Synchronous Motors.pptx
- 1. UNIT – II SYNCHRONOUS MOTOR Principle of operation – Equivalent circuit and phasor diagram – Starting methods – V and inverted V Curves – Power transforming and capability curves – Effect of varying excitation – Hunting – Synchronous condenser – Speed control of synchronous motor.
- 2. Features • Operates at constant speed • Construction is similar to AC generator • By changing the field excitation, the PF can be varied from lagging to leading • Its efficiency is high low initial cost • Improve the power factor
- 3. Principle of Operation • When a three phase voltage is applied to a three phase winding, the flux produced will be the resultant of all the three fields has magnitude of 1.5𝜑𝑚. • The field is rotating is space at a speed is given by, 𝑁𝑠 = 120×𝑓 𝑝 synchronous speed. Interlocking between the stator and rotor poles, the motor runs only at one speed.
- 5. Cross sectional view of rotor
- 7. Starting Methods of Synchronous Motor 1. By using DC source: • Synchronous motor is coupled and started by means of a DC compound motor. The speed of DC motor is adjusted by the speed regulator. • The synchronous motor is then excited and synchronized with AC supply mains.
- 8. 2. By means of Damper Grids in the pole faces • The synchronous motor is made self starting by using a special winding on the rotor poles, known as damper winding. The winding S.C copper bars which are placed in the face of field poles. • At the starting, motor provided with damper starts as a squirrel cage induction motor, when AC supply is given to the stator. • Once the motor is given excitation, the motor starts to runs at sub synchronous speed • DC suppy is given to the field winding, motor gets locked into synchronism and runs at synchronous speed. • The relative speed between damper winding and RMF is zero. So, it disconnected from the circuit.
- 9. 3. As a Slip Ring Induction Motor: • In order to obtain high starting torque, Syn. motor can be stat as SRIM. • By forming damper winding as star or delta connected instead of short circuited. • other end of the winding if connected via slip rings also, external resistance can added in series with rotor circuit • When three phase supply is given, the motor runs as SRIM and high starting torque can be achieved by adding external resistance. • The DC excitation is given to rotor when the motor attains neared to synchronous speed.
- 10. By means of Pony Motor(Small AC Motor) • A small direct coupled induction motor is called pony motor used for starting. • Before switching on the AC supply it must be synchronized by AC bus bars. • Once rotor is attains synchronous speed, DC excitation is given.
- 11. Effect of Load on Synchronous Motor • Synchronous motor, the speed is always constant irrespective of the load, by increasing the load, to supply extra torque and power, synchronous motor draws more current. • The method of drawing increased current from mains on syn. Motor can not be same as that of a DC shunt motor. • The motor can be started by using some external arrangement, then the motor is synchronized with supply. • At the instant of synchorozing an emf is induced in the stator.
- 12. • This emf is called back emf.
- 13. • The angular displacement between the stator and rotor poles is called torque angle or load angle or coupling angle. • Under light load condition, the torque angle is less • The resultant voltage 𝐸𝑅 causes 𝐼𝑎 = 𝐸𝑅 𝑍𝑠 . • 𝐼𝑎 lags 𝐸𝑅 by an angle 𝜃 = tan−1 𝑋𝑠 𝑅𝑎 internal angle.
