Principles of Electro-Mechanical Energy Conversion
Compare magnetic and electric circuits. (AU 2022, GTU 2021)
Explain Faraday's laws of electromagnetic induction and Lenz's law. (AKTU 2018)
First Law
Whenever the flux linking with a coil or closed circuit changes, a static emf is induced in it and such an emf lasts only for the time the change is taking place.
Faraday’s first law can be well understood from the figure below. As long as the magnet is moved to and fro, the flux linked with the coil changes and hence an emf is induced in the coil, whose presence is indicated by the galvanometer.
As soon as the magnet becomes stationary, the changes in flux is stopped and hence emf in the coil also dies.
Second Law
According to this law, the magnitude of induced emf is equal to the rate of change of flux linked with the closed circuit or coil.
Suppose that a coil with N turns is subjected to a change in flux dφ in time dt, then the average emf induced between two terminals of the coil is given by
e = -N(dφ/dt)
here emf is in volts, and dφ/dt is in Wb/s.
The minus sign indicates that the induced emf opposes the changed which produces it, as stated generally in Lenz’s law.
Lenz’s Law
This law gives the direction of induced e.m.f. and hence
current. According to this law the direction of induced emf(and hence
current) is such that it opposes the cause producing it.
In Figure above, if the N-pole is moved towards the coil, then the current in the coil would be induced in such direction, that the portion of coil facing the N-pole of magnet, would start acting like N-pole and thus it would tend to repel the approaching N-pole of magnet(hence trying to decrease the increasing magnetic flux, i.e. cause).
Again, if the N-pole of the magnet is moved away from the coil, then the portion of coil facing the magnet would become S-pole and try to attract the magnet. Hence, the coil tends to increase the decreasing magnetic flux(cause).
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State Bit-Savart law. Write its applications. (AKUB 2020, RTU 2018, 2023)
Biot-Savart law states that at any point P the magnitude of the magnetic field intensity produced by the differential element is proportional to the product of current, magnitude of differential length, and the singe of angle lying between the filament and a line connoting the filament to the point P where the field is desired.
The basic relation for the magnetic flux density, dB at a point P as produced by a current carrying element (See fig.) is
dB = [μ/4πr2](Idlsinθ)
μ is permeability of medium. The permeability of vacuum is designated as μ0 and is equal to 4π x 10-7 H/m.
State Ampere law. (AKUB 2020, RTU 2023)
Suppose, a coil having N turns is wrapped around the rectangular core of some ferromagnetic material and a current i amperes is passed through it. Due to this current, a magnetic field is produced around that coil. If it is assumed that all the flux produced by the core remains in the core and there is no leakage, its Ampere’s law can be given as:
Hl = Ni or H = Ni/l
where H is the magnetic field intensity, and l is the mean length of the core.
The relation between flux density produced in the core and the field intensity is given as:
B = μH, where μ is the permeability of the core material.
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Explain Fleming’s right-hand rule. (PTU 2018, RTU 2023)
The direction of the induced emf can be found out by Fleming's right-hand rule, as shown in the given figure.
Derive the expression for the force and torque on a Current Carrying Conductor. (AU 2022)
The value of the torque is given by
F = BIl
where B is the flux density, i the current and l is the length of the conductor.
If the conductor moves in the magnetic field in such a way that its axis makes an angle with the magnetic field, the force is given by
F = BIl sinθ
Torque, T = wBIlsinθ = (A/l)Bilsinθ = ABIsinθ
where w is width of current carrying loop and "l" is its length.
If we have a multiple loop of N turns, then
T = NABIsinθ
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What is mmf and flux? (RTU 2018)
MMF. A current-carrying conductor produces a magnetic field around it. The coil should have the correct number of turns in order to produce the required flux density. Magnetomotive force (mmf) is defined as the product of current and number of turns.
Magnetomotive force = Current × Turns
Let I be the current through the coil (A) and N be the number of turns in the coil.
F = IN
The unit of mmf is ampere-turns (AT). Since N is dimensionless, its unit is amperes (A) also.
Magnetic Flux. A current-carrying conductor produces a magnetic field around it. The magnetic field is measured in terms of flux lines (or simply flux) f and flux density B, the number of flux lines per unit area.
B = φ/A
The density of flux is greatest near the conductor and tapers off inversely with distance.
B = μ0I/2πr
The unit of flux φ is Weber (Wb) and the unit of magnetic intensity (B) is Weber per metre square (Wb/m2) or tesla (T).
What is an air gap flux? (RTU 2023)
Magnetic flux which does not follow the intended port in a magnetic circuit is called leakage flux. As against this, there is flux(in the air gap) which is utilized for useful purposes. Such flux is called useful flux. Both types of fluxes have been shown in Figure.
Give name of sources that store energy in magnetic circuits. (RTU 2023)
The inductor stores energy in the form of the magnetic field. The capacitor stores energy in the form of the electric field. So that transients are present in the inductor & capacitor.
Give example of linear and non-linear magnetic circuits. (RTU 2023)
- Linear circuits follow the principles of linearity, which means that the output of the circuit is directly proportional to its input. The examples of linear circuits are resistance and resistive circuit, inductor and inductive circuit and capacitor and capacitive circuit.
- Nonlinear circuits, on the other hand, do not follow the principles of linearity. In these circuits, the output is not directly proportional to the input, and the properties of homogeneity and superposition do not hold. Some of the examples of nonlinear circuit of nonlinear elements are diode, transformer, iron core, inductor, transistor etc.
