advertisement

Introduction to Electrical Machines CHAPTER ONE ELECTROMAGNETIC PRINCIPLES 1.1. INTRODUCTION Magnetism plays an integral part in almost every electrical device used today in industry, research, or the home. Generators, motors, transformers, circuit breakers, televisions, computers, tape recorders, and telephones all employ magnetic effects to perform a variety of important tasks. The dynamic age of electricity began with the work of Hans Christian Oersted (17711851), who demonstrated in the year 1819 that a current-carrying conductor produced a magnetic field. This was the first time that a relationship was shown to exist between electricity and magnetism. His discovery set off a chain of experiments all across Europe which culminated in the discovery by Michael Faraday (1791-1867) of his law of electromagnetic induction in 1831. Faraday showed that it was possible to produce an electric current by means of a magnetic field. This led; in a very short time, to the development of electrical generators, motors, and transformers, and opened up our modern electrical era. All electromagnetic devices make use of magnetic fields in their operation. These magnetic fields may be produced by permanent magnets or electromagnets. Magnetic fields are created by alternating- and direct-current sources to provide the necessary medium for developing generator action and motor action. Throughout this book we will be studying the application of magnetic fields to electromechanical energy conversion processes as demonstrated in rotating electric machinery. Also, transformers provide energy transfer from one electric circuit to another via the changing magnetic field. It will become apparent that there is both transfer and storage of energy in the magnetic fields of the various electromagnetic devices. Hence all electromagnetic devices are constructed with appropriate magnetic circuits. 1.2. MAGNETIC FIELDS The oldest magnetic instrument is a suspended permanent magnet, called a compass. We can define a magnetic field as a region in space in which a compass needle is acted upon. In a region where there are no large magnetic objects, the compass needle points in a general north-south longitudinal direction, with the "north" pole of the compass pointing to the earth's north magnetic pole. However, we know that similar to the law of electric charges, unlike magnetic poles attract and like magnetic poles repel. In spite of the fact that the attracting poles of the compass and earth must be of opposite magnetic polarity, this north-seeking pole of the compass is defined as the north pole. Similarly, it would be correct to describe the other (unmarked) pole of the compass as the south-seeking pole. For brevity this pole is called the south pole. It is well known that a bar of iron can be magnetized by placing it in contact with a strong magnet. By observing the direction of the compass needle at many points around the magnetized bar, a map of the magnetic field can be traced. A map of these lines 1 Chapter One: Electromagnetic Principles can be obtained by the familiar method of sprinkling iron filings on a sheet of paper held over the magnetized bar. When this is done, the pattern of Fig, 1-1 is produced. The map of Figure1-1 should not be interpreted too literally. The iron filings are just a local manifestation of the direction of the magnetic field at that point in apace. Each particle of iron has in effect become a small magnet and is aligned with the magnetic field of the larger magnet (the magnetized bar). Although this map seems to show "lines of force." the lines do not actually exist in space. They can, however, be conceptualized and treated as if they had physical reality. This visualization of magnetic lines of force which was developed by Faraday will be of great value in our understanding of electromagnetic principles. Properties of Magnetic Lines of Force The following properties may be ascribed to magnetic lines of force: Property 1. Magnetic lines of force are directed from north to south outside a magnet. The direction is determined by the north pole of a small magnet held in the field. Figure 1.1 Magnetic field pattern near a magnet Figure 1.2 Magnetic field distortion Property 2. Magnetic lines of force are continious. Property 3. Magnetic lines of force enter or leave a magnetic surface at right angles. Property 4. Magnetic lines of force cannot cross each other. Property 5. Magnetic lines of force in the same direction tend to repel each other. Property 6. Magnetic lines of force tend to be as short as possible. Property 7. Magnetic lines of force occupy three-dimensional space extending (theoretically) to infinity. 2 Introduction to Electrical Machines These properties can be seen in the field map of Figure 1.2. Because of properties 1 and 2, we must assume that there is a magnetic field within the bar; the direction of the field is south to north. The lines are perpendicular to the magnetic surface because of property 3. The lines spread out because of properties 4 and 5, and they assume their shape because of the interaction of properties 4, 5, 6, and 7. Because of property 6, if the lines "find" an easy magnetic path (e.g., through iron), they will prefer this to a more difficult path-through air, as seen in Figure 1.2, which is the field map around a magnet when a piece of iron is brought near it. The iron bar is an "easier" path than the air, hence the lines tend to concentrate around this part of the circuit. We say that the reluctance of the iron is less than the reluctance of air, hence the iron is an easier path for the flux lines. Reluctance of a magnetic circuit may be described as magnetic resistance which tends to oppose the establishment of magnetic flux lines. Magnetic Field Produced by Current-Carrying Conductor A magnetic field is always associated with a current-carrying conductor, as illustrated in Figure 1.3. Exploring the magnetic field by means of a compass, we observe the following: 1. The magnetic field is strongest perpendicular to the current direction. Figure 1.3 Direction of magnetic field around a currcnt-carrying conductor. 2. As we traverse a path around the conductor, we find that the magnetic field is always tangent to the direction of current flow. We can trace a path around the conductor so that continuous magnetic lines of force surround the conductor. 3. If we reverse the direction of current flow, the direction of the magnetic field also changes. 4. The field is strongest near the wire and decreases as we move farther from it. (We can ortain a measure of field strength by trying to deflect the magnet needle from the position it has assumed in the field. At a point where the field is strong, it will be more difficult to deflect it than at a point where it is weak.) 5. If we look at a single current-carrying conductor end on, and draw it as in Figure 1.3, where the symbol ⊕ indicates current flowing into the page, it is easier to draw the magnetic field. If we reverse the current, we have the symbol for current coming out of the page, and we have the situation depicted in Figure 1.3. The dot and cross symbols, respectively, represent the head and tail of an arrow. 6. If we grasp the conductor with our right hand, the thumb pointing in the direction of the current, our fingers will point in the same direction as the north pole of the 3 Chapter One: Electromagnetic Principles compass. This method of determining the directions of current flow in a conductor and the surrounding lines of force is called Ampere's right-hand rule as illustrated in Figure 1.4. Field or flux line Current-carrying conductor Figure 1.4 Ampere’s right hand rule showing the direction of field Practical Magnetic Circuits If we construct a coil of many turns, we can increase the magnetic field strength very greatly, as shown in Figure 1.5. We can also increase the magnetic field strength by increasing the magnitude of current in the coil. A cylindrical coil closely wound with a large number of turns of insulated wire is called solenoid . Thus we see that the magnetic field strength is proportional to both the number of turns and the current. We can determine the direction of the magnetic field in a cylindrical coil of many turns of insulated wire by using our right hand. If we grasp the coil with our right hand with the fingers pointing in the direction of the current, the thumb will point in the direction of the north pole. This method of determining directions of current flow in a coil and magnetic fields of force is another form of Ampere's right-hand rule. Andre Marie Ampere (1775-1836), pursuant to the experimental work of Oersted, developed extensively the foundations of electromagnetic theory. Refer to Figure 1.5. Several practical magnetic circuits are illustrated in Figure 1.6. Figure 1.5 Magnetic field inside a long solenoid 4 Introduction to Electrical Machines (a) permanent magnet (b) Lift electromagnet (c) Vertical contactor (e) Watt-hour meter (d) Transformer (f) Synchronous machine Figure 1.6 Practical magnetic circuits: (a) permanent magnet; (b) lift electromagnet; (c) vertical contactor; (d) transfonner; (e) watthour meter; (f) synchronous machine 1.3. 1.3.1. ELECTROMAGNETIC RELATIONSHIPS Magnetic Lines of Force The "quantity of magnetism" which exists in a magnetic field is the magnetic line of force, or more simply, the magnetic flux. In the SI system magnetic flux is measured in units called webers, abbreviated Wb, and its symbol is φ ( (the Greek lowercase letter phi). The weber is defined in terms of an induced voltage, so that the definition of the unit will be postponed until we study electromagnetic induction. Although there is no actual flow of magnetic flux, we will consider flux to be analogous to current in electric circuits. 5 Chapter One: Electromagnetic Principles 3.2. Magnetic Flux Density The total magnetic flux that comes out of the magnet is not uniformly distributed, as can be seen in Figure 1.2. A more useful measure of the magnetic effect is the magnetic flux density, which is the magnetic flux per unit cross-sectional area. We will consider two equal areas through which the magnetic flux penetrates at right angles near one end of the permanent magnet along its centerline. From the illustration it becomes apparent that there is a greater amount of magnetic flux passing through an area that is nearer the magnet pole. In other words, the magnetic flux density increases as we approach closer to the end of the magnet. However, it must be noted that the magnetic flux density inside the magnet is uniformly constant. Magnetic flux density is measured in units of tesla (T) and is given the symbol B. One tesla is equal to 1 weber of magnetic flux per square meter of area.We can state that B= Φ A 1.1 where B = magnetic flux density, T Φ = magnetic flux, Wb A = area through which Φ penetrates perpendirularly, m2 Example 1.1 The total magnetic flux out of a cylindrical permanent magnet is found to be 0.032 mWb. If the magnet has a circular cross section and a diameter of 1 cm, what is the magnetic flux density at the end of the magnet? Solution The total flux = 0.032 x 10-3 Wb, cross-sectional area of magnet: A= πD 2 π(0.01) 2 = = 78.53 × 10− 6 m 2 4 4 Φ 0.032 × 10 −3 B= = = 0.407 T A 78.53 × 10 − 6 Note that this magnetic flux density exists only at the immediate end of the magnet. As we move away from the end of the magnet, the magnetic flux spreads out, and therefore the magnet flux density decreases. 3.3. Magnetomotive Force We have seen that an increase in the magnitude of current in a coil or a single conductor results in an increase in the magnetic flux. If the number of turns in a coil are increased (with the current remaining constant), there is an increase in magnetic flux. Therefore, the magnetic flux is proportional to the products of amperes and turns. This ability of a coil to produce magnetic flux is called the magnetomotive force. Magnetomotive force is abbreviated MMF and has the units of ampere-turns (At). The magnetomotive force is given the symbol Fm. Strictly speaking, the units of MMF are amperes because turns are 6 Introduction to Electrical Machines dimensionless quantities. However, from a pedagogical standpoint, we prefer and shall use throughout this book the units of ampere-turns (At) for MMF. We may write Fm = NI 1.2 where Fm = magnetomotive force (MMF), At N = number of turns of coil I = excitation current in coil, A Magnetomotive force in the magnetic circuit is analogous to electromotive force in an electric circuit. Example 1.2 The coil in Figure 1.7 has 1000 turns wound on a cardboard toroid. The mean (or average) diameter D of the toroid is 10 cm, and the cross section is 1 cm. The total magnetic flux in the toroid is 3µWb when there is an excitation current of 10 mA in the coil. (a) What is the magnetic flux when the current is increased to 20 mA? (b) What is the magnetic flux density within the coil when the current is 20 mA? Solution (a) If we double the current to 20 mA. then Fm = NI = 1000 × 20 × 10−3 = 20At and Φ must double to 6 µWb. (b) For a toroid. the magnetic flux is assumed to be uniform across the interior crosssectional area of the coil. From Eq. (1.1), B= Φ 6 × 10 −6 = = 76 mT A ( π / 4)(1 × 10 − 2 )2 Figure 1.7 Toroid coil. 7 Chapter One: Electromagnetic Principles 3.4. Magnetic Reluctance In Example 1.2 we have seen that doubling the driving force (MMF) in the circuit results in a doubling of the output quantity (magnetic flux). We consider this ratio of MMF to Fm magnetic flux: = ℜm 1.3 Φ where Fm = NI , the MMF, At Φ = magnetic flux, Wb ℜm= reluctance of the magnetic circuit. At/Wb Transposing, we have Fm = ℜ m Φ which shows us that the magnetic flux is directlv proportional m the magnetomotive force. This equation represents Ohm's law of magnetic circuits. The proportionality factor ℜm, is called the reluctance of the magnetic circuit and is obviouslv, analogous to resistance in an electric circuit. Assuming that a coil has fixed turns and a constant excitation current, the amount of magnetic flux produced will depend on the material used in the core of the coil. A much larger amount of flux can be produced in an ironcore coil than in an air-core coil. Thus we see that the reluctance of the magnetic circuit depends on the material properties of the magnetic circuit. For our purposes, the materials are classified as either magnetic or nonmagnetic. Only the ferrous (irons and steels) group of metals, including cobalt and nickel, are magnetic materials. All other materials, such as air, insulators, wood, paper, plastic, brass, and bronze, including vacuum, are nonmagnetic materials. The strength and pattern of the magnetic field in nonmagnetic materials would be identical to that of air or vacuum (free space). In our discussions we will assume that the magnetic properties of air and vacuum are the same. We consider some of the peculiar characteristics of magnetic materials in subsequent sections. The reluctance of a homogeneous magnetic circuit may be expressed in terms of its physical dimensions and magnetic property as follows: ℜm = l µA 1.4 where ℜm = reluctance of the magnetic circuit, At/Wb l = average or mean length of the magnetic path, m A = cross-sectional area of the magnetic path, m2 µ = µ0×µr , absolute (or total) permeability of the magnetic path, H/m Reluctance is in essence magnetic resistance, that is, the property of a magnetic circuit which is reluctant or unwilling to set up magnetic flux. The reciprocal of reluctance is termed as permeance, which is anologous to conductance in electric circuits. 8 Introduction to Electrical Machines 1.3.5. Permeability Permeability is the magnetic property that determines the characteristics of magnetic materials and nonmagnetic materials. The permeability of free space and nonmagnetic materials has the following symbol and constant value in SI units: µ 0 = 4 π × 10 −7 H / m As we can see, the reluctance of magnetic materials µr is much lower than that of air or nonmagnetic materials µ0 . From the inverse relationship of reluctance and permeability, we determine that the total permeability of magnetic materials is much greater than that of air. However, the value of permeability varies with the degree of magnetization of the magnetic material and, of course, the type of material. Since the permeability of magnetic materials µr is variable, we must employ magnetic saturation (B-H) curves to perform magnetic circuit calculations. Permeability in magnetic circuits is somewhat analogous to conductivity in electric circuits. Example 1.3 In Figure 1.7 we assume that the magnetic flux is practically uniform in the cross-sectional area of the toroid. The mean path length is 0.314 m and the crosssectional area through which the flux exists is 78.5 x 10-6 m2. Calculate the number of ampere-turns required to set up magnetic flux of 1 Wb. Solution The reluctance of the homogeneous magnetic circuit is ℜm = l 0 .314 = = 3 .18 × 10 9 At / Wb − 7 µA 4 π × 10 × 78 .5 × 10 − 6 F = ℜmΦ = 3.18 × 109 ×1.0 = 3.18 ×109 At This is obviously a very large number and we may conclude that the path reluctance is very high. This means that it is comparatively difficult to establish a large magnetic flux in air. For this reason, when we need high flux densities, it becomes necessary to use materials having high values of permeability (such as iron or steel) for large portions of the magnetic paths. 1.3.6. Magnetic Field Intensity One other important magnetic quantity is the magnetomotive force gradient per unit length of magnetic circuit, or more commonly, the magnetic field intensity. Its symbol is H and from the definition, F H= m l 1.5 the unit is ampere-turns per meter (At/m). The former name for magnetic field intensity was magnetizing force. We have seen that more ampere-turns (MMF) are required to set up the same magnetic flux in magnetic circuits of air than in iron of similar 9 Chapter One: Electromagnetic Principles configuration. Hence the magnetic field intensity for the air path is much larger than for the iron path. In the toroid of Figure 1.7, a magnetomotive force of 10 At acts along the mean path of 0.314 m. The magnetic field intensity is H= NI 10 = = 31 .8 At / m l 0 .314 Equation (1.5) transposed, Hl = NI is one form of Ampere's circuital law applied to a simple magnetic circuit. Magnetic field intensity in maonetic circuits is analogous to potential or voltage gradient in electric circuits. We can derive a useful relationship for magnetic circuits by summarizing the equations developed so far. Φ A B= Fm = NI Φ= Fm ℜm ℜm = H= l µA Fm l Thus B= Φ F HlµA = m = = µH A ℜm A lA B = µH or 1.6 Equation (1.6) shows that the magnetic flux density is directly dependent on both permeability and magnetic field intensity. Only in air or free space is the permeability (µ0) constant, and thus a linear relationship between B and H exists. In the next section we consider ferromagnetic materials in which the absolute permeability is not a constant but depends on the degree of magnetization. 1.4. MAGNETIC CIRCUITS A toroid of homogeneous magnetic material, such as iron or steel, is wound with a fixed number of turns of insulated wire as shown in Figure 1.7. The magnetic flux (Φ) and the excitation current (I) are related by Eq. (1.6): B = µH = Φ A Thus 10 Introduction to Electrical Machines Φ NI =µ A l that is, φ = (constant) × µI where the constant is NI / l . At the outset, the sample of ferromagnetic material in the toroid was totally demagnetized. In experimental measurements, the excitation current is varied and the corresponding values of magnetic flux recorded. Then the calculated values of B and H are plotted on linear scales as illustrated in Figure 1.8. 1.4.1. Magnetization (B-H) Curve Typical magnetization or B-H curves for sheet steel, cast steel, and cast iron are plotted in Figure 1.8. The nonlinear relationship between magnetic flux density B (teslas) and magnetic field intensity H (ampere-turns per meter) is illustrated. It is observed that the magnetic flux density increases almost linearly with an increase in the magnetic field intensity up to the knee of the magnetization curve. Beyond the knee, a continued increase in the magnetic field intensity results in a relatively small increase in the magnetic flux density. When ferromagnetic materials experience only a slight increase in magnetic flux density for a relatively large increase in magnetic field intensity, the materials are said to be saturated. Magnetic saturation occurs beyond the knee of the magnetization curve. Figure 1.8 Typical Magnitizations curves. The characteristic of saturation is present only in ferromagnetic materials. An explanation of magnetic saturation is based on the theory that magnetic materials are composed of very many tiny magnets (magnetic domains) that are randomly positioned when the material is totally demagnetized. Upon application of a magnetizing force (H), the tiny magnets will tend to align themselves in the direction of this force. In the lower part of the magnetizing curve, the alignment of the randomly positioned tiny magnets increases proportionately to the magnetic field intensity until the knee of the curve is reached. Beyond the knee of the curve, fewer tiny magnets remain to be aligned, and therefore large increases in the magnetic field intensity result in only small increases 11 Chapter One: Electromagnetic Principles in magnetic flux density. When there are no more tiny magnets to be aligned, the ferromagnetic material is completely saturated. In the saturation region of the curve, the magnetic flux density increases linearly with magnetic field intensity, just as it does for free space or nonmagnetic materials. From the origin of the B-H curve there is a slight concave curvature beyond which is the essentially linear region. We shall see that the nonlinear characteristics of the magnetization curve have practical implications in the operation of electrical machines. Hysteresis 4.2. Hysteresis is the name given to the "lagging" of flux density B behind the magnetizing force H. when a specimen of ferromagnetic material is taken through a cycle of magnetization.If the specimen has been completely demagnetized and the magnetizing force H is increased in steps from zero, the relationship between flux density B and H is represented by the curve OAC (Figure 1.9) which is the normal magnetization curve. If the value of H is now decreased, the trace of B is higher than OC and follows the curve CD until H is reduced to zero. Thus when H reaches zero, there is a residual flux density referred to as remnant flux density denoted by Br. In order to reduce B to zero, a negative field strength OE must be applied. The magnetic field intensity OE required to wipe out the residual magnetism Br is called coercive force. As H is further increased in the negative direction, the specimen becomes magnetized with the opposite polarity as shown by the curve EF. If H is varied backwards from LO to OK, the flux density curve follows a path FGC, which is similar to the curve CDEF. The closed loop CDEFGC thus traced out is called the hysteresis loop of the specimen. The term remnant flux density Br is also called retentivity and the term coercive force is often called coercivity. The shape of the hysteresis loop will depend upon the nature of magnetic material. Steel alloyed with 4 % silicon has a very narrow hysteresis loop. Hysteresis in magnetic materials results in dissipation of energy, which is proportional to the area of the hysteresis loop. Hence the following conclusions can be drawn: 1. Flux density B always lags with respect to the magnetizing force H. 2. An expenditure of energy is essential to carry the specimen through a complete cycle of magnetization. 3. Energy loss is proportional to the area of hysteresis loop and depends upon the quality of the magnetic material. Figure 1.9 Hysteresis loop 12 Introduction to Electrical Machines 1.4.3. Relative Permeability Transposition of Eq. (1.6) gives the absolute permeability as the ratio of the magnetic flux density to the corresponding magnetic field intensity: µ= B H Thus we can obtain the values of absolute permeability of ferromagnetic materials from the magnetization (B-H) curves. Another method of obtaining the absolute permeability would be to take the slope (differential) of the curve at various points. Although the differential method may be more realistic, for our purposes in this book, the simpler method of ratios to obtain the absolute permeability will be acceptable. If we wish to compare the permeability of magnetic materials with that of air, we may use the relative permeability µr, which is defined by the equation µr = µ µ0 Where µ = absolute permeability of the material. H/m µ0 = 4π×10-7H/m = permeability of free space µr = relative permeability From the typical magnetization curves of Figure 1.8, we can calculate the value of absolute and relative permeabilities for any magnetic operating condition. When we do this we observe that the value of relative permeability is not a constant but obtains a maximum value at about the knee of the B-H curve. Example 1.4 Calculate the absolute and relative permeabilities of cast steel operating at magnetic flux densities of 0.7 T and 1.0 T. Solution From the saturation curve for cast steel, the values of H are 400 At/m and 800 At/m. respectively. The absolute permeabilities are: For 0.7 T: µ= B 0 .7 = = 1.75 × 10 −3 H / m or T / At / m H 400 For 1.0 T: µ= B 1 .0 = = 1.25 × 10 − 3 H / m H 800 The relative permeabilities are For 0.7 T: 13 Chapter One: Electromagnetic Principles µr = µ 1.75 × 10−3 = = 1392.61 µ0 4π × 10− 7 µr = µ 1.25 × 10−3 = = 994.72 µ0 4π × 10− 7 For 1.0 T: Thus we see that cast steel has at least 1000 times more ability to set up magnetic flux lines than do nonmagnetic materials. Series and Parallel Magnet Circuits By definition, a series magnetic circuit contains magnetic flux, which is common throughout the series magnetic elements. These series magnetic elements may consist of composite sectors of ferromagnetic materials of different lengths and cross-sectional areas, and of air gaps. The simplest series magnetic circuit would be of a toroid of homogeneous material and the steel core of a transformer. More complex series circuits which contain air gaps are illustrated in Figure 1.6. Parallel magnetic circuits are defined by the number of paths that the magnetic flux may follow. Any of these paths or branches may consist of composite sectors of magnetic materials, including air gaps. A detailed calculation for a typical parallel magnet circuit is demonstrated in Section 1.5. Electric Circuit Analogs In our discussion so far, we note the following analogous relationships between magnetic quantities and electric quantities: Electric circuit E (volts) I (amperes) R (ohms) 1 ρ = (conductvity ) σ Magnetic circuit Fm (NI ampere-turns) φ (webers) ℜm (ampere-turns/weber) µ (henries/meter) We can draw useful electrical analogs for the solution of magnetic circuit problems. In an electrical circuit the driving force is the voltage, the output is the current, and the opposition to establishing current is the resistance. In the same way, the driving force in the magnetic circuit is the magnetomotive force, the output is the magnetic flux, and opposition to establishing the flux is the reluctance. Thus we have for the magnetic circuit of Figure 1.9a the analogous electric circuit and the analogous magnetic circuit in Figure 1.9b and c, respectively. The iron and air portions of the magnetic circuit are analogous to the two series resistors of the electric circuit. Analogous to the electric circuit, the magnetomotive force must overcome the magnetic potential drops of the two series reluctances in accordance with Kirchhoff's voltage law applied to magnetic circuits. 14 Introduction to Electrical Machines Therefore, Fm = ℜ miron φ + ℜ mag φ 1.8 is the equivalent magnetic-potential-drop equation. Since the permeability of ferromagnetic materials (iron) is a variable depending on the state of magnetization, we must use the B-H curves to obtain the magnetic field intensity if the magnetic flux density is available. Hence we can calculate the MMF drop for the iron from Eq. (1.5) as follows: Fmiron = H iron l iron At 1.9 Finally, the general MMF-drop equation for series magnetic circuits is modified for calculation purposes to the following form: Fm = H iron l iron + l ag φ µ 0 A ag 1.10 (a) (b) φ ℜ1 ℜag (c) Figure 1.9 Iron-core toroid with air gap: (a) Magnetic circuit; (b) analogous electric circuit; (c) analogous magnetic circuit. Given the physical parameters of the series magnetic circuit and the value of magnetic flux or magnetic flux density, the required magnetomotive force can be calculated in a straightforward manner using Eq. (1.10). 15 Chapter One: Electromagnetic Principles The general principles of electric circuits embodied in Ohm's and Kirchhotf's laws are applied as analogous equivalents to parallel magnetic circuits. With the presence of air gaps, most complex magnetic circuits are solved using the seriesparallel equivalent analogs.In analogous equivalents, Kirchhoff's current law for magnetic circuits states that the sum of magnetic fluxes entering a junction or node is equal to the sum of magnetic fluxes leaving the junction or node. Needless to a say, magnetic flux must not be perceived as flowing. Fringing and Leakage Flux In a series magnetic circuit containing an air gap, there is a tendency for the airgap flux to spread out (i.e., to create a bulge) as shown in Figure 1.10. This spreading effect, termed fringing, reduces the net flux density in the air gap. Figure 1.10 Fringing flux Leakage flux is that flux in a magnetic circuit which is not useful or effective. Since a large amount of leakage flux requires a greater magnetomotive force, the designer of electromagnetic devices must minimize this ineffective flux. Magnetic Core (Iron) Losses It will be shown later that the magnetic flux within the armature of dc machines changes direction as rotation occurs past the magnetic field poles. This change in direction of the armature magnetic flux is effectivelv an alternating flux. This results in core losses, which are treated in more detail in forthcoming chapters . Magnetic core losses consist of hysteresis losses and eddy-current losses. 1.5. MAGNETIC CIRCUIT CALCULATIONS We have seen that magnetic circuits may be represented by electric circuit analogs. Thus the methods of solution for series and parallel electric circuits may be applied to magnetic circuit problems. Typically, we will be required to calculate the magnetomotive force, flux, or permeability for some given conditions. The major difference between the two types of circuits is the nonlinear characteristics of ferrous magnetic materials. Thus it is necessary to make use of B-H curves and graphical methods. Calculation of Ampere Turns Ampere turns for various parts of the magnetic circuit will be calculated separately. To calculate the ampere turns for a particular part, the following procedure is followed in general: 16 Introduction to Electrical Machines 1. The reluctance of the part is calculated using Eq. (1.4) as the case may be. 2. The magnetic flux φ established in that part is calculated using Eq. (5.10). 3. Cross-sectional area of the part is calculated from the given dimensions. 4. 5. 6. 7. Magnetic flux density is found by dividing the flux by the cross-sectional area, i.e. φ B= A Ampere turns per meter of the magnetic flux path length in that part at the flux density calculated above is found by using the magnetization curve for the magnetic material of that part. Length of the magnetic flux path in that part is estimated from the given dimensions. Total ampere turns for the part are obtained by multiplying ampere turns per meter by the length of the flux path. 8. 9. General procedure is now applied to various parts of the magnetic circuit. Total ampere turns for the complete magnetic circuit can now be found by adding algebraically the ampere turns needed by the various parts of the magnetic circuit. Calculation of Ampere Turns for the Air Gap Total ampere turns for the air gap is given by, Fag = φ (Flux) × ℜag (reluctance) Reluctance for air gap, ℜag (for which µr=1) = Thus Fag = φ × lg µ0 × A g lg µ0 × A g lg φ × Ag µ0 1 = × Bg l g µ0 1 = × Bg l g 4 π × 10 − 7 = (1.11) Hence to calculate the ampere turns for the air gap, the following general procedure may be followed: 1. calculate the magnetic flux in the air gap, 2. calculate the gap area from the given data. 3. calculate gap density, Bg = φ , and Ag using Eq. (1.11), calculate the ampere turns needed for the air gap. 17 Chapter One: Electromagnetic Principles Series Magnetic Circuits Example 1.5 Illustrates the method of solution for a simple one-material series circuit. The circuit of Figure 1.11 is a magnetic core made of cast steel. A coil of N turns is wound on it. For a flux of 560 µWb, calculate the necessary current, neglecting any fringing effects.The cross-sectional area A is constant. Solution N = 550 turns l1 = 20 cm = 20 ×10-2 m l2 = 12 cm = 12 ×10-2 m A = 4 cm2 = 4 ×10-4 m2 φ =560 ×10-6 Wb B= 560 × 10−6 4 × 10 −4 = 140 × 10 − 2 = 1.4T For B = 1.4 T, H = 2200 At/m (from the B-H curve of Figure 1.7). The average or mean length of the magnetic path is 20 + 12 + 20 + 12 cm = 64 = 0.64 m. Therefore, Hl=NI=2000×0.64 At I= 2200 × 0.64 = 2.56 A 550 Figure 1.11 Magnetic circuit for Example 1.5 Example 1.6 The magnetic frame shown in Figure 1.12 is built up of iron of square cross-section, 3 cm side. Each air gap is 2 mm wide. Each of the coils is wound with 1000 turns and the exciting current is 1.0 A. The relative permeability of part A and part B may be taken as 1000 and 1200 respectively. Calculate the following: (i) reluctance of part A, (ii) reluctance of part B,(iii) reluctance of two air gaps, (iv) total reluctance of the 18 Introduction to Electrical Machines complete magnetic circuit, (v) the mmf, (vi) total flux, and (vii) flux density. Leakage and fringing may be neglected. φ Figure 1.12 magnetic circuit of example Solution Figure 1.12 shows the given magnetic frame consisting of two iron parts A and B separated by two air gaps of 2 mm each. Dotted line shows the mean magnetic circuit set up in this frame, when the coils wound over it carries current. The magnetic circuit consists of four portions connected in aeries, i.e. (i) magnetic flux path in part B, (ii) magnetic flux path in air gap, (iii) magnetic flux path in part A and (iv) flux path in air gap. Hence the total reluctance of this magnetic circuit will be equal to the sum of the reluctances of these four parts calculated separately. i) Reluctance offered by a magnetic path is given by, lg ℜ= Reluctance for part A, ℜA = µ0 µ r A lA µ0 µ r A Cross-sectional area of part A, whose cross-section is a square with 3 cm side, AA = 3 × 3 = 9 cm2 = 9 × 10-4 m2. Length of mean path of flux in part A, lA = 20 − (1.5+1.5) = 17 cm = 0.17m Permeability, µ 0 = 4π × 10 −7 Relative permeability µr for part A = 1000 Thus reluctance of part A, ℜA = 0.17 4π × 10 × 1000 × 9 × 10 − 4 = 15.03 × 104 At / Wb ii) Reluctance of part B, ℜ B = −7 lB µ0 µ r A B Cross-sectional area of part B, AB = 9 × 10-4 m2 . 19 Chapter One: Electromagnetic Principles Relative permeability µr , of part B = 1200 Length of mean path of flux in part B, lB = 17 + 8.5 + 8.5 = 34 cm = 0.34 m Thus reluctance of part B, 0.34 ℜB = 4π × 10 × 1200 × 9 × 10 − 4 = 25.04 × 10 4 At / Wb iii) Reluctance of air gaps, ℜag = −7 lg µ0 A Length of mean path of flux in the two air gaps, lg = 2 + 2 = 4 mm = 0.004 m. 0.004 ℜag = Hence reluctance of two air gaps, 4π × 10 − 7 × 9 × 10 − 4 = 353.5 × 10 4 At / Wb iv) Total reluctance of the magnetic circuit, ℜ = ℜ A + ℜ B + ℜag = (15.03 + 25.04 + 353.5) × 10 4 = 393.57 × 10 4 At / Wb v) The magnetomotive force produced by two coils on iron part B, each having 1000 turns and carrying a current of 1 A = (2 × 1000) × 1 ∵ (mmf = NI) = 2000 At. vi) As per Ohm's law for magnetic circuits, mmf = Flux × Reluctance φ= = NI ℜ 2000 393.57 × 10 − 4 = 5.08 × 10 − 4 Wb vii) Flux density, B= = φ A 5.08 × 10 − 4 −4 = 0.564 Wb / m 2 [Tesla ] 9 × 10 Example 1.7 An electromagnet is of the form and dimensions as shown in Figure 1.13. It is made of iron of square section 4 cm side. A flux of 1.1 mWb is required in the air gap. Neglecting leakage and fringing, calculate the number of ampere turns required. Take the relative permeability to be 2000 at this flux density. Figure 1.13 magnetic circuit of example 20 Introduction to Electrical Machines Solution The magnetic circuit of the electromagnet shown in Figure 1.13 is completed by four parts connected in series, viz. (i) iron portion C (ii) air gap, (iii) iron portion D, and (iv) air gap. Total ampere turns required for this magnetic circuit, FT = ampere turns required for iron portion C , FC + ampere turns required for air gap, Fag + ampere turns for iron portion D, FD + ampere turns of air gap, Fag. Or FT = FC + Fag + FD + Fag Or FT = FC + FD + 2Fag i) Ampere turns required for iron portion C Flux, φ = 1.1 × 10-3 Wb Cross-sectional area, = 4 × 4 = 16 cm2 = 16×10-4 m2 Flux density, B= = φ A 1.1 × 10 − 3 16 × 10 −4 = 0.6875 Wb / m 2 Relative permeability of iron = 2000 Thus ampere turns per meter of flux path length, H = Or HC = B µ0 µ r 0.6875 2π × 10 − 7 × 2000 = 272 At / m Length of mean flux path in the iron portion C, lC = 25 cm = 0.25 m Ampere turns required for iron portion C, FC = HC× lC = 272 × 0.25 = 68 At. ii) The materials for portions C and D are the same. Section is also the same. Thus ampere turns per meter of flux path length, HD = HC = H = 272 At/m. Length of mean flux path in iron portion D, lD = 30 cm = 0.3 m Thus, Ampere turns required for iron portion D, FD = H × lD = 272 × 0.3 = 82 At. iii) Ampere turns required for air gap, 21 Chapter One: Electromagnetic Principles Fag = = 1 × Bg l g µ0 1 4π × 10 − 7 = 1094 At × 0.6875 × 0.002 Ampere turns required for two air gaps = 2 × 1094 = 2188 At Total ampere turns required, FT = 68+ 82+ 2188 = 2318 At Parallel Magnetic Circuits Figure 1.14a shows a parallel magnetic circuit.There are NI ampere-turns on the center leg.The flux that is produced by the MMF in the center leg exists in the center leg and then divides into two parts, one going in the path afe and the other in the path bcd. If we assume for simplicitv that afe = bcd, the flux is distributed evenly between the two paths. Now φg = φafe + φbcd Where 1.11 φg = flux in portion g φafe = flux in portion afe φbcd = flux in portion bcd Equation (1.11) is actually the analog of Kirchhoff's current law, but now we can say that the amount of flux entering a junction is equal to the amount of flux leaving the junction. Another observation that we may make on this circuit is that the MMF drops around a circuit are the same no matter what path we take. Thus the MMF drop around afe must be equal to the MMF drop around bcd. This can be stated more precisely as Hala + Hflf + Hele = Hblb + Hclc + Hdld 1.12 (a) 22 Introduction to Electrical Machines ℜma φafe φbcd ℜmb ℜag ℜmc ℜmf φg ℜme ℜmd (b) (c) Figure 1-14 Magnetic circuit with center leg: (a) Magnetic circuit; (b) equivalent magnetic circuit; (c) analogous electric circuit. The drop in MMF around either path afe or bcd must also be equal to the MMF drop along path g. But g also has an "active source," the NI ampere-turns of the coil. The actual MMF existing between X and Y is the driving force NI minus the drop Hglg in path g. Then we can write (NI - Hglg) = Hala + Hflf + Hele 1.13 = Hblb + Hclc + Hdld Again we can draw analogous magnetic and electrical circuits as in Figure 1.14b and c. For Figure 1.l4b we may write NI - ℜmgφg = φbcd (ℜmb + ℜmc + ℜmd ) 1.14 = φafe (ℜma + ℜmf + ℜme ) and in Figure 1.l0c we may write E - RgIg = Ibcd (Rb + Rc + Rd ) 1.15 = Iafe (Ra + Rf + Re ) In the analogous magnetic circuit, note that NI is drawn in series with Rmg, although physically the coil surrounds the central magnetic path. 23 Chapter One: Electromagnetic Principles Example 1.8 In Figure 1-14a, the following dimensions apply: lg = lf = lc = 12 cm la = lb = le = ld = 14 cm Aa = Ab = Ac = Ad = Ae = A = 1 cm2 Ag = 3 cm2 The material is sheet steel. The flux densitv in the center leg is 0.9 T. Calculate the MMF required to produce this flux density. Solution The total flux in the center leg is 0.9 × 3 × 10-4 = 2.7 × 10-4 Wb. The flux divides into two parts, the left-hand path through afe and the right-hand path through bcd. The flux density in path g is Bg = 0.9 T and therefore Hg = 320 At/m. The flux density in section a is Ba = 2.7 × 10 −4 2 × 1 × 10 − 4 = 1.35 T and therefore Ha = 950 At/rn Ha = Hb = Hc =Hd =He= Hf Therefore, NI = Hglg + Ha (la + lf + le ) = 320 × 12 × 10-2 + 950 ( 14 + 12 + 14 ) × 10-2 = 38.4 + 380 = 418.4 At Example 1.9 We can add one more degree of complexity to the circuit of Figure 1.14a. In Figure 1.15 we cut an air gap in the center leg, and the air gap is 1.5 mm wide. All other dimensions remain unchanged and the flux density in the center leg is still 0.9 T. Find the number of ampere-turns on the center leg required to produce this flux density. Solution We can still use the equivalent-circuit concept as shown in Figure 1.14b, the only difference being that NI is now in series with two reluctances in the center path, the air gap and the steel in leg g. (NI) - (MMF drop in air gap) - (MMF drop in section g ) = MMF drop in section b + c + d = MMF drop in section a + f + e In the center leg, the flux density is still fixed at 0.9 T. Therefore. Bg = 0.9 T 24 Introduction to Electrical Machines The MMF drop per unit length in the center steel section is still Hg = 320 At/m, as before. Therefore, MMF drop in leg g = 320 (12 - 0.15) × 10 -2 = 37.92 At The MMF drop across the air gap is found from Fmgap = Hgaplgap For air µ = µ0 = 4π×10-7 Wb/(At/m) or H/m Therefore, H gap = 0 .9 4π × 10 −7 = 7.16 × 105 At / m Figure 1.15 Parallel magnetic circuit with air gap. The MMF drop across the gap Hgap lgap= 7.16 ×105 ×1. 