Unit 1-1: Electromagnetism and Transformers PDF

Introduction to Electrical Engineering F.Y. B. TECH. _________________________________________________________________________________________ Unit I: Electromagnetism and Transformers Syllabus Fundamentals of...

Introduction to Electrical Engineering F.Y. B. TECH. _________________________________________________________________________________________ Unit I: Electromagnetism and Transformers Syllabus Fundamentals of Magnetic Circuits, B-H characteristics. Faraday‟s Laws of Electromagnetic Induction and types of EMF‟s. Transformer: Construction, Working principle, Ideal and practical transformer. Equivalent circuit, losses in transformers, Regulation and efficiency. Auto- transformer. 1.1. Introduction A conductor carrying current is always surround all along its length by a magnetic field. This important relationship between magnet and electric current forms the basis of electromagnetism. Electromagnetism deals with magnetic effects of electric current. Magnetism and its Effects A. Properties of a Magnet (i) It attracts small pieces of iron (ii) When freely suspended by a piece of silk fiber, its north pole points the north direction and the South Pole the south direction. (iii) Like magnetic poles repel and unlike poles attracts each other. B. Laws of Magnetism : There are two fundamentals of laws of magnetism: First law: It state that like magnetic poles repel and unlike poles attracts each other. Second law: (Coulomb’s Law) It state that force (F) exerted by one pole on another pole is directly proportional to the product of the pole strength and inversely proportional to the square of the distance (d) between them. It also depends on the nature of the medium surrounding the poles. Mathematically, we can write: ( ) ( ) where m1 and m2 are pole strength and K is a constant whose value depends on the nature of the surrounding medium. 1.1.1. Magnetic Field The region around a magnet within which the effect or influence of magnet can be experienced is called magnetic field. JSPM’s RAJARSHI SHAHU COLLEGE OF ENGINEERING, PUNE 1 Introduction to Electrical Engineering F.Y. B. TECH. _________________________________________________________________________________________ 1.1.2. Magnetic lines of force or Lines of Magnetic Flux : These are the imaginary lines around a magnet used to represent its magnetic field. A lines of force may be defined as a line along which an isolated N- pole would travel if free to move in a magnetic field and it is such that the tangent at any point gives the direction of the resultant force at that point. Properties of lines of force : 1. Lines of force are always in the form of closed curves originating from N – pole and terminating on a S – pole outside the magnet and S – pole to N – pole inside the magnet 2. Lines of force never cross one another. 3. Parallel lines of force acting in the same direction repel on another. 4. Lines of force always try to contract and thus behave like stretched elastic threads. 5. Magnetic lines of force always take the path of least reluctance (opposition). 1.1.3. Pole Strength The magnitude of fore exerted by one magnet on another magnet gives rise to the idea of pole strength. If under exactly identical conditions two magnetic poles are exerting equal force on another poles, then they said to have equal pole strengths. Unit of pole strength is weber (also called as a unit pole). 1.1.4. Magnetic Flux (ф) The total number of magnetic lines of force in any particular magnetic field is called the magnetic flux. The unit of magnetic flux is weber (108 lines). The flux which the unit north pole radiates out is known as one weber. 1.1.5. Magnetic Flux Density (B) The flux per unit area (A) in a phase at right angles to the flux is known as flux density. Its unit is weber/ meter2 or Tesla (T), 1.1.6. Magnetic Field Strength (H) The force experienced by a unit North Pole (i.e. N – pole with one weber pole strength) placed at any point in a magnetic field is known as magnetic field strength (H) at that point. Higher the value of this force, stronger the field. Its unit is Newton / weber (N / wb) or Ampere Turns per meter. The mathematical expression for calculating magnetic field strength is, JSPM’s RAJARSHI SHAHU COLLEGE OF ENGINEERING, PUNE 2 Introduction to Electrical Engineering F.Y. B. TECH. _________________________________________________________________________________________ 1.2. Magnetic Field Due to Straight Conductor We know that a straight conductor will have a magnetic field all along its length when it is carrying a current. The lines of force are in the form of concentric circles in planes at right angles to the conductor and their direction is dependent on the direction of the current producing. This can be very well demonstrated by the straight conductor passed through a card as shown in the figure (a). When iron filling is sprinkled on the card board, They are themselves in centric circles around the current carrying conductor. The direction of the field can be found by replacing the fillings by compass needles. Figure (b) illustrates the conventional representation of current carrying conductor along with the direction of current flowing through it and the direction of magnetic field around it. Here the small circle represents the conductor in section. The cross on it represents the rear view of the feathered end of an arrow and indicates a current flowing in to the plane of the paper (i.e. away from the observer) in a conventional manner, on the other hand, the dot represents the tip of an arrow and indicates a current flowing out of the plane of the paper (i.e. towards the observer). The direction of the field may be quickly determined by means of the following rules: The Right Hand Gripping Rule Grip the current carrying conductor in the right hand with the thumb outstretched parallel to the conductor and pointing in the direction of the current as shown in the figure. The curled fingers then point in the direction of the magnetic field