UNIT I DC MACHINES

Three phase circuits, a review. Construction of DC machines – Theory of operation of DC generators – Characteristics of DC generators- Operating principle of DC motors – Types of DC motors and their characteristics – Speed control of DC motors- Applications

DC GENERATOR

A dc generator is an electrical machine which converts mechanical energy into direct current electricity. This energy conversion is based on the principle of production of dynamically induced emf.

CONSTRUCTION:

Cross Sectional View of DC Generator

Above figure shows the constructional details of a simple 4-pole DC generator. A DC generator consists of two basic parts, stator and rotor.

Basic constructional parts of a DC generator are described below:

Yoke : The outer frame of a generator or motor is called as yoke. Yoke is made up of cast iron or steel. Yoke provides mechanical strength for whole assembly of the generator (or motor). It also carries the magnetic flux produced by the poles.

Poles : Poles are joined to the yoke with the help of screws or welding. Poles are to support field windings. Field winding is wound on poles and connected in series or parallel with armature winding or sometimes separately.

Pole shoe : Pole shoe is an extended part of the pole which serves two purposes, (i) to prevent field coils from slipping and (ii) to spread out the flux in air gap uniformly.

Armature core : Armature core is the rotor of a generator. Armature core is cylindrical in shape on which slots are provided to carry armature windings .

Commutator and brushes : As emf is generated in the armature terminals, it must be taken out to make use of generated emf. But if we can't directly solder wires to Commutator conductors as they rotate. Thus commutator is connected to the armature conductors and mounted on the same shaft as that of armature core. Conducting brushes rest on commutator and they slides over when rotor (hence commutator) rotates. Thus brushes are physically in contact with armature conductors hence wires can be connected to brushes.

WORKING PRINCIPLE OF A DC GENERATOR :

According to Faraday's law of electromagnetic induction , when a conductor moves in a magnetic field (thereby cutting the magnetic flux lines), a dynamically induced emf is produced in the conductor. The magnitude of generated emf can be given by emf equation of DC generator . If a closed path is provided to the moving conductor then generated emf causes a current to flow in the circuit.

Thus in DC generators, when armature is rotated with the help of a prime mover and field windings are excited (there may be permanent field magnets also), emf is induced in armature conductors. This induced emf is taken out via commutator-brush arrangement.

EMF EQUATION OF A DC GENERATOR:

Let Ø = flux/pole in Wb (weber)

Z = total no. of armature conductors

P = no. of generator poles

A = no. of parallel paths in armature

N = rotational speed of armature in revolutions per min. (rpm)

E = emf induced in any parallel path in armature

Now,

Generated e.m.f Eg = e.m.f generated in any one of the parallel paths i.e E.Average e.m.f geneated /conductor = dΦ/dt volt (n=1)Now, flux cut/conductor in one revolution dΦ = ΦP WbNo.of revolutions/second = N/60

Time for one revolution, dt = 60/N secondHence, according to Faraday's Laws of Electroagnetic Induction,E.M.F generated/conductor is

For a simplex wave-wound generator, No.of parallel paths = 2No.of conductors (in series) in one path = Z/2

E.M.F. generated/path is

For a simplex lap-wound generatorNo.of parallel paths = PNo.of conductors (in series) in one path = Z/P

E.M.F.generated/path

In general generated e.m.f

where A = 2 - for simplex wave-winding

A= P - for simplex lap-winding

CHARACTERISTICS OF DC GENERATOR :

In a separately excited DC generator, the field winding is excited by an external independent source. There are generally three most important characteristic of DC generator:

Magnetic or Open Circuit Characteristic of Separately Excited DC Generator

The curve which gives the relation between field current (If) and the generated voltage (E0) in the armature on no load is called magnetic or open circuit characteristic of a DC generator. The plot of this curve is practically same for all types of generators, whether they are separately excited or self-excited. This curve is also known as no load saturation characteristic curve of DC generator. Here in this figure below we can see the variation of generated emf on no load with field current for different fixed speeds of the armature. For higher value of constant speed, the steepness of the curve is more. When the field current is zero, for the effect residual magnetism in the poles, there will be a small initial emf (OA) as show in figure.

Let us consider a separately excited DC generator giving its no load voltage E0 for a constant field current. If there is no armature reaction and armature voltage drop in the machine then the voltage will remain constant. Therefore, if we plot the rated voltage on the Y axis and load current on the X axis then the curve will be a straight line and parallel to X-axis as shown in figure below. Here, AB line indicating the no load voltage (E0).

When the generator is loaded then the voltage drops due to two main reasons-

1) Due to armature reaction,

2) Due to ohmic drop ( IaRa ).

Internal or Total Characteristic of Separately Excited DC Generator

The internal characteristic of the separately excited DC generator is obtained by subtracting the drops due to armature reaction from no load voltage. This curve of actually generated voltage ( Eg ) will be slightly dropping. Here, AC line in the diagram indicating the actually generated voltage (E_g ) with respect to load current. This curve is also called total characteristic of separately excited DC generator.

External Characteristic of Separately Excited DC Generator

The external characteristic of the separately excited DC generator is obtained by subtracting the drops due to ohmic loss ( Ia Ra ) in the armature from generated voltage ( Eg ).

Terminal voltage(V) = Eg - Ia Ra.

This curve gives the relation between the terminal voltage (V) and load current. The external characteristic curve lies below the internal characteristic curve. Here, AD line in the diagram below is indicating the change in terminal voltage(V) with increasing load current. It can be seen from figure that when load current increases then the terminal voltage decreases slightly. This decrease in terminal voltage can be maintained easily by increasing the field

current and thus increasing the generated voltage. Therefore, we can get constant terminal voltage.

Separately excited DC generators have many advantages over self-excited DC generators. It can operate in stable condition with any field excitation and gives wide range of output voltage.

The main disadvantage of these kinds of generators is that it is very expensive of providing a separate excitation source.

CHARACTERISTICS OF SELF EXCITED DC GENERATOR :

In shunt wound DC generators the field windings are connected in parallel with armature conductors as shown in figure below. In these types of generators the armature current Ia divides in two parts. One part is the shunt field current Ish flows through shunt field winding and the other part is the load current IL goes through the external load.

Three most important characteristic of shunt wound dc generators are discussed below:

Magnetic or Open Circuit Characteristic of Shunt Wound DC Generator

This curve is drawn between shunt field current(Ish) and the no load voltage (E0). For a given excitation current or field current, the emf generated at no load E0 varies in proportionally with the rotational speed of the armature. Here in the diagram the magnetic characteristic curve for various speeds are drawn. Due to residual magnetism the curves start from a point A slightly up from the origin O. The upper portions of the curves are bend due to saturation. The external load resistance of the machine needs to be maintained greater than its critical value otherwise the machine will not excite or will stop running if it is already in motion. AB, AC and AD are the slops which give critical resistances at speeds N1, N2 and N3. Here, N1 > N2 > N3.

Critical Load Resistance of Shunt Wound DC Generator

This is the minimum external load resistance which is required to excite the shunt wound generator.

Internal Characteristic of Shunt Wound DC Generator

The internal characteristic curve represents the relation between the generated voltage Eg and the load current IL. When the generator is loaded then the generated voltage is decreased due to armature reaction. So, generated voltage will be lower than the emf generated at no load. Here in the figure below AD curve is showing the no load voltage curve and AB is the internal characteristic curve.

External Characteristic of Shunt Wound DC Generator

AC curve is showing the external characteristic of the shunt wound DC generator. It is showing the variation of terminal voltage with the load current. Ohmic drop due to armature resistance gives lesser terminal voltage the generated voltage. That is why the curve lies below the internal characteristic curve.

The terminal voltage can always be maintained constant by adjusting the of the load terminal.

When the load resistance of a shunt wound DC generator is decreased, then load current of the generator increased as shown in above figure. But the load current can be increased to a certain limit with (up to point C) the decrease of load resistance. Beyond this point, it shows a reversal in the characteristic. Any decrease of load resistance, results in current reduction and consequently, the external characteristic curve turns back as shown in the dotted line and ultimately the terminal voltage becomes zero. Though there is some voltage due to residual magnetism.

Now, when IL increased, then terminal voltage decreased. After a certain limit, due to heavy load current and increased ohmic drop, the terminal voltage is reduced drastically. This drastic reduction of terminal voltage across the load, results the drop in the load current although at that time load is high or load resistance is low.

That is why the load resistance of the machine must be maintained properly. The point in which the machine gives maximum current output is called breakdown point (point C in the picture).

CHARACTERISTICS OF DC SERIES GENERATOR:

In these types of generators the field windings, armature windings and external load circuit all are connected in series as shown in figure below.

Therefore, the same current flows through armature winding, field winding and the load.

Let, I = Ia = Isc = IL

Here, Ia = armature current

Isc = series field current

IL = load current

There are generally three most important characteristics of series wound DC generator which show the relation between various quantities such as series field current or excitation current, generated voltage, terminal voltage and load current.

Magnetic or Open Circuit Characteristic of Series Wound DC Generator

The curve which shows the relation between no load voltage and the field excitation current is called magnetic or open circuit characteristic curve. As during no load, the load terminals are open circuited, there will be no field current in the field since, the armature, field and load are series connected and these three make a closed loop of circuit. So, this curve can be obtained practically be separating the field winding and exciting the DC generator by an external source. Here in the diagram below AB curve is showing the magnetic characteristic of series wound DC generator. The linearity of the curve will continue till the saturation of the poles. After that there will be no further significant change of terminal voltage of DC generator for increasing field current. Due to residual magnetism there will be a small initial voltage across the armature that is why the curve started from a point A which is a little way up to the origin O.

Internal Characteristic of Series Wound DC Generator

The internal characteristic curve gives the relation between voltage generated in the armature and the load current. This curve is obtained by subtracting the drop due to the demagnetizing effect of armature reaction from the no load voltage. So, the actual generated voltage ( Eg) will be less than the no load voltage (E0). That is why the curve is slightly dropping from the open circuit characteristic curve. Here in the diagram below OC curve is showing the internal characteristic or total characteristic of the series wound DC generator.

External Characteristic of Series Wound DC Generator

The external characteristic curve shows the variation of terminal voltage (V) with the load current ( IL). Terminal voltage of this type of generator is obtained by subtracting the ohomic drop due to armature resistance (Ra) and series field resistance ( Rsc) from the actually generated voltage ( Eg).

Terminal voltage V = Eg - I(Ra + Rsc)

The external characteristic curve lies below the internal characteristic curve because the value of terminal voltage is less than the generated voltage. Here in the figure OD curve is showing the external characteristic of the series wound DC generator.

It can be observed from the characteristics of series wound DC generator, that with the increase in load (load is increased when load current increases) the terminal voltage of the machine increases. But after reaching its maximum value it starts to decrease due to excessive demagnetizing effect of armature reaction. This phenomenon is shown in the figure by the dotted line. Dotted portion of the characteristic gives approximately constant current irrespective of the external load resistance. This is because if load is increased, the field current is increased as field is series connected with load. Similarly if load is increased, armature current is increased as the armature is also series connected with load. But due to saturation, there will be no further significance raises of magnetic field strength hence any further increase in induced voltage. But due to increased armature current, the effect of armature reaction increases significantly which causes significant fall in load voltage. If load voltage falls, the load current is also decreased proportionally since current is proportional to voltage as per Ohm's law. So, increasing load, tends to increase the load current, but decreasing load voltage, tends to decrease load current. Due these two simultaneous effects, there will be no significant change in load current in dotted portion of external characteristics of series wound DC generator. That is why series DC generator is called constant current DC generator .

CHARACTERISTICS OF DC COMPOUND GENERATOR:

In compound wound DC generators both the field windings are combined (series and shunt). This type of generators can be used as either long shunt or short shunt compound wound generators as shown in the diagram below. In both the cases the external characteristic of the generator will be nearly same. The compound wound generators may be cumulatively compounded or differentially compounded. Differentially compound wound generators are very rarely used. So, here we mainly concentrate upon the characteristic of cumulatively compound wound generators.

In series wound DC generators, the output voltage is directly proportional with load current and in shunt wound DC generators, output voltage is inversely proportional with load current.

The electric current in the shunt field winding produces a flux which causes a fall in terminal voltage due to armature reaction and ohmic drop in the circuit. But the current in the

series field also produces a flux which opposes the shunt field flux and compensates the drop in the terminal voltage and try to operate the machine at constant voltage.

