Q1. State Faraday’s law of electromagnetic induction [2002, 2004, 2005, 2006, 2007, 2008].

Faraday’s law of Electromagnetic Induction: These are the following laws which are known as the faradays laws of electromagnetic induction.

1 st law: According to the first law of electromagnetic induction,.’ when ever the flux linking with a coil or circuit changes, an emf. Induced in it.’’

This linkage can be obtained either by rotating the conductor in the magnetic field or by rotating magnetic field keeping the conductor stationary.

2 nd law : According to the second law of electromagnetic induction,’ the magnitude of the induced emf. In coil is directly proportional to the rate of changed of flux linkage.’’e rate of change of flux linkage.

Let there be a coil having ‘N’ the initial flux being 1 and the final flux value after time‘t’ is 2.

The net change in flux= 2 - 1

And the rate of change of linkage

=

Nφ 2 -N φ 1

tNow according to faradays laws of electromagnetic induction. The emf rate of change of flux linkage.

Nφ 2 -N φ 1

t=k (φ 2-φ 1)

tx N volts

where k is the constant of proportionality and here being unity

So, e = (φ 2−φ 1)

t x N volts

Induced emf.. = rate of change of flux x no. of conductors and it can be otherwise stated as

e = d/dt x N volts

Q2. Define Fleming’s right hand rule [2002, 2004, 2005, 2007].

Fleming’s Right Hand Rule: The direction of induced electromagnetic e.m.f. in a coil can be found out by applying Fleming's Right Hand rule. As shown in Fig. "Stretch first finger, second finger and thumb of right hand mutually perpendicular to each other. If first finger indicates the direction of main magnetic field, thumb indicates the direction of motion of conductor, then 2nd finger will indicate the direction of induced e.m.f. in the conductor.

Q3. State Lenz’s law [2002, 2004, 2005, 2008].

Lenz’s law: The e.mf. induced, according to the faradays laws of electromagnetic induction, has got only the strength of the magnitude and is silent over the direction. The direction was stated by Lenz’s laws in 1835.

According to Lenz’s laws, the direction of induced e.g. because of the electromagnetic induction is such a way as to the oppose the caused which is responsible for the production of this e.g. the equation can be given as

e = -d/dt x N volts.The minus sign indicates the direction of the induced e.g. i.e. opposing.

Q4. Derive emf equation of DC generator [2002, 2004, 2005, 2006, 2007, and 2008].

EMF Equation of DC Generator: According to faradays laws of electromagnetic induction, if a conductor cuts the magnetic flux or it experience a rate of change of flux, an e.m.f. is generated in that conductor, this e.m.f. can be given as.

e = d/dt V

Now here in case of D.C. generator, having the following data:

= the flux per pole in WbZ = number of conductorsN = speed in r.p.m.P = number of polesa = number of parallel paths

and E = the induced e.m.f. in volts.

Now the total flux produced= (P x ) Wb

Now the time required in one revolution when running at N.rpm.

= 1/N min= 60/N sec

Now e.m.f. induced = rate of change of flux

= P/60/N V= PN/60 V

It is the voltage induced per conductor. But there are Z conductors so the e.m.f.

= ZNP/60 V

In D.C. armature winding the closed type winding is done so the total armature conductors (Z) are divided into the number of parallel paths depend upon the type of winding. The voltage taken out will be the voltage induced per parallel path. Here let ‘a, be the parallel paths, so the e.m.f. generated can be given as

E = ZNP/60a V

The above equation is said the e.m.f equation of the generator.

Q5. Enlist essential parts of a DC machine and explain [2004, 2005, 2006, 2007, 2008].

Parts of a DC Machine: The parts of D.C. generator can be broadly divided into the followings:

(a) Stationary parts (b) Rotating parts Stationary parts:The parts of D.C. generator which remains stationary during the working of the generator, are known as stationary parts. These are followings: Eye bolt, body yoke, poles, side cover, brushes, and rocker, bearings and legs and bed sheet.

Rotating parts:The parts of D.C. generators which rotates during the working of d.c generators, are known as rotating parts. These are as follows:

Armature, commentator, fan and shaft.

The brief description of these parts is as follows:

Stationary parts:

(a) Eye bolt: It is a stationary part and fixed on the top of the body or yoke. It is used for lifting the machine. These are one of two depending upon the frame size of the machine.

(b) Body or yoke: It is the outer frames of the machine. It accommodates all parts of the generator. It is made of forged steel or cast steel or cast iron. The cast steel is used for the machines of large capacity because of the good magnetic properties. The cast iron frames are used for the machines of low capacity: because the magnetic properties are not so good

It is experienced that the frames size of cast iron or cast steel differ for the same capacity, cast steel requires half the size. Nowadays the yokes of large machines are almost invariably fabricated steel because this material has good magnetic properties, for example the permeability is twice that the cast iron; and has hence the weight. (fig. 12.1. parts of D.C. generator).

The yoke provides the magnetic path for the magnetic flux.The flux in yoke is half than that of the flux pear pole, because the flux passing diverts into the paths causing half the flux through the yoke. Some times the body and yoke of the machine are different but in most of the cases these are same.

(3) Poles: The main magnetic field is produced by the poles excited by the Field coils. The poles are made either of the cast iron or soft steel or the laminations of silicone steel. In small machines the poles are casted with the body. In some construction the pole faces are separate and attached to the

mountings. Nowadays the complete poles are made by laminations of silicone steel which are pressed hydraulically and riveted together. The poles are attached with the yoke or body by means of bolts etc.The field coils are wound with the different number off conductors (shunt field coil- having more number of turns of thin conductors series field coil- less number of turns thick conductor) are place over the poles, in such away so that it may not come out. The pole faces are made circular as to provide uniform air gap around the armature and uniform flux density also. The poles are always in pair; i.e. 2, 4, 6 and 8 etc.

(4) Brushes and brush gears: The main function of brushes is to collect the current from the commutate and supply to the external load circuit. These are housed in rectangular chamber. The brushes are provided with a spring as to offer some pressure which could easily be adjusted by the spring loading finger. The assembly is called brush holder, generally made a brass. These brush holders are mounted over the round construction known as the “rocker” by the displacement of rocker over the commentator, the brush position can be changed.

(5) Bearings: The bearings are fitted in the cover, so as to have minimum friction between the rotating and stationary portions (shaft and side cover). The bearings also help in keeping the armature in the centre for smooth running.

Rotating or rotary part:The part which revolves during operation is called the rotating part.

(1) Armature: The armature rotates in the magnetic field. The conductors in which the e.m.f is induced, are housed in the slots of the armature .the armature is made of laminations of silicone steel to reduce the eddy current and hysteresis loss. The laminations are assemble and riveted under hydraulic

pressure to avoid any air gap between the laminations. The lamination are insulted from each other by means of the varnish or sometimes thin insulating paper. By adding the silicone in steel, the resistance is increased and thus decreases the eddy currents which in other words decrease the eddy current losses, the lamination instead of solid block, caused the further reduction in current and ultimately reducing both the eddy current and hysteresis losses. Generally the thickness off each lamination is 0.4 to 0.6 mm.

In order to dissipate the heat, thus produced, some ventilating ducts are provided. The air circulation through these ducts increases the heat dissipation and keeps the machine temperature under space limits.

(2) Commutator: It is made of the hard drawn copper segments which are insulated from each other and from shaft by means of mica or minacity. It is mounted on the shaft of the machine. The copper segments are tapered and there is a rises on one side of each segment. These Is a space provided to solder the conductor with the riser or segment as shown in fig 12.3. It is round in shape to facilitate the collection of current (in case of generator) from the armature.

The construction of the segment is such as, that it has “v,, groove on both the sides to protect the segment from coming out because of the centrifugal forces. Every segment is properly insulated from every side i.e. from segment to segment, from segment to shaft from segment to the sleeve and ‘v’ checks nuts by means of mica or minacity.

(3) Fan and shaft: a fan is mounted over the shaft in opposite direction of commentator. It is made of cast iron or thick mild steel sheets. It circulates the air through armature, armature winding etc. to keep the temperature down.

A shaft generally of mild steel is used, which carries the armature, commentator, fan and bearings. The pulley is also mounted on the shaft after the side cover to enable the mechanical energy to the load.

Q6. Explain the working principle of a DC generator [2002, 2004, 2006, 2008, 2009].

Working Principle of a DC Generator: The D.C. generator works on the principal of faraday’s law of electromagnetic induction and that to the dynamic induction. The e.m.f. thus induction or produced is known as the dynamically induced e.m.f.

According to that “principal”, when ever the conductor cuts the magnetic lines of force, an e.m.f. is generated in that conductor. The e.m.f generated is directly proportional to the rate of change of flux i.e. the angle of flux linkage. The total

e.m.f generated in the armature is also proportional to the number off conductors.

Q7. Draw a block diagram showing a power stages of DC generator [2002, 2006, 2007, 2009].

Power Stages in D.C. GeneratorThe various power stages in case of a D.C. generator are given below:

Iron and friction losses= A-B

Copper losses = B - C

Q8. State the different losses of a DC generator [2002, 2004, 2006, 2007, 2008, 2009].

Losses in DC Generator: The losses in d.c. machines may be divided into two groups: (1) Variable losses, and (2) Constant losses.

1. Variable losses are proportional to the square of the load current in the armature and the series and interpole windings when employed. Due to the contact resistance between the brushes and the commutator, there is amore variable loss.

2. Constant losses are assumed to be constant over the load range and comprise:(a) Iron loss (due to hysteresis and eddy currents)(b) Windage and friction loss, and(c) Shunt field excitation loss.

The above losses can be calculated or obtained separately from test results. The eddy currents losses due to change in flux with load are additional to above losses and known as stray losses.

The above two groups of losses are discussed below:

(i) Armature losses: thee losses are of two types:(a) I2 R loss in the windings of the armature; and(b) Iron loss in the core of the armature, due to the hysteresis and eddy currents. Hysteresis loss depends upon the quality of iron. It is a function of frequency f and approximately proportionate to the square of the flux (2). The eddy current losses are proportional to the square of frequency and the square of flux. Moreover the eddy current loss is proportional to the square of the thickness of the laminations.

(ii) Commutator losses: Commutator losses occur due to the contact resistance and friction between the brushes and the commutator segments.(a) Contact resistance losses are dependent upon the quality of the brushes.(b) The frictional losses in the commutator segments and brushes depend upon the brush pressure on the segments, the coefficient of friction and the peripheral speed of the commutator.

