electrical drives ans controls

8/3/2019 Electrical Drives Ans Controls

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SYED AMMAL ENGINEERING COLLEGE, RAMANATHAPURAM

III Semester.SUBJECT CODE: 10122ME306

SUBJECT NAME: ELECTRICAL DRIVES AND CONTROLS MECHANICAL ENGINEERING BRANCH

Regulation 2010

ANSWER KEY

1. List out some advantages of electric drives .

i. Availability of electric drives over a wide range of power a fewwatts to mega watts.

ii. Ability to provide a wide range of torques over wide range of speeds.

iii. Electric motors are available in a variety of design in order tomake them compatible to any type of load.

2. What is a Group Electric Drive (Shaft Drive)?

This drive consists of single motor, which drives one or moreline shafts supported on bearings.

The line shaft may be fitted with either pulleys & belts or gears,by means of which a group of machines or mechanisms may beoperated.

3. Give an expression for the losses occurring in a machine.

The losses occurring in a machine is given byW = W c + x 2 WvWhere Wc = Constant lossesWv = Variable losses at full loadX = load on the motor expressed as a function of rated load.

4. How heating occurs in motor drives? The heating of motor due to losses occurring inside the motor while

converting the electrical power into mechanical power and these lossesoccur in steel core, motor winding & bearing friction.


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5. Define heating time constant & Cooling time constant?

Heating time constant

Time required to heat the machine to 0.632 times of its final steady state

value.Cooling time constant

Time required to cool the machine to 0.367 times of the initialtemperature rise above the ambient temperature.

PART B

6. (a) (i) Explain the factors governing the selection of motors.

1. Shaft power & speed 11. Speed range2. Power range 12. Efficiency3. Starting torque 13. Influence on the supply network4. Maintenance 14. Special competence5. Total purchase cost 15. Cost of energy losses6. Influence on power supply 16. Environment7. Availability 17. Accessibility8. Nature of electric supply 18. Nature of load9. Types of drive 19. Electrical Characteristics10. Service cost 20. Service capacity & rating

(b) (i) Discuss in detail the determination of power rating of motors

1. Method of Average losses2. Method of equivalent power3. Method of equivalent current 4. Method of equivalent Torque

7. (a) (i) Explain the different types of loading of drives. (8 )

(i)Continuous or contant loads

In this type load occurs for a long time under tha same condition(e.g) fan loads ,paper makig machines etc

(ii)Continous variable loads

The load is variable over a period of time but it occurs repeatetively forlonger duration.


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(e.g) metal cutting lathes , conveyors ,etc

(iii) Pulsating loads:

Torque which exhibits combination of constant load torque super

imposed by pulsations.(e.g) Reciprocating pump, compressors, all load having crank shafts

(iv) Impact load:

There are peak loads occur at regular intervals of the time .

(e.g) Rolling mill, presses, sharing machines, forging hammersMotor for such loads are provided with heavy loads

(v) Short time intermittent loads : The load appears periodically identical duty cycles, each consisting of aperiod of application of load and one or rest.(e.g) all forms of cranes, hoists, elevators

(vi) Short time loads :

A constant load appears on the drive and the system rests for remainingpriod of cycle.

(e.g) motor generator set for charging batteries, household equipments

(ii) Explain the choice of selection of the motor for differentloads (8 )

Ans: Description of few points in 6 th question

(OR)

(b) (i) Describe the simplifications based on which the heating andcooling calculations of an electric motor are made.

i. The machine is considered to be a homogeneous body havinga uniform temperature gradient. All the points at which heatgenerated have the same temperature. All the points atwhich heat is dissipated are also at same temperature.

ii. Heat dissipation taking place is proportional to the differenceof temperature of the body and surrounding medium. Noheat is radiated.


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iii. The rate of dissipation of heat is constant at alltemperatures.

(3 )(ii) Establish the heating time constant and the heating curves.

(13)

8. (a) (i) Compare the D.C and A.C drives.

DC Drives

Advantage:

Simple torque and speed control without sophisticated electronics

Limitations:

Regular Maintenance Expensive motor Heavy motor Sparking

AC Drives

Induction motors are the most common motors used for variousequipments in industry.

Their popularity is due to their simple design, they are inexpensive (half or less of the cost of a DC motor) High power to weight ratio (about twice that of a DC motor) easy to maintain can be directly connected to an AC power source

(ii) A 100 kW motor, having rated temperature rise of 60C, has full-load efficiency of 80% and the maximum efficiency occurs at 85%full load. It has thermal time constants of 80 minutes and 65minutes. It is cyclically loaded, 120% of full load for one hour and50% of full load for the next hour. Find the temperature rise after 3hour (8)

K=(0.8)2=0.64, =0.9, P 0 =100 kW

WL (full load loss)= P0 ((1/ )-1)= 100 ((1/ 0.9 )-1)=11.11 kW

WL= (k+1)Wcu =1.64 Wcu


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Wcu =11.11/1.64 =6.775 kWWconst =WL -Wcu=11.11-6.775=4.336 kWAt 120 % of full load, the permissible (steady state temperature rise is ss at this load