Draw magnetic lines of a magnet showing magnetic flux lines clearly. (AKUB 2020)
Answer: Magnetic Field lines are imaginary lines along which the North Magnetic Pole would move. In a bar magnet, the magnetic field lines look like.
Which material is suitable for making permanent magnet? Give reason. (PTU 2020)
Usually ferromagnetic or ferrimagnetic materials are used for making permanent magnets these materials includes iron, nickel, cobalt and some rare earth metals. These materials were exposed to strong magnetic fields until it retains its magnetic field.
What is the importance of electromagnet in machine? (RTU 2018)
Electromagnetism has important scientific and technological applications. It is used in many electrical appliances to generate desired magnetic fields. It is even used in an electric generator to produce magnetic fields for electromagnetic induction to occur. It has many more technological applications, including MRI scanning (magnetic resonance imaging) and electric bells.
Uses of Electromagnets: Generators, motors, and transformers, Electric buzzers and bells, Headphones and loudspeakers, Relays and valves, Data storage devices like VCRs, tape recorders, hard discs, etc., Induction cooker, Magnetic locks, MRI machines, Particle accelerators, Mass spectrometers.
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Why laminated core in electrical machines are used. (PTU 2020)
Eddy current losses can be reduced by splitting the solid core into thin sheets called laminations, in the plane parallel to the magnetic field. Each lamination is insulated from each other by a thin layer of coating of varnish or oxide film. By laminating the core, the area of each section is reduced and hence the induced emf also reduces. As the area through which the current is passed is smaller, the resistance of the eddy current path increases.
While comparing magnetic and electric circuit, the flux of magnetic circuit is compared with which parameter of the electric circuit. (PTU 2020)
Current
On the same platform, draw the B-H curve for soft as well as hard magnetic material. (AKUB 2020, RTU 2018, 2023)
Explain the magnetization curve of ferromagnetic material. (GTU 2020)
Describe multiple excited magnetic field system. (AU 2022)
Answer: In doubly excited magnetic field system, there are two independent sources of excitation such as DC separately excited generator, synchronous motor, loud speakers, tachometers, etc.
Differentiate statically induced EMF from dynamically induced EMF. Give one example for each. (AU 2023)
- Statically Induced EMF is produced when there is no relative movement between conductor and the Source of the magnetic field. Rather, magnetic field itself varies in magnitude. this rate of change of magnetic field around the conductor causes the EMF induction in the conductor. Example of such EMF is the self and mutually induced EMFs in inductors when supplied with Alternating Current.
- Dynamically Induced EMF is produced when there exists relative movement between conductor and the Source of the Magnetic field. Here Magnetic field lines does not change in magnitude rather it moves with respect to the conductor, and when it cuts the conductor an EMF is induced in the conductor. Example of such EMF is the EMF induced in the armature of a DC Generator.
With the help of neat and labelled diagram, show energy and co-energy. (AKUB 2020)
Energy stored in the magnetic field is called field energy.
The electrical energy which is converted into magnetic energy is called co-energy. From Figure, the area above the magnetization curve (or λ - i) gives the field energy while the area below the magnetization curve gives the co-energy which is denoted by W’field.
Write the energy balance equation for the generator and motor. (AKTU 2019)
The energy balance equation is an expression which shows the complete process of energy conversion. In an electromechanical energy conversion device, the total input energy is equal to the sum of three components :
Energy dissipated or lost
Energy stored
Useful output energy
Therefore, for an electric motor, the energy balance equation can be written as,
Electrical energy input = Energy dissipated + Energy stored + Mechanical energy output
Where,
The electrical energy input is the electricity supplied from the main supply.
Energy stored is equal to sum of the energy stored in the magnetic field and in the mechanical system in the form of potential and kinetic energies.
The energy dissipated is equal to sum of energy loss in electric resistance, energy loss in magnetic core (hysteresis loss + eddy current loss) and mechanical losses (windage and friction losses).
For an electric generator, the energy balance equation can be written as,
Mechanical energy input = Electrical energy output + Energy stored + Energy dissipated
Where, the mechanical energy input is the mechanical energy obtained from a turbine, engine, etc. to turn the shaft of the generator.
Why do all practical energy conversion devices make use of a magnetic field as a coupling medium rather than an electric field? Also, write the energy balance equation. (AKTU 2022, 2023)
The magnetic field is used as the coupling medium between electrical and mechanical medium because the energy storing capacity of the magnetic field is much higher than the electric field.
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Develop a block diagram indicating the process of electromechanical energy conversion. (AU 2022, AKTU 2023, GTU 2020)
Electromechanical energy conversion is the transformation of energy between an electrical and a mechanical system. If energy is converted from mechanical to electrical form, the device is called a generator, while if electrical energy is converted into mechanical energy, the device is known as a motor. Both these effects are shown in following figure.
Single Phase Transformers
How are the transformer losses affected by the power factor of the connected load? (AKTU 2019)
Power factor also plays a key role in transformer losses. If load is having a low lagging power factor, it may lead to high load current increasing I2R losses. Whereas any reduction in current for the same kW load by improvement in power factor reduces load losses (I2R losses).
What is Hysteresis loss in transformer? Explain in detail how eddy current and hysteresis losses of a transformer can be minimised. (JNTUH 2023, MDU 2018)
Hysteresis Losses: During each A.C. cycle, current flowing in the forward and reverse directions magnetizes and demagnetizes the core alternatively. Energy is lost in each hysteresis cycle within the magnetic core. Energy loss is dependent on the properties (e.g. coercivity) of particular core material and is proportional to the area of the hysteresis loop (B-H curve). The iron losses are minimised by using high grade core material like silicon steel having very low hysteresis loop and by manufacturing the core in the form of laminations.