The combination of a series generator and a shunt generator gives the characteristic of a cumulative compound wound generator.

At no load condition there is no current in the series field because the load terminals are open circuited. But the shunt field current helps to produce field flux and excite the machine. When the dc generator supplies load, the load current increases and current flows through the series field. Therefore, series field also provides some field flux and emf is also increased. The voltage drop in the shunt machine is therefore compensated by the voltage rise in the series machine.

Characteristics of DC Compound Generator:

For small distance operation the flat compounded generators are generally used because the length of the feeder is negligible. But to maintain constant voltage over a long period, the over compounded generators are used. It works as a generator and a booster (boost the terminal voltage).

External characteristic of DC compound wound generator is drawn between the terminal voltage and the load current. By adjusting the no. of amp-turns in the series field winding we can get following external characteristics:

If the series turns are so adjusted that with the increase in load current the terminal voltage also increases, then the generator is called over compounded. The curve AB in the figure showing this characteristic. When the load current increases then the flux provides by the series field also increases. It gives the additional generated voltage. If the increase in generated voltage is greater than the voltage drops due to armature reaction and ohmic drop then, terminal voltage of the generator is increased.

If the series turns are so adjusted that with the increase in load current the terminal voltage remains constant, then the generator is called flat compounded. The curve AC in the figure showing this characteristic. When the load current increases then the flux provides by the series field also increases and gives the additional generated voltage. If the increase in generated voltage is equal to the voltage drops due to armature reaction and ohmic drop then, rated terminal voltage of the generator remains same as no load voltage.

If the series field winding has lesser no. of turns then the rated terminal voltage becomes less than the no load voltage, then the generator is called under compounded. Because, the increase in generated voltage is lesser than the voltage drops due to armature reaction and ohmic drop. Curve AD in the figure is showing this characteristic.

DC MOTOR: A motor is a device which converts an electrical energy into the mechanical energy . The energy conversion process is exactly opposite to that involved in a d.c. generator. In a generator the input mechanical energy is supplied by a prime mover while in a d.c. motor, input electrical energy is supplied by a d.c. supply. The construction of a d.c. machine is same whether it is a motor or a generator.Principle of Operation of a D.C. Motor The principle of operation of a d.c. motor can be stated in a single statement as 'when a current carrying conductor is placed in a magnetic field' it experiences a mechanical force'. In a practical d.c. motor, field winding produces a required magnetic field while armature conductors play a role of a current carrying conductors and hence armature conductors experience a force. As a conductors are placed in the slots which are in the periphery, the individual force experienced by the conductors acts as a twisting or turning force on the armature which is called a torque. The torque is the product of force and the radius at which this force acts. So overall armature experiences a torque and starts rotating. Let us study this motoring action in detail. Consider a single conductor placed in a magnetic field as shown in the Fig .1(a). The magnetic field is produced by a permanent magnet but in a practical d.c. motor it is produced by the field winding when it carries a current.

Fig. 1 Now this conductor is excited by a separate supply so that it carries a current in a particular direction. Consider that it carries a current away from an observe as shown in the Fig. 1(b). Any

current carrying conductor produces its own magnetic field around it. hence this conductor also produces its own flux, around. The direction of this flux can be determined by right hand thumb rule. For direction of current considered, the direction of flux around a conductor is clockwise. For simplicity of understanding, the main flux produced by the permanent magnet is not shown in the Fig. 1(b). Now there are two fluxes present,1. The flux produced by the permanent magnet called flux.2. The flux produced by the current carrying conductor. There are shown in the Fig.2(a). Form this, it is clear that on one side of the conductor, both the fluxes are in same direction. In this case, on the left of the conductor there is gathering of the flux lines as two fluxes help each other. As against this, on the right of the conductor, the two fluxes are in opposite direction and hence try to cancel each other. Due to this, the density of the flux lines in this area gets weakened. So on the left, there exists high flux density area while on the right of the conductor there exists low flux density area as shown in the Fig. 2(b).

Fig. 2 This flux distribution around the conductors acts like a stretched rubber band under tension. This exerts a mechanical force on the conductor which acts from high flux density area towards low flux density area. i.e. from left to right for the case considered as shown in the Fig. 2(b).

Fig. 3

Key point : In the practical d.c. motor, the permanent magnet is replaced by a field winding which produces the required flux called main flux and all the armature conductors, mounted on the periphery of the armature drum, get subjected to the mechanical force. Due to this, overall armature experiences a twisting force called torque and armature of the motor starts rotating.

1. Direction of Rotation of Motor The magnitude of the force experienced by the conductor in a motor is given by, F = B l I Newton (N)B = Flux density due to the flux produced by the field winding.l = Active length of the conductor.I = Magnitude of the current passing through the conductor. The direction of such force i.e. the direction of rotation of a motor can be determined by Fleming's left hand. So Fleming's right hand rule is to determine direction of induced e.m.f. i.e. for generating action while Fleming's left hand rule is to determine direction of force experienced i.e. for motoring action.1.1 Fleming's left hand rule The rule states that, 'Outstretch the three fingers of the left hand namely the first finger, middle finger and thumb such that they are mutually perpendicular to each other. Now point the first finger in the direction of magnetic field and the middle finger in the direction of the current then the thumb gives the direction of the force experienced by the conductor'. The Fleming's left hand rule can be diagramatically shown as in the Fig. 1.

Fig. 1 Apply the rule to crosscheck the direction of force experienced by a single conductor, placed in the magnetic field, shown in the Fig. 2(a), (b), (c) and (d).

Fig. 2 It can be seen from the Fig. 2 that if the direction of the main field in which current carrying conductor is placed, is reversed, force experienced by the conductor reverses its direction. Similarly keeping main flux direction unchanged, the direction of current passing through the conductor is reversed. The force experienced by the conductor reverses its direction. However if both the directions are reversed, the direction of the force experienced remains the same.

Key point : So in a practical motor, to reverse its direction of rotation, either direction of main field produced by the field winding is reversed or direction of the current passing through the armature is reversed. The direction of the main field can be reversed by changing the direction of current passing through the field winding, which is possible by interchanging the polarities of supply which is given to the field winding . In short, to have a motoring action two fluxes must exist, the interaction of which produces a torque.

Significance of Back E.M.F.

It is seen in the generation action, that when a conductor cuts the lines of flux, e.m.f. gets induced in the conductor. The question is obvious that in a d.c. motor, after a motoring action, armature starts rotating and armature conductors cut the main flux. So there is a generating action existing in a motor. After a motoring action, there exists a generating action. There is an induced e.m.f. in the rotating armature conductors according to Faraday's law of electromagnetic induction. This induced e.m.f. in the armature always acts in the opposite direction of the supply voltage. This is according to the Lenz's law which states that the direction of the induced e.m.f. is always so as to oppose the cause producing it. In a d.c. motor, electrical input i.e. the supply voltage is the cause and hence this induced e.m.f. opposes the supply voltage. This e.m.f. tries to set up a current through the armature which is in the opposite direction to that, which supply voltage is forcing through the conductor. So as this e.m.f. always opposes the supply voltage, it is called back e.m.f. and denoted as Eb. Though it is obtained as Eb, basically it gets generated by the generation action which we have seen earlier in case of generation. So its magnitude can be determined by the e.m.f. equation which is derived earlier. So,

where all symbols carry the same meaning as seen earlier in case of generators.

Fig. 1

This e.m.f. is shown schematically in the Fig. 1(a). So if V is supply voltage in volts and R a

is the value of the armature resistance, the equivalent electric circuit can be shown as in the Fig. 1(b).

TYPES OF D.C. MOTORSSimilar to the d.c. generators, the d.c. motors are classified depending upon the way of connecting the field winding with the armature winding. The difference types of d.c. motors are ;1. Shunt motor2. Series motors3. Compound motors The compound motors are further classified as ;

1. Short shunt compound2. Long shunt compound

DC SHUNT MOTOR:

In this type, the field winding is connected across the armature winding and the combination is connected across the supply, as shown in the Fig. 1.

Fig. 1 D.C. shunt motor

Let Rsh be the resistance of shunt field winding. Ra be the resistance of armature winding. The value of Ra is very small while Rsh is quite large. Hence shunt field winding has more number of turns with less cross-sectional area.1.1 Voltage and Current Relationship The voltage across armature and field winding is same equal to the supply voltage V.The total current drawn from the supply is denoted as line current IL. IL = Ia + Ish

Ish = V/Rsh

and V = Eb + Ia Ra + Vbrush

Vbrush is generally neglected. Now flux produced by the field winding is proportional to the current passing through it i.e. Ish.

Note : As long as supply voltage is constant, which is generally so in practice, the flux produced is constant. Hence d.c. shunt motor is called constant flux motor.

DC SERIES MOTOR: In this type of motor, the series field winding is connected in series with the armature and the supply, as shown in the Fig. 1.

Fig. 1 D.C. series motor

Let Rse be the resistance of the series field winding. The value of Rse is very small and it is made of small number of turns having large cross-sectional area.1.1 Voltage and Current Relationship Let IL be the total current drawn from the supply. So IL = Ise = Ia

and V = Eb + Ia Ra + Ise Rse + Vbrush

V = Eb + Ia (Ra + Rse) + Vbrush

Supply voltage has to overcome the drop across series field winding in addition to Eb and drop across armature winding.Note : In series motor, entire armature current is passing through the series field winding. So flux produced is proportional to the armature current.

for series motor

DC COMPOUND MOTOR:The compound motor consists of part of the field winding connected in series and part of

the field winding connected in parallel with armature. It is further classified as long shunt compound and short shunt compound motor.1.1 Long Shunt Compound Motor In this type, the shunt field winding is connected across the combination of armature and the series field winding as shown in the Fig. 1.

Fig. 1 Long shunt compound motor

Let Rse be the resistance of series field and Rsh be the resistance of shunt field winding. The total current drawn from supply is IL. So IL = Ise + Ish

But Ise = Ia

... IL = Ia+ Ish

And Ish = V/Rsh And V = Eb + Ia Ra + Ise Rse + Vbrush

But as Ise = Ia ,... V = Eb + Ia (Ra + Rse) + Vbrush

1.2 Short Shunt Compound Motor In this type, the shunt field is connected purely in parallel with armature and the series field is connected in series with this combination shown in the Fig. 2.

Fig. 2 Short shunt compound motor

IL = Ise The entire line current is passing through the series field winding. and IL = Ia + Ish

Now the drop across the shunt field winding is to be calculated from the voltage equation. So V = Eb + Ise Rse + Ia Ra + Vbrush but Ise = IL

... V = Eb + IL Rse + Ia Ra + Vbrush

... Drop across shunt field winding is, = V - IL Rse = Eb + Ia Ra + Vbrush

Apart from these two, compound motor can be classified into two more types,i) Cumulatively compound motors and ii) Differential compound motors.Note : If the two field windings are wound in such a manner that the fluxes produced by the two always help each other, the motor is called cumulatively compound. If the fluxes produced by the two field windings are trying to cancel each other i.e. they are in opposite direction, the motor is called differential compound. A long shunt compound motor can be of cumulative or differential type. Similarly short shunt compound motor can be cumulative or differential type.

TORQUE EQUATION OF A D.C. MOTOR It is seen that the turning or twisting force about an axis is called torque. Consider a wheel of radius R meters acted upon by a circumferential force F newtons as shown in the Fig. 1.

Fig. 1 The wheel is rotating at a speed of N r.p.m. Then angular speed of the wheel is, ω = (2πN)/60 rad/sec So workdone in one revolution is, W = F x distance travelled in one revolution = F x 2 R joules And P = Power developed = Workdone/Time = (F x 2πR) / (Time for 1 rev) = (F x 2πR) / (60/N) = (F x R) x (2πN/60)... P = T x ω watts Where T = Torque in N - m ω = Angular speed in rad/sec. Let Ta be the gross torque developed by the armature of the motor. It is also called armature torque. The gross mechanical power developed in the armature is Eb Ia, as seen from the power equation. So if speed of the motor is N r.p.m. then, Power in armature = Armature torque x ω

... Eb Ia = x (2N/60) but Eb in a motor is given by, Eb = (ΦPNZ) / (60A)... (ΦPNZ / 60A) x Ia = Ta x (2πN/60)

This is the torque equation of a d.c. motor.TORQUE SPEED EQUATIONS:

Before analysing the various characteristics of motors, let us revise the torque and speed equations are applied to various types of motors.... T α Φ Ia from torque equation. This is because, 0.159(PZ)/A is a constant for a given motor. Now Φ is the flux produced by the field winding and is proportional to the current passing through the field winding. Φ α Ifield

But for various types of motors, current through the field winding is different. Accordingly torque equation must be modified.