- 15. Expression for Power and Torque developed in Synchronous Motor • OL – Supply voltage / Phase • Ia – Armature current • LM – Back emf at a load angle of 𝛿 • OM – Resultant Voltage 𝐸𝑅 • 𝐸𝑟 − 𝐼𝑎𝑍𝑠 • 𝐼𝑎 lags or leads V by an angle 𝜑 and lags behind 𝐸𝑟 by an angle 𝜃 𝜃 = tan−1 𝑋𝑠 𝑅𝑎 Line NS is drawn at an angle 𝜃 to LM
- 16. • LN and QS are perpendicular to NS and QL • Mechanical power developed per phase in the rotor 𝑃𝑚𝑒𝑐ℎ = 𝐸𝑏𝐼𝑎 cos ψ ------------(1) Consider ΔOMS, MS = 𝐼𝑎𝑍𝑠 cos ψ MS = NS –NM = LQ –NM 𝐼𝑎𝑍𝑠 cos ψ = V cos 𝜃 − 𝛿 − 𝐸𝑏 cos 𝜃 𝐼𝑎 cos ψ = 𝑉 𝑍𝑠 cos 𝜃 − 𝛿 − 𝐸𝑏 𝑍𝑠 cos 𝜃 -------(2) Substitute (2) in (1) 𝑃𝑚𝑒𝑐ℎ/𝑝ℎ = 𝐸𝑏 𝑉 𝑍𝑠 cos 𝜃 − 𝛿 − 𝐸𝑏 𝑍𝑠 cos 𝜃 𝑃𝑚𝑒𝑐ℎ/𝑝ℎ = 𝐸𝑏𝑉 𝑍𝑠 cos 𝜃 − 𝛿 − 𝐸𝑏 2 𝑍𝑠 cos 𝜃
- 17. This is the expression for mechanical power developed in terms of load angle 𝛿 and the internal angle 𝜃 of the motor for a constant voltage V &𝐸𝑏 If, 𝑍𝑠 = 𝑋𝑠, Ra is negligible, 𝜃 = 90° 𝑃𝑚𝑒𝑐ℎ/𝑝ℎ = 𝐸𝑏𝑉 𝑋𝑠 cos 90° − 𝛿 𝑃𝑚𝑒𝑐ℎ/𝑝ℎ = 𝐸𝑏𝑉 𝑋𝑠 sin 𝛿 If T is the gross torque developed by the motor, then, 𝑃𝑚𝑒𝑐ℎ = 𝑇 × 𝜔𝑛 T = 𝑃𝑚𝑒𝑐ℎ 2𝜋𝑁 60 = 9.55𝑃𝑚𝑒𝑐ℎ 𝑁
- 18. Condition for maximum power developed Differentiating equ(4) w.r.t load angle 𝛿 and then equating in to zero 𝑑𝑃𝑚𝑒𝑐ℎ 𝑑𝛿 = − 𝐸𝑏𝑉 𝑋𝑠 sin 𝜃 − 𝛿 = 0 sin 𝜃 − 𝛿 = 0 𝜃 = 𝛿 Value of maximum power, 𝑃𝑚𝑒𝑐ℎ max = 𝐸𝑏𝑉 𝑍𝑠 − 𝐸𝑏 2 𝑍𝑠 cos 𝜃 Since running at constant speed, the maximum power and hence torque depends on V and excitation voltage 𝐸𝑏
- 19. • Maximum value of 𝜃 is 90°. For all values of V and 𝐸𝑏, maximum torque will be proportional to the maximum power developed, 𝑃𝑚𝑒𝑐ℎ max = 𝐸𝑏𝑉 𝑍𝑠 when (𝛿 = 90°) This corresponds to pull out torque,
- 20. • To determine the value of excitation or induced emf𝐸𝑏 to give maximum power developed possible, diff, w.r.t 𝐸𝑏 and equate to zero. 𝑑𝑃𝑚𝑒𝑐ℎ 𝑑𝐸𝑏 = 𝑉 𝑍𝑠 − 2𝐸𝑏 𝑍𝑠 cos 𝜃 = 0 𝐸𝑏 = 𝑉 2 cos 𝜃 Substitute 𝐸𝑏 in 𝑃𝑚𝑒𝑐ℎ max, 𝑃𝑚𝑒𝑐ℎ max = 𝑉2 2𝑍𝑠 cos 𝜃 − 𝑉2 4𝑍𝑠 cos 𝜃 = 𝑉2 4𝑍𝑠 cos 𝜃 = 𝑉2 4𝑅𝑠
- 21. Torque Developed in Synchronous Motor: Starting Torque: The torque developed by the motor at starting condition while the rated voltage is applied to the stator. It is also sometimes called “break away torque”. Running Torque: Running torque is the torque developed by the motor under running condition. It is determined by the output power rating and speed of the driven machine. Peak output power determines the maximum torque that would be required . The breakdown or maximum running torque of a motor must be greater than this value in order to avoid stalling of the machine.