Wx=Wcu[k+(Px/P nom]2]= 6.775[0.64+(1.2]

2]=14.092 kW

ss/ ss= Wx/ WL=14.092/11.11=1.268

ss=101.46 C

lly At 50 % of full load ,the steady state temperature rise is . At this load ss=43.42 C

th/T h=60/60=1

h1 =temperature rise after 1 hr = ss(1-e(-th/Th) )=101.46(1-e -t)=64.135 Cc1 =temperature rise after 2 hrs = ss(1-e(-th/Th) )+ h1 e(-th/Th) =51.05 Ch2 =temperature rise after 3 hrs = ss(1-e(-th/Th) )+ c1 e(-th/Th) =82.195 Cc1 =temperature rise after 4 hrs = ss(1-e(-th/Th) )+ h2 e(-th/Th) =57.95 C

Or(b) (i) Write a brief note on classes of duty for an electric motor. (8)

1. Continuous duty2. Short time duty operation of motor Main classes of duties3. Intermittent periodic duty4. Intermittent periodic duty with starting5. Intermittent periodic duty with starting & braking6. Continuous duty with intermittent periodic loading7. Continuous duty with starting & braking8. Continuous duty with periodic load changes

(ii) The thermal time constant and final steady temperature of amotor on continuous running is 30 minutes and 60C. Find out thetemperature.

i) After 15 minutes at this load.ii) After 1 hour at this load.iii) If temperature rise at 1 hour rating is 60C, find the

maximum steady temperature.iv) What will be the time required to increase the

temperature from 40C to 60C at 1 hour rating. (8)


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PERIODICALTEST II, July 2011

1. Why DC series motor should never be started on no-load?

However, at light or no-load, the motor tends to attain dangerouslyhigh speed. So it should not be started without load.

2. State the condition at which the starting toque developed in aslip-ring induction motor is maximum.

R 2 = s X 2Slip corresponding to maximum torque, s = R 2/X2.It can be shown that:

3. What are the advantages and disadvantages of Electrical Braking?

1. To augment the brake power of the mechanical brakes.2. To save the life of the mechanical brakes.3. To regenerate the electrical power and improve the energy

efficiency.4. In the case of emergencies to step the machine instantly.5. To improve the through put in many production process by

reducing the stopping time.

4. What is meant by Regenerative braking in DC Motor?

In the regenerative braking operation, the motor operates as agenerator, while it is still connected to the supply here; the motor speedis grater that the synchronous speed. Mechanical energy is converter intoelectrical energy, part of which is returned to the supply and rest as heatin the winding and bearing.

5. What are the factors influencing the selection of starters ?


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PART B ---- (2 1/2 X 16 = 40 marks)

6. (a) List out the advantages and disadvantages of electrical

braking over mechanical braking. (8 )

1. To augment the brake power of the mechanical brakes.2. To save the life of the mechanical brakes.3. To regenerate the electrical power and improve the energy

efficiency.4. In the case of emergencies to step the machine instantly.5. To improve the through put in many production process by

reducing the stopping time.

Or

(b) Discuss any one method of electrical braking of DC Machines.(8 )

Dynamic braking

In dynamic braking the motor is disconnected from the supply andconnected to a dynamic braking resistance RDB.

In and Fig. 1 this is done by changing the switch from position 1 to2 .

Fig .1 Connections

The supply to the field should not be removed.


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Due to the rotation of the armature during motoring mode anddue to the inertia, the armature continues to rotate.

An emf is induced due to the presence of the field and the rotation. This voltage drives a current through the braking resistance.

The direction of this current is opposite to the one which wasflowing before change in the connection.

Therefore, torque developed also gets reversed. The machine actslike a brake.

The torque speed characteristics separate by excited shunt of themachine under dynamic braking mode is as shown in Fig. 2 for aparticular value of RDB.

Fig 2: Characteristics

The positive torque corresponds to the motoring operation.

Fig. 3 shows the dynamic braking of a shunt excited motor and the

corresponding torque-speed curve.


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Fig 3: Dynamic braking

Here the machine behaves as a self excited generator.

Below a certain speed the self-excitation collapses and the braking actionbecomes Zero.

7. (a) Explain about the speed-torque characteristics of a DC ShuntMotor with suitable graph and equations.

Fig. (1) shows the connections of a d.c. shunt motor. The field

current Ish is constant since the field winding is directly connected to thesupply voltage V which is assumed to be constant. Hence, the flux in ashunt motor is approximately constant.

Fig :1 Fig:2

(i)Ta/Ia Characteristic.


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We know that in a d.c. motor, T a Ia

Since the motor is operating from a constant supply voltage, flux f is constant (neglecting armature reaction).

