ELECTROMECHANICAL ENERGY CONVERSION AND TRANSFORMERS.pdf

HO TECHNICAL UNIVERSITY FACULTY OF ENGINEERING DEPARTMENT OF ELECTRICAL AND ELECTRONIC ENGINEERING Lecture Note Electromechanical Energy Conversion and Transformers (BEEE 104 & EEE 132) Level 10...

HO TECHNICAL UNIVERSITY FACULTY OF ENGINEERING DEPARTMENT OF ELECTRICAL AND ELECTRONIC ENGINEERING Lecture Note Electromechanical Energy Conversion and Transformers (BEEE 104 & EEE 132) Level 100  Table of Contents Contents Pages CHAPTER 1 MAGNETIC CIRCUITS AND MAGNETIC MATERIALS 1 1.1 Introduction 1 1.2 Classifications of Magnetic Materials 1 1.3 Magnetic Field and its Significance 2 1.4 Magnetic Circuit and its Analysis 2 1.5 Important Terms Associated with Magnetic Circuit 3 1.6 The differences between a magnetic circuit and an electric circuit 4 1.7 Comparison between Magnetic and Electric Circuits 4 1.8 Series Magnetic Circuits 6 1.9 Parallel Magnetic Circuits 6 1.10 Magnetization or B-H Curve 14 1.11 Magnetic Hysteresis 15 1.12 Importance of Hysteresis Loop 17 1.13 Application of Ferromagnetic Material 18 CHAPTER 2 PRINCIPLES OF ELECTROMECHANICAL ENERGY CONVERSION 19 2.1 Introduction 19 2.2 Forces and Torques in Magnetic Fields 19 2.3 Force Between Two Current-Carrying Wires 20 2.4 Force on a Current-Carrying Wire in Magnetic Field 21 2.5 Generated Voltage in Magnetic Systems 22 2.6 Voltage Induced on a Moving Conductor 24 2.7 Force Developed in an Electromagnetic System 25 2.7.1 Force of Alignment 25 i  2.7.2 Force of Interaction 27 2.8 Basic Structure of Rotating Electrical Machines 27 2.9 Structure Types of Rotating Electrical Machines 28 2.10 Energy Balance Method 32 2.11 Energy in Singly-Excited Magnetic Field Systems 33 2.11.1 Derivation of force from energy 33 2.11.2 Derivation of force from co-energy 35 2.12 Force of Alignment Between Parallel Magnetized Surfaces 38 2.13 Lateral Force of Alignment Between Parallel Magnetized Surfaces 38 2.14 Alternative Force Calculation in A Linear System 39 2.15 Rotating Electrical Machines 41 2.16 Basics of Motors and Generator 43 CHAPTER 3 TRANSFORMER 45 3.1 Introduction 45 3.2 Principle of Operation of a Transformer 45 3.3 Constructional features of two-winding transformers 47 3.3.1 Comparison of Core and Shell Type Transforms 48 3.4 EMF Equation of a Transformer 48 3.5 Concept of Ideal Transformer 51 3.5.1 Ideal Transformer on No Load 52 3.5.2 Ideal Transformer on Load 53 3.6 Practical Transformer 54 3.6.1 Practical Transformer on No Load 55 3.6.2 Practical Transformer on Load 56 3.7 Transformer with resistance and leakage reactance 58 3.7.1 Shifting Impedances in a Transformer 60 3.8 Voltage Regulation, Testing, and Efficiency of a transformer 68 3.8.1 Voltage Regulation of Transformer 68 3.8.2 Losses in a Transformer 68 3.8.3 Efficiency of a Transformer 69 3.8.4 All-Day Efficiency 71 ii  3.8.5 Testing of Transformers 72 3.9 Parallel Operation of Single-Phase Transformer 79 3.10 Auto Transformer 81 3.11 Components of a Typical Power Transformer 84 3.12 COOLING OF TRANSFORMER 87 3.13 Thres-Phase Transformers 90 3.13.1 Principle of Operation of Three Phase Transformer 90 3.13.2 Construction of Three-Phase Transformers 91 3.13.3 Three-Phase Transformer Connection 92 3.14 Transformer Tap Changer 98 iii  CHAPTER 1 MAGNETIC CIRCUITS AND MAGNETIC MATERIALS 1.1 Introduction It is always advantageous to utilise electrical energy since it is cheaper, can be easily transmitted, easy to control and more efficient. The electrical energy is generally generated from natural resources such as water, coal, diesel, wind, atomic energy, etc. From these sources, first mechanical energy is produced by one way or the other and then that mechanical energy is converted into electrical energy by suitable machines. For the utilisation of electrical energy, it is again converted into other forms of energy such as mechanical, heat, light etc. It is a well- known fact that the electric drives have been universally adopted by the industry due to their inherent advantages. The energy conversion devices are always required at both ends of a typical electrical system. The devices or machines which convert mechanical energy into electrical energy and vice-versa are called electro–mechanical energy conversion devices. The operation of all the electrical machines such as DC machines, transformers, synchronous machines, induction motors, etc., rely upon their magnetic circuits. The closed path followed by the magnetic lines of force is called a magnetic circuit. The operation of all the electrical devices (e.g., transformers, generators, motors, etc.) depends upon the magnetism produced by their magnetic circuits. Therefore, to obtain the required characteristics of these devices, their magnetic circuits have to be designed carefully. 1.2 Classifications of Magnetic Materials All the materials in this universe may be classified as either magnetic or non-magnetic. Magnetic materials are those which are affected by magnetic field and non-magnetic materials are those which are not affected or slightly affected by magnetic field. Non-magnetic material can further be classified as diamagnetic and paramagnetic materials. Materials respond differently to the force of a magnetic field. A magnet will strongly attract ferromagnetic materials, weakly attract paramagnetic materials and weakly repel diamagnetic materials. Diamagnetic; these materials are slightly repelled by a magnetic field and the material does not retain the magnetic properties when the external field is removed. In diamagnetic materials all the electron are paired so there is no permanent net magnetic moment per atom. Diamagnetic properties arise from the realignment of the electron paths under the influence of an external magnetic field. Most elements in the periodic table, including copper, silver, and gold, are diamagnetic. Paramagnetic; These materials are slightly attracted by a magnetic field and the material does not retain the magnetic properties when the external field is removed. Paramagnetic properties are due to the presence of some unpaired electrons, and from the realignment of the electron paths caused by the external magnetic field. Paramagnetic materials include magnesium, lithium, and tantalum. 1 Ferromagnetic; These materials exhibit a strong attraction to magnetic fields and are able to retain their magnetic properties after the external field has been removed. Ferromagnetic materials have some unpaired electrons so their atoms have a net magnetic moment. They get their strong magnetic properties due to the presence of magnetic domains. When a ferromagnetic material is in the demagnetized state, the domains are nearly randomly organized and the net magnetic field for the part as a whole is zero. When a magnetizing force is applied, the domains become aligned to produce a strong magnetic field within the part. Iron, nickel, and cobalt are examples of ferromagnetic materials. 1.3 Magnetic Field and its Significance The region around a magnet where its poles exhibit a force of attraction or repulsion is called magnetic field. The existence of the magnetic field at a point around the magnet can also be determined by placing a magnetic needle at that point as shown in Fig. 1.1. Although magnetic lines of force have no real existence and are purely imaginary, yet their concept is very useful to understand various magnetic effects. It is assumed (because of their effects) that the magnetic lines of force possess the following important properties: (i) The direction of magnetic lines of force is from N-pole to the S-pole outside the magnet. But inside the magnet their direction is from S-pole to N-pole. (ii) They form a closed loop. (iii) Their tendency is to follow the least reluctance path. (iv) They act like stretched cords, always trying to shorten themselves. (v) They never intersect each other. (vi) They repel each other when they are parallel and are in the same direction. (vii) They remain unaffected by non-magnetic materials. 1.4 Magnetic Circuit and its Analysis The closed path followed by magnetic flux is called magnetic circuit. A magnetic circuit usually consists of magnetic materials having high permeability (e.g., iron, soft steel, etc.). In this circuit, magnetic flux starts from a point and finishes at the same point after completing its path. Figure 1.2 shows a solenoid having N turns wound on an iron core (ring). When current I ampere is passed through the solenoid, magnetic flux Φ Wb is set-up in the core. Let l = mean length of magnetic circuit in m; a = area of cross-section of core in m2; µr = relative permeability of core material.  Flux density in the core material, B = Wb/m 2 a Magnetising force in the core material. B  H= = = AT / m o  r a  o  r According to work law, the work done in moving a unit pole once round the magnetic circuit (or path) is equal to the ampere-turns enclosed by the magnetic circuit.  NI i.e., Hl = NI or  l = NI or  = Wb o  r ( l / a o  r ) The above expression reveals that the amount of flux set-up in the core is (i) directly proportional to N and I i.e., NI, called magnetomotive force (mmf). It shows that the flux increases if either of the two increases and vice-versa. 