For a d.c. shunt motor, Ish is constant as long as supply voltage is constant. Hence Φ flux is also constant.... T α Ia for shunt motors For a d.c. series motor, Ise is same as Ia. Hence flux Φ is proportional to the armature current Ia.... T α Ia α Ia

2 for series motors.

Similarly as Eb = (ΦPNZ)(60A), we can write the speed equation as,

Eb α Φ N ... N α Eb/Φ But V = Eb + Ia Ra neglecting brush drop... Eb = V - Ia Ra

... Speed equation becomes, N α (V-Ia Ra)/Φ So for shunt motor as flux is constant,... N α V - Ia Ra

While for series motor, flux Φ is proportional to Ia.

FACTORS AFFECTING THE SPEED OF A D.C. MOTOR According to the speed equation of a d.c. motor we can write,

The factors Z, P, A are constants for a d.c. motor. But as the value of armature resistance Ra and series field resistance Rse is very small, the drop Ia Ra and (Ra + Rse) is very small compared to applied voltage V. Hence neglecting these voltage drops the speed equation can be modified as,

Thus the factors affecting the speed of a d.c. motor are,1. The flux Φ2. The voltage across the armature3. The applied voltage V depending upon these factors the various methods of speed control are,1. Changing the flux Φ by controlling the current through the field winding called flux control methods.2. Changing the armature path resistance which in turn changes the voltage applied across the armature called rheostatic control.3. Changing the applied voltage called voltage control method.

D.C. Motor CharacteristicsThe performance of a d.c. motor under various conditions can be judged by the following characteristics

i) Torque - Armature current characteristics (T Vs Ia ) : The graph showing the relationship between the torque and the armature current is called a torque-armature current characteristic. These are also called electrical characteristics.ii) Speed - Armature current characteristics(N Vs Ia ) : The graph showing the relationship between the speed and armature current characteristic.iii) Speed - Torque characteristics(N Vs T) : The graph showing the relationship between the speed and the torque of the motor is called speed-torque characteristics of the motor. These are also called mechanical characteristic. The nature of these characteristics can easily be obtained by using speed and torque equations derived in previous post. These characteristics play a very important role in selecting a type of motor for a particular application.

Characteristics of dc shunt motor:i) Torque - Armature current characteristics For a d.c. motor T α Φ Ia

For a constant values of Rsh and supply voltage V, Ish is also constant and hence flux is also constant.... Ta α Φ Ia

The equation represents a straight line, passing through the origin, as shown in the Fig. 1. Torque increases linearly with armature current. It is seen earlier that armature current is decided by the load. So as load increases, armature current increases, increasing the torque developed linearly.

Fig. 1 T Vs Ia for shunt motor Now if shaft torque is plotted against armature current, it is known that shaft torque is less than the armature torque and the difference between the two is loss torque T f as shown. On no load Tsh = 0 but armature torque is present which is just enough to overcome stray losses shown as Ta0. The current required is Ia0 on no load to produce Ta0 and hence Tsh graph has an intercept of Ia0 on the current axis. To generate high starting torque, this type of motor requires a large value of armature current at start. This may damage the motor hence d.c. shunt motors can develop moderate starting torque and hence suitable for such applications where starting torque requirement is moderate.ii) Speed - Armature current characteristics From the speed equation we get, N α (V - Ia Ra)/Φ

αV - Ia Ra as Φ is constant So as load increases, the armature current increases and hence drop Ia Ra also increases. Hence for constant supply voltage, V - Ia Ra decreases and hence speed reduces. But as Ra is very small, for change in Ia from no load to full load, drop Ia Ra is very small and hence drop in speed is also not significant from no load to full load.

Fig. 2 N Vs Ia for shunt motor

So the characteristics is slightly droping as shown in the Fig. 2. But for all practical purposes these type of motors are considered to be a constant speed motors.iii) Speed - Torque characteristics These characteristics can be derived from the above two characteristics. This graph is similar to speed-armature current characteristics as torque is proportional to the armature current. This curve shows that the speed almost remains constant through torque changes from no load to full load conditions. This is shown in the Fig. 3.

Fig. 3 N Vs T for shunt motor

Characteristics of dc series motor:i) Torque - Armature current characteristics In case of series motor the series field winding is carrying the entire armature current. So flux produced is proportional to the armature current.... Φ α Ia Hence Ta α Φ Ia α Ia

2

Thus torque in case of series motor is proportional to the square of the armature current. This relation is parabolic in nature as shown in the Fig. 1.

As load increases, armature current increases and torque produced increases proportional to the square of the armature current upto a certain limit. As the entire Ia passes through the series field, there is a property of an electromagnet called saturation, may occur. Saturation means though the current through the winding increases, the flux produced remains constant. Hence after saturation the characteristics take the place of straight line as flux becomes constant, as shown. The difference between Ta and Tsh is loss torque Tf which is also shown in the Fig. 2. At start as Ta αIa

2 , these types of motors can produce high torque for small amount of

armature current hence the series motors are suitable for the applications which demand high starting torque.ii) Speed - Armature current characteristics From the speed equation we get, N α (Eb/Φ) ) αV - Ia Ra - Ia Rse)/ Ia as Φ α Ia in case of series motor Now the values of Ra and Rse are so small that the effect of change in Ia on speed overrides the effect of change in V - Ia Ra - Ia Rse on the speed. Hence in the speed equation, Eb ≈ Vand can be assumed constant. So speed equation reduced to, N α 1/Ia

So speed-armature current characteristics is rectangular hyperbola type as shown in the Fig. 2.iii) Speed - Torque characteristics In case of series motors, T α Ia

2 and N α 1/Ia

Hence we can write, N α 1/√T Thus as torque increases when load increases, the speed decreases. On no load, torque is very less and hence speed increases to dangerously high value. Thus the nature of the speed-torque characteristics is similar to the nature of the speed-armature current characteristics. The speed-torque characteristics of a series motor is shown in the Fig. 3.

Fig. 3 N Vs T for series motor

DC Compound Motor Characteristics:

Compound motor characteristics basically depends on the fact whether the motor is cumulatively compound or differential compound. All the characteristics of the compound motor are the combination of the shunt and series characteristic. Cumulative compound motor is capable of developing large amount of torque at low speeds just like series motor. However it is not having a disadvantages of series motor even at light or no load. The shunt field winding produces the definite flux and series flux helps the shunt field flux to increase the total flux level. So cumulative compound motor can run at reasonable speed and will not run with dangerously high speed like series motor, on light or no load condition. In differential compound motor, as two fluxes oppose each other, the resultant flux decreases as load increases, thus the machine runs at a higher speed with increase in the load. This property is dangerous as on full load, the motor may try to run with dangerously high speed. So differential compound motor is generally not used in practice. The various characteristics of both the types of compound motors cumulative and the differential are shown in the Fig. 1(a), (b) and (c).

Fig. 1 characteristics of d.c. compound motor

The exact shape of these characteristics depends on the relative contribution of series and shunt field windings. If the shunt field winding is more dominant then the characteristics take the shape of the shunt motor characteristics. While if the series field winding is more dominant then the characteristics take the shape of the series characteristics.

Why Series Motor is Never Started on No Load?It is seen earlier that motor armature current is decided by the load. On light load or no load, the armature current drawn by the motor is very small. In case of a d.c. series motor, Φ α Ia and on no load as Ia is small hence flux produced is also very small. According to speed equation, N α 1/Φ as Eb is almost constant. So on very light load or no load as flux is very small, the motor tries to run at dangerously high speed which may damage the motor mechanically. This can be seen from the speed-armature current and the speed-torque characteristics that on low armature current and low torque condition motor shows a tendency to rotate with dangerously high speed. This is the reason why series motor should never be started on light loads or no load conditions. For this reason it is not selected for belt drives as breaking or slipping of belt causes to throw the entire load off on the motor and made to run motor with no load which is dangerous.STARTING METHODS OF A DC MOTORBasic operational voltage equation of a DC motor is given as E = Eb + I aRa and hence I a = (E - E b) / Ra

Now, when the motor is at rest, obviously, there is no back emf E b, hence armature current will be high at starting.This excessive current will-

1. Blow out the fuses and may damage the armature winding and/or commutator brush arrangement.

2. Produce very high starting torque (as torque is directly proportional to armature current), and this high starting toque will produce huge centrifugal force which may throw off the armature windings.

Thus to avoid above two drawbacks, starters are used for starting of DC machine.

Starting Methods of a DC MotorThus, to avoid the above dangers while starting a DC motor , it is necessary to limit the starting current. For that purpose, starters are used to start a DC motor. There are various starters like, 3 point starter, 4 point starter, No load release coil starter, thyristor starter etc.The main concept behind every DC motor starter is, adding external resistance to the armature winding at starting.

3 POINT STARTER:The internal wiring of a 3 point starter is as shown in the figure.

When motor is to be started, the lever is turned gradually to the right. When lever touches point 1, the field winding gets directly connected across the supply, and the armature winding gets connected with resistances R1 to R5 in series. Hence at starting full resistance is added in series with armature. Then as the lever is moved further, the resistance is gradually is cut out from the armature circuit. Now, as the lever reaches to position 6, all the resistance is cut out from the armature circuit and armature gets directly connected across the supply. The electromagnet E (no voltage coil) holds the lever at this position. This electromagnet releases the lever when there is no (or low) supply voltage.

When the motor is overloaded beyond a predefined value, over current release electromagnet D gets activated, which short circuits electromagnet E , and hence releases the lever and motor is turned off.

4 POINT STARTER:

The main difference between a 3 point starter and a 4 point starter is that the no voltage coil is not connected in series with field coil. The field gets directly connected to the supply, as the lever moves touching the brass arc. The no voltage coil (or Hold on coil) is connected with a current limiting resistance Rh. This arrangement ensures that any change of current in the shunt field does not affect the current through hold on coil at all. This means that electromagnet pull of the hold-on coil will always be sufficient so that the spring does not unnecessarily restore the lever to the off position.

This starter is used where field current is to be adjusted by means of a field rheostat.

2 POINT STARTER : (DC Series motor starter)

Construction of DC series motor starters is very basic as shown in the figure. A start arm is simply moved towards right to start the motor. Thus at first maximum resistance is connected in series with the armature and then gradually decreased as the start arm moves towards right. The no load release coil holds the start arm to the run position and leaves it at no load.

SPEED CONTROL METHODS OF DC MOTOR:

We know, back emf of a DC motor E b is the induced emf due to rotation of the armature in magnetic field. Thus value of the Eb can be given by the EMF equation of a DC generator .

Eb = PØNZ / 60A

(where, P= no. of poles, Ø=flux/pole, N=speed in rpm, Z=no. of armature conductors , A=parallel paths)

Eb can also be given as, E b = V- I aRa

thus from above equations

N = E b 60A / PØZ

but, for a DC motor A, P and Z are constant

N α K E b/Ø (where, K=constant)

thus, it shows speed is directly proportional to back emf and inversely proportional to the flux per pole.

SPEED CONTROL METHODS OF DC MOTOR Speed Control of Shunt Motor

1. Flux Control Method

It is seen that speed of the motor is inversely proportional to flux. Thus by decreasing flux speed can be increased and vice versa.

To control the flux, a rheostat is added in series with the field winding, as shown in the circuit diagram. Adding more resistance in series with field winding will increase the speed, as it will decrease the flux. Field current is relatively small and hence I 2 R loss is small, hence this method is quiet efficient. Though speed can be increased by reducing flux with this method, it puts a limit to maximum speed as weakening of flux beyond the limit will adversely affect the commutation.

2. Armature Control Method

Speed of the motor is directly proportional to the back emf E b and E b = V- I aRa.That is when supply voltage V and armature resistance Ra are kept constant, speed is

directly proportional to armature current Ia. Thus if we add resistance in series with armature, Ia decreases and hence speed decreases.Greater the resistance in series with armature, greater the decrease in speed.

