors are often used as a power factor correction device, phase advancers and phase modifiers for voltage regulation of the transmission lines. This is possible because the excitation of the synchronous motor can be adjusted as per the requirement.

       The disadvantages of synchronous motor are their higher cost, necessity of frequent maintenance and a need of d.c. excitation source, auxiliary device or additional winding provision to make it self starting. Overall their initial cost is very high.

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 (E_b > V) then power factor angle increases almost upto 90^o. 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.

Synchronization With Infinite Bus Bar

There is a specific procedure of connecting synchronous machine to infinite bus bars. Infinite bus bar is one which keeps constant voltage and frequency although load varies. The Fig. 1 shows a synchronous machine which is to be connected to the bus bars with the help of switch K.

        If the synchronous machine is running as a generator then its phase sequence should be some as that of bus bars. The machine speed and field current is adjusted in such a way so as to have the machine voltage same as that of bus bar voltage. The machine frequency should be nearly equal to bus bar frequency so that the machine speed is nearer to synchronous speed.

Fig. 1

        When the above conditions are satisfied, the instant of switching for synchronization should be determined. This can be determined by lamps dark method, Lamps bright and dark method or by using synchroscope.

 

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.

     

Salient Pole Synchronous Motor

The analysis of salient pole synchronous motor is based on the Blondel's two reaction. The direct and quadrature axis components of current and reactance are same as defined earlier for the synchronous generators. Thus,

                      X_d = Direct axis reactance

                      X_q  = Quadrature axis reactance

                      I_d  = Direct axis component of I_a

                       I_q = Quadrature axis component of I_a

       The complete phasor diagram of lagging p.f. is shown in the Fig. 1.

Fig. 1 Phasor diagram for lagging p.f.

       From the phasor diagram it can be derived that,

Note : For the proof of above results refer example 2.

       The complete phasor diagram of leading p.f. is shown in the Fig. 2.

Fig. 2  Phasor diagram for leading p.f.

       For this leading p.f. case,

Note : Φ should be taken negative for the leading power factor for calculating other parameters.

       While the mechanical power developed per phase is given by,

       Total P_m  = 3 x P_m   

Blondel Diagram { Constant Power Circle)

The Blondel diagram of a synchronous motor is an extension of a simple phasor diagram of a synchronous motor.

       For a synchronous motor, the power input to the motor per phase is given by,

                    P_in  = V_ph I_ph cosΦ                                               ............ per phase

       The gross mechanical power developed per phase will be equal to the difference between  P_in  per phase and the per phase copper losses of the winding.

                   Copper loss per phase = (I_aph)²  R_a

.^..                 P_m   = V_ph I_ph cosΦ - (I_aph)²  R_a                       .......... per phase

       For mathematical convenience let V_ph = V and I_aph = I,

.^..                  P_m   = VI cos - I² R_a

.^..                  I² R_a  - VI cos + P_m  = 0

       Now consider the phasor diagram as shown in the Fig. 1.

Fig. 1

       The equation (1) represents polar equation to a circle. To obtain this circle in a phasor diagram, draw a line OY at an angle θ with respect to OA.

Fig. 2   Blondel diagram

       The circle represented by equation (1) has a centre at some point O' on the line OY. The circle drawn with centre as O' and radius as O'B represents circle of constant power. This is called Blondel diagram, shown in the Fig. 2.

       Thus if excitation is varied while the power is kept constant, then working point B while move along the circle of constant power.

       Let   O'B = Radius of circle = r

       OO' = Distant d

       Applying cosine rule to triangle OBO',

       Now OB represents resultant E_R   which is I_a  Z_s. Thus OB is proportional to current and when referred to OY represents the current in both magnitude and phase.

                             OB = I_a  = I say

       Substituting various values in equation (2) we get,

                              r² = I² + d² - 2dI cosΦ

       Comparing equations (1) and (3) we get,

       Thus the point O' is independent of power P_m  and is a constant for a give motor operating at a fixed applied voltage V.

       Comparing last term of equations (1) and (3),

       The equation shows that as power P_m  must be real, then 4P_m Ra  _≥ V² . The maximum possible power per phase is,

       And the radius of the circle for maximum power is zero. Thus at the time of maximum power, the circles becomes a point O'.

While when the power   P_m   = O, then

                                        r = V/2Ra    = OO'

       This shows that the circle of zero power passes through the points O and A.

       The radius for any power P_m  is given by,

 This is generalised expression for the radius for any power.

Condition for Maximum Power Developed

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

Note : Thus when R_a is negligible, θ = 90^o 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 P_m.

