UNIT II SYNCHRONOUS MOTOR

Ideal Condition on No Load The ideal condition on no load can be assumed by neglecting various losses in the motor.        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. Eph Vph As magnitude of Ebph  and Vph  is same and opposes the phasor diagram for this condition

Synchronous Motor on No Load (With Losses) 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 C B δ is very small Eph Vph – Eph = IaZs IaZs OB = IaZs = ERph δ δ A Eph O Vph

Synchronous Motor on Load

C B ER1ph OB = ER1ph Ebph LIGHT LOAD δ1 Vph A Eph O B C δ2 > δ1 OB = ER2ph ER2ph ER2ph > ER1ph Ebph HEAVY LOAD δ2 Vph A Eph O

Operation of Synchronous motor at constant load Variable Excitation Field Excitation (Field Current) is changed Keeping Load Constant, The Synchronous motor reacts by changing its power factor of operation Normal Excitation Excitation is adjusted to get Eb = Vph Motor draws a certain current Ia from supply PF of Motor is lagging If changes Eb changes If changes Eb also changes Eb α If

B C Constant Ia Cos ф Normal Excitation Excitation is adjusted Eb = Vph ER Ebph Constant Vph ϴ δ A O Vph ф Ia B Under Excitation Excitation is adjusted Eb < Vph Ebph ER δ ϴ A Vph ф Ia B Ebph ER Ia Over Excitation Excitation is adjusted Eb > Vph ϴ δ ф A Vph B Critical Excitation Excitation is adjusted Eb > Vph Ebph ER ϴ δ A Ia Vph

Synchronous condensers A synchronous motor takes a leading current when over-excited and, therefore, behaves as a capacitor. An over-excited synchronous motor running on no-load in known as Synchronous condenser. Synchronous Motor is over-excited it takes leading P.F. current If Synchronous Motor is no NO-LOAD, where load angle δ Over Excitation Excitation is adjusted Eb > Vph Ia Ia Ebph ER Ф = 90 δ 90 Vph

Advantages By varying the field excitation, the magnitude of current drawn by the motor can be changed by any amount. This helps in achieving stepless control of power factor. The motor windings have high thermal stability to short circuit currents. The faults can be removed easily. Disadvantages (i) There are considerable losses in the motor. (ii) The maintenance cost is high.

Hunting When Synchronous motor is on NO LOAD The stator and rotor pole axis coincide with each other. When Synchronous motor is LOAD the rotor pole fall back with respect to stator. ER2 ER Ebph ER1 R δ Load angle by which the Synchronous motor rotor falls back from the aim of stator is called load angle(δ) O Vph Ia1 Ia Ia2

Causes of Hunting in Synchronous Motor Sudden change in load. Sudden change in field current. A load containing harmonic torque. Fault in supply system Effects of Hunting in Synchronous Motor It may lead to loss of synchronism. Produces mechanical stresses. Increases machine losses and cause temperature rise. Cause greater surges in current and power flow

Reduction of Hunting in Synchronous Motor Two techniques should be used to reduce hunting. These are – Use of Damper Winding  It consists of low electrical resistance copper / aluminium brush embedded in slots of pole faces in salient pole machine. Damper winding damps out hunting by producing torque opposite to slip of rotor. The magnitude of damping torque is proportional to the slip speed. Use of Flywheels : The prime mover is provided with a large and heavy flywheel. This increases the inertia of prime mover and helps in maintaining the rotor speed constant. Designing synchronous machine with suitable synchronizing power coefficients

V curves and Inverted V curves

V curves and Inverted V curves Unity PF Armature Current I PF Lagging PF Leading PF A Normal Excitation If

Synchronous Machines on Infinite Bus-Bars   Infinite bus-bars are those whose frequency and the phase and magnitude of potential differences are not affected by changes in the condition of any one machine connected to it. The variation in excitation of a synchronous machine connected to an infinite bus-bar will cause a large change in the reactive component supplied by the alternator and so change its power. The increase in torque of the prime-mover of one alternator, it is further loaded and equivalent load is removed from the other unit (s) with which the machine is paralleled. If the output of the alternator, whose prime-mover torque has been increased, becomes more than the total load being supplied, then the other machine (s) will operate as synchronous motor (s). The active and reactive power loading of an alternator operating on an infinite bus-bar is controlled by controlling the input power to it and ecitation respectively.