3. Voltage Control Method

a) Multiple voltage control: In this method the, shunt filed is connected to a fixed exciting voltage, and armature is supplied with different voltages. Voltage across armature is changed with the help of a suitable switchgear. The speed is approximately proportional to the voltage across the armature.

b) WARD-LEONARD SYSTEM:

Ward Leonard control system is introduced by Henry Ward Leonard in 1891. Ward Leonard method of speed control is used for controlling the speed of a DC motor. It is a basic armature control method. This control system is consisting of a dc motor M_1 and powered by a DC generator G. In this method the speed of the dc motor (M_1) is controlled by applying variable voltage across its armature. This variable voltage is obtained using a motor-generator set which consists of a motor M_2(either ac or dc motor) directly coupled with the generator G. It is a very widely used method of speed control of DC motor.

Principle of Ward Leonard Method

Basic connection diagram of the Ward Leonard speed control system is shown in the figure below.

The speed of motor M1 is to be controlled which is powered by the generator G. The shunt field of the motor M1 is connected across the dc supply lines. Now, generator G is driven by the motor M2. The speed of the motor M2 is constant. When the output voltage of the generator is fed to the motor M1 then the motor starts to rotate. When the output voltage of the generator varies then the speed of the motor also varies. Now controlling the output voltage of the generator the speed of motor can also be controlled. For this purpose of controlling the output voltage, a field regulator is connected across the generator with the dc supply lines to control the field excitation. The direction of rotation of the motor M1 can be reversed by excitation current of the generator and it can be done with the help of the reversing switch R.S. But the motor-generator set must run in the same direction.

Advantages of Ward Leonard System

1. It is a very smooth speed control system over a very wide range (from zero to normal speed of the motor).

2. The speed can be controlled in both the direction of rotation of the motor easily.3. The motor can run with a uniform acceleration.4. Speed regulation of DC motor in this ward Leonard system is very good.

Disadvantages of Ward Leonard System

1. The system is very costly because two extra machines (motor-generator set) are required.2. Overall efficiency of the system is not sufficient especially it is lightly loaded.

Application of Ward Leonard System

This Ward Leonard method of speed control system is used where a very wide and very sensitive speed control is of a DC motor in both the direction of rotation is required. This speed control system is mainly used in colliery winders, cranes, electric excavators, mine hoists, elevators, steel rolling mills and paper machines etc.

Speed Control Of Series Motor

1. Flux Control Method

a) Field diverter :

A variable resistance is connected parallel to the series field as shown in fig. This variable resistor is called as diverter, as desired amount of current can be diverted through this resistor and hence current through field coil can be decreased. Hence flux can be decreased to desired amount and speed can be increased.

b) Armature divertor:

Diverter is connected across the armature as in fig .For a given constant load torque, if armature current is reduced then flux must increase. As, Ta α ØIa This will result in increase in current taken from the supply and hence flux Ø will increase and subsequently speed of the motor will decrease.

c) Tapped field control:

As shown in fig field coil is tapped dividing number of turns. Thus we can select different value of Ø by selecting different number of turns.

d) Paralleling field coils:

In this method, several speeds can be obtained by regrouping coils as shown in fig

2. Variable Resistance In Series With ArmatureBy introducing resistance in series with armature, voltage across the armature can be

reduced. And hence, speed reduces in proportion with it.

3. Series-Parallel ControlThis system is widely used in electric traction, where two or more mechanically coupled

series motors are employed. For low speeds, motors are joined in series, and for higher speeds motors are joined in parallel.

When in series, the motors have the same current passing through them, although voltage across each motor is divided. When in parallel, voltage across each motor is same although current gets divided.

REVIEW QUESTIONS

Construction details

1. Explain the constructional details of a DC Generator. Describe its working principle and derive its emf equation. (16) (N/D-12)

2. Explain in detail, the functions of various parts of a DC motor. (16)

Emf equation

1. A six pole lap connected DC generator has a useful flux per pole of 0.045 Wb. If the no-load voltage at 400 rpm is 300 V, find the conductors on the armature periphery. (8) (N/D-09)

2. Derive the emf equation from fundamentals and also explain why the lap winding is preferred compared to wave winding for low voltage high current machines. (8)

3. Derive an expression for emf generated in a DC machine. (8)

Self and separately excited generators

1. A separately excited generator, when running at 1000 rpm supplies 200 A at 125 V. What will be the load current when the speed drops to 800 rpm, if the field current is unchanged? Assume the armature resistance as 0.04 and the brush drop is 2 V. (16) (N/D-09)

Characteristics of series, shunt and compound generators

1. Explain in detail, the characteristics of a DC generator with neat diagrams. (16) (N/D-09)2. A DC shunt generator has a terminal voltage of 160 V and a no-load induced emf of 168 V.

The resistances of armature and field are 0.03 Ω and 20 Ω respectively. Find the armature current, field current and load current. Neglect armature reaction. (6)

Principle of operation of a DC motor

1. Explain the principle of operation of a DC motor with relevant diagrams. (8) (N/D-11)

Back emf and torque equation

1. Explain the significance of back emf. (6)

2. Prove that the torque developed by a DC motor is proportional to flux and armature current. (8)

Characteristics of series, shunt and compound motors

1. A 25 kW, 250 V, DC shunt generator has an armature and a field resistance of 0.06 and 100 respectively. Determine the total armature power developed during operating condition. (6) (A/M-10)

2. A 4 pole, 220 V DC shunt motor has 540 lap-wound conductors. It takes 32 A from the supply mains and develops output power of 5.595 kW. The field winding takes 1 A. The armature resistance is 0.9 and the flux per pole is 30 mWb. Calculate the speed in rpm and the torque developed in Nm. (6) (A/M-10)

3. Explain in detail, the different classifications of DC motors with respect to their equivalent circuits. (8) (N/D-09)

4. Estimate the percentage reduction in speed of a separately excited DC generator working with constant excitation of 400 V, bus bar voltage is decreasing as its load varies from 600 kW to 400 kW. The machine resistance is 0.02 . Neglect the armature reaction. (8) (A/M-11)

5. The armature winding of a 200 V, 4 pole series motor is lap connected. There are 280 slots with 4 conductors per slot. The current is 45 A and the flux per pole is 18 mWb. The field resistance is 0.3 Ω, armature resistance 0.5 Ω and the total iron and the friction losses are 800 W. The pulley diameter is 0.41 m. Find the pull in newton at the rim of the pulley. (10)

6. Explain the speed-current, speed-torque and torque-current characteristics of a DC shunt and series motor. (8)

Types of starters

1. Draw the schematic diagram of a 3 point starter and explain its working principle. (8) (M/J-12)

2. Explain the operation of a 4 point starter with a neat diagram and also write its advantages over a three point starter. (10) (A/M-10)

Speed control of DC shunt motors

1. (i) A 4 pole DC motor runs at 600 rpm on full load and takes 25 A , 450 V the armature is lap wound with conductors and flux per pole is given by the equation = (1.7 X 10-2) x I0.5 Wb where I is the motor current. If the supply voltage and torque are halved, calculate the speed at which the motor will run. Neglect stray losses. (8)

(ii) Explain the various types of speed control techniques of DC Machines. (8) (M/J-12)

2. Explain the procedure to predetermine the efficiencies of a DC Machine as a motor and as a generator. (8) (N/D-12)

3. Explain in detail, the Ward-Leonard system of speed control of a DC motor.4. Explain in detail, the different methods of speed control of a DC motor. (16) (N/D-12)

UNIT II TRANSFORMER

Introduction – Single phase transformer construction and principle of operation – EMF equation of transformer-Transformer no–load phasor diagram –– Transformer on–load phasor diagram –– Equivalent circuit of transformer – Regulation of transformer –Transformer losses and efficiency-All day efficiency –auto transformers.

SINGLE PHASE TRANSFORMERWorking Principle of 1-Phase Transformer1. Introduction The main advantage of alternating currents over direct current is that, the alternating currents can be easily transferable from low voltage to high voltage or high voltage to low. Alternating voltages can be raised or lowered as per requirements in the different stages of electrical network as generation, transmission, distribution and utilization. This is possible with a static device called transformer. The transformer works on the principle of mutual induction. It transfer an electric energy from one circuit to other when there is no electrical connection between the tow circuits. Thus we can define transformer as below :Key point : The transformer is a static piece of apparatus by means of which an electrical power is transformed from one alternating current circuit to another with the desired change in voltage and current, without any change in the frequency. The use of transformers in transmission system is shown in the Fig 1.1.

Fig. 1.1 Use of transformer in transmission system

2. Principle of working The principle of mutual induction states that when tow coils are inductively coupled and if current in one coil is changed uniformly then an e.m.f. gets induced in the other coil. This e.m.f can drive a current, when a closed path is provided to it. The transformer works on the same principle. In its elementary form, it consists of tow inductive coils which are electrically separated but linked through a common magnetic circuit. The two coils have high mutual inductance. The basic transformer is shown in the Fig 1.2. One of the two coils is connected to source of alternating voltage. This coil in which electrical energy is fed with the help of source called primary winding (P). The other winding is connected to load. The electrical energy transformed to this winding is drawn out to the load.

Fig.1.2 Basic transformer

Fig 1.3 Symbolic representation This winding is called secondary winding (S). The primary winding has N1number of turns while the secondary winding has N2 number of turns. Symbolically the transformer is indicated as shown in the Fig 1.3. When primary winding is excited by an alternating voltage, it circulates an alternating current. This current produces an alternating flux (Φ)which completes its path through common magnetic core as shown dotted in the Fig 1.2. Thus an alternating, flux links with the secondary winding. As the flux is alternating, according to Faraday's law of an electromagnetic induction, mutually induced e.m.f. gets developed in the secondary winding. If now load is connected to the secondary winding, this e.m.f. drives a current through it. Thus through there is no electrical contact between the two windings, an electrical energy gets transferred from primary to the secondary.Key point : The frequency of the mutual induced e.m.f. is same as that of the alternating source which is supplying energy to the primary winding.

3. Can D.C. Supply be used for Transformer? The d.c. supply can not be used for the transformers.

The transformer works on the principle of mutual induction, for which current in one coil must change uniformly. If d.c. supply is given, the current will not change due to constant supply and transformer will not work. Practically winding resistance is very small. For d.c., the inductive reactance XL is zero as d.c. has no frequency. So total impedance of winding is very low for d.c. Thus winding will draw very high current if d.c. supply is given to it. This may cause the burning of windings due to extra heat generated and may cause permanent damage to the transformer. There can be saturation of the core due to which transformer draws very large current from the supply when connected to d.c. Thus d.c. supply should not be connected to the transformers.

CONSTRUCTION OF TRANSFORMER

There are two basic parts of a transformer i) Magnetic Core ii) Winding or Coils.The core of the transformer is either square or rectangular in size. It is further divided into tow parts. The vertical position on which coils are wound is called limb while the top and bottom horizontal portion is called yoke of the core. These parts are shown in the Fig.1(a). Core is made up of lamination. Because of laminated type of construction, eddy current losses get minimised. Generally high grade silicon steel laminations (0.3 to 0.5 mm thick) are used. These laminations are insulated from each other by using insulation like varnish. All laminations are varnished. Laminations are overlapped so that to avoid the air gap at joints. For this generally 'L' shaped or 'I' shaped laminations are used which are shown in the Fig 1(b).

Fig. 1 Construction of transformer

The cross-section of the limb depends on the type of coil to be used either circular or rectangular. The different cross-section of limbs, practically used are shown in the Fig. 2.

Fig. 2 Different cross-sections

Types of Windings The coils used are wound on the limbs and are insulated from each other. In the basic transformer shown in the Fig 1.2 (see post : Working Principle of 1-Phase Transformer ) the two windings wound are shown on two different limbs i.e. primary on one limb while secondary on other limb. But due to this leakage flux increases which effects the transformer performance badly. Similarly it is necessary that the windings should be very closes to each other to have high mutual inductance. To achieve this, the tow windings are split into number of coils and are wound adjacent to each other on the same limb. A very common arrangement is cylindrical coils as shown in the Fig. 3.