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

.^..               R_a = Z_s cosθ   and X_s = Z_s sinθ

       Substituting   cosθ = R_a/Z_s in equation (6b) we get,

        Solving the above quadratic in E_b we get,

       As E_b 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 (P_m ) 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 P_m = (P_m)_max

         Substitute            cosθ = R_a/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,

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 = E_bph and I_aph is also ψ.

      

Power Flow in Synchronous Motor

Net input to the synchronous motor is the three phase input to the stator.

.^..                      P_in = √3 V_L I_L cosΦ W 

       where         V_L = Applied Line Voltage

                          I_L = Line current drawn by the motor

                          cosΦ = operating p.f. of synchronous motor

       or                P_in  = 3 ([er phase power)

                                 = 3 x V_ph I_aph cosΦ W

       Now in stator, due to its resistance R_a per phase there are stator copper losses.
       Total stator copper losses = 3 x (I_aph)² x R_a W

.^..     The remaining power is converted to the mechanical power, called gross mechanical power developed by the motor denoted as P_m.

.^..                P_m = P_in - Stator copper losses
       Now    P = T x ω
.^..                P_m  = T_g  x (2πN_s/60)  as speed is always N_s

     

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

Case i) Under excitation, E_bph < V_ph .

       Z_s = R_a + j X_s = | Z_s |  ∟θ Ω

       θ = tan^-1(X_s/R_a)

       E_Rph  ^ I_aph = θ, I_a lags always by angle θ.

       V_ph = Phase voltage applied

       E_Rph = Back e.m.f. induced per phase

       E_Rph = I_a x Z_s V            ... per phase

       Let p.f. be cosΦ, lagging as under excited,

       V_ph  ^ I_aph = Φ

       Phasor diagram is shown in the Fig. 1.

Fig. 1 Phasor diagram for under excited condition

       Applying cosine rule to Δ OAB, 

       (E_bph)² = (V_ph)² + (E_Rph)² - 2V_ph E_Rph x (V_ph ^ E_Rph)

       but V_ph ^ E_Rph = x = θ - Φ

       (E_bph)² = (V_ph)² + (E_Rph)² - 2V_ph E_Rph x (θ - Φ)                   ......(1)

       where E_Rph =  I_aph x Z_s

       Applying sine rule to Δ OAB,

       E_bph/sinx = E_Rph/sinδ

        So once E_bph is calculated, load angle δ can be determined by using sine rule.

Case ii) Over excitation, E_bph > V_ph

       p.f. is leading in nature.

       E_Rph  ^ I_aph = θ

       V_ph  ^ I_aph = Φ

       The phasor diagram is shown in the Fig. 2.

Fig.2  Phasor diagram for overexcited condition

       Applying cosine rule to Δ OAB,

       (E_bph)² = (V_ph)² + (E_Rph)² - 2V_ph  E_Rph x cos(V_ph ^ E_Rph)

       V_ph ^ E_Rph = θ + Φ

.^..    (E_bph)² = (V_ph)² + (E_Rph)² - 2 V_ph   E_Rph cos(θ + Φ) .......(3)

       But θ + Φ is generally greater than  90^o

.^..    cos (θ + Φ) becomes negative, hence for leading p.f., E_bph > V_ph .

       Applying sine rule to Δ OAB,

       E_bph/sin( E_Rph ^ V_ph) = E_Rph/sinδ

       Hence load angle δ can be calculated once E_bph is known.

Case iii) Critical excitation

       In this case E_bph ≈ V_ph, but p.f. of synchronous motor is unity.

.^..         cos = 1   .^..    Φ = 0^o

       i.e. V_ph and I_aph are in phase

       and  E_Rph ^ I_aph = θ

       Phasor diagram is shown in the Fig. 3.

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

       Applying cosine rule to OAB,

       (E_bph)² = (V_ph)² + (E_Rph)² - 2V_ph E_Rph cos θ            ............(5)

       Applying sine rule to OAB,

       E_bph/sinθ = E_Rph/sinδ

       where   E_Rph = I_aph x Zs V

V-Curves and Inverted V-Curves

it is clear that if excitation is varied from very low (under excitation) to very high (over excitation) value, then current I_a 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 (I_a) against field current (I_f) 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 I_L = I_a, per phase value can be determined, from the stator winding connections. 

       I_L  = I_aph for stator connection

       I_L/√3 = I_aph 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) I_a Vs I_f  → V-curve

2) cosΦ Vs I_f  → Inverted V-curve

       The entire procedure can be repeated for various load conditions to obtain family of V-curves and Inverted V-curves.