T a Ia

Hence Ta/Ia characteristic is a straight line passing through theorigin as shown in Fig. (2). The shaft torque (T sh) is less than Ta and isshown by a dotted line. It is clear from the curve that a very large currentis required to start a heavy load. Therefore, a shunt motor should not bestarted on heavy load.

(ii)N/Ia Characteristic.

The speed N of a. d.c. motor is given by;N Eb

The flux and back e.m.f. E b in a shunt motor are almost constantunder normal conditions. Therefore, speed of a shunt motor will remainconstant as the armature current varies (dotted line AB in Fig. 3). Strictlyspeaking, when load is increased, E b (= V- IaRa) and f decrease due to thearmature resistance drop and armature reaction respectively. However,Eb decreases slightly more than f so that the speed of the motordecreases slightly with load (line AC).

(iii)N/Ta Characteristic .

The curve is obtained by plotting the values of N and Ta for variousarmature currents (See Fig. 3). It may be seen that speed falls somewhatas the load torque increases.

Fig :3 Fig:4


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Conclusions

Following two important conclusions are drawn from the abovecharacteristics:

(i) There is slight change in the speed of a shunt motor from no-load tofullload.Hence; it is essentially a constant-speed motor.

(ii) The starting torque is not high because Ta Ia.

Or

(b) Explain how an induction motor is brought to stop by (i) Pluggingand (ii) dynamic braking. (16)

(i) Rheostat or Dynamic braking

In this method, the armature of the running motor is disconnectedfrom the supply and is connected across a variable resistance R.However, the field winding is left connected to the supply. The armature,while slowing down, rotates in a strong magnetic field and, therefore,operates as a generator, sending a large current through resistance R.

This causes the energy possessed by the rotating armature to bedissipated quickly as heat in the resistance. As a result, the motor isbrought to standstill quickly.

Fig: 1Fig. (1] show dynamic braking of a shunt motor. The braking

torque can be controlled by varying the resistance R. If the value of R isdecreased as the motor speed decreases, the braking torque may bemaintained at a high value. At a low value of speed, the braking torque


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becomes small and the final stopping of the motor is due to friction. Thistype of braking is used extensively in connection with the control of elevators and hoists and in other applications in which motors must bestarted, stopped and reversed frequently.

We now investigate how braking torque depends upon the speed of themotor.

Referring to Fig. (1),Armature current,

Braking torque

where k2 and k3 are constantsFor a shunt motor, is constant.Braking torque, T B N

Therefore, braking torque decreases as the motor speed decreases.

(ii) Plugging

In this method, connections to the armature are reversed so thatmotor tends to rotate in the opposite direction, thus providing thenecessary braking effect. When the motor comes to rest, the supply mustbe cut off otherwise the motor will start rotating in the opposite direction.


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Fig: 2Fig. (2) shows plugging of a d.c. shunt motor. Note that armature

connections are reversed while the connections of the field winding arekept the same. As a result the current in the armature reverses. Duringthe normal running of the motor [See Fig. 2], the back e.m.f. E b opposesthe applied voltage V. However, when armature connections are reversed,back e.m.f. E b and V act in the same direction around the circuit.

Therefore, a voltage equal to V + E b is impressed across the armaturecircuit. Since E b ~ V, the impressed voltage is approximately 2V. In order10 limit the current to safe value, a variable resistance R is inserted inthe circuit at the time of changing armature connections.

We now investigate how braking torque depends upon the speed of themotor. Referring to Fig. (2),

Armature current,

Braking torque,

For a shunt motor, f is constant.


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Braking torque, T B = k5 + k6N

Thus braking torque decreases as the motor slows down. Note that thereis some braking torque (TB = k5) even when the motor speed is zero.

8. (a) Draw a neat schematic diagram of a three point starter andexplain its working. (16)

Three-Point Starter

This type of starter is widely used for starting shunt andcompound motors.Schematic diagram

Fig. (1) shows the schematic diagram of a three-point starter for a

shunt motor with protective devices. It is so called because it has threeterminals L, Z and A. The starter consists of starting resistance dividedinto several sections and connected in series with the armature. Thetapping points of the starting resistance are brought out to a number of studs. The three terminals L, Z and A of the starter are connectedrespectively to the positive line terminal, shunt field terminal andarmature terminal. The other terminals of the armature and shunt fieldwindings are connected to the negative terminal of the supply. The no-volt release coil is connected in the shunt field circuit. One end of thehandle is connected to the terminal L through the over-load release coil.

The other end of the handle moves against a spiral spring and makescontact with each stud during starting operation, cutting out more andmore starting resistance as it passes over each stud in clockwisedirection.


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Fig: 1


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Operation

(i) To start with, the d.c. supply is switched on with handle in the OFFposition.

(ii) The handle is now moved clockwise to the first stud. As soon as itcomes in contact with the first stud, the shunt field winding is directlyconnected across the supply, while the whole starting resistance isinserted in series with the armature circuit.