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Why is transformer not used on dc supply? (AKTU 2022)
This is because a transformer works on the principle of electromagnetic induction, which requires a voltage source that changes with time (alternating source). Thus since a dc supply is unidirectional, the transformer cannot work. Simple, no voltage or output.
Discuss the significance of voltage regulation in transformers. How the same is arrived at? (AKTU 2023)
Voltage regulation is the measure of how well a power transformer can maintain constant secondary voltage given a constant primary voltage and wide variance in load current. The lower the percentage (closer to zero), the more stable the secondary voltage and the better the regulation it will provide.
These answers are taken from study material for B. Tech. exams by amiestudycircle.com. With our study material which is prepared by IIT, Roorkee faculty (Retd.), no text book study is required.
Define all day efficiency of a transformer. (AU 2022, GTU 2021)
All day efficiency means the power consumed by the transformer throughout the day. It is defined as the ratio of output power to the input power in kWh or Wh of the transformer over 24 hours.
What is the difference between commercial efficiency and all day efficiency? (KTU 2020)
All-day efficiency of the transformer depends on their load cycle. The load cycle of the transformer means the repetitions of load on it for a specific period. The ordinary or commercial efficiency of a transformer is defined as the ratio of the output power to the input power.
Write the working principle of a step down auto transformer with a single diagram. (AU 2022)
Answer: Described in this module.
What is transformer ratio? (JNTUH 2023)
EMF on primary side is
V1 = 4.44φmaxfN1volts
On secondary side
V2 = 4.44φmaxfN2volts
Hence, V2/V1 = N2/N1= K (say)
Where K is known as the transformation ratio of the transformer.
Now, if I1 is the primary current and I2 is the secondary current, then
V1I1 = V2I2
(Since there is no power loss in as ideal transformer)
V2/V1 = N2/N1 = I1/I2 = K
In other words, ratio of secondary current and primary current is inversely proportional to the transformation ratio ‘K’.
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Why stepped core section is preferred to a square section for transformers. (PTU 2018)
As the rating are increased stepped or multi stepped cores are preferred, which improves mechanical stability, reduce the length of mean turn and results savings in copper besides increasing efficiency and improves regulation.
Justify following statements:
(i) Transformer core is laminated.
(ii) Transformer rating is in KVA. (GTU 2020, 2021, HPTU 2022, JNTUH 2023, PTU 2020)
(i) The transformer's core is laminated to reduce eddy current and improve efficiency.
(ii) Iron loss on voltage and copper loss of a transformer depends on current. Hence, total transformer heat loss depends on volt-ampere (VA) and independent of phase angle among voltage and current i.e., it is independent of power factor. That is why the rating of the transformer is in kVA and not in kW.
Why Secondary of current transformer should not be open? (GTU 2020)
The secondary side of a current transformer should never be kept in open condition because, when kept open, there is a very high voltage found across the secondary side. This high voltage causes a high magnetizing current to build up on the secondary side which in turn causes high flux and makes the core to saturate.
What is the order of the efficiency of a general purpose transformer? State reason. (AKUB 2020)
The efficiencies of power transformers normally vary from 97 to 99 percent. The power supplied to the load plus resistive, eddy current, hysteresis, and flux losses must equal the input power. The input power is always greater than the output power.
These answers are taken from study material for B. Tech. exams by amiestudycircle.com. With our study material which is prepared by IIT, Roorkee faculty (Retd.), no text book study is required.
Draw phasor diagram of transformer at no load condition. (AKTU 2019, AU 2023, RGPV 2020
The emf per turn of a single-phase transformer 6.6 kV/440 V, 50 Hz transformer is approximately 12 V. Calculate the number of turns in the HV and LV windings. (AKTU 2022)
Explain polarity test of single phase transformer. (GTU 2020, 2022)
Polarities of a transformer can be checked by a simple test requiring only voltage measurements with a transformer on no-load. Polarity test is required for parallel operation of transformers.
In this test, the rated voltage is applied on one winding and one terminal of the primary winding is connected to one terminal of the secondary winding through a voltmeter. The other two terminals of the windings are short circuited as shown in following figure.
Suppose the voltages induced in the primary and secondary windings are E1 and E2 respectively. If the voltmeter reading is equal to E1 – E2 then the terminals A and C are of same polarity at any one instant.
Hence,
terminals B and D will be of opposite polarities. A and C are like
terminals and B and D are unlike terminals. If the voltmeter reading is
equal to E1 + E2 then A and C are of opposite
polarities. In that case if terminal A is positive at any instant then
terminal D will be positive at that instant. At the same instant,
terminals B and C will have negative polarities.
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Explain why parallel operation of transformer is necessary? (HPTU 2022)
The transformers arc connected in parallel when the load on them is more than the rating of the individual transformers. Several smaller units are operated in parallel which share a common load. Thus it is avoided that the total load is supplied by single unit due to use of parallel operation. For operating transformers in parallel, their primaries are connected to the supply bus and the secondaries are connected together to form a secondary bus to feed the common load. This is shown in following figure.
State the conditions for parallel operations of transformers. (AKTU 2023, AU 2022, GTU 2021, HPTU 2022, KTU 2020, MDU 2018, RGPV 2019)
- The transformers must be connected in the same polarity.
- The transformation ratios of the transformers should be same.
- Equivalent impedances of the transformers should be divided in inverse proportion to the current ratings, that is the internal impedance drop of the transformers should be equal.
