MAHALAKSHMI ENGINEERING COLLEGE TIRUCHIRAPALLI

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MAHALAKSHMI ENGINEERING COLLEGE TIRUCHIRAPALLI 621213 QUESTION BANK WITH ANSWER --------------------------------------------------------------------------------------------------------------- Sub.

1 MAHALAKSHMI ENGINEERING COLLEGE TIRUCHIRAPALLI QUESTION BANK WITH ANSWER Sub. Code : EE2251 Semester : IV Subject : ELECTRICAL MACHINES I Unit : V PART-A 1. What are the losses occurring in a dc machine? (AU2013) The losses occurring in a dc machine are i. Copper losses ii. Iron losses iii. Mechanical losses 2. What is the advantages of swinburne s test? (AU2009) 1.This method is economical since power required to test a large machine is very small(i.e.,)no load input power. 2.This method enables us to determine thye losses and efficiency without actually loading the machine. 3. What are the advantages and dis-advantages of hopkinsons test? (AU2011) Advantages : 1. Power required for the test is small as compared to full load powers of the two machines. SETHU MADHAVAN R S AP/EEE Page 1

2 2. Since the machines can be tested under full load conditions for long duration, the performance of the machines regarding commutation and temperature rise can be studied. Dis-advantage : 1. Two identical machines are required. 4. Why brake test is not suitable for large size machines? (AU2008) The brake test is not suitable for large dc motors owing to yhe difficulty of dissipating the heat generated at the pulley. 5. Mention the factors on which hysteresis loss depends. (AU2008) 1. Hystersis coefficient 2. Maximum flux density 3. Frequency 4. Volume of the core 6. Hopkinson s test is also called a regenerative test. Justify. (AU2008) Hopkinson s test is also called a regenerative test. Because the machine 1running as a motor drives machine 2 as a generator. Since both the machines are coupled electrically, the power output of generator 2 is fed to the motor 1.Due to this reason, Hopkinson s test is also called a regenerative test. 7. What is a regenerative test? (AU2013) Regenerative test is a hopkinson s test. In this method two identical d.c machines are coupled, both mechanically and electrically and are tested simultaneously.one of the machines is made to run as a motor and it drives the other machine as a generator. 8. Name the two methods of improving commutation. (AU2011) (i) Emf commutation. (ii)resistance commutation SETHU MADHAVAN R S AP/EEE Page 2

3 9. What are the losses in dc motor? (AU2010) 1.Copper losses 2.Iron losses 3.Mechanical losses 10. What are the drawbacks of brake test on DC machines? (AU2011) 1.The brake test can be used for small motors only, because incase of large motors, it is difficult to dissipate the large amount of heat generated at the brake. 2. This method cannot be used for determining the internal losses. 3. The output of the motor cannot be measured directly. PART-B 1.what are the various starting method of DC motor? Explain any one method. (or).with a neat diagram explain the principle of operation of 3 point starter which is used for DC shunt motor? (AU MAY/JUN 11) Introduction: There are three types of DC motor starters namely, Two point starter- (used for dc series motor) Three point starter Four point starter When the handle is moved ON position the soft iron,which is attracted by the electromagnet. When the handle is in ON position,the motor achieves its full speed,which develop the back emf.this back emf regulates the armature current. The starting resistance is connected in series with the armature of a dc motor. SETHU MADHAVAN R S AP/EEE Page 3

4 A handle,which can be moved over the starting resistance against the spring. A no voltage release (NVR) coil is connected in series with the field winding. An OLR coil is connected in series with the field winding. A movable arm is placed near the OLR coil. Operation: To start the motor,the DC supply is given and the main switch is closed. The handle is now slowly moved clockwise to the stud 1. When the handle touches the stud 1,the full resistance is connected in series with the armature.but the shunt field winding is directly connected across the supply voltage. As the handle is gradually moved over to the final stud,the starting reistances is cut out of the armature circuit in steps.the handle move against the spring force. When the handle reaches the final stud the soft iron piece is attracted by the electromagnet. Protective devices used in starters: (i) No-Volt Release Coil(NVR): When the handle is ON position,the no volt coil is magnetised and attracts the soft iron and keeps the handle in ON position against the spring tension. SETHU MADHAVAN R S AP/EEE Page 4

