Electrical Machine - II -English-QB Compose By Lalit Chaudhary

Synchronous Motor 51 But half a period later, the stator poles interchange their polarity being the supply AC and under these condition the south pole and the rotor will fail under north pole of stator and north pole of rotor under south pole of stator. Being opposite poles they are attracted and the rotor tends to rotate in clockwise direction. Thus, they find that due to continuous and rapid rotation of stator poles the rotor is subject to a torque which is rapidly reversing. Owing to its large inertia, the rotor does not responded to much quick reversing torque, with the result the rotor remains stationary that is why the synchronous motor is not a self-starting motor. In case arrangement is made that rotor poles also shift their positions with the stator poles, then they will continuously experience a unidirectional torque. It is only required at the time of starting, once the rotor comes in the magnetic coupling, it will go on speed. Q.2. Explain the construction and principle of operation of 3-φ synchronous motor. (S/2010) Explain the construction of synchronous motor. (2017) Ans. Principle of Operation: Refer Q.1. Construction: A synchronous motor has the same relationship with an alternator like a DC motor have a DC generator i.e., if an AC generator (alternator) supplied AC power it is capable of rotating as a motor and doing mechanical work. In case of a DC motor the field and armature, both are supplied from DC supply mains, since both require direct current. But in case of a synchronous motor the field structure is to be energized by direct current as in the alternator, where as armature winding is connected to a three phase AC supply mains. Such a machine, therefore, requires two sources of supply, an AC source to supply the power for driving the armature and a DC source to excited the field. The essential parts of a three phase synchronous motor are as follows: (i) laminated stator core with three phase winding, (ii) rotating field structure complete with damper winding and slip rings, (iii) brushes and brush holders, (iv) two end shields to house the bearings to support the rotor shaft. The stator core and windings of a synchronous motor are similar to those of a three phase alternator. The leads for the stator winding terminate in a terminal box usually mounted on the side of the motor frame. The rotor of the synchronous motor usually has salient field poles connected to give alternate polarity. The number of rotor field poles must be the same as that of stator field poles. The field circuit leads are brought out to two slip rings mounted on the shaft. A squirrel cage or damper winding is provided for making the motor self starting. Carbon brushes mounted in brushes holders make contact with the

52 Electrical Machines–II two slip rings. The terminals of the field circuit are brought out from the brush holders to a second terminal box mounted on the motor frame. Q.3. Explain the analysis of phasor diagram of synchronous motor under normal working condition. Ans. Consider a phasor diagram with normal excitation i.e., such a current through field winding will produce flux that will adjust magnitude of Ebph and Vph. Let, δ be the load angle corresponding to the load on the motor. So from the exact opposing position of Ebph with respect to Vph. Ebph gets displaced by angle δ. Vector difference of Ebph and Vph, gives the phasor which represents IaZs called Ebph. Now, Zs = Ra + jXs Ω Where, Ra = resistance of stator per phase Xs = synchronous reactance of stator per phase i.e. θ = tan–1 Xs Ra and |Zs| = √Ra2 + Xs2 Ω This angle θ is called internal machine angle or an impedance angle. The significance of θ will tell that phasor Iaph lags behind Erph i.e., IaZs by angle θ. Current always lags in case of inductive impedance with respect to voltage drop across that impedance. So, phasor Iaph can be shown lagging with respect to Erph by angle θ. Practically, Ra is very small as compared to Xa and hence θ tends to 900. Fig. 2.2 Phasor Diagram Under Normal Working Condition

Synchronous Motor 53 ∴ φ = Vph Iaph = power factor angle and cosφ = power factor at which motor is working. The nature of this pf is lagging if Iaph lags Vph by angle φ. While, it is leading if Iaph leads Vph by angle φ. Phasor diagram indicating all the details is shown in fig. 2.2. Q.4. How pf of synchronous machine is dependent on its excitation? Explain. (S/2006) Explain the effect of changing excitation when the load is constant on a synchronous motor. (2008) With the help of the vector diagrams, explain the effect of excitation on the power factor of the synchronous motor. (S/2009) Describe the effect of varying the excitation of synchronous motor upon armature current and pf when input is constant. Draw the vector diagram. (2011, 14, S/17) Explain the effect of change in excitation of synchronous motor on its armature current and power factor. (S/2012) With the help of the suitable vector diagrams, explain how pf of synchronous motor changes with change in excitation. (2012, 14, S/15) Explain the effect of changing excitation on armature current and power factor of synchronous motor with the help of vector diagram. (S/2016) Draw phasor diagram of a synchronous motor when operating at unity, lagging and leading power factor for showing the effect of varying field excitation. (2017) Ans. The change in field excitation does not effect the motor speed but effect the power factor and armature current, considering the supply voltage and input to motor constant, but the effect occurs in synchronous motor during excitation are as follows: (i) Fig. 2.3 shows the case for 100% excitation, i.e., E = V. Here, armature current I is in phase with the applied voltage V. So, current drawn by motor would be minimum and the motor runs at a unity power factor. In the vector diagram, Fig. 2.3 Excitation is 100%

54 Electrical Machines–II α = angle between V and E φ = angle between V and I. (ii) Fig. 2.4 shows the excitation is less then 100%., i.e., E < V. So, when the excitation is reduced, the induced emf decreases in magnitude to the value which shifts the resultant voltage E in clock-wise direction and since, the phase angle between the resultant voltage Er and current I is constant (900), so current vector I is also shifted in clockwise direction. Therefore, phase φ between supply voltage V and current I increases. Due to increase of phase angle φ, the power factor cos f Fig. 2.4 Excitation is Less than 100% decreases and current I increases since I cos φ is constant. Hence, with reduction in excitation, synchronous motor draws more current from the supply at lower power factors and the motor operates at lagging pf. (iii) Fig. 2.5 represents the condition for over-excited (excitation more than 100% motor) i.e., E > V. So, the resultant voltage vector Er is pulled anti-clockwise and so is I. It is seen that now motor operates at a leading power factor and will draw a larger value of armature current than it does when normally excited. Fig. 2.5 Excitation is above 100% Therefore, the power factor of a synchronous motor can be controlled through adjustment of its field current. A certain value of field current

Synchronous Motor 55 (normal excitation) for any particular load on the motor will give unity power factor operation. As the field current is reduced from the normal excitation value, the power factor will become more and more lagging. As the field current is increased above the normal excitation value, the power factor will become more and more leading. For any particular load the armature current will vary with the excitation. The armature current will have its smallest value when the motor is normally excited so that its power factor is unity. Q.5. Derive expression for power developed by a synchronous motor. Ans. Except for very small machines, the armature resistance of a synchronous motor is negligible as compared to its synchronous reactance. Hence, the equivalent circuit for the motor becomes as shown in fig. 2.6 (a). From the phasor diagram of fig. 2.6 (b). It is seen that, (a) (b) Fig. 2.6 AB = Eb sin α = IaXs cos φ or VI cos φ = Eb V sin α Now, VIa cos φ = motor power input/phase ∴ Pin = EbV sin α per phase Xs Pin = 3 EbV sin α for three phase Xs Since, stator Cu losses have been neglected, Pin also represents the gross mechanical powers (Pm) developed by the motor. ∴ Pm = 3EbV sin α Xs The gross torque developed by the motor is, Tg = 9.55 Pm N-m Ns Where, Ns = 120f in rpm P

56 Electrical Machines–II Q.6. Write a short note on comparison between 3-φ induction motor and synchronous motor. (S/2010) Compare induction motor and synchronous motor. (S/2011, S/17) Compare 3-φ induction motor with 3-φ synchronous motor. (S/2012) Compare synchronous motor and induction motor. (2015, 17) Ans. The differentiation/comparison between synchronous motor and induction motor are as follows: S.No. Synchronous Motor Induction Motor (i) It has no self-starting torque It is self-starting and does (ii) and requires some external not require any external (iii) means for its starting. means for its starting. (iv) (v) Its average speed is constant Its average speed is not (vi) (vii) and is dependent on load. constant but falls, depending (viii) on load. (ix) It requires DC excitation. It does not require DC (x) excitation. (xi) Its speed can’t be controlled. Its speed can be controlled to some extent. Its torque is sensitive to Its torque is more sensitive change in supply voltage. to change in supply voltage. It is complicated and costly. It is simple and cheaper. It can be operated under It operates only at lagging wide range of pf both lagging pf. and leading. It can be used for supply It is used for supply mechanical load as well as for mechanical load only. power factor improvement. As load increases, load angle As load increases, the speed increases, keeping speed keeps on decreasing. constant as synchronous. Motor is sensitive to sudden Phenomenon of hunting is load changes and hunting absent. results. Breakdown torque is Breakdown torque is proportional to the supply proportional to the square of voltage. the supply voltage. Q.7. Write the advantages, disadvantages and applications of synchronous motor. Ans. Advantages of Synchronous Motor: The advantages of synchronous motor are as follows:

Synchronous Motor 57 (i) It gives constant speed from no load to full load. (ii) Electromagnetic power varies linearly with the voltage. (iii) These motors usually operate at higher efficiencies, especially in the low speed which make them better mechanically. (iv) These motor can be constructed with wider air gap, than induction motor, which make them better mechanically. (v) These motor can be made to operate at a leading pf and thereby improve the pf of an industrial plant from one that is normally lagging to one that is close to unity. Disadvantages of Synchronous Motor: The disadvantages of synchronous motor are as follows: (i) It cannot be used for variable speed jobs as there is no possibility of speed adjustment. (ii) It requires DC excitation which must be supplied from external source. (iii) It cannot be started under load. Its starting torque is zero. (iv) It has a tendency to hunt. (v) It may fall out synchronism and stop when over loaded. (vi) Collector rings and brushes are required. (vii) The cost per KW is generally higher than induction motor. Applications of Synchronous Motor: The applications of synchronous motor are as follows: (i) It is used in power houses and substations in parallel to the bus bars to improve the pf. For this purpose, its run without mechanical load on it and over excited. (ii) In factories it is used with a large number of induction motors or other power apparatus operating at lagging pf, it is used for improving the pf. (iii) It is used to control the voltage at the end of transmission line by varying its excitation. (iv) It is also used in rubber mills, textile mills, cemment factories, mining industries and other big industries for power applications. Q.8. Write a short note on V-curves. (2010, S/11, S/15) What is the effect of excitation on pf of synchronous machine? Draw V-curve. (2011) Write short note on V-curves of synchronous motor. (S/2012) With the help of suitable diagram explain V-curve. (2015) Draw V-curve. (S/2017) Draw and explain V-curve of synchronous motor. (2017) Ans. Effect of Excitation on PF of Synchronous Machine: Refer Q.4. V-Curve: V-curves of synchronous motor are the curves which shows the relationship between field current and armature current for constant mechanical output. The current has large values both for low and high

58 Electrical Machines–II values of excitation (though it is lagging for low excitation and leading for higher excitation). In between, it has minimum value corresponding to a certain excitation. V-curves are shown in fig. 2.7 (a). Fig. 2.7 For the same input, armature current varies over a wide range and so cause the power factor also to vary accordingly. When over- excited, motor runs with leading pf and with lagging pf when under- excited. In between the pf is unity. The variation of pf with excitation are shown in fig. 2.7 (b). The curves for pf looks like inverted V-curves. It would be noted minimum armature current corresponds to unity pf. Q.9. Write the characteristics of synchronous motor. Ans. The characteristics of synchronous motor are as follows: (i) It runs either at synchronous speed or nor at all i.e. while running it maintains a constant speed. The only way to change its speed is to vary the supply frequency (because Ns = 120 f/P). (ii) It is not inherently self starting. It has to be run upto synchronous speed by some means, before it can be synchronised to the supply.

Synchronous Motor 59 (iii) It is capable of being operated under a wide range of power factors both lagging and leading. Hence, it can be used for power correction purposes, in addition to supplying torque to drive loads. (iv) If the excitation is increased the stator flux is neutralized by rotor flux and therefore, power factor becomes unity. (v) On no load synchronous motor have very little current from the supply mains to meet internal losses. With fixed excitation the input current increases with increases in load. As, input current reaches maximum, no further increases in load possible. If the motor will overloaded then the motor will stop. Q.10. Explain the following torques associated with a synchronous motor : (i) starting torque, (ii) running torque, (iii) pull-in torque, (iv) pull-out torque. (S/2009) Define in a synchronous motor : (i) starting torque, (ii) running torque, (iii) pull-in torque, (iv) pull-out torque. (2012, S/15, 15, 17) Define starting torque and running torque in a synchronous motor. (S/2017) Ans. (i) Starting Torque: When full voltage is applied to stator (armature) winding then torque is developed by motor. It is also called as break away torque. Its value can be as high as 200 to 250% as in case of loaded reciprocating two cylinder compressors and as low as 10% as in case of centrifugal pumps of full load torque. (ii) Running Torque: This torque is developed by the motor under running conditions. It is defined by the horse power and speed of the driven motor. The peak horse power indicates the maximum torque that should be required by the motor. The motor should have a maximum running torque greater than this value in order to avoid stalling. (iii) Pull-in Torque: When synchronous motor is started as an induction motor till it runs 2 to 5% below the synchronous speed. Then excitation is switched on and the rotor pulls in step with the synchronously rotating stator field. This quantity of torque at which the motor will pull into step is called the pull-in torque. (iv) Pull-out Torque: This is the maximum torque when the motor can develop without pulling out of step or synchronism is called the pull-out torque. When the load on motor is increased, the process of rotor tends to fall back in phase by some angle (called load angle) behind the synchronously revolving stator magnetic field through it runs

60 Electrical Machines–II synchronously. Motor develops maximum torque when its rotor is retarded by an angle of 900. If there is any increment in load will cause the motor to pull-out of step and stop. Q.11. Write a short note on synchronous capacitor. (S/2012) Write a short note on synchronous condenser. (S/2013,13) Ans. In large industrial plants, which have a low lagging power factor load, it is often found economical to install a synchronous motor, even though the motor is not required to drive a load. The motor is operated over excited at no-load so that the current drawn by it leads the voltage by nearly 900. A synchronous motor used in this way is said to float on the line because it has no mechanical output. Since, the motor operating at no-load acts in the same manner as a static capacitor and when operated in this manner is called synchronous condenser or a synchronous capacitor. As compared with a synchronous motor with equal armature voltage and current ratings, a synchronous condenser requires more copper in the field winding to carry the increased field current. The synchronous condenser does not require so large shaft and bearings as the synchronous motor because no shaft torque is required. Synchronous condenser are some times operated at power factor ranging from lagging through unity to leading for voltage control. When operated in this manner a synchronous condenser is called as synchronous reactor. Q.12. What are the advantages and disadvantages of synchronous condenser? Ans. Advantages: The advantages of synchronous condenser are as follows: (i) By varying the field excitation of current drawn by the motor which can changed by any amount. This helps in for achieving less control of power factor. (ii) It have high thermal stability to short circuit currents. (iii) The faults can be removed easily. Disadvantages: The disadvantages of synchronous condenser are as follows: (i) The maintenance cost is high. (ii) It produces noise. (iii) It have considerable losses in the motor. (iv) The cost is greater than that of static capacitors of same rating. Q.13. Write a short note on damper winding. (2011) Ans. Synchronous motor are not self-starting machines. These machines are made self-starting by providing a special winding in the rotor poles, known as damper winding or squirrel cage windings. The damper winding consists of short circuited copper bars embedded in the face of the rotor poles.

Synchronous Motor 61 When an AC supply is provided to stator in a three phase synchronous motor, stator winding produces rotating magnetic field. Due to the damper winding present in the rotor winding of the synchronous motor, machine start as induction motor. The exciter for synchronous motor moves along with rotor. When the motor attains about 95% of the synchronous speed, the rotor windings is connected to exciter terminals and the rotor is magnetically locked by the rotating magnetic field of stator and it runs as a synchronous motor. Functions of Damper Winding: The main functions of damper winding are as follows: (i) Damper winding helps the synchronous motor to start on its own (self-starting machine) by providing starting torque. (ii) By providing damper winding in the rotor of synchronous motor hunting of machine can be suppressed. When there is change in load, excitation or change in other conditions of the system rotor of the synchronous motor will oscillate and for about an equilibrium position. At times these oscillations becomes more violent and resulting in loss of synchronism of the motor and comes to halt. Q.14. Explain why synchronous motor is not self-starting? Write starting methods of synchronous motor and explain any one method. (2003) What are the methods of starting synchronous motor? Explain in brief. (2013, S/17) What are the different methods of starting synchronous motor explain? (2017) Ans. Reason for not Self-Starting Synchronous Motor: Refer Q.1. Starting Methods of Synchronous Motor: The starting methods of synchronous motor are as follows: (i) From DC Source: In this method the synchronous motor is coupled with a small DC compound motor whose speed is control through a speed regulation. The DC supply is given to the rotor of the synchronous motor and the motor is rotated at synchronous speed, when it starts building up voltage equal to supply voltage the synchronising switch is closed and the DC motor is disconnected. (ii) By Means of AC Motor: In this method a small induction motor is coupled with the synchronous motor. The DC supply from the exciter is given to the rotor of the synchronous motor. Before switch on, AC supply to the synchronous motor, it is synchronised with the bus-bar. The induction motor used for this purpose is generally of less number of poles than the poles of synchronous motor so that synchronous speed can be achieved.