around the conductor. The Corkscrew Rule Imaging a right handed corkscrew or wood screw as shown in the figure below, placed alongside of the current carrying conductor with its axis parallel to the conductor and tip pointed in the direction of the current flow. Then the direction of the magnetic field is given by the direction of in which the screw must be turned to advance it in the direction of the current. JSPM’s RAJARSHI SHAHU COLLEGE OF ENGINEERING, PUNE 3 Introduction to Electrical Engineering F.Y. B. TECH. _________________________________________________________________________________________ 1.2.1 Magnetic field strength Due to long straight conductors: The magnetic field strength at a point „r‟ meter from the group of N long conductors each carrying current I amperes in same direction as shown in the figure here, is 1.2.2 Magnetic field strength Due to solenoid: The magnetic field strength along the axis of the solenoid as shown in the figure below is Where: N=Number of turns of the solenoid, I= current flowing in the coil around solenoid in Ampere, L= Length of solenoid in meters. 1.3.1. Permeability or Absolute Permeability (µ) It is defined as the prosperity of the magnetic circuit to allow the magnetic lines of force to pass through it. Also it can be stated as It is a property of magnetic material which decides how much magnetic line of forces required allow to pass through it or it is degree of measure by which the magnetic line of forces can penetrate the magnetic material. It is denoted by µ and measured in H/m. 1.3.2. Permeability of Free Space (µ0) It is defined as the ratio of flux density to magnetic field strength producing that flux density. it is degree of measure by which the magnetic line of forces can penetrate the free space. It is denoted by µ0 and measured in H/m. µ0 = B / H = 4π x 10-7 H / m 1.3.3. Relative Permeability (µr ) It is defined as the ratio of the flux density produced in a material to the flux density produced in the vacuum (or free space ) by the same magnetic field strength under identical conditions it is denoted by µr. It is unit less quantity. for air and nonmagnetic material µr = 1 1.4.1. Magneto Motive Force (MMF) It is the work done in joules on unit magnetic poles in taking it once around a closed magnetic circuit. MMF = N.I = ф.s = H.I Thus MMF is given by the product of the number turns „N‟ and current „I‟ flowing in the coil of the solenoid. JSPM’s RAJARSHI SHAHU COLLEGE OF ENGINEERING, PUNE 4 Introduction to Electrical Engineering F.Y. B. TECH. _________________________________________________________________________________________ 1.4.2. Reluctance (S) Reluctance is the resistance offered by the material to the passage of the magnetic flux through it. It is given by the following equations Where, L is the length of the magnetic circuit in meter, a is the cross – sectional area in m2, µ0 is permeability of free space and µr is relative permeability. ф is the flux in the magnetic circuit in weber. 1.4.3. Permeance It is the reciprocal of reluctance and it is defined as the property of the magnetic circuit due to which it permits the passage of magnetic flux through it Permeance = 1 / Reluctance (wb / A) 1.4.4. Relation between B and H Everywhere within the magnetic field B is proportional to H. The relation between B and H is depends on the type of material within which the relation is required to be find. B=µ H 1.5. Comparison between Magnetic and Electric Circuits Similarities Sr. No. Magnetic Circuits Electric circuit 1. Magnetic circuit is the path of Electric circuit is the path of magnetic flux. electric current. 2. MMF is responsible to produce EMF is responsible to produce current in magnetic circuit. current in electric circuit. 3. Ohms low for Magnetic circuit Ohms low for Electric circuit F=ØS V = IR 4. Opposition to the flow of flux is Opposition to the flow of current called as reluctance. is called as resistance. 5. Flux = MMF / Reluctance Current = EMF / Resistance Ф = NI /s I = V/R 6. Flux density B = ф /a Current density, s = l / a amp /m2 7. Permeance = 1/ s Conductance = 1/ R 8. Kirchhoff‟s laws for electric circuit Kirchhoff‟s laws for electric circuit 1) Σ Ø =0 1) ΣI =0 2) ΣØS + ΣF =0 2) ΣIR + ΣE =0 JSPM’s RAJARSHI SHAHU COLLEGE OF ENGINEERING, PUNE 5 Introduction to Electrical Engineering F.Y. B. TECH. _________________________________________________________________________________________ Dissimilarities Sr. No. Magnetic Circuit Electric circuit 1. Flow of flux in magnetic circuit is Current in electric circuit is actual just imagination; actually nothing flow of electrons. is flowing in magnetic circuit. 2. Energy is required to create the Energy is required to create and magnetic flux in magnetic circuit maintain electric current in but do not required maintain it electric circuit. 3. Magnetic line of forces always Electric line of forces never form form closed loops closed loops 4. Magnetic flux can flow through Electric current never flows in free free space. space 5. There is no any insulator which There are electric insulators to avoid flow of magnetic flux avoid flow of electric current Numerical Ex. 1) An iron ring has circular cross section of 8cm in diameter and average circumference of 100 cm. The ring is uniformly wound with a coil of 700 turns. Find current required in coil to produce 2 mWb flux inside the ring, if relative permeability of the iron is 900. Solution: d = 8 cm = 8 x 10-2 m, l = 100 cm =1 m, Ø = 2 mWb = 2 x 10-3 Wb, N = 700, µr = 900 A= π/4 x (d)2 = π/4 x (8 x 10-2)2 = 50.2654 x 10-4 m2 F= NI = ØS I = ØS/N I= Ø I= Ø I = 2 x 10-3 x I = 0.5025 A Ex. 2) An iron ring has inner diameter of 30 cm and outer diameter of 34 cm. the ring has uniform square cross section of side 4 cm. The ring is uniformly wound with coil carrying current of 2 A. If flux inside the ring is 2 mWb, then find number of turns of the coil. Assume relative permeability of iron as 900. JSPM’s RAJARSHI SHAHU COLLEGE OF ENGINEERING, PUNE 6 Introduction to Electrical Engineering