5 In case of failure or disconnection of the supply or a break in the field circuit,the NVR coil is denergised therby releasing the Arm, which is pulled back by the spring to the OFF position. (ii) OLR(Over Load Release) If the motor becomes over loaded beyond a certain predetermined value,line current (or) armature current increases and hence the attracting power of the electromagnet increases,then the movable arm is lifted and the short circuits the electromagnet.hence the arm is released and retuns to OFF position. Demerits of 3-Point Starter: While employing this method,the field current is decreased to achieve the speeds above the rated speed. This low value of the current also passes through NVR,which is unablie to create enough electromagnetic pull to overcome the spring tension.hence the arm is pulled back to OFF position. (i).with a neat sketch,explain the working of four point starter for DC shunt motor (AU NOV/DEC11) In the four point starter,the no volt release coil is connected in series directly across the line through a protective distance,r p. The no-volt release coil is independent of the shunt field current.therefore proper speed control can be excited without affecting the operation of no-volt release coil. Operation: To start the motor,the dc supply is given and the main switch is closed. The handle is now slowly moved clockwise to the stud 1. When the handle touches the stud 1,the the line current divides into three parts (a)one part passes through the Starting resistance and motor armature. (b)the second part passes through the shunt field winding and the field rheostat. (c)the third part passes through the no volt coil release coil and current protecting resistance R p. SETHU MADHAVAN R S AP/EEE Page 5

6 Therefore,the electromagnet pull produced by the hold on coil will always be the same and sufficient to hold the handle in ON position. Disadvantages: It wil not protect the motor from high protection.during running condition,if the field gets opened,the field current reduces. (ii)with a neat sketch,explain the working of Two point starter for DC shunt motor In two point starter,the starting resistance is connected in series with the armature of series motor.the no volt release coil is connected in series with the armature.after closing the supply,the handle is moved from OFF position. SETHU MADHAVAN R S AP/EEE Page 6

7 Then the full starting resistance is included.therefore the starting current is reduced.then the starting resistance is gradually cut down and the motor gathers speed, which will then develop back emf. this NVC is also called as hold on coil. Disadvantages: The main problem of dc series motor is its overspeeding action,when the load is less.this problem can be presented by using two poin starter. 2.A 230 V shunt motor has an armature resistance of 0.2 ohm.the starting armature current must not exceed 50 A.If the number of sections of reistances in the starter is 5,find the value of resistance in each section.(au NOV/DEC 11) Given: V=230 V R a =0.2Ω, I max =50 A SETHU MADHAVAN R S AP/EEE Page 7

8 Number of sections =5.Hence there are 6 studs,i.e., n=6 Total resistance from armature terminal to stud 1 is R 1 =230/50 =4.6 Ω R 6 =R 0 =0.2 Ω R 1 /R a =K n-1, 4.6/0.2 =K 5 Hence K= R 2 =R 1 /K =4.6/1.8721= Ω, R 3 =R 2 /K =2.4571/1.8721= Ω, R 4 =R 3 /K =1.3125/1.8721=0.701 Ω, R 5 =R 4 /K =0.701/ = Ω, r 1 = R 1 - R 2 = =2.14 Ω, r 2 = R 2 - R 3 = =1.466 Ω, r 3 = R 3 - R 4 = = Ω, r 4 = R 4 - R 5 = = Ω, r 5 = R 5 - R a = =0.1744Ω, 3.A starter required for a 220 V shunt motor.the maximum allowable current is 55 A and the minimum current is about 35 A.Find the number of starter resistance required and the resistance of each section.the armature resistance of the motor is 0.4 ohm. ( AU NOV/DEC 11) Given: V= 220 V R a = 0.4 ohm I 1 = I max =55 A, I 2 = I min =35 A (I 1 / I 2 ) n-1 =R 1 /R a (55 / 35 ) n-1 =R 1 /0.4 now R 1 = V/I 1 =220/55 =4 ohm SETHU MADHAVAN R S AP/EEE Page 8

9 ( ) n-1 =4/0.4 =10 (n-1)log =log 10 (n-1) ( )=1 i.e., n-1 =5.09 n=6.09 =6 Thus there are 6 studs and 5 sections. I 1 / I 2 =K=55/35= R 1 =K R 2 i.e., R 2 =2.545 Ω R 2 =K R 3 i.e., R 3 = Ω R 3 =K R 4 i.e., R 4 = Ω R 4 =K R 5 i.e., R 5 =0.656 Ω R 5 =K R a i.e., R a =0.41 Ω=0.4 Ω (given) The resistances of various sections are section 1= R 1 - R 2 = Ω section 2= R 2 - R 3 = Ω section 3= R 3 - R 4 = Ω section 4= R 4 - R 5 =0.3748Ω section 5= R 5 - R a = Ω 4.Explain the Ward Leonard system for the speed control of dc motors? Mention its merits and demerits.(au NOV/DEC 11) This system is used where a very sensitive speed control is required. Examples:elevators,paper mills,etc. M 1 is the main motor whose speed control is required.the field of this motor is permanently connected across the dc supply lines. By applying the a variable voltage across the armature,any desired speed can be obtained. SETHU MADHAVAN R S AP/EEE Page 9

This variable voltage is supplied by a motor set which of either a dc or ac motor M 2.The motor M 2 is directly coupled to the generator G. The motor M 2 runs at an approximately constant speed.