2 (ii) inversely proportional to l/a μ0 μr called reluctance of the magnetic path. In fact, reluctance is the opposition offered to the magnetic flux by the magnetic path. The lower is the reluctance, the higher will be the flux and vice-versa. m.m.f Thus, FLux = reluctance It may be noted that the above expression has a strong resemblance to Ohm’s law for electric current (I = emf/resistance). The mmf is analogous to emf in electric circuit, reluctance is analogous to resistance and flux is analogous to current. Because of this similarity, the above expression is sometimes referred to as Ohm’s law of magnetic circuits. 1.5 Important Terms Associated with Magnetic Circuit While studying magnetic circuits, generally, we come across the following terms: 1. Magnetic field: The region around a magnet where its poles exhibit a force of attraction or repulsion is called magnetic field. 2. Magnetic flux (Φ): The number of magnetic lines of force set-up in a magnetic circuit is called magnetic flux. Its unit is weber (Wb). It is analogous to electric current I in electric circuit. 3. The magnetic flux density at a point is the flux per unit area at right angles to the flux at that point. It is, generally, represented by letter ‘B’. Its unit is Wb/m2 or Tesla, i.e.,  B= Wb/m 2 or T (1 Wb/m2 = 1 x 104 Wb/cm2) A 4. Permeability: The ability of a material to conduct magnetic lines of force through it is called the permeability of that material. It is generally represented by μ (mu, a Greek letter). The greater the permeability of a material, the greater is its conductivity for the magnetic lines of force and vice-versa. The permeability of air or vacuum is the poorest and is represented as o (where o = 4π × 10–7 H/m). Relative permeability: The absolute (or actual) permeability μ of a magnetic material is much greater than absolute permeability of air o. The relative permeability of a magnetic material is given in comparison with air or vacuum. Hence, the ratio of the permeability of material μ to the permeability of air or vacuum o is called the relative permeability  r of the material.  e.i., r = or  = o r o Obviously, the relative permeability of air would be o o = 1. The value of relative permeability of all the non-magnetic materials is also 1. However, its value is as high as 8000 for soft iron, whereas, its value for mumetal (iron 22% and nickel 78%) is as high as 1,20,000. 5. Magnetic field intensity: The force acting on a unit north pole (1 Wb) when placed at a point in the magnetic field is called the magnetic intensity of the field at that point. It is denoted by H. In magnetic circuits, it is defined as mmf per unit length of the magnetic path. It is denoted by H, mathematically m.m.f NI H= = AT / m length of magnetic path l 3 6. Magnetomotive force (mmf): The magnetic pressure which sets-up or tends to set-up magnetic flux in a magnetic circuit is called magnetomotive force. As per work law it may be defined as under: The work done in moving a unit magnetic pole (1 Wb) once round the magnetic circuit is called magnetomotive force. In general, mmf = NI ampere-turns (or AT) It is analogous to emf in an electric circuit. 7. Reluctance (S): The opposition offered to the magnetic flux by a magnetic circuit is called its reluctance. It depends upon length (l), area of cross-section (a) and permeability (  = o r of the material that makes up the magnetic circuit. It is measured in AT/Wb. l Reluctance, S = aμ oμ r It is analogous to resistance in an electric circuit. 8. Permeance: It is a measure of the ease with which flux can be set-up in the material. It is just reciprocal of reluctance of the material and is measured in Wb/AT or henry. 1 aμ oμ r Permeance = = Wb/AT or H reluctance l It is analogous to conductance in an electric circuit. 9. Reluctivity: It is specific reluctance and analogous to resistivity in electric circuit. 1.6 The differences between a magnetic circuit and an electric circuit 1. Electric current in an electric circuit is the flow of electric charge whereas flux does not actually flow in a magnetic circuit. 2. Conductivity of different materials are different that means it greatly varies from material to material but permeability of any magnetic material does not largely vary from material to material. 3. There are many materials which are fairly bad conductor of electricity but there is hardly any material which can be considered as insulator of magnetic flux. 4. For a particular temperature the permeability depends upon the flux density or total amount of flux but in case of conductivity, it is fixed and independent of current density at a particular temperature. 5. Air or free space is very good insulator for electric current whereas either of them cannot resist the flux to pass. 6. There is no loss of energy due to reluctance in magnetic circuit but electric current through the resistance causes energy loss in form of heat. 7. Energy is required to establish flux in a magnetic circuit but energy is not required to maintain the flux in magnetic circuit but in case of electric circuit energy is continuously required to maintain the flow of current in it. 1.7 Comparison between Magnetic and Electric Circuits Although magnetic and electric circuits have many points of similarity but still, they are not analogous in all respects. A comparison of the two circuits is given below: 4  Dissimilarities 1. In fact, the magnetic flux does not flow but 1. The electric current (electrons) actually it sets up in the magnetic circuit (basically flows in an electric circuit. molecular poles are aligned). 2. For magnetic flux, there is no perfect 2. For electric current, there are large number insulator. It can be set-up even in the non- of perfect insulators like glass, air, rubber, magnetic materials like air, rubber, glass etc. etc., which do not allow it to follow through with reasonable mmf them under normal conditions. 3. The reluctance (S) of a magnetic circuit is 3. The resistance (R) of an electric circuit is not constant rather it varies with the value of almost constant as its value depends upon the B. It is because the value of μr changes value of ρ which is almost constant. considerably with the change in B. However, the value of ρ and R may vary slightly if temperature changes. 4. Once the magnetic flux is set-up in a 4. Energy is expanded continuously, so long magnetic circuit, no energy is expanded. as the current flows through an electric However, a small amount of energy is circuit. This energy is dissipated in the form required at the start to create flux in the of heat. circuit. 5 Ampere-turns Calculations In a magnetic circuit, flux produced. m.m.f NI  = = reluctance l / a o  r l B or AT, NI = = l = Hl a o  r o  r 1.8 Series Magnetic Circuits A magnetic circuit that has a number of parts of different dimensions and materials carrying the same magnetic field is called a series magnetic circuit. Such as series magnetic circuit (composite circuit) is shown in Fig. 1.1. Figure 1.1 Series magnetic circuit Total reluctance of the magnetic circuit S = S1 + S2 + S3 + Sg l1 l2 l3 lg = + + + a1o  r1 a2 o  r 2 a3 o  r 3 a g o Total mmf = ΦS  l1 l2 l3 lg  =  + + +   a1o r1 a2 o r 2 a3 o r 3 ag o   B1l1 B2l2 B3l3 Bg l g = + + +  o  r1 o  r 2 o  r 3 o = H1l1 + H 2l2 + H 3l3 + H g l g 1.9 Parallel Magnetic Circuits A magnetic circuit which has two or more than two paths for the magnetic flux is called parallel magnetic circuit. Its behaviour can be just compared to a parallel electric circuit. Figure 1.2 shows a parallel magnetic circuit. A current carrying coil is wound on the central limb AB. This 6 coil sets-up a magnetic flux Φ1 in the central limb which is further divided into two paths i.e., (i) path ADCB which carries flux Φ2 and (ii) path AFEB which carries flux Φ3. Figure 1.2 Parallel Magnetic Circuit It is clear that Φ1 = Φ2 + Φ3 The two magnetic paths ADCB and AFEB are in parallel. The ATs required for this parallel circuit is equal to the ATs required for any one of the paths. If S1 = reluctance of path BA i.e., l1 a1o r1 S2 = reluctance of path ADCB i.e., l2 a2 o r 2 S3 = reluctance of path AFEB i.e., l3 a3 o r 3 Therefore, total mmf required = mmf required for path BA + mmf required path ADCB or path AFEB. i.e., Total mmf or AT = Φ1S1 + Φ2S2 = Φ1S1 + Φ3S3 Leakage Flux The magnetic flux which does not follow the intended path in a magnetic circuit is called leakage flux. When some current is passed through a solenoid, as shown in Fig. 1.3, magnetic flux is produced by it. Most of this flux is set-up in the magnetic core and passes through the air gap (an intended path). This flux is known as useful flux u. However, some of the flux is just set-up around the coil and is not utilised for any work. This flux is called leakage flux l. Total flux produced by the solenoid.  = u + l Leakage co-efficient or leakage factor: The ratio of total flux (  ) produced by the solenoid to the useful flux ( u ) set-up in the air gap is known as leakage co-efficient. It is generally represented by letter ‘  ’.   