(Ebph )2 = (Vph )2 + (ERph )2 - 2Vph ERph * Cos (ϴ - ф) Expression for Back EMF or Induced EMF per phase in Synchronous Motor Case i : Under Excitation Eb < Vph Power factor is Lagging Zs = Ra + jXs ERph = IaZs Ebph B = │Zs│∟ ϴ Ebph x δ ϴ   δ A ф Vph O ERph ᴧ Iaph = ϴ Ia Vph ᴧ Iaph = ф X = ϴ - ф Δ OAB ERph = Iaph * Zs (Ebph )2 = (Vph )2 + (ERph )2 - 2Vph ERph * Cos (ϴ - ф) Ebph = ERph Sin x Sin δ Sin δ = ERph Sin (ϴ - ф) E bph

(Ebph )2 = (Vph )2 + (ERph )2 - 2Vph ERph * Cos (ϴ + ф) Expression for Back EMF or Induced EMF per phase in Synchronous Motor Case ii : Over Excitation Eb > Vph Power factor is Leading Zs = Ra + jXs ERph Ebph Ia B = │Zs│∟ ϴ Ebph ϴ   δ ф δ A Vph O ERph ᴧ Iaph = ϴ Vph ᴧ Iaph = ф X = ϴ + ф Δ OAB ERph = Iaph * Zs (Ebph )2 = (Vph )2 + (ERph )2 - 2Vph ERph * Cos (ϴ + ф) Sin δ = ERph Sin (ϴ + ф) E bph

(Ebph )2 = (Vph )2 + (ERph )2 - 2Vph ERph * Cos (ϴ + ф) Expression for Back EMF or Induced EMF per phase in Synchronous Motor Case iii : Critical Excitation Eb ≡ Vph Power factor is Unity Zs = Ra + jXs ERph Ebph B = │Zs│∟ ϴ Ebph ϴ   δ ф δ A Vph O ERph ᴧ Iaph = ϴ Ia Vph ᴧ Iaph = ф X = ϴ + ф Δ OAB ERph = Iaph * Zs (Ebph )2 = (Vph )2 + (ERph )2 - 2Vph ERph * Cos (ϴ + ф) Ebph = ERph Sin ϴ Sin δ Sin δ = ERph Sin ϴ E bph

(Ebph )2 = (Vph )2 + (ERph )2 - 2Vph ERph * Cos (ϴ ± ф) ERph = Iaph * Zs + sign for lagging Zs = Ra + jXs - sign for leading  

A 400 V, 3 phase, star connected synchronous motor has an armature resistance of 0.2 ohms per phase and synchronous reactance of 2 ohms per phase. While driving a certain load, it takes 25 A from the supply. Calculate the back EMF induced in the motor if it is working with 0.8 Lagging 0.8 leading Unity power factor conditions

(Ebph )2 = (Vph )2 + (ERph )2 - 2Vph ERph * Cos (ϴ ± ф) Given Data VL = 400 V Star connected ERph = Iaph * Zs Ra = 0.2 Ohms Zs = Ra + jXs Xs = 2 Ohms Zs = 0.2 + j 2 Ω IL = Iaph = 25 A Vph = VL / √3 = 2.001 ∟ 84.29 ERph = Iaph * Zs Vph = 400 / √3 ERph = 25 * 2.001 Vph = 230.94 V ERph = 50.025 V (i) Cos ф = 0.8 Lagging Ф = 36.869 ° ϴ = 84.29° (Ebph )2 = (Vph )2 + (ERph )2 - 2Vph ERph * Cos (ϴ ± ф) (Ebph )2 = (230.94)2 + (50.025)2 - 2x(230.94)x(50.025) * Cos (84.29 – 36.86) Ebph = 200.505 V Ebph < Vph 200.50 < 230.94

(Ebph )2 = (Vph )2 + (ERph )2 - 2Vph ERph * Cos (ϴ ± ф) Vph = 230.94 V ERph = 50.025 V ϴ = 84.29° (ii) Cos ф = 0.8 Leading Ф = 36.869 ° (Ebph )2 = (Vph )2 + (ERph )2 - 2Vph ERph * Cos (ϴ + ф) (Ebph )2 = (230.94)2 + (50.025)2 - 2x(230.94)x(50.025) * Cos (84.29 +36.86) Ebph >Vph 252.56 < 230.94 Ebph = 252.56V (iii) Cos ф = 1Unity PF Ф = 0° (Ebph )2 = (Vph )2 + (ERph )2 - 2Vph ERph * Cos (ϴ + ф) (Ebph )2 = (230.94)2 + (50.025)2 - 2x(230.94)x(50.025) * Cos (84.29 +0) Ebph = 231.38V Ebph ≡ Vph 252.56 ≡ 230.94

A 3 phase, 500 V, synchronous motor draws a current of 50 A from the supply while driving a certain load. The stator is star connected with armature resistance of 0.4 ohms per phase and a synchronous reactance of 4 ohms per phase. Find the power factor at which the motor would operate when the field current is adjusted to give the line values of generated emf as (i) 600 V (ii) 380 V