5 ×10-3 = 1.074×103 At Noting that the MMF drop across the path afe is still 380 At, as before, NI − (37.92 + 1074) = 380 At NI = 1491.91 At By adding a very small air gap, the MMF required has increased by a factor of 3.57. This is because the reluctance of the air is so high and the reluctance of unsaturated steel is very low. This. in turn, is the reason why the largest part of a magnetic circuit is usually in iron and only a small portion is in air. 1.6. PERMANENT MAGNETS Permanent magnets are communlv used as compasses and magnetic lifts. Today, there is a substantial increase in the application of permanent magnets for electromagnetic devices such as instruments. magnetic clutches and brakes, loudspeakers and relays, as well as small generators and motors. 25 Chapter One: Electromagnetic Principles Modern Permanent magnets materials are alloys composed of nickel, aluminum, and iron, described by the trade name Alnico. Current research has developed rare-earth materials for permanent magnets having extremely high values of residual flux density. A wide variety of powdered-composition permanent magnets called ferrites are useful for relatively high-frequency applications. The composition materials of ferrites are usually barium and ceramic. 1.7. ELECTROMAGNETIC INDUCTION Oersted at Copenhagen in 1820 discovered a very important phenomenon giving the relationship between magnetism and electricity. As per this relationship, a conductor carrying a current I is surrounded all along its length by a magnetic field, the lines of magnetic flux being concentric circles in planes at right angles to the conductor. This phenomenon of a magnetic field being associated with a current carrying conductor lead to the question whether the converse of the above is possible, i.e. can a magnetic field generate a current? Michael Faraday, on 29 Aug. 1831, succeeded in generating an electric current with the aid of magnetic flux. From his experiments, Faraday concluded that a current was generated in a coil so long as the lines of force bearing through the conductor changed. The current thus generated is called the induced current and the emf that gives rise to this induced current is called the induced emf. This phenomenon of generating an induced current in a closed circuit by changing the magnetic field through it, is called electromagnetic induction. The operation of electrical equipments like motors, generators, transformers, etc. is mainly based upon the laws formulated by Faraday. 1.7.1. Faraday's Laws of Electromagnetic Induction Faraday conducted the following experiment to obtain an electric current with the aid of magnetic flux. (a) (b) (c) Figure1.16 Faraday's experiment Figure 1.16 shows a coil connected to a galvanometer G. When the magnet was kept inside the coil nothing happened as shown in Figure 1.16 (a). But when the north pole of the magnet was inserted in the coil as shown in Figure 1.16 (b), the galvanometer pointer was deflected momentarily on one side and the direction of current was found to be anticlockwise. When the magnet was withdrawn as shown in Figure 1.16 (c), the pointer of the galvanometer deflect on the other side and the direction of current was found to be clockwise. Similar results were obtained when the south pole of the magnet was inserted 26 Introduction to Electrical Machines or withdrawn, but the direction of current in this case was reverse to that obtained with the North Pole. Faraday summed up the results of the experiments described above in the form of following two laws, known as Faraday's laws of electromagnetic induction. Faraday's first law states that whenever the magnetic flux associated or linked with a closed circuit is changed, or alternatively, when a conductor cuts or is cut by the magnetic flux, an emf is induced in the circuit resulting in an induced current. This emf is induced so long as the magnetic flux changes. Faraday's second law states that the magnitude of the induced emf generated in a coil is directly proportional to the rate of change of magnetic flux. These two basic laws discovered by Faraday changed the course of electrical engineering and led to the development of generators, transformers, etc. The change of flux as discussed in the Faraday's laws can be produced in two different ways: (i) by the motion of the conductor or the coil in a magnetic field, i.e. the magnetic field is stationary and the moving conductors cut across it. The emf generated in this way is normally called dynamically induced emf; (ii) by changing the current (either increasing or decreasing) in a circuit. thereby changing the flux linked with stationary conductors, i.e. the conductors or coils remain stationary and the flux linking these conductors is changed. The emf is termed statically induced emf. Statically induced emf can be further subdivided into (a) self-induced emf and (b) mutually induced emf. The concept of dynamically induced emf gave rise to the development of generators, whereas statically induced emf was helpful in developing transformers. 1.7.2. Direction of Induced emf The direction of induced emf can be determined by two methods namely (a) Fleming's right hand rule and (b) Lenz's law. In case of dynamically induced emf, Fleming's right hand rule is used to obtain the direction of induced emf , whereas Lenz's law is normally used to fix the direction of statically induced emf. (a) Fleming's Right Hand Rule: Stretch the forefinger, the middle finger and the thumb of the right hand in three mutually perpendicular directions as shown in Figure 1.17. If the forefinger points in the direction of the magnetic flux, the thumb points in the direction of motion of the conductor relative to the magnetic field, then the middle finger represents the direction of the induced emf. (b) Lenz's Law: The direction of statically induced emf can be obtained with the help of Lenz's law which states: "the direction of the induced emf is always such that it tends to set up a current opposing the change of flux responsible for producing that emf. Lenz’s law is further clarified by using it with reference to Figure 1.16. When the north pole of the magnet is inserted in the coil, an emf is induced in it due to the motion of the magnet, thereby generating induced current. According to Lenz's law, the direction of this induced current generated in the coil should be such that the motion of the magnet is opposed, which is possible only when the upper end of the coil behaves as a north pole. For this to happen, the current generated in the coil should be in the anticlockwise 27 Chapter One: Electromagnetic Principles direction as was observed by Faraday. In a similar way, the direction of the induced emf can be determined for any case utilizing Lenz's law. ν ν φ S N Motion e Flux EMF Figure 1.17 direction of induced emf 1.7.3. Magnitude of Induced Emf in a Coil Let a coil consist of N number of turns over it. Assume that the flux through the coil changes from its initial value φ1 to φ2 in an interval t second. Initial value of' flux linkages = Nφ1 Value of flux linkages after t s = Nφ2 Change of flux linkages in time t s == Nφ1− Nφ2 The term flux linkages used over here simply means the product of flux in Weber and the number of turns with which the flux is linked. Now as per Faraday's laws of electromagnetic induction, induced emf in the above coil due to a change of flux is given by, Induced emf = N (φ1 − φ2 ) V t Based on the above, the instantaneous value of emf induced in the coil can be represented as, e=− Or d ( Nφ) dt e = −N dφ dt 1.16 The negative sign in the Eq. (1.16) above equation signifies that the induced emf generates a current tending to oppose the increase of flux through the coil. The relation expressed by the above equation can be called Faraday's law. 1.7.3.1. Dynamically Induced emf 28 Introduction to Electrical Machines Dynamically induced emf is produced by the movement of the conductor in a magnetic field. Figure 1.18 shows a uniform magnetic field of flux density B tesla, in which the conductor is moving in the direction shown and cuts the flux at right angles. If l = length of the conductor in meter cutting the field v = velocity of motion of conductor in m/s dx = distance moved by the conductor in time dt ν Figure 1.18 Dynamically induced emf Then area swept by the moving conductor = l× dx Hence change in flux, when the conductor moves a distance dx in time dt, dφ = B × l dx Wb The dynamically induced emf is the rate of change of flux linkages, i.e. Dynamically induced emf = Bldx dt = Bl As Thus dynamically induced emf 1.7.3.2. dx dt dx = ν (velocity) dt = Blν V 1.17 Statically Induced emf When the conductor or coil remains stationary and the flux linking with these conductors or coil undergo a change, an emf is induced in the conductors. Such an induced emf is termed as statically induced emf. Statically induced emf can be further classified as (i) self-induced emf and (ii) mutual induced emf. Self-induced emf Any electrical circuit in which the change of current is accompanied by the change of flux, and therefore by an induced emf, is said to be inductive or to possess self inductance. Thus the property of the coil which enables to induce an emf in it whenever the current changes is called self-induction. 29 Chapter One: Electromagnetic Principles φ I Figure 1.19 Self-induced emf Consider a coil of N turns carrying a current of I amperes and let φ be the resulting flux linking the coil. The magnetic flux forms complete loops as shown in Figure 1.19. The product Nφ is normally termed as flux linkages. Now if the current flowing in the coil is changed, then the number of lines linking the coil also changes. As such emf is induced in the coil according to Faraday's laws of electromagnetic induction. This emf is termed as statically self-induced emf or the emf of self-induction. The phenomenon of selfinduction is felt only when the current is changing, either increasing or decreasing. As per Faraday's laws of electromagnetic induction, this induced emf is given by, e = −N dφ , dt V 1.18 The coil in question is wound on an iron core, whose permeability is constant. Thus flux is proportional to the currcut through the coil, i.e. φ ∝I Or φ = cons tan t I Now flux can also be written as, Flux = i.e. φ= Flux × current current φ ×I I Now if current is changed at a certain rate, the flux also changes at the same rate. Thus, φ φ the rate of change of flux = × rate of change of current = cons tan t . I I Substituting this in Eq. (1.18), φ e = − N × rate of change of current I Or e=− Nφ di × I dt 1.19 30 Introduction to Electrical Machines Nφ i.e. flux linkages/ampere is generally called the self-inductance of the coil I or the coefficient of self-induction and is denoted by a symbol L. With this replacement, Eq, (1.19) becomes, The term e = −L Where L=− di dt 1.20 Nφ henry I The negative sign in Eq. (1.20) indicates that it is an emf opposing the change, i.e. if the current is increasing, this emf will oppose the increase in current (emf will be opposite to the applied voltage), in case the current is decreasing, the induced emf tends to prevent the decrease of current and its direction is therefore the same as that of current or the applied voltage. It also indicates that the energy is being absorbed from the electric circuit and stored as magnetic energy in the coil. The coefficient of self-induction L of the circuit is thus defined as the magnetic flux linked with the coil when a unit current flows through it. It is also numerically equal to the induced emf due to unit rate of change of current in the coil. The practical unit of inductance is henry. Mutually Induced emf The phenomenon of generation of induced emf in a circuit by changing the current in a neighbouring circuit is called mutual induction. Consider two coils P and S such that P is connected to a cell through switch K and S to a galvanometer as shown in Figure 1.20. When the switch K is closed suddenly to start current in the coil P, the galvanometer gives a sudden "kick" in one direction. Now when K is opened, the galvanometer again shows a deflection but in the oppoaite direction. The above cbservations indicate clearly that an induced current is set up in the coil S when the current is changed in the coil P, though the coil S is not connected physically to coil P. Two coils possessing this property are said to have mutual inductance. The unit of mutual inductance is also henry. It is denoted by M. Two coils are said to possess a mutual inductance of 1 henry when current changing at the rate of l ampere per second in one coil induces an emf of 1 volt in the other. Figure 1.20 Mutually induced emf Let φ1 be the flux in coil P due to curent I flowing in it and φ2 the flux induce in S due to φ flux φ1 in coil P. The ratio 2 is denoted by K. Thus φ1 31 Chapter One: Electromagnetic Principles φ2 =K φ1 Or φ2 = Kφ1 Also φ ∝I Or φ = cons tan t I Now φ φ2 = 2 × I I Or φ2 = Kφ1 × current I When current is changed at a certain rate, φ2 also changes at the same rate. Thus, Rate of change of φ2 = Kφ1 × rate of change of current I 1.21 According to Faraday's laws of electromagnetic induction the emf induced in S is given by, eS = N 2 × (rate of change of flux φ2 ) 1.22 where N2 is the number of turns in coil S Combining Eqs (1.21) and (1.22), Or eS = N 2 × Kφ1 × rate of change of current I eS = N 2 × Kφ1 di × I dt = M× Where M= = di dt N 2 Kφ1 N 2 φ2 = I I flux linkage of coil S current in coil P The constant M in the above equation , which is equal to the flux linkages of coil S per ampere of current in coil P, is called the coefficient of mutual induction or mutual inductance. Hence the coeffiaicnt of mutual induction is defined as the number of lines of force passing through the secondary coil S when unit current changes in the primary coil P. It 32 Introduction to Electrical Machines is also numerically equal to the induced emf in one circuit due to a unit rate of change of current in the other circuit. 1.7.4. Inductance Having introduced the necessary electromagnetic background, we can now address inductance. Inductance is, in some sense, a mirror image of capacitance. While capacitors store energy in an electric field, inductors store energy in a magnetic field. While capacitors prevent voltage from changing instantaneously, inductors, as we shall see, prevent current from changing instantaneously. Consider a coil of wire carrying some current creating a magnetic field within the coil. As shown in Figure 1.21, if the coil has an air core, the flux can pretty much go where it wants to, which leads to the possibility that much of the flux will not link all of the turns of the coil. To help guide the flux through the coil, so that flux leakage is minimized, the coil might be wrapped around a ferromagnetic bar or ferromagnetic core as shown in Figure 1.22. The lower reluctance path provided by the ferromagnetic material also greatly increases the flux φ. We can easily analyze the magnetic circuit in which the coil is wrapped around the ferromagnetic core in Figure 1.22(a). Assume that all of the flux stays within the low-reluctance pathway provided by the core, and apply (1.3): φ= F Ni = ℜ ℜ 1.23 Figure 1.21 A coil with an air core will have considerable leakage flux. φ N i e _ + (a) (b) Figure 1.22 Flux can be increased and leakage reduced by wrapping the coils around a ferromagnetic material that provides a lower reluctance path. The flux will be much higher using the core (a) rather than the rod (b). 33 Chapter One: Electromagnetic Principles From Faraday’s law , changes in magnetic flux create a voltage e, called the electromotive force (emf), across the coil equal to e=N dφ dt 1.24 Substituting (1.23) into (1.24) gives e=N d Ni N 2 di di =L = dt ℜ ℜ dt dt where inductance L has been introduced and defined as Inductance, L= N2 heneries ℜ 1.25 Notice in Figure 1.22 (a) that a distinction has been made between e, the emf voltage induced across the coil, and V, a voltage that may have been applied to the circuit to cause the flux in the first place. If there are no losses in the connecting wires between the source voltage and the coil, then e = v and we have the final defining relationship for an inductor: v=L di dt As given in (1.25), inductance is inversely proportional to reluctance ℜ. Recall that the reluctance of a flux path through air is much greater than the reluctance if it passes through a ferromagnetic material. That tells us if we want a large inductance, the flux needs to pass through materials with high permeability (not air). Example 1.10 Inductance of a Core-and-Coil. Find the inductance of a core with effective length l = 0.1 m, cross-sectional area A = 0.001 m2, and relative permeability µr somewhere between 15,000 and 25,000. It is wrapped with N = 10 turns of wire. What is the range of inductance for the core? Solution When the core’s permeability is 15,000 times that of free space, it is µcore = µrµ0 = 15,000 × 4π × 10−7 = 0.01885 Wb/At-m So its reluctance is ℜ= l µ core A = 0 .1 = 5305 At / Wb 0.01885 × 0.001 and its inductance is 34 Introduction to Electrical Machines L= N2 102 = = 0.0188 H = 18.8 mH ℜ 5305 Similarly, when the relative permeability is 25,000 the inductance is N 2 N 2µ r µ 0 A 10 2 × 25,000 × 4π × 10 −7 × 0.001 L= = = ℜ l 0 .1 = 0.0314 H = 31.4 mH The point of Example 1.9 is that the inductance of a coil of wire wrapped around a solid core can be quite variable given the imprecise value of the core’s permeability. Its permeability depends on how hard the coil is driven by mmf so you can’t just pick up an off-the-shelf inductor like this and know what its inductance is likely to be. The trick to getting a more precise value of inductance given the uncertainty in permeability is to sacrifice some amount of inductance by building into the core a small air gap. Another approach is to get the equivalent of an air gap by using a powdered ferromagnetic material in which the spaces between particles of material act as the air gap. The air gap reluctance, which is determined strictly by geometry, is large compared to the core reluctance so the impact of core permeability changes is minimized. The following example illustrates the advantage of using an air gap to minimize the uncertainty in inductance. It also demonstrates something called Ampere’s circuital law, which is the magnetic analogy to Kirchhoff’s voltage law. That is, the rise in magnetomotive force (mmf) provided by N turns of wire carrying current i is equal to the sum of the mmf drops R φ around the magnetic loop. Example 1.11 An Air Gap to Minimize Inductance Uncertainty. Suppose the core of Example 1.10 is built with a 0.001 m air gap. Find the range of inductances when the core’s relative permeability varies between 15,000 and 25,000. Figure 1.23 for Example 1.10 Solution The reluctance of the ferromagnetic portion of the core when its relative permeability is 15,000 is ℜcore = l core 0.099 = = 5252 At / Wb µ core A 15,000 × 4π × 10 − 7 × 0.001 And the air gap reluctance is 35 Chapter One: Electromagnetic Principles ℜag = l ag µ0A = 0.001 4π × 10 − 7 × 0.001 = 795.775 At / Wb So the total reluctance of the series path consisting or core and air gap is ℜTotal = 5252 + 795,775 = 801,027 At/Wb And the inductance is L= N2 10 2 = = 0.0001248 H = 0.1248 mH ℜ 801,027 When the core’s relative permeability is 25,000, its reluctance is ℜcore = l core 0.099 = = 3151 At / Wb µ core A 25,000 × 4π × 10 − 7 × 0.001 And the new total inductance is N2 10 2 L= = = 0.0001251 H = 0.1251 mH ℜ 3151 + 795,775 This is an insignificant change in inductance. A very precise inductance has been achieved at the expense of a sizable decrease in inductance compared to the core without an air gap. 1.7.5. Energy Stored In Magnetic Field Consider a coil having a constant inductance of L Henry, in which the current increases by di in dt seconds, then induced emf in the coil , e becomes e = −L di dt The applied voltage must balance the voltage drop across resistor R and neutralize the above induced emf, thus, V = iR + L di dt 1.26 Multiplying Eq.(1.26) throughout by i.dt V i dt = i 2 Rdt + L i di dt 1.27 where, V i dt is the energy supplied by the source in time dt i2 R dt the energy dissipated in the form of heat 36 Introduction to Electrical Machines Li di the energy absorbed by the inductance of the coil in building up the magnetic field. Thus energy absorbed by the magnetic field during the time dt second = L i dt Joules Hence total energy absorbed by the magnetic field when the current increases from zero to I amperes I = ∫ Lidi 0 I = L ∫ idi 0 = Energy stored 1 2 LI 2 J 1.28 Energy Stored in Magnetic Field in Terms of Volume of Field Energy stored in Magnetic field = Self inductance of the coil, L = Thus energy 1 2 LI 2 Nφ I stored 1 Nφ 2 ×I 2 I 1 = N φI 2 = J Total ampere turns on the coil, NI = Hl Also flux, φ = BA Thus energy stored 1 Hl × BA 2 1 = BH × (l × A ) 2 = But A × 1 = volume of the magnetic field Hence energy stored = 1 BH × (volume of the field ) 2 37 Chapter One: Electromagnetic Principles Or 1 BH 2 1.29 Energy stored/cubic meter = 1 µ 0µ r H 2 2 1.30 Energy stored/cubic meter = 1 B2 2 µ 0µ r 1.31 Energy stored/cubic meter = Since B = µ 0µ r H Or 1.8. ELECTROMAGNETIC FORCES By the interaction of magnetic fields produced in electromagnetic devices, mechanical forces are developed which may do useful work. Electromagnetic forces fall into two general classifications: (1) the magnetic tractive force, and (2) thc force on a conductor. There are many examples of forces acting in electromagnetic fields. An electromagnet used to separate ferrous from nonferrous material is one, the deflection of an electron beam in a cathode-ray tube is another, and the action of an electric motor is a third. A fourth example is the attraction of an armature to an electromagnet, such as in relays, contactors, and lift magnets. Lifting Power of Magnet (Magnetic Tractive Force) We will consider the forces of attraction acting in an air gap between parallel surfaces Referring to Figure 1.24, let F be the force in Newton between the poles of the magnets and the pole cross-sectional area A in square meter having a flux density of B tesla. If the upper pole is pulled through a small distance δx against the force F, then Work done = Fδx J 1.32 Work done as given by above equation is equal to the increase in energy stored in the magnetic field. Energy stored per cubic meter of magnetic field = 1 BH 2 = 1 B2 2 µ 0µ r If the field is in air, µr = 1 Thus energy stored per cubic meter = 1 B2 = 2 µ0 where B is the flux density of the field Additional volume of magnetic field = Aδx m3 Thus increase in energy stored in the magnetic field 38 Introduction to Electrical Machines = 1 B2 A δx 2 µ0 Now by equating the above to equations Fδx = Or Pull F = 1 B2 A 2 µ0 1 B2 A δx 2 µ0 N 1.33 Pull in Kg wt F= B2 A kg wt 9.81 × 2µ 0 Or F= B2 A kg wt 19.62 × 2µ 0 1.34 N δx F N Figure 1.24 lifting power of magnet Example 1.12 Let us calculate the current required to lift a large cast-iron plate using the electromagnet of Figure 1.25. We will assume that since the magnet and the plate are both rough surfaces, we have an equivalent air gap of about 1.5 mm. The plate has a mass of 400 kg. Solution The force required to lift the magnet is Total force = 2 × force per pole = 2× B2 A B2 A = 2µ 0 µ0 newtons ( N ) The force weight is F = ma = 400 x 9.80 = 3920 N Therefore, 39 Chapter One: Electromagnetic Principles 3920 = B 2 ( 0.10 × 0.20 ) A 4 π × 10 − 7 and 4 π × 10 −7 × 3920 = 0.25 0.02 B = 0 .5 B2 = NI = Hlcs + HlCI + B( 2lag ) µ0 = 250 × 0.6 + 1950 × 0.35 + = 150 + 682.5 + 119.4 = 952 At 0.5 × 3.0 × 10 − 3 4 π × 10 − 7 Figure 1.25 Lifting electromagnet for Example 1.8. Thus if N = 1000 turns, I = 0.95 A. In general we observe that almost all of the ampere-turns are usually consumed by the relatively small air gap. In a practical case, since leakage flux and fringing have been neglected, increasing the value of the current by about 20% would probably yield a satisfactory solution. Example 1.13 A solenoid 80 cm in length and 8 cm in diameter has 4000 turns uniformly wound over it. Calculate (i) the inductance and (ii) the energy stored in the magnetic field when a current of 2 A flows in the solenoid. Solution i). Inductance of the solenoid is given by, L= Nφ I Flux, 40 Introduction to Electrical Machines d2 φ = B ⋅ A = (µ 0 H ) ⋅ π = 4 d2 NI = µ0 = ×π l 4 4000 × 2 π 2 = 4π × 10 −7 × × × 8 × 10 −4 0 .8 4 −3 = 0.06322 × 10 Wb Inductance, 4000 × 0.06322 × 10 −3 L= 2 = 0.126 H Energy stored in the magnetic field = ii). 1 2 LI 2 1 × 0.216 × 2 2 = 2 = 0.252 J = Force on a Conductor Ampere demonstrated in 1820 that there is a magnetic field associated with a conductor carrying current. When placed in a transverse magnetic field, this conductor experiences a force that is proportional to (a) the strength of the magnetic field, (b) the magnitude of current in the conductor, and (c) the length of the conductor in, and perpendicular to, the magnetic field. In SI units, the electromagnetic force developed on the conductor carrying current in a magnetic field B is given by F=BlI newtons 1.35 Much use will be made of this important equation in subsequent chapter. 41 Chapter One: Electromagnetic Principles PROBLEMS 1.1. A coils of 200 turns is wound uniformly over a wooden ring having a mean circumference of 600 mm and a uniform cross sectional area of 500 mm2. If the current through the coil is 4 A, calculate: (a) the magnetic field strength, (b) the flux density, and (c) the total flux Ans.: 1333 A/m, 1675×10 -6 T, 0.8375 µWb 1.2. A coil of 1000 turns is wound on air-core toroid as shown in Figure 1.26. The current in the coil is 5A. Do=7 cm, and Di =5 cm. calculate the flux density inside the coil, assuming that it is uniformly distributed over the coil cross section. 1.3. A coil of 1000 turns is wound on an air-core toroid as shown in Figure 1.26, where Di=4 cm and Ds= 0.5 cm. If the flux in the cross section is 0.8 µWb, calculate I. Assume that the flux is confined to the inside of the coil and is uniformly distributed across the cross section. Ans.: 4.58 A Figure 1.26 magnetic circuit for problem 1.3. 1.4. Consider the toroid shown in Figure 1.26, given that Di= 13 cm, and Ds=1.5 cm. the cast steel material has the magnetic characteristics of the B-H curve shown in Figure 1.7. if the coil has 150 turns, calculate I so that the magnetic flux density is 1.5 T. 1.5. The toroid shown in Figure 1.26 has circular cross section and a cast iron core. The inner diameter is 4 cm, the outer diameter is 6 cm. If the flux in the core is 0.059 mWb, find the current in coil. Ans.: 1.38 A 1.6. A coil having 200 turns is wound on the toroid of Figure 1.26. A 1.5mm air gap is cut in the cast steel and a current of 2 A is passed through the coil. Di = 13 cm and Ds= 1.5 cm. Assuming no leakage flux, calculate the flux density in the air gap. 1.7. An air-gap 1 mm wide is cut in the toroid of Problem 1.4. What is the current now required to produce flux density of 1.5 T ? Ans.: 17.0 A 1.8. Calculate the MMF needed to produce a flux of 5.3× 10 -3 Wb in the air gap of Figure 1.27. Ans.: 1801 At 1.9. A magnetic circuit has the dimensions shown in Figure 1.28 (all dimensions in centimeters except as indicated). Find I if the air-gap flux is 645 µWb.Ans.1.82 A 42 Introduction to Electrical Machines Figure 1.28 magnetic circuit for problem 1.9. Figure 1.27 magnetic circuit for problem 1.8. 1.10. A magnetic core is made up of two parts, a sheet steel portion plus a cast steel portion (Figure 1.29). The effective length of the cast steel is 8 cm, and of the sheet steel, 20 cm. The core has a uniform cross section of 15 cm2. Find the number of ampere-turns of the coil required to produce a flux of 2.1 mWb in the core. Figure 1.29 magnetic circuit for problem 1.10. 1.11. A magnetic circuit made of cast steel has the dimensions shown in Figure 1.30. The cross-sectional area is 4 cm2 for all parts in the magnetic path. The air gap is 1.5 mm wide. Find the number of ampere-turns required on the center leg to produce a flux of 0.52 mWb in the center leg. Ans.: 1942 At Figure 1.30 magnetic circuit for problem 1.11. 1.12. A magnetic circuit consists of a cast steel yoke which has a cross-sectional area of 2 cm2 and a mean length of 12 cm. There are two air gaps, each 0.2 mm long. Calculate the ampere-turns required to produce a magnetic flux of 50 µWb in the air gaps. 1.13. A cast steel electromagnet has an iron path of mean length 40 cm and an air gap of 2 mm. it is desired to produce a flux density of 1.0 Wb/m2 in the air gap by putting a coil on electromagnet. Assuming negligible leakage and fringing, find the total ampere turns required. B-H curve may be plotted as per the data given below: Ans.: 1940 At 43 Chapter One: Electromagnetic Principles H (At/m) 1.14. 500 1000 2000 3000 4000 B (Wb/m2) 0.6 1.05 1.38 1.5 1.58 The shunt-field winding of a dc two-pole machine has 1200 turns shown in Figure 1.19. The magnetic flux path has a net cross-sectional area of 200 cm2. The iron portion has a mean length of 50 cm, and there are two air gaps, each 0.1 cm in length. The magnetization curve for the iron in the circuit is: H (At/m) 350 650 1250 B (Wb/m2) 1.0 1.2 1.4 Draw the magnetization curve, for the two-pole machine shown in Figure 1.31, find the shunt-field current required to set up a flux of 0.02 Wb in each air group. Neglect all leakage and fringing effects. Figure 1.31 magnetic circuit for problem 1.13. 1.15. A horse shoe type relay needs an excitation of 2000 ampere turns to raise the armature, when the equivalent gap is 1.5 mm. Each pole shoe has an area of 3 cm 2 and the length of iron path is 50 cm. Find (i) the pull on the armature, assuming the relative permeability of iron to be 300. (ii) if the gap closes to 0.2 mm, find the force needed to pull the armature away. the excitation remaining the same. Ans.: 150.4 N ; 216.6 N 44 Introduction to Electrical Machines CHAPTER TWO TRANSFORMERS 2.1. INTRODUCTION The transformer is a static device that transfers electrical energy from one electrical circuit to another electrical circuit through the medium of magnetic field and without a change in the frequency. The electric circuit which receives energy from the supply mains is called primary winding and the other circuit which delivers electrical energy to the load is called secondary winding. Actually the transformer is an electric energy conversion device, since the energy received by the primary is converted to useful electrical energy in the other circuits (secondary winding circuit). If the secondary winding has more turns than the primary winding, then the secondary voltage is higher than the primary voltage and the transformer is called a step-up transformer. When the secondary winding has less turns than the primary windings then the secondary voltage is lower than the primary voltage and the transformer is called step down transformer. Note that a step-up transformer can be used as a step-down transformer, in which the secondary of step-up transformer becomes the primary of the step-down transformer. Actually a transformer can be termed a step-up or step-down transformer only after it has been put into service. The most important tasks performed by transformers are:i) ii) iii) Changing voltage and current levels in electrical power systems Matching source and load impedances for maximum power transfer in electronic and control circuit and Electrical isolation (isolating one circuit from another ) Transformers are used extensively in ac power systems. AC electrical power can be generated at one central location, its voltage stepped up for transmission over long distances at very low losses and its voltage stepped down again for final use. 2.2. CONSTRUCTION OF TRANSFORMER There are basically two types of transformer, the core-type and the shell-type. The two types differ from each other by the manner in which the windings are wound around the magnetic core. The magnetic core is a stock of thin silicon-steel laminations about 0.35mm thick for 50Hz transformers. In order to reduce the eddy current losses, these laminations are insulated from one another by thin layer of varnish. In the core-type, the windings surround a considerable part of steel core as shown in Figure 2.1(a). In shell-type the steel core surrounds a major part of the windings as shown in Figure 2.1(b). For a given 45 Chapter One: Electromagnetic Principles output and voltage rating, core-type transformer requires less iron but more conductor material as compared to a shell-type transformer. The vertical portions of the core are usually called limbs or legs and the top and bottom portions are called yoke. This means that for single-phase transformers, core-type has two-legged core where as shell type has three-legged core. In core-type transformers, most of the flux is confined to high permeability core. However, some of the flux leaks through the core legs and non-magnetic material surrounding the core. The flux called leakage flux, links one winding and not the other. A reduction in this leakage flux is desirable as it improves the transformer performance considerably. Consequently, an effort is always made to reduce it. In the core-type transformer, this is achieved by placing half of the low voltage (LV) winding over one leg and the other half over the second leg or limb. For the high voltage (HV) winding also , half of the winding is over one leg and the other half over the second leg, as shown in Figure 2.1. φ (a) φ 2 φ 2 φ 2 φ 2 (b) Figure 2.1 Constructional details of single-phase (a) core-type transformer (b) Shell-type transformer Low voltage winding is placed adjacent to the steel core and high voltage winding outside, in order to minimize the amount of insulation required. In shell-type transformer the low voltage and high voltage windings are wound over the central limb and are interleaved or sandwiched as shown in Figure2.1(b). Note that the bottom and top are low voltage coils. In core-type transformer, the flux has a single path around the legs or yokes. Figure 2.1(a). In shell-type transformer, the flux in the central limb divides equally and returns through the outer two legs as shown in Figure 2.1(b). There are two types of windings employed in transformers. The concentric coils are used for core-type transformer as shown in Figure 2.1(a) and interleaved (or sandwiched) coils for shell- type transformers as shown in Figure 2.2(b). One type of laminations for the core and shell type of transformers is illustrated in Figure 2.2 (a) and (b) respectively. In both core and shell-type transformers, the individual laminations are cut in the form of long strips of L's, E’s and I's as shown in Figure 2.3 . 46 Introduction to Electrical Machines ) (b) Figure 2.2 two adjacent layers for (a) core and (b) shell type transformers Figure 2.3 long strips of E’s, L’s and I’s laminations In order to avoid high reluctance at the joints where the laminations are butted against each other, the alternative layers are stacked as shown in Figure 2.4. Figure 2.4 Arrangement of butt joints in magnetic core 47 Chapter One: Electromagnetic Principles During the transformer construction first the primary and secondary winding are wound, then the laminations are pushed through the coil openings, layer by layer and the steel core is placed. The laminations are then tightened by means of clamps and bolts. Low-power transformers are air cooled whereas larger power transformers are immersed in oil for better cooling. In oil-cooled transformer, the coil serves as a coolant and also as an insulation medium. 2.3. PRINCIPLE OF TRANSFORMER ACTION φ V1 P N1 N2 S Figure 2.5 Schematic diagram of a two-winding transformer The primary winding P is connected to an alternating voltage source, therefore, an alternating current Im starts flowing through N1 turns. The alternating mmf N1Im sets up an alternating flux φ which is confined to the high permeability iron path as indicated in Figure 2.5. The alternating flux induces voltage E1 in the primary P and E2 in secondary S. If a load is connected across the secondary, load current starts flowing. 2.4. IDEAL TWO-WINDING TRANSFORMER For a transformer to be an ideal one, the various assumptions are as follows 1. Winding resistances are negligible. 2. All the flux set up by the primary links the secondary windings i.e. all of the flux is confined to the magnetic core. 3. The core losses (hysteresis and eddy current losses) are negligible. 4. The core has constant permeability, i.e. the magnetization curve for the core is linear. 2.5.1. EMF Equation of A Transformer Let the voltage V1 applied voltage primary be sinusoidal (or sine wave). Then the current Im and, therefore, the flux φ will flow with the variations of Im . That is, the flux φ is in time phase with the current Im and varies sinusoidally. Let sinusoidal variation of flux φ be expressed as φ = φm Sin ωt Where φm is maximum of the magnetic flux in Weber and ω = 2πf is the angular frequency in rad/sec and f is the supply frequency in Hz. 48 Introduction to Electrical Machines The emf e1 in volt, induced in the primary of N1 turns by the alternating flux is given by dφ dt = − N1ωφmCos ω t e 1 = − N1 = N1ωφm sin( ωt − π ) 2 π Its maximum value, E1max occurs when Sin ωt − is equal to 1. 