leakage co-efficient,  = u 7  Figure 1.3 Fringing: It may be seen in Fig. 1.3 that the useful flux when sets-up in the air gap, it tends to bulge outwards at b and b’ since the magnetic lines set-up in the same direction repel each other. This increases the effective area in the air gap and decreases the flux density. This effect is known as fringing. The fringing is directly proportional to the length of the air gap. Work Examples 1.1 An iron ring of 400 cm mean circumference is made from round iron of cross-section 20 cm2. Its permeability is 500. If it is wound with 400 turns, what current would be required to produce a flux of 0·001 Wb? The magnetic circuit is shown in Figure 1.4. Figure 1.4 Solution Mean length of magnetic path, lm = 400 cm = 4 m Area of X-section of iron ring, a = 20 × 10–4 m2 Absolute permeability, μ0 = 4π × 10–7 Now mmf = flux × reluctance lm NI =   a o  r 4 400  I = 0.001  −4 20  10  4  10−7  500 0.001  4 Current, I = = 7.958 A 20  10  4  10−7  500  400 −4 8 Example 1.2 A coil of insulated wire of 500 turns and of resistance 4Ω is closely wound on iron ring. The ring has a mean diameter of 0·25 m and a uniform cross-sectional area of 700 mm2. Calculate the total flux in the ring when a DC supply of 6V is applied to the ends of the winding. Assume a relative permeability of 550. The magnetic circuit is shown in Figure 1.5. Figure 1.5 Magnetic Circuit Solution Mean length of iron ring, l = πD = π × 0·25 = 0·25π m Area of cross-section, a = 700 mm2 = 700 × 10–6 m2 Current flowing through the coil, voltage applied across coil 6 I = = = 1.5 A Resistance of coil 5 NI NI  ao r Total flux in the ring,  = = l / a o  r l 500  1.5  700  10−6  4  10 −7  550 = = 0.462 mWb 0.25 Example 1.3 What are the similarities between electrical circuits and magnetic circuits? An iron ring of mean length 50 cm and relative permeability 300 has an air gap of 1 mm. If the ring is provided with winding of 200 turns and a current of 1 A is allowed to flow through, find the flux density across the airgap. Solution: Here, li = 50 cm = 0·5 m; μr = 300; lg = 1mm = 0·001 m; N = 200 turns; I = 1 A B Ampere turns required for air gap = lg o B Ampere-turns required for iron, = li o  r B B Or Total ampere-turns required = lg + li o o  r 9 Ampere-turns provided by the coil = N I = 200 × 1 = 200 B  li  B  0.5  Thus, 200 =  lg + =  0.01 +  o   r  o  300  B B = ( 0.001 + 0.00167 ) =  0.00267 o o 200  o 200  4  10−7 Hence, Flux density, B = = = 0.09425 T 0.00267 0.00267 Example 1.4 A coil of 1000 turns is wound on a laminated core of steel having a cross-section of 5 cm2. The core has an air gap of 2 mm cut at right angle. What value of current is required to have an air gap flux density of 0·5 T? Permeability of steel may be taken as infinity. Determine the coil inductance. Solution: Here, N = 1000 turns; a = 5 cm2 = 5 × 10–4 m2; lg = 2 mm = 2 × 10–3 m; B = 0·5 T; r =  Total ampere-turn required, B B 0.5 AT = lg + li = −7  2  10−3 + 0 = 796 r o  r 4  10  B   As r = ;  li = 0   o  r  AT 796 Current required, I = = = 0.796 A N 1000 N N  B  a 1000  0.5  5  10−4 Inductance of Coil, L= = = = 0.314 H I I 0.796 Examples 1.5 A flux density of 1·2 Wb/m2 is required in 2 mm air gap of an electro-magnet having an iron path 1 metre long. Calculate the magnetising force and current required if the electro magnet has 1273 turns. Assume relative permeability of iron to be 1500. Solution: Flux density, B = 1·2 Wb/m2 Relative permeability of iron, μr = 1500 No. of turns, N = 1273 Length of iron path, li = 1m Length of air gap, lg = 2mm = 0·002 m B 1.2 Magnetising force for iron, H i = = = 636.6 AT/m o  r 4  10−7  1500 10  B 1.2 Magnetising force for air gap, H i = = = 954900 AT/m o 4  10−7 ATs required for iron path = Hili = 636·6 × 1 = 636·6 ATs required for air gap = Hglg = 954900 × 0·002 = 1909·8 Total ampere-turns = 636·6 + 1909·8 = 2546·4 Total ATs 2546.4 Current required, I = = =2A N 1273 Example 1.5 A rectangular magnetic core shown in Fig. 1.6 (a). has square cross-section of area 16 cm2. An air-gap of 2 mm is cut across one of its limbs. Find the exciting current needed in the coil having 1000 turns wound on the core to create an air-gap flux of 4 m Wb. The relative permeability of the core is 2000. Figure 1.6 Magnetic Circuit Solution: Here, Area of x-section, a = 16 cm2 = 16 x 10–4 m2; lg = 2 mm = 2 x 10–3 m No. of turns, N = 1000; Flux, Φ = 4 m Wb = 4 x 10–3 Wb; μr = 2000  4  10−3 Flux density required, B = = = 2.5 T a 16  10−4 Each side of the cross-section = 16 = 4 cm  4 4 Length of iron-path, li =  25 - 2  + 20 - 2    2 - 0.2  2 2 = 73.8 cm = 0.738 m 11  B B Total ampere-turns required = lg + li o o  r 2.5  2  10−3 2.5  0.738 = −7 + 4  10 4  10−7  2000 = 3979 + 734 = 4713 Total ampere-turns 4713 Exciting current required, I = = = 4.713 A N 1000 Examples An iron ring is made up of three parts; l1 = 10 cm, a1 = 5 cm2; l2 = 8 cm, a2 = 3 cm2; l3 = 6 cm, a3 = 2·5 cm2. It is wound with a 250 turns coil. Calculate current required to produce flux of 0·4 mWb. 1 = 2670,  2 = 1050, 3 = 600. Figure 1.7 shows the series magnetic circuit as per data. Figure 1.7 Series magnetic Solution  0.4  10−3 Flux density, B1 = = a1 5  10−4 = 0.8 Wb/m2  0.4  10 −3 Flux density, B2 = = a2 3  10−4 = 1.33 Wb/m2  0.4  10 −3 Flux density, B3 = = a3 2.5  10 −4 = 1.6 Wb/m2 Total Ampere-turns required 12  B1 B2 B3 AT = l1 + l2 + l3 o  r o  r o  r 1  0.8 1.33 1.6  = −7   0.10 +  0.08 +  0.06  4  10  2670 1050 600  = 231.92 Current required to produce given flux AT 231.92 I = = = 0.928 A N 250 Examples A magnetic core made of annealed sheet steel has the dimensions as shown in Fig. 1.8. The cross-section everywhere is 25 cm2. The flux in branches A and B is 3500 m Wb, but that in the branch C is zero. Find the required ampere-turns for coil A and for coil C. Relative permeability of sheet steel is 1000. Figure 1.8 Given parallel magnetic circuit Figure 1.9 Flux distribution in the parallel magnetic circuit Solution: The given magnetic circuit is a parallel circuit. To determine the ATs for coil ‘A’, the flux distribution is shown in Fig. 1.9. Since path ‘B’ and ‘C’ are in parallel with each other w.r.t. path ‘A’, Therefore, mmf for path ‘B’ = mmf for path C, Thus, 1S1 = 2S2 3500  10−6  30 10 −2 80  10 −2 Thus, Thus, = 2  a o  r a o  r  2 = 1312.5  10−6 Wb Total flux in the path ‘A’, Φ = Φ1 + Φ2 = (3500 + 1312·5) × 10–6 = 4812·5 × 10–6 Wb Actual (resultant) flux in path ‘A’ = Φ – Φ2 = 3500 × 10–6 Wb ATs required for coil ‘A’ = ATs for path ‘A’ + ATs for path ‘B’ or ‘C’ 13  3500  10−6 =  ( 0.8 + 0.3) 4  10−7  1000  25  10 −4 = 1225.5 To neutralise the flux in section ‘C’, the coil produces flux of 1312·5 μWb in opposite direction. Therefore, ATs required for coil ‘C’ = ATs for path ‘C’ only 1312.5  10−6  0.8 = = 334.22 4  10−7  1000  25  10−4 1. Estimate the number of ampere-turns necessary to produce a flux of 100000 lines round an iron ring of 6 cm2 cross section and 20 cm mean diameter having an air gap 2 mm wide across it Permeability of the iron may be taken 1200. Neglect the leakage flux outside the 2 mm air gap. The magnetic circuit is shown in figure 1.10 [ Ans. 3344.79 AT] Figure 1.10 Magnetic circuit A circular ring 20 cm in diameter has an air gap 1 mm wide cut in it. The area of a cross-section of the ring is 3·6 cm2. Calculate the value of direct current needed in a coil of 1000 turns uniformly wound round the ring to create a flux of 0·5 mWb in the air gap. Neglect fringing and assume relative permeability for the iron as 650. [ Ans. 2.1735 A] 1.10 Magnetization or B-H Curve The graph plotted between flux density B and magnetising force H of a material is called the magnetisation or B–H Curve of that material. The general shape of the B–H Curve of a magnetic material is shown in Fig. 1.11 The shape of the curve is non-linear This indicates that the relative permeability (μr = B/μ0 H) of a magnetic material is not constant but it varies. The value of μr largely depends upon the value of flux density. Its shape is shown in Fig. 1.12 (for cast steel). 14 Figure 1.11 B-H curve of a magnetic material Figure 1.12 µr-B Curve The B–H curves of some of the common magnetic materials are shown in Fig. 1.13. The B–H curve for a non-magnetic material is shown in Fig. 1.14. It is a straight-line curve since B = o H or B  H as the value of µ0 is constant. Figure 1.13 B-H curve for different magnetic Figure 1.14 B-H curve for a non-magnetic materials materials 1.11 Magnetic Hysteresis When a magnetic material is magnetised first in one direction and then in the other (i.e., one cycle of magnetisation), it is found that flux density B in the material lags behind the applied magnetising force H. This phenomenon is known as magnetic hysteresis. Hence, the phenomenon of flux density B lagging behind the magnetising force H in a magnetic material is called magnetic hysteresis. ‘Hysteresis’ is the term derived from the Greek word hysterein meaning to lag behind. To understand the complete phenomenon of magnetic hysteresis, consider a ring of magnetic material on which a solenoid is wound uniformly as shown in Fig. 1.15. The solenoid in connected to a DC source through a double pole double throw reversible switch (position ‘1’). 