(iii) Excitation losses: These include the loss in a shunt circuit (if so used) which is equal to the product of the shunt current and the terminal voltage. In a shunt-wound generator, this loss varies between no-load and full-load, whereas in a level compound generator it is almost constant. Excitation losses also occur in commutating poles or interpoles series, and compensating windings (if used), and are proportional to the square of the armature current.

(iv) Losses due to bearing friction and windage: The losses due to bearing friction are roughly proportional to the speed. Windage loss is ordinarily very small unless the generator is fitted with a cooling fan device, and in that case, windage loss is proportional to the cube of the speed.

(v) Losses due to stray load: The effect of armature reaction to distort to flux. The flux densities in certain regions of the armature are increased and in certain other regions they are decreased. Hence the iron loss depends upon the square of the flux density, Hence the iron loss is increased due to armature reaction. Such losses are known as stray losses.

Q8. Explain in detail the parallel operation of DC Shunt generator with the help of diagram [2002, 2004, 2005, 2006, 2009].

Parallel Operation of Shunt Generators: When two shunt generators are connected in parallel, they function together to supply power to a common load and they are completely stable whether or not their exact voltage versus current characteristics are identical. This is because of the drooping voltage-current characteristics of shunt generators.

Fig. shows the circuit diagram for the operation of two shunt generators in parallel. Assume that generator A is in operation with switch SA closed and that the load it isdelivering is about to be increased sufficiently to require the use of a machine of larger rating than A or theconnection of a second generator to operate in parallel with A. The procedure for connection of two shunt generators in parallel is as follows:

1. Second generator B is brought up to rated speed by its prime mover.

2. The field switch SB2 is closed whereupon the voltage will build, assuring of course that all conditions for build up are satisfied.

3. The voltage of generator B2 is adjusted until it is equal to a slightly higher than that of generator A, the field rheostat of the incoming machine being manipulated for this purpose; special care should be taken that polarity of B is exactly same as that for A with respect to the load, i.e., the plus and minus terminals of the linking machine must be traced to the corresponding bus polarities across open switch SB.

4. With adjustment made and precautions taken as indicated, main switch SB is closed; this places generator B in parallel with A but the later still supplying the entire load and machine B running idle, i.e. floating.

5. To shift the load from A to B it is merely necessary to adjust the field rheostat of two generators simultaneously, cutting in resistance in the field circuit of A and at the same time cutting out resistance in the field circuit of B.

Any desired load shifting may be readily accomplished in this way; in fact the entire load may be transferred to B, after which

the main switch SA can be opened to disconnect generator A from the line. When it is necessary to shunt down any of the generators, it is not convenient to open its main switch suddenly. This would cause severe sparking at the contacts and would cause a sudden increase in the loads on the other machine. Instead it is better to decrease its field current till the current supplied by this generator becomes almost zero. Then its main switch can be safely opened and its prime mover can be shut down.

Now let us designate generator A and generator B as generator 1 and generator 2 respectively and these are connected in parallel. These generators have identical voltages and the characteristic curves of both the generators can be plotted back to back with terminal voltage in common on the vertical axis. When generators of this sort are connected in parallel, the sum of their line current must equal the current supplied to the load on the system.

IL1 + IL2 = Iload

The terminal voltage at which the power system operates will be exactly the line voltage required to make the sum of the two generator's currents equal to the current required for the loads.

If the speed or field current of generator 1 is increased in the system, the entire characteristic curve of generator 1 moves upward and the voltage at which the sum of two machine’ currents adds up to equal the load current rises. In addition, generator 1 now assumes a large fraction of the total load on the system. Similar adjustment of either generator 1 or generator 2 can permit the system voltage and the power sharing between the generators to be freely adjusted.

Q10. Explain the working principle of DC motor with diagram [2002].

Working Principle of DC Motor: In a D.C. motor, when the terminals are connected to external D.C. supply, the field magnets on excitation become alternate N and S poles and develop magnetic field which extends from North Pole to South Pole. Also the armature conductors of themotor carry currents. All the conductors under South Pole carry currents in one direction while under North Pole; those carry currents in the other direction. As explained,the mechanical force acts on the conductors (placed in the slots)and the armature of the D.C. motor tends to rotate in the anti-clockwise direction as shown. The direction of motion of the motor is determined by the Fleming’s left hand rule.

It is the arrangement of placing the conductors under the poles such that the direction of the current flowing through these armature conductors is as shown in Fig. or vice-versa. If the di-rection of current in the conductors is otherwise reversed from that shown in Fig. The motor will run in the clockwise direction. It is thus a matter of convention we use. The direction of motion of the motor is always determined by the Fleming's left hand rule.

The force on conductors gives a combined effect to produce a driving torque which makes the armature to continue rotating.

To understand the difference between motor and generator action let we discuss there two separately Fig. (b) shows a generator action where a mechanical force moves a conductor in upward direction inducing an e.m.f. in the direction shown by Fleming's Right hand rule. When a current flows as a result of this e.m.f., there is a current carrying conductor existing in a magnetic field: hence motor action takes place shown as a dotted line in Fig. (a) the force developed as a result of motor action opposes the motion which produced it.

Q11. State back emf of DC motor [2002, 2004, 2005].

Back EMF: In a D.C. motor when the armature rotates, the conductors on it cut the lines of force of magnetic field in which they revolve, so that an e.m.f. is induced in the armature as in a generator. The induced e.m.f. acts in opposition to the current in the machine and, therefore, to the applied voltage, so that it is customary to refer to this voltage as the "back e.m.f. 'That this is so can be deduced by Lenz's law, which states that the direction of an induced e.m.f. is such as to oppose the change causing it, which is, of course, the applied voltage.

The magnitude of the back or counter e.m.f. can be calculated by using formula for the induced e.m.f. in a generator, and it is important in the case of the motor, to appreciate that this is proportional to the product of the flux and the speed. Thus if Eb

denotes the back e.m.f., φ the flux and N the speed, we may write,where k is a number depending on nature of armature winding.

The value of back e.m.f. (Eb) is always less than the applied voltage, although difference is small when the machine is running under normal conditions. It is the difference between these two quantities which actually drives current through the resistance of the armature circuit. If this resistance is represented by Ra, the back e.m.f. by Eb and the applied voltage by V, then we have

V = Eb + IaRa

Where Ia is the current in the armature circuit.

Q12. Draw block diagram showing the power stages of dc motor [2002, 2004].

Block Diagram of Power Stages IN D.C. Motor

Q13. Differentiate between secondary cell and primary cell [2002, 2005].

Difference between Primary & Secondary CellsPrimary Cells Secondary Cells

1. Primary cells cannot be recharged.

Secondary cells can be recharged.

2. One may change the parts of the cells.

Plates cannot be changed they to be thrown out.

3. Primary cell is light weight. Secondary cell is heavy weight.4. Primary cell has high

internal resistance.Secondary cell has low internal resistance.

5. Primary cells have short life. Secondary cells have long life.6. Primary cells can be used

for low energy applications.Secondary cells can be used for high energy applications.

Q14. List parts of DC generator [2004].

Parts of DC generator: The basic essential parts of a DC generator are:

(i) A magnetic field(ii) A conductor which can so move to cut the flux

Q15. Explain types of DC generator [2004, 2006, 2007, 2008].

Types of DC Generator: When the field coils are excited by the exciting current supplied by the generator itself, it is said to be self-excited generator.

In a self-excited generator, the field coils may be connected in series with the armature coils (series) or in parallel with the armature coils (shunt) or partly in series and partly in parallel with armature coils (compound). According to the type of connections of the field coils with the armature coils the self-excited generators may be classified in to:(i) Series wound generator. (ii) Shunt wound generator. (ii) Compound wound generator.

(i) Series Wound Generator: Fig. shows a series excited generator in which the field coils are in series with the arma-ture and carry the total current of the generator. Fig. ( i ) shows the connections of a series wound generator where as Fig. (ii) shows its conventional diagram. Since the field coils carries the full load current therefore, it has a few turns of wire of large cross-sectional area having low resistance. Before the machine will excite, the external circuit must be closed. Series generators are scarcely ever used except for special purposes e.g., boosters.

(i) Shunt Wound Generator : In a shunt wound generator the field coils are connected in parallel (or shunt) across the armature terminals as shown in Fig. ( i ) & (ii). These coils have many turns of relatively small wire having

high resistance and carry a current IF or Ish =

VR sh , the terminal

voltage divided by the resistance of the field coil circuit. This exciting current varies from about 0-5 percent of the full load current of the machine to about 5 percent, depending chiefly on the size of the machine, the larger proportion of current being taken by the smaller machines. Fig. ( i ) & (i i ) shows the connections and conventional diagrams of a shunt wound generator.

(iii) Compound Wound Generators: Fig. ( i ) & ( i i ) shows compound wound generators in which there are both shunt and series field coils on each pole-One is in series and the other in parallel with the armature ; thus the compound wound generator may be:

(a) Short shunt. In which the shunt coils are connected in-side the series coils or in which the shunt field coils are only in parallel with armature, the connection is said to be short shunt as shown in Fig. (i).

(i)

(b) Long shunt. When the shunt coils are connected outside the series coils, the machine is said to have a long-shunt connection as shown in Fig. (ii).

(ii)It makes no appreciable difference in the operating charac-teristics which way the shunt is connected and the choice is determined by mechanical considerations of connections or reversing switches.

Q16. Explain the process of armature reaction with diagram in a DC generator [2002, 2004, 2005, 2007, 2008, 2009].

Armature Reaction: The armature reaction is basically the effect of the flux produced by the current carrying armature conductors on the main magnetic flux. So,” the armature reaction is defined as the effect of the magnetic field produced

by the armature conductors on the distribution of the flux under the main poles.

Now consider an armature rotating in the magnetic field produced by the two poles machine. When ever load is not connected on the armature so no current is flowing through the armature conductors, the magnetic field produced is merely because of the main magnetic field as shown in the following fig.

According to Fleming’s right hand ruler the conductors under North Pole are carrying current in such a direction as shown. Now the magnetic field produced by armature conductors is shown in fig. (b) .the magnetic flux will contribute the magnetic belt in the both sides of MNA and hence the direction of armature flux is upwards. Now two magnetic fluxes, because of main magnetic field and because of the armature conductors are working at right angle to each other when these are energized as shown in fig. (c). As a result, the main flux will be distorted and the flux will be strong and TPT and weak at LPT. Hence the brushes are to be given the forward lead in the direction of motion of the armature, i.e. the new neutral axis, which is at right angle to the resultant magnetic field. The angle lead depends upon the load on the armature more load, more angle and vice versa.