Fig. 3 Cylindrical concentric coils

Such cylindrical coils are used in the core type transformer. Theses coils are mechanically strong. These are wound in the helical layers. The different layers are insulated from each other by paper, cloth or mica. The low voltage winding is placed near the core from ease of insulating it from the core. The high voltage is placed after it. The other type of coils which is very commonly used for the shell type of transformer is sandwiching coils. Each high voltage portion lies between the two low voltage portion sandwiching the high voltage portion. Such subdivision of windings into small portion reduces the leakage flux. Higher the degree of subdivision, smaller is the reactance. The sandwich coil is shown in the Fig. 4. The top and bottom coils are low voltage coils. All the portion are insulated from each other by paper.

Fig. 4 Sandwich coils

CONSTRUCTION OF SINGLE PHASE TRANSFORMERThe various constructions used for the single phase transformers are,1. Core type 2. shell type and 3. Berry type1. Core Type Transformer It has a single magnetic circuit. The core rectangular having two limbs. The winding encircles the core. The coils used are of cylindrical type. As mentioned earlier, the coils are wound in helical layers with different layers insulated from each other by paper or mica. Both the coils are placed on both the limbs. The low voltage coil is placed inside near the core while high voltage coil surrounds the low voltage coil. Core is made up of large number of thin laminations. As The windings are uniformly distributed over the two limbs, the natural cooling is more effective. The coils can be easily removed by removing the laminations of the top yoke, for maintenance. The Fig. 1(a) shows the schematic representation of the core type transformer while the Fig 1(b) shows the view of actual construction of the core type transformer.

Fig. 1 Core type transformer2. Shell Type Transformer It has a double magnetic circuit. The core has three limbs. Both the windings are placed on the central limb. The core encircles most part of the windings. The coils used are generally multilayer disc type or sandwich coils. As mentioned earlier, each high voltage coil is in between tow low voltage coils and low voltage coils are nearest to top and bottom of the yokes. The core is laminated. While arranging the laminations of the core, the care is taken that all the joints at alternate layers are staggered. This is done to avoid narrow air gap at the joint, right through the cross-section of the core. Such joints are called over lapped or imbricated joint. Generally for very high voltage transformers, the shell type construction is preferred. As the windings are surrounded by the core, the natural cooling does not exist. For removing any winding for maintenance, large number of laimnations are required to be removed. The Fig. 2(a) shows the schematic representation while the Fig. 2(b) shows the outaway view of the construction of the shell type transformer.

Fig 2 Shell type transformer

3. Berry Type Transformer This has distributed magnetic circuit. The number of independent magnetic circuits are more than 2. Its core construction is like spokes of a wheel. Otherwise it is symmetrical to that of shell type. Diagramatically it can be shown as in the Fug. 3.

Fig. 3 Berry type transformer The transformers are generally kept in tightly fitted sheet metal tanks. The tanks are constructed of specified high quality steel plate cut, formed and welded into the rigid structures. All the joints are painted with a solution of light blue chalk which turns dark in the presence of oil, disclosing even the minutes leaks. The tanks are filled with the special insulating oil. The entire transformer assembly is immersed in the oil. Oil serves two functions :i) Keeps the coil cool by circulation and ii) Provides the transformers an additional insulation. The oil should be absolutely free from alkalies, sulphur and specially from moisture. Presence of very small moisture lowers the dielectric strength of oil, affecting its performance badly. Hence the tanks are sealed air tight to avoid the contact of oil with atmospheric air and moisture. In large transformers, the chambers called breather are provided. The breathers prevent the atmospheric moisture to pass on to the oil. The breathers contain the silica gel crystal which immediately absorb the atmospheric moisture. Due to long and continuous use, the sludge is formed in the oil which can contaminate the oil. Hence to keep such sludge separate from the oil in main tank, an air tight metal drum is provided, which is placed on the top of tank. This is called conservator.

4. Comparison of Core and Shell Type Transformers

E.M.F EQUATION OF A TRANSFORMERWhen the primary winding is excited by an alternating voltage V1, it circulates alternating

current, producing an alternating flux Φ. The primary winding has N1number of turns. The alternating flux Φ linking with the primary winding itself induces an e.m.f in it denoted as E1. The flux links with secondary winding through the common magnetic core. It produces induced e.m.f. E2 in the secondary winding. This is mutually induced e.m.f. Let us derive the equations for E1 and E2. The primary winding is excited by purely sinusoidal alternating voltage. Hence the flux produced is also sinusoidal in nature having maximum value of Φm as show in the Fig. 1.

Fig. 1 Sinusoidal flux The various quantities which affect the magnitude of the induced e.m.f. are : Φ = Flux Φm = Maximum value of flux N1 = Number of primary winding turns

N2 = Number of secondary winding turns f = Frequency of the supply voltage E1 = R.M.S. value of the primary induced e.m.f. E2 = R.M.S. value of the secondary induced e.m.f. From Faraday's law of electromagnetic induction the voltage e.m.f. induced in each turn is proportional to the average rate of change of flux.... averagee.m.f. per turn = average rate of change of flux... averagee.m.f. per turn = dΦ/dtNow dΦ/dt = Change in flux/Time required for change in flux Consider the 1/4 th cycle of the flux as shown in the Fig.1. Complete cycle gets completed in 1/f seconds. In 1/4 th time period, the change in flux is from 0 to Φm.... dΦ/dt = (Φm - 0)/(1/4f) as dt for 1/4 th time period is 1/4f seconds = 4 f Φm Wb/sec ... Average e.m.f. per turn = 4 f Φm volts As is sinusoidal, the induced e.m.f. in each turn of both the windings is also sinusoidal in nature. For sinusoidal quantity, From factor = R.M.S. value/Average value = 1.11... R.M.S. value of induced e.m.f. per turn = 1.11 x 4 f Φm = 4.44 f Φm

There are number of primary turns hence the R.M.S value of induced e.m.f. of primary denoted as is E1, E1 = N1 x 4.44 f Φm volts While as there are number of secondary turns the R.M.S values of induced e.m.f. of secondary denoted is E2 is, E2 = N2 x 4.44 f Φm volts The expression of E1 and E2 are called e.m.f. equation of a transformer. Thus e.m.f. equations are, E1 = 4.44 f Φm N1 volts ............(1) E2 = 4.44 f Φm N2 volts .............(2)

Ratios of a Transformer

Consider a transformer shown in Fig.1 indicating various voltages and currents.

Fig. 1 Ratios of transformer

1. Voltage Ratio We known from the e.m.f. equations of a transformer that E1 = 4.44 f Φm N1 and E2 = 4.44 f Φm N2

Taking ratio of the two equations we get,

This ratio of secondary induced e.m.f. to primary induced e.m.f. is known as voltage transformation ratio denoted as K, Thus,

1. If N2 > N1 i.e. K > 1, E2 > E1 we get then the transformer is called step-up transformer.2. If N2 < N1 i.e. K < 1, we get E2 < E1 then the transformer is called step-down transformer.3. If = i.e. K= 1, we get E2 = E1 then the transformer is called isolation transformer or 1:1

transformer.2. Concept of Ideal Transformer A transformer is said to be ideal if it satisfies following properties :i) It has no losses.ii) Its windings have zero resistance.iii) Leakage flux is zero i.e. 100% flux produced by primary links with the secondary.iv) Permeability of core is so high that negligible current is required to establish the flux in it.Key point : For an ideal transformer, the primary applied voltage V1 is same as the primary induced e.m.f. V2 as there are no voltage drops. Similarly the secondary induced e.m.f. E2 is also same as the terminal voltage V2 across the load. Hence for an ideal transformer we can write,

No transformer is ideal in practice but the value of E1 is almost equal to V1 for properly designed transformer.

3. Current ratio For an ideal transformer there are no losses. Hence the product of primary voltage V1 and primary current I1, is same as the product of secondary voltage V2 and the secondary current I2.So V1 I1 = input VA and V2 I2 = output VA For an ideal transformer, V1 I1 = V2 I2

Key point : Hence the currents are in the inverse ratio of the voltage transformation ratio.

4. Voltage ampere rating When electrical power is transferred from primary winding to secondary there are few power losses in between. These power losses appear in the form of heat which increase the temperature of the device.Now this temperature must be maintained below certain limiting values as it is always harmful from insulation point of view. As current is the main cause in producing heat, the output maximum rating is generally specified as the product of output voltage and output current i.e.V2 I2. This always indicates that when transformer is operated under this specified rating, its temperature rise will not be excessive. The copper loss (I2R) in the transformer depends on the current 'I' through the winding while the iron or core loss depends on the voltage 'V' as frequency of operation is constant. None of these losses depend on the power factor (cos Φ) of the load. Hence losses decide the temperature and hence the rating of the transformer. As losses depend on V and I only, the rating of the transformer is specified as a product of these two parameters VxI.

Key point : Thus the transformer rating is specified as the product of voltage and current called VA rating. On both sides, primary and secondary VA rating remains same. This rating is generally expresses in KVA (kilo volt amperes rating).Now V1 /V2 = I2 /I1 = K... V1 I1 = V2 I2

If V1 and V2 are the terminal voltages of primary and secondary then from specified KVA rating we can decide full load currents of primary and secondary, I1 and I2. This is the safe maximum current limit which may carry, keeping temperature rise below its limiting value.

Key point : The full load primary and secondary currents indicate the safe maximum values of currents which transformer windings can carry.

Ideal Transformer on No Load

Consider an ideal transformer on no load as shown in the Fig. 3. The supply voltage is and as it is V1 an no load the secondary current I2 = 0. The primary draws a current I1 which is just necessary to produce flux in the core. As it magnetising the core, it is called magnetising current denoted as Im. As the transformer is ideal, the winding resistance is zero and it is purely inductive in nature. The magnetising current is Im is

very small and lags V1 by 30o as the winding is purely inductive. This Im produces an alternating flux Φ which is in phase with Im.

Fig. 1 Ideal transformer on no load

The flux links with both the winding producing the induced e.m.f.s E1 and E2 , in the primary and secondary windings respectively. According to Lenz's law, the induced e.m.f. opposes the cause producing it which is supply voltage V1. Hence E1 is in antiphase with V1 but equal in magnitude. The induced E2 also opposes V1 hence in antiphase with V1 but its magnitude depends on N2. Thus E1 and E2 are in phase. The phasor diagram for the ideal transformer on no load is shown in the Fig. .2.

Fig. 2 Phasor diagram for ideal transformer on no load It can be seen that flux Φ is reference. Im produces Φ hence in phase with Φ. V1leads Im by 90o as winding is purely inductive so current has to lag voltage by 90o. E1 and E2 are in phase and both opposing supply voltage . The power input to the transformer is V1 I1 cos (V1 ^ I1 ) i.e. V1 Im cos(90o) i.e. zero. This is because on no load output power is zero and for ideal transformer there are no losses hence input power is also zero. Ideal no load p.f. of transformer is zero lagging.

Practical Transformer on No Load

Actually in practical transformer iron core causes hysteresis and eddy current losses as it is subjected to alternating flux. While designing the transformer the efforts are made to keep these losses minimum by,

1. Using high grade material as silicon steel to reduce hysteresis loss.2. Manufacturing core in the form of laminations or stacks of thin lamination to reduce eddy

current loss.

Apart from this there are iron losses in the practical transformer. Practically primary winding has certain resistance hence there are small primary copper loss present. Thus the primary current under no load condition has to supply the iron losses i.e. hysteresis loss and eddy current loss and a small amount of primary copper loss. This current is denoted as Io.Now the no load input current Io has two components :

1. A purely reactive component Im called magnetising component of no load current required to produce the flux. This is also called wattless component.

2. An active component Ic which supplies total losses under no load condition called power component of no load current. This also called wattful component or core loss component of Io. Thtotal no load current Io is the vector addition of Im and Ic.

In practical transformer, due to winding resistance, no load current Io is no longer at 90o with respect to V1. But it lags V1 by angle Φo which is less than 90o . Thus cos Φo is called no load power factor of practical transformer.

Fig 1. Practical transformer on no load

The phasor diagram is shown in the Fig. 1. It can be seen that the two components Io are,

This is magnetising component lagging V1 exactly by 90o .

This is core loss component which is in phase withV1. The magnitude of the no load current is given by,

While Φo = no load primary power factor angle The total power input on no load is denoted as Wo and is given by,

It may be denoted that the current is very small, about 3 to 5% of the full load rated current. Hence the primary copper loss is negligibly small hence Ic is called core loss or iron loss component. Hence power input Wo on no load always represent the iron losses, as copper loss is negligibly small. The iron losses are denoted as Pi and are constant for all load conditions.