62 Electrical Machines–II (iii) By Means of Damper Winding in the Pole Shoes of Rotor: In this method, the field magnets of rotor are provided with an additional winding known as damper winding (similar to squirrel cage winding) in shown in fig. 2.8. The damper winding consists of copper bars (short circuited to ends) embedded in pole shoes/faces of the rotor. To start the motor three phase supply is given to the stator winding of the motor, and the rotor field winding is left unenergised. The rotating magnetic field produced in the stator which induces current in the damper winding and the motor accelerates like a squirrel cage induction motor. When motor reaches nearly the synchronous speed, then the rotor field winding is energised. This results in magnetically interlocking or rotor poles with stator and the motor continues to run at synchronous speed. Fig. 2.8 (iv) By Using Pony Induction Motor: In this method, the rotor is brought to the synchronous speed, with the help of some external device like small induction motor. Such an external device is called pony induction motor. Once the rotor attains the synchronous speed, the DC excitation to the rotor is switched on and the synchronism is established pony motor is decoupled. The motor then continuous to rotate as synchronous motor. Q. 15. Write a short note on hunting in synchronous machine. (S/2002, S/09, S/11) Write the effect of hunting in synchronous motor. Also write methods to reduce hunting? (S/2010) Explain hunting produced in synchronous motor. How is hunting prevented? (2011) Write a short note on hunting and its prevention. (2012, S/15, S/16)

Synchronous Motor 63 Explain hunting phenomenon in synchronous motor and how it can be prevented. (S/2017) What is hunting? How it is prevented? (2017) Ans. When a synchronous motor is used for driving a varying load, then a condition known as hunting is produced. Hunting may also be caused if supply frequency is pulsating. When a synchronous motor is loaded, its rotor falls back in phase by the coupling angle α. As load is progressively increased, this angle also increases so as to produce more torque for coping with the increased load. If now, there is sudden decrease in the motor load, the motor is immediately pulled up or advanced to a new value of α corresponding to the new load. But in this process, the rotor overshoots and hence, is again pulled back. In this way, the rotor starts oscillating about its new position of equilibrium corresponding to the new load. The amplitude of these oscillations is built upto a large value and may eventually become so great so as to throw the machine out of synchronism. This condition is known as hunting. Hunting is objectionable because it produces severe mechanical stresses and great variations in current and power drawn by the motor. Prevention of Hunting: The methods which are used to reduced or prevent the hunting are as follows: (i) To stop the build up of these oscillations, damper windings (also known as squirrel cage winding) are employed. These damper winding consists of short circuited Cu bars embedded in the faces of the field poles of the motor. The oscillatory motion of the rotor sets up eddy current in the damper windings which flow in such a way as to suppress these oscillations. (ii) It can also be reduced by making the flywheels of motor in a big size. (iii) It is also reduced by using the dashpot on governor of the motor. Q.16. Write short note on alternator cooling methods. (2005) Explain different methods of cooling synchronous machine. (S/2017, 17) Ans. By improving the cooling system of a machine, motor can increase the current carrying capacity of the machine and hence its output. The temperature of various parts of a machine is raised above the room temperature due to losses of energy in its different parts like core, conductors and bearings which appear in the form of heat. Various methods of cooling are adopted to dissipate the heat to the atmosphere and lower the temperature of the machine. Methods of Cooling: There are different methods of cooling synchronous motor which are as follows: (i) Small machines are cooled by providing ventilating ducts.

64 Electrical Machines–II (ii) Bigger machines are cooled by providing ventilating ducts and also fans on the rotor shaft, which induce a stream of air flow into machine. (iii) In still bigger machines, a separate fan driven by an independent motor may be used. (iv) The stator winding are cooled by passing water or oil through the hollow stator conductors. This type of cooling is called closed circuit cooling. (v) Hydrogen cooling is adopted for the cooling of rotor conductors of turbo-alternators. The rotor conductors are made hollow through the hydrogen gas at a pressure more than atmospheric pressure is allowed to flow through them. The stator cooling is done by making a provision of a number of radical inlet passages for the gas to flow in. Since, hydrogen forms an explosive mixture with air therefore, first carbon dioxide is filled in the casing and then hydrogen is introduced by replacing carbon dioxide. Q.17. Write the advantages of doing distributed winding in synchronous machine. (S/2016) What are the advantages of distributed winding in synchronous motor? (2017) Ans. For obtaining smooth sinusoidal emf wave form conductors are placed in several slot under a single pole, this armature winding is known as distributed winding. Advantages: The advantages of doing distributed winding in synchronous machine are as follows: (i) It reduces harmonic emf and so wave form is improved. (ii) It also diminishes armature reaction. (iii) Even distribution of conductors, help for better cooling. (iv) The core is fully utilized as the conductors are distributed over the slots on the armature periphery. NUMERICAL PROBLEMS AND SOLUTIONS Prob.1. A 3-phase 400V star connected synchronous motor is taking 50A at 0.8 pf. Motor impedance is (0.25 + j3.2) Ω per phase. Calculate input power and back emf of motor. (S/2007, 09) A 3-phase 400V star connected synchronous motor is taking 50A at 0.8 pf leading. Impedance of motor is (0.25 + j3.2) per phase. Calculate the input power of the motor. (2010) Sol. Given, Voltage, VL = 400 V Current, IL = 50A cos φ = 0.8 Input power, P2 = √3 VL IL cosφ

Synchronous Motor 65 = √3 × 400 × 50 × 0.8 1000 = 27.71 kW Ans. Impedance per phase, Zs = √(R)2 + (Xs)2 = √(0.25)2 + (3.2)2 = 3.21 Ω θ = tan–1 Xs = 3.21 R 0.25 = tan–1 12.8 = 85.50 φ = cos–1 (0.8) = 36.80 Voltage per phase, V = 400 √3 = 231 V In fig. 2.9, shows the vector diagram. AB line shows induced or back emf Eb and OB line shows the resultant voltage. ∴ Er = IZs Fig. 2.9 = 50 × 3.21 = 160.5 V ∴ (θ + φ) = 85.50 + 36.80 = 122.30 Now, cos φ = cos (1800 – 122.30) = cos 57.70 = 0.5344 From triangle OAB, Eb = √V2 + Er2 + (2VEr cosφ) = √(231)2 + (160.5)2 + (2×231×160.5×.5344) = 344.60 volt/phase Voltage, EL = √3 Eb EL = √3 × 344.60 Ans. = 596.86 V Prob.2. A 400 V, 5 HP, 3-phase synchronous motor has negligible armature resistance and a synchronous reactance of 12 ohm / phase. Determine the minimum current and the corresponding induced emf for full-load condition. Assume an efficiency of 80%. (1985) Sol. Motor output = 5 HP = 5 × 746 = 3730 W Efficiency, η = 0.80

66 Electrical Machines–II Output Motor input = η = 3730 0.8 = 4662.5 W Because motor input = √3 VL IL cos φ IL cos φ = Motor input in watts √3 VL = 4662.5 √3 × 400 = 6.73 A The current will be minimum when power factor, cos φ is unity and in such condition, I = IL cos φ Ans. = 6.73 A Impedance drop, Er = I × Zs = 6.73 × 12 = 80.76 V Supply voltage per phase, V = 400 = 231 V √3 Induced emf/phase, E = √V2 + E2r = √(231)2 + (80.76)2 = 244.71 V Line induced emf = √3 × 244.71 Ans. = 423.85 V Prob.3. The input an 11000 V, 3-φ star connected synchronous motor is 60 A. The effective resistance and synchronous reactance / phase are respectively 1 Ω and 30 Ω Find: (i) the power supplied to the motor, (ii) mechanical power developed, (iii) induced emf for a pf of a 0.8 leading. Sol. (i) Motor power input = √3 × 11000 × 60 × 0.8 Ans. = 915 kW (ii) Stator Cu loss/phase = 602 × 1 = 3600 W Cu loss for three phases = 3 × 3600 = 10.8 kW Pm = P2 – rotor Cu loss = 915 – 10.8 = 904.2 kW Ans.

Synchronous Motor 67 (iii) VP = 11000 √3 = 6350 V φ = cos–1 0.8 = 36.90 ( )30 θ = tan–1 1 = 88.10 cos (θ + φ) = cos (88.10 + 36.90) = – 0.573 Zs = 30 Ω Stator impedance drop/phase = IaZs Er = 60 × 30 = 1800 V E2b = (Vp)2 + (Er)2 – 2VpEr × cos (θ + φ) E2b = 63502 + 18002 – 2 × 6350 × 1800 × (– 0.573) Eb = 7527 V Ans. Line value of Eb = 7527 × √3 = 13037 V Prob.4. A 75 kW, 50 Hz, 440 V, 4 pole star connected cylindrical rotor synchronous motor operates at rated condition with 0.8 pf leading. The motor efficiency excluding field and stator losses is 95% and Xs = 3.5 Ω. Calculate : (i) mechanical power developed, (ii) armature current, (iii) power angle, (iv) maximum torque of the motor. Sol. Given, Output of motor, Pout = 75 kW = 75 × 103 W Frequency, f = 50 Hz Voltage, V = 440 V Power factor, cosφ = 0.8 Efficiency, η = 95% or 0.95 Synchronous speed, Ns = 120 f = 120 × 50 P4 = 1500 rpm (i) Mechanical power developed, Pm = Pout = Pin η

68 Electrical Machines–II = 75 × 103 0.95 = 78947.37 W Ans. (ii) Since, power input is known as armature current (Ia) is, √3 × V × Ia × cosφ = Pin ∴ √3 × 440 × Ia × 0.8 = 78947.37 Ia = 78947.37 √3 × 440 × 0.8 = 129.49 A Ans. (iii) Voltage/phase = 440 = 254V √3 Let, V = 254 ∠00 as shown in fig. 2.10, then, Where, V = Eb + j Ia Xs Eb = V – j Ia Xs Ia = 129.49 ∠ cos–10.8 = 129.49 ∠ 36.90 tanθ = Xs Fig. 2.10 Ra If, Ra is common then θ = 900 Xs = 2.5 ∠ 900 Eb = 254 ∠00 – 129 ∠ 36.90 × 2.5 ∠ 900 = 254 ∠00 – 322 ∠126.90 = 254 – 322 (cos 126.90 + j sin 126.90) = 254 – 322 (– 0.6 + j 0.8) = 516 ∠– 300 Ans. (iv) Maximum torque developed at α = 900 Maximum mechanical power, Pm = 3EbV sin δ Xs = 3 × 254 × 516 × sin 900 2.5 = 157276.8 W Maximum torque = 9.55 × 157276.8 1500 = 1001.3 N-m Ans. Prob.5. A 3-phase, 3300 V, 1.5 MW star connected synchronous motor has Xd = 4Ω/phase and Xq = 3Ω/phase. Neglecting all losses.