- The ratio of equivalent leakage reactance to equivalent resistance should be equal for all the transformers. This condition ensures that the transformers share active and reactive power according to their ratings.
What is an auto transformer and its applications? (HPTU 2022, RTU 2018)
In two winding transformers, the windings are electrically isolated and the e.m.f. in secondary gets induced due to induction.
In practice, it is possible to use only one winding for the transformer so that part of this winding is common to the primary and secondary. Such a special type of transformer having only one winding such that part of the winding is common to the primary and secondary is called autotransformer.
Obviously the two windings are electrically connected and it works on the principle of conduction as well as induction. Such an autotransformer is very much economical where the voltage ratio is less than 2 and the electrical isolation of the two windings is not necessary.
The power transfer in 2 winding transformer is fully inductively while in autotransformer the power is transferred from primary to secondary by both inductively as well as conductively.
Applications of auto-transformer
- Auto-transformers with a number of tappings are used for starting induction motors and synchronous motors.
- Auto-transformers are used as variable a.c. voltage source (variac).
- Auto-transformers are used as boosters to give a small boost to a distribution cable for compensating the voltage drop.
- They are used as interconnections of power systems of different voltage levels.
- They can be used as furnace transformers to supply power to the furnaces at the required supply voltage.
- No isolation between the primary and secondary winding The main disadvantage of the autotransformer is that it does not have electrical isolation between primary and secondary windings.
- More current during fault As an autotransformer has only one winding impedance, so, the impedance is less ( in an Isolation type transformer both primary and secondary windings will have impedance). Due to less impedance, the current flow during an output short circuit will be more. Hence more short circuit current problems.
- Full voltage at the output during wire break in the winding Full voltage appears at the output during wire damage. Now the load connected to the output will get high voltage and will get damaged.
- Not economical for less voltage ratio The autotransformer is not economical for less voltage ratio between output and input. The size and cost of the autotransformer will increase with less output-to-input voltage ratio (fixed ratio type).
Give a comparison of an auto transformer with a two-winding transformer. (RTU 2018, GTU 2021, 2022)
Auto-transformer
- In autotransformer, a part of the winding is common for primary as well secondary.
- In autotransformer, there are movable contacts are available.
- The autotransformer is operateas a step-down and step-up transformer.
- In autotransformer, copper saving is more because one winding is used as primary as well as the secondary winding.In autotransformer, we get variable secondary voltage.
- In an autotransformer, there is no electrical isolation between the primary and secondary windings.
- The size of the autotransformer is very small because one single winding is used.
- In autotransformer number of winding is one.
- The cost of an autotransformer is less.
- In autotransformer losses in winding is low.
- In autotransformers, efficiency is very high.
- In autotransformer voltage regulation is better.
- The autotransformer works on the self-induction principle.
2-winding transformer
- In two winding transformers separate primary and secondary windings are used.
- In two winding there are no movable contacts. Two winding transformers are static devices.
- The two winding transformers are also operated as a step-up as well as a step-down transformer.
- In two winding transformer copper saving is none because two separate windings are used.
- In two winding transformers, we can not get any variable secondary voltage.
- In two winding transformers, electrical isolation exists between the primary and secondary windings.
- The size of two winding transformers is very large because two separate winding are used.
- In two winding transformer number of windings are two.
- The cost of two winding transformers is very large.
- In two winding transformer losses in winding is high.
- In two winding transformer efficiency is very low.
- In two winding transformer efficiency is poor.
- The two winding transformer works on the mutual induction principle.
An auto transformer has primary voltage V1 and secondary voltage V2, where V1 > V2. Calculate the fraction of power transferred inductively. (AKTU 2019)
Since the primary and secondary windings of an autotransformer are connected magnetically as well as electrically, the power from primary is transferred to the secondary inductively (transformer action) as well as conductively (i.e., directly as windings are electrically connected).
Output apparent power = V2I2
Apparent power transferred inductively = V2(I2 - I1) = V2(I2 - KI2)
= V2I2 (1 - K) = V1I1 (1 - K)
Hence, Power transferred inductively = input x (1 - K)
Power transferred conductively = input - input x (1 - K) = K x input
Suppose the input power to an ideal autotransformer is 500 W and its voltage transformation ratio K = 1/4. Then,
Power transferred inductively = input x (1 - K) = 500(1 - 1/4) = 375W
Power transferred conductively =K x input = 1/4 x 500 = 125W
Note that input power to the autotransformer is 500 W. Out of this, 375 W is transferred to the secondary by transformer action (inductively) while 125 W is conducted directly from the source to the load (i.e., it is transferred conductively to the load).
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Three Phase Transformers
Write the advantages of three phase transformer. (AU 2023)
- A three phase transformer occupies less space for same rating, compared to a bank of three single phase transformers.
- It weighs less.
- Its cost is less.
- Only one unit is required to be handled which makes it easy for the operator.
- It can be transported easily.
- The core will be of smaller size and the material required for the core is less.
- Single three phase unit is more efficient.
- In case of three single phase units, six terminals are required to be brought out while in case of one three phase unit, only three terminals are required to be brought out.
- The overall busbar structure, switchgear and installation of single three phase unit is simpler.
What are the basic units of a 3-phase transformer? (RTU 2023)
3-phase transformers typically have at least 6 windings- 3 primary and 3 secondary. The primary and secondary windings can be connected in different configurations to meet different requirements. In common applications, the windings are usually connected in one of two popular configurations: Delta or Wye.