10 This variable voltage is supplied by a motor set which of either a dc or ac motor M 2.The motor M 2 is directly coupled to the generator G. The motor M 2 runs at an approximately constant speed.the output voltage of 'G' is directly fed to the main motor M 1. The voltage of the generator can be varied from zero to its maximum value by means of its field generator. The field current of the generator can be reversing switch R s.therefore the generated voltage can be reversed and hence the direction of the relation of M 1 is also reversed. It should be remembered that motor generator set always in the same direction. As this method can give the unlimited speed control in either directions,this system is commonly employed for elevators,hoists and main drive in steel mills.also this system is ideal in applications where frequent starting,stopping and reversals are required. As the generator voltage can be raised from zero,the motor starts up smoothly without any extra starting equipment. Although this system is advantageous as it giving wide range of speeds it requires two extra machines which involves high capital cost. In modern days SCRs are used for obtaining variable DC voltage which will take power from AC mains through a transformer. Though it is not less expensive,the arrangement is neat and free from maintainence problems.it will give automatic control of speed. The modified Ward Leonard system is called Ward Leonard -ILgner system in which a flywheel is used in addition to motor generator set.it is used to reduce the fluctuations in SETHU MADHAVAN R S AP/EEE Page 10

11 power demand from the supply. When load on main motor is suddenly increased,the driving motor M 2 from the motor generator slows down.thus inertia of flywheel is used to supply part of the overload. However when load is suddenly decreased from motor M 1,then the motor M 2 from set speeds up which allows energy to store in the flywheel. Merits: This method is very effective and a wide range of speed control is obtained without resistance losses. The motor can be brought to standstill quickly,simply by reducing the voltage of the generator G. This method is used for speed control,of large motors when a dc supply is not available. Demerits: The capital cost of such system is high,since three machines are employed. 5.Sketch the necessary schematic circuits for the following methods of controlling the speed of the DC series motor. (AU NOV/DEC 11) (i)armature diverter control (ii)tapped field control. Mention clearly the speed will increase or decrease, in each case with reasons. Armature field control: In this method,a variable resistance is connected in parallel with the armature. The armature current can be varied by adjusting the diverter resistance. For a constant load, if I a is decreased using the armature diverter,then the flux Φ must increase to produce the same torque as Tα I a. To satisfy the condition current drawn from the supply will be more to increase flux. Since the Nα1/Φ,the motor speed is decreased. By adjusting the armature diverter,the speed of the motor can be varied. This method is suitable to obtain the speed below the normal speed. SETHU MADHAVAN R S AP/EEE Page 11

12 (ii)tapped Field Control: In this method the number of tappings are provided in the series winding The number of series field turns in the circuit can be changed by tappings.here the flux is reduced by decreasing the number of turns with full turns of the field winding,the motor runs at normal speed and as the field turns are cut out, the speed higher than normal speed is achieved. 7.Explain armature reaction and commutation in details. SETHU MADHAVAN R S AP/EEE Page 12

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16 8.Explain the characteristic of DC motor. a)electrical characteristics: i) Torque and armature current characteristics. (Ta/Ia) b) Mechanical characteristic: ii)speed and torque characteristics (N/T) c)speed and armature current characteristics (N/Ia) CHARACTERISTICS OF DC SERIES MOTOR: i)t/ia characteristics: we know that, T α ǿ Ia in this case, as field windings also carry the armature current upto magnetic saturation, ǿ α Ia, so that T α Ia 2 after magnetic saturation, ǿ is constant, so that T α Ia. Thus up to magnetic saturation, the armature torque is proportional to the square of the armature current. If Ia increases, T increases as the square of the current. Hence T/Ia curve is a parabola, after magnetic saturation, torque is directly proportional to the armature current. Therefore, T/Ia curve after magnetic saturation is a straight line. T α Ia (after saturation) TαIa 2 (upto saturation) Ia then T SETHU MADHAVAN R S AP/EEE Page 16

17 ii)n/ia characteristics: The speed N of a series DC motor, where, Eb= V-Ia (Ra + Ise) When Ia increases, the back emf Eb decreases due to Ia (Ra+Rse)drop, also the flux ǿ increases however, Ia(Ra+Rse) drop is smaller under normal conditions and may be neglected With increased Ia, ǿ also increases. Nα 1/Ia. If Ia, then N. N/T (OR) MECHANICAL CHARACTERISTICS: versa. It is found from the characteristic curves that when speed is high, torque is low and vice The relationship between N/T is shown. SETHU MADHAVAN R S AP/EEE Page 17