It is observed that the flux through the armature is no longer uniform and symmetrical about the poles axis. The resultant armature flux as shown in fig. has two components: the d = the demagnetizing component and c = the cross-magnetizing component. The demagnetizing the main flux and the cross-magnetizing component is at right angle to the main magnetic flux.

Q17. Explain the necessity of starter for DC motor [2004].

Necessity of a starter: When the motor is at rest the speed of the motor is zero, there fore back emf Eb is zero and if a dc motor is connected directly tot eh supply mains, a heavy current will flow through the armature conductors because from the emf equation for armature circuit (Eb= V- IaRa) armature current Ia is given as and armature resistance Ra is very small. For example consider a 400 V, 20 kW dc motor having a total resistance of 0.5 Ω. If switched directly on to the supply, it would draw a current of a while the full-load current would probably be about 64 ampere. The starting current thus would be i.e.12.5 times the full- load values. When running, of course, the applied voltage V is opposed by the induced back emf and, there fore, a much smaller current flows.

Heavy inrush of current at the stating instant may cause(i) Heavy sparking at the commutator and even flash-

over’s.(ii) Damaged to the armature winding, either by the heat

developed in the windings, or by the mechanical forces set up by electro- magnetic action.

(iii) Damage to the rotating parts of the motor and load to development of large starting torque an quick acceleration and

(iv) Large dip in the supply voltage.

Hence for the protection of the motor against the flow of excessive current during starting period (say 5 to 10 seconds),

it is necessary that a high resistance be connected in series with the armature of the motor at the instant of starting and gradually cut in steps as the motor gains speed, the additional resistance from the armature circuit it totally disconnected. If this additional loss of energy resulting in reduced operating efficiency and (ii) reduction in operating speed of the motor.

However, a very small motor (frictional kW motor) may be started simply by closing the switch which connects it to the supply mains. The reasons for it are given below.

(i) The resistance and inductance of the armature winding in case of small motors are generally sufficiently large to limit the initial of current to values that are not particularly serious.

(ii) The inertia of a small armature is generally so low that it comes upto speed very quickly, there by minimizing the detrimental effects that might otherwise result from the excessive sustained current.

Q18. Define the operation and protection offered by 4 point starter [2004, 2009].

4-Point Starter: The three point starter cannot be used to advantage on variable-speed motors having field control. Such motors normally have a speed variation of 5 to 1. This results in the field current having approximately the same range. The hold up magnet may be too strong, therefore, at the higher values of field current and too weak at the lower values. This difficulty can be overcome by a four point starter as shown in Fig. (b)., except that the hold up coil is of high resistance and is connected directly across the line. The only difference in the connection is that the "line" terminal must be connected to the side of the line which runs directly to the common armature and field terminals. When the supply voltage is shut down, the hold up coil becomes de-energized and allows the arm to spring back to the starting position.

In stopping a motor, the line switch should always be opened rather than throwing back the starting arm. With shunt motors the line switch can be opened with no appreciable arc, since the motor develops a counter e.m.f. nearly equal to the line voltage and the net voltage across the switch contacts is small. The electro-magnetic energy stored in the field does not appear at the switch but is discharged gradually through the armature. On the other-hand, if the starting arm is thrown back, the field circuit is broken at the last contact button. Owing to the inductive nature of the field, this results in a hot arc which burns the contact. To prevent the contact from being burned, a small finger can be introduced to break the arc.

Q19. State types of DC motors [2004, 2005].

Types of DC motors: A dc motor must receive their excitations from an outside source, therefore, they are separately excited, their field and armature windings are connected, however, in one of the three different ways employed for self-excited dc generators, and so according to the field arrangement there are three types of dc motors

namely: (i) series wound (ii) shunt wound and (iii) compound wound.

1. Series Wound Motor: A series motor is one in which the field winding consisting of few turns of thick wire is connected in series with the armature so that the whole current drawn by the motor passes through the field winding as well as armature. Connection diagram is shown in fig.

In a dc series motor,Armature current, la = Series field current, lse

= Line current, IL = I( say)Back emf developed, Eb = V-I(Ra + Rse)Power drawn from supply mains = VI wattswhere V is the supply voltage and I is the input current.Power developed = Power input-losses in armature and field

= VI – I2 (Ra + Rse)= I [V – I (Ra + Rse)] = Eb I watts.

2. Shunt Wound Motor: A shunt wound motor is one in which the field winding consisting of large number of turns of comparatively fine wire is connected in parallel

with armature, as illustrated in fig.. The current supplied to the motor is divided into two paths, one through the shunt field winding and second through the armature.

i.e. input line current, IL = Ia + Ish where Ia is the armature current and Ish is the shunt field current given by the expression

Ish =

VR sh R sh being the shunt fiend resistance and V the supply

voltage.Back emf developed, Eb = V – Ia Ra

Power input = VIL

Power developed = Power input - losses in armature and shunt field

= V IL –VIsh - Ia2 Ra

= V (IL –Ish ) - Ia2 Ra

= V Ia – Ia2 Ra = Ia (V- Ia Ra)

= Eb la

3. Compound Wound Motors: Compound wound motors are of two types namely cumulative compound wound and differential compound wound.

Cumulative compound wound motor is one in which the field windings are connected in such a way that the direction of flow

of current is same in both of the field windings, as illustrated in fig:. In the motor of this type the flux due to series field winding strengthens the field due to the shunt field winding.

Differential compound wound motor is one in which the field windings are connected in such a way that the direction of flow of current is opposite to each other in the two field windings, as illustrated in fig.. In this type of motor the flux due to series field winding weakens the field due to shunt field winding.

Q20. Explain with diagram, how the torque is developed in a DC motor [2004].

TORQUE DEVELOPED IN A MOTOR: When the field of a machine (of the type described as generator) is excited and a potential difference is impressed upon the machine terminals, the current in the armature winding reacts with the air-gap flux to produce a turning moment or torque which tends to cause the armature to revolve. Fig. illustrates production of torque in a motor.

When the brushes are on the neutral axis, all the armature conductors lying under the North Pole carry currents in a given direction, while those lying under South Pole carry currents in the reverse direction. The commutator (just as in a generator) serves to reverse the current in each armature coil at the instant it passes through the neutral axis, so the above relation is always maintained as the armature rotates.

All conductors under the North Pole carry inward-flowing currents which react with the air gap flux to produce down-ward acting forces and a counter clockwise torque. Similarly the conductors under the South Pole carry outward-flowing currents which produce upward-acting forces. These forces also give rise to counter clockwise torques. If the air-gap flux is assumed to be radially directed at all points, each of the force acts tangentially and produces a turning moment equal to the force multiplied by its lever arm — the radial distance from the centre of the conductor to the centre of the shaft.

Magnitude of torque developed by each conductor= Bllr Nm

If the motor contains Z conductors, the total torque developed by the armature

Ta = BllrZ NmWhere B = gap density, T (Wb/m2)I = armature current in a conductor, A I = active length of each conductor, mr = average lever arm of a conductor or the average radius at which conductors are placed, m Z = total number of armature conductors.It is more convenient to express Ta in terms of armature current Ia, total flux per pole φ and number of poles p.

I =

I a

a

And B =

φA

Where a = number of parallel paths,and A = the cross-sectional area of flux path at radius r.

A =

2 π rl p

Then B =

Z φlI a r2 π rl x

pa

=

Z φI a p2 π a Nm

Ta = 0.159Zφ p x

I a

a

Or Ta = kφ la Nm

Where k =

Zp2π a is a constant for any machine.

Q21. State the applications of DC motors depending upon their characteristics [2004].

Applications of motor depending upon their characteristic

1. Shunt motors: The characteristics of a shunt motor reveal that it is an approximately constant speed motor. It is therefore, used:

(i) Where the speed is required to remain almost constant from no load to full load.

(ii) Where the load has to be driven at a number of speeds and any one of which is required to remain nearly constant.

Industrial Use: Lathes, drills, boring mills, shapers, spinning and weaving machines etc.

2. Series motors: It is variable speed motors i.e. speed is low at high torque and vice versa. However, at light or no load, the motor tends to attain dangerously high speed. The motor has a high starting torque. It is therefore, used:

(i) Where large starting torque is required e.g. in elevators and electric traction.

(ii) Where the load is subjected to heavy fluctuations and the speed is automatically required to reduce at high torques and vice versa.

Industrial Use: Electric traction, cranes, elevators, air compressors, vacuum cleaners, hair dry, sewing machines etc.

3. Compound motors: Differential compound motors are rarely used because of their poor torque characteristics. However, cumulative compound motors are used where a fairly constant speed is required with irregular loads or suddenly applied heavy loads.

Industrial Use: Presses, shears, reciprocating machines etc.

Q22. Define commutation [2004, 2006].

Commutation: Commutation means the process of current collection by the brush or the changes which take place in a coil during the period of short circuit by a brush. The period during which this happens is called the commutation period.

Commutation is considered to be good when the changes of the current in the coil sections are not accompanied by sparking between the brushes and the commutator, and the surface of the latter remains clean and undamaged during continuous

operation of the machine. Conversely, commutation is considered to be poor if there is sparking at the brushes and appreciable damage to the commutator surface hindering reliable operation of the machine.

One of the main tasks in the analysis of commutation process is to find the causes of sparking between the brush and the commutator. The sparking may be called forth by causes differing in their physical nature, i.e. mechanical and electrical.

Q23. State function of lead acid battery [2004, 2005, 2007, 2008, 2009].

Lead acid battery: Lead acid battery cells are used where large currents and energy storage is needed. Lead acid cells the most commonly used secondary cell and it is the power source for the electric system of the most cars, trucks and tractors. It can provide large current needed to crank internal combustion engine. A lead acid cell produces about 2.1V. Higher voltages are obtained by connecting cells together to form batteries. A 12 V automobile battery actually has a nominal voltage of 12.6 volts, because it contains six cells. A lead acid battery consists of 3 cells, 6 cells and 12 cells. Now we are going to study in detail construction, action and characteristics of, lead acid battery one by one.

Q24. State the different methods of speed control of DC motor and explain any one in brief [2004, 2007, 2009].

Speed control of d.c. series motor: The speed control of D.C. series motors can be obtained by (i) Flux control method (ii) Armature resistance control method. The latter method is mostly used.

