*1.1 Lorentz force on a current carrying conductorThe force is greatest when the conductor is perpendicular to the field (Fig. 1).When a current-carrying conductor is placed in a magnetic field, it is subjected to a force, known as electromagnetic force, or Lorentz force.This force is of fundamental importance because it constitutes the basis of operation of motors, of generators, and of many electrical instruments.The magnitude of the force depends upon the orientation of the conductor with respect to the direction of the field.

*The force acting on a straight conductor is given by F=BIl sin The force is zero when it is parallel to it (Fig. 2). Between these two extremes, the force has intermediate values. where F = force acting on the conductor [Newton, N] B =flux density of the field [Tesla, T] Tesla=Weber/m2 l = active length of the conductor [meter, m] I = current in the conductor [Ampere, A] = angle between the conductor and the flux

*1.2 Direction of Lorentz forceThe direction of Lorentz force is given by Flemings left hand rule.Flemings left hand ruleStretch the thumb, first (fore) and second (middle) fingers of the left hand so that they are mutually perpendicular to each other. If the first finger represents the direction of field (B) and the second finger the direction of current (I), the thumb points in the direction of force (F) on the conductor.

*1.3 Voltage induced in a conductorIn many motors and generators, the coils move with respect to a flux that is fixed in space. The relative motion produces a change in the flux linking the coils and, consequently, a voltage is induced according to Faraday's law. The value of the induced voltage is given by E = Blv sin where E = induced voltage [Volt, V] B = flux density [T] l = active length of the conductor in the magnetic field [meter, m] v = relative speed of the conductor [meter/second, m/s] = angle between the conductor and the flux

*Induced EMF

*Magnet:It is a substance having the properties of attracting iron and its alloys. Magnetic field: It is defined as the region of space around a magnet in which there exists a magnetic force (either force of attraction or force of repulsion). Magnetic lines of force: These are the imaginary lines of force in a magnetic field which start from the North pole, pass through air or magnetic medium, and end at South pole. Magnetic circuit: It is a continuous path occupied by magnetic lines of force.

*Magnetic flux: It is the number of lines of magnetic force crossing the space occupied by a magnetic field().Magneto-motive-force (m.m.f): It is the force which drives or tries to drive magnetic flux to flow through magnetic circuit and is the analogy to e.m.f. in electric circuit. It is equal to the effective current flow applied to the core (Ni). Its unit is Ampere-turn. Reluctance (S or ) : This is the term given to the material characteristics that resist the generation of magnetic flux inside it. It is the opposition offered by a magnetic path to the establishment of a magnetic flux and is the analogy to resistance in electric circuit. (AT/Wb) Reluctance, S =Resistance R =

*1.4 Direction of induced voltageThe direction of induced voltage is given by Flemings right hand rule.Flemings right hand ruleExtend the thumb, first and second fingers of the right hand so that they are mutually perpendicular to each other. If the thumb represents the direction of motion and the first finger the direction of B, then the second finger represents the direction of the induced voltage E.

*1.5 Direction of magnetic field produced by currentThe direction of magnetic field due to a current can be obtained by applying right hand rule.Magnetic field due to current in a coilIf the thumb of the right hand points in the direction of magnetic field, the other fingers curl in the direction of current.Maxwell's Cork- Screw Rule: Imagine a right-handed corkscrew being rotated along the wire in the direction of the current. The direction of rotation of the thumb gives the direction of the magnetic lines of force.

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Faradays lawLenzs lawThe polarity of the voltage induced by a changing flux tends to oppose the change in flux that produced the induced voltage. , the negative sign is due to Lenzs lawCombining Faradays law and Lenzs law we have:

*1.6 Magnetic circuitsMagnetic circuits consist predominantly of iron paths of specific geometry which serves to confine the flux; air gaps may be included.The magnetomotive force (MMF) of the coil produces a flux which is confined to the iron and to that part of the air having effectively the same cross sectional area as the iron.

The electrical equivalent of this magnetic circuit is shown in Fig. 3 (b).

*Fig. 3(c) shows two more magnetic circuits.

*For a magnetic circuit: F = R (similar to E=IR in electrical circuits) where F is the magnetomotive force (MMF) in Ampereturns (At) is flux in Weber (Wb) R is reluctance in At/Wb

*1.7 Magnetization curve or B-H CurveWhenever a magnetic flux exists in a body or component, it is due to the presence of a magnetic field intensity, H or magnetic field strength (It is proportional to the length of a conductor and the amount of electrical current passing through the conductor).