What is harmonics in transformer? (RTU 2018)
If the flux in a transformer core is sinusoidal, the induced emf is also sinusoidal but the magnetizing current is not sinusoidal due to non-linear B-H curve for core material. The magnetizing current contains third and higher harmonics necessary to produce sinusoidal flux. The amplitude of these harmonics increases with increase in maximum flux density in the core.
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List the problems associated with harmonics in 3- transformers. (HPTU 2022)
- Source voltage wave shape distortion
- Efficiency loss in transmission and distribution depending on RMS current increase on lines
- Failures in compensation systems
- Overheating in electrical engines and transformers
- Failures in sensitive electronic devices, PLC and CNC devices
- Difficulties and abrasions in insulation levels of equipments
- Increased losses in the system
- Having faulty works in protection and control systems
- Increased voltage reductions
- Resonance risks at high frequencies in systems
What is tap changing? Why is it required? (GTU 2020, RGPV 2019)
The tap changer is connected where the voltage to the neutral is minimum. In a star-connected transformer, the tapped end of the windings are connected to form the star point though physically the tapped coils are placed in the middle of the winding. In a delta-connected transformer, it is essential to provide the tapped coils in the middle so that the tap-changing gear is far removed from the line and lighting surges.
Depending on whether the tap changer is designed to operate when the transformer is out of service or when the transformer is in operation, it can be of two types: off-load tap changer and on-load tap changer.
These answers are taken from study material for B. Tech. exams by amiestudycircle.com. With our study material which is prepared by IIT, Roorkee faculty (Retd.), no text book study is required.
Explain the purposes of a third winding in three winding transformers. (KTU 2021, 2022)
Three-winding transformers contain an additional winding, other than primary and secondary windings, called tertiary winding.
Tertiary winding is provided in electrical power transformer to meet one or more of the following requirements: It reduces the unbalancing in the primary due to unbalancing in the three-phase load. It removes harmonic components in voltage.
Unlike two-winding transformers, the kVA rating of the three windings are different. The voltage rating of the primary is the highest and that of the tertiary is lowest. The voltage rating of the secondary winding lies in between primary and tertiary.
Compare core type and shell type 3-phase transformers. (AKTU 2022)
Core type
- The winding encircles the core.
- The cylindrical type of coils are used.
- As windings are distributed, the natural cooling is more effective.
- The coils can be easily removed from maintenance point of view.
- The construction is preferred for low voltage transformers.
- It has a single magnetic circuit.
- In a single phase type, the core has two limbs.
Shell type
- The core encircles most part of the windings.
- Generally, multilayer disc type or sandwich coils are used.
- As windings are surrounded by the core, the natural cooling does not exist.
- For removing any winding for the maintenance, large number of laminations are required to be removed. This is difficult.
- The construction is used for very high voltage transformers.
- It has a double magnetic circuit.
- In a single phase type, the core has three limbs.
State the condition for which a 3 phase - 4 wire distribution transformer will give maximum efficiency and the range of loading for maximum efficiency. (AU 2022)
The transformer will give the maximum efficiency when their copper loss is equal to the iron loss.
What are vector groups in 3-phase transformer? (AKTU 2022, KTU 2021)
Three phase transformers can be connected in variety of different ways such as in star, delta, V or zigzag. In many cases, vector diagrams of the primary and secondary e.m.f.s are useful to describe the characteristics, advantages and disadvantages of a given type of connection. A vector diagram can be constructed based on following principles,
- The voltages of primary and secondary windings on the same limb are in phase opposition and the two induced e.m.f.s are in phase.
- The e.m.f. s induced in the three phases are equal, balanced and displaced mutually by one third period in time and have a definite sequence.
In polyphase transformers, polarity alone is insufficient to represent a definite relation between the high voltage and low voltage windings. Thus we make use of vector diagrams in addition to lead markings to represent angular phase displacement between the high voltage and low voltage windings and the time order of phase sequence.
DC Generators
Give classification of DC generators with neat connection diagram. (GTU 2021)
- Separately excited D.C. Generator
- Self excited (a) D.C. Series Generator (b) D.C. Shunt Generator (c) D.C. Compound Generator (i) Long shunt (ii) Short shunt
Mention the parts of a DC machine. Explain the use of any one of them. (GTU 2021, JNTUH 2023)
See following figure.
The basic nature of e.m.f. induced in the armature conductors is alternating. This needs rectification in case of d.c. generator, which is possible by a device called commutator.
Functions :
- To facilitate the collection of current from the armature conductors.
- To convert internally developed alternating e.m.f. to unidirectional (d.c.) e.m.f.
- To produce unidirectional torque in case of motors.
It is cylindrical in shape and is made up of wedge shaped segments of hard drawn, high conductivity copper. These segments are insulated from each other by thin layer of mica. Each commutator segment is connected to the armature conductor by means of copper lug or strip. This connection is shown in the following figure.
What are the material used for brushes in de machines. Also give reason. (PTU 2020)
Brushes are stationary and resting on the surface of the commutator. Brushes are usually made from carbon or graphite, but copper brushes are preferred in low voltage high current requirements.
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What are the losses occur in DC Machines? (JNTUH 2023, KU 2021, RTU 2018)
- Armature copper losses
- Copper losses in the field winding
- Magnetic losses or Iron losses
- Mechanical losses
Total Losses in D.C. Generator
= Armature copper losses + Field copper losses + Iron losses + Mechanical losses.
Draw the internal and external load characteristics of a DC shunt generator. (AU 2022)
Internal characteristics
Ideally the induced e.m.f. is not dependent on the load current IL or armature current Ia But as load current increases, the armature current Ia increases to supply load demand. As Ia increases, armature flux increases.