15 Figure 1.15 Circuit to trance hysteresis loop Figure 1.16 Hysteresis loop When the field intensity H is increased gradually by increasing current in the solenoid (by decreasing the value of R), the flux density B also increases until saturation point a is reached and curve so obtained is oa. If now the magnetising force is gradually reduced to zero by decreasing current in the solenoid to zero. The flux density does not become zero and the curve so obtained is ab as shown in Fig. 1.16. When magnetising force H is zero, the flux density still has value ob. Residual Magnetism and Retentivity This value of flux density ‘ob’ retained by the magnetic material is called residual magnetism and the power of retaining this residual magnetism is called retentivity of the material. To demagnetise the magnetic ring, the magnetising force H is reversed by reversing the direction of flow of current in the solenoid. This is achieved by changing the position of double pole, double throw switch (i.e., position ‘2’). When H is increased in reverse direction, the flux density starts decreasing and becomes zero and curve follows the path bc. Thus residual magnetism of the magnetic material is wiped off by applying magnetising force oc in opposite direction. Coercive Force This value of magnetising force oc required to wipe off the residual magnetism is called coercive force. To complete the loop, the magnetising force H is increased further in reverse direction till saturation reaches (point ‘d’) and the curve follows the path cd. Again, H is reduced to zero and the curve follows the path de. Where oe represents the residual magnetism. Then H is increased in the positive direction by changing the position of reversible switch to position ‘1’ and increasing the flow of current in the solenoid. The curve follows the path of efa and the loop is completed. Again, of is the magnetising force utilised to wipe off the residual magnetism oe. Hence, cf is the total coercive force required in one cycle of magnetisation to wipe off the residual magnetism. Since the meaning of hysteresis is lagging behind, and in this case flux density B always lags behind the magnetising force, H, therefore, loop (abcdefa) so obtained is called hysteresis loop. 16 Hysteresis Loss When a magnetising force is applied, the magnetic material is magnetised and the molecular magnets are lined up in a particular direction. However, when the magnetising force in a magnetic material is reversed, the internal friction of the molecular magnets opposes the reversal of magnetism, resulting in hysteresis. To overcome this internal friction of the molecular magnets (or to wipe off the residual magnetism), a part of the magnetising force is used. The work done by the magnetising force against this internal friction of molecular magnets produces heat. This energy, which is wasted in the form of heat due to hysteresis, is called hysteresis loss. Hysteresis loss occurs in all the magnetic parts of electrical machines where there is reversal of magnetisation. This loss results in wastage of energy in the form of heat. Consequently, it increases the temperature of the machine which is undesirable. Therefore, a suitable magnetic material is selected for the construction of such parts, e.g., silicon steel is most suitable in which hysteresis loss is minimum. 1.12 Importance of Hysteresis Loop The shape and size of hysteresis loop of a magnetic material largely depends upon its nature. For a particular location, the choice of the magnetic material depends upon the shape and size (i.e., area) of its hysteresis loop. The hysteresis loops of some of the common magnetic materials are shown in Fig. 1.17. (i) Hard steel: The hysteresis loop for hard steel is shown in Fig. 1.17 (a). This loop has larger area which indicates that this material will have more hysteresis loss. Therefore, it is never used for the construction of machine parts. However, its loop shows that the material has high retentivity and coercivity. Therefore, it is more suitable for making permanent magnets. Figure 1.17 Hysteresis loop for different magnetic materials (ii) Silicon steel: The hysteresis loop for silicon steel is shown in Fig. 1.17 (b). This loop has smallest area which indicates that this material will have small hysteresis loss. Therefore, it is most suitable for the construction of those parts of electrical machines in which reversal of magnetisation is very quick e.g., armature of DC machines, transformer core, starter of induction motors etc. (iii) Wrought iron: Figure 1.17 (c) shows the hysteresis loop for wrought iron. This loop shows that this material has fairly good residual magnetism and coercivity. Therefore, it is best suited for making cores of electromagnets. 17 1.13 Application of Ferromagnetic Material Ferromagnetic material (e.g. iron, steel, nickel, cobalt etc) are widely used in a number of applications. The choice of a ferromagnetic material for a particular application depends upon its magnetic properties. Such as, retentivity, coercivity and are of the hysteresis loop. Ferromagnetic materials are classified as being either soft (soft iron) or hard (steel). As shown in the figure 1.18. This shows the hysteresis loop for soft and hard ferromagnetic materials. The table below also gives the magnetic properties of hard and soft ferromagnetic materials. Figure 1.18 Magnetic Property Soft Iron Hard Steel Hysteresis Loop Narrow Large area Retentivity High High Coercivity Low High Saturation flux density High Good (i) The permanent magnets are made from hard ferromagnetic materials (steel, cobalt steel, carbon steel etc). Since these materials have high retentivity, the magnet is quite strong. Due to their high coercivity, they are unlikely to be demagnetized by stray magnetic fields. (ii) The electromagnets or temporary magnets are made from soft ferromagnetic materials (e.g. soft iron). Since these materials have low coercivity, they can be easily demagnetized. Due to high saturation flux density, they make strong magnets. (iii) The transformer cores are made from soft ferromagnetic materials. When a transformer is in use, its core is taken through many cycles of magnetization. Energy is dissipated in the core in the form of heat during each cycle. The energy dissipated is known as Hysteresis loss and is proportional to the area of hysteresis loop. Since the soft ferromagnetic materials have narrow hysteresis loop (i.e. smaller hysteresis loop area), they are used for making transformer cores. 18  CHAPTER 2 PRINCIPLES OF ELECTROMECHANICAL ENERGY CONVERSION 2.1 Introduction Electromechanical devices convert electrical energy into mechanical energy and vice versa. Energy conversion takes place through the medium of magnetic field. Electromechanical energy conversion is a reversible process except for the losses in the system. We can categorise the electromechanical energy conversion devices in the following manner: 1. Transducers: These are the first category of devices that involve low-energy conversion signals from electrical to mechanical or vice versa. Microphones, sensors, and loudspeakers are typical examples of this category of devices which operate generally under linear conditions. 2. Force-producing devices: These are the second category of devices that consist of force- or torque-producing devices with limited mechanical motion. Examples are solenoids, relays, and electromagnets. 3. Continuous energy conversion equipment: Examples are motors and generators that are used for bulk energy conversion and utilisation. 2.2 Forces and Torques in Magnetic Fields According to Lorentz’s force law, in pure magnetic field systems, the force on a moving particle of charge q is indicated by the following vector cross product: ( F =q v  B ) where v is the velocity in m/s of the particle of charge q moving in a magnetic field density of B Tesla (Figure 3.1). The magnitude of this vector cross product is l F = qvBsin = q B sin  = ilB sin  t Figure 2.1 Direction of force (right hand rule II) A 6 m long wire carries a current of 100 A in a magnetic field density of 0.8 T as shown in Figure 2.2. The angle between the magnetic field and the current flow is 110°. What is the force on the wire, and what is its direction? 19  Figure 2.2 Solution F =Bil sin θ = 0.8 ×100 6 sin1100 = 451N 2.3 Force Between Two Current-Carrying Wires As illustrated in Figure 2.3, when a current-carrying conductor is brought near another current- carrying conductor, a magnetic force is experienced between the wires. If we have two parallel wires of length l separated by a distance d currents i1 and i2 flowing in opposite directions in these parallel wires from our previous chapters, we know that a magnetic field intensity H due to current i2 on conductor carrying current i1 at distance d would be (Figures 2.3 and 2.4) i2 Hd = 2 d The force exerted on i1 will be in an outward direction (right-hand rule II) (Figure 2.3) with a magnitude of i2 i1i2l F = Bil sin  = i1  o  l = o N 2 d 2 d where μ0 represents the magnetic permeability of the media which happens to be air in our example. 