B= 0 H where, B = flux density [T] H = magnetic field intensity [At/m] 0 = magnetic constant or permeability of vacuum B-H curve of vacuumIn vacuum, the magnetic flux density B is directly proportional to the magnetic field intensity H, and is expressed by the equation

*Nonmagnetic materials such as copper, paper, rubber, and air have B-H curves almost identical to that of vacuum.The B-H curve of vacuum is a straight line. A vacuum never saturates, no matter how great the flux density may be (Fig. 4).

*B-H curve of a magnetic material The flux density in a magnetic material also depends upon the magnetic field intensity to which it is subjected. Its value is given by

B= 0 r H

where B, 0 and H have the same significance as before, and r is the relative permeability of the material.The value of r is not constant but varies with the flux density in the material. Consequently, the relationship between B and H is not linear, and this makes the equation, B= 0 r H rather impractical to use.

*Fig. 5 shows typical B-H curves of three materials commonly used in electrical machines: silicon iron, cast iron, and cast steel. The curves show that a magnetic field intensity of 2000 At/m produces a flux density of 1.4 T in cast steel but only 0.5 T in cast iron.

Example 1A metal ring has a mean diameter of 21 cm and cross sectional area of 10 cm2. The ring is made up of semicircular sections of cast iron and steel, with each joint having a reluctance equal to an air gap of 0.2 mm. Find the ampere turns required to produce a flux of . The relative permeabilities of steel and cast iron are 800 & 166 respectively.*

Air gapLength (L)Reluctance (R)*Ampere turns required SteelLength (L)Reluctance (R) Ampere turns requiredCast ironProceeding in the same manner as in case of steel the ampere turns required in case of cast iron can be found out as 1265 At. Total ampere-turns required At/Wb At At/Wb At = 255 +263 + 1265 = 1783 At

Example 2A magnetic circuit made of iron is shown below. It has a coil of 1000 turns on its central limb. Find the current that the coil should carry to produce a flux of 2.5 mWb in the air gap. The cross sectional area of the central limb and each of the side limbs is 25 cm2. The magnetization curve for iron is as follows.B 0.2 0.5 0.7 1.0 1.2 (Wb/m2)H 300 540 650 900 1150 (At/m)*

*SolutionCentral limbFlux density in the central limb and air gap

From the magnetization curve of iron for B= 1 Wb/m2, H= 900 At/mAmpere turns for the central limb Air gapH in air gap Ampere turns for air gap

*Side limb (BADC)Flux density in the side limbFrom the magnetization curve of iron for B= 0.5 Wb/m2 H= 540 At/mAmpere turns for the side limb Total ampere turns in the closed loop ABCDACurrent through the coil:

*1.8 Hysteresis loop Transformers and most electric motors operate on alternating current. In such devices the flux in the iron changes continuously both in value and direction. The magnetic domains are therefore oriented first in one direction, then the other, at a rate that depends upon the frequency.

*Thus, if the flux has a frequency of 50 Hz, the domains describe a complete cycle every 1/50 of a second, passing successively through peak flux densities +Bm and -Bm as the peak magnetic field intensity alternates between +Hm and -Hm. If we plot the flux density B as a function of H, we obtain a closed curve called hysteresis loop (Fig. 6).

*1.8.1 Hysteresis lossThe magnetic material absorbs energy during each hysteresis cycle and this energy is dissipated as heat. The amount of heat released per cycle is equal to the area of the hysteresis loop. To reduce hysteresis losses, we select magnetic materials that have a narrow hysteresis loop, such as the grain-oriented silicon steel used in the cores of transformers.

*1.9. Eddy current lossFrom Faradays law we see that time varying magnetic fields give rise to electric fields. In magnetic materials these electric fields result in eddy currents, which circulate in the core material. The I2R loss associated with eddy currents is known as eddy current loss.The eddy current loss can be reduced by splitting the core into thin sections and insulating the sections from one another. The voltage induced in each section is much less than what it was before, with the result that the eddy currents and the corresponding losses are considerably reduced.

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* 1.10 Machine ratingsThe nameplate of an electrical machine indicates the power, voltage, speed, and other details about the machine. These ratings, or nominal characteristics, are the values guaranteed by the manufacturer. For example, the following information is punched on the nameplate of a 100 kW DC generator:Power100 kWSpeed 1200 r/minVoltage250 VTypecompoundExciting current 20 A ClassBTemperature rise500 C

*These specifications tell us that the machine can deliver, continuously, a power of 100 kW at a voltage of 250 V, without exceeding a temperature rise of 50 C. Power100 kWSpeed 1200 r/minVoltage250 VTypecompoundExciting current 20 A ClassBTemperature rise500 CThe class B designation refers to the class of insulation used in the machine. Examples of class B insulators are mica, glass fiber etc. It can therefore supply a load current of 100000/250 = 400 A. It possesses a series winding, and the current in the shunt field is 20 A. In practice, the terminal voltage is adjusted to a value close to its rating of 250 V. We may draw any amount of power from the generator, as long as it does not exceed 100 kW and the current is less than 400 A.