I. Flux Control Method: In this method, the flux produced by the series motor is varied and hence the speed. The variation of flux can be achieved in the following ways:

(i) Field Diverters: In this method, a variable resistance (called field diverter) is connected in parallel with series field winding as shown in figure. Its effect is to shunt some portion of the line current from the series field winding, thus weakening the field and increasing the speed (N = 1/). The lowest speed obtainable is that corresponding to zero current in the diverter (i.e. diverter is open). Obviously, the lowest speed obtainable is the normal speed of the motor.

Consequently, this method can only provide speeds above the normal speed. The series field diverter method is often employed in traction work.

(ii) Armature Diverter: In order to obtain speeds below the normal speed, a variable resistance (called armature diverter) is connected in parallel with the armature as shown in figure. The diverter shunts some of the lien current, thus reducing the armature current. Now for a given load, if Ia is decreased, the flux must increase (T = Ia). Since, N 1/, the motor speed is decreased. By adjusting the armature diverter any speed lower than the normal speed can be obtained.

(iii) Tapped Field Control: In this method, the flux is reduced (and hence speed is increased) by decreasing the number of turns of the series field winding as shown

in figure. The switches can short circuit any part of the field winding, the motor runs at normal speed and as the field turns are cut out, speeds higher than normal speed are achieved.

II. Armature Resistance Control: In this method, a variable resistance is directly connected in series with the supply to the complete motor as shown in figure. This reduces the voltage available across the armature and hence the speed falls. By changing the value of variable resistance, any speed below the normal speed can be obtained. This is the most common method employed to control the speed of d.c. series motors. Although this method has poor speed regulation, this has no significance for series motors because they are used in varying speed applications. The loss of power in the series resistance for many applications of series motors is not too serious since in these applications, the control is utilized for a large portion of the time for reducing the speed under light load conditions and is only used intermittently when the motor is carrying full load.

Q25. Enumerate types of cell. Explain working of nickel iron battery [2004, 2006, 2007].

Types of cells

1. Primary Cell: The basic working principle of cell is .that an emf can be obtained by two dissimilar metal when acted upon by an electrolyte. This emf will produced until the chemical action has changed all of the electrolyte or electrode in some other chemical form, then the cell is said to the discharged. A primary cell is one which cannot be recharged. In a primary cell it is not possible to reverse the chemical action to bring the material into their original condition. The carbon-Zinc, mercuric oxide silver oxide and alkaline cells are the most common examples of primary cell.

2. Secondary Cell: Secondary cells work on the same principles as primary cells but differ in the process in which they may be renewed. In a secondary cell the chemical reaction which takes between the electrolyte and electrode on discharge can be completely reversed by sending a current through the cell in the opposite direction from an external source of emf. This is called charging. A secondary cell is one which can be charged. In a secondary cell it is possible to reverse the chemical action to bring the material into their original condition. The lead acid nickel cadmium, silver Zinc and Edison cells are the most common examples of secondary cell. Difference between a

primarv cell and a secondary, cell is that the primary cell and a secondary cell is that the primary cell cannot be recharged after use while the secondary cell can be recharged after use.

Nickel iron battery (Edison cell): The nickel-iron battery is a storage battery having a nickel (III) oxide-hydroxide cathode and an iron anode, with an electrolyte of potassium hydroxide. The active materials are held in nickel-plated steel tubes or perforated pockets. The nominal cell voltage is 1.2V. It is a very robust -battery which is tolerant of abuse, (overcharge, over-discharge, short-circuiting and thermal shock) and can have very long life even if so treated. It is often used in backup situations where it can be continuously charged and can last for more than 20 years. Its limitations, namely, low specific energy, poor charge retention, and poor low-temperature performance, and its high cost of manufacture compared with the lead-acid battery led to a decline in usage along with it having the lowest energy-to-weight ratio.

The ability of these batteries to survive frequent cycling is due to the low solubility, of the reactants in the electrolyte. The formation of metallic iron during charge is slow because of the low solubility of the Fe3O4 which is good and bad. It is good because the slow formation of iron .crystals preserves the electrodes; bad because it limits the high rate performance: these cells charge slowly, and are only able to discharge slowly.

Nickel-iron batteries have long been used in European, mining operations because of their ability to withstand vibrations, high temperatures and other physical stress. They are being examined again for use in wind and solar power systems and for modern electric vehicle applications.

In many respects the Nickel/Iron battery was almost ''too good". A battery that lasts for decades in many cases can outlast the equipment that is was originally designed to power.

So from an economic standpoint lead acid, NiCd and other technologies have been deemed "good enough" and are the predominant technologies in use today even though they do not last as long as a Nickel/Iron counterpart.

Q26. Define preparation of electrolyte of a required specific gravity [2004, 2006, 2007, 2009].

Method of preparation of electrolyte for a given specific gravity

Electrolyte: A substance through which an electric current can pass in solution form or in molten form is called electrolyte.

Example: Nacl, KBr, H2SO4, NaOH, etc.

Types of Electrolyte: There are three types of electrolytes:(i) Strong electrolytes

(ii) Weak electrolyte(iii) Non-electrolytes(i) Strong Electrolytes: The compounds which ionize to a

large extent in dilute solution and conduct electric current to a large extent are called strong electrolytes.

Examples: HCl, NaOH, NaCl, etc.

(ii) Weak Electrolytes: The compound which ionize to a small extent in dilute aqueous solution and conduct electric current to small extent are called weak electrolytes.

Examples: CH3COOH, benzoic acid, carbonic acid, etc.

(iii) Non-Electrolytes: The compounds, which do not ionize in aqueous solution and do not conduct electric current, are called non-electrolytes.

Examples: Sugar, urea, glucose, etc.

An electrolyte can be prepared by adding pure distilled water in dilute sulphuric acid at specific gravity.

During charging process the amount of H2SO4 increased while water content increased during discharging. The value of specific gravity is never constant in electrolyte during charging and discharging. Battery condition can be easily understand by given specific gravity charge with the help of this chart we can understand the battery condition and relevant specific gravity of electrolyte.

Battery condition Relevant Specific gravityFully charged 1.26 to 1.30

About ¾ charged 1.23 to 1.26Half charged 1.20 to 1.23

About ¼ charged 1.17 to 1.2Discharged 1.11 to 1.14

The another factor for charging of specific gravity is temperature of battery. At high temperature battery is fully charged at comparatively low specific gravity as compared to low temperature.

It should be noted that amount of heat is produced when acid and water are mixed further if water added into acid it is harmful for worker so first acid is added drop by drop into distilled water for dilution.

Q27. Explain purpose of interpoles [2004, 2005, 2006, 2009].

Purpose of inter-poles : These are the poles mounded in between the main poles. These are small in size than the main poles. Generally these are 1/3 to 3/4 th of the main poles as shown in Fig. The winding is done with thick conductor. It is connected in series with the armature. The magnetic field produced by these poles is proportional to the load current.

In case of D.C. motor, the polarity of inter-poles will be opposite to the main poles ahead in the direction of rotation of the armature.

Q28. Explain with neat diagram the Ward Leonard system of DC motor speed control [2006].

Ward Leonard System: This method of control not only gives a wide range of operating speeds, but reduces to the very minimum the wastage of energy that ay take place at starting and stopping. Fig shows the schematic arrangement of Ward Leonard method.

M = main motor whose speed is to be controlledG = separately excited generator which feeds the armature of the motor ME = an exciter (a small shunt generator) which provides field excitation to the generator G and motor MM’ = driving motor – a constant speed motor which drives G and ER = a potentiometer rheostatS = a double throw switch

The working of this system is as follows: The motor M’ drives the generator G and excitor E at

constant speed. The voltage fed to motor M can be controlled by varying the setting of R. A change in voltage applied to motor M changes its speed. The speed can be adjusted to any value from zero to maximum in either direction by means of a rheostat R and switch S.

When the sliding contact of R is at extreme right, the motor is running at full speed in one direction. To decrease the speed the sliding contact is moved to the

left. When the sliding contact is at the left extreme position, the speed of motor M is zero. In order to reverse the speed of the motor, the sliding contact is shifted to the extreme left, the switch S is reversed and the sliding contact shifted to right again.

A modification of the Ward Leonard system is known as Ward Leonard Ilgner system, which uses a small motor generator set with the addition of a flywheel whose function is to reduce fluctuations in the power demand from the supply circuit. When the main motor M becomes suddenly overloaded, the driving motor M’ slows down, thus allowing the inertia of the flywheel to supply a part of the overload. However, when the load is suddenly thrown off the main motor M, then M’ speeds up thereby again storing energy in the flywheel. When Ilgner system is driven by means of an AC motor (whether induction or synchronous) another refinement in the form of a slip regulator can be usefully employed thus giving an additional control.

One important feature of the Ward Leonard system is its regenerative action. When a locomotive, fitted with this system, is descending a slope, its speeds up due to the action of gravity. The speed of motor M increases until its back emf exceeds the applied voltage motor M then runs as generator and feeds the machine G which now works as a generator and feeds electrical energy back into the trolley wire. This results in salvaging of considerable amount of energy and a superior and smooth braking action. Such an action in known as regenerative braking.

Q29. What are the necessary requirements for parallel operation of DC generator [2008].

Requirements for Parallel Operation of DC Generator: The following are the principle types of situations where paralleling of DC generators is required:

Paralleling shunt generators of the same or varying sizes

Paralleling compound generators of the same or varying sizes

There are certain requirements that must be met for successful electrical paralleling in all different situations. A parallel circuit is defined as one in which the same voltage exists across each unit as the paralleling point.

This is absolutely required by Kirchhoff’s voltage law.

The following three conditions may be met if the generated voltages of the individual generators are not all the same, and they are paralleled:

(i) If a generator is developing an internally generated voltage Eg that is appreciably above the voltage at the paralleling point, generator action is taking place and the unit is delivering current to the load.

(ii) When a generator is producing the same voltage as that existing at the paralleling point, no effective generating action is taking place and no current is flowing to the load. The generator is said to the floating on the line. It is neither contributing nor drawing current and is still being rotated by its own prime mover.

(iii) If the setting of the generator is so made that it develops less internal Eg than voltage at the paralleling terminal, it will draw current from the paralleling point and will be operating as motor.

The above three situations are in entire agreement with Kirchhoff’s current law, as any parallel circuit must be.

The following are the requirements or conditions of paralleling DC generators:

(i) The polarities of the generators must be the same or the connections must be interchanged until they are.

(ii) The voltages should be nearly if not exactly identical so that each machine will contribute.

(iii) The change of voltage with change of load should be of the same character.

(iv) The prime mover that drive the generators should have similar and stable rotational speed characteristics.