Synchronous Motor 69 Calculate the excitation emf. When motor supplies rated load at unity pf. Calculate the maximum mechanical power which the motor would develop for this field excitation. Sol. Given, Input power, Pi = 1.5 MW Pf, cos φ = 1.5 × 106 W sin φ φ =1 Voltage, V =0 Voltage per phase = 00 = 3300V = 3300 = 1905V √3 Input power, Pi = √3 VIa cos φ Ia = 1.5 × 106 √3 × 3300 × 1 = 262 A Since, tan ψ = V sin φ – IaXq V cos φ ∴ ψ ∴ Power angle α = 1905 × 0 – 262 × 3 1905 = – 0.4125 = – 22.40 = φ – ψ = 0 – (– 22.4) = 22.40 Since, armature current, Ia has two components, Id and Iq, then Id = 262 × sin (– 22.40) = –100A Iq = 262 × cos (– 22.40) = 242A Back emf, Eb = Vcos α – IdXd = 1905 cos (– 22.40) – (– 100 × 4) = 2160V Mechanical power per phase, Pm = EbV sin α + V2(Xd– Xq) sin 2 α Pm kW/phase Xd 2XdXq = 2160 × 1905 × sin α+ (1905)2 (4 1–030)0× sin 2 α 4 × 1000 2 ×4×3× = 1029 sin α + 151 sin 2 α If developed power has to reach maximum value, then, dPm = 1029 cos α + 2 × 151 cos 2 α = 0 dα

70 Electrical Machines–II 1029 cos α + 302 (2 cos2α –1) = 0 604 cos2α + 1029 cos α – 302 = 0 Solve this quadratic equation as, cos α = –1029 + √(1029)2 + 4 × 604 × 302 2 × 604 = 0.255 α = 75.20 ∴ Maximum, Pm = 1029 sin 75.20 + 151 cos 2 × 75.20 = 12343 kW/phase Hence, Maximum power developed for three phases = 3 × 12343 = 37029 kW Ans. Prob.6. A 2300, 3-phase star connected synchronous motor has a resistance of 0.2Ω per phase and a synchronous reactance of 2.2 Ω per phase. The motor is operating at 0.5 pf leading with a line current of 200 A. Determine the value of the generated emf per phase. Sol. Given, Voltage, V = 2300 V Resistance, R = 0.2 Ω/phase Synchronous reactance, Xs = 2.2 Ω/phase Power factor, cos φ = 0.5 Then, φ = cos–1 (0.5) = 600 (leading) Fig. 2.11 θ = tan–1 Xs = tan–1 2.2 R 0.2 = 84.80 (θ + φ) = 84.80 + 600 = 144.80 cos 144.80 = – cos 35.20 V = 2300 = 1328V √3 Zs = √R2 + Xs2 = √(0.2)2 + (2.2)2 = 2.209 Ω IZs = 200 × 2.209 = 442 V The phasor diagram is shown in fig. 2.11, Eb = √V2 + E 2 + 2VEr cos (θ + φ) r = √(1328)2 + (442)2 + 2 × 1328 × 442 × √cos 35.20 = 1708 V/phase Ans.

Synchronous Motor 71 Prob.7. A 3-phase, 6.6 kV, 50 Hz, star-connected synchronous motor takes 50 A current. The resistance and synchronous reactance per phase are 1ohm and 20 ohm respectively. Find the power supplied to the motor and induced emf for a pf of: (i) 0.8 lagging, (ii) 0.8 leading. Sol. Given, Voltage, V = 6.6 kV = 6600 V Frequency, f = 50 Hz Current, Ia = 50 A Resistance per phase, Ra = 1Ω Synchronous reactance per phase, Xs = 20Ω (i) Power applied to the motor = √3 × 6600 × 50 × 0.8 = 457261 W Supply voltage/phase = 6600 = 3810 W √3 φ = cos–1 (0.8) = 360 52' According to fig. 2.12(a), ( ) ( )θ = tan–1 Xs = tan–1 20 Ra 1 = 870.8' Zs = √(Xs)2 + (Ra)2 = √(20)2 + (1)2 = 20Ω Impedance drop, Er = IaZs = 50 × 20 = 1000 V/phase Eb2 = V2 + Er2 – 2VEr cos (θ + φ) Eb2 = (3810)2 + (1000)2 – 2 × 3810 × 1000 × cos5103' Eb = 3278 V/phase Line induced emf = 3278 √3 = 1892 V Ans. (ii) Power supplied to the motor will same. The value of current is shown in fig. 2.12 (b) at a leading angle of Ω = 36052'

72 Electrical Machines–II (a) (b) Fig. 2.12 Now, (θ + φ) = 8708' + 36052' ∴ = 1240 cos 1240 = – cos 560 Eb2 = (3810)2 + (1000)2 – 2 × 3810 × 1000 × (– cos 560) Eb = 4447 V/phase Ans. Line induced emf = √3 × 4447 = 7702 V FF

73 3 SYNCHRONOUS GENERATOR Chapter SUBJECTIVE QUESTIONS AND ANSWERS Q.1. Explain the principle of working of three phase alternator. Derive its emf equation. (S/2005) Derive an emf equation of an alternator. (S/2008, S/13, 13) Derive an emf expression for an alternator. (2009) Derive the emf equation of synchronous generator. (2011) Write the working principle of 3 alternator and derive the equation of electromotive force of an alternnator. (2015) Derive an induced emf equation for an alternator. (S/2016) Write emf equation of an alternator. (S/2017) Write application of synchronous generator (alternator). (S/2017) Explain working principle of 3-φ alternator. Derive emf equation of an alternator. (2017) Ans. Principle of Working of an Alternator: AC generators or alternators operate on the same fundamental principles of electromagnetic induction as DC generators, which states that when a rotating coil cuts the magnetic lines of force, an emf is induced in it. The direction of which at any instant is given by Fleming’s Right Hand Rule. The magnitude of induced emf at any instant is given by BLv sin θ volts. Where, B = magnetic flux density (in wb/m2), L = length of conductor (in m), v = velocity (in m/s) and θ = the angle through which the plane of the coil rotates from the magnetic field axis. When the rotor rotated by the prime mover, the stator winding (being stationary) are cut by the magnetic flux of rotor poles. Consequently, an emf is induced in the stator conductors. Since, the magnetic poles in the rotor are alternating N and S, so they induce an alternating emf (and hence alternating current) in the stator conductors, the frequency of which depends upon the number of N and S poles moving past a stator conductor per seconds. The frequency f of generated alternating emf is given by, f = PN …(i) 120 Where, P = total number of magnetic poles, N = speed of rotor in rpm, f = frequency of generated emf in Hz. (73)

74 Electrical Machines–II The generating emf in the stator conductor is taken out for supply by connecting three leads to the stator windings. EMF Equation of an Alternator: The expression or equation of an induced emf of an alternator is to be derived as follows. Let, Z = number of conductors or coil sides in series/phase = 2T, where T is the number of coils or turns/phase P = number of poles f = frequency of induced emf in Hz φ = flux/pole in webers Kd = distribution or breadth factor = sin βm /2 m sin β/2 Kc or Kp = pitch or coil span factor cos α 2 Kf = form factor = 1.11 (if emf is assumed sinusoidal) N = rotor speed in rpm In one revolution of the rotor (i.e. in 60/N seconds) each stator conductor is cut by a flux of φP webers. ∴ dφ = φP and dt = 60 seconds N Average emf induced/conductor = dφ = φP dt 60/N = φPN V …(i) 60 Now, user know that, f = PN 120 or N = 120 f …(ii) P Substituting this value of N from equation (ii) to equation (i) they get, Average emf induced/conductor, = φP × 120 f 60 P = 2fφ volts If there are Z conductors in series/phase, then, Average emf/phase = 2fφZ volts (... Z = 2T) = 4f φT volts RMS value of emf/phase = 1.11 × 4fφT E = 4.44fφT volts …(iii) This would have been the actual value of the induced voltage if all the coils in a phase were full pitched and concentrated or bunched in