- 22. Pull in torque The amount of torque developed by synchronous motor to pull into synchronism when changing from induction to synchronous motor operation. Pull out Torque It is the maximum torque that the synchronous motor will develop without pulling out of synchronism. Its value ranges from 1.25 to 3.5 times the full load torque. At pull load torque, load angle 𝛿is equal to 90°. The motor runs out of synchronism for 𝛿 greater than 90°
- 23. Effect of Varying Excitation on Armature current and power factor
- 25. V Curve and Inverted V Curves
- 30. Synchronous Condenser • An over excited synchronous motor running on no load. • It takes a leading current when over excited and therefore behaves as a capacitor • When such a machine is connected in parallel with induction motor or other devices that operate at lagging PF, the leading kVAR supplied by the synchronous motor partially neutralizes the lagging kVAR of the loads. • Power factor of the system is improved
- 31. • Impacts of Power Factor P = VIcos ∅ I = 𝑃 V cos ∅ If PF gets lowered then, current will increase dramatically. • For withstanding high value of current, large size of armature conductor must be used • Cost of the machine is high • More losses will occur hence efficiency gets lowered • Poor voltage regulation Synchronous capacitors are installed in substations solely for power factor improvement. They are economical in large sizes than the static capacitors
- 32. V – applied voltage/ph ∅1 − angle by which the current 𝐼𝑎1 lags behind V In order to improve the PF, the synchronous motor is connected across the AC mains. After that, the motor will draw the current 𝐼𝑎2 which leads the applied voltage V by an angle of ∅𝑏. The net effective current is the phasor sum of 𝐼𝑎1 and 𝐼𝑎2. This total current 𝐼𝑒𝑓𝑓 lags V by an angle of very low value and the power factor improved.
- 34. Current locus for constant power lines: In synchronous motor, the input power per phase is, 𝑃𝑖𝑛 3 If V be the input voltage/phase, the power component of the current is, Icos ∅ = 𝑃𝑖𝑛 3 × 1 𝑉 A vertical is first erected representing the axis for the voltage V. A horizontal line is drawn at a point A such that OA = 𝑃𝑖𝑛 3𝑉 The current locus is varies on the straight line for various powers. For under excitation, it takes a lagging current i.e., the current vector lying on the right side of the voltage vector V. For over excitation, it takes a leading current, the current vector lying on the left side of the voltage vector V.
- 36. Current locus for Constant Power Developed P = VIcos ∅ - 𝐼𝑎 2 𝑅𝑎 𝐼𝑎 2 - 𝑉 𝑅𝑎 Icos ∅ = - 𝑃 𝑅𝑎 𝐼𝑎 2 + 𝑉 2𝑅𝑎 2 - 2 𝑉 2𝑅𝑎 Icos ∅ = 𝑉 2𝑅𝑎 2 - 𝑃𝑚 𝑅𝑎 𝑉 2𝑅𝑎 and 𝐼𝑎 are taken as two sides of a triangle, shown by OA and OB respectively. The third side is given by 𝑉 2𝑅𝑎 2 − 𝑃𝑚 𝑅𝑎 . When Pm remains constant both 𝑉 2𝑅𝑎 and 𝑉 2𝑅𝑎 2 − 𝑃𝑚 𝑅𝑎 remains constant.