*The power rating of a transformer is equal to the product of the nominal voltage times the nominal current of the primary or secondary winding. However, the result is not expressed in watts, because the phase angle between the voltage and current may have any value at all, depending on the nature of the load. Consequently, the power-handling capacity of a transformer is expressed in voltamperes (VA), in kilovoltamperes (kVA) or in megavolamperes (MVA), depending on the size of the transformer. The temperature rise of a transformer is directly related to the apparent power that flows through it. This means that a 500 kVA transformer will get just as hot feeding a 500 kvar inductive load as a 500 kW resistive load. The rated kVA, frequency, and voltage are always shown on the nameplate. In large transformers the corresponding rated currents are also shown.

*1.11 LossesWhenever a machine transforms energy from one form to another, there is always a certain loss. The loss takes place in the machine itself, causing (1) an increase in temperature and (2) a reduction in efficiency.

From the standpoint of losses, electrical machines may be divided into two groups: those having revolving parts (motors, generators, etc.) and those do not (transformers, reactors, etc.). Electrical and mechanical losses are produced in rotating machines, while only electrical losses are produced in stationary machines.Here we shall analyze the losses in DC machines, but the same losses are also found in most machines operating on alternating current. The study of power losses is important because it gives us a clue as to how they may be reduced.

*1.11.1 Mechanical lossesMechanical losses are due to bearing friction, brush friction, and windage. The friction losses depend upon the speed of the machine and upon the design of the bearings, brushes, commutator, and slip rings. Windage losses depend on the speed and design of the cooling fan and on the turbulence produced by the revolving parts. It draws in cool air from the surroundings, blows it over the windings, and expels it again through suitable vents. Rotating machines are usually cooled by an internal fan mounted on the motor shaft. In the absence of prior information, we usually conduct tests on the machine itself to determine the value of these mechanical losses.

*1.11.2 Electrical lossesElectrical losses are composed of the following:1. Conductor I2R losses (sometimes called copper losses)2. Brush losses 3. Iron losses

*Conductor losses: The losses in a conductor depend upon its resistance and the square of the current it carries. in whichR= resistance of conductor []L= length of conductor [m]A = cross section of conductor [m2] = resistivity of conductor at temperature t [-m]0 = resistivity of conductor at 0 C [-m] = temperature coefficient of resistance at 0C [C] The following equations enable us to determine the resistance at any temperature and for any material:The resistance, in turn, depends upon the length, cross section, resistivity, and temperature of the conductor.

*In DC motors and generators, copper losses occur in the armature, the series field, the shunt field, the commutating poles, and the compensating winding. These I2R losses show up as heat, causing the conductor temperatures to rise above ambient temperature.

*Brush losses: The I2R losses in the brushes are negligible because the current density is only about 0.1 A/mm2, which is far less than that used in copper. However, the contact voltage drop between the brushes and commutator may produce significant losses. The drop varies from 0.8 V to 1.3 V, depending on the type of brush, the applied pressure, and the brush current (Fig. 7).

*Iron losses: Iron losses are produced in the core of a machine. 1.12 Temperature riseThe temperature rise of a machine or device is the difference between the temperature of its warmest accessible part and the ambient temperature.

They are due to hysteresis and eddy currents.

Iron losses depend upon the magnetic flux density, the speed of rotation, the quality of the steel, and the size of the armature.

Consequently, temperature rise is a very important quantity.It also has a direct bearing on its useful service life.

Temperature rise has a direct bearing on the power rating of a machine or device.

*1.13 Life expectancy of electric equipmentApart from accidental, electrical and mechanical failures, the life expectancy of electrical apparatus is limited by the temperature of its insulation: the higher the temperature, the shorter its life.

Tests made on many insulating materials have shown that the service life of electrical apparatus diminishes approximately by half every time the temperature increases by 10C.

This means that if a motor has a normal life expectancy of eight years at a temperature of 105 C, it will have a service life of only four years at a temperature of 115C, of two years at 125C, and of only one year at 135C.

The factors that contribute most to the deterioration of insulators are (1) heat, (2) humidity, (3) vibration, (4) acidity, (5) oxidation, and (6) time.