20  Figure 2.3 Force on parallel current carrying wires Figure 2.4 Magnetic flux lines around current carrying conductor 2.4 Force on a Current-Carrying Wire in Magnetic Field Now we will look at the magnetic force from a different point of view of magnetic flux lines. Let us consider two permanent magnets with their N and S poles facing each other as shown in Figure 2.5. Let us also consider a piece of wire running midway and parallel to the pole faces and right through and perpendicular to the paper. Assuming the wire is thin, the magnetic field and flux lines between the two pole faces will be as plotted in Figure 2.5. If a current flow in the wire and into the paper (Figure 2.6), it will cause a magnetic field of its own which will help the magnetic flux lines in the upper part and oppose the magnetic flux lines in the lower part. As a result, the distorted magnetic flux lines will be as illustrated in Figure 2.6. The distorted lines of flux can be visualised as acting like stretched elastic cords. They will try to return to the shortest paths between pole faces, thereby exerting a force pushing the conductor out of the way. The resultant effect is to strengthen the magnetic field on the upper side and weaken that on the lower side, thus giving the distribution shown in Figure 2.6. This is a good way to visualise how the forces appear on the current carrying wires in magnetic fields. 21  Figure 2.5 Magnetic flux lines without current flowing Figure 2.6 Magnetic flux lines with current flowing Example The two conductors are 2 cm apart, and both carry 2000 A of current in opposite directions as shown in the figure. What is the force on each conductor if both wires are 1 m long? Solution ii l 2000  2000 1 F = o 1 2 = 4  10−7  = 40 N 2 d 2  0.02 2.5 Generated Voltage in Magnetic Systems As stated earlier, we stated Faraday’s law as the voltage induced (e) in an N-turn coil when the flux Φ linking or threading through it changes. d d  e=N = dt dt Now consider the magnetic circuit of Figure 2.7 where a changing voltage v(t) is applied to a coil of N turns wound on a magnetic material to produce a changing magnetic flux of Φ(t) passing through the coil. According to Faraday’s law, changing magnetic flux Φ(t) will induce a voltage e(t) on the coil: d (t ) d  (t ) e(t ) = N = V dt dt According to Lenz’s law, the direction of the induced voltage e(t) on the coil will be such as to cause a current flow in the “closed coil” which will produce a flux opposing the original flux in the core. In more general terms, Φ(t) in the above expression is defined with the surface integral of the dot product B.ds : Φ(t) =  B.ds =Total flux passing through the coil where “s” is the area enclosed by the coil. Therefore, for our case, in scalar terms, Φ(t) = B.s.sin θ 22 Here, θ is the angle between the plane of the area s and the magnetic flux lines. The induced voltage according to Faraday’s law is also known as the “transformer voltage.” If we consider the static single loop and a changing magnetic flux going through it as shown in Figure 2.8, according to Faraday’s law the induced transformer voltage would be d d  dB  etransformer = =N = N. A.(sin  ).   dt dt  dt  Here, θ = Angle between the plane of the area enclosed (A) and flux (Φ) Figure 2.7 Induced voltage in a coil with changing magnetic field Figure 2.8 Generation of induced voltage In accordance with Lenz’s law, the direction of this induced voltage would be so as to make a current flow from a to b outwards to oppose the original flux change that is creating this induced voltage. 23 We can apply the “right-hand rule” to this loop to find the direction of the current due to the induced voltage that would have to flow from a to b (outwards) [If Φ(t) is increasing, the induced voltage is positive]. 2.6 Voltage Induced on a Moving Conductor For a straight conductor with length l moving with velocity v in a uniform magnetic field B which is not changing with time (Figure 2.9), the induced motion voltage on the conductor emotional would be d Induced voltage on the conductor, em = dt If the distance covered by the conductor in Δt seconds is Δx, then d Bldx.sin  em = = = v.l.B.sin  dt dt Figure 2.9 Moving conductor at an angle in a magnetic field Figure 2.10 Moving conductive bar on rails in a magnetic field 24 Here, θ is the angle between velocity v and magnetic field B (from v to B). Direction of this induced voltage can be found by Lenz’s law or by the right-hand rule of Fleming which can be summarised as “If the thumb, forefinger, and middle finger of the right hand are held at right angles to one another with the thumb pointing in the direction of the motion of the conductor the forefinger pointing in the direction of the magnetic field then the middle finger will point in the direction of the induced current. If the angle θ is 90°, then the voltage induced is a maximum: d Bldx.sin  em = = = v.l.B dt dt If in Figure 2.10 magnetic field B is in the direction shown (into the paper) and the conductor l is moving to the right with a velocity of u, then the induced voltage would be d em = = u.l.B dt According to Fleming’s right-hand rule or Lenz’ law, this voltage would be in the direction Vab. 2.7 Force Developed in an Electromagnetic System An electromagnetic system can develop a mechanical force in two ways: 1. by alignment 2. by interaction. 2.7.1 Force of Alignment To be able to understand the force of alignment, consider the two illustrations shown in Figure 2.11 and 2.12, in which two poles are opposite to each other; each is made of a ferromagnetic material. In Figure 3.16 a flux passes from one to the other and the two magnetised surfaces are attracted towards one another. Magnetic attraction forces will try to bring the poles together since this decreases the reluctance of the air gap in the magnetic circuit. Hence, this will increase the flux and consequently the stored energy. In the second case shown in Figure 2.12, forces try to achieve greater stored magnetic energy by two component actions: 1.by attraction of the poles 2.by aligning the poles laterally. If the poles move laterally, the cross-sectional area of the air gap is increased and the reluctance is reduced. This will again increase the flux and consequently the stored magnetic energy as before. In both of the above cases, forces (force of alignment) occur so as to maximise the magnetic energy stored in the magnetic circuit. 25  Figure 2.11 Force of attraction Figure 2.12 Force of attraction and alignment The arrangement (relay) shown in Figure 2.13 is an example of an electromechanical energy conversion device which utilises force of alignment giving rise to linear motion. When the coil is energised, a flux is set up in the relay core and air gap. The air gap surfaces become magnetised and are attracted. Hence, armature plate is pulled in the direction indicated. This relay is used extensively in old type of telephone exchanges as a simple switch. Recently, they have been replaced by digital switches. The force of alignment can also be used to produce rotary motion as in the reluctance motor shown in Figure 1.14. The rotating part (piece) experiences a torque due to the magnetised rotor and pole surfaces attempting to align themselves. Figure 2.13 Telephone exchange relay utilises force of alignment 26  2.14 Reluctance motor utilises force of alignment 2.7.2 Force of Interaction Force of interaction is utilised in the principles of operation of the electrical machines. By passing a current through the coil, a force is experienced on each of the coil sides resulting in a torque about the axis of rotation. A practical machine requires many turns of conductors in order to develop a sufficient torque. Depending on how the conductors are arranged, various electrical machines are created. 2.8 Basic Structure of Rotating Electrical Machines Structure wise electrical machines have two major parts, stator and rotor, separated by an air gap, as shown in Figures 2.15. Stator: This part of the machine does not move and normally is the outer frame of the machine. Rotor: This part of the machine is free to move and normally is the inner part of the machine. Figure 2.15 Typical electrical machine. Magnetic materials of electrical machines require materials of high permeability and low resistivity. Proper mechanical strength is another requirement. In electrical machines, the coil in which the magnetic field is created is called the “field winding.” the coil to which the voltage is applied (or received) is called the “armature winding.” Whether the field or the armature winding is on the stator or rotor winding varies according to the type of the machine. 27 Machine windings are inserted in the slots of the machine (Figure 2.16) or wound on the protruding parts of the relevant structure. Figure 2.16 Rotor, Stator and slots of electrical machines The part of the machine (stator and/or rotor) in which AC voltages and currents are involved, must be made in laminated form for smaller iron losses (Figure 2.17). Laminated structure is not required if the part is not subjected to AC currents. Figure 2.17 Stator and rotor lamination 2.9 Structure Types of Rotating Electrical Machines In “salient pole structure” coil is wound around protruding poles (Figures 2.18). Rotating machines which have a salient pole rotor or stator are called salient pole machines. In salient pole machines, there is a nonuniform air gap. 28  Figure 2.18 Salient pole machine with slots in stator In cylindrical (non-salient) pole structure, coil is wound in slots cut into a cylindrical magnetic structure (Figure 2.19). Rotating machines which have a cylindrical rotor and stator (Figure 2.19) are called “cylindrical” or “non-salient pole” machines. In these machines, there is a uniform air gap. Depending on the machine type, the field and the armature coils can be on the stator or rotor of the machine. Figure 2.19 Cylindrical machines with slots in rotor and stator a. DC machines: Rotor is cylindrical and stator is salient pole (Figure 2.20) field is on stator. armature is on rotor. b. Induction machines: Both rotor and stator are cylindrical (Figure 2.21). Field winding is on the stator. Armature winding is on the rotor. c. Synchronous machines: Stator is cylindrical, and rotor is cylindrical or salient pole (Figure 2.22). Field is on rotor. Armature is on stator. 