Transformer on Load (M.M.F Balancing on Load)

When the transformer is loaded, the current I2 flows through the secondary winding. The magnetic and phase of I2 is determined by the load. If load is inductive, I2 lags V2. If load is capacitive, I2 leads V2 while for resistive load, I2 is in phase withV2. There exists a secondary m.m.f. N2 I2 due to which secondary current sets up its own flux Φ2. This flux opposes the main flux Φ which is produced in the core due to magnetising component of no load current. Hence the m.m.f. is N2 I2 called demagnetising ampere-turns. This is shown in the Fig.1(a). The flux Φ2 momentarily reduces the main flux Φ, due to which the primary induced e.m.f. also E1 reduces.

This additional current drawn by primary is due to the load hence called load component of primary current denoted as I2' as shown in the Fig.1(b).

Fig. 1 Transformer on load

This current I2' is in antiphase with I2. The current sets up its own flux Φ2' which opposes the flux Φ2 and helps the main fluxΦ. This flux Φ2' neutralises the flux Φ2produced by I2. The m.m.f. i.e. ampere turns N2 I2' balances the ampere turns N2 I2. Hence the net flux in the core is again maintained at constant level.Key point : Thus for any load condition, no load to full load the flux in the core is practically constant.

The load component current I2' always neutralises the changes in the loads. Hence the transformer is called constant flux machine. As the ampere turns are balanced we can write, N2 I2=N2 I2' ... I2' =(N2/N1) = K I2 ..................(1) Thus when transformer is loaded, the primary current I1 has two components :1. The no load current Io which lags V1 by angle Φo. It has two components Im and Ic.2. The load component I2' which in antiphase with I2. And phase of I2 is decided by the load. Hence primary current I1 is vector sum of Io and I2'.... Ī1 = Īo + Ī2 ...............(2) Assume inductive load, I2 lags E2 by Φ2, the phasor diagram is shown in the Fig. 2(a). Assume purely resistive load, I2 in phase with E2, the phasor diagram is shown in the Fig.2(b). Assume capacitive load, I2 leads E2 by Φ2, the phasor diagram is shown in the Fig. 2(c). Note that I2' is always in antiphase with I2.

Fig. 2 Actually the phase of I2 is with respect to V2 i.e. angle Φ2 is angle between I2 and V2. For the ideal case, E2 is assumed equal to V2 neglecting various drops. The current ratio can be verified from this discussion. As the no load current I o is very small, neglecting Io we can write, I1 ~ I2' Balancing the ampere turns, N1 I1 = N1 I1 = N2 I2

... N2 /N1 = I1 /I2 = K Under full load conditions when Io is very small compared to full load currents, the ratio of primary and secondary current is constant..

Effect of Winding Resistances

1. Effect OF Winding Resistances A practical transformer windings process some resistances which not only cause the power

losses but also the voltage drops. Let us see what is the effect of winding resistance on the performance of the transformer. Let R1 = primary winding resistance in ohms R2 = secondary winding resistance in ohms Now when current I1 flows through primary, there is voltage drop I1 R1 across the winding. The supply voltage V1 has to supply this drop. Hence primary induced e.m.f. E1 is the vector difference between V1 and I1 R1.

Similarly the induced e.m.f. in secondary is E2. When load is connected, current I2 flows and there is voltage drop I2 R2. The e.m.f. E2 has to supply this drop. The vector difference between E2 and I2 R2 is available to the load as a terminal voltage.

The drops I1 R1 and I2 R2 are purely resistive drops hence are always in phase with the respective currents I1 and I2.1.1 Equivalent Resistance The resistance of the two windings can be transferred to any one side either primary or secondary without affecting the performance of the transformer. The transfer of the resistances on any one side is advantageous as it makes the calculations very easy. Let us see how to transfer the resistances on any one side. The total copper loss due to both the resistances can be obtained as, total copper loss = I1

2 R1 + I22 R2

= I12 R1 +(I2

2/I12) R2

= I12R1 + (1/K2) R2 .......(3)

Where I2/I1 = 1/K neglecting no load current. Now the expression (3) indicates that the total copper loss can be expressed as I1

2 R1 + I1

2 .R2/K2. This means R2/K2 is the resistance value of R2 shifted to primary side which causes same copper loss with I1 as R2 causes with. This value of resistance which R2 /K2 is the value of R2 referred to primary is called equivalent resistance of secondary referred to primary. It is denoted as R2'. R2' = R2 /K2 ........(4) Hence the total resistance referred to primary is the addition of R1 and R2' called equivalent resistance of transformer referred to primary and denoted as R1e. = R1 + R2'= R1 + R2 /K2 .........(5) This resistance R1e causes same copper loss with I1 as the total copper loss due to the individual windings. total copper loss = I1

2 R1e = I12 R1 + I2

2 R2 ......(6) So equivalent resistance simplifies the calculations as we have to calculate parameters on one side only. Similarly it is possible to refer the equivalent resistance to secondary winding. total copper loss = I1

2 R1 + I22 R2

= I22 (I1

2/I22) R1 + R2

= I22 ( K2 R1 + R2) ........(7)

Thus the resistance K2 R1 is primary resistance referred to secondary denoted as R1'. R1' = K2 R1 .......(8)

Hence the total resistance referred to secondary is the addition of R2 and R1'called equivalent resistance of transformer referred to secondary and denoted as R2e. R2e = R2 + R1' = R2 + K2 R1 .........(9) total copper loss = I2

2 R2e ........(10) The concept of equivalent resistance is shown in the Fig. 1(a), (b) and (c).

Fig. 1 Equivalent resistanceKey Point : When resistance are transferred to primary, the secondary winding becomes zero resistance winding for calculation purpose. The entire copper loss occurs due to R1e. Similarly when resistances are referred to secondary, the primary becomes resistanceless for calculation purpose. The entire copper loss occurs due to R2e.Important Note : When a resistance is to be transferred from the primary to secondary, it must be multiplied by K2. When a resistance is to be transferred from the secondary to primary, it must be divided by K2. Remember that K is N1 /N2. The result can be cross-checked by another approach. The high voltage winding is always low current winding and hence the resistance of high voltage side is high. The low voltage side is high current side and hence resistance of low voltage side is low. So while transferring resistance from low voltage side to high voltage side, its value must increase while transferring resistance from high voltage side to low voltage side, its value must decrease.

Key point :High voltage side → Low current side → High resistance sideLow voltage side → High current side → Low resistance side

Effect of Leakage Reactance1. Effect of Leakage Reactance Uptill now it is assumed that the entire flux produced by the primary links with the secondary winding. But in practice it is not possible. Part of the primary flux as well as the secondary flux completes the path through air and links with the respecting winding only. Such a flux is called leakage flux. Thus there are two leakage fluxes present as shown in the Fig. 1.

Fig .1 Individual impedance

The flux ΦL1 is the primary leakage flux which is produced due to primary current I1. It is in phase with I1 and links with primary only. The flux ΦL2 is the secondary leakage flux which is produced due to current I2. It is in phase with I2 and links with the secondary winding only. Due to the leakage flux ΦL1 there is self inducede.m.f. eL1 in primary. While due to leakage flux ΦL2 there is self inducede.m.f. eL2 in secondary. The primary voltage V1 has to overcome this voltage eL1 to produce E1 while induced e.m.f. E2 has to overcome eL2 to produce terminal voltage V2. Thus the self inducede.m.f.sare treated as the voltage drops across the fictitious reactance placed in series with the windings. These reactances are called leakage reactance of the winding. So X1 = Leakage reacatnce of primary winding. and X2 = Leakage reactance of secondary winding. The value of X1 is such that the drop I1 X1 is nothing but the self inducede.m.f. eL1 due to fluxΦL1. The value of X2 is such that the drop I2 X2 is equal to the self inducede.m.f. eL2 due to flux ΦL1. Leakage fluxes with the respective windings only and not to both the windings. To reduce the leakage, as mentioned, int eh construction both the winding's are placed on same limb rather than on separate limbs.1.1 Equivalent Leakage Reactance Similar to the resistances, the leakage reactances also can be transferred from primary to secondary or viceversa. The relation through K2 remains same for the transfer of recatnaces as it is studied earlier for the resistances. Let X1 is leakage reactance of primary and X2 is leakage reactance of secondary. Then the total leakage reacatance referred to primary is X1e given by, X1e = X1 + X2' where X2' = X2/K2

While the total leakage reacatnce referred to secondary is given by , X2e = X2 + X1' where X1' = K2 X1 And K = N2/N1 =transformation ratio

Equivalent Impedance The transformer primary has resistance R1 and reactance X1. While the transformer secondary has resistance R2 and reacatnce X2. Thus we can say that the total impedance of primary winding is Z1 which is, Z1 = R1 + j X1 Ω ..........(1)

And the total impedance of the secondary winding is which is , Z2 = R2 + j X2 Ω ...........(2) This is shown in the Fig. 1.

Fig. 1 Individual impedance

The individual magnitudes of and are, Z1 = √(R1

2 + X12) ...........(3)

and Z2 = √(R22 + X2

2) .........(4) Similar to resistance and reactance, the impedance also can be referred to any one side.Let Z1e = total equivalent impedance referred to primarythen Z1e = R1e + j X1e

Z1e = Z1 + Z2' = Z1 + Z2/K2 ............(5)Similarly Z2e = total equivalent impedance referred to secondarythen Z2e = R2e + j X2e

Z2e = Z2 + Z1' = Z2 + K2 Z1 ............(6) The magnitude of Z1e and Z2e are, Z1e = √(R1e

2 + X1e2) ........(7)

and Z2e = √(R2e2 + X2e

2) ...........(8) It can be denoted that, Z2e = K2 Z1e and Z1e = Z2e /K2

The concept of equivalent impedance is shown in the Fig. 2.

Fig 2 Equivalent impedance

Phasor Diagrams for Transformer on LoadPhasor Diagrams for Transformer on Load Consider a transformer supplying the load as shown in the Fig. 1.

Fig. 1 The various transformer parameters are, R1 = Primary winding resistance X1 = Primary leakage reactance R2 = Secondary winding resistance X2 = Secondary leakage reactance ZL = Load impedance I1= Primary current I2 = Secondary current = IL = Load current now Ī1 = Īo + Ī2'where Io = No load current I2'= Load component of current decided by the load = K I2 where K is transformer component The primary voltage V1 has now three components,1. -E1, the induced e.m.f. which opposes V1

2. I1 R1, the drop across the resistance, in phase with I1 3. I1 X1, the drop across the reactance, leading I1 by 90o

The secondary induced e.m.f. has also three components, 1. V2, the terminal voltage across the load2.I2 R2, the drop across the resistance, in phase with I2

3. I2 X2, the drop across the reactance, leading I2 by 90o

The phasor diagram for the transformer on load depends on the nature of the load power factor. Let us consider the various cases of the load power factor.

1.1 Unity power factor load, cosΦ2 = 1 As load power factor is unity, the voltage V2 and I2 are in phase. Steps to draw the phasor diagram are,1. Consider flux Φ as reference2. E1 lags Φ by 90o. Reverse E1 to get -E1.3. E1 and E2 are inphase4. Assume V2 in a particular direction5. I2 is in phase with V2.6. Add I2 R2 and I2 X2 to to get E2.7. Reverse I2 to get I2'.8. Add Io and I2' to get I1.9. Add I1 R1 and to -E1 to get V1. Angle between V1 and I1 is Φ1 and cosΦ1 is primary power factor. Remember that I1X1 leads I1 direction by 90o and I2 X2 leads I2 by 90o as current through inductance lags voltage across inductance by 90o. The phasor diagram is shown in the Fig.2

Fig. 2 Phasor diagram for unity power factor load

Lagging Power Factor Load, cos Φ2

As load power factor is lagging cosΦ2, the current I2 lags V2 by angle Φ2. So only changes in drawing the phasor diagram is to draw I2 lagging V2 by Φ2 in step 5 discussed earlier. Accordingly direction of I2 R2, I2 X2, I2', I1, I1 R1 and I1X1 will change. Remember that whatever may be the power factor of load, I2X2 leads I2 by 90o and I1X1 leads I1 by 90o. The complete phasor diagram is shown in the Fig. 3.