(Ebph )2 = (Vph )2 + (ERph )2 - 2Vph ERph * Cos (ϴ ± ф) Given Data VL = 500 V Star connected ERph = Iaph * Zs Ra = 0.4 Ohms Zs = Ra + jXs Xs = 4 Ohms Zs = 0.4 + j 4 Ω IL = Iaph = 50 A Vph = VL / √3 = 4.019 ∟ 84.89 ϴ = 84.89° ERph = Iaph * Zs Vph = 500 / √3 (i) Cos ф = ? Ф = ? Vph = 288.67 V ERph = 50 * 4.019 (i) Eb LINE = 600 v ERph = 200.95 V Ebph = Eb Line / √3 Ebph = 600 / √3 = 346.410 V (Ebph )2 = (Vph )2 + (ERph )2 - 2Vph ERph * Cos (ϴ ± ф) (346.410 )2 = (288.67)2 + (200.95)2 - 2x(288.67)x(200.95) * Cos (84.89 + Ф ) (i) Cos ф = 0.9977 Leading Ф = 3.877 Ф = 3.877 Ebph > Vph 346.410 > 288.67

(Ebph )2 = (Vph )2 + (ERph )2 - 2Vph ERph * Cos (ϴ ± ф) Given Data ERph = Iaph * Zs Vph= VL / √3 Vph = 500 / √3 Zs = Ra + jXs Vph = 288.67 V Zs = 0.4 + j 4 Ω = 4.019 ∟ 84.89 (ii) Eb LINE = 380 v ϴ = 84.89° ERph = Iaph * Zs Cos ф = ? Ф = ? ERph = 50 * 4.019 ERph = 200.95 V Ebph = Eb Line / √3 Ebph = 380/ √3 = 219.393 V (Ebph )2 = (Vph )2 + (ERph )2 - 2Vph ERph * Cos (ϴ ± ф) (219.393 )2 = (288.67)2 + (200.95)2 - 2x(288.67)x(200.95) * Cos (84.89 - Ф ) Cos ф = 0.819 Lagging Ф = 34.93 Ф = 34.93 Ebph < Vph 219.393 < 288.67

A 3 phase, 6600 V, star connected synchronous motor delivers 500 kW power to the full load efficiency is 83%. Its armature resistance of 0.3 ohms per phase and a synchronous reactance of 3.2 ohms per phase. Its works with 0.8 leading power factor on full load. Calculate (i) Generated EMF on full load (ii) The load angle Given Data VL = 6600 V Star connected Ra = 0.3Ohms Ƞ = Output Input Xs = 3.2 Ohms Efficiency = 83 % Pin = VLILCOS ф Pout = 500 kW Cos ф = 0.8 leading

(Ebph )2 = (Vph )2 + (ERph )2 - 2Vph ERph * Cos (ϴ + ф) Sin δ = ERph Sin ϴ E bph (Ebph )2 = (3810.52)2 + (211.71)2 -2x(3810.52)x(211.71)*Cos (84.96 + 36.86) Ebph = 2925.31 V Sin δ = ERph Sin ϴ E bph Sin δ = (211.71)2 Sin (84.96 + 36.86) 3925.31 Sin δ = 2.63

Equivalent Circuit and torque equation Zs Equivalent circuit for one armature phase Eb δ V Eb O δ IaZs -Eb IaXs IaRa Ia V is the Vector sum of Reversed back EMF and Impedance drop V = -Eb +Ia Ra Load Angle δ

V O A δ 90 - ф ф ф IaXs Ia Eb 90 B C AB = Eb Sin δ Since Stator copper losses neglected Pin also represents gross mechanical Power(Pm) Cos ф = AB / IaXs AB = IaXs Cos ф Pm= (3 V Eb Sin δ ) / Xs AB = Eb Sin δ T = Pm / ɷm ɷm = (2πN )/ 60 IaXs Cos ф = Eb Sin δ Ia Cos ф = Eb Sin δ / Xs T = (3 V Eb Sin δ ) ((2πN ) / 60 )Xs P = V Ia Cos ф T = 9.55 Pm / N P = V Eb Sin δ / Xs Pin =(3 V Eb Sin δ ) / Xs For 3 phase

Torque in Synchronous Motor Starting Torque : This is the torque developed by the synchronous motor at start when rated voltage is applied to the stator. It is also called BREAK AWAY torque. It is necessary to overcome friction and inertia Running Torque : It is the torque developed by the synchronous motor under running conditions. It is decided by the OUTPUT rating of the motor and speed Pull in Torque : Initially Synchronous motor is rotated at a speed slightly less than the synchronous speed. The amount of torque developed by the motor at the time of pulling into synchronous speed is called PULL in Torque Pull our Torque : When the Synchronous motor is loaded, the rotor falls back with respect to stator by an angle called load angle δ. As δ increases magnetic locking between stator and rotor decreases. The maximum torque developed by the synchronous motor without pulling out of synchronous speed is called Pull out Torque

Applications of Synchronous Motors Synchronous motor are used drive load which are at constant speed applications in industries. They are pumps, compressor, blowers, rolling mills, paper mills, crushers etc. Synchronous motor are used to drive an AC generator to generate electricity at frequency other than supply frequency Synchronous motor are used to improve power factor (lagging) Synchronous motor are used to regulate the Transmission line voltage