External Characteristics
For d.c. shunt generator we know that, E = Vt + IaRa neglecting other drops. So as load current IL increases, Ia increases. Thus the drop IaRa increases and terminal voltage Vt = E - IaRa decreases. But the value of armature resistance is very small, the drop in terminal voltage as Il changes from no load to full load is very small. This is shown in the given figure. Hence d.c. shunt generator is called constant voltage generator.
What is the necessity of commutating poles (interpoles) in DC machines? (HPTU 2022, AU 2023)
Interpoles (commutating poles or compoles) are provided in between the main poles of DC machine and are energised to such an extent that they must neutralise the armature field produced by the armature winding when machine is loaded. At the same time they must neutralise the emf induced due to inductance in the coil undergoing commutation.
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Explain with the help of an example why in an electrical machine the number of stator poles should be equal to the number of rotor poles. (AKTU 2018)
The number of stator and rotor poles must be equal for the production of net Electromagnetic Torque. If the number of poles in the stator and rotor are different then the net average torque produced will be zero because their magnetic fields will rotate at different speeds.
Draw the pictorial view of armature flux and field flux positions in the air gap during demagnetizing and cross magnetizing situations of DC machine. (AU 2023)
List the important conditions for exciting a self-excited DC generator. (AU 2022)
- There must be some residual magnetism in the field system.
- The residual magnetism must be in proper direction. The field coils should be connected with the armature in such a way that current flowing through them should increased the emf induced by the residual magnetism.
- For a series wound generator, the resistance of the external circuit should be less than the critical resistance.
- For a shunt wound generator, resistance in the field circuit must be less than critical resistance for field circuit and resistance in the load circuit must be greater than critical resistance for the load circuit.
What are the advantages and disadvantages of separately exited DC machines? (KU 2022)
Separately excited DC generators have distinct 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 more expensive to provide a separate excitation source.
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Define back pitch and front pitch of armature winding in dc machine. (AKTU 2022, 2 marks)
Back Pitch (YB)
The distance between the armature conductors in a coil on the back side of the armature is called as back pitch. It is denoted by This pitch expresses the distance at the non commutator end or back side hence it is called as back pitch.
Front Pitch (YF)
It is given by the distance between the second conductor of one coil and the first conductor of the next coil whose ends are connected to same commutator bar. Front pitch is denoted by YF. This pitch expresses the distance at the front, hence is called front pitch. Both the front pitch and back pitch for both lap and wave windings are shown in given figure.
Lap Winding
- The lap winding can be defined as a coil which can be lap back toward the succeeding coil.
- The connection of the lap winding is, the armature coil end is connected to the nearby section on the commutators.
- The numbers of the parallel path are equal to the total of number poles.
- Another name of lap winding is multiple winding otherwise Parallel Winding.
- The e.m.f of lap winding is Less.
- The no. of brushes in lap winding is Equivalent to the no. of parallel paths.
- The types of lap winding are Simplex lap winding & Duplex lap winding.
- The efficiency of the lap winding is Less.
- The additional coil used in the lap winding is Equalizer Ring.
- The winding cost of the lap winding is High.
- The lap winding used for high current, low voltage machines.
Wave Winding
- The wave winding can be defined as the loop of the winding can form the signal shape.
- The connection of the wave winding is, the armature coil end is connected to commutator sections at some distance apart.
- The number of parallel paths is equal to two.
- Another name of wave winding is Series Winding otherwise Two-circuit.
- The e.m.f of wave winding is More.
- The no. of brushes in wave winding is Equivalent toTwo.
- The types of wave winding are Progressive & Retrogressive.
- The efficiency of the wave winding is High.
- The additional coil used in the wave winding is Dummy coil.
- The winding cost of the wave winding is Low.
- The applications of wave winding include low current and high voltage machines.
Why is the wave winding useful for high voltage low current DC machines? (AKTU 2019)
For a given number of poles and armature conductors it gives more emf than that of lap winding. Hence wave winding is used in high voltage and low current machines.
Explain why equalizer connections are used in lap-winding and dummy coils are sometimes used in wave-windings. (AKTU 2018)
The function of the equalizer ring is to cause the circulating current to flow within the armature winding itself, without letting them pass through the brushes.
Wave winding is actually called as incomplete winding because after completion of winding some slots are left empty. These empty slots are generally filled with some extra coils namely called as dummy coils. These dummy coils are used for providing mechanical balance for the rotor only.
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Draw sample waveform of wave winding and lap winding. (AKUB 2020)
Q.25. (AKTU 2019, 2 marks): Why is the wave winding useful for high voltage low current DC machines?
For a given number of poles and armature conductors it gives more emf than that of lap winding. Hence wave winding is used in high voltage and low current machines.
Q.26. (AKTU 2018, 2 marks): Explain why equalizer connections are used in lap-winding and dummy coils are sometimes used in wave-windings.
The function of the equalizer ring is to cause the circulating current to flow within the armature winding itself, without letting them pass through the brushes.
Wave winding is actually called as incomplete winding because after completion of winding some slots are left empty. These empty slots are generally filled with some extra coils namely called as dummy coils. These dummy coils are used for providing mechanical balance for the rotor only.
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Explain clearly, the necessity for introducing dummy coils in a 4-pole, double layer, wave wound armature of a DC machine, having 24 slots, with 2 coil sides per slot. (KU 2021)
The dummy coils are similar to other coils except their ends are cut, short and taped. They do not connect with the commutator bars. The dummy coils are simply to provide mechanical balance for the armature. As they do not connect with commutator bars, they do not affect the electrical characteristics of the winding.