Synchronous Generator 75 one slot. But this not being so, the actually available voltage is reduced in the ratio of above mentioned factors (Kd and Kc). ∴Actually available voltage/phase = 4.44 KcKdfφT volts …(iv) = 4 KcKdKffφT volts …(v) If the alternator is star connected then the line voltage is √3 times the phase voltage. Application of Synchronous Generator (Alternator): The application of synchronous generator (alternator) are as follows: (i) used in automobile, (ii) used in diesel electric multiple unit, (iii) used in marine, (iv) used in electrical power generator, (v) used in radio alternator like radio frequency transmission. Q.2. Which type of alternator is suitable for coupling with steam turbines and why? (2010, 15, 17) Write a short note on salient and cylindrical rotors. (2012, 14) Explain different types of rotors of an alternators and also write their uses. (S/2016) How does a rotor of a synchronous motor rotate? (2016) Define construction of an alternator. (S/2017) Explain different types of rotors of an alternator. Write their applications. (2017) Ans. Alternator is just like a DC generator, essentially consist of an armature winding and a magnetic field. But there is one important difference between them. Whereas in DC generators, the armature rotates and the field system is stationary, the arrangement in the alternators is just the reverse of it. In their case, standard construction consists of armature winding mounted on a stationary element called stator and field windings on a rotating element called rotor. Construction: The construction of an alternator consists in two ways which are as follows: (i) Stator: Stator is a stationary part build up of sheet steel lamination, having slots on its inner pheriphery. A three phase star or delta winding is placed in these slots and the whole assembly is held in a cast iron frame. In order to minimise the eddy-current losses, the stator core is laminated. The stator core acts as return path for the magnetic flux. The core laminations are insulated from each other and have space between them for allowing the cooling air to pass through the motor. (ii) Rotor: The rotor is like a fly wheel having N and S poles fixed to its outer rim. The magnetic poles are excited by providing DC supply at 125 to 600 volts from a small DC generators (usually shunt generators), which is mounted on the shaft of the alternator itself. Because the field

76 Electrical Machines–II magnets are rotating, this current is supplied through two slip rings. Rotor construction consists of two types which are as follows: (a) Salient or Projected Pole Type: It has a large number of projecting (salient) poles, having their cores bolted or dovetailed into a heavy magnetic wheel or cast iron or steel of good magnetic quality. The pole faces are usually provided with slots for damper winding. The field coils are placed on the pole pieces and connected in series. The ends of the field windings are connected to DC source, through slip ring (carrying brushes) and mounted on the shaft. This is shown in fig. 3.1 (a). Such generators are characterised by their large diameter and short axial lengths. They are used for low and medium speeds (120 to 400 rpm) when the prime-mover is a water turbine or a reciprocating engine. This is because of the following reasons: (i) At high speed, such poles would cause an excessive windage losses and at the same time produce noise. (ii) Their construction cannot be made strong enough to withstand the mechanical stresses encountered at high speeds. (b) Non Salient or Cylindrical Pole Type: It has a smooth, radial cylinder rotor (made of forged steel) having a number of slots along its outer periphery in which field winding are connected in series, are embeded. The ends of the field windings are connected to DC source through slip ring (carrying brushes) and mounted on the shaft. This is shown in fig. 3.1 (b). Such generators are characterised by small diameters and very long axial (or rotor) lengths to avoid excessive peripheral velocity. These are used for steam-driven alternators i.e., (a) Salient Pole (b) Non Salient Pole Fig. 3.1

Synchronous Generator 77 turbo alternators, which run at very high speeds. Such rotors are designed mostly for 2-pole or 4-pole turbo-generators running at 3600 rpm or 1800 rpm. It should be noted that in this case, poles are non-salient i.e., they do not project out from the surface of the rotor. In this type of construction a better emf waveform is obtained, since the flux distribution around the periphery is nearly a sine wave and windage losses are also less. They can run at high speed because as: (i) Its mechanical robustness in construction. (ii) This type of construction gives better balance and quites operation at high speeds. Q. 3. Write comparison between salient and cylindrical rotor synchronous generators. (S/2004) Differentiate between salient pole and non salient pole alternators. (2013, 17) Compare between salient pole and cylindrical rotor synchronous machine. (2015) Draw diagram of salient pole and cylindrical rotor of synchronous motor and compare then. (2016) Ans. Diagram of Salient Pole and Cylindrical Rotor: Refer Q.2. Difference/Compare Between Salient Pole and Non Salient/ Cylindrical Rotor: The difference/compare between salient pole and non salient pole alternators are as follows: S.No. Salient Rotor Non Salient/Cylindrical Rotor (i) It has 6 to 40 large number It has 2 to 4 number of poles of poles. only. It has short diameters and (ii) It has large diameters and long axial lengths. short axial lengths. It has speed upto 1000 to 3000 rpm. (iii) It has speed upto 150 It is driven by steam turbine. to 750 rpm. They do not uses damper (iv) It is driven by water turbine winding. or diesel engine Air gap is uniform due to as a alternator. smooth cylindrical periphery. This rotor is mechanically (v) They uses damper windings. robust. For same size, rating is (vi) It has non uniform air gap. higher than salient pole type. (vii) This rotor is mechanically (viii) weak. For same size, the rating is smaller than cylindrical type.

78 Electrical Machines–II Q.4. Derive the relation among the number of poles, synchronous speed and the frequency of an alternator. (2001) Ans. Let, P = total number of magnetic poles N = rotational speed of the rotor in rpm f = frequency of generated emf in Hz Since, one cycle of emf is produced when a pair of poles passes past a conductor, the number of cycles of emf produced in one revolution of the rotor is equal to the number of pair of poles. Frequency of one cycle, f = P 2 Synchronous speed Ns of one cycle per second, then the frequency, f = Ns × P 2 If speed of synchronous is per minute then frequency, f = Ns × P 60 × 2 = Ns × P 120 Synchronous speed, Ns = 120 × f P Q.5. Write advantages of stationary armature in an alternator. (S/2003) Write advantages of stationary armature or rotating magnetic system in alternator. (S/2012) Ans. The advantages of stationary armature or rotating magnetic system in alternator are as follows: (i) It is easy to insulate stationary armature winding for high AC voltage, upto 33 kV. (ii) The output current may be load directly from fixed terminals on the stator or armature winding to the load circuit without passing through contacts. (iii) The sliding contacts can be easily insulated due to slip ring are transferred to the low voltage, low power DC field circuit. (iv) The armature winding can be more easily braked to prevent any type of deformation, which may be produced by the mechanical stresses setup as a result of short circuit current and high centrifugal forces. Q.6. Define pitch factor and distribution factor for an alternator. (2004) Write a short note on distribution and pitch factor of an alternator. (S/2010)

Synchronous Generator 79 Define distribution factor and pitch factor. (2011, 17) Explain pitch factor and distribution factor with the help of a vector diagram. (S/2013, 16) Write short note on pitch factor and distribution factor. (2015) Explain distribution factor. (S/2017) Define pitch factor. (S/2017) Ans. Pitch Factor or Coil Span: In short pitched coil, the induced emf of two coil sides is vectorically added to get resultant emf of the coil. In short pitched coil, the phase angle between the emf’s induced in two opposite coil sides is less than 1800 (electrical). But, in full pitched coil, the phase angle between the emf induced in two coil sides is exactly 1800 (electrical). Hence, the resultant emf of a foil pitched coil is just arithmetic sum of the emf induced in both sides of the coil. It is denoted by Kc or Kp defined as, it is the ratio of the vector sum of induced emf per coil to the arithmetic sum of induced emf per coil. It is always less than unity. The pitch factor is shown in fig. 3.2. EMF with short pitched winding Kc = EMF with full pitched winding = Vector sum of the induced emf/coil Arithmetic sum of the induced emf/coil Kc = cos α 2 Where, α = Number of slots by which × 1800 winding is short pitched Number of slots/pole Fig. 3.2 Distribution or Breadth Factor: If all the coil sides of any one phase under one pole are bunched in one slot, the winding obtained is known as concentrated winding and the total emf induced is equal to

80 Electrical Machines–II arithmetic sum of the emf induced in all the coils of one phase under one pole. But in practical cases, for obtained smooth sinusoidal voltage wave form, armature winding of alternator is not concentrated but distributed among the different slots to form polar groups under each pole. In distributed winding, coil sides per phase are displaced from each other by an angle equal to the angular displacement of the adjacent slots. Hence, the induced emf per coil side are not an angle equal to the angular displacement of the slots. It is denoted by Kd and is defined as it is the ratio of vector sum of the emf’s of distributed winding to the arithmetic sum of the emf’s concentrated winding. The distribution factor is shown in fig. 3.3. Fig. 3.3 Kd = EMF induced in distributed winding EMF induced in the concentrated winding Vector sum of the emf’s of distribution = winding Arithmetic sum of the emf’s of concentrated winding or Kd = sin m β/2 Where, m sin β/2 m = Number of slots/pole/phase β = 1800/number of slots/pole Q.7. Explain the equivalent circuit of an alternator. (S/2010) Draw and explain the equivalent circuit of an alternator. (2010, 15, 17)