Q30. State the applications of DC Series, Shunt and Compound motors [2008].

Applications of DC Motors

1. DC Shunt Motor(i) Drills and milling machines(ii) Line shaft drives(iii) Boring mills(iv) Grinders and shapers(v) Spinning and weaving machines(vi) Wood working machines(vii) Small printing presses(viii) Light machine tools generally

2. DC Series Motor(i) Traction drives generally(ii) Tram cars and railway cars(iii) Cranes, derricks, hoists, elevators and winches(iv) Fans and air compressors(v) Vacuum cleaners, hair driers, sewing machines(vi) Universal machines generally

3. Cumulative Compound Motor(i) Punching, shearing and planning machines(ii) Lifts, haulage gears and mine hoists(iii) Pumps and power fans(iv) Rolling mills, stamping presses and large printing

presses

(v) Trolley buses

4. Differential Compound Motor(i) Battery boosters(ii) Experimental and research work

Q31. What is the importance of DC machine testing? Why it is necessary [2008, 2009].

Testing of d.c. machinesThe following important performance tests are

conducted on D.C. machines:1. The magnetization or open circuit test.2. The load characteristics.3. The determination of efficiency curve.4. The temperature rise test.

The procedure to conduct the magnetization or open circuit test and load characteristic (external characteristic) tests

The methods for determining efficiency can be divided into following three methods:

(i) Direct method (ii) Indirect method(iii) Regenerative method.

Direct method: In direct test the generator or motor is put on full-load and whole of the power developed by it is wasted. Brake test is a typical example of direct test. The direct tests can be used only on small machines.

Though this method is simple in looking, but involves complication in the measurement of mechanical power input in the case of a generator and output in case of a motor.

If brake is to be applied to a series motor, the brake must be tight before the motor is started, otherwise the armature may get damaged and fly to pieces.

Indirect method: This method consists in measuring the losses and then calculating the efficiency. The simplest of the indirect tests is Swinburne test.

This method enables the determination of losses without actually loading the machine. The power is required to supply the losses only, so there is no difficulty in applying this method even to very large machines.

The disadvantage of this method is that machine is run light during the test which gives no indication as to the temperature rise on load or to the commutating qualities of the machine.

Regenerative method: This method requires two identical machines; one of them works as a motor and drives the other, which is mechanically coupled to it. The other machine works as a generator and feed back power into the supply. Thus the total power drawn from the supply is only for supplying internal losses of the two machines. Thus even very large machines may be tested as the power required is small.

Hopkinson test is a regenerative test for determining efficiency of D.C. machines.

Q32. Explain the Hopkinson’s test [2004, 2008, 2009].

Hopkinson testThe back to back test for dc machines was devised in its original form by Hopkinson and is similar to the Sumpner's test for transformers. In practice the modification of the original

Hopkinson test by Kapp is employed and it is usually called the Hopkinson test. The connection diagram for the Hopkinson test is shown in Fig.. Two dc shunt machines are mechanically coupled and losses are supplied electrically.

Machine M is started with the help of a starter as a dc shunt motor and brought to speed with switch SW open. Since both the machines M and G are mechanically coupled, both the machines run at the same speed. The field current of the machine G is adjusted till the voltmeter VI reads zero and the switch is then closed. Under this condition the generator will float neither taking nor giving current to supply. Any desired load can be put on the generator by increasing the value of its induced emf; this in turn can be accomplished either by increasing the generator exciting current or by decreasing the exciting current of the motor or both adjustments may be made. The machine with lower field current will act as motor and the other will work as generator. The power supplied from the mains is only the power required to supply the losses of the two machines. At the end of the test, the resistances ra1 and ra2 of the armature circuit are measured by passing full load current through them and measuring voltage drops with field circuit open. The total current 11 is supplied from the mains to two dc machines M and G.

Armature current of generator is

lg = I2 + I4

Armature current of motor islm = I1 - I4 + I2

Armature copper loss of generator= I2

g ra2

Armature copper loss of motor= I2

mra1

Field copper loss of generator= VI4

Armature copper loss of motor= VI3

Total iron + friction losses for the two machines= Vl1= - (I2

mra1+ I2g ra2 + VI3 + VI4))

Total iron + friction loss per machines P0 = 0.5 (Vl1- (I2

mra1+ I2g ra2 + VI3 + VI4))

Motor power loss P11 = P0 + I2

mra1+ VI3

Motor input powerP1 = V (I1 + I2)

Motor output power = P1 - P11

Motor efficiency is given by

Efficiency =

P1 - P11

P 1=

V ( I 1+ V2 )−(P0+ I2m r a1+Vl3)

V ( I 1 +12 ) Generator power loss P12 = P0 + I2

gra2+ VI4

Generator output power P2 = VI2

Generator input power P2 = P12

Generator efficiency is given by

Efficiency =

P2

P 2 +P12= V 12

VI2+P0+ I2m r a2+Vl4

Q33. Explain the methods of charging of battery and explain in brief any one of them [2001, 2004, 2008].

Charging of battery: For charging a battery a d.c. supply of voltage little higher than the battery voltage is applied across the battery. The care should be taken that positive terminal of battery should be connected to positive terminal of supply and negative plate to negative terminal of supply.

The current passes from positive plate to the negative plate inside the cell the chemical reaction is an follows.

At Negative plate: PbSO4 + 2H+ = Pb +

H2SO4

(Lead sulphate) (Hydrogen) (Lead) (Sulphuric acid)

At Positive plate: PbO2 + SO-

4 + 2H2O = PBO2 + 2H2SO4

(Lead sulphate) (Sulphate) (Water) (Lead peroxide) (Sulphuric acid)

The changes occurs during charging is as below: (i) The positive plate changes to lead peroxide (PbO2),

which is dark chocolate brown in colour. (ii) The negative plate changes to spongy lead which is

slate grey in colour.

(iii) The voltage per cell increases from 1.8 volt to 2.1 volt.

(iv) The specific gravity of each cell rises from 1.180 to 1.280.

The cell stores electrical energy in form of chemical energy. Fig. illustrate charging of battery.

Q34. How the battery is discharged, give its process in brief [2004, 2008].

Discharging of Battery: When the cell is fully charged the positive plate is of nickel hydroxide [(Ni (OH4)] and the negative plate, is of iron (Fe). When load is connected to the terminal of the cell, current flow from positive plate outside the cell and from negative to positive inside cell. The current through the electrodes breaks it into potassium ions (K+) and hydroxide ions (OH-). The K+ moves toward the anode and OH-

towards cathode the chemical reactions is as given below:At Negative plate:Fe++ + 2OH - Fe (OH)2(Iron) (Hydroxide of potassium) (Ferrous hydroxide)

At Positive plate:

Ni (OH)4 + 2K+ = Ni (OH), + 2KOH (Nickel hydroxide) Potassium (Nickel of lower oxide) (Potassium hydroxide)

The following-changes occur during discharging:(i) The negative plate turns into ferrous oxide,(ii) The positive plate turns into the lower hydroxide of nickel,(iii) The strength of electrolyte remains constant.

Q35. Explain Swinburn’s and Dynamometer test [2009].

Swinburne's test: This is an indirect method of determining the efficiency of a motor by measuring the losses. In this method the iron and friction losses are determined by measuring the input to the machine on no-load, the machine being run as a motor at normal voltage and speed. Copper losses are calculated from measured values of the various resistances. This test is carried out for shunt machines. The connection diagram for Swinburne's test is shown in Fig.

In the case of shunt motor, let I0 = the total current on no-load, V being the supply voltage. Then the total losses on no load are Vlo. A small and normally negligible portion of these losses consists of ohmic losses in the armature, brush contacts and inter-pole endings. The armature copper loss is I2

a0 ra where ra

is the armature resistance and the armature current is Ia0 = I0 - If. The total iron, friction and windage losses are

Po=VI0-I2a0 ra

For an input current I, Ia = I- If and line voltage V, the efficiency of the dc machine as a motor is given by

Efficiency = VI- I

2a r2 - P 0 - VIf

VI

Knowing the losses the efficiency of the machine as a generator can be determined from the relation

Efficiency =

VI

VI- I2a r2 - P 0 - VIf

The armature resistance is measured by a dc test by passing a rated current from a battery supply.

Dynamometer test: The dyno must be able to operate at any speed, and load the prime mover to any level of torque that the test requires. A dynamometer is usually equipped with some means of measuring the operating torque and speed.

Dynamometers can be equipped with a variety of control systems. If the dynamometer has a torque regulator, it operates at a set torque while the prime mover operates at whatever speed it can attain while developing the torque that has been set. If the dynamometer has a speed regulator, it develops whatever torque is necessary to force the prime mover to operate at the set speed.

A motoring dynamometer acts as a motor that drives the equipment under test. It must be able to drive the equipment at any speed and develop any level of torque that the test requires.

Only torque and speed can be measured; Power must be calculated from the torque and speed figures according to the formula:

Where K is determined by the units of measure used as can be seen below:

To calculate power in horsepower (hp) use:

where:

Torque is in pound-feet (lbf·ft)

Rotational speed is in revolutions per minute (rpm)

To calculate power in kilowatts use:

where:

Torque is in newton-metres (N·m)

Rotational speed is in revolutions per minute (rpm)

(On graphs of torque vs. rpm the numerical values of torque and power are always equal when the rpm value is equal to the constant, K. The numerical values of horsepower and lbf·ft of torque are always equal at 5252 rpm because 5252 rpm in the numerator cancels out the constant, 5252 in the denominator leaving only the torque figure equal to the power, Fig.)

Detailed dynamometer description

Electrical dynamometer setup showing engine, torque measurement arrangement and tachometer

A dynamometer consists of an absorption (or absorber/driver) unit, and usually includes a means for measuring torque and rotational speed. An absorption unit consists of some type of rotor in a housing. The rotor is coupled to the engine or other equipment under test and is free to rotate at whatever speed is required for the test. Some means is provided to develop a braking torque between dynamometer's rotor and housing. The means for developing torque can be frictional, hydraulic, electromagnetic etc. according to the type of absorption/driver unit.

One means for measuring torque is to mount the dynamometer housing so that it is free to turn except that it is restrained by a torque arm. The housing can be made free to rotate by using trunnions connected to each end of the housing to support the dyno in pedestal mounted trunnion bearings. The torque arm is connected to the dyno housing and a weighing scale is positioned so that it measures the force exerted by the dyno housing in attempting to rotate. The torque is the force indicated by the scales multiplied by the length of the torque arm measured from the center of the dynamometer. A load cell

transducer can be substituted for the scales in order to provide an electrical signal that is proportional to torque.