Fig. 3 Phasor diagram for lagging power factor

Loading Power Factor Load, cos Φ2

As load power factor is leading, the current I2 leads V2 by angle Φ2. So change is to draw I2 leading I2 by angle Φ2. All other steps remain same as before. The complete phasor diagram is shown in the Fig. 4

Fig. 4 Phasor diagram for leading power factor

Equivalent circuit of Transformer1. Equivalent circuit of Transformer

The term equivalent circuit of a machine means the combination of fixed and variable resistances and reactances, which exactly simulates performance and working of the machine. For a transformer, no load primary current has two components, Im = Io sinΦo = Magnetizing component Ic = Io cosΦo = Active component Im produces the flux and is assumed to flow through reactance Xo called no load reractance while Ic is active component representing core losses hence is assumed to flow through the reactance Ro. Hence equivalent circuit on no load can be shown as in the Fig. 1. This circuit consisting of Ro and Xo in parallel is called exciting circuit. From the equivalent circuit we can write,Ro = V1/Ic

and Xo= V1/Im

Fig. 1 No load equivalent circuit

When the is connected to the transformer then secondary current I2 flows. This causes voltage drop across R2 and R2. Due to I2, primary draws an additional current I2' = I2/ K. Now I1 is the phasor addition of Io and I2'. This I1 causes the voltage drop across primary resistance R1 and reactance X1. Hence the equivalent circuit can be shown as in the Fig. 2.

Fig. 2 But in the equivalent circuit, windings are not shown and it is further simplified by transferring all the values to the primary or secondary. This makes the transformer calculation much easy.

So transferring secondary parameters to primary we get, R2'= R2/K2 , X2' = X2/K2' , Z2' = Z2/K2

While E2' = E2/K' I2' = K I2

Where K = N2 /N1

While transferring the values remember the rule that Low voltage winding High current Low impedance High voltage winding Low current High impedance Thus the exact equivalent circuit referred to primary can be shown as in the Fig. 3.

Fig. 3 Exact equivalent circuit referred to primary

Similarly all the primary value can be referred to secondary and we can obtain the equivalent circuit referred to secondary. R1' = K2 R1 , X1' = K2 X1, Z1' = K2 Z1 E1'= K E1, Io' = I1 /K' Io' = Io /K Similarly the exciting circuit parameters also gets transferred to secondary as R o'and Xo '. The circuit is shown in the Fig.4.

Fig. 4 Exact equivalent circuit referred to secondary

Now as long as no load branch i.e. exciting branch is in between Z1 and Z2', the impedances can not be combined. So further simplification of the circuit can be done. Such circuit is called approximate equivalent circuit.1.1 Approximate Equivalent Circuit To get approximate equivalent circuit, shift the no load branch containing Ro and Xo to the left of R1 and X1. By doing this we are creating an error that the drop across R1 and X1due to Io is neglected. Hence such an equivalent circuit is called approximate equivalent circuit. So approximate equivalent circuit referred to primary can be as shown in the Fig. 5.

Fig. 5 Approximate equivalent circuit referred to primary

In this circuit now R1 and R2' can be combined to get equivalent resistance referred to primary R1e as discussed earlier. Similarly X1and X1' can be combined to get X1e. And equivalent circuit can be simplified as shown in the Fig. 6.

Fig. 6

We know that, R1e = R1 + R2'= R1 + R2/K2

X1e = X1 + X2' = X1 + X2/K2

Z1e = R1e + j X1e

Ro = V1 /Ic and Xo = V1 /Im

Ic = Io cosΦo and Im = Io sinΦo In the similar fashion, the approximate equivalent circuit referred to secondary also can be obtained.

Voltage Regulation of Transformer1. Voltage Regulation of Transformer Because of the voltage drop across the primary and secondary impedances it is observed that the secondary terminal voltage drops from its no load value (E2) to load value (V2) as load and load current increases.

This decrease in the secondary terminal voltage expressed as a fraction of the no load secondary terminal voltage is called regulation of a transformer.

The regulation is defined as change in the magnitude of the secondary terminal voltage, when full load i.e. rated load of specified power factor supplied at rated voltage is reduced to no load, with primary voltage maintained constant expressed as the percentage of the rated terminal voltage.

Let E2 = Secondary terminal voltage on no load V2 = Secondary terminal voltage on given load then mathematically voltage regulation at given load can be expressed as,

The ratio (E2 - V2 / V2 ) is called per unit regulation.

The secondary terminal voltage does not depend only on the magnitude of the load current but also on the nature of the power factor of the load. If V2 is determined for full load and specified power factor condition the regulation is called full load regulation.

As load current increases, the voltage drops tend to increase V2 and drops more and more. In case of lagging power factor V2 < E2 and we get positive voltage regulation, while for leading power factor E2 < V2 and we get negative voltage regulation.

The voltage drop should be as small as possible hence less the regulation better is the performance of a transformer.

1.1 Expression for Voltage Regulation

The voltage regulation is defined as, %R = (E2 - V2 /V2 ) x 100 = (Total voltage drop/V2) x 100 The expression for the total approximate voltage drop is already derived. Total voltage drop = I2 R2e cos Φ ± I2 X2e sin Φ Hence the regulation can be expressed as,

'+' sing for lagging power factor while '-' sing for leading power factor loads. The regulation van be further expressed interms of I1 , V1, R1e and X1e. V2 /V1 =I1 /I2 = K... V2 = KV1 , I2 = I1/Kwhile R1e =R2e/K2 , X1e = X2e /K2 Substituting in the regulation expression we get,

1.2 Zero Voltage Regulation

We have seen that for lagging power factor and unity power factor condition V2 < E2 and we get positive regulation. But as load becomes capacitive, V2 starts increasing as load increase. At a certain leading power factor we get E2 = V2 and the regulation becomes zero. If the load is increased further, E2 becomes less than V2and we get negative regulation.

... for zero voltage regulation, E2 = V2

... E2 - V2 = 0or VR cos Φ - Vx sin Φ = 0 .......... -ve sing as leading power factor where VR = I2 R2e /V2 = I1 R1e /V1 and Vx = I2 X2e /V2 = I1 X1e /V1

... VR cos Φ = Vx sin Φ

... tan Φ = VR /Vx

... cos Φ = cos tan-1(VR /Vx) This is the leading p.f. at which voltage regulation becomes zero while supplying the load.

1.3 Constants of a Transformer

From the regulation expression we can define constants of a transformer. %R= (( I2 R2e cos Φ ± I2 X2e sin Φ )/ E2) x 100 = (I2 R2e /E2) cos Φ ± (I2 X2e/E2 ) sin Φ x 100 The ratio (I2 R2e /E2) or (I1 R1e /E1) is called per unit resistive drop and denoted as VR. The ratio (I2 X2e/E2 ) or (I1 X1e/E1) is called per unit reactive drop and is denoted as Vx.

The terms VR and Vx are called constants of a transformer because for the rated output I2, E2, R1e, X1e, R2e , X2e are constants. The regulation can be expressed interms of VR and Vx as,

%R = (VR cos Φiuu ±Vx sin Φ ) x 100 On load condition, E2 = V2 and E1= V1

where V1 and V2 are the given voltage ratings of a transformer. Hence VR and Vx can be expressed as, VR = I2 R2e/ V2 = I1 R1e/ V1

and Vx =I2 R2e/ V2 = I1 X1e/ V1

where V1and V2 are no load primary and secondary voltages, VR and Vx can be expressed on percentage basis as, Percentage resistive drop = VR x 100 Percentage reactive drop = Vx x 100Key Point : Note that and are also called per unit resistance and reactance respectively.

LOSSES IN A TRANSFORMER1. Losses in a Transformer In a transformer, there exists two types of losses.i) The core gets subjected to an alternating flux, causing core losses.ii) The windings carry currents when transformer is loaded, causing copper losses.1.1 Core or Iron Losses Due to alternating flux set up in the magnetic core of the transformer, it undergoes a cycle of magnetisation and demagnetisation. Due to hysteresis effect there is loss of energy in this process which is called hysteresis loss. It is given by, hysteresis loss = Kh

Bm1.67 f v watts

where Kh = Hysteresis constant depends on material.

Bm = Maximum flux density. f = Frequency. v = Volume of the core. The induced e.m.f. in the core tries to set up eddy currents in the core and hence responsible for the eddy current losses. The eddy current loss is given by, Eddy current loss = Ke

Bm2 f2 t2 watts/ unit volume

where Ke = Eddy current constant

t = Thickness of the core As seen earlier, the flux in the core is almost constant as supply voltage V1 at rated frequency f is always constant. Hence the flux density Bm in the core and hence both hysteresis and eddy current losses are constants at all the loads. Hence the core or iron losses are also called constant losses. The iron losses are denoted as Pi.

The iron losses are minimized by using high grade core material like silicon steel having very low hysteresis loop by manufacturing the core in the form of laminations.1.2 Copper Losses The copper losses are due to the power wasted in the form of I2

R loss due to the resistances of the primary and secondary windings. The copper loss depends on the magnitude of the currents flowing through the windings. Total Cu loss = I1

2 R1 + I2

2 R2 = I1

2 ( R1 + R2' )= I2

2 ( R2 +R1' )

= I12 R1e = I2

2 R2e

The copper looses are denoted as. If the current through the windings is full load current, we get copper losses at full load. If the load on transformer is half then we get copper losses at half load which are less than full load copper losses. Thus copper losses are called variable losses. For transformer VA rating is or. As is constant, we can say that copper losses are proportional to the square of the KVA rating. So, Pcu α I2α (KVA)2

Thus for a transformer, Total losses = Iron losses + Copper losses = Pi +Pcu

Key point : It is seen that the iron losses depend on the supply voltage while the copper losses depend on the current. The losses are not dependent on the phase angle between voltage and current. Hence the rating of the transformer is expressed as a product of voltage and current and called VA rating of transformer. It is not expressed in watts or kilo watts. Most of the times, rating is expressed in KVA.

EFFICIENCY OF A TRANSFORMER

Efficiency of a Transformer Due to the losses in a transformer, the output power of a transformer is less than the input power supplied.... Power output = Power input - Total losses... Power input = Power output + Total losses = Power output + Pi +Pcu

The efficiency of any device is defined as the ratio of the power output to power input. So for a transformer the efficiency can be expresses as, η = Power output/power input... η = Power output/(power output + Pi + Pcu ) Now power output = V2 I2 cos Φwhere cos Φ = Load power factorThe transformer supplies full load of current I2 and with terminal voltage V2. Pcu = Copper losses on full load = I2

2 R2e

... η = (V2 I2 cos Φ2 )/(V2 I2 cos Φ2 + Pi + I22 R2e)

But V2 I2 = VA rating of a transformer... η = (VA rating x cos Φ) / (VA rating x cos Φ + Pi + I2

2 R2e)

This is full load percentage efficiency with, I2 = Full load secondary current But if the transformer is subjected to fractional load then using the appropriate values of various quantities, the efficiency can be obtained. Let n =Fraction by which load is less than full load = Actual load/Full load For example, if transformer is subjected to half load then, n = Half load/Full load = (1/2)/2 = 0.5 when load changes, the load current changes by same proportion.... new I2 = n (I2) F.L. Similarly the output V2 I2 cosΦ2 also reduces by the same fraction. Thus fraction of VA rating is available at the output. Similarly as copper losses are proportional to square of current then, new Pcu = n2

(Pcu ) F.L.Key Point : So copper losses get reduced by n2. In general for fractional load the efficiency is given by,

where n = Fraction by which load power factor lagging, leading and unity the efficiency expression does not change, and remains same.