Synchronous Generator 81 Ans. The equivalent circuit of loaded alternator is shown in fig. 3.4. The direction of current is positive in this circuit. This means it comes from positive terminal. Fig. 3.4 Equivalent Circuit of an Alternator If there is number of load on alternator then open circuit voltage E will receive at its terminals. When load current I flows then it receives voltage E' which is equal to equivalent voltage drop of armature reaction less from E. The leakage reactance in armature winding will voltage drop jIXL and in resistance winding voltage drop equals to IRa which equals to lesses from E', load connected to alternator’s terminal voltage V. If voltage drop IRa, jXL and jXa are connected to terminal voltage V then open circuit voltage, E of alternator will received. E = V + IRa + jI (XL + Xa) E = V + IRa + jIXs Here, Xs is an equivalent reactance And, E = V + I Zs Here, Zs = synchronous impedance (a) On Lagging PF (b) On Leading PF Fig. 3.5 Vector Diagram of an Alternator Fig. 3.5 (a) and (b) shows the vector diagram of alternator at leading and lagging pf. From fig. 3.5, it is clear that in terminal voltage V, on adding voltage drops IaRa, jIXc and jIXa gives open circuit voltage E. S is the phase angle between E and V and is known as power angle of machine and φ is the phase angle between V and I is known as load pf angle.

82 Electrical Machines–II Q.8. Write short note on armature reaction. (2000) Ans. When a synchronous generator (alternator) is running on no- load there will be no current flowing through the armature windings. The flux produced in the air gap will be due to the field ampere-turns only. When load is connected across the armature terminals, current will flow through the armature windings. This three phase current will produce a rotating magnetic field in the air gap. The effect of armature flux on the flux produced by the field ampere-turns is called as armature reaction. The armature flux will distort, oppose or help the field flux causing reduction or increase in the air gap flux depending upon the power factor of the load. In the case of alternators, the pf of the load has a considerable effect on the armature reaction. This consists of three cases which are as follows: (i) Unity Power Factor: Fig. 3.6 In this case the armature flux is cross-magnetising. The result is that the flux at the leading tips of the pole is reduced while it is increased at the trailing tips. However, these two effects nearly offset each other leaving the average field strength constant. In other words, armature reaction for unity pf is distortional, this is shown in fig. 3.6. (ii) Zero Power Factor Fig. 3.7 Lagging: As seen from fig. 3.7, here the armature flux is in direct opposition to the main flux. Hence, the main flux is decreased. Therefore, it is found that armature reaction in this case, is wholly demagnetising with the result, it is due to weakening of the main flux less emf is generated. To keep the value of generated emf the same, field excitation will have to be increased to compensate for this weakening.

Synchronous Generator 83 (iii) Zero Power Factor Leading: This case as shown in fig. 3.8 (a) armature flux has moved forward by 900 so that it is in phase with the main flux wave. This result added with the main flux. Hence, in this case, armature reaction is wholly magnetising, which results in greater induced emf. To keep the value of generated emf the same field excitation will have to be reduced somewhat. Fig. 3.8 Armature Relation at Various pf For intermediate pf as shown in fig. 3.8 (b), the effect is partly distortional and partly demagnetising (because pf is lagging). Q. 9. What do you understand by synchronous reactance and synchronous impedance of an alternator? Ans. For the same field excitation, terminal voltage is decreased from its no-load value E0 to V (for a lagging pf). This is because the reason are as follows: (i) drop due to armature resistance IRa, (ii) drop due to leakage reactance IXL, (iii) drop due to armature reaction. Fig. 3.9 The drop voltage is due to armature reaction may be accounted by assuming the presence of a fictions reactance Xa in the armature winding. The value of Xa is such that IXa represents the voltage drop due to armature reaction.

84 Electrical Machines–II The leakage reactance XL and the armature reactance Xa may be combined to give synchronous reactance Xs. Hence, Xs = XL + Xa Therefore, total voltage drop in an alternator under load Xs = IRa + jIXs = I (Ra + jXs) = IZs Where, Zs is known as synchronous impedance of the armature, the word synchronous being used merely as an indication that it refers to the working conditions. When the synchronous reactance is combined vectorially with armature effective resistance, the quantity obtained is called synchronous impedance i.e., Zs = Ra + jXs Hence, conclude that the vector difference between no-load voltage E0 and terminal voltage V is equal to IZs as shown in fig. 3.9. Q.10. Draw phasor diagram of an alternator at lagging and leading pf. (S/2012) Ans. While drawing phasor diagram of an alternator at lagging and leading pf some symbols are used which are as follows: E0 = no-load emf. This being the voltage induced in armature in the absence of three factors. Hence, it represents the maximum value of the induced emf. E = load induced emf. It is the induced emf after allowing for armature reaction. E is vectorially less than E0 by IXa. Sometimes, it is written as Ea. V = terminal voltage. It is vectorially less than E0 by IZs or it is vectorially less than E by Iz where, Z = √(R2a + XL2). It may also be written as Za. I = armature current/phase. φ = load pf angle. Hence, according to phasor diagram, E0 = √(Vcosφ + IRa)2 + (Vsinφ + IXs)2 Fig. 3.10

Synchronous Generator 85 In fig. 3.10 (a) is shown the case for unity pf, in fig. 3.10 (b) for lagging pf and in fig. 3.10 (c) for leading pf. All these diagrams apply to one phase of a three phase machine diagrams for the other phases can also be drawn similarly. Q.11. Write a short note on voltage regulation. (2012, 14) Ans. It is clear that, with change in load, there is a change in terminal voltage of an alternator. The magnitude of this change depends not only on the load but also on the load pf. The voltage regulation of an alternator is defined as, the rise in voltage when full-load is removed (field excitation and speed remaining the same) divided by the rated terminal voltage. ∴ % regulation up = E0 – V × 100 V It is to be note down clearly as follows: (i) E0 – V is the arithmetical difference and not the vectorial one. (ii) In the case of leading load pf terminal voltage will fall on removing the full-load. Hence, regulation is negative in this case. (iii) The rise in voltage when full-load is thrown off is not the same as the fall in voltage when full-load is applied. In the case of small machines, the regulation may be found by direct loading. The procedure is as follows: The alternator is driven at synchronous speed and the terminal voltage is adjusted to its rated value V. The load is varied until the wattmeter and ammeter indicate the rated values at desired pf. Then the entire load is thrown-off while the speed and field excitation are kept constant. The open circuit or no-load voltage E0 is read. Hence, regulation can be found from, % regulation = E0 – V × 100 V In the case of large machines, the cost of finding the regulation by direct loading becomes prohibited. Hence, other indirect methods are used are as follows: (i) synchronous impedance or emf method, (ii) the ampere-turn or mmf method, (iii) zero power factor or potier method. Q.12. Explain the short circuit test of an alternator. (S/2010) Explain the open circuit test of an alternator. Ans. Open Circuit Test: Like the magnetism curve for a DC machine, the open circuit test or characteristic of an alternator is the curve between armature terminal voltage (phase value) on open circuit and the field current when the alternator is running at rated speed (synchronous speed).

86 Electrical Machines–II Fig. 3.11 Fig. 3.11 shows the circuit for determining the OCC of an alternator. The alternator is run on no load at the rated speed. The field current If is gradually increased from zero (by adjusting field rheostat) until open circuit voltage = E0 (phase value) is about 59% greater than the rated phase voltage. Short Circuit Test: In a short circuit test, the alternator is run at rated speed and the armature terminals are short circuited through identical ammeters. Only one ammeter need be read, but three are used for balance. The field current If is gradually increased from zero until the short circuited armature current Isc is about twice the rated current. The short circuit test is shown in fig. 3.12. Fig. 3.12 The graph is drawn between open circuit voltage values and the corresponding values of If and also the graph between short circuit armature current and field current are shown in fig. 3.13.