What is the purpose of providing compensating winding in DC machines? In which part of the machine is the compensating winding embedded? (GTU 2020, KU 2021, PTU 2018)
The compensating windings are basically used to neutralise the armature flux in the pole are region which will otherwise cause severe distortion of main field flux. These windings are of concentric type and are placed in axial slots in the pole faces as shown in the figure.
Define commutation in DC machine.(AKUB 2020, KU 2022)
Commutation is a process of producing a unidirectional or direct current from the alternating current generated in the armature coils.
The current generated in the armature conductors of a dc generator are alternating. These currents flow in one direction when the armature conductors are under north pole and the opposite direction when they are under south pole.
As conductors move out of the influence of the north pole and enter south pole, the currents in them are reversed. When a brush spans two commutator segments, the winding element connected to those segments is short-circuited. During the period of short circuit of an armature coil by a brush, the current in the coil must be reversed and also brought up to its full value in the reversed direction. The time of short circuit is called the period of commutation.
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Explain the function of a commutator in a DC machine for motoring and generating action. (AKTU 2018)
Explained in this post.
The role of the commutator in the DC machine is that of a mechanical rectifier. Justify the statement. (AKTU 2023)
A commutator works like a rectifier that changes AC voltage to DC voltage within the armature winding. It is designed with a copper segment, and each copper segment is protected from the other with the help of mica sheets. It is located on the shaft of the machine.
Sketch the OCC of a DC shunt generator and explain the possible causes for the failure of the machine to build up voltage. (KU 2021, MDU 2018)
The shunt generator is allowed to build-up at first before loading it.
Due to the presence of residual magnetism in the poles, a small emf is generated which circulates a small current in the field circuit. If the field is properly connected to the armature, this circulation of current further increases the pole flux which in turn increases the generated emf and this continues until the full open circuit emf is generated.
Any improper connection of field coil to armature may destroy the residual magnetism.
The generated emf in the armature supplies the ohmic drop and overcomes the self-induced emf in the field coil. For the current OE shown in following figure, EF goes to supply the ohmic drop and the rest FG overcomes the self-induced emf.
For the field current OM. the generated emf is fully utilised to supply the ohmic drop and cannot overcome the self-induced emf.
There is no energy stored in field poles and no further increased interpole flux and generated emf.
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Explain the conditions for parallel operation of DC Generators. (GTU 2022)
The general conditions for operating two dc generators in parallel are the following:
- Same polarities of both the machines must be connected together.
- The emf generated should be approximately the same.
- The load sharing depends upon their emf generated and internal resistances.
- Equalizer bars must be connected in series or compound generators for stability of operation.
A 220 V DC shunt machine has a armature resistance of 0.5 Ω. If the full load armature current is 20, calculate the induced emf when the machine act as a (a) generator (b) motor. (AU 2022)
Armature resistance (Ra) = 0.5 ohm
Voltage (V) = 220 V
Armature current (Ia) = 20 A
Back emf (Eb) = 220 - 20 (0.5) = 210 V
Generated emf (Eg) = 220 + 20 (0.5) = 230 V
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DC Motors
Give name of losses related to DC motors. (RTU 2023, KTU 2022)
- Mechanical losses. As their name suggests, mechanical losses are caused by movement of the motor.
- Magnetic losses (core losses or iron losses). These losses are associated with magnetic paths of the motor. Magnetic losses include hysteresis losses, caused by the changing polarity of the flux in the core, and eddy currents, which are induced in the steel core by the changes in flux polarity.
- Copper losses (electrical losses or winding losses). These losses can be referred to by many names, including the term I2R losses,” since they’re caused by the resistance of the field and armature windings.
- Brush losses. During commutation, some losses occur between the commutator and the brushes.
- Stray load losses. Losses that cannot be easily accounted for are generally lumped into the “stray load” category. Common examples are short-circuit currents during commutation and flux distortion due to armature reaction.
State the application of various types of motors. (GTU 2022, JNTUH 2023)
- Shunt DC motor. For driving constant speed line shafting lathes, centrifugal pumps, machine tools, blowers and fans and reciprocating pumps
- Series DC motor. For traction work, i.e. electric locomotives, rapid transit systems, trolley cars etc., cranes and hoists conveyors.
- Compound DC motor. used in elevators, shears and punches, steel rolling mills, printing press and cutting machines, stamping presses, mixers.
Write the applications of series and shunt DC motors. (HPTU 2022)
Already answered in this post.
Write voltage equation of dc motor. (AKUB 2020)
The voltage equation of a d.c. motor can be written as,
V = Eb + IaRa + brush drop
Neglecting the brush drop, the generalised voltage equation is,
V = Eb + Ia Ra
The back e.m.f. is always less than supply voltage (Eb < V).
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How the direction of rotation of the DC shunt motor can be changed. (AKTU 2019)
The direction of rotation of DC shunt motor can be reversed by reversing either field terminals or armature terminals but not both.
Enumerate the types of speed control in DC machines. (AKTU 2023)
Speed control of a motor means the intentional variation of speed according to the requirement of the workload connected with the motor.
Various methods of speed control of dc motors can be understood by examining the expression for speed of a dc motor. The speed of a dc motor can be expressed as
N = (V - IaRa)/Kφ
Speed N can be varied by varying the field flux φ, by adding some extra resistance in the armature circuit and by varying the supply voltage V.
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Why the speed of a DC shunt motor is practically constant? (KTU 2022)
We know that N ∝ Eb/φ. As the shunt field resistance is constant, the shunt field current is constant. Accordingly, the flux per pole is practically constant. The back emf is also practically constant. Hence, by speed relation, it is seen that the speed of a DC shunt motor is constant.