(iii) As the handle is gradually moved over to the final stud, the startingresistance is cut out of the armature circuit in steps. The handle is nowheld magnetically by the no-volt release coil which is energized by shuntfield current.

(iv) If the supply voltage is suddenly interrupted or if the field excitationis accidentally cut, the no-volt release coil is demagnetized and thehandle goes back to the OFF position under the pull of the spring. If no-volt release coil were not used, then in case of failure of supply, thehandle would remain on the final stud. If then supply is restored, themotor will be directly connected across the supply, resulting in anexcessive armature current.

(v) If the motor is over-loaded (or a fault occurs), it will draw excessivecurrent from the supply. This current will increase the ampere-turns of the over-load release coil and pull the armature C, thus short-circuitingthe novolt release coil. The no-volt coil is demagnetized and the handle ispulled to the OFF position by the spring. Thus, the motor isautomatically disconnected from the supply.

[Or]

(b) Draw a neat schematic diagram of a four point starter andexplain its working. (16)

In a four-point starter, the no-volt release coil is connected directlyacross the supply line through a protective resistance R. Fig. 1 shows theschematic diagram of a 4-point starter for a shunt motor (over-loadrelease coil omitted for clarity of the figure). Now the no-volt release coilcircuit is independent of the shunt field circuit. Therefore, proper speedcontrol can be exercised without affecting the operation of no volt releasecoil.

Note that the only difference between a three-point starter and afour-point starter is the manner in which no-volt release coil is


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connected. However, the working of the two starters is the same. It maybe noted that the three point starter also provides protection against anopen field circuit. This protection is not provided by the four-pointstarter.

Fig: 1

Operation

(i) To start with, the d.c. supply is switched on with handle in the OFFposition.

(ii) The handle is now moved clockwise to the first stud. As soon as itcomes in contact with the first stud, the shunt field winding is directlyconnected across the supply, while the whole starting resistance isinserted in series with the armature circuit.

(iii) As the handle is gradually moved over to the final stud, the starting


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resistance is cut out of the armature circuit in steps. The handle is nowheld magnetically by the no-volt release coil which is energized by shuntfield current.

(iv) If the supply voltage is suddenly interrupted or if the field excitation

is accidentally cut, the no-volt release coil is demagnetized and thehandle goes back to the OFF position under the pull of the spring. If no-volt release coil were not used, then in case of failure of supply, thehandle would remain on the final stud. If then supply is restored, themotor will be directly connected across the supply, resulting in anexcessive armature current.

(v) If the motor is over-loaded (or a fault occurs), it will draw excessivecurrent from the supply. This current will increase the ampere-turns of the over-load release coil and pull the armature C, thus short-circuitingthe novolt release coil. The no-volt coil is demagnetized and the handle ispulled to the OFF position by the spring. Thus, the motor isautomatically disconnected from the supply.


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MODEL TEST

1. What are the factors influencing the choice of electric drives?

1) Steady state operation requirements.2) Transient operation requirements.3) Requirements related to the source.4) Capital and running cost, maintenance needs life.5) Space and weight restriction.6) Environment and location.7) Reliability.

2. What is meant by continuous duty?

Operation with a constant load state as shown in Figure1 with a

duration sufficient to reach thermal equilibrium. The load period tB ismuch greater than the thermal time constant T.

Figure2

3. Why differential compound motors are not used in practical?

If the load increases the speed also increasing which will cause thedamage.

4. Give the application where DC Shunt, DC Series and DCCompound motors are used.

Shunt: driving constant speed, lathes, centrifugal pumps, machine tools,blowers and fans, reciprocating pumpsSeries: electric locomotives, rapid transit systems, trolley cars, cranesand hoists, conveyorsCompound: elevators, air compressors, rolling mills, heavy planners.


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5. What are the factors influencing the selection of starters?

Cost Type of motorQuickness

PerformanceEfficiency

6. Why stator resistance starter is rarely used?

This method suffers from two drawbacks. First, the reducedvoltage applied to the motor during the starting period lowers the startingtorque and hence increases the accelerating time. Secondly, a lot of power is wasted in the starting resistances.

7. What will be the effect of change in supply voltage on the speed of

DC Shunt Motor?

Speed is directly proportional to the voltage applied to a motor .Decreasein speed will decrease in speed

8. What are the advantages and disadvantage of solid state drivemethods?

9. List the different methods of speed control of three phaseinduction motor.

Stator voltage control. Frequency control Pole changing control. Slip power recovery control.