F.Y. B. TECH. _________________________________________________________________________________________ Solution: D1 = 30 cm = 30 x 10-2 m, D2 = 34 cm = 32 x 10-2 m, d= 4 cm = 4 x 10-2 m, I = 2 A,Ø = 2 mWb = 2 x 10-3 Wb, µr = 900 Mean diameter of the ring D = (D1 + D2 )/2 = 64/2 = 32 cm = 32 x 10-2 m Mean length of the magnetic circuit = π D = π x 32 x 10-2 m = 1 m. Across section area A = (side)2 = (4 x 10-2)2 = 16 x 10-4 m2. F= NI = ØS N = ØS/I N= Ø N= Ø N = 2 x 10-3 x N = 552.9 Turns= 553 Turns Ex. 3) An iron ring of circular cross section 3 cm2 and mean diameter 20cm is wound with 500 turns of wire and carries current of 2.09 A to produce magnetic flux 0f 0.5 mWb in the ring. Find relative permeability of the material. Solution: A= 3 cm2 = 3 x 10-4 m2, D = 20 cm = 20 x 10-2 m, N=500, I= 2.09 A, Ø= 0.5 mWb= 0.5 x 10-3 Wb. Mean length l = π D = π x 20 x 10-2 = 0.6283 m Reluctance S S (1) Also F = NI = ØS S = NI/ Ø = 500 x 2.09/ 0.5 x 10-3 = 2 x 106 AT/Wb (2) From equations (1) and (2) 2 x 106 = = 833.33 Ex.4) An iron ring has8 cm diameter is made up of round iron of diameter 1 cm and permeability 900, has an air gap of 2 mm wide. It consists of winding with 400 turns carrying a current of 3.5 A. Find MMF, total reluctance, flux and flux density in the ring. Solution: D=8 cm = 8 x 10-2 m, d = 1 cm = 10-2 m, µr = 900, lg = 2 mm = 2 x 10- 3 m, N = 400, I= 3.5 A. Mean length of iron material lm = π D = π x 8 x 10-2 = 0.2493 m. Area of cross section A = π/4 x (d)2 = π/4 x (10-2)2 = 0.7853 x 10-4 m2. MMF produced F =NI = 400 x 3.5 = 1400 AT. Reluctance of iron material Sm = 28.06947 x 105 AT/Wb JSPM’s RAJARSHI SHAHU COLLEGE OF ENGINEERING, PUNE 7 Introduction to Electrical Engineering F.Y. B. TECH. _________________________________________________________________________________________ Assuming area of cross section of iron = area of cross section for air gap Reluctance of air gap Sg = 202.667 x 105 AT/Wb. Total reluctance S = Sm + Sg = 28.06947 x 105 + 202.667 x 105 S = 230.737 x 105 AT/Wb Flux Ø = NI/S Ø = 1400/230.737 x 105 Ø = 6.067 x 10-5 Wb Flux Density B = Ø/A B = 6.067 x 10-5/0.7853 x 10-4 B = 0.7725 T Ex. 5) An iron ring has circular cross section 8cm in diameter and average circumference of 100 cm. The ring is uniformly wound with a coil of 700 turns. Find current required in coil to produce 2 mWb flux inside the ring, if relative permeability of the iron is 900. If now a saw cut of 1 mm wide is made in the ring then calculate the current which will give same flux inside the ring. Solution: d = 8 cm = 8 x 10-2 m, l = 100 cm =1 m, Ø = 2 mWb = 2 x 10-3 Wb, N = 700, µr = 900 A= π/4 x (d)2 = π/4 x (8 x 10-2)2 = 50.2654 x 10-4 m2 F= NI = ØS I = ØS/N I= Ø I= Ø I = 2 x 10-3 x I = 0.5025 A Now a saw cut of 1 mm wide made inside ring which reduces length of iron part. Mean length of iron material lm = l - lg lm = 1- 1 x 10-3 = 99.9 x 10-2 m Total reluctance of the magnetic circuit S = Sm + Sg S = 3.34079 x 105 AT/Wb. JSPM’s RAJARSHI SHAHU COLLEGE OF ENGINEERING, PUNE 8 Introduction to Electrical Engineering F.Y. B. TECH. _________________________________________________________________________________________ It is required to maintain same flux inside the ring Ø = NI/S 2 x 10-3 = 700 x I/ 3.34079 x 105 I = 0.9545 A. Ex. 6) A magnetic circuit has mean length of 60 cm and material relative permeability 750. When a coil having 600 turns wound on the magnetic material produces flux density of 0. 5 T. Find current in the coil. Solution: l = 60 cm = 0.6 m, µr = 750, N = 600, B= 0.5 T. Magnetic field intensity H= B/µ H = 0.5 / (4π x 10-7 x 750) H = 530.785 AT/m MMF in magnetic circuit F=NI = Hl I = Hl/N I = 530.785 x 0.6/ 600 I = 0.530 A. Electromagnetic Induction 1.6. Faradays Laws for Electromagnetic Induction Faradays first law: Whenever magnetic flux linking with conductor or coil is changed, an EMF is always inducing in that conductor or coil. Faradays second law: The magnitude of the induced EMF is depends on the rate at which magnetic flux linking with conductor or coil is changed and it is given by Where, N is Number of turns of coil and negative sign is as per Lenzs law, which states that the induced EMF always acts in a direction to opposite to cause responsible to produce it. 1.7. Types of Induced EMF The EMF which is induced in conductor or coil may be a dynamically induced EMF or statically induced EMF 1.7.1. Dynamically Induced EMF The EMF which is induced in a coil or conductor due to actual physical moment between the conductor and magnetic field is called as dynamically induced EMF. Expression for Dynamically induced EMF: Consider a conductor which is placed within the magnetic field. Let, l =Active length of the conductor in meters B = Magnetic flux density where conductor is placed in T. v= Velocity component of motion of the conductor within magnetic field in m/s. JSPM’s RAJARSHI SHAHU COLLEGE OF ENGINEERING, PUNE 9 Introduction to Electrical Engineering F.Y. B. TECH. _________________________________________________________________________________________ If conductor is moving in some inclined direction with angle of inclination Ɵ with respect to the magnetic field, then the induced EMF becomes e = B l v Sin(Ɵ) Direction of this induced EMF is given by Flemings Right hand Rule which state that “Stretch three fingers of right hand mutually perpendicular to each other. If fore finger represent direction of flux, thumb represent direction of motion of conductor then middle finger represent direction of induced EMF”. 