*In crystallizing, organic insulators become hard and brittle.

Because of these factors, the state of the insulation changes gradually; it slowly begins to crystallize and the transformation takes place more rapidly as the temperature rises.

Eventually, the slightest shock or mechanical vibration will cause them to break.

Under normal conditions of operation, most organic insulators have a life expectancy of eight to ten years provided that their temperature does not exceed 100C. Low temperatures are just as harmful as high temperatures are, because the insulation tends to freeze and crack.

Special synthetic organic insulators have been developed which retain their flexibility at temperatures as low as -60C.

*1.14 Cooling To prevent rapid deterioration of the insulating materials inside a transformer, adequate cooling of the winding and core must be provided.

Oil carries the heat away to the tank, where it is dissipated by radiation and convection to the outside air (Oil is a much better insulator than air ).As the power rating increases, external radiators are added to increase the cooling surface of the oil filled tank. Oil circulates and moves through the radiators, where the heat is again released to surrounding air. For still higher ratings, cooling fans blow air over the radiators.

Distribution transformers below 200 kVA are usually immersed in mineral oil and enclosed in a steel tank.

Such dry type transformers are used inside buildings.

Larger transformers can be built the same way, but forced circulation of clear air must be provided.

Indoors transformers below 200 kVA can be directly cooled by the natural flow of the surrounding air.

*Fig. Transformers with radiator tank and cooling fins

*Oil provides electrical insulation as well as cooling in Transformers.

An electrical insulator is a material that does not respond to an electric field and completely resists the flow of electric charge. In practice, perfect insulators do not exist. Therefore, dielectric materials with high dielectric constants are considered insulators.

Pure water is an insulator, but in the "real world", water usually has impurities, that actually make it a good conductor.

A dielectric material is a substance that is a poor conductor of electricity, but an efficient supporter of electrostatic field.

For insulation of heat, thermal insulation is used.

*1.15 Three phase systems Most of the generation, transmission and heavy power utilization of electric energy involves three phase systems i.e. systems in which three sources equal in magnitude but differing in phase from each other are available.

A three phase load is one which can utilize the output of a three phase source.

A three phase source is one which has three equal voltages which are 1200 out of phase with one another.

The three voltages are usually generated in the same machine.

*Note that the phasor sum of the three voltages is zero. Three voltage sources forming a three phase system are shown in Fig. (a). The phasor diagram of these voltages is shown in Fig. (b).

*The instantaneous three phase voltages, shown in Fig. 9, are as follows:

*Delta ConnectionStar Connection1.15.1 Y and connections

*1.15.1 Y and connectionsThere are two possibilities for the utilization of voltages generated in this manner: the six terminals a, a/, b, b/, c, c/ of the winding may be connected to three independent single phase systems or three phases of the winding may be interconnected and used to supply a three phase system.

The three phases of the winding can be interconnected in two possible ways, as shown in Fig 10. Terminals a/, b/ and c/ may be joined to form the neutral o, yielding a Y or star connection, or terminals a and b/, b and c/ and c and a/ may be joined individually, yielding a (delta) connection.

*In Y connection, a neutral connector may or may not be brought out.

If a neutral connector exists, the system is a four wire three phase system; if not it is a three wire three phase system.

In a delta connection no neutral exists and only a three wire three phase system can be formed.

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*.The three phase voltages are equal and phase displaced by 120 electrical degrees, a general characteristic of a balanced three phase system.

Furthermore, the impedance in any one phase is equal to that in any other phase, so that the resulting phase currents are equal and displaced from each other by 120 electrical degrees.

Likewise, equal power and equal reactive power flow in each phase.

* An unbalanced three phase system, on the other hand, may lack any or all of these equalities and 1200 displacements. The phase order or phase sequence is abc.

*1.15.2 Relationship between line voltage and phase voltage

*From the voltage phasor diagram, The current phasor diagram is:

*The voltage and current phasor diagrams for delta connection are shown in Fig. 13 and Fig. 14. It can be shown that for a delta connection:

*1.15.3 Power triangleS = Apparent power (VA) = VI = I2Z = V2/ZP = Real/active power (W) = VI cos = I2R = V2/R Q = Reactive power (var) = VI sin = I2X = V2/X = Power factor angle

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1.15.4 Three phase power3 phase real power (for Y or delta connected system)3 phase reactive power (for Y or delta connected system)3 phase voltamperes (for Y or delta connected system)

*Example 3Three impedances of value as shown below. For balanced line to line voltages of 208 V, find the line current, the power factor, the total power, reactive power and voltamperes.are connected in Y,

*SolutionThe line to neutral voltage across any one phase:Hence, phase current Power factor lagging

*1.16 Rotating magnetic fieldThe ac windings used in polyphase induction and synchronous machines produce magnetic fields of constant amplitude rotating at a uniform speed around the air gap circumference when the windings carry polyphase currents. This fundamental fact can be demonstrated by considering the elementary three phase two pole winding of Fig. It consists of three coils 1200 apart in space, each coil forming one phase of a three phase system.