29  Figure 2.20 Salient pole DC machines (two - pole) Figure 2.21 Induction motor structure Figure 2.22 Salient pole synchronous machine (four – pole) 30 Both stator and rotor are made of ferromagnetic materials. In most machines, slots are cut on the inner periphery of the stator and outer periphery of the rotor structure, as shown in Figure 2.23. Conductors are placed in these slots to form windings. Figure 2.23 Slotted structure of rotor and stator If the stator or rotor (or both) is subjected to a time-varying magnetic flux, the iron core is laminated to reduce eddy current losses. Eddy currents heat up the core and wastes energy. If laminations are provided, the area is reduced and hence resistance is very high which limits the current to a minimum value. The winding in which voltage is induced is called the armature winding. The winding through which a current is passed to produce the primary source of flux in the machine is called the field winding. Permanent magnets are used in some machines to provide the major source of flux in the machine. The three basic and common rotating electrical machines are DC machines induction machines synchronous machines. To understand the behaviour of rotating machinery, we must realise that the rotor structure is a magnetic field, and the stator structure is another magnetic field. In a similar manner, a compass needle tries to align with the Earth’s magnetic field, these two sets of fields attempt to align, and a torque is therefore produced with their displacement from alignment. Thus, in a motor, the stator magnetic field rotates ahead of that of the rotor, pulling on it and performing work. The opposite is true for a generator in which the rotor does the work on the stator. The following formula F = q.v.B.sinθ = i.l.B.sinθ is the simplest and easiest way to calculate the forces acting on systems. Unfortunately, it is not useful apart from some practical situations. A more general method called “energy method” is used to calculate the net forces in the electromechanical energy conversion devices. 31 2.10 Energy Balance Method Basics of the energy balance method rely on the principle of conservation of energy. The simple force-producing device shown in Figure 2.24 is schematically modelled in Figure 2.25 as a lossless, magnetic field-based electromechanical energy conversion device with an electrical input and mechanical output. In between is a lossless magnetic energy storage system which acts as a reservoir between the electrical input and mechanical output terminals. Figure 2.24 Magnetic field-based electromechanical energy conversion device Figure 2.25 Model of the device in figure 2.24 The electrical input has two variables: a voltage e a current i. The mechanical output also has two variables: a force fx a position x. The electromechanical energy conversion will occur through the medium of the magnetic energy. The time rate of change of the stored energy in the magnetic field (Wf) is equal to the electric power input minus the mechanical power output. dW f dx = ei − f x (2.1) dt dt 32 From Farday’s Law, we have d e= (2.2) dt Inserting equation (2.2) in equation (2.1) and multiplying both sides of equation (2.1) with dt, we get dW f = id  − f x dx (2.3) This is the basic equation for the energy method, and it permits us to solve for the force as a function of the flux λ the mechanical output position x. The equation above holds for motor action. It applies equally well to generator action; right- hand terms would then have negative values. 2.11 Energy in Singly-Excited Magnetic Field Systems 2.11.1 Derivation of force from energy So far, we have mainly dealt with fixed-geometry magnetic circuits. The electromechanical energy conversion device shown in Figure 2.26 constitutes a lossless magnetic energy storage system with its magnetic core and armature, and it has an air gap which can change in dimensions only in the direction of displacement x. The system can schematically be shown as in Figure 2.25. We will assume that magnetic energy stored in the system is totally in the air gap. The air gap in the electromechanical energy conversion device therefore acts like a buffer energy between the electrical and mechanical systems. Since the system is lossless, the magnetic energy stored in the system (Wf ) is a state function of the two independent variables λ and x. As a property of state functions, the value of Wf is determined by the final values of x and λ, and it is irrelevant of how they are brought to these values. The case resembles the potential energy of a particle which is independent of the path through which the particle has come to its final height (PE = mgh ). Therefore, as illustrated in Figure 2.27, in order to reach to a certain value of stored magnetic energy Wf 1 (x1, λ1) in the system we can choose any path which starts from the origin in Figure 2.27 and ends at point Wf1 (x1, λ1). Three such paths a (a= a1 + a2), b, and c are shown in Figure 2.27. The magnetic energy stored (Wf ) in a lossless electromechanical energy conversion system such as the one in Figure 2.26 can be obtained by the integration of the energy balance equation (2.3). dW f = id  − f x dx → W f =  id  −  f x dx Because of the properties of the state functions, the result of the above integral is irrelevant of how we arrived at this value of Wf. We can go along any of the paths (a, b, or c) shown in Figure 2.27. They all ought to give the same result. 33  Figure 2.26 Singly excited electromechanical system Figure 2.27 Energy stored in magnetic field Wf (state function). Let us assume that we have chosen path “a” which consists of the addition of two paths a1 and a2 (a = a1 +a2). Along path a1, we have kept λ as zero. Hence, there will be no magnetic field, and obviously there will be no force fx, and therefore there will be no magnetic energy stored in the system. Hence, energy balance equation along this path (a1) reduces to dW fa1 = 0 → W fa1 = 0 So, no contribution to stored magnetic energy comes from this path a1. In the second part of the path a (a2), we keep x constant at x1 and increase λ until we reach to λ = λ1. In this case, the term dx will be zero. Therefore, the energy balance equation reduces to W f dW f = id  → i = (2.4)  x where the partial derivative is taken while holding x constant. We must keep in mind that the above is a mathematical approach and is not in any way related to whether x is held constant or not during the operation of the actual device. In actual fact, this is the case of the singly excited magnetic energy storage system with a static air gap. The stored energy in the gap of these devices does not change, and there is no electromechanical energy conversion. 34 Now going back to Figure 2.27, let us assume that this time we have chosen path “c” which consists of the addition of two paths c1 and c2 (c = c1 +c2). Along path c1, we have kept x as zero. Hence, there will be no air gap in the system, and as the permeability of the core is assumed to be infinity, there will be no magnetic energy stored in the system. Hence, energy balance equation along this path (c1) reduces to dW fc1 = 0 → W fc1 = 0 So, no contribution to stored magnetic energy comes from this path c1. In the second part of the path c(c2), we keep λ constant at λ1 and increase x until we reach to x = x1. In this case, the term dλ will be zero. Therefore, the energy balance equation reduces to W f ( , x) dW f = − fdx → f = − (2.5) x  where the partial derivative is taken while holding λ constant. If Wx(x, λ) is known as a function of x and λ, force fx can be calculated from the above equation. This is the equation we need for determining the force in an electromechanical energy conversion configuration. We must keep in mind that the above is purely a mathematical approach and is not in any way related to whether λ is held constant or not during the operation of the actual device. For a linear system where L(x) is the self-inductance of the configuration, we have  = i.L( x) Integrating equation (2.5) while keeping x a constant and inserting the above linear relation in the result, we get    2 1 W f =  id  =  d = = i 2.L( x) (2.6) 0 0 L( x) 2 L( x) 2 Now let us insert this in equation (2.5) W f ( , x)   2   2 dL( x ) f =− =−   = x  x  2 L( x )  2 L( x) 2 dx  Using this expression, we get i 2 dL( x)  = L( x)i → f x = (2.7) 2 dx 2.11.2 Derivation of force from co-energy According to the energy equation (2.3) for a particular value of the air gap length shown in Figure 2.26, the energy stored in the magnetic field Wf is given by dW f = id  which means that the area between the λ axis and the λ− i characteristic represents the energy stored in the magnetic field in the air gap. The area between the i axis and the λ− i characteristic is known as the co-energy Wf ′ and is defined as 35  i dW f' =   di 0 From figure 2.28 W f' + W f = i Coenergy has no physical significance. However, it can be used to derive expressions for force (or torque) developed in an electromagnetic system. Let us remember that if λ− i curve is nonlinear → Wf ′ > Wf if this curve is linear → Wf ′= Wf Now we shall find the expression for force fx from another approach. Without changing anything else, let us slowly push the movable part in Figure 2.26 from position x1 to another position x2 so that air gap decreases. The current i before and after the movement will be the same. Since Wf ′ is a state function, what happens to current i during the motion has no importance. We can assume the movable part has moved slowly and the current i has remained the same. The λ-i characteristics of the system for these two positions are shown in Figure 2.29 as a and b. Figure 2.28 Stored magnetic energy and co-energy Figure 2.29 Energy and co-energy of magnetic fields 36 The operating point has therefore moved upward from point a to b as shown in Figure 2.29. 