1.7.2. Statically Induced EMF: The EMF which is induced in a conductor or coil without any physical moment between the conductor and magnetic field is called statically induced EMF. The statically induced EMF is further classified into Self- Induced EMF and Mutually-Induced EMF. 1.7.2.1. Self-Induced EMF: The EMF which induced in a coil due change in its own magnetic field or due to change in its own current is called as self-induced EMF. Expression for Self-Induced EMF: Consider a coil connected across a source of EMF. Let, I= Current in coil Ø =Flux produced by coil. N = Number of turns of coil. Where, L: Inductance or Self-inductance of the coil. Self-inductance of the coil is defined as it is flux linkage to the coil with respect to change in current in a coil or it is property of a coil which opposes change in current through it. It is measured in H (Henry). Expression for self-Inductance: But, Ø = NI/S JSPM’s RAJARSHI SHAHU COLLEGE OF ENGINEERING, PUNE 10 Introduction to Electrical Engineering F.Y. B. TECH. _________________________________________________________________________________________ S Therefore, Factors affecting self-inductance of the coil: Following are the factors which affect value of self-inductance of coil.  Self-inductance of the coil decrease with increase in length of coil.  Self-inductance of the coil increase with increase in diameter (cross section) of coil.  Self-inductance of the coil increase with increase in number of turns of coil.  Self-inductance of the coil increase with increase in relative permeability of magnetic material used as core for the coil.  Air core coil as least self-inductance. 1.7.2.2. Mutually Induced EMF Consider two coils placed near to each other as shown in figure. Terminals of first coil are connected across a source of EMF whereas terminals of second coil are connected to galvanometer. When current in a first coin increased from its zero value up to it maximum value then, magnetic flux produced by this coil increases. The increasing magnetic flux, first of all, links with the same coil and hence according to Faradays law an EMF gets induced in same coil called as self-induced EMF. At the same time increasing flux produced by first coil links with second coil and an EMM get also induced in second coil. When current in a first coil decreases from its maximum value up to its zero value then flux produced by this coil decreases. The decreasing magnetic flux again links with first and second coils and hence EMF gets induced in both the coils. The EMF induced in second coil due to change in current in first coil is called as mutually induced EMF. The EMF which induced in one coil due change in magnetic field produced by another coil or due to change in current in another coil is called as mutually induced EMF. Mutually induced EMF. JSPM’s RAJARSHI SHAHU COLLEGE OF ENGINEERING, PUNE 11 Introduction to Electrical Engineering F.Y. B. TECH. _________________________________________________________________________________________ Expression for Mutually Induced EMF: Consider two coils placed near to each other. Let,I1= Current in first coil. I2= Current in second coil. N1 = Number of turns of first coil. N1 = Number of turns of second coil. Ø1 =Flux produced by first coil (self-flux). Ø2 =Flux produced by second coil (self-flux). Ø12 = Amount of flux linking with first coil from second coil (mutual flux). Ø21 = Amount of flux linking with second coil from first coil (mutual flux). K1 = Amount of flux produced by first coil linking with second coil. Where, Ø21 = K1 Ø1 : Mutual inductance between two coils. Mutual inductance between two coils is defined as it is flux linkage to the one coil with respect to change in current in another coil. It is measured in H (Henry). If source of EMF is connected to second coil and galvanometer is connected to first coil then, xpressions for mutually induced EMF and mutual inductance can be written as. Numerical Ex.1) A electric conductor of effective length 0.3 m is moving with velocity 5 m/s in a magnetic field having density 0.5 T. Find EMF induced in a conductor if conductor is moving (a) Perpendicular to the field (b) In a direction 450 with respect to magnetic field. (c) In a direction 300 with respect to magnetic field. Solution: l= 0.3 m, v = 5 m/s, B=0.5 T. Conductor is moving perpendicular to field e = Blv e = 0.5 x 0.3 x 5 e = 0.75V Conductor is moving in direction 450 with respect to magnetic field e = BlvSin(Ɵ) JSPM’s RAJARSHI SHAHU COLLEGE OF ENGINEERING, PUNE 12 Introduction to Electrical Engineering F.Y. B. TECH. _________________________________________________________________________________________ e = 0.5 x 0.3 x 5 x Sin (45) e = 0.530 V Conductor is moving in direction 300 with respect to magnetic field e = BlvSin(Ɵ) e = 0.5 x 0.3 x 5 x Sin (30) e = 0.375 V Ex. 2) Find the inductance of air core coil having length 25 cm, radius 5 cm and 200 turns. Solution: l= 25cm = 0.25 m, r =5 cm 0.05 m, N= 200. Area of cross section A= π r2 = π x (0.05)2 = 78.53 x 10-4 m2 For air core coil S S S = 2.533 x 107 AT/Wb. Inductance L = N2/S = 2002/2.533 x 107 L = 1.579 x 10-3 H. Ex. 3) A coil of 450 turns is uniformly wound around a ring of an iron alloy of mean circumference 100 cm and cross sectional area 1.125 cm2. When a current of 0.5 A is linearly reduced to zero in 0.01 second, the EMF induced in the coil is 2V. Find relative permeability of iron alloy. Solution: l= 100 cm = 1m, A = 1.125 cm2 = 1.125 x 10-4 m2, N = 450. dI = change in current = Final value – Initial Value dI = 0 – 5 dI = – 5 A = - 500 L = - 2/(-500) L = 0.04H S S L = N2/S putting values of L, N and S µr =1398 Ex. 4) Two windings connected in series are wound on a ferromagnetic ring having cross sectional area of 750 mm2 and mean diameter of 175 mm. The two windings have 250 and 750 turns, while the relative permeability of the material is 1500. Find self-inductance of each coil and mutual inductance JSPM’s RAJARSHI SHAHU COLLEGE OF ENGINEERING, PUNE 13 Introduction to Electrical Engineering F.Y. B. TECH. _________________________________________________________________________________________ between them. Also find EMF induced in coil 2 if current in coil 1 is uniformly increased from zero to 5A in 0.01 sec. Solution: A = 750 mm2 = 7.5 x 10-4 m2, D = 175 mm = 0.175 m, N1= 250, N2 = 750. Mean length of circuit l = π D = π x 0.175 = 0.5497 m. Reluctance of the circuit S S = 3.88833 x 105 AT/Wb Self-Inductances of the coils L1 = N12/S L1 = 2502/3.88833 x 105 L1 =0.1607 H L2 = N22/S L1 = 7502/3.88833 x 105 L1 =1.4466 H. Mutual inductance M = N1 