2 ∴ E1m = N1ωφm and π e 1 = E1m sin ωt − 2 ∴ The RMS value of the induced emf E1 in the primary winding is E 2π E1 = 1m = fN1ωφm 2 2 = 2 πfN1φm (2.1) = 4.44 fN1φm Since the primary winding resistance is negligible hence e1, at every instant, must be equal and opposite of V1. That is, v1 = −e1 = − N1 or dφ dt V1 = − E1 The emf induced in the secondary is dφ = − N 2 ωφm cos ωt dt π = N 2 ωφ s in ( ω t − ) 2 π = Em 2 sin ( ω t − ) 2 e2 = − N 2 ∴ Rms value of emf E2 induced in secondary winding is given by E E 2 = 2m = 2 πfN 2 φm 2 = 4.44 fN 2 φm (2.2) 49 Chapter One: Electromagnetic Principles 2.5.2. Voltage Transformation Ratio From Eqs. (2.1) and (2.2), we get E1 N1 = =k E2 N 2 (2.3) The ratio is known as voltage transformation ratio. i) If N2 > N1 i.e., K<1, then the transformer is called a step-up transformer. ii) If N2 < N1 i.e., K>1, then the transformer is known as a step-down transformer. Again in an ideal transformer Input VA = Output VA V1I1 = V2 I 2 and I1 V2 1 = = I 2 V1 k Hence, the currents are in the inverse ratio of the (voltage) transformation ratio of Eq. (2.3). E E Also, the ratio of 1 = 2 = 2 π f φm and this shows that the emf per turn in each of N1 N 2 the windings is the same. Example 2.1 A single phase transformer has 350 primary and 1050 secondary turns. The net cross-sectional area of the core is 55 cm2. If the primary winding be connected to a 400 V, 50 Hz single phase supply, calculate (i) the maximum value of flux density in the core and (ii) the voltage induced in the secondary winding. Solution Voltage applied to the primary = 400 V Induced emf in the primary, E1 ≈ voltage applied to the primary, V1 = 400 V Number of turns in the primary N1 = 350 Net cross-sectional area Ai = 55 cm2 = 55 ×10-4 m2 Frequency of the supply f = 50 Hz 50 ) ) Introduction to Electrical Machines Induced emf in the primary is given by E1 = 4.44f φm N1 = 4.44f Bm A i N1 E1 = Maximum value of flux density in the core, Bm = 400 4.44 × 50 × 55 × 10 − 4 × 350 = 0.93 T ( Wb / m 2 ) Number of turns in the secondary winding, N2 = 1050 For an ideal transformer, E1 N1 = E2 N 2 Voltage induced in the secondary winding, N2 N1 1050 = 400 × 350 = 1200 V E 2 = E1 × Example 2.2 The required no-load voltage ratio in a single phase 50 Hz, core type transformer is 6600/500. Find the number of turns in each winding, if the flux is to be 0.06 Wb. Solution No-load voltage ratio = 6000 500 No-load voltage of low voltage winding = 500 V Flux φ= 0.06 Wb Frequency f = 50Hz Induced emf in the low voltage winding (secondary) of the transformer is given by, E 2 = 4.44f φm N 2 or 500 = 4.44 × 50 × 0.06 × N 2 Number ut turns in the low voltage, 500 N2 = 4.44 × 50 × 0.06 = 37.5 ( not possible) The number of turns in each winding should be a whole number, moreover each winding in the core type transformer is accommodated on both the limbs. i.e. half number of turns of each winding on one limb. As such the number of turns in each winding should be even. 51 Chapter One: Electromagnetic Principles Considering these facts , the number of turns in low voltage winding, N2 = 38 Number of turns in high voltage winding V1 6600 = 38 × V2 500 = 501.6 ( not possible) N1 = N 2 × Considering all the factors mentioned above, the number of turns in the high voltage winding N1=500. Here the number of turns finally taken is 500 and not 502, because the high voltage winding will be split up into a number of coils. With 250 turns on each limb, high voltage winding on one limb can be split into 5 coils of 50 turns each. 2.5. EQUIVALENT CIRCUIT OF A TRANSFORMER The equivalent circuit for any electrical engineering devices can be drawn if the equations describing its behavior are known. If any electrical device is to be analyzed and investigated further for suitable modifications, its appropriate equivalent circuit is necessary. The equivalent circuit for electromagnetic devices consists of a combination o1 resistances, inductances, capacitances, voltages etc. Such an equivalent circuit (or circuit model) can, therefore, be analyzed and studied easily by the direct application of electric circuit theory. As stated above equivalent circuit is simply a circuit representation of the equations describing the performance of the device. In the equivalent circuit of Figure 2.6(a) (rl +jx1) and (r2 + jx2) are the leakage impedances of the primary and secondary windings respectively. The voltage V1' is treated as a voltage drop in the direction of I1. Recall that the magnitude of V1' does not change appreciably from no load to full load in large transformers. The magnitude of V1' depends on f ,N1 and φm, since V1' = E1 . The primary current I1 consists of two components. One component I1' is the load component and counteracts the secondary m.m.f. I2N2 completely. The other component is exciting current Ie which is composed of Ic and Im. The current Ic is in phase with V1' and product V1' Ic gives core loss. The resistance Rc parallel with V1' represents the core loss Pc, such that. Pc = And I c2 R c = V1' I c 2 ( V1' ) = Rc V' Rc = 1 Ic The current Im lags V1' by 90° and this can, therefore , be represented in the equivalent circuit by a reactance Xm, such that V' Xm = 1 Im 52 Introduction to Electrical Machines Rc and Xm are shown in Figure 2.6 (b), which is the exact equivalent circuit of a transformer. The resistance Rc and reactance Xm are called core-loss resistance and magnetizing reactance, respectively. For minor changes in supply voltage and frequency, which is common under normal operation, Rc and Xm are treated constant. In Figure 2.6 (a) and (b) , the ideal transformer has been introduced to show the transformation of voltage and current between primary and secondary windings . Even at this stage the transformer magnetization curve is assumed linear, since the effect of higher order harmonic can't be represented in the equivalent circuit. E2 E1 (a) Exciting current neglected I1' E1 (b) Exact equivalent circuit jx'2 I1' r2' N I1' = I 2 2 N1 E1 N E 2 1 N2 N V2 1 N2 (c) Referred to primary 53 Chapter One: Electromagnetic Principles r1' N V1 2 N1 jx1' N I1 1 N2 N I e 1 N2 I'm I'c N E1 2 N1 I2 V2 E2 (d) referred to secondary Figure 2.6 Development of the exact equivalent circuit of a transformer In transformer analysis, it is usual to transfer the secondary quantities to primary side or primary quantities to secondary side. Secondary resistance drop I2 r2 when transferred to primary side must be multiplied by the turns ratio N1 /N2. N ∴ Secondary resistance drop, when transferred to primary = (I 2 r2 ) 1 N2 N N = I1 ⋅ 1 ⋅ r2 1 N2 N2 N putting I 2 = I1 ⋅ 1 N2 N 2 = I1 1 r2 = I1r2' N 2 Where N r2' = r2 ⋅ ⋅ 1 N2 2 If resistance r2' is placed in the primary circuit, then the relation between voltage V1 and V2 is unaffected. This resistance r2' is called the secondary resistance referred to primary. Therefore, the total resistance in the primary circuit is 2 N re1 = r1 + r2 ⋅ 1 = r1 + r2' N2 Hence re1 is called the transformer equivalent (or total) resistance referred to primary 2 N winding. Similarly the primary resistance referred to secondary is r1 ⋅ ⋅ 2 and the N1 equivalent ( or total) resistance referred to secondary is 54 Introduction to Electrical Machines 2 N re 2 = r2 + r1 ⋅ ⋅ 2 = r2 + r1' N1 Figure 2.6 (e) equivalent circuit in general form. Secondary leakage reactance drop I2 x2, when transferred to primary is N 2 N1 = I1 1 x 2 = I1x '2 I 2 x 2 N N 2 2 The quantity x '2 is called the secondary leakage reactance referred to primary. Total primary leakage reactance is 2 N x e1 = x1 + x 2 ⋅ 1 = x1 + x '2 N2 Where xe1, is called the equivalent or total leakage reactance referred to primary. Likewise, the equivalent or total leakage reactance referred to secondary is 2 N xe 2 = x2 + x1 ⋅ 2 = x2 + x1' N1 The equivalent (or total) leakage impedance referred to primary is z e1 = re1 + jx e1 The equivalent (or total) leakage impedance referred to secondary is z e2 = re2 + jx e2 Following the above procedure, it can be shown that 2 2 N N z e1 = 1 z e2 and z e2 = 2 z e1 N2 N1 55 Chapter One: Electromagnetic Principles In general, when values are referred to either circuit the following conditions should be kept in mind The energy condition (i.e. the active and reactive) power should be remain unchanged The phase angle between voltage and current i.e. power factor, should be remain the same and The referring factor must be the same for all values of the same type. Simplification of the exact equivalent circuit: The equivalent circuit of Figure 2.6 (b) can be simplified by referring all the quantities to primary or secondary and at the same time, moving the ideal transformer to one side. If the secondary quantities are referred to primary, the equivalent circuit of Figure2.6 (c) is obtained. Since it is usual to omit the ideal transformer, it is shown dotted for the sake of completeness. When the primary quantities are referred to the secondary side, the equivalent circuit of Figure 2.6(d) is 2 2 N N obtained. Note = R c 2 and X 'm = X m 2 . The exact equivalent circuits N1 N1 of Figure 2.6(c) and (d) are known as T-circuits for a transformer, referred to primary and secondary windings respectively. that R 'c In the equivalent circuits of Figure 2.6 (c) and (d), the referred quantities with suitable notation , have been used. A more general equivalent circuit can be drawn as shown in Figure 2.6(e), where for simplicity (i) a particular notation for referred-quantities has been dropped (ii) the complex notation (bar over I, j with reactances etc.) has been given up and (iii) the ideal transformer is not shown. If the general equivalent circuit refers to the primary, one has to keep in mind that the secondary quantities have been referred to the primary side. On the other hand, if the general equivalent circuit refers to the secondary, then the primary quantities must be referred to the secondary side. Thus in the general equivalent circuit of a transformer, one has merely to keep in mind about the side to which all the quantities have been referred. It may be interesting at this stage to draw the phasor diagram for the equivalent circuit of Figure 2.6(e) from a knowledge of the electric circuit theory. Assume that the secondary load voltage V2 load current I2 and angle θ2, by which I2 lags V2 are known. First of all draw I2 lagging V2 by an angle θ2 and then add I2 (r2 + jx2) to V2 to obtain E2, Figure 2. It is obvious from Figure (le) that current Im due to voltage E2, must lag it by 90° and further Ic must be in phase with E2. The phasor sum of Ic and Im gives Ie and phasor sum of I2 and Ie gives I1. The voltage drop I1 (r1 + jx1) is now added to E2 to obtain V1 as shown in Figure 2. The secondary p.f. is cosθ2 lagging and the primary p.f. is cos θ1 lagging . The voltage drops I1 (r1 + jx1) and I2 (r2 + jx2) have been drawn to a much larger scale, in comparison-with V1 or V2 for the sake of clarity. 56 Introduction to Electrical Machines θ1 θ2 Figure 2.7 Phasor diagram for equivalent circuit of Figure 2.6 (e) Approximate Equivalent Circuit: Approximate equivalent circuit is obtained from the exact equivalent circuit Figure 2.6(e), if the shunt branch (Rc and Xm in parallel) is moved to the .primary or secondary terminals as shown in Figure 2.8(a) and (b) respectively. It may be seen from Figure 2.8 (a) that the exciting current Ie does not flow through rl and x1, whereas Ie does flow through r1 and x1 in the exact equivalent circuit. Thus the primary leakage impedance drop due to the exciting current, i.e. Ie( r1 +jx1) has been neglected in Figure 2.8 (a), though it is not so actually. It may also be seen from Figure 2.8 (b) that Ie flows through r2 and x2, whereas Ie does not flow through r2 and x2 in the exact equivalent circuit. Thus the secondary leakage impedance drop due to Ie, i.e. Ie (r2 + jx2) has been included, though Ie (r2 + jx2) is actually zero. (a) (b) 57 Chapter One: Electromagnetic Principles r1+r2=re x1+x2=xe + + I1=I2 V1 V2 _ _ (c) (d) Figure 2.8 (a) and (b) Approximate equivalent circuits of a transformer (c) and (d) Simplified forms of the approximate equivalent circuit. Since the exciting current is only about 2 to 6 per cent of the rated winding current in power and distribution transformers, the error introduced by neglecting Ie( r1 +jx1) or including Ie (r2 + jx2) is insignificant. However, the computational labor involved is reduced considerably by the use of approximate equivalent circuits of Figure 2.8(a) and (b). As before, one must keep in mind about the side to which all the equivalent-circuit quantities have been referred. Still further simplification is achieved by neglecting the shunt branch Rc and Xm in Figure 2.8 (a) and (b) and this leads to equivalent circuit of Figure 2.8(c). This simplification is tantamount to neglecting exciting current Ie in comparison with rated currents, which is almost justifiable in large transformers, say over 100 KVA or so. For transformers having ratings near 500 KVA or more, the equivalent resistance re is quite small as compared with equivalent leakage reactance xe. Consequently re may be neglected, leading to the equivalent circuit of Figure3 (d). Thus, when a large power system is studied, a transformer is usually replaced by its equivalent circuit of the form shown in Figure 2.8(d). The equivalent circuit Figure 2.6(e) should be used only when the exciting current is a large percentage of the rated current e.g., in audio-frequency transformers used in electronic circuits, in transformers used for relaying and measurement purposes etc. For high voltage surge investigations, the transformer equivalent circuit must be modified to include the effects of inter-turn and turn to earth capacitances. 2.6. OPEN-CIRCUIT AND SHORT-CIRCUIT TESTS These two tests on a transformer help to determine (i) The parameters of the equivalent circuit of Figure 2.8 (ii) the voltage regulation and (iii) efficiency The equivalent circuit parameters can also be obtained from the physical dimensions of the transformer core and its winding details. Complete analysis of the transformer can be carried out, once its equivalent circuit parameters are known. The power required during these two tests is equal to the appropriate power loss occurring in the transformer. 58 Introduction to Electrical Machines 2.6.1. Open Circuit (or No-Load) Test The circuit diagram for performing open circuit test on a single phase transformer is given in Figure 2.9 (a). In this diagram, a voltmeter, wattmeter and an ammeter are shown connected on the low voltage side of the transformer. The high voltage side is left open circuited. The rated frequency voltage applied to the primary, i.e. low voltage side, is varied with the help of a variable ratio auto-transformer. When the voltmeter reading is equal to the rated voltage of the L.V. winding , all three instrument readings are recorded. (a) (b) Figure 2.9 (a) Circuit diagram for open-circuit test on a transformer and (b) approximate equivalent circuit at no load The-ammeter records the no-load current or exciting current Ie. Since Ie is quite small (2 to 6%) of rated current), the primary leakage impedance drop is almost negligible, and for all practical purposes, the applied voltage V1 is equal to the induced emf E1. Consequently, the equivalent circuit of Figure 2.6 (e) gets modified to that shown in Figure 2.9( b). The input power given by the wattmeter reading consists of core loss and ohmic loss. The exciting current being about 2 to 6 percent of the full load current, the ohmic loss in 2 2 the primary( = I e2 r1 ) varies from 0.04 percent to 0.36 percent of the ⋅ × 100 100 100 full-load primary ohmic loss . In view of this fact, the ohmic loss during open circuit test is negligible in comparison with the normal core loss (approximately proportional to the square of the applied voltage). Hence the wattmeter reading can be taken as equal to transformer core loss. V1 = Applied rated voltage on L.V. side, Ie = exciting current ( or no-load current) and Pc = core loss Then Pc = V1I e cos θo 59 Chapter One: Electromagnetic Principles ∴ No load p.f. = cos θo = Pc V1I e From phase diagram of Figure 2.7, it follows that I c = I e cos θo and I m = I e sin θo P From Figure 4( b), I c = c V1 V V1 R CL = 1 = I c I e cos θo ∴ Core loss resistant V2 V12 = = 11 V1I e cos θo Pc o Also I c2 R CL = Pc ∴ P Pc R CL = c = I c2 (I e cos θo )2 Magnetizing reactance X mL = Pc V1 = I m I e sin θo The subscript L with Rc and Xm is used merely to emphasize that theses values are for the L.V. side. It must be kept is mind that the values of Rc and Xm, in general, refer to the side, in which the instruments are placed (the L.V. side in the present case). A voltmeter is sometimes, used at the open-circuited secondary terminals, in order to determine the turns ratio. Thus the open-circuit test gives the following information: (i) core loss at rated voltage and frequency, (ii) the shunt branch parameters of the equivalent circuit, i.e. Rc and Xm and (iii) turns ratio of the transformer. Short-Circuit Test The low voltage-side of the transformer is short-circuited and the instruments are placed on the high voltage side, as illustrated in Figure 2.10 (a). Short circuit 2.6.2. Figure 2.10 (a) connection diagram for short circuit test on a transformer 60 Introduction to Electrical Machines r1 x1 x2 Rc r2 Xm Figure 2.10 (b) Equivalent circuit with short-circuit on the secondary side r1 x1 x2 r2 Isc Vsc Figure 2.10 (c) Transformer equivalent circuit with secondary short-circuited The applied voltage is adjusted by auto-transformer, to circulate rated current in the high voltage side. In a transformer, the primary m.m.f. is almost equal to the secondary m.m.f., therefore, a rated current in the H.V. winding causes rated current to flow in the L.V. winding. A primary voltage of 2 to 12% of its rated value is sufficient to circulate rated currents in both primary and secondary windings. From Figure 2.10 (b) , it is clear that the secondary leakage impedance drop appears across the exciting branch (RC and Xm in parallel). About half (1 to 6%) of the applied voltage appears across the secondary leakage impedance and, therefore, across the exciting branch. The core flux induces the voltage across the exciting branch and since the latter is 1 to 6% of rated voltage, the core flux is also 1 to 6% of its rated value. Hence the core loss, being approximately 1 1 proportional to the square of the core flux, is 0.01 percent = × × 100 to 0.36 100 100 6 6 percent = × × 100 of its value at rated voltage. The wattmeter, in short circuit 100 100 test, records the core loss and the ohmic loss in both windings. Since the core loss has been proved to be almost negligible in comparison with the rated voltage core loss, the wattmeter can be taken to register only the ohmic losses in both windings. At rated-voltage, the exciting Current is 2 to 6% of full load current. When the voltage across the exciting branch is 1 to 6% of rated voltage, the exciting current may be 0.02 1 6 2 6 percent = × × 100 to 0.36% percent = × × 100 of its full-load 100 100 100 100 current and can, therefore, be safely ignored. As a result of this the equivalent circuit of Figure2.6(e), with the secondary short-circuited, gets modified to that shown in Figure 2.10 (c) 61 Chapter One: Electromagnetic Principles Let VSC, ISC and PSC be the voltmeter, ammeter and wattmeter readings; then from Figure 2.10 (c), equivalent leakage impedance referred to H.V. side, V Z eH = SC I SC P equivalent resistance referred to H.V. side, reH = SC and equivalent leakage reactance 2 I SC 2 2 referred to H.V. side, X eH = Z eH − reH In reH, XeH and ZeH„, the subscript H is used to indicate that these quantities are referred to H.V. side. These parameters can however, be referred to the L.V. side, if required. In the analysis of transformer equivalent circuit, the values of equivalent resistance and equivalent leakage reactance referred to either side are used. However, if the leakage impedance parameters for both primary and secondary are required separately, then it is usual to take r1 = r2=½ re) and x1 = x2=½ xe, referred to the same side. Thus, the short-circuit test gives the following information: (i) ohmic loss at rated current and frequency and (ii) the equivalent resistance and equivalent leakage reactance. Voltage regulation of a transformer can be determined from the data obtained from short-circuit test. Data of both open-circuit and short-circuit tests is necessary (i) for, obtaining all the parameters of exact equivalent circuit and (ii) for calculating the transformer efficiency. How can a wattmeter connected on the H.V. side, record the ohmic in the L.V. winding also? When rated current is made to flow in the H.V. winding, the 1.v. winding must also carry rated current, because the transformer action requires I1N1= I2N2. The flow of rated current in the L.V. winding causes ohmic loss, which must be supplied from somewhere. The only way to provide L.V. winding loss is from the input to H.V. side. But the entire input power to H.V. side is recorded by the wattmeter, therefore, the ohmic losses in both windings are given by the wattmeter reading. It has already been stated that open-circuit and short-circuit tests should be performed on the L.V. side and H.V. side respectively only for the sake of convenience. This can he illustrated by considering a 3300/220V, 33KVA, single-phase transformer. For open-circuit test on low voltage side, the ranges of voltmeter, ammeter and wattmeter are 220V (rated value), 6A ( 2 to 6% of rated current of 150A) and 6A, 220V respectively. These are the standard ranges for ordinary instruments and therefore, more accurate readings can be obtained. If the open circuit test is performed on the H.V. side, a source of 3300V may not be readily available. At the same time, the instrument ranges are 3300V, 0.4A and 0.4A , 3300V which are which are not within the range of ordinary instruments and the results obtained may not be so accurate . Also it may not be safe to work on the high voltage side. For a short-circuit test on the H.V. side, the instrument ranges are 165V (2 to 12% of rated voltage of 3300V), l0A (rated current) and 10A, 165V, which are well within the 62 Introduction to Electrical Machines range of the ordinary instruments. On the other hand, instrument ranges, for a shortcircuit test on L.V. side are 11V, 150A, and 150A. 11V. Instruments of such ranges and auto-transformer capable of handling 150A may not be readily available and at the same time, the results may not be so accurate. It is for these reasons that the open-circuit and short-circuit tests are conducted on L.V. and H.V. sides respectively. 2.6.3. Polarity Test On the primary side of a two-winding transformer, one terminal is positive with respect to the other terminal at any one instant. At the same instant, one terminal of the secondary winding is positive with respect to the other terminal. These relative polarities of the primary and secondary terminals at any instant must be known if the transformers are to be operated in parallel or are to be used in a polyphase circuit. E1- E2 E1+ E2 V A2 A1 V - - E1 E2 + + (a) a2 A2 - + E1 a1 A1 a1 E2 - + a2 (b) Figure 2.11 Polarity test on a two winding transformer (a) subtractive polarity and (b) additive polarity When viewed from the H.V. side, the terminals are marked A1 and A2, the former, i.e. A1 being on the extreme right. Terminals A1 and A2 marked plus and minus arbitrarily in Figure 6. Now terminal A1 is connected to one end of the secondary winding and a voltmeter is connected between A2 and the other end of the secondary winding. A voltage of suitable value is now applied to the H.V. winding. Let E1 and E2 be the e.m.fs induced on H.V. and L.V. sides respectively. If the voltmeter reading is equal to E1–E2 then secondary terminal connected to A1 is positive and is marked a1, the L.V. terminal connected to A2 through the voltmeter is negative and is marked a2 as shown in Figure 2.11(a). If voltmeter reading is equal to E1+E2, then the terminals connected to A1 and A2 are negative and positive and are marked a2 and a1 respectively as shown in Figure 2.11(b). The subscript numbers 1,2 on the H.V. and L.V. windings are so arranged that when A2 is negative with respect to A1. a2 is also negative with respect to a1 at the same instant. In other words, if the instantaneous emf is directed from A2 to A1 in H.V. winding, it is at the same time directed from a2 to a1 in the L.V. winding. When the voltmeter reads the difference E1–E2, the transformer is said to possess a subtractive polarity and when voltmeter reads E1+E2 the transformer has additive polarity. In subtractive polarity, the voltage between A2 and a2 (or A1 and a1) is reduced. The leads connected to these terminals and the two windings are, therefore, not subjected to high voltage stress. In additive polarity the windings and the leads connected to A1, 63 Chapter One: Electromagnetic Principles A2, a1 and a2 are subjected to high voltage stresses. On account of these reasons, subtractive polarity is preferable to additive polarity. Example 2.3 A 20 kVA, 2500/250 V, 50 Hz, single-phase transformer gave the following test result Open-circuit test (on L.V. side):250 V, 1.4 A, 105 W. Short-circuit test (on H.V. side): 104 V, 8 A, 320 watts. Compute the parameters of the approximate equivalent circuit referred to high-voltage and low-voltage sides. Also draw the exact equivalent circuit referred to the low -voltage side. Solution From open-circuit test: 105 = 0 .3 250 × 1.4 θ0 = 72.55° and sin θ0 =0.954 Ic = Ie cos θ0 = 1.4×0.3 = 0.42 A Im = Ie sin θ0 = 1.4×0.954 = 1.336 A No-load power factor, cos θ0 = and V 250 R cL = 1 = = 595 Ω I c 0.42 V 250 X mL = 1 = = 187 Ω I m 1.336 Alternatively, the value of RcL and XmL can be determined as follows: hence, Now V 2 (250)2 R cL = 1 = = 595 Ω Pc 105 V 250 Ic = 1 = = 0.42 A R cL 595 I m = I e2 − I c2 = 1.4 2 − 0.42 2 = 1.336 A V 250 X mL = 1 = = 187 Ω I m 1.336 From short circuit test: V 104 Z eH = sc = = 13 Ω I sc 8 P 320 reH = sc = = 8Ω 2 I sc 82 and 2 2 ∴ x eH = Z eH − reH = 132 − 52 = 12 Ω Equivalent circuit parameters referred to L.V. side are: RcL = 595 Ω XmL = 187 Ω 64 Introduction to Electrical Machines 2 2 2 2 1 1 1 1 reL = reH × = 5 = 0.05 Ω ; x eL = x eH × = 12 = 0.12 Ω k 10 k 10 This equivalent circuit is shown in Figure below (a) x eL = 0.12 Ω R cL = 595Ω 0.025 Ω reL = 0.05 Ω X cL = 187 Ω 0.06 Ω 595Ω 0.06 Ω 0.025 Ω 187 Ω (a) (b) Figure (a) approximate equivalent circuit referred to L.V. side and exact equivalent circuit referred to L.V. side. Equivalent circuit parameters referred to H.V. side are: R cH = R cL × (k )2 = 595(10 )2 = 59,500 Ω X mH = X mL × (k )2 = 187(10)2 = 18,700 Ω reH = 5Ω ; xeH = 12Ω An equivalent circuit showing these parameters can easily be drawn. Exact equivalent circuit parameters referred to L.V. side are: r1L = r1' H = 1 1 reL = 0.05 = 0.025 Ω 2 2 x1L = x1' H = 1 1 x eL = 0.12 = 0.06 Ω 2 2 RcL = 595 Ω and XmL = 187 Ω The exact equivalent circuit is shown in Figure (b) 65 Chapter One: Electromagnetic Principles 2.7. TRANSFORMER PHASOR DIAGRAMS The purpose of first considering an ideal transformer, i.e. a transformer with no core losses, no winding resistance, no magnetic leakage and constant permeability, is merely to highlight the most important aspect of transformer action. Such transformer never exists and now the phasor diagrams of real transformer will be considered. 2.6.4. a) Transformer Phasor Diagram at no-load The magnetic flux φm being common to both the primary and secondary is drawn first. The induced emf E1 and E2 lag φm by 900 and are shown accordingly in Figure 2.12. The voltage -E1 is being replaced by V′1 just for convenience. Effect of transformer core loss The core loss (or iron loss) consists of hysteresis loss and eddy current loss. These losses are always present in the ferromagnetic core of the transformer, since the transformer is an ac operated magnetic device. The hysteresis loss in the core is minimized by using high grade material such as cold-rolled-grain-oriented (CRGO) steel and the eddy current loss is minimized by using thin lamination for the core. The current in the primary is alternating, therefore, the magnetizing force H is cyclically varying from one positive value say Hl to a corresponding negative value −Hl, Figure 2.12 (a). When the magnetizing force is - Hl, the flux density is maximum negative equal to OM. As the magnetizing force decreases from - Hl, the current Ie decreases and becomes zero for a flux density, or flux, equal to ON. When the current Ie becomes positive and equal to OP, the flux is reduced to zero but it is going to become positive. The traverse of the loop along the arrows involves time. When Ie is crossing zero positive (passing through zero and becoming positive), the core flux is negative and is equal to ON in Figure 2.12(a). This is shown in Figure 2.12(b) at instant ωtl, where waveforms are assumed sine waves. When Ie is positive and equal to OP, Figure 2.12(a), the flux is crossing zero and becoming positive; this is shown in Figure 2.12(b) at instant ωt2. It is seen from Figure 2.12(b) that exciting current Ie leads the magnetic flux φ (or φ lags Ie) by some time angle α. This angle of lead, or lag, being dependent on the hysteresis loop, is called the hysteretic angle. In Figure 2.12 (c), Ie is shown leading φ, or φ is shown lagging Ie, by hysteretic angle α. Β or Φ φ ,φ ωt α ωt1 (a) ωt 2 (b) 66 Introduction to Electrical Machines θ0 φl1 φ (c) Figure 2.12 (a) Hysteresis loop for transformer core (b) exciting current and core flux waveforms and (c) no-load phasor diagram of a transformer. The no-load primary current Ie is called the exciting current of the transformer and can be resolved into two components. The component Im along φm is called the reactive or magnetizing current , since its function is to provide the required magnetic flux φm. The second component along V'1 is Ic and this component is called the core- loss component. When multiplied by V'1 gives the total core loss Pc. P V ' 1 I c = Pc or I c = c Amp . V' 1 From Figure 2.12 (c), it is seen that I e = I m2 + I 2c Note that in an ideal transformer, core-loss current I c = 0 and therefore exciting current Ie equals to magnetizing current Im i.e. I e = I m . b) Effect of transformer resistance The effect of primary resistance r1 can be accounted for, by adding to V′1, a voltage drop equal to Ie r1 as shown in Figure 2.12 (c). Note that Ier1 is in phase with Ie and is drawn parallel to Ie in the phasor diagram. c) Effect of leakage flux The existence of electrical potential difference is essential for the establishment of current in an electric circuit. Similarly the magnetic potential difference is necessary to establish flux in a magnetic circuit. This magnetic potential difference establishes: i) ii) the mutual flux φm linking both windings and the primary leakage flux φl1 which links only the primary winding. The distinctive behavior of the mutual flux φm and the primary leakage flux φl1The mutual flux φm exists entirely in the ferromagnetic core and, therefore, involves hysteresis loop. The current Ie that establishes φm must lead it by some hysteresis angle . 67 Chapter One: Electromagnetic Principles On the other hand, the primary leakage flux φl1 exists largely in air. Although φl1 does pass through some iron, the reluctance offered to φl1 is mainly due to air. Consequently φl1 does not involve any hysteresis loop and it can be taken to be in phase with the current Ie that produces it, Figure2.12(c). In the primary winding,φ induces an emf E1 lagging it by 90°; similarly the primary leakage flux φl1 induces an emf Ex1 in the primary winding and lagging it (i.e. φl1) by 90°. Since Ie leads Ex1 by 90°, it is possible to write EX1= -jIexl. The primary applied voltage Vl must have a component jIexl, equal and opposite to Exl. Here xl has the nature of reactance and is referred to as the primary leakage reactance in ohms. It may be noted that x1 is a fictitious quantity merely introduced to represent the effects of primary leakage flux. Figure 2.13 Transformer at no load. The total voltage drop in primary at no-load is Ie (r1+jx1) = Iez1 where z1 is the primary leakage impedance. Therefore Figure 2.12(c) gives the phasor diagram of transformer at no-load, where Nl is assumed to be equal to N2. The primary voltage equation at no-load can be written as: V1 = V1' + I e ( r1 + jx1 ) The primary leakage impedance drop shown in Figure 2.12(c), is drawn to a larger scale, in comparison with Vl' or Vl, just for the sake of clarity. At no-load and V'1 and V1 are very nearly equal. Even at full load primary leakage impedance drop in power transformer is about 2 to 5% of V1, so that the magnitude of V'1 or E1 does not change appreciably from no-load to full load. 2.6.5. Transformer Phasor Diagram Under Load The secondary circuit of transformer is considered first and then the primary circuit, for developing the phasor diagram of a transformer under load. Figure 2.14 Transformer under load When the switch S is closed, secondary current I2 starts flowing from terminal n to the load. Assume the load to have a lagging power factor so that I2 lags secondary load voltage V2 by an angle θ2. At first V2 is drawn with I2 lagging V2 by the secondary p.f. angle θ2, Figure 1.11 (a). The secondary resistance drop is accounted for, by drawing I2r2 parallel to I2. The secondary m.m.f. I2N2 gives rise to a leakage flux φl2 which links only 68 Introduction to Electrical Machines the secondary and not the primary. The flux φl2 is called the secondary leakage flux and is in phase with I2, for the same reason that φl1 is in phase with Ie in Figure 2.12(c). The secondary leakage flux induces emf Ex2 in the secondary winding, lagging φl2 by 90°. The secondary no load voltage E2 must have a component equal and opposite to –jx2I2. Thus the phasor sum of V2, I2r2 and jx2I2 gives the secondary induced emf E2 as shown in Figure 2.15(a). The voltage equation for secondary circuit can now be written as E2 = V2 + I 2 ( r2 + jx2 ) = V2 + I 2 z 2 where z2 is the secondary leakage impedance of the transformer. Further the mutual flux φ is drawn leading E2 by 90° and exciting current Ie is drawn leading φ by the hysteretic angle α. Note that the phasor V2 has purposely been taken to the left of vertical line, so that E2 is vertically downward and the mutual flux φ is horizontal. The component of the primary current which neutralizes the demagnetizing effect of I2 is I'1 (I′1N1 = I2N2) and drawn opposite to I2. The phasor sun of I'1 and Ie gives the total primary current I1 taken from the supply mains . The primary leakage impedance drop I1(r1+jx1) is depicted as explained earlier. The voltage equation for primary circuit under load can be written as V1 = V1' + I1( r1 + jx2 ) = V1' + I1 z1 where z1 is the primary leakage impedance of transformer. Note that the angle θ1 between V1 and I1 is the primary power factor angle under load. If the secondary load current I2 leads the voltage V2 such that the load p.f. is leading, then the phasor diagram for the transformer is as shown in Figure 2.15 (b). The entire procedure for drawing the phasor diagram is the same as explained for Figure 2.15 (a). x1 jI 1 θ1 θ1 φl 2 α φ α φ θ2 θ2 x2 jI 2 x2 jI 2 Figure 2.15 Transformer phasor diagram for (a) lagging p.f. load and (b) leading p.f. load 69 Chapter One: Electromagnetic Principles 2.8. VOLTAGE REGULATION OF A TRANSFORMER Constant voltage is the characteristics of most domestic, commercial and industrial loads. It is therefore, necessary that the output voltage of a transformer must remain within narrow limits as the load and its power factor vary. This requirement is more stringent in distribution transformers as these directly feed the load centers. The voltage drop in a transformer on load is chiefly determined by its leakage reactance which must be kept as low as design and manufacturing techniques would permit. The voltage regulation is defined as voltage in secondary terminal voltage, expressed as a percentage (per unit) of secondary rated voltage i.e. Voltage regulation = E2 − V2 in p .u sec ondary rated voltage where E2 = Secondary terminal voltage at no load V2 = Secondary terminal voltage at any load It is stipulated that the secondary rated voltage of a transformer is equal to the secondary terminal voltage at no load, i.e. E2. E − V2 E − V2 Voltage regulation = 2 in p .u = 2 × 100 in percentage E2 E2 ∴ At no-load, the primary leakage impedance drop is almost negligible, therefore, the N secondary no-load voltage E 2 = V1 2 . The expression for voltage regulation can also N1 be written as V1 N2 N − V2 V1 − V2 1 N1 N2 × 100 in percentage = × 100 in percentage N2 V 1 V1 N1 Here V1 is the primary applied voltage. The change in secondary terminal voltage with load current is due to the primary and secondary leakage impedances of the transformer. The magnitude of this change depends on the load power factor, load current, total resistance and leakage reactance of a transformer. A distribution transformer should have a small value of voltage regulation (i.e. good voltage regulation) so that the terminal voltage at the consumers does not vary widely as the load changes. For a transformer of large voltage regulation (i.e. poor voltage regulation), the voltage at the consumers' terminals will fall appreciably with increase in load. This has a detrimental effect on the operation of fluorescent tubes, T.V. sets, refrigeration motors, etc since these are designed to operate satisfactorily at a constant 70 Introduction to Electrical Machines voltage. Thus distribution transformer should be designed to have a low value of leakage impedances. The voltage regulation of a transformer can be obtained form its approximate equivalent circuit referred to primary or secondary. Figure 2.16 (a) illustrates the approximate equivalent circuit of a transformer referred to the secondary side and the phasor diagram for this circuit is drawn in Figure 2.16 (b) for a lagging power factor load. For the calculation of voltage regulation, draw an arc of radius OD meeting the extension of line OA in F. It may be seen from Figure 2.16 (b) that OF (= E2) is approximately equal to OC. ∴ E2 = OC = OA + AB + BC (or B' C' ) = OA + AB' cos θ2 + DB' sin θ2 = V2 + I 2 re 2 cos θ2 + I 2 xe 2 sin θ2 θ2 θ2 θ2 θ2 Figure 2.16 (a) approximate equivalent circuit of a 2-winding transformer, referred to secondary; (b) the phasor diagram of the circuit of Figure 2.16(a) for lagging power factor load. Thus the voltage drop in the secondary terminal voltage E2 − V2 = I 2 re 2 cos θ2 + I 2 xe 2 sin θ2 2.1 Note carefully that E2-V2 is not equal to AD i.e. I2ze2. The change in secondary terminal voltage is equal to the magnitude of E2 minus the magnitude of V2. In Eq.(2.1), per unit voltage regulation for any load current I2 is I x E 2 − V2 I 2 re 2 = cos θ2 + 2 e 2 sin θ2 E2 E2 E2 2.2 In case I2, is rated current, then I 2 re 2 voltage drop across re 2 at rated current = E2 Rated ( = base ) voltage E 2 = p .u . equivalent resis tan ce or p .u . resis tan ce drop = ε r ( say ) 71 Chapter One: Electromagnetic Principles Also I r I2 r Ohmic loss at rated current ε r = 2 e2 = 2r e2 = E2 E2 I 2 r Rated VA I ⋅x Similarly, for rated current I2, 2 e 2 = ε x E2 From Eq. (2.2), the per unit voltage regulation at rated current is given by ε r cos θ2 + ε x sin θ2 (2.3a) Percentage voltage regulation at rated load (ε r cos θ2 + ε x sin θ2 ) × 100 (2.3b) It should be noted that Eqs. (2.1) to (2.3) are valid for lagging power factors only. For leading power factor loads, the phasor diagram of Figure reveals that E2 ≅ Oc = V2 + I 2 re 2 cos θ2 − I 2 xe 2 sin θ2 Therefore , secondary terminal voltage drop, for any load current I2, is E2 − V2 = I 2 re 2 cos θ2 − I 2 xe 2 sin θ2 ∴ p.u. voltage regulation at any load current I2 is given by I 2 re 2 I x cos θ2 − 2 e 2 sin θ2 E2 E2 In case I2 is the rated ( or full-load)current, then p.u. voltage regulation is given by ε r cos θ2 − ε x sin θ2 Condition for zero voltage regulation: It can be seen from Eq. (2.3) that voltage regulation varies with load power factor. If load power factor is varied with constant values of load current and secondary emf, then zero voltage regulation will occur when ε r cos θ2 + ε x sin θ2 = 0 I 2 re 2 r ε tan θ2 = − r = − = − e2 I x εx xe2 E2 2 e 2 E2 x ∴ magnitude of the load p.f. , cos θ2 = e 2 re 2 The negative value of tanθ2 indicates a leading power factor. Therefore, zero voltage x regulation occurs when load power factor is e 2 leading . For leading p.f.s. greater than ze2 xe 2 , the voltage regulation will be negative, i.e. the voltage will rise from its no load ze2 value, as the transformer load is increased. 72 Introduction to Electrical Machines Condition for maximum voltage regulation: P.u. voltage regulation = ε r cos θ2 + ε x sin θ2 . The condition for maximum voltage regulation is obtained by dedifferentiating the above expression with respect to θ2 and equating the results to zero. Here again the load current and secondary emf are assumed to remain constant. d ( p .u . regulation ) = −ε r sin θ2 + ε x cos θ2 = 0 dθ 2 ε x tan θ2 = x = e 2 εr re 2 r cos θ2 = e 2 ze2 ∴ Or Any Here tanθ2 is positive, therefore, maximum voltage regulation occurs at lagging load p.f. r equal to e 2 . In other words, maximum voltage regulation occurs when load powerze2 factor angle θ2 is equal to the leakage impedance angle φ of the transformer. ∴ the magnitude of maximum voltage regulation is: r x I r r I x x = ε r e2 + ε x e2 = 2 e2 ⋅ e2 + 2 e2 ⋅ e2 ze2 re 2 E2 ze2 E2 re 2 I z I2 = re22 + xe22 = 2 e 2 = z e 2 pu E2 ze2 E2 ) ( θ2 θ2 θ2 = φ I 2 re2 cos θ2 I 2 x e2 cos θ2 θ2 θ2 θ2 90° (a) (b) (c) Figure 2.17 Phasor diagram for 1-phase transformer for (a) negative voltage regulation (V.R);(b) zero V.R and (c) maximum V.R Thus the magnitude of maximum voltage regulation is equal to the p.u value equivalent leakage impedance of the transformer. For example, if a transformer has ze2 = 0.054. then magnitude of maximum possible voltage regulation is 5.4%. Phasor diagrams for a single-phase transformer for different operating power factors are illustrated in Figure 2. In Figure 2.17 (a), E2 < V2 voltage regulation (V.R.) is therefore 73 Chapter One: Electromagnetic Principles negative . In Figure 2.17 (b) E2 = V2. V.R is zero. Figure 2.17 (c) is drawn under the condition of maximum V.R, because here load power-factor angle θ2 = leakagex = cos −1 re 2 . impedance angle φ of the transformer where φ = tan −1 e 2 r z e2 e2 Example 2.4 A 6600/440 V, single-phase transformer has an equivalent resistance of 0.02 p.u. and an equivalent reactance of 0.05 p.u. Find the full-load voltage regulation at 0.8 pf lagging, if the primary voltage is 6600 V. Find also the secondary terminal voltage at full load. Solution = ε r cos θ2 + ε x sin θ2 = (0.02 )(0.8) + (0.05)(0.6 ) = 0.046 E 2 − V2 ∴ = 0.046 E2 For a primary voltage of 6600 V, the secondary no load voltage E2 is 440 V. ∴ The change in the secondary terminal voltage E2 − V2 = 440 (0.046) = 20.25 V and secondary terminal voltage V2 = 440 + 20.25 V P.u. voltage regulation Example 2.5 A short-circuit test, when performed on the H.V. side of a 10 kVA, 2000/400 V, single-phase transformer gave the following data: 60 V, 4 A, 100 W If the L.V. side is delivering full load (or rated) current at 0.8 p.f. lag and at 400 V, find the voltage applied to H.V. side. Solution From short circuit data P 100 reH = sc = = 6.25 Ω 2 I sc 42 V 60 Z eH = sc = = 15 Ω I sc 4 ∴ 2 2 x eH = Z eH − reH = 152 − 6.252 = 13.61 Ω For the L.V. side, the parameters are 2 2 1 1 reL = reH × = 6.25 = 0.25 Ω k 5 74 Introduction to Electrical Machines 2 2 1 1 x eL = x eH × = 13.61 = 0.544 Ω k 5 Full load secondary current I 2L = Now S 10,000 = = 25 A V2 400 E 2 − V2 = I 2 L reL cos θ2 + I 2 L x eL sin θ2 = (25 × 0.25 × 0.8) + (25 × 0.544 × 0.6 ) = 13.16 V For V2 = 400 V, E2 = 400 + 13.16 = 413.16 V ∴ The voltage applied to the primary is = 413.16 2000 = 2065.8 V . 