Synchronous Generator 87 Fig. 3.13 Open Circuit and Short Circuit Characteristics of an Alternator There is no need to take more than one reading because SCC is a straight line passing through the origin. The reason is simple. Since, armature resistance is much smaller than the synchronous reactance, the short circuit armature current lags the induced voltage by very nearly 900. Consequently, the armature flux and field flux are in direct opposition and the resultant flux in small. Since, the resultant flux is small, the saturation effects will be negligible and the short circuit armature current, therefore, is directly proportional to the field current over the range from zero to well above the rated armature current. Q.13. Explain how will you calculate regulation of an alternator by synchronous impedance method? Ans. For calculating the voltage regulation of an alternator by synchronous impedence method are as follows: (i) According to base draw an open circuit characteristics (OCC) and short circuit characteristics (SCC) as shown in fig. 3.14. (ii) The SCC is a straight line which passes through the main point while OCC is a curved line, both line meet at a common line, If , and Ei is open circuit voltage. As the winding is shorted at that time voltage terminal is zero. (a) Fig. 3.14 (b)

88 Electrical Machines–II ∴ Zs = E1 (open circuit) I1 (short circuit) or Xs = √Z2s – R2a Known Ra and Xs vector diagram can be drawn for any load and power factor as shown in fig. 3.14 (b). Here, OD = E0 ∴ E0 = √OB2 + BD2 or E0 = √[(V cos φ + IRa)2 + (V sin φ + IXs)2] ∴ % regulation up = E0 – V × 100 V Q.14. What is the error in determining voltage regulation by synchronous impedance method? Ans. The regulation obtained by this method is found higher since the synchronous reactance calculated by this method is in the straight part of its magnetisation curve whereas the value of the synchronous reactance varies with the degree of saturation. Hence, in the actual the degree of saturation will be higher than the straight part of the magnetisation curve. This method is theoretically accurate for non salient pole machines with distributed field winding when saturation is not considered. The reason of error is that in this method synchronous impedance or reactance is assumed to remain constant while it is not constant. The value of synchronous impedance varies with the saturation. Q.15. Draw the load characteristics of an alternator at various power factor (pf). Ans. It has been observed that the change of terminal voltage of a synchronous generator depends on load and load pf. A series of load characteristics is as shown in fig. 3.15, which is drawn for different pf Fig. 3.15 Load Characteristics of an Alternator

Synchronous Generator 89 loads. When a synchronous generator is independely supplying a load, increasing its excitation will increases the no-load voltage. This will raise the level of the whole load characteristics and increase the value of the terminal voltage. Q.16. Write a short note on parallel operation of alternators. (2004, S/06, S/07, 09, S/13) Describe the necessity of parallel operation of alternators and explain the conditions of parallel operation. (S/2010) Explain the necessity of parallel operation of alternators. (2010) Write necessity and important conditions for parallel operation of three phase alternator. (2012, 14, S/15) What are the conditions of parallel operation of alternators? Explain the necessity of parallel operation of alternators. (2013) Write conditions of parallel operation of alternators. Explain the necessity of parallel operation of alternators. (2017) Ans. Parallel Operation of Alternators: The operation of connecting an alternator in parallel with another alternator or with common bus-bars is called synchronizing. Necessity of Parallel Operation: The necessity of parallel operation of alternators are as follows: (i) Interconnections of the several alternators of different power stations gives better continuity of supply to consumers. In case fault any unit can be shunt down for repairs and supply can still be maintained to consumers. (ii) Interconnections of several power stations by a grid is economical and advantageous. (iii) For routine maintenance and inspection at power stations, a single unit should be shunt down for a certain period. It can be done without switch off the supply. (iv) With several units in parallel, the number of alternator operating at a time can be changed as demand. In this way it is possible to operate alternators at full load capacity and it reduces the operating cost. Conditions for Parallel Operation: Before a synchronous generator can be put to share the load, it should be properly connected in parallel with a common bus-bar. The operation of connecting an alternator in parallel with common bus-bar is known as synchronising. For proper synchronisation of alternators there are three conditions which are as follows: (i) The terminal voltage (effective) of the incoming alternator must be the same as bus-bar voltage. (ii) The speed of the incoming machine must be such that its frequency equals to the bus-bar frequency.

90 Electrical Machines–II (iii) Phase sequence of the voltages of the incoming alternator should be the same as that of the bus-bar. Q.17. Write the advantages of parallel operation of alternators. Write the conditions necessary for parallel operation. (2015, S/17) Ans. Conditions Necessary for Parallel Operation: Refer Q.16. Advantages of Parallel Operation of Alternators: The advantages of parallel operation of alternators are as follows: (i) Continuity of Supply and Maintenance: This connection is quite fessible when the repair and periodical maintenance work is concerned. It is easier for maintenance when smaller individual generating units are used since, it can schedule the maintenance of each unit one after the other without affecting the continuity of power generation. If we are using a single unit then the whole system has to be shut down for its maintenance work, this will definitely effect the power demand during the duration. (ii) Efficiency: The efficiency versus load current curve of the generator, the generator unit must run on full load, since efficiency is maximum when the load is 100%. If we are using a single large unit then it is uneconomical to use it on lower load, and while using small unit it can add up or switch off the generator units as per the load requirements to meet the maximum efficiency. (iii) Expansion Plans: The current capacity of the power plan is 500 MW, if the plan is expand for a 2 year then the capacity will expand to 700 MW then it will costlier to replace the whole single generator setup and buy a bigger unit. (iv) Size of Alternator: More the rating of generator, it is huge size of setup. It is very difficulty to manage a single large alternator which may range around 1000 MVA or more. Rather if we are using small individual units connected in parallel, it will easier to manage its size. Q.18. What do you mean by synchronizing of alternators? Describe any one method of synchronizing. (S/2010) Write a short note on method of synchronization of alternators. (2012, 14, S/15) Ans. Synchronization or parallel connection of alternators can be achieved by the two method which are as follows: (i) Synchronizing by Three Lamp Method: In this method of synchronizing an alternator, three lamps are connected as shown in fig. 3.16. Two lamps are cross connected with the bus-bar. In this method the brightness of the lamps will vary in sequence. A particular sequence will indicate if the incoming alternator is running too fast or too slow. Perfect synchronizing will occur lamp L1 is dark while lamps L2 and L3 are equally bright.

Synchronous Generator 91 Fig. 3.16 Synchronization of an Alternator with the Bus-Bar When the speed and voltage have been adjusted, the switch of the incoming synchronous machine can be closed only when lamp L1 is dark while lamps L2 and L3 are equally bright. If the frequency of the incoming alternator is higher than the bus-bar frequency, the phasor R2 - Y2 - B2 representing the alternator voltage will be rotating faster than the phasors R1 - Y1 - B1 representing the bus-bar voltages. At the instant when R1 is in phase with R2, lamp L1 will be dark and the other two lamps will be equally bright. After one-third of the cycle, B2 will be in phase with Y2. Since, the lamp L2 is connected across B2 and Y2, it will be dark. After another one-third of a cycle, lamp L3 will be dark. Thus, if the frequency of the incoming alternator is higher the lamps will become dark in the sequence L1 - L2 - L3. Similarly, if the frequency of the incoming alternator is lower, the lamps will become dark in the sequence L1 - L3 - L2. The speed of the alternator will, therefore, have to be slowly adjusted so that the lamp L1 is dark and lamps L2 and L3 are bright. At this instant, the switch can be closed. The incoming machine thus get connected in parallel with the bus-bar. (ii) Synchronizing by Using a Synchroscope: A synchroscope determines the instant of synchronizm more accurately than the three lamp method. A synchroscope consists of a rotor and a stator; one of which is connected to the incoming alternator and the other to the bus- bar as shown in fig. 3.17. A pointer is connected to the rotor which will rotate if there is a difference in frequencies of the incoming alternator and bus-bar. Anticlockwise rotation of the rotor pointer indicates that the frequency of the incoming alternator is slower, whereas clockwise rotation of the pointer indicates that the frequency is higher than the bus-bar frequency. The speed of the prime-mover driving the alternator will therefore, have to be adjusted such that, when the frequencies are equal

92 Electrical Machines–II Fig. 3.17 Sychronizing of Alternator using Synchroscope the pointer is stationary. The alternator can be switched on to the bus- bar by closing the switch, s at this instant. Q.19. Draw and explain power angle characteristics of synchronous generator. (S/2011, 14, S/15, S/17) Draw and explain power angle characteristics of an alternator. (2017) Ans. Electrical Power Output of Salient Pole Generator: The one-line diagram of a salient pole synchronous machine connected to an infinite bus-bars of voltage Vb through a line of series reactance Xext (per phase) is shown in fig. 3.18. The total d and q-axis reactances are then. Xd = xd + xext Xq = xq + xext Fig. 3.18

Synchronous Generator 93 The resistances of the machine armature and line are assumed negligible. Fig. 3.19 below gives the phasor diagram when the machine is generating. It is easy to see that real power delivered to the bus-bar. Fig. 3.19 Phasor-Diagram of a Salient Pole Synchronous Machine when Generating Pa = IdVb sin δ + IqVb cos δ Now, Id = Ef – Vb cos δ and Xd Iq = Vb sin δ Xq Pa = EfVb sin δ + Vb2 Xd – Xq sin 2δ...(i) Xd 2XdXq } Reluctance power This equation gives the electrical power output of a salient-pole generator and electrical power input of a motoring machine where Ef lags Vb by angle δ. The second term of equation (i) is known as reluctance power or torque. It varies as sin 2δ with a maximum value at δ=450. Power Angle Characteristics: The power angle plots of both the terms of equation (i) from the above expression along with the form of resultant power angle characteristic as shown in fig. 3.20. The resultant or total characteristics clearly indicates, that the maximum power output occurs, when the power angle is less than π/2. Hence, the power angle, δ in alternators is a very important parameter, because this angle decides the power transfer between the alternator and the constant voltage bus. When the alternator is feeding power to the bus. Power angle, δ of the alternator also plays an important role in the stability of the alternator operation.