10. What is VVVF control?

Variable voltage variable frequency (V/f) constant


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PART B ---- (5X 16 = 80 marks)

11.(a) Draw the typical temperature rise-time curve and derivethe equation for temperature rise in an electric drive. (16)

[Or]

(b) The thermal time constant and final steady temperatureof a motor on continuous running is 30 minutes and 60C. Find outthe temperature.

i) After 15 minutes at this load.ii) After 1 hour at this load.iii) If temperature rise at 1 hour rating is 60C, find the

maximum steady temperature.iv) What will be the time required to increase the

temperature from 40C to 60C at 1 hour rating. (16)

K=(0.8)2=0.64, =0.9, P 0 =100 kW

WL (full load loss)= P0 ((1/ )-1)= 100 ((1/ 0.9 )-1)=11.11 kW

WL= (k+1)Wcu =1.64 Wcu

Wcu =11.11/1.64 =6.775 kWWconst =WL -Wcu=11.11-6.775=4.336 kWAt 120 % of full load, the permissible (steady state temperature rise is ss at this load

Wx=Wcu[k+(Px/P nom]2]= 6.775[0.64+(1.2]2]=14.092 kW

ss/ ss= Wx/ WL=14.092/11.11=1.268

ss=101.46 C

lly At 50 % of full load ,the steady state temperature rise is . At this load ss=43.42 C

th/T h=60/60=1

h1 =temperature rise after 1 hr = ss(1-e(-th/Th) )=101.46(1-e -t)=64.135 Cc1 =temperature rise after 2 hrs = ss(1-e(-th/Th) )+ h1 e(-th/Th) =51.05 Ch2 =temperature rise after 3 hrs = ss(1-e(-th/Th) )+ c1 e(-th/Th) =82.195 Cc1 =temperature rise after 4 hrs = ss(1-e(-th/Th) )+ h2 e(-th/Th) =57.95 C


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12 .(a) Explain the Speed-Torque characteristics of three phaseinduction motor with neat diagrams. (16 )

1. A torque speed curve shows how a motor's torque productionvaries throughout the different phases of its operation.

2. Starting torque, also called locked rotor torque, is produced by amotor when it is initially turned on. Starting torque is the amountrequired to overcome the inertia of a standstill.

3. Pull-up torque is the minimum torque generated by a motor as itaccelerates from standstill to operating speed. If a motor's pull-uptorque is less than that required by its application load, the motorwill overheat and eventually stall.

4. Breakdown torque is the greatest amount of torque a motor canattain without stalling. High breakdown torque is necessary forapplications that may undergo frequent overloading. One suchapplication is a conveyor belt. Often, conveyor belts have moreproduct placed upon them than their rating allows. Highbreakdown torque enables the conveyor to continue operatingunder these conditions without causing heat damage to the motor.


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5. Full load torque is produced by a motor functioning at a ratedspeed and horsepower. The operating life is significantlydiminished in motors continually run at levels exceeding full loadtorque.

6. Synchronous speed is the speed at which no torque is generated bya motor. This occurs in motors that run while not connected to aload. At synchronous speed, the rotor turns at exactly the samerate as the stator's rotating magnetic field. Since there is no slip,there is no torque produced.

[Or]

(b) (i) Explain the speed torque curve of single phase inductionmotors in detail . (8)

Depending on the various start techniques, Single-Phase ACInduction Motor is classified: Split-Phase AC Induction Motor, CapacitorStart AC Induction Motor, Permanent Split Capacitor (Capacitor Run) ACInduction Motor, Capacitor Start/Capacitor Run AC Induction Motor,Shaded-Pole AC Induction Motor.

Below is the figure showing the torque-speed curves of variouskinds of single-phase AC induction motors. From this figure, you willknow more about different types of single phase ac induction motors.

Figure. Torque-speed curve of different types of Single-phaseinduction motor

(ii) Explain the method of regenerative braking employed in DCMotors. (8)

In the regenerative braking, the motor is run as a generator. As aresult, the kinetic energy of the motor is converted into electrical energyand returned to the supply. Fig. (1) Shows two methods of regenerativebraking for a shunt motor.


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(a) In one method, field winding is disconnected from the supply and fieldcurrent is increased by exciting it from another source [See Fig. 1 (i)].As a result, induced e.m.f. E exceeds the supply voltage V and themachine feeds energy into the supply. Thus braking torque is providedupto the speed at which induced e.m.f. and supply voltage are equal. As

the machine slows down, it is not possible to maintain induced e.m.f. ata higher value than the supply voltage. Therefore, this method is possibleonly for a limited range of speed.

(b) In a second method, the field excitation does not change but the loadcauses the motor to run above the normal speed (e.g., descending loadon a crane).As a result, the induced e.m.f. E becomes greater than thesupply voltage V [See Fig. 1 (ii)]. The direction of armature current I,therefore, reverses but the direction of shunt field current If remainsunaltered. Hence the torque is reversed and the speed falls until Ebecomes less than V.

13. (a) Draw a neat schematic diagram of a four point starter andexplain its working(16)

In a four-point starter, the no-volt release coil is connected directlyacross the supply line through a protective resistance R. Fig. 1 shows theschematic diagram of a 4-point starter for a shunt motor (over-loadrelease coil omitted for clarity of the figure). Now the no-volt release coilcircuit is independent of the shunt field circuit. Therefore, proper speedcontrol can be exercised without affecting the operation of no volt releasecoil.