Q36. Explain parallel operation of DC Compound Generator [2009].

Parallel Operation of Compound Generator: Since these machines possess winding in series with the armature and usually having a rising characteristics, it is again necessary to equalize their field currents. This is usually affected by means of an auxiliary bar on the main switch board, the bar being called the "equalizer bus bar". Essential connections for parallel operation of compound generators are shown in Fig. In absence of the equalizing bar, an accidental increase in the generated voltage of one machine causes it to supply more load current which in time strengthens its series field. This increases the generated emf further, with still further increase in its load current.

When the equalizing conductor is used, two series field coil windings are connected in parallel and this cumulative action can not take place. An increase in generated emf of one machine increases its armature current but series field is not strengthened so that successful parallel operation is possible.

The equalizer should be connected at the junction of inter-pole and series field windings. This is necessary so that the inter-pole winding carries the armature current. For satisfactory operation, the resistance of the equalizing connector should be very small in comparison with that of the field coils.

Q37. Write short notes on the following [2002, 2004, 2005, 2006, 2007, 2008, 2009].

Dry Cell: It is a portable cell. Dry cell is a modification of Lachlanche cell. In this cell the electrolyte in the paste shape. It is a mixture of FeSO4, 7H2O, NH4C1 and H2O. The depolarizer is a mixture of crystals of carbon, MnO2, H2O etc. A Zn cylinder is used to house all the chemicals and other components; it works as negative terminal. A carbon rod is placed in centre and it works as positive terminal of the cell. The paste is surrounding the carbon rod in a muslin bag which works as the porous pot. A vent plug is provided to escape the NH3 gas. Figure shows the simple construction of the dry cell.

3-Point DC Motor Starter: The starter shown in Fig. is of more practical arrangement. This is a shunt motor starter. The field has been permanently connected to the first stud of the resistance and the risk of opening the shunt field circuit is thus

avoided. An electromagnet, known as "no-volt release" is arranged to hold the arm in "on" position against the pull of the spring provided in the starter arm. When the supply fails then the arm is released and is pulled back by the spring to the 'off position.

An 'over load release' shown in Fig. provides an overload protection. When the current is high enough to attract the ar-mature of the overload release, then a pair of contacts is bridged. These contacts are connected to the ends of the coil of the no-volt release which, consequently is released and return to the "off position. This overload release is instantaneous in action. In practice overloads of large magnitude but of short duration usually occur, thus such an instantaneous 'overload release' is sometimes suitable.

Voltage Build up in DC Generator: One of the simplest forms of self-excited generator is the shunt wound machine, the

connection diagram (without load) of which is shown in figure. The manner in which a self-excited generator manages to excite its own field and build a D.C. voltage across its armature is described with reference to figure in the following steps:

(i) Assume that the generator starts from rest, i.e., prime-mover speed is zero. Despite a residual magnetism, the generated e.m.f. E, is zero.

(ii) As the prime-mover rotates the generator armature and the speed approaches rated speed, the voltage due to residual magnetism and speed increases.

(iii) At rated speed, the voltage across the armature due to residual magnetism is small, E1 as shown in the figure. But this voltage is also across the field circuit whose resistance is Rf. Thus, the current which flows in the field circuit I1, is also small.

(iv) When I1 flows in the field circuit of the generator an increases in mmf results due to IfTf, Tf being field turns, which aids the residual magnetism in increasing the induced voltage to E2 as shown in figure.

(v) Voltage E2 is now impressed across the field, causing a large current I2 to flow in the field circuit. I2Tf is an increased mmf which produces generated voltage E3.

(vi) E3 yields I3 in the field circuit, producing E4. But E4

causes I4 to flow in the field producing E5; and so on, up to E8, the maximum value.

(vii) The process continues until that point where the field resistance line crosses the magnetization curve in

figure. Here the process stops. The induced voltage produced, when impressed across the field circuit, produces a current flow that in turn produces an induced voltage of the same magnitude, E8, as shown in the figure.

Chemical Action during Charging and Discharging of Lead Acid Battery:

Chemical Process during Discharging: By discharging of a cell we mean that it is delivering current to the external circuit. Consider a charged lead acid cell with anode of PbO2 and cathode of Pb; the electrolyte being dilute H2SO4 (See figure). Sulphuric acid splits up into hydrogen ions (H+H+) and sulphate ions (SO4

--). The sulphate ions move towards the cathode and hydrogen ions move towards the anode causing the following chemical actions:

At cathode: On reaching the cathode, a sulphate ion (SO4--)

gives up its two extra electrons to become sulphate radical. These electrons given up at the cathode move through the external circuit to the anode where they are available to neutralize the positive ions (H+H+) arriving there. Since sulphate radical cannot exist, it enters into chemical action with cathode material (Pb) to form lead sulphate (PbSO4).

SO4-- - 2e SO4 (radical)

Pb + SO4 PbSO4

At anode: On reaching the anode, each hydrogen ion takes one electron from it to become hydrogen gas. This electron is given by the sulphate ion at the cathode and has come to the anode via the external circuit.

H+H+ + 2e 2H

The hydrogen gas liberated at the anode acts chemically on the anode material (PbO2) and reduces it to lead oxide (PbO).

PbO2 + 2H PbO + H2O

Sulphuric acid reacts with PbO to form PbSO4

PbO + H2SO4 PbSO4 + H2O

The chemical changes that take place during discharging can be summed up as under:(i) Both the plates are converted into lead sulphate

(PbSO4) which is whitish in colour.(ii) Water is formed which lowers the specific gravity of

the electrolyte (H2SO4). When the cell is fully discharged, the specific gravity of H2SO4 falls to about 1.18.

(iii) The e.m.f. of the cell falls. The lead acid cell should not be discharged beyond the point where its e.m.f. falls to about 1.8 volts.

(iv) The chemical energy stored in the cell is converted into electrical energy.

It is important to note that e.m.f. of the cell provides little indication to the state of discharge of the cell since it remains close to 2V for 90% of the discharge period. In practice, specific gravity of the electrolyte (H2SO4) is used to know the

state of discharge. The cell should be recharged when specific gravity of H2SO4 falls to 1.18.

Chemical Process during ChargingConsider a discharged lead acid cell having both the plates converted to lead sulphate (PbSO4). In order to recharge the cell, direct current is passed through the cell in the reverse direction to that in which the cell provided current. To do so, the anode is connected to the positive terminal of d.c. source and cathode to the negative terminal of the source as shown in figure. The electrolyte (H2SO4) breaks up into hydrogen ions (H+H+) and sulphate ions (SO4

--). Hydrogen ions move towards cathode and sulphate ions move towards anode causing the following chemical reactions:

At anode: On reaching the anode, a sulphate ion (SO4--) gives

up its two extra electrons to become sulphate radical. These electrons given up at the anode move through the external circuit to the cathode where they are available to neutralize the positive ions (H+H+) arriving there. Since sulphate radical cannot exist, it enters into chemical reaction with water as under:

SO4-- - 2e SO4 (radical)

SO4 + H2O H2SO4 + O

The oxygen in the atomic state (i.e. O) is very active and reacts chemically with anode material (PbSO4) to produce the following chemical change.

PbSO4 + O + H2O PbO2 + H2SO4

At cathode: On reaching the cathode, each hydrogen ion (H+) takes one electron from it to become hydrogen gas. This electron is given up by the sulphate ion at anode and has come to the cathode via the external circuit.

H+H+ + 2e 2H

The hydrogen gas liberated at the cathode reacts with cathode material (PbSO4) to reduce it to lead (Pb) as under:

PbSO4 + 2H Pb + H2SO4

As the charging process goes on, the anode is converted into PbO2 and cathode into Pb. The H2SO4 produced in the chemical reactions above increases the specific gravity of the electrolyte. The chemical changes that occur during recharging can be summed up as under:

(i) The positive plate (anode) is converted into PbO2 and the negative plate (cathode) into Pb.

(ii) H2SO4 is formed in the reactions. Therefore, the specific gravity of the electrolyte (H2SO4) is raised.

When the cell is fully charged, the specific gravity of H2SO4 rises to about 1.28.

(iii) The e.m.f. of the cell rises. The e.m.f. of a fully charged lead acid cell is about 2 volts.

(iv) Electrical energy supplied is converted into chemical energy which is stored in the cell.

Nickel Cadmium Battery: Active material used for positive plate and the electrolyte used in nickel cadmium cells are the same as that used for nickel iron cell. For negative plate a mixture of cadmium and iron is used.

The number of positive plates used in nickel-cadmium cell is one more than the number of negative plates.

The working of nickel-cadmium cell is almost similar to nickel iron cell. When current passes through the cell, the electrolyte KOH breaks up into K+ and OH- ions. During discharging the K+ ions move to anode and OH- ions move to cathode. Following chemical changes take place when the cell is supplying current to the external circuit.

At Anode: Ni (OH)2 + 2OH Ni (OH)4

And at Cathode: Cd + 2OH Cd (OH)2

When the cell is charged by connecting the anode to the positive terminal of supply and cathode with the negative terminal of supply the direction of current flow is opposite to that during discharging. Therefore, K+ ions move to cathode and OH- ions move to anode. Following chemical reactions take place during charging.

At Anode: Ni (OH)2 + 2OH Ni (OH)4

At Cathode: Cd (OH)2 + 2K Cd + 2KOH

Nickel cadmium batteries are more suitable than nickel iron batteries for floating duties in conjunction with charging

dynamo. Open circuit losses in these batteries are also low as compared to nickel iron batteries.

Efficiencies of DC Generator

1. Electrical efficiency:

λ m=

Generatorou output in wattsElect . power developed in armature

=

CB Ish =

V I L

E g I a

2. Mechanical efficiency:

λ m=

Elect . power developed in armature in watts .Mechanical power input

=

BA

=

E g I a

B . H . P . x 735.5

3. Overall or Commercial efficiency:

λ c =

Generator out put .Mech . power input

=

CA =

V I L

B . H . P . x 735. 5Unless otherwise told, commercial efficiency is always understood.

Shaft Torque: The torque developed by the armature is the gross torque. Whole of this torque is not available at the pulley, since certain percentage of torque developed by the armature is lost to overcome the iron and friction losses. The torque which is available for useful work is known as shaft torque Tsh. It is so called because it is available at the shaft. The horse power obtained by using shaft torque is called brake horse power (B.H.P.).