N2/S M = 250 x 750/3.88833 x 105 M = 0.4822 H. EMF induced in coil 2 e2 = - M dI/dt e2 = - 0.4822 x (5-0)/0.01 e2 = - 241.1 V Single Phase Transformer 1.8 Introduction We know that the a.c. electrical energy is generated at low or economical voltage. It has to be transmitted at high voltages for better efficiency. For distribution purpose again , the voltage needs to be reduced. The rising or lowering of voltages can be done very easily with the help of a static device, the transformer. The transformer is the static device which transforms electrical energy from one alternating current circuit to another with the desired change in voltage or current and without any change in frequency. The transformers are design to operate either on single phase or three phase supply our discussion in this unit is restricted to single phase transformer only. 1.9 Constructional Features The transformer mainly consists of the core and the windings. However, as the size (capacity) and the operating voltages increase, some other parts such as a suitable tank, the bushings, the conservator, the breather, the explosion vent, the Buchholz relay, etc. are needed. The various parts of it are briefly described below JSPM’s RAJARSHI SHAHU COLLEGE OF ENGINEERING, PUNE 14 Introduction to Electrical Engineering F.Y. B. TECH. _________________________________________________________________________________________ (a) Magnetic Core : It usually made up of magnetic material such as high grade silicon steel and is limited with a thickness of 0.35 mm. limbs are vertical portions of the core on which the windings are placed where as yokes the top and the bottom portions. Use of high grade of silicon steel is used to reduced hysteresis losses where as laminations are done to reduce eddy current losses. Yoke and limb laminations are usually interleaved to reduced reluctance and energy losses. (b) Windings : In the elementary transformer, the primary and secondary windings are wound on the separate limbs. These are insulated from each other. These windings are made up of copper. However, in actual practice primary and secondary windings are placed on the same limb. This is done so that there will not be any appreciable flux leakage between the primary and secondary windings. Concentric windings, and (ii) sandwiched windings.Concentric windings are used in the core type transformer where as sandwiched windings in the shell type transformers. (c) Transformer Tank : The whole of the transformer assembly is placed in a fabricated sheet metal tank and immersed in the oil which serves both the purpose of providing insulation and cooling. The coil rencolves the heat generated in the winding and the core. The tank is usually used for higher capacity transformer. (d) Terminal Bushings : The leads of the transformer brought out from the tank are insulated from it with the help of porcelain bushings. These bushings are fitted to the tank. (e) Conservator : The conservator is an air tight metal drum supported on the transformer tank. This drum is connected by pipe to the transformer tank and is always partly filled with oil. The expansion and contraction of the oil in the tank with the change of temperature is now taken up by the conservator. With this arrangement since the main tank is always full with the oil, the surface of the oil is not directly exposed to air. The conservator takes care of the breathing of the transformer where on, when the transformer becomes warm, the oil expands and the air at the top of the oil in the conservator is expelled. When the transformer cools, the oil cons tracts and outside air is drawn into the transformer. (f) Breather : The breather is device connected with the conservator. It consists of a drying agent such as calcium chloride or silica gel. It absorbs the moisture from the atmospheric air coming into the conservator and hence into the transformer there by allowing the dry air into the transformer. If breather were not there, air from the atmosphere entering conservator with moisture in it will start reducing its JSPM’s RAJARSHI SHAHU COLLEGE OF ENGINEERING, PUNE 15 Introduction to Electrical Engineering F.Y. B. TECH. _________________________________________________________________________________________ insulating value. This slow detraction of oil insulation is presented by the breather. (g) Buchholz Relay : It is a protective device connected in between the main tank and the conservator. It has an alarm circuit and a tripping circuit. When the fault in the transformer (called incipient fault) is in the initial stage it gives an alarm so that proper preventive action is taken. If the fault condition is serious, it gives tripping command and isolates the transformer from the supply mains to prevent further damage. (h) Explosion Vent : The bent up pipe fitted on the upper surface of the tank is known as explosion vent or relief value. It is made out of glass sheet, and aluminum foil or a Bakelite sheet. In the event of fault, if the excessive pressure is developed inside the tank due to liberated gases, the diaphragm in the explosion vent bursts and releases the pressure, thus avoiding damage to the transformer. 