- 37. The locus of the current phasor B, as q changes, is a circle. The radius of the circle is 𝑉 2𝑅𝑎 2 − 𝑃𝑚 𝑅𝑎 remains constant. and the co-ordinate of the center is given by, 𝑉 2𝑅𝑎 , 0
- 38. The power developed by the motor is increases, the radius AB on increasing. Hence, the current locus becomes a circle of small diameter. For a development of maximum power, the radius of the circle must be zero. 𝑉 2𝑅𝑎 2 − 𝑃𝑚 𝑅𝑎 = 0 𝑃𝑚,𝑚𝑎𝑥 𝑅𝑎 = 𝑉2 4𝑅𝑎 2 𝑃𝑚,𝑚𝑎𝑥 = 𝑉2 4𝑅𝑎 When the mechanical power developed in the motor is maximum, the current vector OB falls on the voltage vector having magnitude of 𝑉 2𝑅𝑎 i.e., 𝐼𝑎 = 𝑉 2𝑅𝑎
- 39. Power input/ph = V × 𝑉 2𝑅𝑎 = 𝑉2 2𝑅𝑎 Armature copper loss/ph = 𝐼𝑎 2 𝑅𝑎 = 𝑉2 4𝑅𝑎 Power developed 𝑃𝑚 = 𝑃𝑖𝑛 - Armature copper loss Pm = 𝑉2 2𝑅𝑎 - 𝑉2 4𝑅𝑎 = 𝑉2 4𝑅𝑎 The motor efficiency at maximum output becomes 50%. Losses become the half of the input power
- 40. Current locus for constant Excitation Consider the sides of the triangle to the left of voltage vector V, impedance drop 𝐼𝑎𝑍𝑠 and the induced emf E. All the three sides of this triangle are divided by 𝑍𝑠 , we get triangle with sided, 𝑉 𝑍𝑠 , 𝐼𝑎 , 𝐸 𝑍𝑠 . If the field excitation remains constant 𝑉 𝑍𝑠 , 𝐸 𝑍𝑠 remains constant and the current locus 𝐼𝑎 is a circle drawn for a radius 𝐸 𝑍𝑠 from the point A.
- 42. Form fig., 𝐼𝑎 2 = 𝑉 𝑍𝑠 2 + 𝐸𝑓 𝑍𝑠 2 - 2 𝐸𝑓 𝑍𝑠 𝑉 𝑍𝑠 cos 𝛿 𝐼𝑎 consists of two values. One value gives stable operation and another value gives unstable operation. The unstable operating point is represented by 𝐼𝑎. The angle 𝛿 between 𝐸𝑓 𝑍𝑠 and 𝑉 𝑍𝑠 for unstable operating point is more than ∅. The maximum value of angle 𝛿 for the stable operation is impedance angle 𝜑. When the synchronous motor is loaded condition, the rotor field falls back from the stator RMF by an angle 𝛿 and continuous to run in synchronous speed. the motor becomes unstable it may not be able to rotate along with the rotating field if 𝛿 goes beyond ∅.
- 43. A six pole synchronuous motor has synchronous impedance of 10 ohm and a resistance of 0.5ohm.when running on 2kV, 50Hz supply mains. Its field excitation is such that the induced emf in the machine is 1.6kV. Calculate the maximum torque developed by the machine •
- 44. A 5kW, three phase star-connected 50HZ, 400V, cylindrical rotor synchronous motor operates at rated condition with 0.8pf lagging. The motor efficiency excluding field and stator copper losses is 95% and Xs = 2.5Ohm. Calculate (i) Mechanical power developed (ii) Armature current (iii) Back emf (iv) Power angl (v) Maximum torque
- 48. A 1000kVA, 11000V, 3-ph, star connected motor has armature resistance and synchronous reactance per phase are 35ohm and 40 ohm respectively. Determine the induced emf and angular retardation of the rotor when full load 0.8PF lagging and 0.8p.f leading condition
- 52. 1. A Star connected synchronous motor rated at 187KVA, 3-Ph, 300V, 47A, 50Hz, 187.5RPM has an effective resistance of 1.5 Ohm and a synchronous reactance of 20 ohm per phase. Determine the internal power developed by the motor when it is operating at rated current and 0.8 power factor leading.(i/p = 149.78kW, Pcu=9.94kW, Pdev = 139.84kW) 2. A 2.2kV, 3-ph, star connected synchronus motor has Zs=0.2+j2.2 ohms/ph. The motor is operating at 0.5 power factor leading with a line current of 200A. Determine the generated emf per phase.