B.H.P. (metric) =

T sh x 2 πN735 .5

∴ T sh =B . H . P .(metric ) x 735 .5¿

2 πN60

¿

¿¿ ..(5.6)

Where N = speed of armature in r.p.m.The difference Ta - Tsh is known as lost torque (i.e. torque lost in iron and friction losses)

= 0.159 x

i ron and friction losses N

60 Nm

No-Load Saturation Characteristic (or O.C.C.) If the generator is run at constant speed with the main

switch open, and the terminal voltage is noted at various values of exciting or field current then the O.C.C. shown in

Fig. can be plotted. This is also referred to as the 'magnetization curve' since the same graph shows, to a suitably chosen scale, the amount of magnetic flux, there being a constant relationship (depending upon speed of rotation) between flux and induced voltage.

It will be noticed that a small voltage is produced when the field current is zero, this being due to a small amount of permanent magnetism in the field poles. This is called residual magnetism and is usually sufficient to produce 2 or 3 per cent of normal terminal voltage, although in some special cases it is purposely increased to 10 per cent or more.

The first part of the curve is approximately straight and shows that the flux produced is proportional to the exciting current; but after a certain point,

saturation of the iron becomes perceptible as the curve departs from straight line form.

NUMERICALSQ1) A six pole wave wound armature has 640 conductors

the flux / pole is 0.06 wb. Emf generated as 256 V. Find the speed in rpm.P = 6A = 2Z = 640 = 0.06 wbE = 256 VoltsN = ?

E = × N = N = N = 133.33 rpm

Q2) A eight pole wound armature has 960 conductors and it runs at a speed of 800 rpm, the emf is 320 V, for flux / pole.A = P = 8Z = 960N = 500 rpmE = 320 = ?

E = = = = 0.04 wb

Q3) Calculate the e.m.f. generated by a 4-pole wave wound armature having 45 slots with 18 conductor per slot when driven at 1200 r.p.m. The flux per pole is 0.016 weber.

P = 4A = 2Z = 45 × 18N = 1200 = 0.016 wb.emf = ?

emf = × = × = 518.4 volts.

Q4) A conductor of length 3 meter moves at 60 degrees to a uniform magnetic field of flux density 5 web/m2

with a velocity of 45 m/sec. Calculate the e.m.f. induced in it.

B = 5 wb/m2

l = 3 m = 60° = 45 m/scemf = ?

emf = BL Sinemf = 45 × 5 × 3 × Sin 60°emf =

Q5) In a 4 pole dynams, the flux per pole is 30 m web. Calculate the average e.m.f. induced in one of the armature conductor if the armature is driven at 600 r.p.m.

= 30 mwbN = 600 rpmP = 4A = 2emf = ?emf = × emf /Z = ×

= × emf /Z = 0.6 volts

Q6) A conductor of length 10 meter moves at an angle of 45° to uniform magnetic field of flux density 1.5 wb/m2 with a velocity of 25 m/s. Calculate the e.m.f. induced in it.

L = 10 m = 45°B = 1.5 wb/m2

= 25 m/semf = ?

e = Bl Si e = 1.5 × 10 × 25 × Si 45°e =

Q7) A coil is would with 500 turns, a current of 5 Amperes flowing in the coil. Produces a flux of 60 micro, webers. Find the inductance of the coil.

N = 500 tonnesd = 60 wbdi = 5 Amp.L = ?

e = – N e = – L – L = – N L = N L = L = 6 × 10–3 H

Q8) A coil having an inductance of 75 milli Henries is carrying a current of 100 A. Calculate the self induced e.m.f. when the current is

(i) Reduced to zero in 0.02 seconds(ii) Reversed in 0.02 secondsL = 75 mHI = 100 ASelf induced e.m.f. = ?

(a) When current charges by 100 Amps in 0.02 sec.Average rate of change of current =

= 5000 Amp. / sec Self induced e.m.f. = L

= 75 × 10–3 ×

= × = 375 Volts

(b) Current charges by 200 Amp. in 0.02 sec.Average rate of charge of current =

= 10000 Amp. / sec. Self induced e.m.f. = L

= × Q1) A shunt generator supplies 195 Ampere at 220 volt.

Armature resistance 0.02 ohms, shunt field resistance 44 ohm. If the Iron and Friction loss 1600 watt. Find: (a) E.m.f. generated, (b) copper loss, (c) Commercial and electrical efficiencies.IL = 195 AV = 220 voltsRa = 0.02 __.Rsn = 44 __.Iron and Friction loss = 1600 watt.a) Emf generated = ?b) Copper loss = ?c) Commercial efficiency = ?

a) Isn = Isn = Isn = 5AIa = 195 + 5Ia = 200 AArmature drop = Ia Ra

= 200 (0.02) = 4 volts

Generated e.m.f. = 220 + 4 = 224 volts

b) Cu lossArmature Cu loss = Ia

2 Ra

= (200)2 (0.02) = 800 watt.

Shunt Cu loss = V Ish.= 220 × 5= 1,100 watt

Total Cu loss = 800 + 1,100= 1800 watt

c) Commercial efficiency:Total loss = 1900 + Iron & Friction

= 1900 + 1600= 3500 watt

c = × 100% c = × 100% c = 92.45%

Q2) The armature of a 16 pole d.c. generator is required to generate an e.m.f. of 600 volts on open circuit when revolving at a speed of 900 r.p.m. Calculate the magnetic flux per pole required if the armature has 144 slots with 2 conductors per slot. The armature is wave wound.P = 16A = 2Emf = 600 voltsN = 900 rpmZ = 144 × 2 = ?

e = ×

= = = 0.0173611 wb.

Q3) A shunt generator delivers 450 A at 230 volts and the resistances of the shunt field and armature are 50 __ and 0.03 __ respectively. Calculate the generated e.m.f.IL = 450 AV = 230 volts.Rsh = 50 __Rsn = 0.03 __General emf = ?

Ish = = 4.6 AIa = 450 + 4.6 = 454.6 AGeneral emf = V + Ia Ra

= 230 + (454.6) (0.03)= 243.638 volts

Q4) A short shunt compound generator supplies a load of 15 kW at 300 volts through a pair of feeders of total resistance 0.1 __ armature resistance .04 __, series field resistance .03 __, shunt field resistance 80 __. What is the e.m.f. generated.P = 15 kWVl = 300 voltsRa = 0.04 __Rs = 0.03 __Rsh = 80 __Eg = ?

Load current = 15,000 / 300 = 50 A

Voltage drop in series winding = 50 × 0.03 = 1.5 VVoltage drop in series shunt winding = 300 + 1.5 = 301.5 V Ish =

Isn = 3.76875 AIa = 50 + 3.76875Ia = 53. 76875 AIa Ra = 53.76875 × 0.04 = 2.15075 VGenerated emf= 300 + 1.5 + 2.15075

= 303.65075 volts

Q5) A 4 pole DC Generator having wave wound armature has 51 slots, each with 25 conductors. Find emf generated in the machine when driven at 1500 rpm with 67 mwb flux / pole.P = 4A = 2Z = 25 × 51N = 1500 rpm = 6.7 × 10–3 wbE = ?

E = × E = × E = 427.125 volts.

Q6) A short shunt compound generator supplies a current of 50 amperes at 200 volts. The resistance of the shunt field is 50 ohms, of series winding is 0.025 ohms, and of armature 0.05 ohm. Iron and friction losses amount to 1 kW. Find: a) e.m.f. generated, b) The copper losses, c) The b.h.p. of the prime mover, d) The commercial efficiency.

Load current = 50 AVt = 200 voltsRsn = 50 __Rs = 0.025 __Ra = 0.05 __Iron and friction loss = 1 kWa) Emf generated = ?b) Cu loss = ?c) Output of prime moverd) Commercial efficiency

a) Ish = 200 / 50 = 4 AIa = 50 + 4 = 54 AIa Ra = 54 × 0.05 = 2.7 volts

Voltage drop across series winding = 50 × 0.025 = 1.25 voltsEmf generated = 200 + 2.7 + 1.25

= 203.95 voltsb) Cu loss:Armature Cu loss = Ia

2 Ra = 542 × 0.05 = 145 WattShunt Cu loss = V. Ish = 200 × 4 = 800 WattTotal = 800 + 145 = 945 Watt

c) Stray losses = 1000 wattTotal loss = 1000 + 945 = 1945 WattOutput = 200 × 50 = 10000Output of Prime mover = 10000 + 1945

= 11945 Wattd) c = × 100%

c = × 100%c = 83.717%

Q7) A four pole generator, having wave wound armature winding has 51 slots, each slot containing in the machine when driven at 1500 rpm. Assuring the flux / pole to be 70 mwb.P = 4A = 2Z = 51 × 20E = ?N = 1500 rpm = 7 × 10–3 wb.

E = × E = E = 357 volts.

Q8) A 4 pole wave wound armature of a dc generator has 122 conductors and runs at 1200 rpm if the flux / pole is 25 mwb. Calculate the emf generated.P = 4A = 2Z = 122N = 1200 rpm = 25 × 10–3 wbE = ?

E = E = E = 122 volts.

Q9) A short shunt compound generator supplies a load at 200 volts through a pair of feeders of total resistance 0.1 Ohms. The load consists of 4 motors each taking 20 ampere and a lighting load of 20, 60

watt lamps. Armature resistance 0.02 , series field resistance of .05 , shunt field resistance 50 .

V = 200 VoltRT = 0.1 __4 motors = each taking 20 A20 lamps = each 60 WRa = 0.02 __Rse = 0.05 __Rsh = 50 __Load current = ?Terminal voltage = ?E.m.f. generated = ?

Solution:(i) Each motor taking 20 Amp.

So total motor current = 20 × 4= 80 Amp.

Current taken by one lamp of 60 WW = VII = I = = I = 0.3 Amp.Total current of 20 lamps = 20 × 0.3

= 6 Amps. Now total current supplied = 80 + 6

= 86 Amp.(ii) Now current 86 Amp is flowing through the

feeders of having resistance 0.1 __So voltage drip in feeder = 86 × 0.1

= 8.6 Volt

Voltage at terminal = Voltage at load + Feeder drop

= 200 + 8.6= 208.6 Volt Ans.

(iii) The current 86 Amp flowing through series fieldSo series voltage dropVse = Ise × Rse

Vse = 86 × 0.05= 4.3 Volt

Voltage across shunt field = voltage at brush= 208.6 + 4.3= 212.9 Volts

Shunt field current = Ish = Ish = Ish = 4.258 A

Total armature current = Ise + Ish

Ia = 86 + 4.258Ia = 90.258 Amp.