400 2.9. TRANSFORMER LOSSES AND EFFICIENCY Equipment is desired to operate at a high efficiency. Fortunately, losses in transformers are small. Because the transformer is a static device, there are no rotational losses such as windage and friction losses in a rotating machine. In a well-designed transformer the efficiency can be as high as 99%. 2.9.1. Transformer Losses There are mainly two kinds of losses in a transformer, namely i) Core loss and ii) ohmic loss Core loss The core loss Pc occurring in the transformer iron, consists of components, hysteresis loss Ph and eddy current loss Pe i.e. two Pc = Ph + Pe The hysteresis and eddy current losses in the core can be expressed by :- and Ph = K h f Bxm 2 2 Pe = K e f B m Where Kh = proportionality constant which depends upon the volume and quality of the core material and units used. Ke = Proportionality constant whose value depends on the volume and resistivity of the core material, thickness of laminations and the units employed 75 Chapter One: Electromagnetic Principles Bm = maximum flux density in the core and f = frequency of the alternating flux The value of the exponent x (called Steinmetz’s constant) varies from 1.5 to 2.5 depending upon the magnetic properties of the core material. Therefore, the total core loss is Pc = KhfB1.6m + Kef2B2m Ohmic Loss When a transformer is loaded, ohmic loss (I2R) occurs in both the primary and secondary winding resistances. Since the standard operating temperature of electrical machines is 750C. The ohmic loss should be calculated at 750C. 2.9.2. Transformer Efficiency The efficiency of a transformer ( or any other device) is defined as the ratio of the output power to input power. Thus Efficiency η = η= Where Output power Input power V2 I 2 cos θ2 V2 I 2 cos θ2 + Pc + I 2 2 R 2.4 Pc = total core loss I22R = total ohmic losses V2I2 = output VA Cos θ2 = load power factor Since the efficiencies of power and distribution transformers are usually very high, it is therefore, more accurate to determine the efficiency from measurement of losses than from the measurement of output. Condition for Maximum Efficiency. In Eq. (*), Pc is constant and the load voltage V2 remains practically constant. A dη specified values of load p.f. cosθ2, the efficiency will be maximum when = 0. dI 2 dη Therefore, = 0 for Eq.(2.4) is dI 2 ( ) dη V2 I 2 cos θ2 + Pc + I 22 re2 (V2 cos θ2 ) − (V2 I 2 cos θ2 )(V2 cos θ2 + 2 I 2 re2 ) = =0 2 dI 2 2 V2 I 2 cos θ2 + Pc + I 2 re 2 Or ( ) (V2 I 2 cos θ2 + Pc + I 22 re2 ) (V2 cos θ2 ) = (V2 I 2 cos θ2 )(V2 cos θ2 + 2 I 2 re2 ) 76 Introduction to Electrical Machines Or Or I 22 re 2 = Pc 2.5 variable ohmic loss, I 22 re 2 = cons tan t core loss , Pc Hence the maximum efficiency occurs when the variable ohmic loss I 22 re 2 is equal to the fixed core loss Pc. From Eq.(2.5) the load current I2 at which maximum efficiency occurs is given by I2 = Pc Pc = I fl re 2 I 2fl re 2 2.6 If both sides of above equation are multiplied by E2 I 2 E2 I fl = 1000 1000 E2 1000 , we get Pc Full load ohmic losses ∴kVA load for maximum = (rated transformer kVA) × Or (kVA)max⋅η = (kVA) Core loss Ohmic losses at rated current Pc 2 I fl re2 2.7 Thus the maximum efficiency, for a constant load current, occurs at unity power factor (i.e. at purely resistive load). It is seen from Eq. (2.6) that the load current at which maximum efficiency occurs does not depend upon the load power factor because Pc and re2 are almost unaffected by a variation in the load power factor. A reduction in the load power factor reduces the transformer output and therefore the transformer efficiency is also reduced accordingly. Figure 2.18 illustrates the effect of p.f. on efficiency. Note that transformer efficiency is maximum at the same load current regardless of variation in the load power factor. ηmax Figure 2.18 Effect power factor on efficiency 77 Chapter One: Electromagnetic Principles 2.10. RATING OF TRANSFORMER The manufacturer of transformers fixes a name plate on the transformer, on which are recorded the rated output, the rated voltages, the rated frequency etc of a particular transformer. A typical name plate rating of a single phase transformer is as follows: 20 KVA, 3300/220V, 50Hz. Here 20 KVA is rated output at the secondary terminals. Note that the rated output is expressed in kilo-volt-ampere (KVA) rather than in kilowatt (KW). This is due to the fact that rated transformer output is limited by the heating and hence by losses in the transformer. The two types of losses in a transformer are core loss and ohmic ( I2r) loss. The core loss depends on transformer voltage and ohmic loss on transformer current. As these losses depend on transformer voltage (V) and current (I) and are almost unaffected by the load power factor, the transformer rated output is expressed in VA (V×I) or in kVA and are not in kW. For example, a transformer working on rated voltage and rated current with load pf equal to zero has rated losses and rated kVA output but delivers zero power to load. This shows that transformer must be expressed in kVA. For any transformer Rated input in kVA at Rated 0utput in kVA at the primary ter min als (cosθ1 ) = the sec ondary ter min als (cosθ2 ) + Losses Since a transformer operates at very high efficiency, losses may be ignored. Furthermore the primary power factor Cosθ1 and secondary power factor Cosθ2 are nearly equal. Therefore the rated KVA marked on the name plate of a transformer refers to both windings i.e. rated KVA of primary winding and secondary winding are equal. The voltage 3300/220 V refers to the design voltages of the two windings. Either one may serve as primary or secondary. If it is step-down transformer, then 3300 V is rated primary voltage and refers to the voltage applied to the primary winding. The 220V is rated secondary voltage and refers to the voltage developed between output terminals at no-load with rated voltage applied to the primary terminals. Rated primary and secondary currents are calculated from the rated KVA and the corresponding rated voltages. Thus Rated (or full load) primary current = 20,000 = 6.06A 3300 Rated (or full load) secondary current = 20,000 = 90.91A 220 Note that the rated primary and secondary currents refer to the currents for which the windings are designed. Rated frequency refers to the frequency for which the transformer is designed to operate. Example 2.6 A 100 kVA, 1000/10000 V, 50 Hz, single phase transformer has an iron loss of 1100 W. The copper loss with 5 A in the high voltage winding is 400 W. Calculate the efficiencies at (i) 25 %, (ii) 50 % and (iii) 100 % of normal load for power 78 i) ii) iii) Introduction to Electrical Machines factors of (a) 1.0 and (b) 0.8. The output terminal voltage being maintained at 10000 V. Find also the load for maximum efficiency at both power factors. Solution Efficiency at 25% of normal load, unity pf: Iron losses = 1100 W Copper losses with 5 A in secondary = 400 W Secondary full load current, I2 100 × 1000 10000 = 10 A I2 = Current in the secondary at 25 % full load = ¼ × 10 = 2.5 A 2 2.5 Copper losses at 25% full load = × 400 5 = 100 W Output at 25% full load = 0.25 × 100 × 1000 × 1 = 25 000 W Efficiency at 25 % load, unity pf 25000 × 100 25000 + 1100 + 100 = 95.4% η at 25% = Efficiency at 25 % full load, 0.8 pf: 25000 × 0.8 × 100 25000 × 0.8 + 1100 + 100 = 94.34% η at 25% = Efficiency at 50 % full load, unity pf: Copper losses at 25% full load = 400 W Output at 50 % full load, unity pf = 0.5 × 100 × 1000 × 1 = 50 000 W Iron losses = 1100 W 79 Chapter One: Electromagnetic Principles 50000 × 1.0 × 100 50000 × 1.0 + 1100 + 100 = 97.65% η at 50% = Efficiency at 50 % full load, 0.8 pf: 50000 × 0.8 × 100 50000 × 0.8 + 1100 + 400 = 97.1% η at 50% = Efficiency at 100 % full load, unity pf: 2 10 Copper losses at 100% full load = × 400 5 = 1600 W Output = 100 × 1000 × 1 = 100 000 W Iron losses = 1100 W 100000 × 1.0 × 100 100000 × 1.0 + 1100 + 1600 = 97.37% η at 100% = Efficiency at full load, 0.8 pf: 100000 × 0.8 × 100 100000 × 0.8 + 1100 + 1600 = 96.73% η at 100% = Load for maximum efficiency at unity pf: Maximum efficiency occurs when the iron losses equal the copper losses. Let the maximum efficiency occur at x per cent of the full load. 2 Copper losses at x % of full load = x ×1600 Thus x2 ×1600 = 1100 x = 0.829 Load for maximum efficiency = 0.829 × 100 = 82.9 kVA Load for maximum efficiency at 0.8 pf: Load for maximum efficiency will remain the same irrespective of power factor Thus load for maximum efficiency = 82.9 kVA 80 Introduction to Electrical Machines Example 2.7 A single phase transformer working at unity power factor has an efficiency of 90% at both one half load and at the full load of 500 W. Determine the efficiency at 75 % of full load. Solution Efficiency of the transformer at full-load = 0.9 Output at full load = 500 W Let the iron losses of the transformer be = x watts and the total copper losses at full load be = y watts Then, the total losses at full load = x + y Hence, 500 500 + x + y = 10 A 0 .9 = Or 0.9x + 0.9y = 50 (i) Efficiency of the transformer at half of full load = 0.9 2 y 1 Total copper losses at half of full load = × y = 4 2 Output of the transformer = ½ × 500 = 250 W Thus, 0.9 = 250 250 + x + y 4 0.9x + 0.225y = 25 (ii) Solving Eqs. (i) and (ii) y = 37 W and x = 18.53 W i.e. total copper losses at full load = 37 W Iron losses = 18.53 W Output of the transformer at 75 per cent of full load = 0.75 x 500 = 375 W 81 Chapter One: Electromagnetic Principles Total copper losses at 75 per cent of full load = (0.75)2 x 37 = 20.8 W Efficiency at 75 percent of full load 375 × 100 375 + 18.53 + 20.8 = 90.5% η at 75% = 2.11. PARALLEL OPERATION OF SINGLE-PHASE TRANSFORMERS When electric power is supplied to a locality a single transformer, capable of handling the required power demand, is installed. In some cases, it may be preferable to install two or more transformers in parallel, instead of one large unit. Though two or more transformers may be expensive than one large unit, yet this scheme posses certain advantages described below. 1. With two or more transformers, power system becomes more reliable. For instance if one transformer develops a fault, it can be removed and the other transformers can maintain the flow of power , though at reduced load. 2. Transformers can be switched on or off , depending upon the power demand. In this manner, the transformer losses decrease and the system becomes more economical and efficient in operation. 3. The cost of standby (or spare) unit is much less when two or more transformers are installed. In any case, in the long run , electric power demand may become more than rated KVA capacity of already existing transformer or transformers. Under such circumstances, the need for extra transformer arises; the extra unit must be connected in parallel. Note that the parallel operation of transformers requires that their primary windings , as well as secondary windings are connected in parallel . In this section only the parallel operation of single-phase transformers is considered. The various conditions which must be fulfilled for the satisfactory parallel operation of two or more single-phase transformers are as follows: a) The transformer must have the same voltage ratios, i.e with the primaries connected to the same voltage sources, the secondary voltage of all transformers should be equal in magnitude. b) The equivalent leakage impedance in ohms must be inversely proportional to their respective KVA ratings. In other words, per unit (pu) leakage impedance of transformers based on their KVA rating must be equal. 82 Introduction to Electrical Machines c) The ratio of equivalent leakage reactance to equivalent resistance i.e. Xe/re should be the same for all transformers. d) The transformer must be connected properly as far as their polarities are concerned. Out of the conditions listed above, condition(d) must be strictly fulfilled. If the secondary terminals are connected with wrong polarities, large circulating currents will flow and the transformers may get damaged. Condition (a) should be satisfied as accurately as possible ; since different secondary voltages would give rise to undesired circulating currents. For conductions (b) and (c), some deviation is permissible. Thus the fulfillment of condition (d) is essential whereas the fulfillment of the other conditions is desirable. A1 A2 A1 A2 A a1 + B a2 a1 + a2 V Figure 2.19 Two single-phase transformers in parallel Figure 2.19 shows single-phase transformers in parallel, connected to some voltage source on the primary side. Zero voltmeter reading indicates proper polarities. If the voltmeter reads the sum of two secondary voltages, the polarities are improper and can be corrected by reversing the secondary terminals of any one of the transformers. 2.12. THREE-PHASE TRANSFORMERS Generation, transmission and distribution of electric energy is invariably done through the use of three-phase systems because of its several advantages over single-phase systems. As such, a large number of three-phase transformers are inducted in a 3-phase energy system for stepping-up or stepping – down the voltage as required. For 3-phase up or down transformation, three units of 1-phase transformers or one unit of 3-phase transformer may be used. When three identical units of 1-phase transformers are used as shown in Figure 2.20(a), the arrangement is usually called a bank of three transformers or a 3-phase transformer bank. A single 3-phase transformer unit may employ 3–phase core-type construction Figure 2.20(b) or three phase shell type construction. 83 Chapter One: Electromagnetic Principles (a) (b) Figure 2.20 (a)Three-phase transformer bank, both windings in star;(b) three-phase core-type transformer A single-unit 3-phase core-type transformer uses a three-limbed core, one limb for each phase winding as shown in Figure 2.20(b). Actually, each limb has the L.V. winding placed adjacent to the laminated steel core and then H.V. winding is placed over the 1.v. winding. Appropriate insulation is placed in between the core and 1.v. winding and also in between the two windings. A 3-phase core-type transformer costs about 15% less than a bank of three 1-phase transformers. Also, a single unit occupies less floor space than a bank. 2.12.1. Three-Phase Transformer Connections Three-phase transformers may have the following four standard connections (a) Star-Delta ( Y-∆) (b) Delta-Star (∆-Y) (c) Delta-Delta (∆-∆) (d) Star-Star (Y-Y) These connections are shown in Figures 2.21 and 2.22, where V and I are taken as input line voltage and line current respectively. Primary and secondary windings of one phase are drawn parallel to each other. With phase turns ratio from primary to secondary as N1/N2= a, the voltages and current in the windings and lines are shown in Figures 2.21 and 2.22. The various connections are now described briefly. (a) Star-delta (Y-∆) Connection This connection is commonly used for stepping down the voltage from a high level to a medium or low level. The insulation on the h.v. 1 side of the transformer is stressed only to 57.74% = x 100 of line to line voltage 3 For per-phase m.m.f. balance, I2N2 =I1N1 Here primary phase current, I1 = primary line current I 84 Introduction to Electrical Machines ∴Secondary phase current , I 2 = N1 I = aI N2 1 = 3 I 2 = 3 . aI Secondary line current Also, voltage per turn on primary = voltage per turn on secondary 1 V2 = 3 N1 N 2 V . Secondary phase voltage, V2 = N2 V V . = N 1 3 a. 3 Secondary line voltage = secondary phase voltage = Input VA = 3 V 3 . I = output VA = 3. V a. 3 V a. 3 . aI = 3 VI Phase and line values for voltages and currents on both primary and secondary sides of star-delta transformer are shown in Figure 2.21(a) I V aI 3 3aI V 3 I V 3a I aI aI 3 I V V a 3V a 3 (a) (b) Figure 2.21 (a) Star-delta connection and (b) delta-star connection of 3-phase transformers (b) Delta-Star (∆-Y) connection:- This type of connection is used for stepping up the voltage to a high level. For example, these are used in the beginning of h.v. transmission lines so that insulation is stressed to about 57.74% of line voltage Delta-star transformers are also generally used as distribution transformers for providing mixed line to line voltage to high-power equipment and line to neutral voltage to 1-phase low-power equipment. For example, 11kV/400V, delta-star distribution transformer is used to distribute power to consumers by 3-phase four-wire system. Three-phase high– power equipment is connected to 400V, three line wires, whereas 1-phase low-power equipment is energized from 231 V line to neutral circuits. 85 Chapter One: Electromagnetic Principles For per − phase m.m.f . balance , I 2 N 2 = I1 N1 1 Here primary phase current , I1 = ( primary line current I ) 3 N 1 Secondary phase current, I 2 = 1 I1 = a N2 3 V2 V1 Also = N 2 N1 N V Secondary phase voltage, V2 = 2 .V1 = a N 1 V ∴ Secondary line voltage = 3. a I V I Input VA = 3. V . = 3 VI = Output VA = 3. . a = 3 VI a 3 3 Phase and line values for voltages and currents on primary as well as secondary sides of a 3-phase delta-star transformer are shown in Figure 2.21(b). ∆) Connection Delta-Delta (∆-∆ (c) This scheme of connections is used for large 1.v transformers. It is because a deltaconnected winding handles line voltage, so it requires more turns per phase but of smaller cross-sectional area. The absence of star point may be a disadvantage in some applications. In case a bank of three transformers is used, then one transformer can be removed for maintenance purposes while the remaining two transformers (called an open-delta or Vconnection) can still deliver 58% of the power delivered by the original 3-phase transformer bank. For per phase mmf balance, I2N2 = I1N1: primary phase current , I1 = 1 ( primary line current I ) 3 N aI Secondary phase current , I 2 = 1 I1 = N2 3 Secondary line current , aI = aI = 3 3 V2 V = 1 N2 N1 N V Secondary phase voltage , V2 = 2 V1 = ( Here V1 = V) N1 a V Secondary line voltage = V2 = a I V aI Input VA = 3V . = output VA = 3. , = 3 VI a 3 3 Also = 86 Introduction to Electrical Machines Phase and line value for voltages and currents on both primary and secondary sides of a 3-phase delta-delta transformer are shown in Figure 2.22 (a). kI I V I 3 kI V k kI 3 V V 3 I kI V 3k V k (b) (a) Figure 2.22 (a) Delta-delta connection and (b) Star-star connection of three-phase transformers. (d) Star-Star (Y-Y) Connection This connection is used for small h.v transformers . As stated before, with star connection, turns per phase are minimum and the winding insulation is stressed to 57.74% of line voltage. Star-star connection is rarely used in practice because of oscillatory neutral problems. For per − phase m.m.f .balance, Pr imary Phase Current , I 2 N 2 = I1 N1 I1 = primary line current , I N Secondary phase current , I 2 = 1 I1 = aI N2 = Secondary line current N V sec ondary phase voltage , V2 = 2 V1 = N1 3a Secondary line voltage = 3 V2 = 3 . Input VA = 3 . V 3. a = V a V V I = output VA = 3. . aI = 3 VI 3 3. a Phase and line values of voltages and currents on both sides of a star-star transformer are shown in Figure 2.22(b) Example 2.8 A 3-phase transformer is used to step-down the voltage of a 3-phase, 11kV feeder line. Per-phase turns ratio is 12. For a primary line current of 20A, calculate the secondary line voltage, line current and output KVA for the following Connections: (a) star-delta (b) delta-star (c) delta-delta (d) star-star. Neglect losses. Solution (a) Three-phase transformer with star-delta connection is shown in Figure 2.23(a) 87 Chapter One: Electromagnetic Principles V 11000 VP1 = L1 = V 3 3 phase current on primary, I P1 = I L1 = 20A VP1 V Here, = P 2 and I p1 × 12 = I p2 × 1 12 1 11000 ∴ Phase voltage on sec ondary, Vp 2 = = 529.25V 3 x 12 Line voltage on sec ondary , VL 2 = Vp 2 = 529.25V phase voltage on primary, Phase current on sec ondary I p 2 = 12 I p1 = 12 × 20 = 240A Line current on sec ondary I L 2 = 3 I p 2 = 3 × 240 = 415.68A 3Vp2 .I p 2 11000 1 Output KVA = = 3. × 240 x = 381.04 KVA 1000 1000 3 x 12 (b) Delta-star connection of 3-phase transformer is shown in Figure 2.23 (b) Vp1 VL1 11000 = V = 916.67 V 12 12 12 11000 Line voltage on sec ondary VL 2 = 3 Vp2 = 3 × = 1587.67V 12 20 Phase current on primary I p1 = I L1 / 3 = A 3 20 Phase current on sec ondary I p1 = 12 I p1 = 12 × = 138.568A 3 Line current on sec ondary, I L2 = I p2 = 138.568A 11000 12 x 20 Output KVA = 3 × × = 381.04 KVA. 12 1000 x 3 Phase voltage on sec ondary VP 2 = = (c) Delta-delta connection of 3-phase transformer is shown in Figure 2.23(c) Vp1 VL1 11000 = V = 916.7 V 12 12 12 Line voltage on sec ondary, VL2 = VP2 = 916.7 V 20 Phase current on primary, I P1 = A 3 20 Phase current on sec ondary, I P2 = 12 I P1 = 12 × A 3 12 × 20 Line current on sec ondary, I L2 = 3 I P 2 = 3 . = 240A 3 11000 12x 20 1 Output KVA = 3 × × × = 381.04 KVA. 12 1000 3 Phase voltage on sec ondary, VP 2 = = 88 Introduction to Electrical Machines I L1 = 20A VL1 = 11000 V I L2 VP1 VP 2 = VL2 I P1 I L1 = 20 A I P2 VP 2 I P2 I P1 VL1 = 11000 V VL2 (a) (b) I L 2 = 3I P2 I L1 = 20A I L2 VP2 = VL2 VP1 = 11000 V I P1 I P2 VL1 = 11000 V VP1 I P1 I P2 VP 2 VL 2 = 3VP 2 (d) (c) Figure 2.23 for Example 2.8 (d) 3-phase transformer with star-star connection is shown in Figure 2.23(d) V 11000 Phase voltage on sec ondary , VP 2 = P1 = V 12 3 x 12 11000 11000 Line voltage on sec ondary, VL2 = 3 VP 2 = 3 = V 12 3 x 12 Phase current on Pr imary, I P1 = I L1 = 20A Phase current on sec ondary, , I P2 = 12 I P1 = 12 × 20 = 240A Line current on Secondary, I L2 = I P 2 = 240A 3 × 11000 240 Output KVA = × = 381.04 KVA 3 × 12 1000 Example 2.9 An 11000/415V, delta-star transformer feeds power to a 30 kW, 415V, 3phase induction motor having an efficiency of 90% and full-load pf 0.833. Calculate the transformer rating and phase and line currents on both high and low voltage sides. 30 = 40KVA 0.9 x 0.833 Total load in VA 40,000 Solution Line current on l.v. side of transformer = = = 55.65A 3 x line voltage 3 x 415 . Transformer kVA rating = For star connected 1.v. winding, phase current in 1.v. winding = line current on 1.v side = 55.65A. Line current on HV, side of transformer = 40,000 3 x 11000 = 2.1A 89 Chapter One: Electromagnetic Principles For delta connected HV winding, phase current in HV winding: = 1 3 (line current on h.v. side) = 1 3 x 2.1 = 1.212A Phase Shift Some of the three-phase transformer connections will result in a phase shift between the primary and secondary line-voltages. Consider the phase voltages, shown in Figure 2.24, for the Y-∆ connections. The phases VAN and Va are aligned, but line voltage VAB of the primary leads the line voltage Vab of the secondary by 300. It can be shown the ∆-Y connection also provides a 300 phase shift in there line –to-line voltage. This property of phase shift in Y-∆ or ∆-Y connections can be used advantageously in some applications. Figure 2.24 phase shift in line-to-line voltages in a three-phase transformer V-Connection It was stated earlier that in the ∆-∆ connection of three single-phase transformers, one transformer can be removed and the system can still deliver three-phase power to a threephase load. This configuration is known as an open-delta or V connection. It may be employed in an emergency situation when one transformer must be removed for repair and continuity of service is required. (a) 90 Introduction to Electrical Machines 30° φ φ 30° (b) Figure 2.25 V-connection Consider Figure 2.25(a) in which one transformer, shown dotted is removed. For simplicity the load is considered to by Y- connected. Figure 2.25(b) shows the phase diagram for voltages and currents. Here VAB, VBC and VCA represent the line-to line voltage of the primary Vab, Vbc and Vcb secondary and Van, Vbn and Vcn represent the phase voltages of the load. For an inductive load the load currents Ia, Ib and Ic will lag the corresponding voltages Van, Vbn and Vcn by the load phase angle θ. Transformer winding ab and bc deliver power Pab = Vab I a cos (30 + φ) Pbc = Vcb I c cos (30 − φ) Let Vah = Vcb = V, Voltage rating of the transformer secondary winding. Ia = Ic = I current rating of the transformer secondary winding and φ = 0 for a resistive load. Power delivered to the load by the V connection is Pv = Pab + Pbc = 2VI cos 30 2.8 With all three transformers connected in delta, the power delivered is P∆ = 3VI 2.9 From Eqs. (2.8) and (2.9) Pv 2 cos 300 = = 0.58 P∆ 3 The V connection is capable of delivering 58% power without overloading the transformer (i.e., not exceeding the current rating of the transformer winding). 2.12.2. Three-Phase Transformer of A Common Magnetic Core (Three Phase Unit Transformer) A three phase transformer can be constructed by having three primary and three secondary windings on a common magnetic core. Consider three single-phase core-type units as shown in Figure 2.26(a). For simplicity, only the primary winding has been shown. If balanced three-phase sinusoidal voltages are applied to the windings, the fluxes φa, φb, and φc will also be sinusoidal and balanced. If the three legs carrying these 91 Chapter One: Electromagnetic Principles fluxes are merged, the net flux in the merged leg is zero. This leg can therefore be removed as shown in Figure 2.26(b). This structure is not convenient to build. φa φa φc φb φb φc φa + φ b + φc = 0 (a) (b) (c) (d) Figure 2.26 development of a three-phase core-type transformer. However, If section II is pushed in between sections I and III by removing its yokes, a common magnetic structure shown in Figure 2.26(c), is obtained. This core structure can be built using stacked laminations as shown in Figure 2.26(d). Both primary and secondary windings of a phase are placed on the same leg. Note that the magnetic paths of legs a and c are somewhere longer than that of leg b (Figure 2.26 (c). This will result in some imbalance in the magnetizing currents. However, this imbalance is not significant. Figurer 2.27 shows a picture of a three-phase transformer of this type. Such a transformer weight less , costs less, and requires less space than a three-phase transformer bank of the same rating. The disadvantage is that if one phase breaks down, the whole transformer must be removed for repair. 92 Introduction to Electrical Machines Figure 2.27 Cut-away view of three-phase core-type transformer Transformers are usually air-cooled even if placed in metal cases. Larger sizes are placed in tanks with special transformer oil. The oil has a dual function ; it insulates while providing cooling. Still larger sizes have tanks with corrugated sides or cooling fins or radiators to dissipate the heat to the surrounding air. Figure 2.28 shows a typical selfcooled transformer. The oil moves around by natural convection, since warmer oil flows up. It flows down again through the radiator, which gives up this heat to the surrounding air. In larger oil-cooled units, the oil must be pumped around to maintain acceptable temperature levels. No matter what size of transformer is dealt with, they all operate on the same principle. Figure 2.28 large oil-cooled three-phase power transformer (sectional view). 93 Chapter One: Electromagnetic Principles 2.13. AUTOTRANSFORMERS In principle and in general construction, the autotransformer does not differ from the conventional two-winding transformer so far discussed. It does differ from it. however, in the way the primary and secondary windings are interrelated. It will be recalled that in discussing the transformer principles of operation, it was pointed out that a counter emf was induced in the winding, which acted as a primary to establish the excitation ampere turns. The induced voltage per turn was the same in each and every turn linking with the common flux of the transformer. Therefore, fundamentally it makes no difference in the operation whether the secondary induced voltage is obtained from a separate winding linked with the core or from a portion of the primary turns. The same voltage transformation results in the two situations. When the primary and secondary voltage are derived from the same winding. the transformer is called an autotransformer. Load An ordinary two-winding transformer may also be used as an autotransformer by connecting the two windings in series and applying the impressed voltage across the two, or merely to one of the windings. It depends on whether it is desired to step the voltage down or up, respectively. This is shown in Figure 2.29(a) for the step-down connection; the step-up connection is illustrated in Figure 2.29(b). (a) (b) Figure 2.29 Autotransformers: (a) step-down; (b) step-up. In Figure 2.29(a) the input voltage V1 is connected to the complete winding (a-c) and the load RL is connected across a portion of the winding, that is, (b-c).The voltage V2 is related to V1 as in the conventional two-winding transformer, that is, V2 = V1 × N bc N ac 2.10 where Nbc and Nac are the number of turns on the respective windings. The ratio of voltage transformation in an autotransformer is the same as that for an ordinary transformer, namely, k= N ac V1 I 2 = = N bc V2 I1 2.11 with k > 1 for step-down. 94 Introduction to Electrical Machines Assuming a resistive load for convenience, then, V I2 = 2 RL Assume that the transformer is 100% efficient.The power output is P = V2 I 2 2.12 Note that I1 flows in the portion of winding ab, whereas the current (I2 – I1) flows in the remaining portion bc. The resulting current flowing in the winding bc is always the arithmetic difference between I1 and I2 , since they are always in opposite sense. Remember that the induced voltage in the primary opposes the primary voltage. As a result. the current caused by the induced voltage flows opposite to the input current. In an autotransformer, the secondary current is this induced current, that is, I1 + (I 2 − I1 ) = I 2 2.13 Hence the ampere-turns due to section bc, where the substitutions I 2 = kI1 and N N bc = ac are made according to Eq. (2.11), is k ampere-turns due to section bc = (I 2 − I1 )N bc (kI1 − I1 )N bc 1 = I1N ac 1 − = I1N ab k k = ampere − turns due to sec tion ab = Thus the ampere-turns due to sections bc and ab balance each other, a characterstic of all transformer actions. Equation (2.12) gives the power determined by the load. To see how this power is delivered, we can write the equation in a slightly modified forn. By substituting Eq. (2.13) into Eq. (2.12), we obtain P = V2 I 2 = V2 [I1 + (I 2 − I1 )] = V2 I1 + V2 (I 2 − I1 ) W 2.14 This indicates that the load power consists of two parts. The first part is Pc = V2 I1 ≡ conducted power to load through ab 2.15 The second part is Ptr = V2 (I 2 − I1 ) ≡ transformed power to load through bc 2.16 95 Chapter One: Electromagnetic Principles We will see in the following examples that most of the power to the load is directly conducted by winding ab. The remaining power is transferred by the common winding bc. To show these powers Pc and Ptr in terms of the total power P. ue proceed as follows: Pc V2 I1 I1 1 = = = P V2 I 2 I 2 k and Ptr V2 (I 2 − I1 ) (I 2 − I1 ) k − 1 = = = P V2 I 2 I2 k Thus Pc = P P (k − 1) and Ptr = with a > 1 for a step-down autotransformer. k k Example 2.10 A standard 5-kVA 2300/230-V distribution transformer is connected as an autotransformer to step down the voltage from 2530 V to 2300 V. The transformer connection is as shown in Figure 2.29 (a). The 230-V winding is section ab, the 2300-V winding is bc. Compare the kVA rating of the autotransformer with that of the original two-winding transformer. Also calculate Pc, Ptr, and the currents. Solution The rated current in the 230-V winding (or in ab) is 5000 VA = 21.74 A 230 I1 = The rated current in the 2300-V winding (or in bc) is I 2 − I1 = 5000 = 2.174 A 2300 Therefore, I 2 = 2.174 + I1 = 23.914 A The secondary current I2 can also be calculated from I 2 = kI1 = Since the transformation ratio k = 2530 × 21.74 = 23.914 A 2300 2530 = 1 .1 2300 P = V1I1 = V2 I 2 = 2530 × 21.74 = 55.00 kVA The conducted power is Pc = P 55,000 = = 50 kVA k 1 .1 and that transformed is Ptr = P (k − 1) = 55,000 1.1 − 1 = 5.0 kVA k 1 .1 96 Introduction to Electrical Machines Consider now the step-up transformer of Figure 2.29(b). Following reasons similar to those above, it follows that P = V1I1 = V1[I 2 + (I1 − I 2 )] = V1I 2 + V1 (I1 − I 2 ) W 2.17 where we made the substitution of I1 from Eq. (2.13), which really is kirchhoff's current law applied to point b. To show this, note at point b we have I1 + (I 2 − I1 ) = I 2 so that I1 = I 2 − (I 2 − I1 ) = I 2 + (I1 − I 2 ) Again, Eq. (2.17) shows us that the power supplied to the load consists of two parts, Pc = V1I 2 ≡ conducted power to load through ab 2.18 and Ptr = V1 (I1 − I 2 ) ≡ transformed power to load through bc 2.19 In terms of total power, we have Pc V1I 2 I 2 = = =k P V1I1 I1 2.20 Ptr V1 (I1 − I 2 ) (I1 − I 2 ) = = = 1− k P V1I1 I1 2.21 and Thus for the step-up transformer with a < 1, we obtain Pc = kP and Ptr = P (1 − k ) As before, Pc is the power directly conducted to the load and Ptr is the portion that is transformed. Example 2.11 Repeat the problem of Example 2.10 for a 2300 V-to-2530 V step-up connection as shown in Figure 2.29 (b). Solution As calculated in Example 2.10, the current rating of the winding ab is I2 = 21.74 A, which also is the load current. The output voltage is 2530 V; thus the volt-ampere rating of the autotransformer is P = V2 I 2 = 2530 × 21.74 = 55 kVA which is the same as in Example above. The transformer ratio 97 Chapter One: Electromagnetic Principles k= 2300 = 0.909 2530 The conducted power is therefore Pc = aP = 0.909 × 55 kVA = 50 kVA and the transformed power Ptr = P (1 − a ) = 55 kVA(1 − 0.909 ) = 5 kVA The examples given make it clear that an autotranaformer of given physical dimensions can handle much more load power than an equivalent two-winding transformer; in k times its rating as two-winding transformer for the step-down autotransformer fact, 1− k 1 or for the step-up arrangement. A 5-kVA transformer is capable of taking care of k −1 11 times its rating. These great gains are possible since an autotransformer transforms, by transformer action, only a fraction of the total power; the power that is not transformed is conducted directly to the load. It should be noted that an autotransformer is not suitable for large percentage voltage reductions as is a distribution transformer. This is due to the required turns ratio becoming too large; hence the power-handling advantage would be minimal. Furthermore, in the unlikely but possible event that the connections to the lowvoltage secondary were to fail somewhere below point b in Figure 2.29(a), the winding bc would be deleted from the circuit. This implies that the load would see the full high line voltage. Autotransformers are not used for these reasons where large voltage changes are encountered. In situations where autotransformers can be used to their full advantage, it will be found that they are cheaper than a cowentional two-winding transformer of similar rating. They also have better regulation (i.e., the voltage does not drop so much for the same load), and they operate at higher efficiency. In all applications using autotransformer it should be realized that the primary and secondary circuits are not electrically isolated, since one input terminal is common with one output terminal. 98 Introduction to Electrical Machines PROBLEMS 2.1. A 6600/400 V, 50 Hz single-phase core type transformer has a net cross-sectional area of the core of 428 cm2. The maximum flux density in the core is 1.5 T. Calculate the number of turns in the primary and secondary windings. Ans. 462, 28 2.2. A single phase, 50 Hz, 220/3000 V, transformer has a net cross-sectional area of the core 400 cm2. If the peak value of flux density in the core is 1.239 T, calculate the suitable values for the number of turns in primary and secondary windings 2.3. Single-phase 50 Hz transformer has 80 turns on the primary winding and 280 in the secondary winding. The voltage applied across the primary winding is 240 V at 50 Hz. Calculate (i) the maximum flux density in the core and (ii) induced emf in the secondary. The net cross-sectional area of the core can be taken 200 cm2. Ans.(i) 0.675 Wb/m2; (ii) 840 V 2.4. The values of the resistance of the primary and secondary windings of a 2200/200V, 50 Hz single phase transformer are 2.4 and 0.02 Ω respectively. Find (i) equivalent resistance of primary referred to secondary, (ii) equivalent resistance of secondary referred to primary, (iii) total resistance of transformer referred to secondary and (iv) total resistance of transformer referred to primary. 2.5. Short circuit test performed on 20 kVA, 2000/200 V, 50 Hz, single phase transformer gave the following readings: With 100 V applied to the primary, full load current circulated in the short circuited secondary with power drawn of 300 W. Calculate the secondary terminal voltage on fullload (i) at unity power factor, (ii) at pf of 0.75 lagging and (iii) at pf of 0.8 leading. Find also the percentage regulation in each case. Ans.:(i) 197 V, 1.5 % (ii) 191.4 V, 4.28% (iii) 203.3 V, -1.66% 2.6. The iron and full load copper losses in a 40 kVA single phase transformer are 450 and 850 W respectively. Find (i) the efficiency at full load when the power factor of the load is 0.8 lagging, (ii) the maximum efficiency and (iii) the load at which the maximum efficiency occurs. 2.7. Open-circuit and short-circuit tests were conducted on a 50 kVA, 6360/2Q0 V, 50 Hz, single-phase transformer in order to find its efficiency. The observations during these tests are: Open circuit test: Voltage across primary winding 6360 V. Primary current, 1.0A, power input 2 kW. Short circuit test: Voltage across primary 180 V, current in secondary winding 175 A, power input 2 kW. Calculate the efficiency of the transformer when supplying full load at power factor of 0.8 lagging. Ans.:89.2 % 99 Chapter One: Electromagnetic Principles 2.8. Calculate the efficiency at full load, half load and one-fourth load at (i) unity pf and (ii) 0.71 pf lagging, for a 80 kVA, 1100/250 V, 50 Hz, single phase transformer, whose losses are as follows: Iron losses= 800 W Total copper losses with 160 A in the low voltage winding is 200 W. Ans.:(i) 98.04 Y., 97.57 %, 95.92 % (ii) 97.25 %, 96.61 %;, 94.36 %. 