What are the factors that affect the speed of the dc motor? (AKTU 2022, PTU 2018)
- Flux per pole φ;
- Applied voltage V;
- Resistance of armature Ra;
- Armature (or load) current la
Write the reason for higher starting current in DC motors. (AU 2023)
The absence of back emf at the time of starting causes the armature current shoot up to about 20 times the normal current, included, no limiting resistance is included.
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Which dc motor is used to drive cranes? Give reason. (AKTU 2022)
Series wound motor is used because of its very high starting torque. Hence it is used for heavy duty applications such as electric railways, mine hoists, continuous conveyors cranes, rolling mills, metallurgical works etc.
Define back emf in dc motor. What is the significance of back EMF in a DC motor? (GTU 2022, KTU 2020, MDU 2018, PTU 2020)
Due to the presence of back e.m.f. the d.c. motor becomes a regulating machine i.e. motor adjusts itself to draw the armature current just enough to satisfy the load demand. The basic principle of this fact is that the back e.m.f. is proportional to speed, Eb ∝ N.
When load is suddenly put on to the motor, motor tries to slow down. So speed of the motor reduces due to which back e.m.f. also decreases. So the net voltage across the armature (V - Eb) increases and motor draws more armature current. As F = B /I, due to increased current, force experienced by the conductors and hence the torque on the armature increases. The increase in the torque is just sufficient to satisfy increased load demand. The motor speed stops decreasing when the armature current is just enough to produce torque demanded by the new load.
When load on the motor is decreased, the speed of the motor tries to increase. Hence back e.m.f. increases. This causes (V – Eb) to reduce which eventually reduces the current drawn by the armature. The motor speed stops increasing when the armature current is just enough to produce the less torque required by the new load.
So back e.m.f regulates the flow of armature current and it automatically alters the armature current to meet the load requirement. This is the practical significance of the back e.m.f.
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What would happen if the DC Motor is directly switched on to the supply without any starter? (HPTU 2022)
A starter is necessary to start a DC motor because it restricts the initial high armature current that exists, as the value of starting back-EMF is zero. Without a starter high current flow through the armature at starting.
Which arc the different methods of electric braking in DC motors? Describe any one of them. (KTU 2021)
- Dynamic Braking of dc Motors . In this method of braking, the armature terminals of the motor are disconnected from the supply terminals and immediately connected across a resistor (rheostat) while the field winding is kept energised.
- Plugging or Reverse Current Braking. Plugging or plug stopping of a dc shunt or a separately excited motor means bringing the motor to standstill quickly by creating a reverse torque on the armature.
- Regenerative Braking. In both Rheostatic braking and plugging, the energy stored in the rotating system is wasted. In regenerative braking mechanical energy of the rotating system is converted into electrical energy and is returned to the supply source.
Regenerative Braking.
Suppose a dc motor driven car is moving down a load. The speed of the drive motors may become more than its no-load speed due to gravitational force. Under such a condition, the back emf will be more than the supply voltage and hence the dc motor will start working as a generator. The direction of current will reverse because Eb > V and hence the machine will return energy to the supply mains. The direction of electromagnetic torque will reverse and act as a brake to the speeding motor.
Regenerative braking is applicable in holding a descending load with high potential energy, e.g., a metro-train going down the slope; dc motors driving loads such as elevators, cranes, hoists, etc. In all these cases the motor speeds up, Eb becomes greater than V and hence the motor, instead of drawing current from the supply mains returns power to the mains.
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Write the significance and condition for Hopkinson's test. (AU 2022, 2 marks)
This requires two identical machines mechanically coupled to each other. One of the machines will work as a motor and drive the other machine which will work as a generator. The generator will feed back power to the motor. The power to be drawn from the supply is only for supplying the losses in the machines. Large machines can therefore be tested under load conditions without spending much energy from the supply.
The main disadvantage of this method is that two identical machines are required.
Compare brake teat with Swinburne's teat of DC machine. (AU 2023, 2 marks):
Swinburne's test is Indirect method of testing and brake tests are direct method of testing.
Hopkinson’s test conducted on two identical machines gave the following test result: field currents 5 A and 4.2 A, line voltage 250 V, motor armature current 380 A, line currents excluding both the field currents is 50 A. Calculate the efficiency of both machines. Take Ra (each machine) = 0.02 ohms. (AKTU 2022, JNTUH 2022)
See connections.
Generator arm. Cu loss = 3302 × 0.02 = 2,178 W
Power drawn from supply = 250 × 50 = 12,500 W
Stray losses for the two machines = 12,500 − (2,888 + 2,178) = 7.434 W
Stray losses per machine = 7,434/2 = 3,717 W
Motor Efficiency
Arm. Cu loss = 2,888 W
Field Cu loss = 250 × 4.2 = 1050 W
Stray losses = 3,717 W
Total loss = 2,888 + 1050 + 3,717 = 7,655 W
Motor input = 250 × 380 + 250 × 4.2 = 96,050 W
Motor output = 96,050 − 7,655 = 88,395 W
∴ η = 88,395/96,050 = 0.9203 or 92.03%
Generator Efficiency
Arm. Cu loss = 2,178 W
Field Cu loss = 250 × 5 = 1250 W
Stray losses = 3,717 W ; Total losses = 7,145 W
Generator output = 250 × 330 = 82,500 W
Generator input = 82,500 + 7,145 = 89,645 W
∴ η = 82,500/89,645 = 0.9202 or 92.02%
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