Note that the only difference between a three-point starter and afour-point starter is the manner in which no-volt release coil isconnected. However, the working of the two starters is the same. It maybe noted that the three point starter also provides protection against anopen field circuit. This protection is not provided by the four-pointstarter.


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Operation

(i) To start with, the d.c. supply is switched on with handle in the OFF

position.(ii) The handle is now moved clockwise to the first stud. As soon as itcomes in contact with the first stud, the shunt field winding is directlyconnected across the supply, while the whole starting resistance isinserted in series with the armature circuit.

(iii) As the handle is gradually moved over to the final stud, the startingresistance is cut out of the armature circuit in steps. The handle is nowheld magnetically by the no-volt release coil which is energized by shuntfield current.

(iv) If the supply voltage is suddenly interrupted or if the field excitationis accidentally cut, the no-volt release coil is demagnetized and thehandle goes back to the OFF position under the pull of the spring. If no-volt release coil were not used, then in case of failure of supply, thehandle would remain on the final stud. If then supply is restored, themotor will be directly connected across the supply, resulting in anexcessive armature current.


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(v) If the motor is over-loaded (or a fault occurs), it will draw excessivecurrent from the supply. This current will increase the ampere-turns of

the over-load release coil and pull the armature C, thus short-circuitingthe novolt release coil. The no-volt coil is demagnetized and the handle ispulled to the OFF position by the spring. Thus, the motor isautomatically disconnected from the supply.

[Or]

(b) Explain the different starting methods of three phase squirrelcage induction motors with neat sketches. (16)

Except direct-on-line starting, all other methods of starting squirrel-cagemotors employ reduced voltage across motor terminals at starting.(i) Direct-on-line starting

This method of starting in just what the name impliesthe motoris started by connecting it directly to 3-phase supply. The impedance of the motor at standstill is relatively low and when it is directly connectedto the supply system, the starting current will be high (4 to 10 times thefull-load current) and at a low power factor. Consequently, this method of starting is suitable for relatively small (up to 7.5 kW) machines.

Relation between starling and F.L. torques .

We know that:Rotor input = 2p Ns T = kT But Rotor Cu loss = s Rotor inputRelation between starling and F.L. torques .

We know that:Rotor input = 2N s T = kT But Rotor Cu loss = s x Rotor input


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If Ist is the starting current, then starting torque (Tst) is T I2 ( at starting s 1)

If If is the full-load current and sf is the full-load slip, then,

When the motor is started direct-on-line, the starting current is theshort-circuit (blocked-rotor) current I sc .

Let us illustrate the above relation with a numerical example. SupposeIsc = 5 If and full-load slip sf =0.04. Then,

Note that starting current is as large as five times the full-loadcurrent but starting torque is just equal to the full-load torque.

Therefore, starting current is very high and the starting torque iscomparatively low. If this large starting current flows for a long time, itmay overheat the motor and damage the insulation.

(ii) Stator resistance starting

In this method, external resistances are connected in series witheach phase of stator winding during starting. This causes voltage dropacross the resistances so that voltage available across motor terminals isreduced and hence the starting current. The starting resistances are


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gradually cut out in steps (two or more steps) from the stator circuit asthe motor picks up speed. When the motor attains rated speed, theresistances are completely cut out and full line voltage is applied to therotor.

This method suffers from two drawbacks. First, the reducedvoltage applied to the motor during the starting period lowers the startingtorque and hence increases the accelerating time. Secondly, a lot of power is wasted in the starting resistances.

Relation between starting and F.L. torques .

Let V be the rated voltage/phase. If the voltage is reduced by afraction x by the insertion of resistors in the line, then voltage applied tothe motor per phase will be xV.Ist = x Isc Now f


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Thus while the starting current reduces by a fraction x of therated-voltage starting current (Isc), the starting torque is reduced by afraction x2 of that obtained by direct switching. The reduced voltageapplied to the motor during the starting period lowers the startingcurrent but at the same time increases the accelerating time because of

the reduced value of the starting torque. Therefore, this method is usedfor starting small motors only.

(iii) Autotransformer starting

This method also aims at connecting the induction motor to areduced supply at starting and then connecting it to the full voltage asthe motor picks up sufficient speed. Fig. (8.31) shows the circuitarrangement for autotransformer starting.

The tapping on the autotransformer is so set that when it is in thecircuit, 65% to 80% of line voltage is applied to the motor. At the instantof starting, the change-over switch is thrown to start position. Thisputs the autotransformer in the circuit and thus reduced voltage isapplied to the circuit. Consequently, starting current is limited to safevalue. When the motor attains about 80% of normal speed, thechangeover switch is thrown to run position. This takes out theautotransformer from the circuit and puts the motor to full line voltage.Autotransformer starting has several advantages viz low power loss, lowstarting current and less radiated heat. For large machines (over 25H.P.), this method of starting is often used. This method can be used forboth star and delta connected motors.