2.9. The parameters of the equivalent circuit of a 10 kVA, 2000/ 400 V, 50 Hz, single phase transformer are as follows: Primary winding: r1=5.5 Ω; x1=12 Ω Secondary winding: r2 =0.2 Ω; x2 = 0.45Ω If the primary supply voltage is 2000 V, calculate the approximate value of the secondary voltage at full load 0.8 power factor lagging. Ans.:377.6 V 2.10. A 10-kVA 2200/460-V transformer is connected as an autotransformer to step up the voltage from 2200 V to 2660 V. When used to transform 10 kVA, determine the kVA load output. Ans.:57.8 kVA 2.11. Three transformers connected ∆-Y step down the voltage from 12600 to 600 V and deliver a 55-kVA load at a power factor of 0.866 lagging. Calculate : (a) The transformation ratio of each transformer. (b) The kVA and kW load in each transformer. (c) The load currents. (d) The currents in the transformer windings. Ans.: (a) 33k ; (b) 18.3kVA , 15.88kW; (c) 48.1 A ; (d) I2=48.1A, I1=1.46A 2.12. A 50-hp 440-V three-phase motor with an efficiency of 0.88 and a power factor of 0.82 on full load is supplied from a 6600/440-V ∆-Y connected transformer. Calculate the currents in the high- and low-voltage transformer windings when the motor is running at full load. 2.13. A 13,200-V three-phase generator delivers 10000 kVA to a three-phase 66,000-V transmission line through a step-up transformer. Determine the kVA, voltage, and current ratings of each of the single-phase transformers needed if they are connected: (a) ∆-∆ (b)Y -∆ (c) Y –Y (d) ∆-Y Ans.: (a) 3333kVA, V1=13.2kV, V2=66kV, I1=253A, I2=50.5A; (c) 3333kVA, V1=7.62kV, V2=38.1kV, I1=437.5A, I2=87.5A 2.14. A Y-∆ transformer bank supplies a balanced load of 500 kW, 1100 V, 0.85 pf lagging. Determine the primary and secondary voltages and currents . the primary line voltage is 11000 V. 100 Introduction to Electrical Machines CHAPTER THREE INDUCTION MACHINES 3.1. ELECTROMECHANICAL CONVERSION Three electrical Machines (dc, induction & synchronous) are used extensively for electromechanical energy conversion. In these machines, conversion of energy results from the following two electromagnetic phenomena. 1. When a conductor moves in a magnetic field voltage is induced in the conductor: (generator action) 2. When a current –carrying conductor is placed in a magnetic field, the conductor experiences a mechanical force (Motor action) Figure 3.1 Electromechanical energy conversion Note that the two systems in Figure 3.1, electrical and mechanical, are different in nature. In electrical system the primary quantities involved are voltage & current while in mechanical system, the analogous quantities are torque & speed. The coupling medium between these different systems is the magnetic field. The basic electrical machines (induction, dc, and synchronous), which depend on electromagnetic energy conversion, are extensively used in various power ratings. The operation, construction and characteristic features of these machines are discussed in detail in this and other chapters. 3.2. INTRODUCTION The induction machine is the most rugged and the most widely used machine in industry. The induction machine has a stator and a rotor mounted on bearings and separated from the stator by an air gap. However, in the induction machine both stator winding and rotor winding carry alternating current. The alternating current (ac) is supplied to the stator winding machine. The induction machine can operate both as a motor and as a generator. However, it is seldom used as a generator supplying electrical power to a load. The performance characteristics as a generator are not satisfactory for most applications. The induction machine is extensively used as a motor in many applications. 101 Chapter One: Electromagnetic Principles Of all the a.c motors the poly-phase induction motor is the one which is extensively used for various kinds of industrial drives. It has the following main advantages and also some disadvantages. Advantages: 1. 2. 3. It has very simple and extremely rugged, almost unbreakable construction (especially squirrel cage type) Its cost is low and it is very reliable It has sufficiently high efficiency. In normal running condition, no brushes are needed, hence frictional losses are reduced. 4. It has a reasonably good power factor 5. it requires minimum of maintenance 6. It starts up from rest and needs no extra starting motor and has not to be synchronized. Its starting arrangement is simple especially – for squirrel- cage type motor. Disadvantage 1. Its speed cannot be varied without sacrificing some of its efficiency. 2. Just like a d.c. shunt motor, its speed decreases with increase in load 3. Its starting torque is somewhat inferior to that of a d.c shunt motor The induction motor is used in various sizes: • Large three-phase induction motors (in tens or hundreds of horsepower) are used in pumps, fans, compressors, paper mills, textile mills and so forth. • Small single-phase induction motors (in fractional horsepower rating) are used in many household appliances, such as blenders, lawn mowers, juice mixers, washing machines, refrigerators, and stereo turntables. • The linear version of the induction machine has been developed primarily for use in transportation systems. The induction machine is undoubtedly a very useful electrical machine. Two-phase induction motors are used primarily as servomotors in a control system. Three-phase induction motors are the most important ones and are most widely used in industry. In this unit the constructional features, operation, stator windings, characteristic futures, and steady- state performance of the three-phase induction machine are studied in detail. 3. CONSTRUCTION FEATURES Three-phase AC induction motors are commonly used in industrial applications. This type of motor has three main parts, rotor, stator, and enclosure. The stator and rotor do the work, and the enclosure protects the stator and rotor. 102 Introduction to Electrical Machines a) Stator The stator is composed of laminations of high-grade sheet steel and is built up of sheet steel lamination of 0.4 to 0.5mm thickness. Laminations are insulated from each other by means of varnish coating or oxide (Figure 3.2(a)). A three-phase winding is put in slots punched out on the inner surface of the stator frame. It is made up of a number of stampings which are slotted to receive the windings. The stator carries a 3-phase winding and is fed from a 3-phase supply. It is wound for a definite number of poles, the number of poles being determined by the requirements of speed. Greater the number of poles, lesser the speed and vice versa. The stator windings, when supplied with 3-phase currents, produce a magnetic flux which is of constant magnitude but which revolves (or rotates) at synchronous speed (given by N s = 120. f ). This revolving magnetic flux P induces an emf in the rotor by mutual induction. (a) (b) Figure 3.2 induction machine laminations (a)Stator and (b)rotor b) Rotor The rotor also consists of laminated ferromagnetic material, with slots punched out on the outer surface (Figure 3.2 (b). The frequency of the rotor flux is very low; as a result thicker laminations can be used without excessive iron losses. Two types of rotor construction is normally used for three phase induction motor. Bearings, mounted on the shaft, support the rotor and allow it to turn. Some motors, like the one shown in the following illustration, use a fan, also mounted on the rotor shaft, to cool the motor when the shaft is rotating. (i) Squirrel-cage rotor: Motors employing this type of rotor are known as squirrelcage induction motors. (ii) Phase-wound or wound rotor: Motors employing this type of rotor are variously known as phase-wound motors or wound motors or slip-ring motors. Squirrel –Cage Rotor Almost 90 per cent of induction motors are squirrel-cage type, because this type ‘rotor' has the simplest and most rugged construction imaginable and is almost indestructible. The squirrel cage rotor is so called because its construction is reminiscent of the rotating 103 Chapter One: Electromagnetic Principles exercise wheels found in some pet cages. The rotor consists of a cylindrical laminated core with parallel slots for carrying the rotor conductors which, it should be noted clearly, are not wires but consist of heavy bars of copper, aluminum or alloys. One bar is placed in each slot; rather the bars are inserted from the end when semi-closed slots are used. The rotor bars are brazed or electrically welded or bolted to two heavy and stout short circuiting end-rings. It should be noted that the rotor bars are permanently short-circuited on themselves, hence it is not possible to add any external resistance in series with the rotor circuit for starting purposes. Figure 3.3 Cut-away view of squirrel cage IM 1.Shaft ; 2. Ball bearings; 3. Bearings shield; 4. Terminal box ; 5. Fan (ventilator) ; 6. Ball bearings; 7. Bearings shield;8. Ventilator shield ; 9. Rotor core; 10. Stator core; 11. Frame; 12. Basement b) a) Figure 3.4 a) Real squirrel cage rotor and b) Schematic diagram of cage rotor The rotor slots are usually not quit parallel to the shaft but are purposely give a slight skew. This is useful in two ways: i.) it helps to make the motor run quietly by reducing the magnetic hum and ii.) it helps in reducing the locking tendency of the rotor i.e. the tendency of the rotor teeth to remain under the stator teeth due to direct magnetic attraction between the two. 104 Introduction to Electrical Machines In small motors, another method of construction is used. It consists of placing the entire rotor core in a mould and casting all the bars and end-rings in one piece. The metal commonly used is an aluminum alloy. Phase –Wound Rotor This type of rotor is provided with 3-phase, double-layer, distributed winding consisting of coils as used in alternators. The rotor is wound for as many poles as the number of stator poles and is always wound 3-phase even when the stator is wound two-phase. Figure 3.5 Cut-away view of wound-rotor induction machine 1.Ball bearings 2. Bearings shield 3. Enclosure; 4.Stator core with windings 5.Rotor core 6.Bearings shield 7.Ball bearings; 8.Shaft ; 9.Terminal box ; 10. Basement; 11. slip-rings a) b) Figure 3.6 a) schematic diagram of wound-rotor b) real diagram of wound-rotor The three phases are starred internally. The other three winding terminals are brought out and connected to three insulated slip-rings mounted one the shaft with brushes resting on them. These three brushes are further externally connected to a 3-phase star-connected rheostat. This makes possible the introduction of additional resistance in the rotor circuit during the starting period for increasing the starting torque of the motor and for changing 105 Chapter One: Electromagnetic Principles its speed-torque/ current characteristic. When running under normal condition, the sliprings are automatically short-circuited by means of a metal collar which is pushed along the shaft and connect all the rings together (Figure 3.7). Next, the brushes are automatically lifted from the slip-rings to reduce the frictional losses and the wear and tear. Hence, it is seen that under normal running conditions, the wound rotor is short– circuited on itself just like the squirrel-cage rotor. 3-phase supply stator Slip-rings rotor Starting Rheostat Figure 3.7 Three-phase wound –rotor induction motor with external starting rheostat. c) Enclosure The enclosure consists of a frame (or yoke) and two end brackets (or bearing housings). The stator is mounted inside the frame. The rotor fits inside the stator with a slight airgap separating it from the stator. There is no direct physical connection between the rotor and the stator. The enclosure protects the internal parts of the motor from water and other environmental elements. The degree of protection depends upon the type of enclosure (See Appendix D). Comparison of squirrel cage and wound rotors. The squirrel cage motor has the following advantages as compared with the wound rotor machine. i. No slip rings, brush gear, short circuiting devices, rotor terminals for starting rheostats are required. The star delta starter is sufficient for staring. ii. It has slightly higher efficiency. iii. It is cheaper and rugged in construction iv. It has better space factor for rotor slots, a shorter overhang and consequently a smaller copper loss. v. It has bare end rings, a larger space for fans and thus the cooling conditions are better 106 Introduction to Electrical Machines vi. It has smaller rotor overhang leakage which gives a better power factor and greater pull out torque and overload capacity. The greatest disadvantage of squirrel cage rotor is that it is not possible to insert resistance in the rotor circuit for the purpose of increasing the starting torque, reducing the starting current and varying the speed as compared with wound rotor motor. 3.4. ROTATING MAGNETIC FIELD It will now be shown that when three-phase windings displaced in space by 1200 are fed by three-phase current displaced in time by 1200 they produce a resultant magnetic flux which rotated in space as if actual magnetic poles were being rotated mechanically. Let a 3-phase , two-pole stator having three identical winding placed 120 space degrees apart and the flux ( assumed sinusoidal) due to three-phase windings as shown in Figure 3.8(a). The assumed positive directions of the fluxes are shown Figure 3.8(b). φ φB φm 120° φA 120° O 120° θ φC (a) (b) Figure 3.8 (a) flux waveforms due to three-phase windings and (b) positive directions of the fluxes Let the maximum value of flux due to any one of the three phases to φm. The resultant flux φr, at any instant, is given by the vector sum of the individual fluxes φA, φB and φC due to three phases. We will consider values of φr at four instants 1/6 time-period apart corresponding to points marked 0,1,2 and 3 in Figure 3.7(a). i) When θ = 00 i.e. corresponding to point 0 in Figure 3.8 (a). Here φ A = 0, φB = − 3 φm , 2 φC = 3 φm , 2 The vector for Oφ B in Figure 3.9(i) is drawn in a direction opposite to the direction assumed positive in Figure 3.8(b). φr =2× 3 600 3 3 φmCos = 3× φm = φm 2 2 2 2 107 Chapter One: Electromagnetic Principles (ii) When θ = 600 i.e. corresponding to point 1 in Figure 3.8(a). Here φA = 3 φm 2 ………drawn in parallel to OφA of Figure 3.8 (b) as shown in Figure 3.9 (ii) 3 φm 2 φB = − ………drawn in opposition to Oφ B of Figure 3.8(b) as shown in Figure 3.9(ii) φC = 0 φr =2× 3 3 3 φm Cos30° = 3 × φm = φm 2 2 2 3 2 It is found that the resultant flux is again φ m but has rotated clockwise through an angle of 600. (iii) When θ = 1200 i.e. corresponding to point 2 in Figure 3.8(a). Here φA = 3 φm , 2 φ B = 0, φc = − 3 φm 2 3 2 It can be again proved that φ r = φ m .So, the resultant flux is again of the same value but has further rotated clockwise through an angle of 600 [Figure 3.9 (iii)]. (iv) When θ = 1800 i.e. corresponding to point 3 in Figure 3.7a. φ A = 0, The resultant is φB = 3 φm , 2 φC = − 3 φm 2 3 φm and has rotated clockwise through an additional angle of 600 or 2 through an angle of 1800 from the start. 108 Introduction to Electrical Machines ΦA − ΦB ΦC − ΦB Φr = 1.5Φm Φr = 1.5Φm ii) θ = 60° i) θ = 0° − ΦC Φr = 1.5Φm ΦC ΦA Φr = 1.5Φm iii) θ = 120° ΦB iv) θ = 180° Figure 3.9 Resultant flux phasor of 3-phase IM at interval of 60° Hence, we can conclude the above discussion as follow: 1. 3 2 The resultant flux is of constant value = φ m i.e. 1.5 times the maximum value of the flux due to any phase. 2. The resultant flux rotates around the stator at synchronous speed given 120. f by N s = P Figure 3.10 shows the graph of the rotating flux in simple way. As before the positive directions of the flux phasors have been shown separately in Figure 3.10. Arrows on these flux phasors are reversed when each phase passes through zero and becomes negative. φA φ φB φC φm 0° 60° 120° 180° 240° 300° 360° Figure 3.10 a graph of the rotating flux 109 Chapter One: Electromagnetic Principles As seen, positions of the resultant flux phasor have been shown at intervals of 600 only. The resultant flux produces a field rotating in the clockwise direction. N.B. The direction of rotation of a polyphase Induction motor depends on the motor connection to the power lines. Rotation can be readily reversed by interchanging any two input leads. 5. PRINCIPLE OPERATION OF 3-PHASE INDUCTION MOTOR When the 3-phase stator winding are fed by a 3-phase supply then a magnetic flux of constant magnitude but rotating at synchronous speed , is set up. The flux pass through the air gap sweeps past the rotor surface and so cuts the rotor conductors which, as yet stationery. Due to the relative speed between the rotating flux and the stationary conductors, an emf is induced in the latter according to Faraday’s laws of electromagnetic induction. The frequency of the induced e.m.f is the same as the supply frequency. Its magnitude is proportional to the relative speed between the flux and the conductors and its direction is given by Fleming’s Right-hand rule. Since the rotor bars or conductors form a closed circuit, rotor current is produced direction, as given by Lenz’s law is such as to oppose the very cause producing it in this case, the cause which produce the rotor current is relative speed between the rotating flux of the stator and the stationary rotor conductors. Hence to reduce the relative speed, the rotor starts running in the same direction as that of the flux and tries to cutch up with the rotating flux. The setting up of the torque for rotating the rotor is explained below: In Figure 3.11(a) is shown the stator field which is assumed to be rotting clockwise. The relative motion of the rotor with respect to the stator is anticlockwise. By applying Fleming’s right-hand rule, the direction of the induced emf in the rotor is found to be outwards. Hence the direction of the flux due to the rotor current alone is as shown in Figure 3.11 (b). Now by applying the left-hand rule or by the effect of combined field (Figure 3.11c) it is clear the rotor conductors experience a force tending to rotate them in clockwise direction. Hence, the rotor set into rotation in the same direction as the of the stator flux ( or field). stator rotor Relative motion (b) stator F rotor (c) Figure 3.11 pertaining to principle operation of induction motor 110 Introduction to Electrical Machines An induction motor running at no load will have a speed very close to synchronous speed and therefore emf in the rotor winding will be very small. This small emf gives a small current producing a torque just sufficient to overcome the losses such as due to friction and windage and maintain the rotor in rotation. As the mechanical load is applied on the motor shaft, it must slow down because the torque developed at no load will not be sufficient to keep the rotor revolving at the no load speed against the additional opposing torque of load. As the motor slows down, the relative motion between the magnetic field and the rotor is increased. This results in greater rotor emf, rotor current and greater developed torque. Thus, as the load is increased, the motor slows down until the relative motion between the rotor and the rotating magnetic field is just sufficient to result in the development of the torque necessary for that particular load. Slip In practice the rotor never succeeds in catching up with the stator field. If it really did so, then there would be no relative speed between the two hence no rotor emf no rotor current and so no torque to maintain rotation. That is why the rotor runs at a speed, which is always less than the speed of the stator field. The difference in speeds depends upon the load on the motor . The difference between the synchronous speed Ns and the actual speed N of the rotor is known as slip. Though it may be expressed in so many revolutions/ second , yet it is usual to express it as a percentage of the synchronous speed . Actually, the term 'slip' is descriptive of the way in which the rotor 'slips back' from synchronism. S% = Ns − N X 100 Ns Sometimes, Ns – N is called the slip speed. Obviously, rotor (or motor) speed is N = N s (1 − S) It may be kept in mind that revolving flux is rotating synchronously relative to the stator ( i.e. stationary space but at slip-speed relative to the rotor. Frequency of rotor current When the rotor is stationary, the frequency of the rotor current is the same as the supply frequency. But when the rotor starts revolving, then the frequency depends upon the relative speed or on slip-speed. Let at any slip speed, the frequency of the rotor current be fr . Then, 120 f r . p 120 f Also N s = p Ns − N = Dividing one by the other, we get, fr Ns − N = S; = f Ns ∴ f r = sf 111 Chapter One: Electromagnetic Principles As seen, rotor currents have a frequency of fr = sf and when flowing through the individual phases of rotor winding give rise to rotor magnetic fields. These individual rotor magnetic fields produce a combined rotating magnetic field whose speed relative to rotor is = 120 f r 120 sf = = sN s P P However, the rotor itself is running at speed N with respect to space. Hence, speed of rotor field in space = speed of field to rotor + speed of rotor relative to space = SN s + N = N s (1 − S) = N s It means that no matter what the value of slip, rotor currents and stator currents each produce a sinusoidally distributed magnetic field of constant magnitude and constant space speed of Ns. In other words, both the rotor and stator field rotate synchronously which means that they are stationary with respect to each other. These two synchronously rotating magnetic fields in fact, superimpose on each other and given rise to the actually existing rotating field which corresponds to the magnetizing current of the stator winding. Example 3.1 A 3-phase, 50 Hz induction motor has a full-load speed of 1440 r.p. m. For this motor, calculate the following: (a) number of poles ; (b) full-load slip and rotor frequency ;(c) speed of stator field with respect to (i) stator structure and (ii) rotor structure and (d) speed of rotor field with respect to (i) rotor structure (ii) stator structure and (iii) stator field. For parts (c) and (d), answer should be given in rpm and rad /sec. Solution. (a) The use of full-load speed of 1440 rpm as synchronous speed gives NS = 120 × f 1 P Or 1440 = or P= 120 × 50 P 120 × 50 1 = 4 poles 1440 6 Since the number of poles must be even and a whole number, the induction motor must have 4 poles. Note that an induction motor runs at a speed, a little less than synchronous speed. (b) Synchronous speed, NS = 120 × f1 120 × 50 = = 1500 r.p.m. P 4 112 Introduction to Electrical Machines ∴ Slip, S = N s − N 1500 − 1440 = = 0.04 Ns 1500 f 2 = Sf1 = 0.04 × 50 = 2 Hz Rotor frequency, (c) (i) Speed of stator field with respect to stator structure = NS = 1500rpm ∴ ωS = 2 π × N S 2 π × 1500 = = 150.08 rad / s 60 60 (ii) Speed of stator field w.r.t. revolving rotor structure = 1500 − 1440 = 60 rpm = 2π × 60 = 6.283 rad / s 60 (d) (i) Speed of rotor field w.r.t. rotor structure 120 ( rotor frequency ) poles 120 × 2 = = 60 r.p.m. = 6.283 rad / s 4 = (ii) Speed of rotor field w.r.t. stator structure =(Mechanical speed of rotor) + (Speed of rotor field w.r.t rotor structure) = 1440 + 60 =1500 rpm = 150.08 rad/s. (iii) Since both the stator and rotor fields are rotating at synchronous speed of 1500 rpm with respect to stator structure, speed of rotor field with respect to stator field is zero. Thus the stator and rotor fields are stationary with respect to each other. Example 3.2 A properly shunted centre-zero galvanometer is connected in the rotor circuit of a 6-pole, 50 Hz wound-rotor induction motor. If the galvanometer makes 90 complete oscillations in one minute, calculate the rotor speed. Solution. One complete oscillation of galvanometer corresponds to one cycle of rotor frequency. ∴Rotor frequency, f 2 = Sf1 = or Slip, S = Rotor speed, 90 = 1.5 Hz 60 f 2 1 .5 = = 0.03 s = fz fl =501.5 = 0.03 f1 50 N = N S (1 − S) = 120 × 50 (1 − 0.03) = 970 r.p.m. 6 113 Chapter One: Electromagnetic Principles 6. EQUIVALENT CIRCUIT MODEL The preceding sections have provided an appreciation of the physical behavior of the induction machine. We now proceed to develop an equivalent circuit model that can be used to study and predict the performance of the induction machine with reasonable accuracy. In this section a steady-state per-phase equivalent circuit will be derived. For convenience, consider a three-phase wound-rotor induction machine a shown in Figure 3.12. In the case of a squirrel-cage rotor, the rotor circuit can be represented by an equivalent three-phase rotor winding. If currents flow in both stator and rotor windings, rotating magnetic fields will be produced in the air gap. Figure 3.12 Three-phase induction machines equivalent circuit model Because they rotate at the same speed in the air gap, they will produce a resultant air gap field rotating at the synchronous speed. This resultant air gap field will induce voltages in both stator windings (at slip frequency f1) and rotor windings (at slip frequency f2). It appears that the equivalent circuit may assume a form identical to that of a transformer. 3.6.1. Stator Equivalent Circuit The stator winding can be represented as shown in Figure 3.13(a), (a) Where V1 = per-phase terminal voltage R1 = per-phase stator winding resistance X1 = per-phase stator leakage reactance E1 = per-phase induced voltage in the stator winding 114 Introduction to Electrical Machines Xm = per-phase stator magnetizing reactance Rc = per-phase stator core loss resistance Note that there is no difference in form between this equivalent circuit and that of the transformer primary winding. The difference lies only in the magnitude of the parameters. For example, the excitation current Io is considerably large in the induction machine because of the air gap. In induction machines it is as high as 30 to 50 percent of the rated current, depending on the motor size where as it is only 1 to 5 percent in transformers. Moreover, the leakage reactance X1 is large because of the air gap and also because the stator and rotor windings are distributed along the periphery of the air gap rather than concentrated on a core, as in the transformer. 3.6.2. Rotor Equivalent Circuit The rotor equivalent circuit at slip s is shown in Figure 3.13 (b). R2 S (b) (c) Where, E2 = per-phase induced voltage in rotor at standstill (i.e. at stator frequency f1) R2 = per –phase rotor circuit resistance X2 = per –phase rotor leakage reactance Note that this circuit is at frequency f2. The rotor current I2 is I2 = sE 2 R2 + jsX 2 3.1 The power involved in the circuit is P2 = I 2 R2 2 3.2 Which represents the rotor coppers loss per phase Equation 1.1 can be rewritten as I2 = E2 ( R2 / s ) + jX 2 3.3 115 Chapter One: Electromagnetic Principles Equation (3.3) suggests the rotor equivalent circuit of Figure 3.13 (c). Although the magnitude and phase angle of I2 are the same in Eqs.(3.1)and (3.3), there is a significant difference between these two equations and the circuits (Figure 3.13 (b) and (c)) they represent. The current I2 in Eq.(3.1) is at slip frequency f2, where as I2 in Equation (3.3) is at line frequency f1. In Eq.(3.1) the rotor leakage reactance SX2 varies with speed but resistance R2 remains fixed, whereas in Eq.(3.3) the resistance R2/s varies with speed but the leakage reactance X2 remains unaltered. The per-phase power associated with the equivalent circuit of Figure 3.13(c). P = I2 2 R2 P2 = S S 3.4 Because induction machines are operated at low slips (typical values of slip s are 0.01 to 0.05) the power associated with Figure 3.13(c) is considerably larger. Note that the equivalent circuit of Figure 3.13(c) is at the stator frequency, and therefore this is the rotor equivalent circuit as seen from the stator. The power in Eq.(3.4) therefore represents the power that crosses the air gap and thus includes the rotor copper loss as well as the mechanical power developed, Equation (3.4) can be rewritten as . R P = Pag = I 2 2 R2 + 2 ( 1 − s S R = I 22 2 S 3.5 The corresponding equivalent circuit is shown in Figure 3.13(d). R2 (1 − S ) S (d) The speed dependent resistance R2 (1-s)/s represents the mechanical power developed by the induction machine. P mech R 2 ( 1 − S) S = ( 1 − S ) P ag 1− S = .P2 S = I22 and P 2 = I 2 2 R 2 = SP ag Thus P ag : P 2 :P mech = 1:S :1− S This equation indicates that, of the total power input to the rotor (i.e. power crossing the air gap, Pag), a fraction s is dissipated in the resistance of the rotor circuit (known as 116 Introduction to Electrical Machines rotor copper loss) and the fraction 1-s is converted into mechanical power. Therefore, for efficient operation of the induction machine, it should operate at a low slip so that more of the air gap power is converted into mechanical power. Part of the mechanical power will be lost to overcome the windage and friction. The remainder of the mechanical power will be available as out put shaft power. 3.6.3. Complete Equivalent Circuit The stator equivalent circuit, Figure 3.13(a) and the rotor equivalent circuit of Figure 3.13(c) or (d) are at the same line frequency f1 and therefore can be joined together. However, E1 and E2 may be different if the turns in the stator wining and the rotor winding are different. If the turns ratio a = N1 is considered, the equivalent circuit of N2 the induction machine is that shown in Figure 3.13(e). Note that the form of the equivalent circuit is identical to that of a two-winding transformer, as expected. I 2' = I2 a X 2' = a 2 X 2 E2' = aE2 = E1 R2' a 2 R2 = S S (e) Figure 3.13 Development of the induction machine equivalent circuit 3.6.4. Various equivalent circuit The equivalent circuit shown in Figure3.13 (e) is not convenient to use for predicting the performance of the induction machine. As a result, several simplified versions have been proposed in various textbooks on electric machines. There is no general agreement on how to treat the shunt branch (i.e., Rc and Xm), particularly the resistance RC representing the core loss in the machine. Some of the commonly used versions of the equivalent circuit are discussed here. 3.6.5. Approximate Equivalent Circuit If the voltage drop across R1 and X1 is small and the terminal voltage V1 does not appreciably differ from the induced voltage E1, the magnetizing branch (i.e. Rc and Xm), can be moved to the machine terminals as shown in Figure 3.14 (a). 117 Chapter One: Electromagnetic Principles R2' S (a) This approximation of the equivalent circuit will considerably simplify computation, because the excitation current (Io) and the load component (I'2) of the machine current can be directly computed from the terminal voltage V1 by dividing it by the corresponding impedance. Note that if the induction machine is connected to a supply of fixed voltage and frequency the stator core loss is fixed. At no load, the machine will operate close to synchronous speed. Therefore, the rotor frequency f2 is very small and hence rotor core loss is very small. At a lower speed f2 increases and so does the rotor core loss. The total core losses thus increase as the speed falls. On the other hand, at no load, friction and windage losses are maximum and as speed falls these losses decreases. Therefore, if a machine operates from a constant voltage and constant-frequency source, the sum of core losses and friction and windage losses remains essentially constant at all operating speeds. These losses can thus be lumped together and termed the constant rotational losses of the induction machine. If the core loss is lumped with the windage and frication loss Rc can be removed from the equivalent circuit, as shown in Figure 3.14(b). R1 I1 Io V1 X1 X’2 I’2 R2' S Xm Pag (b) Figure 3.14 Approximate equivalent circuit of Induction motor IEEE1 Recommended Equivalent Circuit 3.6.6. In the induction machine, because of its air gap, the exciting current Io is high of the order of 30 to 50 percent of the full-load current. The leakage Reactance X1 is also high . The IEEE recommends that in such situation, the magnetizing reactance Xm not be moved to the machine terminals (as is done in Figure 3.14b), but be retained at its appropriate place, as shown in Figure 3.15. The resistance RC is however, omitted, and the core loss is lumped with the windage and friction losses. This equivalent circuit 1 IEEE-Institute of Electrical and Electronics Engineers 118 Introduction to Electrical Machines (Figure 3.15) is to be preferred for situation in which the induced voltage E1 differs appreciably from the terminal voltage V1. X'2 I'2 R '2 S Figure 3.15 IEEE Recommended Equivalent Circuit 3.6.7. Thevenin’s Equivalent circuit In order to simplify computations V1, R1, X1 and Xm can be replaced by Thevenin's equivalent circuit values Vth, Rth and Xth, as shown in Figure 3.16. R2 S Figure 3.16 Thevenin Equivalent circuit V th = [R X 2 1 + (X m + X 1 m ) 2 ] 1 ⋅ V 2 1 Where If R12 << ( X 1 + X m ) 2 as is usually the case Vth ≈ Xm V1 X1 + X m Vth = K th V1 Where, K th = Xm X1 + X m The Thevenin impedance is Z th = jX m ( R1 + jX 1 ) R1 + j ( X 1 + X m ) = Rth + jX th 119 Chapter One: Electromagnetic Principles If R12 << ( X 1 + X m ) 2 2 Xm R1 R th ≈ X1 + X m = K 2 th R1 and since X1 << Xm Xth ≈ X1 Example 3.3 A 3-phase slip ring induction motor gives a reading of 55V across slip rings on open circuit when at rest with normal stator voltage applied. The rotor is starconnected and has impedance of (0.7 + j 5) per phase. Find the rotor current when the machine is (a) at stand still with the slip rings joined to a star-connected starter with a phase impedance of (4 + j 3) ohms and (b) running normally with a 5 percent slip. Solution (a) At standstill, i.e. S=1 55 = 31.8V 3 Resistance per phase in rotor circuit, r2 = 0.7 + 4 = 4.7 = 4.7 Ω Reactance per phase in rotor circuit, x 2 = 5 + 3 = 8Ω Induced emf per phase in rotor winding, E 2 = Impendence per phase in rotor circuit, z 2 = r22 + x 22 = 4.7 2 + 8 2 = 9.28Ω E 31.8 Rotor current per phase, I 2 = 2 = = 3.425 A z 2 9.28 r 4 .7 Power factor cos θ 2 = 2 = = 0.506 ( lagging ) z 2 9.28 (b) Running normally at 5 Percent slip Induced e.m.f in rotor winding per phase, E2 S = SE2 = 0.05 × 31.8 = 1.59 V Reactance per phase in rotor circuit, x 2 S = Sx2 = 0.05 × 5 = 0.25 Ω Rotor impedance per phase, z 2 S = r22 + x 22S = (0.7 )2 + (0.25 )2 Rotor current per phase, I 2 = = 0.7433Ω E2 S 1.59 = = 2.14 A z 2 S 0.7433 Power factor, r 0 .7 cos θ 2 = 2 = = 0.92 ( lagging ) z 2 S 0.7433 Example 3.4 A 3φ, 15 hp, 460V, 4-pole, 60Hz, 1728 rpm induction motor delivers full output power to a load connected to its shaft. The windage and friction loss of the motor is 750 W. Determine the 120 Introduction to Electrical Machines a) Mechanical power developed. b) Air gap power c) Rotor copper loss. Solution Full-load shft power, Pshaft = 15 × 746 = 11,190 W a) Mechanical power developed, Pmech = Pshaft + Pf&w = 11,190 + 750 = 11,940 W Synchronous speed, N s = b) S= Slip, N s − N 1800 − 1728 = = 0.04 Ns 1800 Air gap power, Pag = c) Pmech 11,940 = = 12 ,437.5 W 1 − S 1 − 0.04 Rotor copper loss, 3.7. 120 × f 120 × 60 = = 1800 rpm P 4 P2 = SPag = 0.04 × 12 ,437.5 = 497.5 W PERFORMANCE CHARACTERISTICS The equivalent circuits derived in the preceding section can be used to predict the performance characteristics of the induction machine. The important performance, characteristics in the steady state are the efficiency, power factor, current, starting torque maximum (or pull-out) torque, and so forth. The mechanical torque developed Tmech per phase is given by Pmech = Tmech . ω = I2 . 2 mech R2 (1 − S ) 33331 S 3.6 Where, ωmech = 2π N 60 3.6 a The mechanical speed ω mech is related to the synchronous speed by ωmech = ( 1 − S )ωsyn = 11 Ns 2π ( 1 − s ) 60 3.7 and 121 Chapter One: Electromagnetic Principles ω syn = 120 f x 2π P .60 3.8 From Eqs. (3.6) and (3.7) 2 I R Tmech .ω syn = 2 2 = pag S T mech = = = 1 ω .Pag 3.10 syn 1 ω syn .I2 1 ω 3.9 2 R2 S 2 I2 syn 3.10a R2 S 3.11 From Thevenin equivalent circuit and Eq.1.11 T mech = 1 ω V th . R + 2 )2 + ( X S ( R th syn 2 . + X th 2 ) 2 R2 S 3.12 Note that if the approximate equivalent circuits (Figure 3.13(b) are used to determine I2 in Eq.(3.12) Vth, Rth and Xth should be replaced by V1, R1 and X1 respectively. The prediction of performance based on the approximate equivalent circuit may differ by 5 percent from those based on the equivalent circuit of Figure 3.15 or Figure 3.16. For a three-phase machine Eq.3.12 should be multiplied by three to obtain the total torque developed by the machine. T 3 φ , mech = 3 ω syn V th . ( R th + 2 R2 2 ) + (X S . th + X 2 )2 R2 S The torque-speed characteristic is shown in Figure 3.17. At low values of slip, (R th + R 2' ) >> ( X S th + X ' 2 ) and R 2' >> S R th And thus 122 Introduction to Electrical Machines Tmech ≈ 1 ω syn V 2 th . .S R' 2 3.13 Figure 3.17 Torque-speed characteristics of an induction motor The linear torque-speed relationship is evident in Figure 3.17 near the synchronous speed. Note that if the approximate equivalent circuits (Figure 3.14) are used in, Eq. (3.13), Vth should replaced by V1. At larger values of slip, R th R 2' + << X S th + X ' 2 and T mech ≈ 1 ω syn . (X th V th R'2 . + X '2 )2 S 3.14 The torque varies almost inversely with slip near S=1, as seen from Figure 3.16. Equation (3.12) also indicates that at a particular speed (i.e. a fixed value of s) the torque varies as the square of the supply voltage Vth (hence V1). Figure 3.18 Torque speed characteristics at different voltages Figure 3.18 shows the T-N profile at various supply voltages. This aspect will be discussed further in a later section on speed control of induction machines by changing the stator voltage. 123 Chapter One: Electromagnetic Principles Maximum Torque dT mech = 0. dS Differentiating Eq. (3.12) with respect to slip s and equating the result to zero gives the following condition for maximum torque. An expression for maximum torque can be obtained by setting [ R 2' = R th2 + ( X th + X 2' ) 2 S T max ] 1 2 3.15 This expression can be also be derived from the fact that the condition for maximum torque corresponds to the condition for maximum air gap power (Eq.3.10). This occurs, by the familiar impedance-matching principle in circuit theory, when the impedance of R'2/s equals in magnitude the impedance between it and the supply voltage V1 (Figure 3.16) as shown in Eq.(3.15). The slip Smax at maximum torque Tmax is ST max = [R R 2' 2 th + ( X th + X ' 2 ) 2 ] 1 3.16 2 The maximum torque per phase from Eqs. (1.12) and (1.16) is Tmax = 1 2ω syn . V th 2 R th + [ R th2 + ( X th + X 2' ) 2 ] 1 3.17 2 Equation (3.17) shows that the maximum torque developed by the induction machine is independent of the rotor circuit resistance. However, from Eq. (3.16) it is evident that the value of the rotor resistance R2 determines the speed at which this maximum torque will occur. The torque-speed characteristics for various values of R2 are shown in Figure.3.19. Figure 3.19 Torque-speed characteristics for varying R2 124 Introduction to Electrical Machines In a wound–rotor induction motor, external resistance is added to the rotor circuit to make the maximum torque occur at standstill so that high starting torque can be obtained. As the motor speeds up, the external resistance is gradually decreased and finally taken out completely. Some induction motors are, in fact, designed so that maximum torque is available at start, that is, at zero speed. If the stator resistance R1 is small (hence Rth is negligibly small), from Eqs. (3.16) and (3.17), S T max R 2' ≈ X th + X 2' T max ≈ 2 1 2ω 3.18 V th X th + X ⋅ syn 3.19 ' 2 Equation (3.19) indicates that the maximum torque developed by induction machine is inversely proportional to the sum of the leakage reactances. From Eq. (3.12), the ratio of the maximum torque developed to the torque developed at any speed is T max T = R R th th R + S T 2 R 2' + S (X + ' 2 max 2 (X + + X th + X th ) 2 ' 2 ' 2 ) ⋅ 2 S S 3.20 T max If R1 (hence Rth) is negligibly small, 2 R 2' + (X S T max = T R 2' ST max + X th ' 2 ) 2 2 + (X th + X ' 2 ) 2 ⋅ S 3.21 S T max From Eqs. (3.18) and (3.21) Tmax = T R 2' S 2 R' + 2 ST max 2 R 2' 2 ST max 2 ⋅ S 3.22 S T max ' Dividing both the numerator and denominator of Eq (3.22) by R2 2 S T max 125 Chapter One: Electromagnetic Principles 2 S T max 1 S T + 1 = 2 S T S T 2 2 1 ST + S S = 2 S2 ST max max max max = +S 1 S 2 S ⋅ ST 2 T max 2 max Hence, S T2 + S T max = T 2ST S 2 3.23 max max Equation (3.23) shows the relationship between torque at any speed and the maximum torque in terms of their slip values. Example 3.5 A 6-pole, 3-phase, 50 Hz, induction motor runs on full load with a slip of 4 percent. Given the rotor standstill impedance per phase as (0.01 + j 0.05)Ω, calculate the available maximum torque in terms of full load torque. Also determine the speed at which the maximum torque occurs. . Solution Rotor resistance per phase, R2=0.01Ω Rotor standstill reactance per phase, X2=0.05Ω Full load slip, S=4%=0.04 Ratio of maximum torque to full load torque 2 R2 + S 2fl X2 = R2 2 ⋅ S fl X2 2 0.01 2 + (0.04 ) 0.0416 0.05 = = 2 .6 0.01 0 . 