2 During the motion, dWe =  ei.dt =  i.d  = area abcd 1 dW f = area obc − oad dWmech = dWe − dW f = area abcd + area 0ad – area 0bc = area 0ab If the motion has occurred under constant-current conditions, the mechanical work done is represented by the shaded area, which, in fact, is the increase in the coenergy (or decrease in the magnetic energy): dWmech = dWf ′ If fx is the mechanical force causing the differential displacement dx, fxdx = dWmech = dW′ W f' (i, x) fx = (2.8) x i = constant We have found another expression for force fx. In fact, this and the expression we found before for the same force fx ought to give the same result. It is up to the user to decide which expression to use. It is a matter of preference of the user. If we assume that the magnetic system in Figure 2.26 is linear, then if L(x) is the self-inductance of the configuration,  = i.L( x) Because the system is linear, we also have Wf′=Wf and hence from equation (3.7), W f' (i, x) W f (i, x) fx = = x x i i Inserting (3.6) in the above equation, W f   i2  i 2 dL( x ) fx = =  L( x)  = (2.9)  x i x  2  i 2 dx which is as expected the same as the result we obtained in equation (3.7) in the above paragraphs. 37 2.12 Force of Alignment Between Parallel Magnetized Surfaces From here onwards, we will be using the following relationship for calculating the force in electromechanical energy conversion arrangements: W f fx = x i The force of alignment between the core members of the arrangement shown in Figure 2.30 is Figure 2.30 Force of alignment dW f Force of alignment, F = dx Wf = (Energy density) x (Volume of the air gap) = wf x v B2 = w f  ( A.x) =  ( A.x) 2 0 As a result dW f B2 A F= = dx 2 o This force will be in the direction to decrease the reluctance of the system by decreasing the distance (shortening the flux path) between the magnetic poles so that flux can travel the path of least resistance. 2.13 Lateral Force of Alignment Between Parallel Magnetized Surfaces Ignoring the effect of leakage flux, let the cross-sectional area of the gap be xl and the gap length be lg. The air gap volume is given by (Figure 2.31) Volume = V = llgx 38  Figure 2.31 Lateral force of alignment B2 W f = w f ll g x = ll g x 2 o dW f B2 F= = ll g dx 2 o 2.14 Alternative Force Calculation in A Linear System In Figure 2.26, if the reluctance of the magnetic core path is negligible compared to that of the air gap path, the λ-i relation becomes linear. For this idealised system where L(x) is the inductance of the coil, whose value depends on the air gap length, we have found the force equation as follows (equations 2.7 and 2.9), 1 dL( x) fm = i2 2 dx We can yet approach the linear system from another point of view to calculate the force produced in the air gap. For the system in Figure 2.26, if the reluctance of the magnetic core path is neglected, the field energy is Bg2 Wf =  Volume of air gap 2 o Bg2 Wf =  Ag.2 g 2 o where Ag is the cross-sectional area of the air gap.   Bg2  fm =   Ag.2 g  g  2 o    Bg2 =  2 Ag 2 o 39 Hence, the force per unit area of air gap, called magnetic pressure Fm, is Bg2 Fm = N/m 2 (2.10) 2 0 In most linear systems, the above approach may seem to be more practical to calculate the force in the air gap. Example 2.1 The magnetic circuit in Figure 2.32 has N (number of turns) = 1,000, i (current) = 1 A, air gap width = 1.0 cm, air gap depth = 2 cm, and air gap length lg = 2 mm. Assuming the permeability of the core is infinity and the leakage and fringing flux effects are negligible, calculate a. the force of attraction on the two faces of the air gap. b. energy stored in the air gap. Figure 2.32 Solution a. Ampere’s circuital law, → Hl = Ni  Bg  Ni 0   lg = Ni → Bg =  0  lg 2  Ni 0     Ag Bg2  lg  N 2i 2 0 Force = f m =  Ag = =  Ag 2 0 2 0 2lg2 10002 12  4 10−7 =  2  1 10−4 = 31.4 N 2  22 10−6 Bg2 b. Wf =  air gap volume 2 0 Bg2 =  Ag  lg = 31.4  2 10−3 = 0.0628 J 2 0 Example 2.2 The electromechanical energy conversion device shown in Figure 2.33 is in equilibrium when the gap distance g is at 0.5 mm and the current intake is i = 1 A. The torque produced by the system on the arm is 30 nm. If the cross-sectional area of the pole faces is 4 × 4 cm2 and the reluctance of the core can be neglected, calculate the length l of the arm. 40  Figure 2.33 Solution Let us write the Ampere’s circuital law by ignoring the core, Bg Ni = H g  2 g =  2g 0 Ni 0 → Bg = 2g Bg W = Magnetic energy stored in the air gap =  (Volume of the air gap) 2 0 2 dW Bg  Ni 0  A F= =  2A =    dx 20  2 g  0 5002 12  4 10−7  4  4  10−4 = = 502.4 N-m 4  0.52 10−6 Torque = 30 = F x l 3000 The length of the arm, l = = 5.97cm 160 2.15 Rotating Electrical Machines Most of the electromechanical energy converters and especially the high-power ones produce rotational motion. In these types of energy converters, the rotor is mounted on a shaft and is free to rotate between the poles of the stator. Let us take a general case in which both stator and rotor have windings carrying currents is and ir, as shown in Figure 2.34. The current is fed into the rotor circuit through fixed brushes and rotor-mounted slip rings. As we saw before, Wf, the magnetic energy deposited in the system, is a state function and how it comes to a certain value is of no consequence. So let us assume the system is static, there is no mechanical output, and Wf is the stored magnetic field energy within the system. Consequently, 41  dW f = es is dt + er ir dt = is d s + ir d r (2.11) For a linear magnetic system, s = Lss is + Lsr ir (2.12) r = Lrs is + Lrr ir (2.13) Here, Lss is the self-inductance of the stator winding. Lrr is the self-inductance of the rotor winding. Lrs and Lsr are mutual inductances of the rotor and stator windings. Figure 2.34 General structure of a rotating machine From equations (4.1), (4.2), and (4.3), dW f = is d ( Lssis + Lsr ir ) + ir d ( Lsr is + Lrr ir ) = Lss is dis + Lrr ir dir + Lsr d (isir ) Total field energy will be is ir is ir W f = Lss  is dis + Lrr  ir dir + Lsr  d (isir ) 0 0 0 1 1 = Lss is2 + Lrr ir2 + Lsr is ir 2 2 Similar to the force equation we obtained in previous chapters, in rotational electromechanical energy conversion systems, the torque developed would be 42  W f' ( i,  m ) T=  m i = constant In a linear magnetic system, the stored magnetic energy Wf would be equal to the coenergy of the system Wf′. Therefore, from the two equations above, we can write 1 dL 1 dL dL T = is2 ss + ir2 rr + is ir sr (2.14) 2 d m 2 d m d The first two terms on the left-hand side of the above equation represent “torques produced in the machine because of variation of self-inductance with rotor position.” This component of torque is called the “reluctance torque.” The third term represents “torque produced by the variation of the mutual inductance” between the stator and rotor windings. Here, we should take notice that the derivation is with respect to mechanical (shaft) angle θm. 2.16 Basics of Motors and Generator Continuous electromechanical energy converters are systems that convert electrical and mechanical energies in continuous form from one to the other. As said before when conveying of energy is considered, electrical energy has economical advantages compared to other forms of energy (e.g., heat energy). Therefore, continuous electromechanical energy converters called electrical machines are important. We divide electrical machines into two groups: Motors that convert electrical energy into mechanical energy in continuous form. Generators that convert mechanical energy into electrical energy in continuous form. An electrical machine can be made to work as a generator or as a motor. Both are inherent in electrical machines. In other words, in these machines, the energy conversion is reversible. In their normal performance, there are voltages that are induced in motors there are forces that are developed in electrical generators Both motors and generators work on two principles: 1. A force (F) is produced on a current-carrying wire in magnetic field and the wire starts accelerating (conversion of electrical energy to mechanical energy). 