1. Condition for Maximum Efficiency When a transformer works on a constant input voltage and frequency then efficiency varies with the load. As load increases, the efficiency increases. At a certain load current, it achieves a maximum value. If the transformer is loaded further the efficiency starts decreasing. The graph of efficiency against load current I2 is shown in the Fig.1

Fig. 1

The load current at which the efficiency attains maximum value is denoted as I 2mand maximum efficiency is denoted as ηmax. Let us determine,1. Condition for maximum efficiency.2. Load current at which ηmax occurs.3. KVA supplied at maximum efficiency. The efficiency is a function of load i.e. load current I2 assuming cos Φ constant. The secondary terminal voltage V2 is also assumed constant. So for maximum efficiency,

dη /d I2 = 0Now η = (V2 I2 cos Φ2 )/(V2 I2 cos Φ2 + Pi + I2

2 R2e)

... (V2 I2 cos Φ2 + Pi + I22 R2e)(V2 cos Φ2) - (V2 I2 cos Φ2)(V2 cos Φ2 + 2I2

R2e) = 0

Cancelling (V2 cos Φ2) from both the terms we get, V2 I2 cos Φ2 + Pi +I2

2 R2e - V2 I2 Φ2 - 2I2

2 R2e = 0

... Pi - I22 R2e= 0

... Pi = I22 R2e = Pcu

So condition to achieve maximum efficiency is that, Copper losses = Iron losses1.1 Load Current I2m at Maximum Efficiency For ηmax, I2

2 R2e = Pi but I2 = I2m

I2m2 R2e = Pi

I2m = √(Pi / R2e) This is the load current at ηmax,Let (I2)F.L. = Full load current... I2m /(I2) F.L.= (1/(I2) F.L.)√(Pi / R2e)... I2m /(I2) F.L.= √(Pi )/((I2) F.L.2

R2e) = √(Pi )/((Pcu) F.L.)... I2m = (I2 )F.L.√(Pi )/((Pcu) F.L.) This is the load current at ηmax interms of full load current.1.2 KVA supplied at maximum Efficiency For constant V2 the KVA supplied is the function of. KVA at ηmax = I2m V2= V2 (I2) F.L. x √(Pi) /((Pcu)F.L.) KVA at ηmax = (KVA rating) x √(Pi) /((Pcu)F.L.) Substituting condition for in the expression of efficiency, we can write expression for ηmax as,

ii) KVA at = KVA rating x √(Pi) /((Pcu)F.L.)

Effect of Power factor on Efficiency The efficiency of a transformer is given by,

Now, input = output + losses = V2 I2 cos Φ + lossesUsing in (1), η = 1 - (losses / (V2 I2 cos Φ + losses))

Let losses/V2 I2 = x and using in (2),

Thus as the power factor of the load is more i.e. cos is higher, x/cosΦ is lesser. Hence the second term in the equation (3) becomes lesser and efficiency will be more.Key Point : As power factor increases, the efficiency increases. Thus the family of efficiency curves are obtained as power factor increases, as shown in the Fig.1.

Fig. 1 Effect of p.f. on efficiency

Effect of Frequency and Supply Voltage on Iron LossesEffect of Frequency and Supply Voltage on Iron Losses The iron losses of a transformer includes two types of losses,1. Hysteresis loss and 2. Eddy current loss For a given volume and thickness of laminations, these losses depend on the operating frequency, maximum flux density and the voltage.

It is known that for a transformer, V = 4.44 f Φm N = 4.44 f Bm A NWhere A = area... Bm α (V/f) .......... For constant A and N Thus as voltage changes, the maximum flux density changes and both eddy current and hysteresis losses also changes. As voltage increases, the maximum flux density in the core increases and total iron loss increases. As frequency increases, the flux density in the core decreases but as the iron loss is directly proportional to the frequency hence effect of increased frequency is to increase the iron losses.Key Point : Thus iron loss increases as the voltage and frequency increases for the transformer.

ALL DAY EFFICIENCY OF A TRANSFORMER For a transformer, the efficiency is defined as the ratio of output power to input power. This is its efficiency. But power efficiency is not the true measure of the performance of some special types of transformers such as distribution transformers. Distribution transformer serve residential and commercial loads. The load on such transformers vary considerably during the period of the day. For most period of the day these transformers are working at 30 to 40 % of full load only or even less than that. But the primary of such transformers is energised at its rated voltage for 24 hours, to provide continuous supply to the consumer. The core loss which depends on voltage, takes place continuously for all the loads. But copper loss depends on the load condition. For no load, copper loss is negligibly small while on full load it is at its rated value. Hence power efficiency can not give the measure of true efficiency of such transformers. in such transformers, the energy output is calculated in kilo watt hour (kWh). Then ratio of total energy output to total energy input (output + losses) is calculated. Such ratio is called energy efficiency or All Day Efficiency of a transformer. Based on this efficiency, the performance of various distribution transformers is compared. All day efficiency is defined as,

While calculating energies, all energies can be expressed in watt hour (Wh) instead of kilo watt hour (kWh). Such distribution transformers are designed to have very low core losses. This is achieved by limiting the core flux density to lower value by using a relative higher core cross-section i.e. larger iron to copper weight ratio. The maximum efficiency in such transformers occurs at about 60-70 % of the full load. So by proper designing, high energy efficiencies can be achieved for distribution transformers. The calculation of all day efficiency for a transformer are illustrated in the Ex. 1.

AUTOTRANSFORMER:

An Autotransformer has only one single voltage winding which is common to both sides. This single winding is “tapped” at various points along its length to provide a percentage of the primary voltage supply across its secondary load. Then the autotransformer has the usual magnetic core but only has one winding, which is common to both the primary and secondary circuits.

Therefore in an autotransformer the primary and secondary windings are linked together both electrically and magnetically. The main advantage of this type of transformer design is that it can be made a lot cheaper for the same VA rating, but the biggest disadvantage of an autotransformer is that it does not have the primary/secondary winding isolation of a conventional double wound transformer.

The section of winding designated as the primary part of the winding is connected to the AC power source with the secondary being part of this primary winding. An autotransformer can also be used to step the supply voltage up or down by reversing the connections. If the primary is the total winding and is connected to a supply, and the secondary circuit is connected across only a portion of the winding, then the secondary voltage is “stepped-down” as shown.

Autotransformer Design

When the primary current IP is flowing through the single winding in the direction of the arrow as shown, the secondary current, IS, flows in the opposite direction. Therefore, in the portion of the winding that generates the secondary voltage, VS the current flowing out of the winding is the difference of IP and IS.

The Autotransformer can also be constructed with more than one single tapping point. Auto-transformers can be used to provide different voltage points along its winding or increase its supply voltage with respect to its supply voltage VP as shown.

Autotransformer with Multiple Tapping Points

The standard method for marking an auto-transformer windings is to label it with capital (upper case) letters. So for example, A, B, Z etc to identify the supply end. Generally the common neutral connection is marked as N or n. For the secondary tapping’s, suffix numbers are used for all tapping points along the auto-transformers primary winding. These numbers generally start at number 1 and continue in ascending order for all tapping points as shown.

Autotransformer Terminal Markings

An autotransformer is used mainly for the adjustments of line voltages to either change its value or to keep it constant. If the voltage adjustment is by a small amount, either up or down, then the transformer ratio is small as VP and VS are nearly equal. Currents IP and IS are also nearly equal.

Therefore, the portion of the winding which carries the difference between the two currents can be made from a much smaller conductor size, since the currents are much smaller saving on the cost of an equivalent double wound transformer.

However, the regulation, leakage inductance and physical size (since there is no second winding) of an autotransformer for a given VA or KVA rating are less than for a double wound transformer.

Autotransformer’s are clearly much cheaper than conventional double wound transformers of the same VA rating. When deciding upon using an autotransformer it is usual to compare its cost with that of an equivalent double wound type.

This is done by comparing the amount of copper saved in the winding. If the ratio “n” is defined as the ratio of the lower voltage to the higher voltage, then it can be shown that the saving in copper is: n.100%. For example, the saving in copper for the two autotransformer’s would be:

The Autotransformer have many advantages over conventional double wound transformers. They are generally more efficient for the same VA rating, are smaller in size, and as they require less copper in their construction, their cost is less compared to double wound transformers of the same VA rating. Also, their core and copper losses, I2R are lower due to less resistance and leakage reactance giving a superior voltage regulation than the equivalent two winding transformer.

Disadvantages of an Autotransformer

The main disadvantage of an autotransformer is that it does not have the primary to secondary winding isolation of a conventional double wound transformer. Then autotransformer’s can not safely be used for stepping down higher voltages to much lower voltages suitable for smaller loads.

If the secondary side winding becomes open-circuited, current stops flowing through the primary winding stopping the transformer action resulting in the full primary voltage being applied to the secondary circuit.

If the secondary circuit suffers a short-circuit condition, the resulting primary current would be much larger than an equivalent double wound transformer due to the increased flux linkage damaging the autotransformer.

Since the neutral connection is common to both the primary and secondary windings, earthing of the secondary winding automatically earths the primary as there is no isolation between the two windings. Double wound transformers are sometimes used to isolate equipment from earth.

The autotransformer has many uses and applications including the starting of induction motors, used to regulate the voltage of transmission lines, and can be used to transform voltages when the primary to secondary ratio is close to unity.An autotransformer can also be made from conventional two-winding transformers by connecting the primary and secondary windings together in series and depending upon how the connection is made, the secondary voltage may add to, or subtract from, the primary voltage.

The Variable Autotransformer

As well as having a fixed or tapped secondary that produces a voltage output at a specific level, there is another useful application of the auto transformer type of arrangement which can be used to produce a variable AC voltage from a fixed voltage AC supply. This type of Variable Autotransformer is generally used in laboratories and science labs in schools and colleges and is known more commonly as the Variac.

The construction of a variable autotransformer, or variac, is the same as for the fixed type. A single primary winding wrapped around a laminated magnetic core is used as in the auto transformer but instead of being fixed at some predetermined tapping point, the secondary voltage is tapped through a carbon brush.This carbon brush is rotated or allowed to slide along an exposed section of the primary winding, making contact with it as it moves supplying the required voltage level.

Then a variable autotransformer contains a variable tap in the form of a carbon brush that slides up and down the primary winding which controls the secondary winding length and hence the secondary output voltage is fully variable from the primary supply voltage value to zero volts.

The variable autotransformer is usually designed with a significant number of primary windings to produce a secondary voltage which can be adjusted from a few volts to fractions of a volt per turn. This is achieved because the carbon brush or slider is always in contact with one or more turns of the primary winding. As the primary coil turns are evenly spaced along its length. Then the output voltage becomes proportional to the angular rotation.

Variable Autotransformer

The variac can adjust the voltage to the load smoothly from zero to the rated supply voltage. If the supply voltage was tapped at some point along the primary winding, then potentially the output secondary voltage could be higher than the actual supply voltage. Variable autotransformer’s can also be used for the dimming of lights and when used in this type of application, they are sometimes called “dimmerstats”.

Variac’s are also very useful in Electrical and Electronics workshops and labs as they can be used to provide a variable AC supply. But caution needs to be taken with suitable fuse protection to ensure that the higher supply voltage is not present at the secondary terminals under fault conditions.

PART B

Construction details

1. Explain in detail, the constructional features of a transformer with neat diagrams. (16)

Principle of operation

1. Explain the working principle of a transformer. (6)

Emf equation

1. Derive the emf equation of a transformer. (6) (N/D-12)

2. Derive the emf equation of a transformer and prove that the number of turns on HV winding and LV winding is in the ratio of their voltages. (8) (A/M-10)

Transformer on no-load

1. Explain in detail, the operation of a transformer on no-load and load. (10) (N/D-12)

2. The required no-load voltage ratio in a 1, 50 Hz, core type transformer is 6000/250 V. Find the number of turns in each winding, if the flux is 0.06 Wb. (10)

3. Draw and explain the no-load vector diagram of an ideal transformer and a practical transformer. (8)

Equivalent circuit

1. Draw and explain the equivalent circuit of a transformer on no-load and load. (8) (N/D-12)

2. Develop the equivalent circuit of a transformer on no-load and load. (16)

3. Calculate in terms of the primary, the equivalent resistance and the leakage reactance of a transformer which gives the following data on test with the secondary terminals short circuited:

Applied voltage = 60 V; Current = 100 A; Power input = 1.2 kW. (10)

Transformer on-load

1. Draw and explain the phasor diagram of a single phase transformer supplying the following:a) a UPF load b) a lagging power factor load (8) (A/M-10)

2. Explain the various losses in a transformer. (8)

3. A single phase transformer is rated at 240/120 V, 50 Hz, find voltage and frequency of secondary at no-load for the following conditions:a) if primary voltage is 120 V at 25 Hz b) if primary voltage is 240 V DC (6)

4. Explain the principle of operation of a single phase transformer and its behaviour on load with phasor diagrams. (16)

Regulation

1. The maximum efficiency of a single phase 11000/400 V, 550 kVA transformer is 97.5% and the efficiency occurs at 80% of full load at unity power factor. The percentage impedance is 3.5% and the load power factor is varied while the load current and the supply voltage are held constant at their rated values. Determine the load power factor, at which the secondary terminal voltage is minimum. Also find the value of the secondary voltage. (16) (M/J-12)