1.10 Principal of Working The operation of the transformer is based on the principal of mutual induction between two circuit linked by a common magnetic field. Let us consider an elementary transformer as shown in the figure (a) its symbol is shown in figure (b). it essentially consists of two windings ( P and S) electrically separated but wound JSPM’s RAJARSHI SHAHU COLLEGE OF ENGINEERING, PUNE 16 Introduction to Electrical Engineering F.Y. B. TECH. _________________________________________________________________________________________ on a common laminated steel core( C ). The vertical portions of the core on which the windings are placed are known as limbs. The top and bottom portions are known as yokes. The winding (P) which is connected to the existing supply system and which receives energy from it is known as the primary winding. The other winding (S) delivering energy to the load at the desired voltage is known as secondary winding. Let N1 & N2 represents the number of turns of primary and secondary windings respectively. When the primary winding is connected to the a.c. supply, the applied voltage circulate an alternating current through it. This current produce an alternating flux (ф ) as shown in figure (a). most of this flux links with the secondary winding through the core and induce an alternating e.m.f. in the winding. This is according to Faraday‟s law of electromagnetic induction and the phenomena of an e.m.f. being induced in the secondary winding due to current in the primary winding is known as mutual induction. The e.m.f. so induced is known as mutually induced e.m.f. or e.m.f. of mutually induction. The frequency of this e.m.f. is same as that of the supply voltage. If a load is connected through the secondary winding, a current will floe through it. Thus there is a transfer of energy from primary to the secondary circuit through the core magnetically. It should be noted that the transformer should not be connected to the direct current (d.c.) source or supply. this is so because, when we try to connected a d.c. supply to the primary winding. There will be a constant flux produced and due to this constant flux through the core, as it is not varying, no e.m.f. will be induced. In fact, with d.c. supply connected, the primary winding will draw very large amount of current from the supply and this may damage the winding. However with a a.c. supply, there will be counter e.m.f. in the primary winding and the primary current is mainly limited by it as it oppose the applied voltage. 1.11 Types of Transformers Based on the arrangement of core and windings, the transformer are classified as follows: 1. Core type transformers 2. Shell type transformer 3. Berry type transformer 1. Core type transformer : The figure here shows the arrangement of a core type transformer. The sailent or main feature of this of transformer is as follows. (i) It has a single magnetic circuit JSPM’s RAJARSHI SHAHU COLLEGE OF ENGINEERING, PUNE 17  Introduction to Electrical Engineering F.Y. B. TECH. _________________________________________________________________________________________ (ii) The winding are cylindrical. (iii) As the windings are placed on the separate limbs of the core, natural cooling is more effective. (iv) The winding surround major portion of the core. (v) Maintenance or repairs is easy as windings can be withdrawn for repairs just by dismantling the laminations of the top yoke. 2. Shell type Transformer : The figure below shows the arrangement of a shell type transformer. The main features of this transformer are explained below: (i) It has a double magnetic circuit (ii) The primary and the secondary windings are wound on the central limb and the outer limb provides the low reactance path. (iii) The voltages and high voltages windings are placed alternatively like a sandwich. Therefore such a winding is also known as sandwich winding. (iv) The core for shell type transformer is made up of U & T shape. (v) Natural cooling is surrounded by a core (vi) It is difficult to maintain and repair as winding cannot be easily dismantled. (vii) The shell type transformer gives better supports against electromotive forces. Comparison between core and shell type transformers: Sr. No. Core type transformer Shell type transformer 1. It has a single magnetic circuit. It has double magnetic circuit. 2. Windings are cylindrical form. Sandwich type winding are used. 3. Core is surrounded by the The winding are surrounded by Winding. the core. 4. Repair and maintenance are Repair and maintenance are difficult. Easy. 5. Natural cooling is good. Natural cooling is poor. JSPM’s RAJARSHI SHAHU COLLEGE OF ENGINEERING, PUNE 18 Introduction to Electrical Engineering F.Y. B. TECH. _________________________________________________________________________________________ 1.12 E.M.F. Equation of A Transformer Let us derive the mathematical expression for the e.m.f. induces in the primary and the secondary windings of a single phase transformer. As soon as the primary windings of the transformer is connected across the a.c. supply an alternating current flows through it. It produces an alternating flux. This flux links, through the core, the secondary winding and induce an e.m.f. in it by mutual induction. This flux also links the turnsof the primary winding and induce an e.m.f. in it. Let N1 = No. of turns in primary N2 = No. of turns in secondary Φ m = Maximum flux in core in webers = Bm × A f = Frequency of a.c. input in Hz As shown in Fig., flux increases from its zero value to maximum value Φm in one quarter of the cycle i.e. in 1/4 f second Average rate of change of flux= = 4f Φ m wb/sec or volts Now, rate of change of flux per turn means induced e.m.f. in volts. Average e. m. f per turn = 4f Φ m volts If flux Φ varies sinusoidally, then r.m.s. value of induced e.m.f. is obtained by multiplying the average value with form factor. Form Factor = RMS value of e. m. f per turn = 1.11 X 4f Φ m volts = 4.44 f Φ m volts Now, r.m.s. value of the induced e.m.f. in the whole of primary winding = (induced e.m.f/turn) × No. of primary turns E1= 4.44 f N1 Φ m =4.44 f N1 B m A Similarly, r.m.s. value of the e.m.f. induced in secondary is, E2= 4.44 f N2 Φ m =4.44 f N2 B m A Again, for an ideal transformer, input VA = output VA. JSPM’s RAJARSHI SHAHU COLLEGE OF ENGINEERING, PUNE 19 Introduction to Electrical Engineering F.Y. B. TECH. _________________________________________________________________________________________ 1.13 Voltage and Current Ratio of a Transformer Refer transformer with its secondary winding connected to the load. Voltage ratio : The ration of primary to secondary terminal voltage is known as the voltage ratio = V1 / V2 Voltage Transformation Ratio (K) We know, Primary induced emf is E1= 4.44 f N1 Φ m =4.44 f N1 B m A Secondary induced emf is E2= 4.44 f N2 Φ m =4.44 f N2 B m A From above equations), we get This constant K is known as transformation ratio. 