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Relation between starting and F.L. torques .

Consider a star-connected squirrel-cage induction motor. If V isthe line voltage, then voltage across motor phase on direct switching is V3 and starting current is Ist = Isc. In case of autotransformer, if a tapping

of transformation ratio K (a fraction) is used, then phase voltage acrossmotor is KV/3 and I st = K Isc,

The current taken from the supply or by autotransformer is I1 =KI2 = K2Isc. Note that motor current is K times, the supply line currentis K2 times and the starting torque is K 2 times the value it would havebeen on direct-on-line starting.

14. (a) Explain how the speed of a DC Shunt Motor can be varied both above and below the speed at which it runs with full fieldcurrent. (16)

The speed of a shunt motor can be changed by (i) flux controlmethod (ii) armature control method (iii) voltage control method. The firstmethod (i.e.flux control method) is frequently used because it is simpleand inexpensive.


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1. Flux control method

It is based on the fact that by varying the flux f, the motor speed (N 1/f) can be changed and hence the name flux control method. In this

method, a variable resistance (known as shunt field rheostat) is placed inseries with shunt field winding as shown in Fig. (1).

Fig :1 Fig:2

The shunt field rheostat reduces the shunt field current Ish andhence the flux f.Therefore, we can only raise the speed of the motor abovethe normal speed (See Fig. 2). Generally, this method permits to increasethe speed in the ratio 3:1.Wider speed ranges tend to produce instabilityand poor commutation.Advantage

(i) This is an easy and convenient method.

(ii) It is an inexpensive method since very little power is wasted in theshunt field rheostat due to relatively small value of I sh .

(iii) The speed control exercised by this method is independent of load onthe machine.

Disadvantages

(i) Only speeds higher than the normal speed can be obtained since thetotal field circuit resistance cannot be reduced below Rshthe shuntfield winding resistance.

(ii) There is a limit to the maximum speed obtainable by this method. It isbecause if the flux is too much weakened, commutation becomes poorer.


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Note . The field of a shunt motor in operation should never be openedbecause its speed will increase to an extremely high value.

2. Armature control method

This method is based on the fact that by varying the voltageavailable across the armature, the back e.m.f and hence the speed of themotor can be changed. This is done by inserting a variable resistance RC(known as controller resistance) in series with the armature as shown inFig. (3).

Fig :3 Fig:4 N V - Ia(Ra + RC)where RC = controller resistance

Due to voltage drop in the controller resistance, the back e.m.f.(Eb) is decreased. Since N Eb, the speed of the motor is reduced. Thehighest speed obtainable is that corresponding to RC = 0 i.e., normalspeed. Hence, this method can only provide speeds below the normalspeed (See Fig. 4).

Disadvantages

(i) A large amount of power is wasted in the controller resistance since itcarries full armature current Ia.

(ii) The speed varies widely with load since the speed depends upon thevoltage drop in the controller resistance and hence on the armaturecurrent demanded by the load.

(iii) The output and efficiency of the motor are reduced.

(iv) This method results in poor speed regulation.


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Due to above disadvantages, this method is seldom used to control tiespeed of shunt motors.Note . The armature control method is a very common method for thespeed control of d.c. series motors. The disadvantage of poor speedregulation is not important in a series motor which is used only where

varying speed service is required.[Or]

(b) Explain in detail the single phase semi-converter speed controlfor DC drive for separately excited motor.. (16)

In these configurations, the armature and / or the field windingvoltage can be varied using single-phase full bridge AC-to-DC convertersas shown in Fig. 1-a. This is a two quadrant converter, as shown in Fig.

1-b, where the load current is always positive while the load voltage maybe positive or negative. The generated waveforms are shown in Fig. 1-c.Single-phase AC-to-DC semi-converters can be also used for DC motorsspeed control as shown in Fig. 2-a. However, this is a first-quadrantconverter, as shown in Fig. 2-b, where the load current and voltage arealways positive. The generated waveforms are shown in Fig. 2-c. Usuallythe same AC supply is used as an input for the two converters. Theaverage value of the armature and / or field voltage (and hence therotational speed) can be varied and controlled by varying the firing angleof the thyristors used. The equation relating the armature and the fieldvoltage with the supply voltage and the firing angle can be expressed asfollows,


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Fig.1 For single-phase full converters connected to the armatureterminals:

The average value (DC) of the armature voltage is,

where Va is the average value of the armature voltage, V m is themaximum value of the supply voltage, and a is the armatureconverters thyristor firing (delay) angle.

For single-phase full converters connected to the field terminals:

The average value (DC) of the field voltage is,

where Vf is the average value of the field voltage, and f is the fieldconverters thyristor firing (delay) angle.