016 2 ⋅ 0.04 0.05 Maximum torque, Tmax = 2.6 T fl (ii) Slip corresponding to maximum torque, ST max = R2 0.01 = = 0 .2 X 2 0.05 Speed corresponding to maximum torque, N = N s ( 1 − ST max ) = 120 × 50 ( 1 − 0.2 ) 6 = 800 r . p .m . Example 3.6 A 4-pole, 50 Hz, 3-phase induction motor has a rotor resistance of 0.024Ω per phase and standstill reactance of 0.6 Ω per phase. Determine the speed at which the maximum torque is developed. 126 Introduction to Electrical Machines Solution Rotor resistance per phase, R2=0.024Ω Rotor standstill reactance per phase, X2=0.6Ω Since the torque under running condition is maximum at that value of the slip which makes rotor reactance per phase equal to the rotor resistance per phase, R 0.024 Slip corresponding to maximum torque, ST max = 2 = = 0.04 X2 0 .6 Speed corresponding to maximum torque, N = N S (1 − S T max ) 120 × 50 (1 − 0.04 ) 4 = 1440 rpm = Example 3.7 The maximum torque of a 3-phase induction motor is twice the full load torque and starting torque is equal to full load torque. Calculate the full load speed and the slip at which maximum torque occurs. Solution Maximum torque, Tmax = 2T fl Starting torque, Tst = T fl Ratio of starting torque to maximum torque, Tst = 0 .5 Tmax Since from Eq.1.23 Tst 2S = 2 T max Tmax ST max + 1 0 .5 = 2 S T max ST2 max + 1 or ST2 max − 4 S T max + 1 = 0 Therefore, slip corresponding to maximum torque, ST max = 4 ± 42 − 4 = 2 − 3 = 0.268 2 rejecting higher value Let the full load slip of Sfl. Since the ratio of full load torque to maximum torque (Eq.3.23) is given by the expression T fl Tmax Or = 2 ST max ⋅ S fl ST2 max + S 2fl 1 2 × 0.268 × S fl = 2 0.268 2 + S 2fl 127 Chapter One: Electromagnetic Principles Or 0268 2 + S 2fl = 4 × 0.268 = 0 Or S fl = 0.072 rejecting higher value. Full load speed, N = N S (1 − S fl ) = N s (1 − 0.072 ) = 0.928 N s i.e. 0.928 times synchronous speed STATOR CURRENT From IEEE recommended equivalent circuit Figure 3.16, the input impedance is Z1 R 2' = R 1 + jX 1 + X m // + jX S = R 1 + jX 1 + X m // Z 2' jX Z 1 = R 1 + jX = 1 + m R 2' + jX S R 2' + j( X S m ' 2 ' 2 + X 2' ) Z1 ∠ θ1 The stator current is I1= V1 = I o + I 2' Z1 At synchronous speed (i.e., S=0), R2'/s is infinite and so I2' = 0. The stator current I1 is the exciting current Io. At larger values of slip S2'(=R2'/s +jX2’) is low and therefore I2'(and hence I1) is large. In fact, the typical starting current (i.e. at S=1) is five to eight times the rated current. The typical stator current variation with speed is shown in Figure 3.20. Figure 3.20 Stator current as a function of speed 128 Introduction to Electrical Machines INPUT POWER FACTOR The supply power factor is given by PF = Cosθ1 where θ1 is the phase angle of the stator current I1. This phase angle θ1 is the same as the impedance angle of the equivalent circuit of Figure 3.15. The typical power factor variation with speed is shown in Figure 3.21. Figure 3.21 Power factor as a function of speed EFFICIENCY In order to determine the efficiency of the induction machine as a power converter, the various losses in the machine are first identified. These losses are illustrated in the power flow diagram of Figure 3.22 For a 3φ machine the power input to the stator is Pin =3V1 I1 Cos θ1 The power loss in the stator windings is P1 = 3 I 12 R 1 Where R1 is the ac resistance (including skin effect) of each phase winding at the operating temperature and frequency. Power is also lost as hysteresis and eddy current loss in the magnetic material of the stator core. The remaining power, Pag , crosses the air gap. Part of it is lost in the resistance of the rotor circuit. P2 = 3I 22 R 2 Where, R2 is the ac resistance of the rotor winding. If it is a wound-rotor machine, R2 also includes any external resistance connected to the rotor circuit through slip rings. Power is also lost in the rotor core. Because the core losses are dependent on the frequency f2 of the rotor, these may be negligible at normal operating speeds, where f2 very low. 129 Chapter One: Electromagnetic Principles Pin = 3V1I1Cosθ1 Figure 3.22 Power flow diagram of induction motor The remaining is converted into mechanical form. Part of this is lost as windage and friction losses, which are dependent on speed. The rest is the mechanical output power Pout which is the useful power output from the machine. The efficiency of the induction motor is η= Pout Pin 3.24 The efficiency is highly dependent on slip. If all losses are neglected except those in the resistance of the rotor circuit, Pag = Pin P2 = sPag Pout = Pmech = Pag (1−S) and the ideal efficiency is η (ideal ) = Pout =1− S Pin 3.25 Sometimes η(ideal) is also called the internal efficiency as it represents the ratio of the power output to the air gap power. The ideal efficiency as a function of speed is shown in Figure 3.23. It indicates that an induction machine must operate near its synchronous speed if high efficiency is desired. This is why the slip is very low for normal operation of the induction machine. If other losses are included, the actual efficiency is lower than the ideal efficiency of Eq.(3.25) as shown in Figure.3.23. The full-load efficiency of a large induction motor may be as high as 95 percent. 130 Introduction to Electrical Machines η . Figure 3.23 Efficiency as a function of speed Example 3.8 A 3φ, 460V, 1740rpm, 60Hz, 4-pole wound-rotor induction motor has the following parameters per phase: R1=0.25Ω R2’=0.2Ω X1=X2’=0.5Ω Xm=30Ω The rotation losses are 1700 w. with the rotor terminals short-circuited, find a.) i) Starting current when started direct on full-voltage ii) Starting torque b.) i) Full-load slip ii) Full-load current iii) Ratio of starting current to full-load current iv) Full-load power factor v) Full-load torque vi) Internal efficiency and the motor efficiency at full-load c.) i) Slip at which maximum torque is developed ii) Maximum torque developed d.) How much external resistance per phase should be connected in the rotor circuit so that the maximum torque occurs at start? Solution a) V1 = 460 = 265.6 V / phase 3 131 Chapter One: Electromagnetic Principles At start S=1. The input impedance is r' jX m 2 + jx '2 s = Z1 = r1 + x1 + ' r2 + j x m + x '2 s ( ) j30(0.2 + j0.5) 0.2 + j(30.5) = 0.25 + j0.5 + = 1.08∠66 Ω I st = 265 .6 = 245 .9 ∠ − 66°A 1.08∠66° 2 πN s 2 π × 1800 = = 188.5 rad / s 60 60 ωsyn = Vth = V1 jX m 266.5 × j 30.0 = = 261.3V r1 + j (xm + x1 ) 0.25 + j 30.5 Z th = jX m ( R1 + jX 1 ) j30( 0.25 + j0.5 ) = = 0.55∠63.9°Ω R1 + j( X 1 + X m ) 0.25 + j30.5 = 0.24 + j0.49 Ω ∴ Rth = 0.24 Ω ; X th = 0.49 Ω T st = 3 ω syn 2 ⋅ ( R th V th ' 2 + R2 ) + ( X th + X 2 )2 . R 2' S 3 261 . 3 2 0 .2 × × = 2 2 188 . 5 ( 0 . 24 + 0 . 2 ) + ( 0 . 49 + 0 . 5 ) 1 = 185 . 2 N .m = b) At full-load S= N s − N 1800 − 1740 = = 0.0333 Ns 1800 R2' 0.2 = = 6.01 Ω s 0.0333 132 Introduction to Electrical Machines j30(6.01 + j0.5) 6.01 + j(30.5) = 0.25 + j0.5 + 5.598 + j1.596 = 6.2123∠19.7 Ω Z1 = 0.25 + j0.5 + I FL = 265.6 = 42.754 ∠ − 19.7°A 6.2123∠19.7 ° iii) I st 245.9 = = 5.75 I FL 42.754 iv) PF= cos(19.7)=0.94 lagging v) T FL vi) Air gap power 3 261 . 3 2 × 6 . 01 188 . 5 ( 0 . 24 + 6 . 01 ) 2 + ( 0 . 49 + 0 . 5 ) 2 = 163 . 11 N . m = Pag = Tωsyn = 163.11 × 188.5 = 30,746.2 W Rotor copper loss: P2 = sPag = 0.0333 × 30,746.2 = 1023.9 W P2 = (1 − s )Pag = (1 − 0.0333) × 30,746.2 = 29,722.3 W Pout = Pmech − Prot = 29 ,722.3 − 1700 = 28 ,022.3W Pin = 3V1 I 1 cos θ1 = 3 × 265.6 × 42.754 × 0.94 = 32,022.4 W ηmotor = Pout 28,022.3 = × 100 = 87.5% Pin 32,022.4 ηint ernal = ( 1 − s ) = 1 − 0.0333 = 0.967 → 96.7% c) i) From Eq. 3.16 S T max = [R 2 th + (X = R ' 2 th + X [0.24 ' 2 )2 ] 1 2 0.2 2 + (0.49 + 0.5) 2 ] 1 = 2 0.2 = 0.1963 1.0187 133 Chapter One: Electromagnetic Principles From Eq.1.17 2 3 Vth . 2 ω syn R + [ R 2 + ( X + X ' ) 2 ] 1 2 th th th 2 2 3 261 . 3 = 2 × 188 .5 0 . 24 + 0 . 24 2 + (0 .49 + 0 .5 )2 = 431 .68 N . m Tmax = [ 1 2 ] Tmax 431.68 = = 2.65 TFL 163.11 d) S T max = 1 = = R ' ext [R [0 . 24 = 1 . 0187 2 R 2 th ' 2 + R ' ext + ( X th + X R '2 + R 'ext ' 2 )2 ] 1 ] + (0 . 49 + 0 . 5 ) 2 − 0 . 2 = 0 . 8187 Ω 2 1 2 = R '2 + R 1 . 0187 ' ext Note that for parts (a) and (b) it is not necessary to use Thevenin’s equivalent circuit. Calculation can be based on the equivalent circuit of Figure. 3.15 as follows: Z1 = r1 + x1 + R e + jX e = 0.25 + j0.5 + 5.598 + j1.596 3 2 3 T= I1 R e = × 42.754 2 × 5.598 ωsyn 188.5 = 163 N.m 8. INDUCTION MOTOR PHASOR DIAGRAM The rotor m.m.f. F2 lagging behind air-gap flux by a space angle of 90 + θ2. The rotor m.m.f. reacts on the stator and calls for a compensating load component of stator current I2' such that load component of stator m.m.f. Fl' = rotor m.m.f. F2 or I '2 N1' = I 2 N '2 N k N '2 ' I2 = I2 = I 2 × 2 w2 ' N1k w1 N1 Here N1' and N2′ are the effective number of stator and rotor turns respectively. In Figure 3.24 load component of m.m.f. F1' = I '2 N1' of the total stator m.m.f. Fl, is shown opposing rotor m.m.f. F2. Similarly, load-component current I2', of the total stator current I1, is 134 Introduction to Electrical Machines shown opposite to rotor current I2. In Figure 3.24, per-phase rotor induced emf E2 lags Φ by a time-phase angle of 90°. φ θ2 Figure 3.24 pertaining to the induction motor phasor diagram If hysteresis is neglected then air-gap flux Φ is in phase with the resultant air-gap m.m.f. Fr. As in a transformer, the no-load magnetizing m.m.f. of the motor does not differ from resultant air-gap m.m.f. Fr. It is because of this reason that motor magnetizing current Im, is shown in phase with Fr, in Figure 3.24. The phasor sum of Fl' and Fr gives the total stator m.m.f. Fl. Similarly the phasor sum of stator-load component of current I2’and motor magnetizing current Im gives the total stator current I1 as illustrated in Figure 3.24. The stator (or the primary) induced emf El and rotor (or secondary) induced emf E2 are shown lagging Φ by 90° as in transformer. Complete induction motor phasor diagram at standstill is drawn in Figure 3.25(a), where mmfs are not shown for the sake of clarity. At standstill, E2 is shown equal to I2 (r2 + jx2). The core-loss component of stator current, i.e. Ic is in phase with Vl' or - E1. At standstill, friction and windage loss is zero. The stator no-load current is I 0 = I m + I c and the stator load current is I1 = I '2 + I o . The stator applied voltage Vl must balance the stator counter emf Vl' (= - E1) and the stator leakage impedance drop I1 (rl +jxl) as shown in Figure 3.25 (a). The power factor angle θ1 (between V1 and Il) at the stator terminals is very high, i.e. stator power factor is very poor at the time of starting a 3-phase induction motor. At normal operating speed, slip s is small. The rotor voltage equation now becomes, sE2 = I2 (r2 + jsx2) and this is illustrated in the phasor diagram of Figure 3.25(b). In this Figure, I 0 = I m + I fc , where Ifc, is the friction, windage and core-loss component of stator current. The rest of the phasor diagram is drawn in the same manner as illustrated in Figure 3.24(a). Figure 3.25(b) reveals that full load power factor at the stator terminals has considerably improved (0.8 to 0.9 lagging) from its power factor at starting. In the phasor diagrams of Figures 3.24 and 3.25, all quantities have perphase values. 135 Chapter One: Electromagnetic Principles θ1 θ1 φ φ θ2 θ2 (b) (a) Figure 3.25 Induction motor Phasor diagram at (a) standstill and (b) a full-load slip s. 3.9. MODES OF OPERATION OF A 3-PHASE INDUCTION MACHINES The three phase induction machine has the following three modes or operating regions of operations depending upon the values of slips: a) Motoring Mode : 1<S>0 Under normal operation, rotor revolves in the direction of rotating field produced by the stator currents. As such, the slip varies from 1 at standstill to zero at synchronous speed, i.e. 1 < S > 0. The corresponding speed values are zero (S=1) and synchronous speed (S=0). b) Generating Mode: S < 0. For this operating modes, slip is negative, i.e. S<0. An induction motor will operate in this region only when its stator terminals are connected to constant-frequency voltage source and its rotor is driven above synchronous speed by prime mover. The connection of stator terminals to voltage source is essential in order to establish the rotating air gap field at synchronous speed. In case stator is disconnected from voltage source and rotor is driven above synchronous speed by the prime mover, no generating action takes place. c) Braking Mode: S > 1. For this mode, slip is greater than 1. a slip more than one can be obtained by driving the rotor , with a prime mover, opposite to the direction of rotating field. But such a use in practice is rare. A practical utility of slip more than 1 is obtained by bringing the rotor to a quick stop by braking action, called plugging. For obtaining S>1, or for obtaining plugging, any two stator terminal leads are interchanged. With this the phase sequence is 136 Introduction to Electrical Machines reversed and, therefore, the direction of rotating magnetic field becomes suddenly opposite to that of the rotor rotation. The electromagnetic torque T, now acting opposite to rotor rotation, produces the braking action. Thus the motor can be quickly brought to rest by plugging, but the stator must be disconnected from the supply before the rotor can start rotate in the other direction. All the three regions of operation (braking, S=2.0 to S=1.0; motor regions, S=1.0 to S=0 and generator region, S=0 to S=-1) are illustrated in Figure 3.26. BRAKING REGION MOTOR REGION GENERATOR REGION Tmax -NS 0 2NS NS Tmax S=2 S=1 S=0 S=-1 Figure 3.26 torque-slip curve of an induction machine showing its braking, motor and generator regions 3.10. OPERATING CHARACTERISTICS OF INDUCTION MOTORS Steady-state operating characteristics of an induction motor show graphically the variation of speed, power factor, stator current and efficiency as the shaft power output is varied from no-load to full load. For a given induction motor, the operating characteristics are governed by its rotor resistance, air-gap length and shape of both stator and rotor slots. The objective of this article is to describe the nature of the steadystate operating characteristics of induction motors and to examine the factors that govern them. Figure 3.27 illustrates typical operating characteristics of an induction motor. The shape of these curves is now explained below. Figure 3.27 operating characteristics of an induction motor 137 Chapter One: Electromagnetic Principles a.) Speed At no load, the rotor speed is near to synchronous speed; therefore, the no-load slip is very small. Also, the no-load torque, sufficient to overcome the loss-torque required by friction and windage, is very low. As a result of small no-load torque, the rotor current or the rotor m.m.f. F2 is also quite small and the load angle, is very near to 90°. As the applied load torque is increased, electromagnetic torque T must increase accordingly-this can happen only if rotor m.m.f. F2 and the load angle increase. In order that rotor m.m.f. F2 or rotor current I2 increase, the rotor emf sE2 must increase or the slip must increase; in other words the rotor speed must decrease as the load torque is increased. Here E2 is the rotor emf per phase at standstill. At no-load, rotor leakage reactance has little effect on the rotor leakage impedance, because rotor frequency and sx2 are very small and load angle 90 + tan −1 sX 2 is very r 2 near to 90°. When the slip increases with an increase in load, the rotor power factor angle sX θ2 = tan −1 2 increases and as a result of it, load angle (90 + θ2) is also increased. r2 This shows that as the applied load torque is increased, the rotor speed falls and both the rotor m.m.f. F2 and load angle increase to supply the required load torque. The fall in speed from no load to full load is usually in the range of 2 to 5 percent of rated speed. In view of this an induction motor can be said to possess shunt characteristics. b.) Power factor: The stator current of a three-phase induction motor is made up of the following three components: (i) Magnetizing current Im: This component lags stator voltage Vl by 90° as in a transformer. Its function is to set up rotating magnetic field. (ii) Stator-loss component Ic: This component supplies the stator iron loss and stator copper loss as in a transformer. (iii) Load component I’2: This component balances the rotor m.m.f. as in a transformer. At no load, stator current Io is shown in Figure 3.28. The function of the third, i.e. load, component of Io at no load is to supply friction and windage (F.W.) loss. (For a transformer at no load, this third component is zero). The rotor power developed at no load, is, therefore, equal to F.W. loss. As this loss is quite small, rotor current and therefore load component is very small. 138 Introduction to Electrical Machines V1 ' θ1' I 2 B θ1 I '2 A I0 O φ I2 I2 E2 E1 Figure 3.28 Improvement of power factor with increase in load So the current due to the combination of second and third components is quite small. But the magnetizing current Im forms a major component of Io due to the presence of air gap in an induction motor. As a result, no-load current Io lags the stator voltage by an angle θ0 in the range of 75° to 85°. Consequently, the stator power factor at no load may be as low as 0.1 to 0.3, the lower values being applicable for large induction motors. As the motor is loaded, the third, i.e. load, component of stator current rises above its noload value. The increment of load component of stator current above its no load value is responsible for supplying the load torque. The stator-load component I2' given by when added to Io, gives the stator current OA at a power factor of cosθ1. With further increase of load on the motor, the rotor current increases and the stator-load component I2' when added to Io gives the stator current OB at a power factor of cos θ1'. It is thus seen from Figure 3.28 that the p.f. angle decreases and, therefore, the stator power factor improves as the load on the motor is increased. The stator power factors of about 0.85 to 0.88 are obtained at 80 to 90% of full-load outputs. If the motor is loaded beyond this load, power factor decreases slightly because of the predominant effect of stator and rotor leakage reactance drops. c.) Efficiency: As in other electrical machines, the induction machine losses are made up of fixed losses (= core loss + friction and windage loss) and variable load losses. At no load, the shaft power output is zero, therefore, efficiency is zero. At lower values of loads, the fixed losses are more as compared with the output; efficiency is, therefore, low. As the load is increased, the efficiency also rises and becomes maximum when fixed losses and variable losses are equal. Maximum efficiency occurs at about 80 to 95% of rated output, the higher values being applicable for larger motors. If the load is increased beyond the load resulting in maximum efficiency, the load losses increase more rapidly than the output, consequently the efficiency decreases. d.) Stator current: The no-load stator current is about 30 to 50% of rated current, the larger values being applicable to smaller sizes. With the increase in load, the current rises correspondingly. 139 Chapter One: Electromagnetic Principles In the phasor diagram, the locus of the tips of stator current with increasing loads follows a semi-circle and this leads to the induction motor circle diagram. e.) Air gap: In induction motors, for constant supply voltage, the air-gap flux remains substantially constant. If the air-gap length is increased, then constant flux requires more magnetizing current. This reduces the no-load power factor as well as the full-load power factor of the induction motor. Therefore, in order that an induction motor operates at a better power factor, the air-gap length is kept as small as is mechanically possible. Small air-gap clearance in induction motors necessitates a heavier shaft and high-grade bearings than are required for other types of rotating machines of the same rating and speed. Open slots in an induction motor has the effect of increasing the air-gap length (1.10 to 1.20 times the actual air-gap length) as a result of which more magnetizing current is required and the operating power factor of the motor is worsened. But an induction motor with open slots has less leakage reactance and, therefore, more Tst, more Tmax etc. Likewise induction motors with semi-closed slots or closed slots requires less magnetizing current and, therefore, better operating power factor, but its Tst, Tmax etc are reduced. In view of this, at the design stage, a compromise has to be made between Tst, Tmax and the operating power factor of the induction motors. 3.11. DETERMINATION PARAMETERS OF EQUIVALENT CIRCUIT The equivalent circuit parameters of poly phase induction motors can be determined from no load test, blocked-rotor test and stator winding dc resistance. The object of this article is to describe the methods of determining the parameters from these tests. 3.11.1. No-load test (or running light test) The induction motor is made to run at no load at rated voltage and frequency. Per phase values of applied stator voltage Vnl, input current Inl and input power Pnl are recorded. Figure 3.29 Circuit diagram for no-load and blocked-rotor Test r2 in Figure 3.15 of IEEE recommended s equivalent circuit, is very large as compared to Xm. In view of this, the resultant of The no-load slip S is very small, therefore 140 Introduction to Electrical Machines r parallel branches jXm and 2 + jx 2 is almost equal to jXm as illustrated in Figure 3.30 s (a). Thus the no-load reactance Xnl seen from the stator terminals is equal to x1 + Xm, i.e. Xnl = x1 + Xm = X1 3.26 Where, X1 is the stator self-reactance. From the instrument readings at no load, stator no-load impedance: Z n1 = Vn 1 I n1 And stator no-load resistance: Rn1 = Pn1 I n1 2 X n1 = Z nl − Rn1 2 2 ∴ The rotational losses Prot (friction, windage loss and core loss) are usually assumed constant and can be obtained from the relation. PRot = m( Pnl − I nl2 R1 ) Where m is the number of stator phases and r1 is the per phase stator resistance Thus the no-load test gives Xn1 and the rotational losses PR. (a) 3.11.2. Blocked-rotor test Blocked-rotor test, similar to the short-circuit test on a transformer, is performed on the induction motor to calculate its leakage impedance. For performing this test, the rotor shaft is blocked by external means, i.e. the rotor shaft is held stationary by belt-pulley arrangement or by hand. Now balanced poly phase voltages at rated frequency are applied to the stator terminals through a poly phase 141 Chapter One: Electromagnetic Principles variac. This applied voltage is adjusted till rated current flows in the stator winding. Per phase values of applied voltage Vbr, input current (=rated current) Ibr and the input power Pbr are recorded. Current Ibr may be affected by rotor position; in view of this the rotor should be held in a position that gives Ibr equal to the mean of maximum and minimum current value. Measure the dc resistance per phase of the stator winding soon after this test and multiply it by 1.1 to 1.3 in order to obtain the per phase effective stator resistance r1. The equivalent circuit under blocked-rotor test is as shown in Figure.3.30 (b). R1 X2 X1 Xm Vbr R2 S (b) Figure 3.30 Induction motor equivalent circuits for (a) no-load test and (b) blocked-rotor test From the instrument readings during blocked-rotor test, the parameters can be obtained as under: The blocked-rotor impedance Z br = Vbr I br and the blocked-rotor resistance, P Rbr = br 2 I br ∴Blocked-rotor reactance X br = Z br 2 − Rbr 2 An examination of Figure 3.30 (b) reveals that the blocked-rotor impedance seen from the stator terminals is given by 142 Introduction to Electrical Machines Z br = Rbr + jX br = r1 + jx1 + jX m ( r2 + jx 2 ) r2 + j( x 2 + X m ) jX m ( r2 + jx 2 ) r2 − j ( x 2 + xm ) . r2 + j ( x 2 + xm ) r2 − j ( x 2 + xm ) = r1 + jx1 + jX m [ r2 + jr2 x 2 − jr2 ( x 2 + X m ) + x 2 ( x 2 + X m ] 2 = r1 + jx1 + r2 + ( x2 + X m )2 2 jX m [ r2 + x 2 X 2 − jr2 X m ] or Rbr + jX br = r1 + jx1 + 2 r2 + X 22 2 3.27 Where X2 = x2 +Xm, is the rotor self-reactance. Comparison of the imaginary components of both sides of Eq. (3.27) gives X m [ r22 + x 2 X 2 ] X br = x1 + r 22 + X 22 r22 Xm + x2 X 2 = x1 + r22 + X2 X2 Since X 2 >> r2 , therefore r may be neglected. With this 2 2 X2 X m x2 X m x2 = x1 + X2 X m + x2 x2 = x1 + x 1+ 2 Xm X br = x1 + x2 Usually the magnetizing reactance Xm >> rotor leakage reactance x2, therefore X is m negligible small and this gives Xbr = x1 +x2 3.28 There is no practical method of separating x1 and x2. For wound rotor machines x1 is assumed equal to x2 i.e. x1 = x2= ½Xbr 143 Chapter One: Electromagnetic Principles For squirrel cage induction machines, total leakage reactance Xbr (=x1 +x2) can be distributed between stator and rotor as per the following table: Table 3.1: Empirical distribution of leakage reactance Xbr Fraction of Xbr X1 X2 0.5 0.5 0.4 0.6 0.3 0.7 0.5 0.5 Class of motor Class A (normal Tst , high Ist and low slip) Class B (normal Tst, low Ist and low slip) Class C (high Tst , low Ist and high slip) Class D (high Tst , low Ist and high slip) Once x1 is known, then from Eq. (3.26), the stator magnetizing reactance is given by Xm= Xnl –x1 Now taking the real components of both sides of Eq. (3.27) blocked-rotor resistance Rbr as seen from the stator terminals is R br = r1 + r2 X 2m r22 + X 22 For machines of normal design, X2>> r2 therefore, r2 in the denominator can be neglected as compared to X2 X R br = r2 + r2 m X2 2 ∴ Per phase rotor resistance X r2 = ( R br − r1 ) 2 Xm 2 3.29 Thus Xm from Eq. (3.26) r2 from Eq. (3.29), r1 form dc resistance per phase of stator winding and x1, x2 from Eq. (3.28) and Table 3.1 can be determined from three tests. The equivalent circuit can now be used for computing the motor performance. Note: For large motors (above 20KW or so), if induction motor characteristics are required near s =1 (e.g. for starting torque etc.); then since rotor frequency f2 is equal to the line frequency, the blocked- rotor test should be carried out at line frequency and with currents equal to those encountered at the time of starting. In case induction motor characteristics are required near synchronous speed (e.g. during normal operation), then rotor frequency is equal to sf1, therefore, the blocked-rotor test should be carried out at reduced frequency and with normal currents. 3.11.3. Separation of friction and windage loss from the no-load test The power input to the induction motor at no-load has to supply the stator copper loss, core loss and friction and windage loss. 144 Introduction to Electrical Machines The dc resistance of the stator winding is measured and its per phase effective value r1 is calculated from the relation. r1 = (1.1 to 1.3) (dc resistance of one phase) For computing the friction and windage loss, the applied voltage to the unloaded induction motor is varied from 1.25 times the rated voltage to about 20% of the rated voltage. The input power, current and voltage are recorded so that a graph can be plotted. The speed, with reduction in voltage, will fall only slightly so that the friction and windage loss remains substantially constant From each of the input-power readings, the corresponding stator ohmic loss is subtracted to obtain the core loss and friction and windage loss, i.e. Prot = m( Pnl + I nl2 r1 ) Where, Pnl is the per phase power input, Inl is the per phase stator current and r1 is the effective per phase stator resistance. The plot of the rotational loss Prot with variable stator voltage is shown in Figure 3.31. The intercept of the extraplotted Prot curve with the ordinate gives the friction and windage loss, because the core loss is zero for zero applied voltage. Figure 3.31 Variation of rotational loss with voltage In order to get a motor accurate value of mechanical loss (friction and windage loss), rotational loss Prot should be plotted against (Voltage) 2. This plot of Prot with (voltage)2 is almost linear and, therefore, the extrapolation is easier. Example 3.9 A 10-Kw , 400-V, 4-pole delta connected squirrel cage induction motor gave the following test results: No-load Test: 400-V , 8-A, 250-W Blocked rotor Test: 90-V, 35-A, 1350-W The dc resistance of the stator winding per phase measured after the blocked –rotor test is 0.6Ω. Calculate the rotational losses and the equivalent circuit parameters. 145 Chapter One: Electromagnetic Principles Solution No-load rotational losses are 2 8 Prot = Pnl − 3I nl R 1 = 250 − 3 × (0.6 × 1.20) = 203.92W 3 2 Note that the effective stator resistance per phase is taken equal to 1.2 times its d.c. value. From no-load Test: Vnl 400 = 86 .6 Ω = I nl 8 3 Pnl 250 = = = 3 .91 2 2 3(I nl ) 3 8 3 Z nl = R nl ∴ X nl = Z 2nl − R 2nl = 86 . 6 2 − 3 .91 2 = 86 .51Ω From blocked-rotor test: Vbr 90 = 4.45 Ω = I br 35 3 Pbr 1350 = = 1.1Ω R br = 2 2 3(I br ) 35 3 3 Z br = ∴ ∴ And X br = Z 2br − R 2br = 4.452 − 1.12 = 4.32 Ω x1 = x 2 = 1 X br = 2.158 Ω 2 X m = X nl − x 1 = 86.51 − 2.158 = 84.352 Ω X 2 = X m + x 2 = 84.352 + 2.158 = 86.51 Ω From eq.(3.29), per phase rotor resistance is given by X r2 = ( R br − r1 ) 2 Xm 2 2 86.51 = (1.102 − 0.6 × 1.2 ) = 0.402 Ω 84.352 Thus the parameters of the induction motor equivalent circuit are: r1= 0.72Ω; r2= 0.402Ω; x1=x2=2.158Ω; and Xm=84.352Ω 146 Introduction to Electrical Machines 3.12. SPEED CONTROL OF INDUCTION MOTORS The synchronous speed Ns of an ac motor is related to supply frequency f and poles P by the equation. Ns = 120 f p As regards induction motor, the rotor is given by N = (1 − s ) N s Where S is the slip It is found from the above two equations that the basic methods of speed control of an induction motor are: a) by changing the number of poles and b) by varying the line (input) frequency. By the above two methods, the synchronous speed of an induction motor can only be changed. These methods are applicable only to cage induction motors The slip can be changed by the following methods. c) by varying the input voltage d) by varying the rotor resistance The methods(c-d) are applicable to slip-ring (wound rotor) induction motors, whereas only the method (c) can be applied to machines with cage rotor. a) by changing the number of poles (Pole changing methods) Four-pole to eight –pole connections In pole changing induction motors, the stator winding of each phase is divided into two equal groups of coils. These coil groups are connected in series and parallel with the current direction being reversed only in one group, to create two different numbers of poles (even) in the ratio 2.1 respectively. When the connection is changed from series to parallel or vice versa, the current in one group of coils is also reversed at the same time. This technique, termed the consequent pole method, is applied to all three windings (phases). This type of induction motor has always the squirrel cage rotor, which can adapt to any number of stator poles. Figure 3.32(a) shows schematically only four coils of one phase of the windings connected in series, along with the direction of current in them, producing eight poles in the stator. If the current in coils 2 and 4 is reversed and the connection is changed to parallel with two coils (1 and 3, and 2 and 4) connected in series for each path, four poles 147 Chapter One: Electromagnetic Principles are formed in the stator (Figure 3.32(b)). It may be noted that the direction of current in coils 1 and 3 remains the same. Only one type of connection is shown. (a) Eight pole (Series) (b) Four pole (parallel) consequent pole Figure 3.32 Stator winding connection for pole changing induction motor Normally, poles are changed in the ratio 2:1.This method provides two synchronous speeds. If two independent sets of polyphase windings are used, each arranged for pole changing, four synchronous speeds can be obtained for the induction motor. In many industrial application induction motor with 4/6/8/12 poles are used to provide 1500/1000/750/500 synchronous speeds respectively. Squirrel-cage motors are invariably used in this method because the rotor can operate with any number of stator poles. It is obvious, however, that speed can be changed only in discrete steps and that the elaborate stator winding makes the motor expensive. Constant torque and constant horsepower operations This type of pole changing in the stator results in constant torque or constant horsepower operations. For constant torque operation, the change of stator winding is made from series-star to parallel- star, while for constant horsepower operation the change is made from series-delta to parallel-star. Regenerative braking takes place during changeover from higher to lower speeds. Constant torque opération: (Y/YY) In any pole changing (P-pole/2P-pole) induction motor, there are two equal parts as stated above. The changeover for constant torque operation takes place as shown in Figure 3.33. 148 Introduction to Electrical Machines (b) (a) Series-star (Y) connection ( 2P-pole connection) Parallel-star (YY) connection ( P-pole connection) Ns 2 Ns (c) Torque-speed characteristics Figure 3.33 stator connections and torque-speed characteristics of induction motors for constant torque operation Let V – Line voltage I – Maximum current that the winding can carry Then the power drawn from the supply is given by:1. For series-star (Y) connection ( Figure 3.33(a)) PY = 3 (VI cos φ Y ) 2. For parallel-star connection (Figure 3.33(b)) PY Y = 2 3 (VI cos φ Y Y ) It is assumed that the power factor remains unchanged and the motor losses are negligible. With the changeover of stator winding from series star to parallel star, the power drawn from the supply is doubled. Simultaneously, the speed is also doubled. So the motor torque remains constant. Constant torque operation is more common. 149 Chapter One: Electromagnetic Principles Constant horsepower operation: (∆ ∆/YY) The power drawn from the supply is given by: 1. For series-delta ( ∆) connection (Figure 3.34(a)) P∆ = 3 (VICosφ ∆ ) 2. For parallel-star connection (Figure (3.34(b)) PY Y = 2 3 (VI Cos φYY ) = 3.46 (VI Cos φYY ) After changeover from series-delta to parallel-star, the power increases slightly (about 15%), if power factor is assumed to remain constant. The constant horsepower connection is the most expensive, because in this case the motor size becomes the largest. (b) Parallel-star (YY) connection (a) Series-delta (∆) connection ( P-pole connection) ( 2P-pole connection) Ns 2 Ns (c) Torque-speed characteristics Figure 3.34 stator connections and torque-speed characteristics of induction motors for constant power operation 150 Introduction to Electrical Machines b) by varying the line (input) frequency) A variable frequency supply is connected as the key factor in speed control of induction motors. Constant Volt/Hz operation The emf per phase of an induction motor is given by E = 4.44 Φ m f T ph K w The induced emf E is nearly equal to the applied voltage V (neglecting drop in stator impedance). Then, we can write V = 4.44 Φ m T ph K f (V Hz ) When the frequency is reduced, the applied voltage also must be reduced proportionally so as to maintain constant flux, otherwise the core will get saturated resulting in excessive iron losses and magnetizing current. The maximum torque also remains constant under this condition. However, the voltage is not varied proportionally in the lower frequency range to account for the voltage drop in the winding resistance. This type of control (constant V/f) is used for speed control below base frequency (line frequency of 50Hz). As the voltage increase above rated value, when the input frequency goes above base frequency, only constant (rated) voltage with variable frequency (frequency control) is used for speed control. Under this condition, both flux and maximum torque decrease as the frequency is increased. Advantages of constant volt/Hz operations are the following: a) Smooth speed control, b) Small input current and improved power factor at low frequency start, and c) Higher starting torque for low cage resistance Maximum Torque Neglecting the stator winding resistance, the maximum torque is Tmax ≈ 3V 2 2ωsyn (X1 + X '2 ) So, the maximum torque remains constant as stated earlier for constant volt/Hz ratio for frequencies below base frequency, except for very low values of frequency (Figure 3.35). This is taken as constant torque control with constant flux or volt/Hz ratio. For input frequency above base frequency, only constant (input) voltage with variable frequency is applied as stated earlier. In this case the maximum torque changes to 151 Chapter One: Electromagnetic Principles T max = Where, α= 2 ω syn 3 V ' (X 1 + X 2 ) α 2 supply frequency base frequency With α >1 as frequency is higher than base frequency, both maximum torque and flux, as given by volt/Hz ratio, decrease as frequency increases (as shown in Figure 3.34). This is taken as constant power control with variable flux. Figure 3.35 Torque-speed characteristics of an IM with variable-voltage, variable –frequency control The torque-speed characteristic of the load is superimposed on the motor torque-speed characteristic. Note that the operating speeds N1…N7 are close to corresponding synchronous speeds. In this method of speed control, therefore, the operating slip is low and the efficiency is high. The operating slip can be changed by c) Varying the line voltage Recall that the torque developed in an induction motor is proportional to the square of the terminal voltage. A set of torque-speed characteristics with various terminal voltages is shown in Figure 3.36. Note that for this method of speed control the slip increase at lower speeds, making the operation inefficient. However, for fans, or, similar centrifugal loads in which torque varies approximately as the square of the speed, the power decreases significantly with decrease in speed. Therefore, although the power lost in the rotor circuit (= sPag) may be a significant portion of the air gap power, the air gap power itself is small and therefore the rotor will not overheat. The voltage controller circuits are simple and, although inefficient, are suitable for fan, pump, and similar centrifugal drives. 152 Introduction to Electrical Machines TL = N 2 Figure 3.36 Torque-speed characteristics with various terminal voltages d) Varying the rotor resistance The speed of a wound-rotor induction machine can be controlled by connecting external resistance in the rotor circuit through slip-rings. The torque- speed characteristics for four external resistances are shown in Figure 3.37. The load torque-speed characteristic is also shown by dashed line. By varying the external resistance 0 < Rex < Rex4, the speed of the load can be controlled in the range of N1 < N < N5. Note that by proper adjustment of the external resistance (Rex=Rex2), maximum starting can be obtained for the load. 3φ Figure 3.37 the torque- speed characteristics for four external resistances The major disadvantage of the rotor resistance control method is that the efficiency is low at reduced speed because of higher slips. However, this control method is often employed because of its simplicity. In application where low-speed operation is only a small proportion of the work, low efficiency is acceptable. A typical application of the rotor, resistance control method is the hoist drive of a shop crane. This method also can be used in fan or pump drives, where speed variation over a small range near the top speed is required. 