18 T α Ia 2 ; N α 1/Ia; if T, then N. EXPLAIN THE CHARACTERISTICS OF DC SHUNT MOTOR. i)t/ia characteristics: we know that in a DC motor, T α ǿ Ia since the motor is operating from a constant supply voltage, flux ǿ is constant. T α Ia T α Ia If Ia then T. Hence T/Ia characteristic curve is a straight line passing through the origin. The shaft torque is a less then Ta and it is shown by a dotted line. It is clear from the curve that a very large current is required to start a heavy load. Therefore, a shunt motor should not be started on load. ii)n/ia characteristics: The speed N of a DC motor is given by, N α Eb/ǿ α (V-Ia Ra) [ǿ-constant] N α (V- Ia Ra) If Ia, then N. SETHU MADHAVAN R S AP/EEE Page 18

19 Thus flux ǿ and back emf Eb in a shunt motor are almost constant under normal conditions. Therefore, speed of a shunt motor will remains constant as the armature current varies. iii)n/t characteristics: The curve is obtained by plotting the values of N and t for various armature currents. It may be seen that speed falls some what as the torque increases. From the characteristic curves. The following points are observed. there is slight change in the speed of a shunt motor from no load to full load. Hence it is essentially constant speed motor. the starting torque is not high because T α Ia N α (V Ia Ra) so if T, then N. SETHU MADHAVAN R S AP/EEE Page 19

20 (9) Explain the working principle of DC motor. While a DC generator converts mechanical energy in the form of rotation of the conductor (armature) into electrical energy, a motor does the opposite. The input to a DC motor is electrical and the output is mechanical rotation or torque. The fundamental principles and construction of the DC motors are identical with DC generators which have the same type of excitation. A DC machine that runs as a motor will also operate as a generator. Principle of operation: the basic principle of operation of d.c. motor is, "whenever a current carrying conductor is placed in a magnetic field, the conductor experiences a force tending to move it." The magnetic field between two poles N and S is shown in figure. A current carrying conductor is shown in figure (a) along with the direction of the flux loops around it. If a current carrying conductor is placed between two magnetic poles as shown in figure (b)both the fields will distorted. Above the conductor, the field is weakened (less flux) and below the conductor, the field is strengthened. Therefore the conductor tends to move upwards. The force exerted upwards depends upon the intensity of the main field flux and the magnitude of the current. Then the direction of the current through the conductor is reversed as shown in figure(c) Here, the field of the below conductor is less (weakened) and field of the above conductor is more (strengthened). Then the conductor tends to move downwards. The magnitude of the force experienced by the conductor in a motor is given by Where F= BIl Newton, B= Magnetic field density in wb/m I = current in amperes l= length of the conductor in meters. The direction of motion is given by Fleming's Left Hand rule,which states that if the thumb,fore finger and middle finger of the left hand are held such that the fingers show three mutually perpendicular directions and if the fore finger indicates direction of the field, and the SETHU MADHAVAN R S AP/EEE Page 20

21 middle finger indicates the direction of the current, then the thumb points the direction of motion of the conductor. In a d.c. motor, a strong electromagnetic field and a large number of conductors housed in an armature and carrying current, make the armature rotate. (10) What is meant by back emf. An interesting aspect of motoring action is detailed below. The conductors are cutting flux and that is exactly what is required for generator action to take place. This means that even when the machine is working as a motor, voltages are induce in the conductors. This emf is called as the back emf or counter emf, since the cause for this is the rotation, which, in turn, is due to the supply voltage. According to Lenz's law, the direction of the back emf opposes the supply voltage. The back emf is given by the equation for induced emf as derived in chapter. where the symbols, P,A,Z and N have the same meaning as in chapter figure 4.20 shows the equivalent circuit of a motor. Here, the armature circuit is equivalent to a source of emf Eb, in series with a resistance of Ra and then a DC supply is applied across these two. The voltage equation of this DC motor is V= Eb + Ia Ra From this equation, armature current where V - applied voltage Eb - back emf Ia - armature current Ra - armature resistance V - Eb - net voltage in the armature circuit. From equations, the induced emf in the armature of a motor Eb depends upon (i)armature speed, (ii) armature current Ia depends upon the back emf Eb for a constant input voltage V and armature resistance Ra. SETHU MADHAVAN R S AP/EEE Page 21

UNIT I D.C. MACHINES PART A. 3. What are factors on which hysteresis loss? It depends on magnetic flux density, frequency & volume of the material.

UNIT I D.C. MACHINES PART A. 3. What are factors on which hysteresis loss? It depends on magnetic flux density, frequency & volume of the material. EE6352-ELECTRICAL ENGINEERING AND INSTRUMENTATION UNIT I D.C. MACHINES PART A 1. What is prime mover? The basic source of mechanical power which drives the armature of the generator is called prime mover.