2. A voltage is induced on an accelerating wire in a magnetic field and the direction of this induced voltage is opposite to the current that is causing the acceleration of the wire (conversion of mechanical energy to electrical energy). In this way, the force on the wire decreases and finally becomes zero, and the wire travels at a constant speed. Conversion of electrical energy to mechanical energy and vice versa takes place simultaneously in these machines and therefore they are reversible There is production of force in generators and production of induced voltage in motors. That is why we call them electrical machines. There are two main parts of electrical machines: 43 1. The winding where we apply (or receive) the voltage (ARMATURE WINDING). 2.The winding that creates the main magnetic field (FIELD WINDING). In electrical machines, the stationary part is called the stator and the rotating part connected to a shaft that couples the machine to its mechanical load is called the rotor. Depending on the type of the motor, field winding and the armature winding can be in the rotor or in the stator. In order for a machine to convert electrical energy to mechanical energy or vice versa a magnetic field must exist within its air gap. This field can be set up electromagnetically or by a permanent magnet. It is usually produced electromagnetically by exciting the field windings of the machine. The machine can convert electrical and mechanical energies into one another using the armature winding. Therefore, the armature winding is where the energy conversion takes place. Due to this reason, the power rating of a machine is actually the power rating of the armature section. 44  CHAPTER 3 TRANSFORMER 3.1 Introduction Electrical power transformer is a static device which transforms electrical energy from one circuit to another without any direct electrical connection and with the help of mutual induction between two windings. It transforms power from one circuit to another without changing its frequency but may be in different voltage level. What is a Transformer? A transformer is a static piece of equipment used either for raising or lowering the voltage of an AC supply with a corresponding decrease or increase in current. The use of transformers in transmission system is shown in the Figure below. Figure 3.1 A simple single phase power system 3.2 Principle of Operation of a Transformer A transformer in its simplest form will consist of a rectangular laminated magnetic structure on which two coils of different number of turns are wound as shown in Figure 3.2. Figure 3.2 A Typical Transformer The winding to which AC voltage is impressed is called the primary of the transformer and the winding across which the load is connected is called the secondary of the transformer. 45  Figure 3.3 Two-winding transformer Depending upon the number of turns of the primary (N1) and secondary (N2), an alternating emf (E2) is induced in the secondary, as shown in figure 3.3. This induced emf (E2) in the secondary causes a secondary current I2. Consequently, terminal voltage V2 will appear across the load. If V2 > V1, it is called a step up-transformer. On the other hand, if V2 < V1, it is called a step-down transformer. When an alternating voltage V1 is applied to the primary, an alternating flux Φ is set up in the core. This alternating flux links both the windings and induces emfs E1 and E2 in them according to Faraday’s laws of electromagnetic induction. The emf E1 is termed as primary emf and emf E2 is termed as Secondary emf. The following points may be noted carefully: (i) The transformer action is based on the laws of electromagnetic induction. (ii) There is no electrical connection between the primary and secondary. (iii) There is no change in frequency i.e., output power has the same frequency as the input power. 46 Can DC Supply be used for Transformers? croThe DC supply cannot be used for the transformers. This is because the transformer works on the principle of mutual induction, for which current in one coil must change uniformly. If DC supply is given, the current will not change due to constant supply and transformer will not work. There can be saturation of the core due to which transformer draws very large current from the supply when connected to DC. Thus, DC supply should not be connected to the transformers. 3.3 Constructional features of two-winding transformers We usually design a power transformer so that it approaches the characteristics of an ideal transformer. To achieve this, following design features are incorporated: (i) The core is made of silicon steel which has low hysteresis loss and high permeability. Further, core is laminated in order to reduce eddy current loss. These features considerably reduce the iron losses and the no-load current. (ii) Instead of placing primary on one limb and secondary on the other, it is a usual practice to wind one-half of each winding on one limb. This ensures tight coupling between the two windings. Consequently, leakage flux is considerably reduced. (iii) The winding resistances are minimized to reduce Copper loss and resulting rise in temperature and to ensure high efficiency. Transformers are of two types: (i) core-type transformer is shown in figure 3.4 whiles the shell- type transformer is shown in figure 3.5. Core-Type Transformer: In a core-type transformer, half of the primary winding and half of the secondary winding are placed round each limb to reduce the leakage flux. Figure 3.4 Core type Transformer Shell-Type Transformer: This method of construction involves the use of a double magnetic circuit. Both the windings are placed round the central limb to ensure a low-reluctance flux path. 47  Figure 3.5 Shell type transformer 3.3.1 Comparison of Core and Shell Type Transforms CORE TYPE SHELL TYPE The winding encircles the core The core encircles most part of the winding It has single magnetic circuit It has double magnetic circuit The core has two limbs The core has three limbs The cylindrical coils are used The multilayer disc or sandwich type coils are used The winding are uniformly distributed on The natural cooling does not exist as the two limbs hence natural cooling is effective windings are surrounded by the core. Preferred for low voltage transformers Preferred for high voltage transformers. 3.4 EMF Equation of a Transformer Figure 3.6 shows the representation of alternating flux, varying sinusoidally, which increases from its zero value to maximum value (Φm) in one-quarter of the cycle, that is in one-fourth of a second where f is the frequency of AC input in hertz. 48  Figure 3.6 Representation of alternating flux m The average rate of change of flux is given by 1 , that is 4 f m Wb/s or V. 4f Therefore, average emf/turn = 4 f m V. Let N1 and N2 be the number of turns in the primary and secondary. RMS value Again, form factor = = 1.11 Average value Rms value/turn = 1.11 x 4fΦm = 4.44fΦm V. The rms value of induced emf in primary winding is given by E1 = (4.44 f m )  N1 = 4.44 f m N1 = 4.44 fBm AN1 (3.1) m where Bm = is the maximum value of flux density having unit Tesla (T) and A is the area of cross-section A Similarly, RMS value of induced emf in secondary winding is E2 = (4.44 f m )  N 2 = 4.44 f m N 2 = 4.44 fBm AN 2 (3.2) From equation (3.1) and (3.2), we have E1 E2 = = 4.44 f m = constant (3.3) N1 N 2 49  E1 N1 = =a (3.4) E2 N 2 where ‘a’ is the turn ratio of the transformer N1 Thus, a = (3.5) N2 Equation (3.3) shows that emf induced per turn in primary and secondary windings are equal. In an ideal transformer at no load, V1 = E1 and V2 = E2, where V2 is the terminal voltage of the transformer. Equation (3.3) becomes V1 N1 = (3.6) V2 N 2 Example: The number of turns in the secondary coil of a 22 kVA, 2200 V/220 V single-phase transformer is 50. Find the (i) number of primary turns, (ii) primary full load current, and (iii) secondary full load current. Neglect all kinds of losses in the transformer. Example: A 25 kVA single-phase transformer has the primary and secondary number of turns of 200 and 400, respectively. The transformer is connected to a 220 V, 50 Hz source. Calculate the (i) turns ratio, and (ii) mutual flux in the core. 50 Example 1. The primary voltage of an iron core single-phase transformer is 220 V. The number of primary and secondary turns of the transformer are 200 and 50, respectively. Calculate the voltage of the secondary coil. Example 2. The number of primary turns of a 30 kVA, 2200/220 V, 50 Hz single-phase transformer is 100. Find the (i) turns ratio, (ii) mutual flux in the core, and (iii) full load primary and secondary currents. 3.5 Concept of Ideal Transformer A transformer is said to be ideal if it satisfies following properties: i) It has no losses. ii) Its windings have zero resistance. iii) Leakage flux is zero i.e. 100 % flux produced by primary links with the secondary. iv) Permeability of core is so high that negligible current is required to establish the flux in it. NOTE: For an ideal transformer, the primary applied voltage V1 is same as the primary induced emf E1 as there are no voltage drops. For ideal transformer: Volt-Ampere Rating Transformer rating is specified as the product of voltage and current and called VA rating. 51 The full load primary and secondary currents which indicate the safe maximum values of currents which transformer windings can carry can be given as: 3.5.1