Armature volt drop = IaRa

= 90.258 × 0.02= 1.81 Volt

Now E.m.f. generatedE.m.f. = load voltage = feeder drop + series

drop + armature dropE.m.f. = 200 + 8.6 + 4.3 +1.81E.m.f. = 214.71 Volts Ans.

Q10) A short shunt compound generator supplies 200 ampere at 220 volts armature resistance 0.02 Ohms,

the iron and friction losses amount to 1600 watt. The shunt field resistance 44 ohms.Find: (i) E.m.f. Generated, (ii) Copper losses, (iii) B.H.P. of engine, (iv) Electrical efficiency, (v) Commercial efficiency.

I = 200 AV = 220 VRa = 0.02 __Rse = 0.01 __Rsh = 44 __Iron friction loss = 1000 W

(i) E.m.f. generated = ?(ii) Copper losses = ?(iii) B.H.P. of engine = ?(iv) Electrical efficiency = ?(v) Commercial efficiency = ?Solution:

Voltage drop in Rse = Vse = IRse

Vse = 200 × 0.01Vse = 2 Volt

Vb = 220 + 2= 222 Volt

Ish = = Ish = 5.05 AIa = I + Ish

Ia = 200 + 5.05Ia = 205.05 Amp.Voltage drop in armature = IaRa

Va = 205.05 × 0.02Va = 4.1 Volt

E.m.f. generated = Vb + Va

= 222 + 4.1= 226.1 Volt Ans.

(c) Electrical power input = V × I= 220 × 200= 44000 Watt

(b) Electrical power developed in armature = E.m.f. × Ia

= 226.1 × 205.05= 46361.805 Watt

Copper losses = b – c= 46360.805 – 44000= 2361.8 Watt Ans.

Iron friction loss = 1600 W Total losses = Current loss + Iron loss

= 2361.8 + 1600= 3961.8 watt

Mechanical power input = Output + total loss= 44000 + 3961.8= 47961.8 Watt

B.H.P. of engine = = = 64.29 h.p.

Electrical efficiency = × 100= 97.3% Ans.

Commercial efficiency = × 100= 94.05% Ans.

Q11) A shunt generator delivers 440 A and 220 V and the resistances of the shunt field and armature are 25

Ohms and 0.025 Ohms respectively, calculate the generated emf.

I = 440 AV = 220 VRsh = 25 __Ra = 0.025e.m.f. = ?

SolutionIsh = Ish = Ish = 8.8 AIa = I + Ish

Ia = 440 + 8.8Ia = 448.8 Ae.m.f. = Voltage at Brush + Armature drop

= 220 + IaRa

= 220 + (448.8 × 0.025)= 220 + 11.22= 231.22 Volt Ans.

Q12) A shunt generator delivers 430 A and 220V and the resistances of the Shunt Field and armature are 30 and 0.02 respectively. Calculate the generated emf.

I = 430 AV = 220 VRsh = 30 __Ra = 0.02 __

e.m.f. = ?Ish = = Ish = 7.3 AIa = I + Ish

Ia = 430 + 7.3Ia = 437.3 Ae.m.f. = Voltage at brush + Voltage drop at

armature= 220 + IaRa

= 220 + (437.3 × 0.02)= 220 + 8.75= 228.75 Volt Ans.

Q13) Two Shunt generators are operating in parallel have each Ra 0.02 ohm and field resistance is 50 each. The combined load current is 2500A, if generated emf’s of the machine are 560V and 550V respectively, calculate the Bus Bar Voltage and output of each machine.

Ra = 0.02 __ eachRsh = 50 __ eachIL = 2500 Ae.m.f. = E1 = 560 Ve.m.f. = E2 = 550 V

G 220 V30 _ _ Rsh

R = 0.02 _ _a

I = 430 A

Bus-bar voltage = V = ?Output in K.W. = ?

Solution:Let V = Bus-bar voltage

I1 = Current output machine 1I2 = Current output machine 2

I1 + I2 = 2500(A)Ish = = Ish =

Bus-bar voltage + Voltage drop in armature = e.m.f. generated

V + (I1 + ) 0.02 = 560 (1)V + (I2 + ) 0.02 = 550 (2)

Subtract equation (2) from equation (1)0.02 I1 – 0.02 I2 = 10 – = 10 – = 10I1 – I2 = 500(B)

We know thatI1 + I2 = 2500(A)

By adding equation (A) and (B), we haveI1 – I2 = 500I1 + I2 = 25002I1 = 3000I1 = I1 = 1500 AI1 + I2 = 2500

1500 + I2 = 2500I2 = 1000 A

Put value of I1 in equation (1)V + (I1 + ) 0.02 = 560V + (1500 + ) 0.02 = 560V + 30 + = 560V + 30 + 0.0004 V = 560V + 0.0004 V = 560 – 30(1 + 0.0004) V = 5301.0004 V = 530V = V = 529.78V = 530 Volt (approx.)

Bus-bar voltage = V = 530 Volt Ans.K.W. Output of machine 1 =

= = 53 × 15= 795 K.W.

K.W. Output of machine 2 = = = 530 K.W. Ans.

Q14) A short shunt compound generator supplies 200 amperes at 220 volts, armature resistance 0.02 ohms, the Iron and Friction losses amount to 1600 watts, the shunt field resistance is 44 Ohms. Calculate (i) electrical Efficiency (ii) EMF generated (iii) B.H.P of engine.

I = 200 AV = 220 V

Ra = 0.02 __Rse = 0.01 __Rsh = 44 __Iron friction losses = 1600 WElectrical efficiency = ?e.m.f. generated = ?B.H.P. of engine = ?

Solution:Voltage drop in series winding = I × Rse

= 200 × 0.01 = 2 Volt Vb = 220 + 2

Vb = 222 VoltIsh = = Ish = 5.05 A

Ia = I + Ish

Ia = 200 + 5.05Ia = 205.05 A

Voltage drop in Armature = IaRa

Va = 205.05 × 0.02Va = 4.1 Volt

E.m.f. generates = Vb + Va

E.m.f. = 222 + 4.1E.m.f. = 226.1 Volt

(c) Electrical power output = V × I= 220 × 200= 44000 watts

(b) Electrical power developed in armature = E.m.f. × Ia

= 226.1 × 205.05

= 46361.805 watts Copper losses = b – c

= 46361.805 – 44000= 2361.8 watts

Iron and friction losses = 1600 watts= 2361.8 + 1600= 3961.8 watts

Mechanical power input = Output + Total loss= 44000 + 3961.8= 47961.8 watts

B.H.P. of Engine = = = 64.29 h.p.

Electrical efficiency = × 100= 97.3% Ans.

Q1) A 100 volt shunt motor is taking a current of 220 Amp armature resistance 0.015 ohm, shunt field resistance 20 ohms. Calculate back e.m.f.

V = 100 volts.I2 = 220 AxpRa = 0.015 __Rsh = 20 __Back emf = ?Ish = Ish = 5 AIa = 220 – 5 = 215 ABack emf = V – Ia Ra

= 100 – 215 (0.015)Back emf = 96.775 volts.

Q2) A shunt motor running at 600 r.p.m. takes 80 amperes at 250 volts. Armature resistance, 0.1 ohm, shunts field resistance 50 ohms. Iron and friction losses 2188 watts: Find (i) Total torque (ii) Copper losses (iii) Commercial efficiency.N = 600 rpmI2 = 80 AV = 250 voltsRa = 0.1 __Rsh = 50 __Iron & friction losses = 2188 watt

Ish = Ish = 5 AIa = 80 – 5Ia = 75 AEb = V – Ia Ra = 250 – 75 × 0.1 = 242.5 volts

a) TorqueTa = Ta = × = 4.8245 N/m.

b) Cu lossArmature Cu loss = Ia

2 Ra = (75)2 × 0.1 = 562.5 wattShunt Cu loss = V. Ish = 250 × 5 = 1250 wattTotal Cu loss = 562.5 + 1250 = 1812.5 wattc) Commercial efficiency

Iron and friction loss = 2188 watts.Total loss = 2188 + 1812.5 = 4000.5 watt

c = × 100% c = × 100% = 83.333%

Q3) A shunt motor, armature resistance 0.2 __, shunt field resistance 55 __ is tahj 50 A at 220 volts. Find back emf.IL = 50 AV = 220 voltsRa = 0.2 __Rsh = 55 __Eb = ?

Ish = Ish = 4 AIa = 50 – 4 = 46 AEb = V – Ia Ra

Eb = 220 – 46 (0.2) = 210.8 volts

Q4) A shunt motor takes 25 A from a 240V main. Its efficiency is 85%, find the output in horse power and total copper losses, when the field resistance is 150 (ohms) and armature resistance is 0.4 (Ohms).

I = 25 AV = 240 VRsh = 150 __Ra = 0.4 __(Efficiency) = 85%Output in hp = ?Total copper losses = ?

Solution:Ish = =

Ish = 1.6AFor armature current

Ia = I – Ish

Ia = 25 – 1.6Ia= = 23.4 A

Now for copper lossesShunt copper loss = Ish

2 × Rsh

= (1.6)2 × 150= 384 W

Armature copper loss Ia2 × Ra

= (23.4)2 × 0.4= 219.1 W

Total copper losses = Shunt Cu loss + Armature Cu loss

= 384 + 219.1= 603.1 Watt Ans.

Output in h.p.Input = V × I

= 240 × 25= 6000 watt

Efficiency = x = Output = x × input

= × 6000= 85 × 60= 5100 watt= 6.8 h.p. Ans.

Q5) A shunt motor takes 30A from a 230 mains, its efficiency is 80%, find the output in horse power

and copper losses when the field resistance is 160 and armature resistance is 0.4.

I = 30 AV = 230 VRsh = 160 __Ra = 0.4 __Efficiency = 80%Total copper loss = ?Output in hp = ?

Solution:Ish = = Ish = Ish = 1.43 AIa = I – Ish

Ia = 30 – 1.43Ia = 28.57 A

Shunt copper loss = Ish2 × Rsh

= (1.43)2 × 160= 327.18 watt

Armature copper loss = Ia2 × Ra

= (28.57)2 × 0.4= 326.5 watt

Total copper loss = Shunt loss + Armature Loss= 327.18 + 326.5= 653.68 watt

Output in h.p.Input = V × I

= 230 × 30= 6800 watt

Efficiency = = Output = × Input

= × 6900= 80 × 69= 5520 watt= 7.4 h.p. Ans.