3.13. STARTING OF IM Most induction motors-Large and Small-are rugged enough that they could be started across the line without incurring any damage to the motor windings, although about five to eight times the rated current flows through the stator at rated voltage at standstill. However, in large induction motors, large starting current are objectionable in two respects: 153 Chapter One: Electromagnetic Principles • First, the mains supplying the induction motor may not be of a sufficiently large capacity. • Second, because of large starting current, the voltage drops in the lines may be excessive, resulting in reduced voltage across the motor. Because the torque varies approximately as the square of the voltage, the starting torque may become small at the reduced line voltage that the motor might not even start on load. Thus we formulate the basic requirement for starting: • The line current should be limited by the capacity of the mains, but only to the extent that the motor can develop sufficient torque to start (on load , if necessary) A number of methods is available of for starting both cage-rotor and wound-rotor motors: .13.1. Starting of Squirrel-cage Motors For cage motors, the choice of any particular method of starting depends on (i) size and design of the motor (ii) capacity of the power lines and (iii) type of the driven load. There are primarily two methods of starting squirrel-cage induction motors: (a) fullvoltage starting and (b) reduced-voltage starting. The full-voltage starting consists of DOL starting only. The reduced-voltage starting has the advantage of reducing the starting current, but it produces an objectionable reduction in the starting torque, on account of the fact that motor torque is proportional to the square of the applied voltage. Despite this, reduced-voltage starting is the most popular method of starting three-phase squirrel-cage induction motors and consists of stator resistor (or reactor) starting, autotransformer starting and star-delta starting. The various methods are now described in what follows. a. Direct-on-line (across-the-line) starting. As the name suggests, this method involves the direct switching of polyphase stator on to the supply mains. The motor takes low-power factor starting current of 5 to 8 times its full-load current, depending upon its size and design. Such large currents of short duration don't harm the rugged squirrel cage motor, but the high currents may cause objectionable voltage drop in the power supply lines feeding the induction motor. These large voltage drops cause undesirable dip in the supply line voltage; consequently the operation of other equipments connected to the same supply line is effected considerably. A common example is the momentary dimming of lamp and tube-lights in the home at the instant a refrigerator motor starts. If the supply system is of sufficient power capacity and the low-power factor starting-current surges don't cause objectionable voltage dips in the supply line voltage, then the direct-on-line starting should be preferred. The relation between starting torque Tst and full-load torque Tfl is now obtained. Let Ist and Ifl be the per-phase stator currents drawn from the supply mains corresponding to starting and full-load conditions respectively. From Torque equation T mech = 1 ω syn .I2 2 R2 S 154 Introduction to Electrical Machines R2 Tst 1 = I 2st = I R Tfl I 22 fl ⋅ 2 2 fl Sfl I 22st ⋅ ∴ 2 ⋅ Sfl 3.30 Eq. (3.30) is valid in case rotor resistance remains constant. Actually, rotor resistance varies with the frequency of rotor current; at starting rotor frequency is 50 Hz and at full load it is only a few hertz. Here I2st and I2fl are the per-phase rotor currents at starting and full-load conditions respectively. If no-load current is neglected, then Ist × effective stator turns = I 2st × effective rotor turns or I st = I 2st × (effective rotor to stator turns ratio ) or I st (Effective rotor to stator turns ratio )I 2st = I fl (Effective rotor to stator turns ratio )I 2fl or I st I 2st = I fl I 2 fl 2 Tst I st = ⋅ Sfl Tfl I fl From equation (1.30), If V1 is the per phase stator voltage and ZSC is the standstill per phase leakage impedance referred to stator, then per phase short-circuit current at standstill (or at starting) is, I sc = Where, V1 Zsc Zsc = (R 1 + R 2 ) + j(X1 + X 2 ) Note that here shunt branch of the induction motor equivalent circuit is neglected. Therefore, for direct switching, I st = I sc = V1 Zsc Equation (3.30) can, therefore, be written as 155 Chapter One: Electromagnetic Principles 2 Tst I sc ⋅ Sfl = Tfl I fl b. 3.31 Stator resistor (or reactor) starting xV 3xV1 3V1 1 In this method, a resistor or a reactor is inserted in between motor terminals and the supply mains, as illustrated in Figure 3.38. Rotor V1 xV1 Figure 3.38 Reactor (or resistor) starting of squirrel cage induction motor At the time of starting, some voltage drop occurs across the starting resistor or reactor and, therefore, only a fraction x (less than 1) of the supply voltage appears across the stator terminals. This reduces the per-phase starting current Ist drawn by the motor from the supply mains. As the motor speeds up, the reactor is cut out in steps and finally shortcircuited when the motor speed is near to its operating speed. Since per phase voltage is reduced to xV1, see Figure 3.38 , the per-phase starting current Ist is given by I st = xV1 = xI SC Z SC 3.32 2 As before Tst I st = ⋅ S fl T fl I fl 2 Or I Tst = x 2 sc ⋅ Sfl Tfl I fl 3.33 In an induction motor, torque ∝ (voltage) 2 2 ∴ Starting torque with reactor or resistor starting xV1 = x 2 = Starting torque with direct starting V1 3.34 156 Introduction to Electrical Machines Series reactor is more costly than the series resistor, but the former has lower energy loss and is more effective in reducing the voltage, because the induction-motor power factor at starting is quite low. c. Autotransformer starting A fraction of xV1 of the supply voltage V1 is applied to the stator terminals at the time of starting, by means of an autotransformer as shown in Figure 3.39. Figure 3.39 pertaining to auto-transformer starting of squirrel cage induction motor This reduces the motor current and also the current drawn from the supply. After the motor has accelerated near to its operating speed, auto-transformer is disconnected and full line voltage is applied to the induction motor by connecting it directly across the supply mains. Note that here x is less than 1. With autotransformer, per phase starting current in motor winding is = xV1 = x I sc Z1 3.35 If no-load current of autotransformer is neglected, then per phase output VA of an autotransformer must be equal to its per phase input VA. That is I stV1 = xV1 (per phase starting current in motor winding) I st V1 = xV1 (x I sc ) Or ∴ Per phase starting current from the supply mains, I st = x 2 .I sc 3.36 Eq. (3.35) shows that the motor starting current per phase is reduced only to x times the direct switching current Isc ; but the per phase starting current from the supply mains is reduced to x2 times the direct switching Isc . 157 Chapter One: Electromagnetic Principles Now Tst (per phase starting current in motor winding) 2 = .S fl Tfl (per phase motor full − load current) 2 From Eq.(1.35) Tst (xIsc )2 = .Sfl 2 Tfl I fl =x 2 (Isc )2 .S 2 I fl 3.37 fl Per phase staring current, Ist, from the supply mains can be calculated in Eq. (3.37) with the help of Eq. (3.36). Substitution of the value of Isc in terms of Ist gives 2 1 I Tst = x 2 2 ⋅ st .S fl x I T fl fl 2 I = x st .S fl I fl 2 3.38 The ratio of the starting to full-load torque in terms of both Ist and Isc can be obtained from Eq. (3.37) as follows. Tst (x 2 I sc )I sc = .S fl T fl I 2fl = I st I sc ⋅ S fl I 2fl 3.39 Per phase motor starting current in terms of Ist, from Eq. 3.35, is given by x I sc = x ⋅ = 1 I st x2 1 I st x 3.40 For an induction motor, torque ∝ (voltage) 2 158 Introduction to Electrical Machines 2 Tst with an autotransformer xV1 = x 2 = Tst with direct switching V1 3.41 It is from above that with an autotransformer, the starting current Ist from the mains and the starting torque Tst are reduced to x2 times their corresponding values with DOL starting, see Eqs (3.36) and (3.41). d. Star-Delta Method A Star-Delta method starting may also be employed to provide reduced voltage of start. In this method, the normal connection of the stator windings is delta while running (Figure 3.40). Figure 3.40 pertaining to the star-delta starting of a 3-phase cage induction motor If these windings are connected in star at start, the phase voltage is reduced, resulting in less current of starting. As the motor approaches its full-speed, the windings will be connected in delta. Thus, the line current under each of these connection are: IY = Vph I∆ = 3 = Z ph VL 3 Z ph with the windings connected in Star ; VL Z ph with the windings connected in Deta So that the ratio of the current is I Y = I ∆ V L ⋅ 3Z ph Z ph 3V = L 1 3 Showing that the current drawn from the line under a star connection is only one-third ( 1 3 ) of that under delta connection. On the other hand, the ratio of the current in the stator windings is 159 Chapter One: Electromagnetic Principles I ph Y I ph ∆ = V ph Z ph 3 V ph Z ph = 1 3 2 VL Tst with Star − Delta Starter 1 3 = = 2 Tst with direct switching in Delta VL 3 Also This shows the star-delta starter also reduce the starting torque to one-third of that produced by direct switching in delta. Example 3.10 A squirrel induction motor has a full-load slip of 0.05. The motor starting current at rated voltage is 6 times its full-load current. Find the tapping on the auto-transformer starter which should give full-load at start. Also find the line current at starting in terms of full-load current. Solution Here , motor starting or short-circuit current Isc is 6Ifl. 2 I Tst = x 2 sc .S fl I T fl fl 1 = x 2 × 6 2 × 0.05 ∴ x= Or or 1 = 0.745 1 .8 74.5% tappings The starting line current is I st = x 2 I sc = 0.745 2 × 6 I fl = 3.33 I fl Example 3.11 A small 3-phase induction motor has a short-circuit current 5 times of full load current and full load slip 5%. Determine the starting torque and starting current if starting resistance starter is used to reduce the impressed voltage to 60% of normal voltage. Solution Starting current, I st = 0.6I SC = 0.6 × 5I fl = 3I fl Starting torque, 2 I 2 Tst = Tfl st ⋅ Sfl = Tfl (3) × 0.05 = 0.45Tfl i.e. 45% of full load torque. Ifl 160 Introduction to Electrical Machines 3.13.2. Starting of Wound-Rotor Motors The methods used for starting squirrel cage motors can also be employed for starting wound-rotor motors, but it is usually not done so because then the advantages of woundrotor induction motors can't be fully realized. The simplest and cheapest method of starting wound-rotor induction motors is by means of added rotor resistance, with fullline voltage across the stator terminals. It has already been discussed that at the time of start, the addition of external resistance in the rotor circuit of a wound-rotor induction motor. i. decreases its starting current ii. increases its starting torque (for a suitable external resistance) and iii. Improves its starting power factor. At the time of start, the entire external resistance is added in the rotor circuit. As the rotor speeds up, the external resistance is decreased in steps so that motor torque tends to remain maximum during the accelerating period. Finally, under normal operation, the external resistance is fully cut off and the slip rings are short-circuited so that motor now develops full-load torque at low value of slip for which it is designed. Calculation of Resistance of elements (or sections) Consider one phase of the rotor of a wound-rotor induction motor, with resistance r2 and standstill leakage reactance x2. Let R1, R2 ,R3 …….Rn be resistance of the n resistance elements (or sections) and R1’ , R2’, R3’,….Rn’, Rn+1’ be the total resistance in each phase of the rotor circuit on 1st ,2nd ,3rd ,…,nth and (n+1)th stud respectively as shown in Figure 3.40(a) , such that R1= R1 + R2 + R3 +R4 +………+ Rn-1+Rn + r2 R2 = R2 + R3 +R4 +………+ Rn-1+Rn + r2 R3’ = R3 +R4 +………+ Rn-1+Rn + r2 : : Rn+1= r2. 161 Chapter One: Electromagnetic Principles (a) (b) Figure 3.41 (a) pertaining to the design of starter for wound-rotor IM; (b) Variation of input current with time Note that Figure 3.41(a) illustrates n-elements starter, n-section starter, (n+1) stud starter or n-step starter. For calculation of the section resistances, the following assumptions are made: i. During starting time , a constant load torque is assumed ii. The stator leakage impedance and its no-load current are neglected iii. and Stator current is taken to fluctuate between fixed limits I1max (maximum value) I1min (minimum value) as shown in Figure 3.41(b). At the time of start, the movable handle is at stud 1 and the rotor circuit resistance is R1’. When the supply is switched on to the stator, the input current shoots to I1max and its value is given by 162 Introduction to Electrical Machines I 1 max = V1 3.42 2 R 1' + x 2 S1 Note that at the time of start, slip S1 = 1 On first stud, R1’ remains in the circuit until the motor has started and the current has fallen from I1max to I1min . At the same time the slip falls from S1 to S2 ∴ I 1 min = V1 3.43 2 R1' + x 2 S2 As soon as I1min is reached at stud 1, resistance R1 is cut out by moving the handle from stud 1 to stud 2. During the notching process (the process of moving the handle from one stud to the next), the speed is assumed to remain constant, i.e. the slip remains as S2 but current at stud 2 becomes I1max as illustrated in Figure 3.41(b). ∴ I 1max = V1 3.44 2 R + x 2 S2 ' 2 At stud 2, the speed rises so that the slip becomes S3 and current decreases to I1min, ∴ I 1 min = V1 3.45 2 R + x 2 S3 ' 2 During the next notching process, i.e., at the third stud when R2 is cut out, ∴ I 1 max = V1 2 R3' + x 2 S3 3.46 and 163 Chapter One: Electromagnetic Principles ∴ V1 I1 min = 2 . 3.47 R '3 + x2 S4 ⋮ ⋮ and so on From Eqs. (3.42), (3.44), (3.46) etc, we get I1 max = = = V1 2 R 1' + x2 S1 V1 2 R '2 + x2 S2 V1 2 R '3 S3 = ................ + x2 From the above it follows that R1' R '2 R '3 = = S1 S2 S3 R' R' r = ......... = n = n +1 = 2 S n Sn +1 Sfl 3.48 Where Sfl = Sn+1 is the slip under normal operating conditions when all external resistance in rotor circuit is reduced to zero and the input current is I1max . From Eqs. (3.43), (3.45), (3.47), etc., we get R1' R2' R3' = = S 2 S3 S4 Rn' −1 Rn' Rn' = ......... = = = S n S n +1 S fl 3.49 From Eq. (3.48), we get 164 Introduction to Electrical Machines S 2 S3 S 4 = = S1 S 2 S3 S S S = ...... = n = n +1 = fl S n −1 Sn Sn = R '2 R '3 = R 1' R '2 = ....... = = R 'n +1 = R 'n −1 R 'n − 2 R '4 R '3 = R 'n R 'n −1 r = 2 = α (say) R 'n R 'n 3.50 Since the slip S1=1, the total resistance in rotor circuit on the first step (or first stud) from Eq. (3.48), is S1 r r2 = 2 S fl S fl R1' = 3.51 From Eq. 3.50, R '2 = R1' α , R '3 = R '2 α = R '2 α 2 R '4 = R '3α = R '2 α 2 = R1' α 3 ⋮ R 'n +1 = R1' α n r2 = R1' α n or Substituting the values of 3.52 R1’ from Eq. (3.51) in Eq. (3.52), we get r r2 = 2 α n Sfl or α n = Sfl or α = (Sfl 1 )n 3.53 Therefore, resistances of the sections are: R 1 = R 1' − R '2 = R 1' (1 − α ) 3.54 R '2 = R '2 − R '3 = R '2 (1 − α) = αR 1' (1 − α ) = αR 1 R 3 = R '3 − R '4 = R '3 (1 − α) = αR '2 (1 − α ) = α 2 R 1 165 Chapter One: Electromagnetic Principles Similarly R 4 = α3R 1 ⋮ R n = αn −1R1 3.55 The slip Sfl provided I1max=Ifl. In case I1max is different from Ifl, slip Sfl should be calculated accordingly and then from Eq. (3.53), α can be obtained. Once R1’is determined by using Eq.(3.51), first the resistance element R1 is obtained by using Eq.(3.54) and then R2, R3, R4……. Can be calculated from Eq. (3.55). Example 3.13 Calculate the values of resistance elements of a 4-step starter for a 3phase, 400-V, wound-rotor induction motor. The full-load slip is 3% and the maximum starting current is limited to its full-load value. Rotor resistance per phase is 0.015-Ω. Solution For a 4-step starter, there are 4-sections i.e. n=4. Here, Sfl=3%=0.03 1 n 1 4 α = ( S fl ) = 0.03 = 0.416 The total resistance of rotor circuit at the moment of starting R1' = r2 0.015 = = 0.5 Ω S fl 0.03 The resistances of various elements are: R1=R1’(1-α)= 0.5(1-0.416)=0.292 Ω R2=αR1=0.416 x 0.292 = 0.121 Ω R3=α2R1=αR2=0.416 x 0.121 = 0.051Ω R4=α3R1=α2R2=αR3=0.416 x 0.051 = 0.021Ω Checking : R’1=R1+R2+R3+R4+r2 =0.292+0.121+0.051+0.021+0.015= 0.5 Ω 3.14. APPLICATIONS OF POLYPHASE INDUCTION MOTORS For loads requiring low starting torques and substantially constant speeds, squirrelcage induction motor is the best choice, because of its ruggedness, simplicity, low cost and reduced maintenance charges. Squirrel cage motor may be designed with low rotor resistance or with high rotor resistance. As stated before, a high rotor resistance gives better starting conditions but poor running performance. On the 166 Introduction to Electrical Machines other hand, a cage motor with low rotor resistance gives poor starting conditions but better running performance. In view of this, the rotor-circuit resistance should be chosen judiciously at the design stage so that there is a compromise between its starting conditions and running performance. Squirrel cage motors with relatively low rotor resistance (full-load slip 3 to 5%) are used for fans, centrifugal pumps, most machinery tools, wood-working tools etc. Cage motors with relatively high rotor resistance (full-load slip 3 to 7%) are used for compressors, crushers, reciprocating pumps. Squirrel cage motors with still higher values of rotor-circuit resistance (full-load slip 7 to 16%) are used for intermittent loads like punching presses, shears, hoists, elevators etc. A wound-rotor induction motor is used for loads requiring severe starting conditions or for loads requiring speed control. A wound rotor induction motor is more expensive than a squirrel cage motor and also it requires more maintenance because of the brushes and slip rings. A wound-rotor motor, also called slip-ring motor, may be used for hoists, cranes, elevators, compressors etc. The relative advantages of cage motor over a wound-rotor motor of the same power rating are given below. i) A cage rotor requires considerably less conductor material than a wound rotor, consequently I2 R loss in cage rotor is less. Therefore, cage motor is a little more efficient than a wound-rotor motor. ii) Wound-motor construction requires slip rings, brushes, short-circuiting devices etc. As a result of it, a wound-rotor motor is costlier than a cage induction motor. iii) A squirrel-cage rotor has very small length of overhang; therefore, it has low rotor overhang leakage flux. This has the effect of resulting in low leakage reactance x2 for a cage rotor than for a wound rotor. Consequently, the diameter of circle diagram for a cage motor is greater than for a wound-rotor motor. This shows that a cage motor has more pull-out torque, greater maximum power output and better operating power factor as compared to a wound-rotor induction motor. iv) Cage motor is more rugged and requires no slip rings, brushes etc. therefore, its maintenance charges are low. v) Cage rotor can be cooled better because of its bare end-rings. The disadvantages of cage motor as compared to a wound rotor motor are its small starting torque for very large starting current and its poor starting power factor. In addition to it, the total energy lost during starting of cage motor is much more than with the wound-rotor motor and this fact is very important where frequent starting of large number of motors is required. 167 Chapter One: Electromagnetic Principles PROBLEMS ON INDUCTION MACHINES 3.1. A 3-phase, 4-pole, 50Hz, 7.5 kw induction motor runs at 1,440 rpm. Determine: i) the slip, ii) the slip-speed. Ans.i) 0.04 ; ii) 60 r.p.m 3.2. A 3-phase, 50 Hz induction motor has a full-load speed of 1440 rpm. For this motor , calculate the following: a) Number of poles b) Full-load slip and rotor frequency c) Speed of stator field with respect to i. Stator structure and ii. Rotor structure and d) Speed of rotor field with respect to i. Rotor structure ii. Stator structure iii. Stator field Ans. a) 4; b) 4% & 2Hz; c)1500rpm; 60rpm; d) 60rpm; 1500rpm; 0 3.3. A 3-phase, 4-pole alternator driven at 1500 rpm is supplying an induction motor which has full load speed of 960 r.p.m: Determine (i) the number of poles and (ii) the percentage slip. Ans.i) 6 ; ii) 4% 3.4. A 4-pole, 50Hz induction motor runs at 415 rpm Deduce the frequency of the curree rotor windings and the slip. Ans. 1.584 Hz ; 3.167% 3.5. A 4-pole. 50Hz induction motor has an emf in the rotor, the frequency of which is 2 Hz. Determine (i) the synchronous speed (ii) the slip iii) the speed of the motor. Ans.i)1500rpm;ii) 4%; iii) 1440 r.p.m 3.6. A 200-Kw, 3300-V, 6-pole, 50-Hz star-connected slip-ring induction motor has a star connected rotor. Stator to rotor turns ratio is 3.2.rotor resistance and leakage reactance are 0.1-Ω and 1-Ω respectively. Neglecting stator impedance ,find a) Current and torque at starting on rated voltage with slip-rings short circuited and b) The external resistance required to reduce the starting current to 50-A with across-the line starting. Compare starting torque under these conditions. Ans. a)185.14A; 1005.5N.m b) 35.681Ω ; 3.484Ω; 2628.8 N.m 3.7. A 3-phase, 4-pole, 1440 rpm, 50 Hz induction motor has star connected rotor winding , having a resistance of 0.2Ω per phase and a standstill leakage resistance of 1Ω 168 Introduction to Electrical Machines per phase. When the stator is energized at rated voltage and frequency, the rotor induced emf at standstill is 120-V per phase. a) Calculate the rotor current, rotor power factor and torque both at starting and at full-load and compare these results. b) If an external resistance of 1Ω per phase is inserted in rotor circuit, calculate rotor current, rotor power factor and torque at the time of starting. Ans. a) at starting:117.67A; 0.196 lagging; 52.36N.m. and at full-load: 23.53 A ; 0.98 lagging; 52.87 N.m. b) at starting: 76.82 A ; 0.768 lagging; 135.25 N.m. 3.8. In s 6-pole, 3-phase. 50Hz induction motor with a star-connected rotor, the rotor resistance per phase is 0.3Ω, the reactance at standstill is 1.5Ω per phase, and the emf between slip rings on open circuit is 175 V. Calculate : (i) slip (ii) rotor emf per phase (iii) rotor reactance per phase if full load speed is 960rpm. Ans.(i) 4% (ii) 4.05 V (iii) 0.06Ω 3.9. A 50 Hz, 8-pole induction motor has a full load slip of 4%. The rotor resistance and standstill reactance are 0-01Ω and 0.1Ω per phase respectively. Find the ratio of maximum to full load torque and the speed at which the maximum torque occurs. Ans.1.45; 675 rpm 3.10. An induction motor has a rotor resistance of 0.02Ω and a standstill reactance of 0.1Ω. How much external resistance must be added in the rotor circuit so as to get the maximum torque at starting. Ans. 0.08 Ω 3.11. For an induction motor, the starting torque is 1.6 times the full load torque and the rnaximum torque is 2 times the full load torque. Determine the percentage reduction in the rotor circuit resistance so that the full load slip is 0.04%. Neglect the stator impedance. Ans.70% 3.12. A 3-phase induction motor is fed with a power of 48 kw. If the stator losses are 1.2 kw, find the mechanical power developed and copper losses in the rotor, when the slip is 3%. Ans. 45.396kW; 1.404kW 3.13. The power input to a 500 V, 50 Hz, 6-pole, 3-phase induction motor running at 975 rpm is 40kW . The stator losses are 2 kw and the friction and windage losses total 2 kw. Calculate (i) the slip (ii) the copper loss (iii) the output power and (iv) the efficiency. Ans. i) 2.5%; ii) 950W; iii) 35.05kW ; iv) 87.6% 3.14. A 20Kw, 6-pole, 400V, 50Hz, 3-phase induction motor has a full-load slip of 0.02. If the torque lost in mechanical (friction & windage) form constitutes about 20 N.m., find the rotor ohmic loss, motor input and efficiency. The total stator losses are 900 Watts. Ans. 450W; 23.4kW; 85.47% 169 Chapter One: Electromagnetic Principles 3.15. A 3-phase , 400 V, 50Hz star-connected induction motor gave the following test results: No load: 400 V, 7.5A, 0.135 power factor Blocked rotor: 150 V, 35A, 0.44 power factor The ratio of standstill leakage reactance of stator and rotor is estimated as 2. If the motor is running at a speed of 900rpm, determine a) Net mechanical power output b) The net torque and c) Efficiency of the motor Assume stator to rotor copper losses to be equal. Ans. Psh=8648.98 W; Tsh=86.033 N.m;η= 86.23% 3.16. Design the 5-sections of a 6-stud starter for a 3-phase slip-ring induction motor. The full-load slip is 2% and the maximum starting current is limited to twice the fullload current. Rotor resistance per phase is 0.03-Ω. Ans. 0.356 Ω; 0.187 Ω; 0.098 Ω; 0.052 Ω;0.027 Ω 3.17. A squirrel induction motor has a full-load slip of 0.05. The motor starting current at rated voltage is 6 times its full-load current. Find the tapping on the auto-transformer starter which should give full-load at start. Also find the line current at starting in terms of full-load current. Ans. 74.5%; 3.33 Ifl 3.18. A 3-phase squirrel cage induction motor has a short-circuit current of 5 times the full-load current. Its full-load slip is 5%. Calculate the starting torque as percentage of full-load torque if the motor is started by a) Direct-on-line starter b) Start-delta and c) d) Auto-transformer starter, limiting the motor starting current to twice the motor full-load current. What is the percentage auto-transformer tapping under (c) case? Ans. a) 1.25%; b) 0.417; c) 0.20; d) 40% 3.19. Calculate the values of resistance elements of a 4-step starter for a 3-phase, 400-V, wound-rotor induction motor. The full-load slip is 3% and the maximum starting current is limited to its full-load value. Rotor resistance per phase is 0.015-Ω. Ans. 0.291 Ω; 0.121 Ω; 0.051 Ω; 0.021 Ω 170 Introduction to Electrical Machines CHAPTER FOUR DC MACHINCES 4.1. INTRODUCTION The dc machines are versatile and extensively used in industry. A wide variety of voltampere or torque-speed characteristics can be obtained from various connections of the field winding. Dc machines can work as generators, motors & brakes. In the generator mode the machine is driven by a prime mover (such as a steam turbine or a diesel engine) with the mechanical power converted into electrical power. In the motor mode, the machine drives a mechanical load with the electrical power supplied converted into mechanical power. In the brake mode, the machine decelerates on account of the power supplied or dissipated by it and, therefore, produces a mechanical braking action. There is almost no modern use of dc machines as generators although in the earlier stages of electrical power generator and distribution. D.C. generators were the principle means of supplying electrical power to industrial and domestic consumers. Presently, all the land based electrical power networks are a.c systems of generation, transmission and distribution. The almost universal use of ac systems is on account of their lower generation and transmission costs, higher efficiency (large bulk of ac power can be transmitted and distributed over wide areas and long distance at much higher voltages that are impossible in dc system), greater reliability on account of interconnection and control. No doubt, application like aerocrafts, ships and road mounted vehicles which are isolated from land based ac networks employ dc sources including dc generators and secondary batteries for power supply but the modern trend is to use ac generators with the dc supply being obtained by rectification with the help of static power rectifiers. D.C. generators are still being used to produce power in small back-up and stand-by generating plants driven by windmill and mountain streams (minihydro-electric plants) to provide uninterrupted power supply. Apart from dc generators, the dc motors are finding increasing applications, especially where large magnitude and precisely controlled torque is required. Such motors are used in rolling mills, in overhead cranes and for traction purpose like in forklift trucks, electric vehicles, and electric trains. They are also used in portable machine tools supplied from batteries, in automotive vehicles as starter motors, blower motors and in many control applications as actuators and as speed and position sensing device (tachogenerators for speed sensing and servomotors for positioning and tracing). 4.2. CONSTRUCTION OF DC MACHINES The dc machines used for industrial applications have essentially three major parts: a) Field system (stator); b) Armature (Rotor) and c) commutator All the components of the dc machine are illustrated in cut-away view of Figure 4.1. 171 . Chapter One: Electromagnetic Principles Figure 4.1 cut-away view of DC machines 1.shaft; 2.end-bearings; 3. Commutator; 4. brushes; 5.armature; 6. main-pole; 7.main-pole field winding; 8.frame; 9.end-shield; 10.ventilator; 11.basement; 12.bearings Field System The field system is located on the stationary part of the machine called stator. The field system is designated for producing magnetic flux and, therefore, provides the necessary excitation for operation of machine. Figure 4.2 shows that the main flux φ paths which starts from a North pole, crosses the air gap and then travels down to the armature core. There, it divides into two equal (φ/2) halves, each half enter the nearby South Pole so as to complete the flux. Each flux line crosses the air-gap twice. Some flux lines may not enter the armature; this flux, called the leakage flux, is not shown in Figure 4.2. Figure 4.1 Flux paths in a 6-pole dc machines 172 Introduction to Electrical Machines The stator of dc machines comprises of 1. Main poles: These poles are designed to produce the main magnetic flux 2. Frame: These provide support for the machine. In many machines the frame is also a part of the magnetic circuit. 3. Interpoles: These poles are designed to improve commutation conditions to ensure sparkles operation of machine. Figure 4.3 Main-pole Main-pole Poles are made of sheet steel laminations of 1.0 to 1.2mm thickness (nowadays the thickness becomes 0.4-0.5mm). The pole shoes support the field coils placed on the pole body and also spread the total flux over a greater area, thereby reduce the air gap reluctance and giving the desired flux distribution to limit saturation in the teeth of the l armature. (ℜ = ) .The poles are secured to the yoke by means of bolts. In small µA machines the pole are built of steel forgings, bolted directly to the yoke. In case of machines having compensating windings, the pole face is slotted to accommodate the windings. Yoke (Frame) The stator of a dc machines consists of a frame or yoke, and poles which support the field windings. The Frame or Yoke in addition to being a part of a magnetic circuit serves as mechanical support for entire assembly. Earlier, cast iron was used for the construction of yoke but it has been replaced by cast steel. This is because cast iron has saturation density of 0.8 Wb/m2 while saturation occurs in cast steel at density of approximately 1.5 Wb/m2. Thus, the cross section of the cast steel frame or yoke is half that of iron cast and hence cast steel is used in case it is desired to reduce the weight of machine. Fabricated steel yokes are commonly used, as they are economical and have consistent magnetic & mechanical properties. For very small sized machines it may still be advantageous to use cost iron frames but for medium and large sizes rolled steel is used. Interlopes In addition to the main poles, modern direct current machines are also provided with interlopes with windings on them in order to improve commutation under loaded 173 . . Chapter One: Electromagnetic Principles conditions. They are arranged midway between the mains poles and are bolted to the yolk. Laminated interlopes are used in machine with sever commutation problems. For small and medium size machines they could be solid. Armature The armature is the rotating part (rotor) of the dc machine where the process of electromechanical energy conversion takes pace. The armature is a cylindrical body, which rotates between the magnetic poles. An isometric view of a small size armature structure is shown in Figure 4.4 (a). The armature and the field system are separated from each other by an air gap. The armature consists of: Armature core with slots and Armature winding accommodated in slots The purpose of the armature is to rotate the conductors in the uniform magnetic field and to induce an alternating e.m.f in its winding. The armature core is normally made from high permeability silicon-steel laminations of 0.4 to 0.5mm thickness, which are insulated from one another by varnish or ceramic insulation. The use of high grade steel is to keep hysteresis loss low, which is due to cyclic change of magnetization caused by rotation of the core in the magnetic field and to reduce the eddy current in the core which are induced by the rotation of the core in the magnetic field. In order to dissipate the heat produced by hysteresis and eddy current losses etc, ventilating ducts are provided. By the fanning action of the armature, air is drawn in through these ducts, thus producing efficient ventilation. In the armature core of small diameters, circular holes are punched in the center of the laminations for the shaft (Figure 4.4(b)). (a) (b) Figure 4.4 (a) Isometric view of armature; (b) armature lamination Commutator It is mounted on the rotor of a dc machine and it performs with help of brushes a mechanical rectification of power: from ac to dc in case of generators and dc to ac in case of motors. The ends of armature coils are connected to the commutator, which together with the brushes rectifies the alternating e.m.f induced in the armature coils and helps in the collection of current. It is cylindrically shaped and is placed at one end of the 174 Introduction to Electrical Machines armature. The construction of the commutator is quite complicated because it involves the combination of copper, iron and insulating materials. The connection of armature conductors to the commutator is made with the help of risers. The risers connecting the segments to the armature coils are made of copper strips for large machines. The outer end of the riser is shaped so as to form clip into which the armature conductors are soldered. The commutator bars are built of a small wedge shaped segments of high conductivity hard drawn copper insulated from each other by mice or micanite of about 0.8mm thickness. The commutator segments are assembled over a steel cylinder. Vshaped grove is provided at each end of the segments to prevent them from flying away under the action of centrifugal force. Threaded steel rings are used to tighten the various components together (see Figure 4.5). The commutator assembly is force and press fitted on the shaft. Satisfactory performance of dc machines is dependent under good mechanically stability of the commutator under all conditions of speed and temperature within the operating range. A mechanically unstable commutator manifests itself in a pool commutation performance and results in unsatisfactory bush life. Riser Commutator segment Insulating V-groove Riser Insulating V ring Thread bolt End ring Copper commutator bars End ring Thread bolt (a) (b) Figure 4.5 (a) cut-away view of commutator; (b) commutator segment 4.2.4. Brushes and Brush Holder Brushes are needed to collect the current from the rotating commutator or to lead the current to it. Normally brushes are made up of carbon and graphite, so that while in contact with the commutator, the commutator surface is not spoiled. The brush is accommodated in the brush holder where a spring presses it against the commutator with pressure of 1.5 to 2.0 Ncm2 (see Figure 4.6). A twisted flexible copper conductor called pigtail securely fixed in to the brush is used to make the connection between the brush and its brush holder. Normally brush holders used in dc machines are of box type. The numbers of brush holders usually equal to the number of main poles in dc machines. Figure 4.6 Brush and brush holder 175 Chapter One: Electromagnetic Principles 3. PRINCIPLE OPERATION OF DC GENERATOR An electrical generator is a machine, which converts mechanical energy into electrical energy. The energy conversion is based on principle of dynamically induced emf, whenever a conductor cuts magnetic flux, dynamically induced emf is produced in it (Faraday’s law). This emf causes a current to flow if the conductor is closed. The basic essential parts of an electrical generator are: A magnetic Field and A conductor or conductors, which can so move as to cut the flux. Figure 4.7 shows the schematic diagram of a simple machine consists of a coil ABCD rotating in the magnetic field of a strong permanent magnet or powerful electromagnet. The magnetic lines in the space between N and S poles are directed from the North Pole N to the South Pole S as shown in Figure 4.7. The ends of the coil ABCD are connected to two copper rings R1 and R2, fixed on the shaft. Two brushes B1 and B2 connected to the external load circuit make contact with the copper rings R1 and R2 respectively. B C N S A D + B1 R2 B C N S D A + R1 B1 _ B2 R2 Load (a) R1 _ B2 Load (b) Figure 4.7 (a) and (b) Schematic diagram of a simple dc generator Let the coil be rotated in an ACW, with constant surface speed v[m/s] in relation to the magnetic field . According to Faraday’s laws of electromagnetic induction, an emf will be induced in the rotating coil and is given by e = Blv volts As l and v are constant for particular case, e = cons tan t × B volts 176 Introduction to Electrical Machines Hence under the given conditions, the change in the magnitude of induced e.m.f with time depends upon the magnetic flux density distribution under the poles. It may be assumed neglecting harmonics it is a sine wave distribution. The direction of the induced emf in this case can be determined by Fleming’s right hand rule as shown in Figure below. Hence the conductor AB of the coil ABCD moves downward and CD moves upward, the direction of the induced emf in the coil is along DCBA as shown in Figure 4.7(a). The current in the external remains the same half a revolution of the coil starting from its vertical position. ν ν φ S N Motion e Flux EMF Figure 4.8 Right-hand rule to determine the direction of induced emf. Similarly, in the next half of the revolution, the direction of the induced emf is reversed and hence the current flows from brush B2 to B1 as shown in Figure 4.7(b).The magnitude of current in the external circuit also varies with time as per sine law; i.e. its magnitude is not constant with time. If the machine has P poles and the armature rotates at N revolutions per minute, then the frequency of the induced emf in the armature is, f = PN , Hz 120 The above discussion clearly indicates that the e.m.f induced in the armature of a dc generator is of alternating nature, alternating with frequency of f hertz depending upon the number of poles in the machine and the speed of the armature. However, the output voltage or the current of dc generator must be unidirectional and that too of a constant value. Thus to compel the above alternating current to flow in one stipulated direction through the external load circuit, the dc machine is furnished with a special device called the commutator. Figure 4.9 shows that the coil ABCD connected to a ring commutator split in two halves R1 and R2 well insulated from each other. The rings of the commutator are so arranged that during half the revolution of the coil, each half ring remain in contact with a particular brush. Figure 4.9(a) wile during the next half revolution, when the current is reversed, the same half ring is in contact with other brush as shown in Figure 4.9 (b). 177 Chapter One: Electromagnetic Principles B C C N S A N B2 R1 R2 S D D B1 B A B1 B2 R2 Load R1 Load (b) Figure 4.9 coil ABCD connected to a ring commutator Current As a result, current in the external load circuit remains in the same direction. The nature of the variation of current in the external load current with the rotation of the coil, i.e. with time, has been shown in Figure 4.10. Such unidirectional current or emf which fluctuates between maximum and zero values is quite inconvenient for practical purposes. 