1. If N2 > N1 i.e. K > 1, then transformer is called step-up transformer. 2. If N2 < N1 i.e. K < 1, then transformer is known as step-down transformer. 3. If N2 = N1 i.e. K = 1, then transformer is known as Isolation transformer or 1: 1 transformer. Turns ratio : The ratio of primary to secondary turns is known as the turns ratio, Thus the turns ratio = N1 / N2. Current ratio : The ratio of primary to secondary current is known as the current ratio = I1 / I2 Relation between V1 , V2, I1 , I2, N1 & N2 : As the transformer transfers electrical power from one circuit to another circuit very efficiently with negligible power loss, power input = power output , V1 I1 cosф1 = V2 I2 cosф2 Where cosф1 and cosф2 are power factor of primary and secondary circuits respectively. These are hereby equal. Hence, V1 I1 = V2 I2 , I1 / I2 = V2/ V1 But we have V2/ V1 = N2 / N1 = K I1 / I2 = V2/ V1 = N2 / N1 = K JSPM’s RAJARSHI SHAHU COLLEGE OF ENGINEERING, PUNE 20 Introduction to Electrical Engineering F.Y. B. TECH. _________________________________________________________________________________________ 1.14 Ratings of a transformer : The primary voltage , secondary voltage, supply frequency and type (single phase or three phase) are mentioned on the transformer name plate. KVA ratings (Kilo volt – Ampere rating) : It is also mentioned on the name plate of a transformer. If I1 &I2 are full load currents and V1& V2 are rated voltages of the primary and secondary , then KVA ratings = V1 , I1 / 1000 = V2 ,I2 / 1000 The transformer is designed for a particular value of operating value output power. The load connected across the secondary side of the transformer may be lagging, leading or unity. Thus for the same operating voltage and current , the output power can be different at different leading conditions. Hence only operating voltage and operating current are specified. These are known as rated voltage and rated current respectively. Their product i.e. the product of rated voltage and rated current are known as volt – ampere rating. This term divided by 1000 gives the kilo volt – ampere rating. The KVA rating of the transformer can be define as the KVA output which it can deliver at rated voltage and rated frequency under usual service conditions with out exceeding the standards limits of temperature rise. 1.15 Concept of an Ideal Transformer An ideal transformer possesses following properties : (i) The permeability of the core is so high that only a negligible current is required to establish the flux in it (ii) There is no magnetic leakage. In other words, all the flux is confined to the iron core only and links both the windings. (iii) Its windings have no ohmic residence. (iv) It has no losses. It is not possible have an ideal transformer in practice. However it is required to have the concept of an ideal transformer to understand the theory of practical transformers. 1.16 Practical Transformers : The practical transformer has core losses, winding resistance and magnetic leakage. Since the core of the practical transformer is subjected to an alternating flux, core losses (i.e. hysteresis and eddy current losses) are always presents to some extent each through the core is laminated to reduce eddy currents and hysteresis is minimize by the use of high grade silicon steel. Follows, therefore, that even under no load condition (i.e. with the secondary winding open – circuited), the source must be supply enough power to the primary winding to JSPM’s RAJARSHI SHAHU COLLEGE OF ENGINEERING, PUNE 21 Introduction to Electrical Engineering F.Y. B. TECH. _________________________________________________________________________________________ over come the core losses. Therefore, in practical transformers, on no load, the primary winding carries a small current I0which has two components. (i) A reactive or wattles magnetizing component Imag. It produces the magnetic flux and therefore is in phase with it. (ii) An active or power or core loss component Ic. it supplies mainly the core losses and the negligible copper loss in the primary winding. This component is in phase with component – E1 of the impressed voltage. The figure (a) below shows the phasor diagram for a practical transformer on no load. From the diagram. We find that, no load current I0 = (Ic2 + Imag )2 And power factor on no load = cosф0= Ic / I0 With Ic being very small compared to Imag, the no load power factor is very low. 1.17 Equivalent Circuit of Single Phase Transformer: The figure below shows a practical transformer on load : Here R1 & X1 represents resistance and reactance of the primary winding respectively, similarly, R2 & X2 represents the resistance and reactance of the secondary winding respectively. Also I1 & I2 are currents in primary and secondary windings respectively. JSPM’s RAJARSHI SHAHU COLLEGE OF ENGINEERING, PUNE 22 Introduction to Electrical Engineering F.Y. B. TECH. _________________________________________________________________________________________ Now, primary copper losses = I12 R1 Similarly, secondary copper losses = I22 R2 Here output is always less then the input power. Also efficiency is less then 100% and regulation is 1 to 5 % 1.18 Transformer Losses The losses that take place in a transformer are of two types: Iron loses or core loses (constant losses ). And copper losses (variable losses) 1. Iron or Core losses : There losses are caused by the alternating flux in the transformer core. These losses consist of hysteresis and eddy current loses. A. Hysteresis loss : This loss takes place in the transformer core because it is continuously subjected to rapid reversals of magnetization by the alternating flux. Hysteresis loss , Ph = Kn Bm1.6 f v watts , where f is the frequency in hertz, Bm is the maximum flux density in tesla, V is the volume of magnetic material in cubic meters and Kn is a constant. B. Eddy current loss : Pe = K e Bm2 f 2 t2 v watts. Where t is the thickness of the laminations, f is the frequency in hertz, Bm is the maximum flux density