94 Electrical Machines–II Fig. 3.20 Power Angle Characteristics for Salient Pole Alternator Q.20. What is hunting in alternators? How it is be overcomed? (2013) Ans. The torque of a reciprocating engine is pulsating and hence, the power delivered to an alternator driven by it is pulsating. If such a power unit is operated in parallel with another power unit, an unstable condition may results. When the torque of the engine is greatest, its alternator takes momentarily more than its average load and the other alternator takes less than its average load. This oscillation of power if too violent, may be cumulative and if the period of oscillation is too near the natural oscillating frequency of the rotor, the alternator may be thrown out of synchronism. Such, oscillation is called hunting. Destructive oscillation is prevented in alternators driven by internal combustion engines by heavy armature windings similar to the starting squirrel cage winding of synchronous motor or by heavy flywheels or their equivalents. Q.21. Write a short note on cooling of alternators. (2010, 13, 16) Explain cooling methods of alternator. (S/2012) Ans. Methods of Cooling of Alternator: Refer Q. 16 in chapter 2. Cooling in Alternators or Synchronous Generators: The system adopted for the alternator cooling are mainly divided into two groups. They are closed system and open system. In open circuit cooling systems, the cool air is taken in the forms of atmosphere with the help of fan and passed through the generator and

Synchronous Generator 95 hot air coming out from the alternator is discharged either into the atmosphere or into the machine hall for station warming. In a closed circuit cooling system, a given volume of air is calculated continuously through the generator. The hot air discharged from the generator which passed through the water cooled heat exchangers where the heat in the air is removed before it is again circulated back through the generator’s. This system is mainly used for large alternators. The cost of closed type of cooling is costlier as compared to normal open cooling of generators. Some of the advantages of closed system of cooling compared to open system of cooling are as follows: (i) The possibility of the dust and impurities deposition in the machine is reduced. (ii) It does not require any filters. (iii) Operation of the generator will be quiet. (iv) It reduces the amount of moisture which might pass through the alternator in case of open system of generator cooling. Q.22. Draw the model and the phasor diagrams of synchronous machine for generating and motoring modes of operation. Also write the voltage equation for both the case. Ans. Generating Mode: When synchronous motor is generating power at that time the mode is called as generating mode. In this mode Vt is terminal voltage is more the Ef voltage, which is shown in fig. 3.21. The voltage equation in generating mode is, Ef = Vt + Iara + jIaXs (a) Generating Mode (b) Phasor Diagram Fig. 3.21 Motoring Mode: When the value of generated mode is greater than terminal voltage, then its gives power instead of taking is called as motoring mode operation which is shown in fig. 3.22.

96 Electrical Machines–II (a) Motoring Mode (b) Phasor Diagram Fig. 3.22 The voltage equation in motoring mode is, Ef = Vt – Iara – jIaXs Q.23. State the advantages of revolving field type synchronous generators. (S/1990) Ans. The advantages of revolving field type synchronous generators are as follows: (i) Superior insulation system are easily employable and therefore, such alternators are available that generate quite high voltages, even upto 33 KV. (ii) More copper can be employed as the stator slot depth can be increased easily. Hence, it has become possible to design large alternators of ratings upto even 1000 MW. (iii) The problem of current collection at high voltage from slip rings is not present. (iv) Only two slip rings are required for receiving DC excitation for the revolving field system. The field excitation employs current low voltages therefore, the problem of insulating field windings is simple. (v) The length of rotor shaft is smaller, as the slip rings for low voltage working are easily insulated from each other and the spacing between them is quite small. (vi) The armature terminals are stationary hence, are easily insulated for greater safety of working people. Q.24. Name the different types of losses occuring in an alternator and the parts where they occur. (2017) Ans. The different losses occurring in an alternator and parts where they occur are as follows: S.No. Losses Parts where Losses Occur (i) Stator (armature) copper losses. In the stator windings. (ii) Field copper loss. In the DC field windings. (iii) Magnetic or iron losses. In armature core and in core of the field poles. (iv) Frictional losses. In bearings and between the slip rings and brushes. (v) Windage losses. Due to fan action of the rotor.

Synchronous Generator 97 Q.25. If the field excitation of a synchronous generator is kept constant, state what will be the effect of lagging, unity and leading factor loads on its terminal voltage? Ans. The terminal voltage of a synchronous generator will be affected by increasing the load which are as follows: (i) The terminal voltage will decrease if the load power factor is lagging. (ii) The terminal voltage will decrease if the load power factor is unity, but not to that extent as for the lagging power factor load. (iii) The terminal voltage will either decrease more slowly than for unity power factor load, or it will increase. NUMERICAL PROBLEMS AND SOLUTIONS Prob.1. A 3-phase star connected alternator has 50 Hz, 90 slots, 12 conductors per slot. The flux per pole is 0.16 webers. Determine the line voltage if the coil span pitch is reduced by 300 and number of poles are 10. (1990, S/11, 14) A 3-phase, 50 Hz, 10 pole, star connected alternator has 90 slots and there are 12 conductors per slot flux per pole is 0.16 wb. Calculate the line voltage. If coil span is 300 short pitched. (S/2016, 17) Sol. Given, P = 10 φ = 0.16 wb f = 50 Hz Z = 90 × 12 = 1080 = 1080 3 = 360 per phase Speed, N = 120 f P = 120 × 50 10 = 600 rpm Number of turns, T = 360 2 = 180 per phase Number of slots/pole/phase, m = 90 =3 10 × 3 Number of slots/pole = 90 = 9 10 Angular displacement between the slots

98 Electrical Machines–II β = 1800 Number of slots/pole = 1800 = 200 9 Distribution factor, Kd = sin m β/2 m sin β/2 sin 3 × 200 2 = 200 3 sin 2 = sin 300 3 sin 100 = 0.96 Coil span factor, Kc = cos α 2 = cos 300 2 = 0.966 Line emf = 4.44 KcKdfφT volts/phase = 4.44 × 0.966 × 0.96 × 50 × 0.16 × 180 = 5929.16 = 5929.16 × √3 volts (line) = 10269.62 volts (line) Ans. Prob.2. A 3-phase, 50 Hz, 16-pole alternator with star connected winding has 144 slots and 10 conductors per slot. The flux per pole is 0.024 Wb simusoidally distributed. Find (i) speed, (ii) line emf. (2011) Sol. Given, Flux, φ = 0.024 Wb Z = 144 × 10 = 1440 = 1440 per phase 3 = 480 P = 16 (i) Speed, N = 120 × f P = 120 × 50 16 = 375 rpm Ans.

Synchronous Generator 99 (ii) Number of turns, T = 480 = 240 per phase 2 Number of slots/phase/pole, m = 144 =3 16 × 3 Number of slots/pole = 144 = 9 16 Angular displacement between the slots, β = 1800 no. of slots/pole = 1800 = 200 9 Distribution factor, Kd = sin m β/2 m sin β/2 sin 3 × 200 2 = 200 3 sin 2 = 0.96 Assuming the coils to be full pitched, Coils span factor, Kc = 1 Line emf = 4.44 KcKdφTf volts/phase = 4.44 × 1 × 0.96 × 50 × 0.024 × 240 = 1227.5 volt/phase = 1227.5 × √3 volts (line) = 2126 Volts Ans. Prob.3. A 16 pole 3-phase alternator has a star connected winding with 144 slots and 10 conductors per slot. The flux per pole is 0.03 Wb distribution sinusoidally and the speed is 375 rpm. Find the line voltage. (2010) Sol. Given, Number of poles, P = 16 Flux, φ = 0.03 Wb Speed, N = 375 rpm Number of conductors = 144 × 10 = 1440 Number of conductors per phase, Zp = 1440 = 480 3 Frequency, f = PN = 16 × 375 120 120 = 50 Hz

100 Electrical Machines–II Synchronous Generator Number of turns per phase, T = Zp = 480 2 2 = 240 Number of slots per phase per pole, m = 144 = 3 16 × 3 Number of slots per pole, η = 144 = 9 16 Angular displacement between slots, β = 1800 η = 1800 = 200 9 sin m β sin 3 × 200 =2 2 Distribution factor, Kd = m sin β 3 sin 200 2 2 = 0.96 Assumed the coil to be full pitched than Pitch factor, Kp = 1 = √3 × 4.44 × Kd × Kp × φ × f × T = √3 × 4.44 × 0.96 × 1 × 0.03 × 50 × 240 = 2658 V Ans. Prob.4. A 3-phase 16 pole alternator has a star connected winding with 144 slots and 10 conductors per slot. The flux per pole is 0.03 Wb simusoidally distributed and the speed is 375 rpm. Find the frequency and the phase and line emf. Assume full pitched coil. (S/2012) Sol. Given, Number of poles, P = 16 Flux, φ = 0.03 Wb Speed, N = 375 rpm Number of conductors, = 144 × 10 = 1440 Number of conductor per phase, Zp = 1440 = 480 3 Frequency, f = PN = 16 × 375 120 120 f = 50 Hz Ans.