*The three coils have been represented as (a, -a), (b, -b) and (c,-c). Under balanced 3-phase conditions the instantaneous three phase currents are:The phase sequence is assumed to be abc.

The instantaneous currents are shown in Fig. 17.

For the balanced three phase currents the resulting rotating MMF can be shown graphically.

*Consider the instant t = 0, the moment when the phase a current is at its maximum value Im. At this moment the currents ib and ic are both Im/2 in the negative direction (Fig.17).

*It represents a sinusoidal space wave with its positive half wave centered on the axis of phase a and having an amplitude 3/2 times that of the phase a contribution alone.

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*At a later instant the currents in phases a and b are a positive half maximum and that in phase c is a negative maximum. The individual MMF components and their resultant are shown in Fig. 18 (b). the resultant has the same amplitude as at now centered on the axis of phase c i.e. it has rotated counterclockwise 60 electrical degrees in space.

but it isSimilarly, at (when the phase b current is a positive maximum and the phase a and c currents are a negative half maximum) the same resultant MMF distribution is again obtained, but it it hasrotated counterclockwise 60 electrical degrees still farther and is now aligned with the magnetic axis of phase b as shown in Fig. 18 (c). As time passes, the resultant MMF wave retains its sinusoidal form and amplitude but shifts progressively around the air gap. The shift corresponds to a field rotating uniformly around the circumference of the air gap.

*The speed of the rotating magnetic field is given by:wherens= speed of rotating magnetic field also called synchronous speed (r.p.m.)f = system frequency (cycles/sec or Hz)p = number of poles of the machine

*1.17 Linear motion, angular motion

Linear motionAngular motionLinear displacement (s)Angular displacement ()

Linear velocity (v)Acceleration (a)Angular acceleration ()Force (F) = mass (m) x Torque (T) = moment of Inertia (I) acceleration (a) x angular acceleration () Angular velocity () Torque (T) = Force (F) x radial distance (r)

If the speed of a machine is n revolutions per second, its angular velocity () will be 2n radians per second.Similarly, if the frequency of voltage or current is f cycles per second, the angular frequency () will be 2f radians per second.Power (P) = T x 2n* Work done (W) = Force (F) Work done (W)= Torque (T) displacement (s) angular displacement() Linear motionAngular motion

*Summary of the two Systems (Star & Delta)

StarDeltaSimilar coil ends connectedDissimilar ends connectedFor balanced or unbalanced loadsUsually used on balanced loadsVL = 3 VP, IL = IPVL = VP, IL = 3 IPTwo values of voltage possibleOnly one voltageStar point for earthingNo earth pointVL leads VP by 30oIL lags IP by 30o

*Additional examplesProblem 1 A conductor 4 m long carrying a current of 100 A is placed in a magnetic field whose density is 0.5 T. Calculate the force on the conductor (a) if it is perpendicular to the lines of force (b) if it makes an angle of 450 with the lines of force.Solution(a) F = B I l sin= 0.5 100 4 1 = 200 N(b) = 450F=B I l sin450 = 0.5 100 4sin450 = 0.5 100 4 0.707 = 141.4 N

*Problem 2 The stationary conductors of a large generator have an active length of 3 m and are cut by a field of 0.6 teslas, moving at a speed of 200 m/s. Calculate the maximum voltage induced in each conductor.Solution Emax = Blv =0.6 x 3 x 200= 360V

Problem 3A cast steel magnetic structure made of a bar of section 2cm 2cm is shown. Find the current that the 500 turn coil on the left limb should carry so that a flux of 2 mWb is produced in the right limb. Given as

*Solution

or,Ampere-turns for the left limb of 25 cm lengthAmpere-turns for the central left limb of 15 cm lengthAmpere-turns for the closed path ABCD

Therefore, current through the magnetizing coil is,

*Problem 4Three impedances of values are connected in delta, as shown in Fig. For balanced line to line voltages of 208 V, find the line current, the power factor, the total power, reactive power and voltamperes.SolutionThe voltage across any one phase is equal to line to line voltage. Consequently,Power factor = lagging

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