Fig. 2 DC motor speed control using single-phase semi-converters

For single-phase semi-converters connected to the armatureterminals:

The average value (DC) of the armature voltage is,


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For single-phase semi-converters connected to the field terminals:

The average value (DC) of the field voltage is,

15. (a) Explain the static Kramer method and static scherbiusmethod of speed control of three phase induction motor. (16)

In a cage-type induction motor, the rotor current at slip frequencyreacting with the airgap flux develops the torque. The corresponding slippower is dissipated in the rotor resistance. In a wound rotor inductionmotor, the slip power can be controlled to control the torque and speed of a machine. Figure 1 shows a popular slip power-controlled drive, knownas a static Kramer drive. The slip power is rectified to dc with a dioderectifier and is then pumped back to an ac line through a thyristorphase-controlled inverter.

Fig 1: Static Kramer drive system


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The method permits speed control in the subsynchronous speed

range. It can be shown that the developed machine torque is proportionalto the dc link current I d and the voltage V d varies directly with speeddeviation from the synchronous speed. The current I d is controlled by the

firing angle of the inverter. Since V d and V I voltages balance at steadystate, at synchronous speed the voltage V d is zero and the firing angle is90 degrees. The firing angle increases as the speed falls, and at 50%synchronous speed the firing angle is near 180 degrees. This ispractically the lowest speed in static Kramer drive. The transformer stepsdown the inverter input voltage to get a 180-degree firing angle at lowestspeed. The advantage of this drive is that the converter rating is lowcompared with the machine rating. Disadvantages are that the line powerfactor is low and the machine is expensive. For limited speed rangeapplications, this drive has been popular.

Static Scherbius drive

Fig 2: Static Scherbius drive using DC link thyristor converter


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Fig 3: Static Scherbius drive using cycloconverter

The scheme has applications in large power fan and pump driveswhich requires speed control in anrrow range only.

If max. slip is denoted by S max, then power rating of diode, inverterand transformer can be just S max times motor power rating resulting in alow cost drive.

This drive provides a constant torque control.Or

(b) Explain in detail about the various methods of solid statespeed control techniques by using inverters. (16)

Voltage-Fed Inverter Induction Motor

A simple and popular converter system for speed control of aninduction motor is shown in Fig. 1. The front-end diode rectifier converts60 Hz ac to dc, which is then filtered to remove the ripple. The dc voltageis then converted to variable-frequency, variable-voltage output for themachine through a PWM bridge inverter. Among a number of PWMtechniques, the sinusoidal PWM is common, and it is illustrated in Fig. 2or one phase only.

The stator sinusoidal reference phase voltage signal is comparedwith a high-frequency carrier wave, and the comparator logic outputcontrols switching of the upper and lower transistors in a phase leg. The


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phase voltage wave shown refers to the fictitious center tap of the filtercapacitor. With the PWM technique, the fundamental voltage andfrequency can be easily varied. The stator voltage wave contains high-frequency ripple, which is easily filtered by the machine leakageinductance. The voltage-to-frequency ratio is kept constant to provide

constant airgap flux in the machine. The machine voltage-frequencyrelation, and the corresponding torque, stator current, and slip, areshown in Fig. 2. Up to the base or rated frequency b , the machine candevelop constant torque. Then, the field flux weakens as the frequency isincreased at constant voltage. The speed of the machine can becontrolled in a simple open-loop manner by controlling the frequency andmaintaining the proportionality between the voltage and frequency.During acceleration, machine-developed torque should be limited so thatthe inverter current rating is not exceeded.

By controlling the frequency, the operation can be extended in thefield weakening region. If the supply frequency is controlled to be lowerthan the machine speed (equivalent frequency), the motor will act as agenerator and the inverter will act as a rectifier, and energy from themotor will be pumped back to the dc link. The dynamic brake shown isnothing hut a buck converter with resistive load that dissipates excesspower to maintain the dc bus voltage constant. When the motor speed isreduced to zero, the phase sequence of the inverter can be reversed forspeed reversal. Therefore, the machine speed can be easily controlled inall four quadrants.


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Fig.1.Sinusoidal pulse width modulation principle.

Fig.2 Voltage-frequency relation of an induction motor.

Current-Fed Inverter Induction Motor Drive

The speed of a machine can be controlled by a current-fed inverteras shown in Fig. 3.The front-end thvristor rectifier generates a variabledc current source in the dc link inductor. The dc current is thenconverted to six-step machine current wave through the inverter. Thebasic mode of operation of the inverter is the same as that of the rectifier,except that it is force-commutated, that is, the capacitors and seriesdiodes help commutation of the thyristors.


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Figure 3. Force-commuted current-fed inverter control of aninduction motor.

FIGURE 4 PWM current-fed inverter control of an induction motor.

One advantage of the drive is that regenerative braking is easybecause the rectifier and inverter can reverse their operation modes. Six-step machine current, however, causes large harmonic heating andtorque pulsation, which may be quite harmful at low-speed operation.Another disadvantage is that the converter system cannot be controlledin open loop like a voltage-fed inverter.

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