- 53. A 6600V, 3-ph, star connected synchronous motor draws a full load current of 80A at 0.8pf leading. The armature resistance is 2.2 ohm and reactance 22ohm per phase. If the stray losses of the machine are 3200W find a) Induced emf ii) Output power iii) Efficiency of the machine
- 54. A 400 V, 6-pole, 3-Ph, 50Hz, star connected synchronous motor has a resistance and synchronous impedance of 0.5 ohm and 4 ohm respectively. It takes a current of 15A at UPF when operating with a certain field current. If the load torque is increased untill the line current Is increased to 60A, the field current remain unchanged, calculate the gross torque developed and new power factor.
- 55. Advantages: 1. The speed is constant and independent of the load 2. Operate at higher efficiencies 3. Electro magneitc power varies linearly with the voltage 4. These motors can be constructed with wider air gaps than induction motor, which make them mechanically strong 5. An over excited synchronous motor having a leading power factor can be operated in parallel with induction motors Disadvantages: 1. It cannot be started under load 2. Requires DC excitation which must be supplied from the external source 3. It has a tendency to hunt 4. It cannot be used for variable speed jobs, as there is no possibility of speed adjustment 5. Collector rings and brushes are required Applications: 1. They are used in power houses and substations in parallel to the bus-bars to improve power factor 2. Synchronous motors are also used to regulate the voltage at the end of transmission lines 3. Fans, blowers, pumps, compressor, rupper mills and paper mills
- 59. A 12MVA, 5000V, 3-Ph, 4-Pole 50HZ alternator is connected to infinite bus bar. The short circuit current Is 4 times the normal full load current. And the moment of inertia of the rotating system is 22000 kg-m2.determine the normal period of oscillation
- 64. A 6-pole, 3-ph, IM deverlopes a power of 22.38kW including mech losses.which total of 1.492kw at a speed of 950RPM on 550V, 50Hz supply with a PF of 0.88.caculate i) Slip ii) Rotor Copper loss iii) total i/p if stator losses are 2000W iv) Effi v)Line current vi) number of cycles completed by the rotor frequency per minite
- 65. A 6-pole, 3-ph, 50Hz IM has a slip of 1% at no load and 3% full load. Find i) synchronous speed ii) no load speed iii) full load speed iv) frequency of rotor current at stand still v) frequency of rotor current at full load.
- 66. Types of Torque Starting Torque: The torque developed by the motor at starting condition while the rated voltage is applied to the stator. It is also sometimes called “break away torque”. Running Torque: Running torque is the torque developed by the motor under running condition. It is determined by the output power rating and speed of the driven machine. Peak output power determines the maximum torque that would be required . The breakdown or maximum running torque of a motor must be greater than this value in order to avoid stalling of the machine.
- 67. Pull in torque The amount of torque developed by synchronous motor to pull into synchronism when changing from induction to synchronous motor operation. Pull out Torque It is the maximum torque that the synchronous motor will develop without pulling out of synchronism. Its value ranges from 1.25 to 3.5 times the full load torque. At pull load torque, load angle 𝛿is equal to 90°. The motor runs out of synchronism for 𝛿 greater than 90°
- 68. Hunting: The oscillation of rotor about its final equilibrium position is called hunting. Causes: 1. Due to fault occurrence in the system that is supplied by the generator. 2. Sudden changes of load 3. Sudden changes of field current 4. Cycle variations of load torque
- 69. • Motor is loaded, rotor falls back in phase by the coupling angle or load angle • Load increase- angle increase, more torque • Sudden decrease in load- motor immediately pulled up- new value of load angle
- 70. Effects of Hunting: 1. Variation in supply voltage 2. Increases the chance of loosing synchronism 3. Increases the possibility of resonance 4. Huge mechanical stress may be developed in the rotor 5. Loss increases in the machine 6. Increase in temperature Reduction of Hunting: 1. By providing damper winding 2. By using flywheels