in tesla, V is the volume of magnetic material and Ke is the constant. These core losses are nearly constant and independent of the magnitude of the current delivered by the transformer. They are reduced by choosing silicon steel having small value of KL and Ke as a core material and by using laminated construction for the core. 2. Copper losses : These losses are caused due to current flowing in the two windings, made up generally of copper material and are due to loss of power caused by the resistance of the windings. These losses are proportional to the square of the current through the windings. If R1 and R2 are the primary and the secondary resistance respectively and I1 and I2 are the corresponding currents, then total copper losses = I12 R1 + I22 R2. These copper losses are proportional to the square of the KVA output. These are reduced by using the material with good conductivity, like copper, for the windings of the transformer. Efficiency of a Transformer It is defined as the ratio of the output power to the input power in a transformer. JSPM’s RAJARSHI SHAHU COLLEGE OF ENGINEERING, PUNE 23 Introduction to Electrical Engineering F.Y. B. TECH. _________________________________________________________________________________________ Transformer efficiency (ɳ) = output power / input power = output power / output power + losses Output power of the transformer = V2I2 cosф2 losses in the transformer = copper losses + core or iron losses = Pcu + Pi Where Pcu = copper loss and Pi = core or iron loss. Efficiency n = V2I2 cosф2 / V2I2 cosф2 + Pcu + Pi If Pcu is the full load copper loss, then efficiency on full load is given by , Full load efficiency = full load output / full load output + Pcu + Pi Full load efficiency = full load rating (in VA) x p.f. / full load rating (in VA) x p.f. x Pcu + Pi Similarly efficiency at any other load, where x is the degree of loading or percent of full load is given by, Efficiency at (X x full load) = X. (full load rating (in VA)) / X x (full load rating (in VA)) x p.f. x Pcu + Pi It should be remembered that at 25 % load, the value of X will be 0.25, at 50% load it will 0.5 1.19 Auto transformers An auto transformer is a transformer having only one winding on laminated magnetic core, the part of this winding common to both the primary and secondary circuits. It can used as step down or step up transformer as shown in the figure. JSPM’s RAJARSHI SHAHU COLLEGE OF ENGINEERING, PUNE 24 Introduction to Electrical Engineering F.Y. B. TECH. _________________________________________________________________________________________ In figure (a), the winding Ac forms the primary and the portion BC forms the secondary. On the other hand, in figure (b) the winding Ac forms the secondary and its portion Bc forms the secondary. Similarly to the two-winding transformer neglecting the losses, the leakage reactance amp the magnetizing current, we have V2 / V1 = I1 / I2 = N2 / N1 =K (with usual notations) Advantages of the auto transformer : 1. As compared to a two winding transformer, an auto transformer is cheaper as lesser copper is required for the windings. 2. The regulation is better as compared to a two winding transformer as the resistance and leakage reactance values are lesser. 3. Efficiency is higher then a two winding transfer as I2R losses being lesser. Disadvantages of the autotransformer : There is always a rise of serious electric state especially when it is used in high voltage circuit, as the low voltage and high voltage sides are not electrically separate. Applications of the auto transformer : 1. For starting of squirrel cage synchronous motor 2. Coupling the a.c. system having nearly the same operating voltage. 3. As various for getting continuous a.c. supply 4. As booster to rise the voltage in a.c. feed back 5. As furnace transformer for getting suitable supply voltage. JSPM’s RAJARSHI SHAHU COLLEGE OF ENGINEERING, PUNE 25 Introduction to Electrical Engineering F.Y. B. TECH. _________________________________________________________________________________________ Numerical Ex. 1) A single phase transformer has 400 turns on primary and 1000 turns on secondary. Voltage per turn on primary winding is 0.5 volts. Find (a) EMF induced in secondary winding. (b) EMF induced in secondary winding. (c) Maximum value of flux density in core if core has cross sectional area of 150 cm2. (d) kVA rating of the transformer if primary full load current is 12A and frequency 50 Hz. Solution: N1 =400, N2 = 1000, A= 150 cm2 = 150 x 10-4 m2, f = 50 Hz. EMF in primary winding E1 = N1 x Voltage per turn E1 = 400 x 0.5 = 200V EMF in secondary winding E2 = E1 x N2/N1 E2 = 200 x 1000/400 =500V. Maximum flux density E2 = 4.44 N2 Øm f 500 = 4.44 x 1000 x Øm x 50 Øm = 2.25 mWb. Bm = Øm/A Bm = 2.25 x 10-3/150 x 10-4 Bm = 0.15 T. kVA rating of transformer kVA rating = V2 I2fl x10-3 = V1 I1fl x 10-3 kVA rating = 200 x 12 x 10-3 kVA rating = 2.4 kVA. Ex.2) A 1.5 kVA, 220/110V, 50Hz, single phase transformer has iron loss of 32 W and full load copper loss of 44 W. Find efficiency of the transformer at (a) Full load and 0.8 p.f lagging. (b) 70 % of the full load and 0.8 p.f. lagging (c) Half load and 0.8 p.f lagging. Solution : E1 = 220 V, E2 = 110 V, Wi =32 W, Wc = 44 W, kVA = 1.5. Full load efficiency = 94.04 % JSPM’s RAJARSHI SHAHU COLLEGE OF ENGINEERING, PUNE 26 Introduction to Electrical Engineering F.Y. B. TECH. _________________________________________________________________________________________ Efficiency at 70% of full load = 94 % Efficiency at half load = 93.31 % Ex. 3) A single phase 100 kVA, 3.3 kV/230 V, 50 Hz transformer has 89.5 % efficiency at 0.85 lag p.f. both at full load and also at half load. Find efficiency of the transformer at 75 % of the full load and 0.9 p.f. leading. Solution: kVA = 100, f=50 Hz. Full load efficiency (1) Half load efficiency (2) Solving equations (1) and (2) Wiron =3.324022 kW. Wcopper = 6.648044 kW. Efficiency at 75 % of full load and 0.9 p.f. leading, JSPM’s RAJARSHI SHAHU COLLEGE OF ENGINEERING, PUNE 27