- 71. Damper windings: • Damper windings are used in salient pole machine to provide damping. • The damper winding consists of low resistance copper or aluminium bars embedded in notches carved in rotor pole shoes and short circuited at both ends by rings. • When the rotor is running at synchronous speed there is no induced emf in the bars. When the speed is different from synchronous speed, there is a relative motion between stator magnetic field and rotor and voltage induced in the bars • The bars are short circuited at the ends, the current flows, and produces magnetic field • The interaction of the two fields produces a torque, which tend to make the rotor speed equal to synchronous speed. • The addition of damper winding helps in damping out the rotor oscillations
- 72. A 10HP, 400V, 3-ph, star connected synchronous motor has Zs=0.35+j2.8 ohm/ph. Find the angle of retard and the voltage at which it must be excited to give a full load output at 0.866pf leading. Assume efficieny as 0.88.(Pin=8.358kW, 𝞱=82.9, Φ=30, V=231V, Er=39.3V, Eb=249V, δ=8.36)
- 73. Speed control • A synchronous motor is constructionally same an alternator • - It runs at synchronous speed or it remains stand still • - Speed can be varied by varying supply frequency because synchronous speed, Ns = (120f/p) • - Due to unavailability of economical variable frequency sources, this method of speed control was not used in past & they were mainly used for constant speed applications • - The development of semiconductor variable frequency sources such as inverter & cycloconverter allowed the use of synchronous motor in variable speed applications
- 74. Types of synchronous motors • Commonly used synchronous motors are 1. Wound field synchronous motor (Cylindrical & salient pole) 2. Permanent magnet synchronous motor 3. Synchronous reluctance motor 4. Hysteresis motor • - All these motors have a stator with 3 phase winding which is connected to an AC source • - Fractional horse power synchronous reluctance & hysteresis motors employ a 1 phase stator
- 76. Speed – Torque Characteristics
- 78. Self control mode using a rotor position sensor or stator voltage sensor
- 79. True synchronous mode where the speed of motor is determined by the external independent oscillator
- 80. Self controlled synchronous motor drive employing a load commutated thyristor inverter
- 84. `
- 87. Capability Curve of Synchronous Generator: • Capability Curve of Synchronous Generator defines the bounds within which it can operate safely. Various bounds imposed on the machine are: • MVA-loading cannot exceed the generator rating. This limit is imposed by the stator heating. • MW-loading cannot exceed the turbine rating which is given by MVA (rating) x pf (rating). • The generator must operate a safe margin away from the steady-state stability limit (δ = 90°). This can be laid down as a maximum allowable value of δ • The maximum field current cannot exceed a specified value imposed by rotor
- 88. • To draw the capability curve of the synchronous generator, its phasor diagram is used which is redrawn in Fig. armature resistance is neglected. After multiplying voltage magnitude of each voltage phasor by (3Vt/Xs), the phasor diagram is redrawn in Fig. 8.49. It is immediately recognized that OMN is the complex power triangle (in 3-phase values) wherein
- 91. Case 1 − Under-Excitation of Synchronous Motor • The synchronous motor is said to be under- excited if the field excitation is such that (Ef<V). Under this condition, the armature current (Ia) lags behind the supply voltage (V) and consumes lagging reactive power (Q) i.e. the motor power factor is lagging as shown in Figure-1.
- 92. Case 2 − Normal Excitation of Synchronous Motor • If (Ef=V), the synchronous motor is said to be normally excited. Under this condition, the reactive power of the motor is zero (i.e., Q = 0), that is the motor is neither absorbing nor delivering reactive power. Thus, the power factor of the motor is unity. For a given load, at unity power factor, the resultant voltage (Er) and hence, the armature current (Ia) are minimum as shown in Figure-2.
- 93. Case 3 − Over-Excitation of Synchronous Motor • The synchronous motor is said to be over- excited if the field excitation is such that (Ef>V). Under such conditions, the armature current (Ia) leads the supply voltage (V) and the motor supplies lagging reactive power to the system. Hence, motor power factor is leading as shown in Figure-3.