0 900 1800 2700 3600 4500 5400 θ Figure 4.10 Unidirectional current wave shape To overcome the above difficulty of the nature of a move shape, consider two coils whose planes are inclined to each other at an angle of 900 and divided the commutator ring mounted on the same shaft into four parts. The leads of each coil are connected to the two diametrically opposite parts of the ring. In such case, the e.m.f or current wave shape due to either coil will be of the same type but 900 out of phase, i.e. when the current in one reaches maximum value, the current in the other coil has zero value as shown in Figure 4.11. The resultant current in the external circuit due to the rotation of the two coils simultaneously at the same speed can be obtained by superimposing the two current waves. Hence, the resultant current wave shape is less fluctuating. Similarly, if a large number of coils are provided on the rotating armature of the machine with double the number of commutator segments, the wave shape of the resultant current or 178 . . Introduction to Electrical Machines the emf will practically be parallel to the time axis and hence constant with respect to time. θ Figure 4.11 Resultant current wave shape 4.4. TYPES OF DC GENERATORS The field winding and the armature winding can be interconnected in various ways to provide a wide variety of performance characteristics. This can be taken as outstanding advantages of a dc machines. A dc machine can work as an electromechanical energy converter only when its field winding is excited with direct current, except for small dc machines employing permanent magnets. According to the method of their field excitation dc generators are classified into the following group: a) Separately excited and b) Self excited DC machines may have one or more field windings and their method of excitation, determines the performance characteristics of the dc machine. Separately Excited Its field winding consists of several hundreds turns of fine wire and is connected to a separate or external dc source i.e. field winding are energized from an independent external sources of dc current. The voltage of the external dc source has no relation with the armature voltage, i.e. the field winding energized from a separate supply, can be designed for any convenient voltage. Important relationships Ia = I L E g = VL + I a Ra Pdev = E g ⋅ I a Pdel = VL ⋅ I L Figure 4.12 separately excited dc machines Self Excitation 179 Chapter One: Electromagnetic Principles When the field winding is excited by its own armature, the machines is said to be a self excited dc machine. In these machines, the field poles must have a residual magnetism, so that when the armature rotates, a residual voltage appears across the brushes. This residual voltage should establish a current in the field winding so as to reinforce the residual flux. According the connection of the field winding with the armature winding, a self-excited dc machine can be sub-divided as follows: Series Excitation The field winding consists of a few turns of thick wire and is connected in series with the armature. In other words, the series field current depends on the armature current and in view of this; a series field may be called a current operated field. Important relationships I a = I se = I L E g = VL + I a (Ra + Rse ) Pdev = E g ⋅ I a Pdel = VL ⋅ I L Figure 4.13 Series excited dc machine Shunt Excitation The field winding consists of a large number of turns of fine wire and is connected in parallel (or in shunt) with the armature. Therefore the voltage across the armature terminals and the shunt field is the same and it is for this reason that a shunt field may be called voltage operated field. Important relationships V I sh = sh Rsh L I a = I sh + I L E g = VL + I a Ra Pdev = E g ⋅ I a Pdel = VL ⋅ I L Figure 4.14 Shunt excited dc machine Remember that series field and shunt field windings are characterized by low and high resistance respectively. In some application, a shunt excited winding may be replaced by a separately excited winding. 180 Introduction to Electrical Machines Compound Excitation A compound excitation involves both series-exited winding and the shunt-excited winding. From the view point of connections, a dc compound machine may have shortshunt connection or a long shunt connection. In short shunt connection of Figure 4.15 (a) the shunt field or voltage excited winding is connected across the armature terminals. In long-shunt connection, the shunt field is connected across the series connection of the armature and series winding or the machine or line terminals as shown in Figure 4.15 (b). However there is appreciable difference in the operating characteristics of short-shunt and long shunt. The choice between the two types depends on mechanical considerations of connections or reversing switches. Important relationships I se = I L E g − I a Ra VL + I se Rse I sh = = Rsh Rsh I a = I sh + I L E g = VL + I a Ra + I L Rse Pdev = E g ⋅ I a (a) Pdel = VL ⋅ I L Important relationships I a = I se E g − I a (Ra + Rse ) VL I sh = = Rsh Rsh I a = I sh + I L E g = VL + I a (Ra + Rse ) (b) Pdev = E g ⋅ I a Pdel = VL ⋅ I L Figure 4.15 DC compound machine connections for a) short-shunt and b) long shunt In a compound machine, the magnetic flux produced by the shunt field is stronger than the series field. When series field aids the shunt field, so that the resultant air gap flux per pole is increases, then the machine is said to be cumulatively compounded. In Figure 4.16 (a) the direction of arrows corresponds to the direction magnetic flux produced by shunt and series field windings. As the two arrows are in the same direction in Figure 4.16 (a), this Figure is for a cumulatively compounded dc machine. 181 Chapter One: Electromagnetic Principles (a) (b) Figure 4.16 compound excited dc machine (a) cumulative and b) differential On the other hand if series field opposes the shunt field so that the resultant air gap flux per pole is decreased, the machine is called a differentially compounded dc machine as shown in Figure 4.16(b). In Figure 4.15(a), each pole of compound machine is shown to possess shunt and series field windings. Figure 4.17(a) illustrated how these windings are arranged on one pole of a dc machine. In Figure 4.17 shunt field coil is placed near yoke and series field coil near the pole shoe just for sake of clarity. (a) (b) Figure 4.17 Series and shunt field windings on one pole of dc compound machine Actually physical arrangement of these coils is shown in Figure 4.17 (b). It is seen that first shunt field coil is wound around the pole body and over it is then wound the series field coil. The reasons for placing the series field coil outside are: i) convenience in the construction and ii) for its better cooling 5. EMF EQUATION OF DC GENERATOR Let φ = flux per pole in Weber Z = total number of armature conductors = Number of slots × Number of conductors per slot 182 ) Introduction to Electrical Machines P = Number of poles a = number of parallel paths in armature N= armature rotation in revolutions per minute (rpm) E = emf induced in any parallel path in armature Generated emf, Eg= emf generated in one of the parallel path dφ Average emf generated / conductor = , volt dt Now, flux cut / conductor in one revolution, dφ = φP ,Wb N Number of revolution / second = , sec ond 60 Hence according to Faraday’s law of electromagnetic induction dφ φPN emf generated / conductor = = , volt dt 60 For wave winding Number of parallel path a = 2 Z 2 φPN Z φZPN emf generated / path = ⋅ = volt 60 2 2 × 60 For lap winding Number of conductors (in series) in one path = ∴ Number of parallel path a = P Number of conductors (in series) in one path = ∴ emf generated / path = φPN Z 60 ⋅ P = φZN 60 Z P volt In general, the Generated emf Eg = Where, a =2 a=P where, K a = φZN P × volt 60 a for wave winding for lap winding Eg = Ka φ N ZP is machine constant. 60 ⋅ a Example 4.1 A dc shunt generator supplies a load of 10 kW at 220 V through feeders of resistance 0.1Ω. The resistance of armature and shunt field windings is 0.05 Ω and 100 Ω respectively. Calculate, (i) terminal voltage, (ii) shunt field current and (iii) generated emf. Solution Load supplied , Pdel = 10 kW 183 Chapter One: Electromagnetic Principles = 10 × 103 W Voltage at the load terminals = 220 V Thus load current, P 10 × 103 I L = del = V 220 = 45.5A Resistance of the feeders = 0.1 Ω Voltage drop in the feeders = IL × 0.1 = 45.5 × 0.1 = 4.55 V Terminal voltage across the armature terminals, V′ = 220 + 4.55 = 224.55 V Shunt field current, I sh = V' 224.55 = R sh 100 = 2.25A Generated emf, Eg = V' + Ia R a = 224.55 + 45.5 × 0.05 = 226.82 V Example 4.2 A 4-pole dc shunt generator with lap-connected armature supplies a load of 100 A at 200 V. The armature resistance is 0.1 Ω and the shunt field resistance is 80 Ω. Find (i) total armature current, (ii) current per armature path, and (iii) emf generated. Assume a brush contact drop of 2V. Solution Terminal voltage across the armature terminals, V = 200 V Shunt field resistance, Rsh = 80 Ω Shunt field current, I sh = V 200 = R sh 80 = 2.5A 184 Introduction to Electrical Machines Load current, IL = 100 A Armature current, Ia = IL + Ish = 100 + 2.5 = 102.5 A ii) Shunt generator is lap-wound, as such the number of parallel circuits in the armature winding is equal to the number of poles. Thus number of parallel circuits a = 4 Total armature current, Ia = 102.5 A Thus the current per armature path, 102.5 4 = 25.625A = iii) Emf generated, E g = V + I a R a + Vbd = 200 + 102.5 × 0.05 + 2 = 212.25 V Example 4.3 A short shunt compound generator supplies 200 A at 100 V. The resistance of armature, series field and shunt field is respectively, 0.04, 0.03 and 60 Ω. Find the emf generated. Solution Terminal voltage across the load, VL = 100 V Load current, IL = 200 A Resistance of series field winding Rse = 0.03 Ω Voltage drop in series field winding = IL Rse = 200 ×0.03 =6V Terminal voltage across the armature, V = VL + IL Rse = 100 + 6 = 106 V Shunt field current, 185 Chapter One: Electromagnetic Principles V 106 = R sh 60 = 1.77A I sh = Armature current, Ia = IL + Ish = 200 + 1.77 = 201.77 A Generated emf, E g = VL + I L R se + I a R a = 100 + 6 + 201.77 × 0.04 = 114.07 V Example 4.4 The armature of a four pole, wave wound shunt generator has 120 slots with 4 conductors per slot. The flux per pole is 0.05 Wb. The armature resistance is 0.05 Ω and the shunt field resistance 50 Ω. Find the speed of the machine when supplying 450 A at a terminal voltage of 250 V. Solution Terminal voltage, VL = 250 V Load current, IL = 450 A Shunt field resistance, Rsh = 50 Ω Shunt field current, VL 250 = R sh 50 I = 5 .0 A I sh = Armature current, Ia = IL + Ish = 450 + 5 = 455 A Armature resistance, Ra = 0.05 Ω Generated emf, E g = VL + I a R a = 250 + 455 × 0.05 = 272.75 V Generated emf, Eg = PφNZ V 60 × a 186 Introduction to Electrical Machines Number of poles, P == 4; Flux per pole, φ = 0.05 Wb; Number of slots on armature = 120; Conductors per slot = 4 Thus total number of conductors on armature = 120 ×4 = 480 As the armature is wave wound, number of parallel paths, a=2 Substituting these in the above equation, 272.75 = 4 × 0.05 × N × 480 60 × 2 Speed of rotation, 272.75 × 60 × 2 4 × 0.05 × 480 = 341 rpm N= Example 4.5 A long-shunt compound generator supplies a load at 110 V through a pair of feeders of total resistance 0.04 Ω. The load consists of five motors, each taking 30 A and a lighting load of 150 bulbs each of 60 W. The armature resistance is 0.03 Ω, series field resistance 0.04 Ω and shunt field resistance, 55 Ω. Find, (i) load current, (ii) terminal voltage, and (iii) emf generated. Solution i) Current drawn by each motor = 30 A Thus current drawn by five motors = 30 × 5 = 150 A Total lighting load = 150 × 60 = 9000 W Current taken by the lighting load = 9000 = 82 A 110 Hence, total load current = 150 + 82 = 232 A ii) Voltage at the terminals of the load = 110 V Total resistance of the feeders = 0.04 Ω Current through the feeders = 232 A Voltage drop in feeders = 232 × 0.04 = 9.28 V Terminal voltage across the generator terminals, V = VL + drop in feeders = 110 + 9.28 = 119.28 V 187 Chapter One: Electromagnetic Principles iii) Resistance of shunt field, Rsh = 55 Ω Current in shunt field winding, I sh = V 119.28 = = 2 .2 A R sh 55 Current in the armature winding, Ia = IL +Ish = 232 + 2.2 = 234.2 A Current in the series field winding, Ise = Ia = 234.2 A Total resistance of armature and series field winding = Ra + Rse = 0.03 + 0.04 = 0.07 Ω Generated emf , E g = V + I a (R a + R se ) = 119.28 + 234.2 × 0.07 = 135.67 A 7. ARMATURE REACTION By armature reaction is meant the effect of magnetic field set up by armature current on the distribution of flux under main poles. In other words armature reaction is meant the effect of armature ampere-turns upon the value and the distribution of the magnetic flux entering and leaving the armature core. The armature magnetic field has two effects: 1. It demagnetizes or weakens the main flux & 2. It cross –magnetizes or distorts it Let us illustrate (demonstrate) these two effects of armature reaction for 2-pole d.c generator. For better understanding let us see three cases. Case-I: Figure 4.18 shows the distribution of magnetic flux when there is no current (Ia=0) in the armature conductors,. For this case a) The distribution magnetic flux symmetrical with respect to the polar axis. b) The magnetic neutral axis or place (M.N.A.) coincides with geometrical neutral axis or plane (G.N.A) M.N.A may be defined as the axis along which no-emf is produced in the armature conductors because they move parallel to the lines of flux or M.N.A. is the axis which is perpendicular to the flux passing through the armature. In this case, brushes are always placed along M.N.A and the mmf (Fm) producing the main flux is directed perpendicular to M.N.A. 188 Introduction to Electrical Machines Figure 4.18 Magnetic flux distribution due to the main field poles only . M.N.A G.N.A Case-II: Figure 4.19, shows the field (or flux) set up by the armature conductors alone, when current carrying the field coils being unexcited (If = 0). The direction of the armature current is the same as it would be when the generator is loaded & determined by Fleming’s Right-hand rule. Under this case, the magnetic fields, which are set up by armature conductor are symmetrical to G.N.A. and the mmf of the armature conductor (depending on the strength of Ia) is shown separately both in magnitude and direction by the Vector OFa which is parallel to G.N.A Figure 4.19 Magnetic flux distribution due to the armature excitation only In the above two cases, we considered the main mmf and armature mmf separately, as if they existed independently, which is not the case in practice under actual load conditions. The two cases exist simultaneously in generator as will be shown in case III. Case-III: Figure 4.21 shows the combination of case I & II. In this case the main flux through the armature is no longer uniform and symmetrical about the pole-axis, rather it has been distorted. The flux is seen to be crowded at the trailing pole tips but weakened or thinned out at the leading pole tips (the pole tip which is first met during rotation by armature conductors are known as the leading pole tip and the other as trailing pole tip). 189 Chapter One: Electromagnetic Principles In Figure 4.20 is shown the resultant mmf OFR which is found by vectorally combining OFm and OFa. θ θ Figure 4.20 combined magnetic flux distribution due to armature and field The new position of M.N.A which is always perpendicular to the resultant mmf vector OFR is shown in Figure 4.20. Due to the shift of M.N.A, say through an angle θ, brushes are also shifted so as to lie along the new positions of M.N.A. Due to this brush shift (or forward, leads), the armature conductors and hence the armature current is redistributed, i.e. some armature conductors, which were earlier under the influence of N-pole, come under the influence of S-pole and vice-versa. Let us see this condition with help of Figure 4.21. Now the armature mmf is now represented by vector Fa that is no vertical but is inclined by angle θ to the left (Figure 4.21). This vector can be resolved into two rectangular components, Fd parallel to polar axis and Fc perpendicular to this axis, we find that 1. Component Fc is at right angle to the vector OFm (Figure 4.18) representing the main mmf it produces distortion in the main field and is hence called the crossmagnetizing or distorting component of the armature Reaction. 2. Component Fd is in direct opposition to OFm, which represents the main mmf. It exerts a demagnetizing influence on the main pole flux. Hence, it is called the demagnetizing or weakening component of the armature reaction. From the above discussion we can conclude that: 1. The flux across the air gap is no longer uniform, but weakens under the leading pole tips and strengthened under the trailing pole tips. (The pole tip which is first met during rotation by armature conductors is known as the leading pole tip and the other as trailing pole tip).Due to this the resultant mmf given rise to decreases flux. So that emf in the armature under loaded conditions is somewhat less than that of under no-load conditions. 190 Introduction to Electrical Machines 2. The brushes should be shifted in the direction of rotation to avoid a heavy shortcircuit current and sparking at brushes. 3. The field distortion cause, an increase in the iron losses as compared its no-load value because of increases peak value of flux density in the tooth. θ Figure 4.21 the demagnetizing and cross-magnetizing components of armature mmf 8. COMMUTATION The armature conductors carry current in one direction when they are under the influence of N-pole and in opposite direction when they are under S-pole. So when the conductors come under the influence of the S-pole from the influence of N-pole, the direction of flow of current in them is reversed. This reversal of current in a coil will take place when the two commutator segments to which the coil is connected are being short circuited by brush. The process of reversal of current in a coil is termed as commutation. The period during which the coil remains short-circuited is called commutation period, Tc. This commutation period is very small of the order of 0.001 to 0.003s. If the current reversal i.e. the changes from+ I to ZERO and then to –I is completed by the end of short circuit or commutation period, the commutation is Ideal. If current reversal is not completed by that time, then sparking is produced between the brush and the commutator, which results in progressive damage to both. Let us discuss the process of commutation in more detail with help of Figure 4.22 where ring winding has been used for simplicity. (a) (b) 191 Chapter One: Electromagnetic Principles (c) Figure 4.22 commutation process In Figure 4.22 (a) Coil B carries current in clock wise direction but it is about to be short circuited, because brush is about to come in touch with commutator segment “a”. Figure 4.22 (b) shows the coil B in the middle of its short-circuited period and it is observed that current can reach the brush with out passing through coil B, so coil B has no current. Figure 4.22(c) depicts the moment when coil B is almost at the end of commutation or short-circuit period and the current in the coil has to be reversed. During the period of short circuit, period of commutation, the current in the shortcircuited coil should be reversed to full value. Rapid reversal of current in the short circuited coil does not attain its full value in the reverse direction by the end of short circuit. The failure of current in the short-circuited coil to reach the full value in reverse direction by the end of short circuit is the basic cause of sparking at the commutator ( as shown in Figure c current jump from commentator segment “ b” to brush in the form of an arc). The reason for sparking at brushes of dc machine is due to reactance voltage (self-inducted emf), which sets-up by rapid reversal of current in the armature coil and tend to delay the current reversal in the coil. Because coil B has some inductance L, the change of current ∆I in a time ∆t induce a voltage L ∆I in the coil. According Lenz’s law, the direction of this voltage is ( ∆t ) opposite to the change ∆I that is causing it. As a result, the current in the coil does not completely reverse by the time the brushes move from segment b to a. Figure 4.23 Commutation in Coil B 192 Introduction to Electrical Machines 4.7.1. Methods of improving commutation There have been adapted two practical ways of improving commutation i.e. of making current reversals in the short-circuited coil as sparkles as possible. The two methods are: i. resistance commutation and ii. emf commutation. This method is achieved by By replacing low-resistance copper brush by comparatively high resistance carbon brush (approximately 12 times that of copper). However , it should be clearly understood that the main causes of the sparking commutation is the self induced emf ,so brushes alone do not give a sparkles commutation, though they do help in obtaining it. ii. By the help of inter poles, neutralize the self- reactance voltage by producing reversing emf. In this method, arrangement is made to neutralize the reactance voltage by producing a reversing emf in the short-circuited coil under commutation. This reversing emf, as the name shows, is an emf in opposition to the reactance voltage and if its value is made up equal to the latter, it will completely wipe it off, thereby producing quick reversal of current in shortcircuited coil which will result in sparkles commutation. Interpoles or Compoles These are small poles fixed to the yoke and spaced in between the main poles. They are wound with comparatively few heavy gauge copper wire turns and are connected in series with the armature so that they carry full armature current. Their polarity, in the case of a generator, is the same as that of the main pole ahead in the direction of rotation as illustrated in Figure 4.24 (a). For a motor, the polarity of the interpole must be the same as that of the main pole behind it in the direction of rotation as shown in Figure 4.24 (b) . S N S N S N 4.7.2. i. N S (a) (b) Figure 4.24 polarity of Interpoles (a) in generator mode; (b) in motor mode The function of interpole is two fold: 193 Chapter One: Electromagnetic Principles i) As their polarity is the same as that of the main pole ahead, the induced an emf in the coil (under commutation) which helps the reversal of current. The emf induced by the compoles is known as commutating or reversing emf. The commutation emf neutralizes the reactance emf thereby making commutation sparkles. As interpoles carry armature current, their commutating emf is proportional to the armature current. This ensures automatic neutralization of the reactance voltage, which is also due to armature current. ii) Another function of the interpoles is to neutralize the cross-magnetize effect of armature reaction. Hence, brushes are not to be shifted from the original position. Neutralization of cross- magnetization is automatic and for all loads because both are produced by the same armature current. 4.7.3. Compensating winding The effect of cross-magnetization can be neutralized using compensating winding. These are conductors embedded in pole faces, connected in series with the armature windings and carrying current in an opposite direction to that flowing in the armature conductors under the pole face. Once cross-magnetization has been neutralized, the M.N.A does not shift with the load and remains coincident with the G.N.A. at all loads. Figure 4.25 compensating windings CHARACTERISTICS OF DC GENERATORS The behavior of various types of dc generators can be studies by their characteristic. The three most important characteristic curves of a dc generator are: 1. Magnetization characteristic or open-circuit characteristic (O.C.C.) shows the relationship between the field current If and the generated emf Eg at no load and at constant given speed. 2. External characteristic-shows the relationship between the terminal voltage V across the load and the current IL flowing in the external load circuit. 3. Internal characteristic –shows the relationship between the emf generated E (after allowing for demagnetizing effect of armature reaction) at load and the armature current Ia. Magnetization characteristic (O.C.C.) 194 Introduction to Electrical Machines The emf generated in the armature winding of a dc machine under no load condition is given by Eg = Pφ NZ 60 a P, Z and a are constants for a particular generator, hence at constant given speed. Eg α φ ∴ The generated emf is directly proportional to the flux per pole (speed being constant), which in turns depends upon the field current If The characteristic curve plotted between generated emf Eg and the field current If at constant speed of rotation is called the magnetization curve or O.C.C. of the dc generator. The magnetization characteristics of a separately excited generator or shunt generator can be obtained as explained below. Figure 4.26 Circuit diagram for determination of magnetization characteristics Figure 4.26 shows the connections of the generator and the field for determination of O.O.C. A potentiometer arrangement has been made to supply the field winding so that the field current can be varied over a wide range by moving the contact K. Ammeter indicate the field current and voltmeter indicate the generated emf. The field current is increased in steps from zero to maximum and the corresponding value of If and Eg are noted down at each step. On plotting these results, a curve of the form shown in Figure 4.27 is obtained. Eg C D OA=Residual emf (due to residual Magnetism) B AB- Unsaturated Region (Straight line) BC- Knee of the curve ( Operating Region) CD- Saturated Region A O Ish Figure 4.27 Magnetization curve or O.C.C. On analyzing the curve in Figure 4.27, it is observed that a small emf OA is generated by the generator, even when the field current is zero. The reason for this generated emf is the residual magnetism in the poles. This emf which is due to residual magnetism is 195 Chapter One: Electromagnetic Principles normally 1 to 5% of the normal voltage of the generator. The magnetization curve of a shunt generator and a series generator can also be obtained in a similar manner. However, a shunt generator differs compared to separately excited one, in the manner that the field current in shunt generator is due to the generated emf only, where as the field current is independent of the generated emf in case of separately. This magnetization curve is of grate importance because it represents the saturation level in the magnetic system of the dc machine for various value of the excitation mmf (current). 10. VOLTAGE BUILD-UP PROCESS IN SHUNT GENERATOR In the shunt or self-excited generator the field is connected across the armature so that the armature voltage can supply the field current. Under certain conditions, to be discussed here, this generator will build up a desired terminal voltage. If the machine is to operate as a self-excited generator, some residual magnetism must exist in the magnetic circuit of the generator. Figure 4.28 shows the magnetization curve of the dc machine. Also shown in this Figure 4.28 is the field resistance line, which is a plot of Rf If versus If. Figure 4.28 voltage build-up process in self excited dg generator A simplistic explanation of the voltage build-up process in the self-excited dc generator is as follows: Assume that the field circuit is initially disconnected from the armature circuit and the armature is driven at a certain speed. A small voltage, Ear will appear across the armature terminals because of the residual magnetism in the machine. If the switch SW is now closed (Figure 4.29) and the field circuit is connected to the armature circuit, a current will flow in the field winding. If the mmf of this field current aids the residual magnetism ,eventually a current If1 will flow in the field circuit. 196 Introduction to Electrical Machines Figure 4.29 schematic diagram of a shunt or self–excited dc generator The buildup of this current will depend on the time constant of the field circuit. With If1 following in the field circuit, the generated voltage is Ea1 (from the magnetization curve) but the terminal voltage is Vt = Ifl RF < Eal. The increased armature voltage Eal will eventually increase the field current to the value If2, which in turn will build up the armature voltage to Ea2. This process of voltage buildup continues. If the voltage drop across Ra is neglected (i.e. Ra << Rf), the voltage builds up to the value given by the crossing point P of the magnetization curve and the field resistance line. At this point Ea = If Rf = Vt (assume Ra is neglected), and no excess voltage is available to further increases the field current. In the actual case, the changes in If and Ea take place simultaneously and the voltage buildup follows approximately the magnetization curve, instead of climbing the flight of stairs. Figure 4.30 shows the voltage buildup in the self-excited dc generator for various field circuit resistances. At some resistance value Rf3, the resistance line is almost coincident with the linear portion of the magnetization curve. This coincidence condition results in an unstable voltage situation. This resistance is known as the critical field circuit resistance. If the resistance is greater than this value, such as Rf4, buildup (Vt4) will be insignificant on the other hand, if the resistance is smaller than this value, such as Rf1 or Rf2, the generator will build up higher voltages (Vt1, Vt2). To sum up, four conditions are to be satisfied for voltage buildup in a self-excited dc generator. 1. Residual magnetism must be present in the magnetic system. 2. Field winding mmf should aid the residual magnetism. 3. Field circuit resistance should be less than the critical field circuit resistance. 4. The speed at which the armature is rotating should be greater than the critical speed. Figure 4.30 effect of field resistances on voltage build-up process 197 Chapter One: Electromagnetic Principles 11. EXTERNAL CHARACTERISTICS The external characteristics of a dc generator express the relationship between the terminal voltage and the load current at a constant speed and with the field current keeping the same as under the no load condition. The shape of this curve depends upon: i. ii. iii. iv. The armature reaction voltage drop in the armature winding, series , inter pole and compensating windings voltage drop at the brush contact( 0.8- 1,0-V per brush ) and The drop in terminal voltage due to (i) and (ii) results in a decreased field current which further reduces the induced emf. Separately Excited Generator In separately excited generators, the field current is independent of the load current, so that if there were no armature reaction and no voltage drop in various windings the terminal voltage will be equal to the generated emf and would be constant for various values of load current as indicated by curve I in Figure 4.31. Figure 4.31 external characteristics of separately excited generator As the generator is separately excited, the armature current is equal to load current. However, the armature reaction will cause a decrease in the voltage, which depends upon the load current. As such considering the effect of armature action only, the curve of terminal voltage Vs armature current will be slightly drooping as shown by curve II in Figure 4.31. Curve II of the generator, which takes into account the effect of armature reaction, gives to a different scale the emf induced in the armature and thus, it is normally called the internal characteristics of the generator. The curve of terminal voltage Vs load current or armature current is obtained by subtracting the holmic drop in the armature winding with respect to the armature current is represented by the straight line passing through the origin as shown Figure 4.31. When the ordinates of straight line representing the voltage drop in the armature winding (IaRa) are deducted from those of curve II, a cure III is obtained, which given the external characteristic of the generator i.e. curve III = curve II - Ia Ra. External characteristics clearly indicate that the terminal voltage falls as load on the generator increase. 198 Introduction to Electrical Machines Shunt-Wound Generator In this type of generator, the field winding is connoted across the armature winding. The generator will therefore build up its own magnetism. The voltage across the shunt field winding is equal to the terminal voltage of the generator as discussed above, the terminal voltage of the generator will fall down due to the armature reaction and the ohmic drop in the armature winding, as the load on the generator increases. Thus the voltage across the field will not remain constant as the load on the shunt generator increases. The voltage across the field winding decreases with an increase in the load current, which causes a decrease in the exciting current. The terminal voltage further falls down incase of a shunt generator because of decreases in excitation current as explained earlier with increasing load current. Hence the total decreases in the voltage in case of shunt generators is mush greater than in separately excited generators. For obtaining the relation between the terminal voltage and load current, the generator is connected as shown in Figure 4.32 (a). Figure 4.32 (b) shows the external characteristics, of a particular generator, when it is run as a separately excited generator (curve IV) and when run as a shunt generator (Curve III). Comparing these two curves for the same generator, it is observed that with self-excitation the external characteristic is lower than that obtained with separate excitation. (a) (b) Figure 4.32 external characteristics of shunt wound generator The basic reason for the difference in the two curves is that, in the former case the shunt field current decreases with decreasing terminal voltage, while in the case of separate excitation the field current remains constant. If the load on the shunt generator is gradually increased by decreasing the resistance in the external circuit, its terminal voltage tends to fall by a process of exactly a reverse nature to that of building up. Up to the normal load current, steady conditions are obtained without a serious fall in the terminal voltage as shown by the thick line of curve III. When the load on the shunt generator increase beyond its full load value, the drop in terminal voltage becomes more appreciable as shown by the dotted line of curve III of Figure 4.32(b). Up to the point D on curve III, the load current increases upon decreasing the external resistance in the load circuit, where the terminal voltage has fallen to an appreciably low value. The current corresponding to this condition is generally termed as critical current Ic. A further decrease in the external load resistance beyond the point D, does not 199 Chapter One: Electromagnetic Principles increase the current in the load circuit, but on the other hand decreases it, because the load resistance shunts the field winding to such an extent the terminal voltage decreases more rapidly than the load resistance. Hence the external characteristic turns back and the terminal voltage is zero when the armature is actually short-circuited. The armature current at this instant is shown by a vale OE that is purely due to residual magnetism of the generator. To obtain the internal characteristics of the dc shunt generator, the sum of the voltage drop in the armature winding including the brush contact drop is added to the external characteristic, thus obtaining curve II representing this characteristic. Figure 4.32 also shows the no load voltage Eo of the generator represented by the dotted line I. The voltage drop between curve II and line I is due to reduction in flux caused by the combined action of armature reaction and the fall caused by the combined action of armature reaction and the fall in the shunt field current. Series Wound Generator In series- wound generators, the field winding is connected in series with the armature winding. Thus, the current in the field winding is the same as the current in the armature winding. If the generator is driven at the constant rated speed, and the armature current is varied by varying the external resistance in the load circuit, a curve III of Figure 4.33 is obtained by plotting the terminal voltage verses the load current or armature current. I II In III Ex te r M ag ne t iz at na ter io n l C na Ch ha l C ar ha ra ac ct ra te er ct ris is er t ic is t ic t ic s s s VL Figure 4.33 external characteristics of series wound generator The internal or total characteristic of the same generator is represented by curve II in Figure 4.33 which can be obtained by adding the voltage drop in the armature circuit including brush contact drop to terminal voltage (curve III). Curve I, in Figure 4.33, shows the magnetization characteristics of the same generator. The voltage drop between the curves I and II is caused by armature reaction. Compound Generator The shunt generator already discussed has a drooping external characteristic, i.e. the terminal voltage falls with load, whereas series generators have an external characteristic, in which the terminal voltage rises with the load. Hence, a series field winding in dc generators can compensate for the tendency of the shunt generator to lose voltage with load, thus maintaining practically a constant voltage at all loads. For this reason, the majority of dc generators in service have both shunt and series windings. Such a dc generator having both shunt and series windings is called a compound generator. 200 Introduction to Electrical Machines Figure 4.34 the external characteristics of dc compound generator. Curve I shows the external characteristic, in which the series excitation is such that the terminal voltage on full load is the same as on no load and the terminal voltage remains practically constant from no load to full load. A dc compound generator giving such an external characteristic is called level-compounded generator. The external characteristic shown by curve II indicates that the terminal voltage rises with the load. Such a compound generator with this external characteristic is said to be over compounded generator. The compound generator having an external characteristic of the nature represented by curve III is called under compounded generator. In all the above three types of compound generators, i.e. level-compounded , overcompounded and under-compounded, the series field aids the shunt field and thus these compound generators can also be called as cumulative compound generator. Cumulative compound generator is most widely used in practice. Their external characteristic can match to all classes of service. These types of generators used for electric railways, for supplying current of incandescent lamps, etc. In case the series field opposes the shunt field, the external characteristic of the generator will be highly drooping with large demagnetizing armature reaction as shown by curve IV in Figure 4.34. Such a compound generator said to be differential-compound generator. Differential compound generators find their field of application in arc welding where a large voltage drop is desirable, when the current increase. Example 4.6 The open circuit characteristic of generator driven at 500 rpw is as follows: Field current, Ish (A) 0.2 0.4 0.6 0.8 Emf , Eg (V) 86 40 66 1.0 1.2 1.4 1.6 101 112 121 128 133 The machine is connected as shunt generator and driven at 500 rpm. Find i) open circuit voltage, when the field circuit resistance is 94 Ω, ii) the additional resistance required in the field circuit to reduce the emf to 110 V and iii) critical value of shunt field resistance. 201 Chapter One: Electromagnetic Principles Solution Figure 4.35 shows the magnetization characteristic drawn as per the given data. Line OA has been drawn as the field resistance line, representing a resistance of 94 Ω. Any point on the field resistance line can be found out corresponding to a particular value of field current, for example, when the field current is 1.0 A, voltage across the shunt field will be Vsh = Ish × Rsh =1.0 × 94 = 94 V, thus establishing a point B on the field resistance line. The field resistance line is drawn joining the point B with the origin O. i) The field resistance line OA cuts the magnetization curve at the point A. Hence the generator will develop an emf corresponding to the operating point A, which is equal to OC or 126 V. ii) Corresponding to the voltage of 110 V, a horizontal line is drawn, which cuts the OCC at the point D. Join the point D with the point O. The line OD represents the field resistance line that would generate a voltage of 110 V. Figure 4.35 Magnetization curve for example 4.6 Hence to generate a voltage of 110 volts, the total resistance of the shunt field circuit should be RF = 70 = 116.7 Ω 0 .6 Resistance of the shunt field winding, Rsh is 94 Ω. Thus additional resistance in the shunt field circuit is Radd = 116.7 - 94 = 22.7 Ω 202 Introduction to Electrical Machines iii) Critical value of shunt field resistance is obtained by drawing a tangent from the origin to the initial portion of the magnetization curve. Line DE represents the critical resistance of the shunt field. Thus critical resistance, 40 0 .2 = 200 Ω R cr = Example 4.7 The open circuit characteristic of a dc generator at rpm is as follows: Field current, Ish (A) 0.5 1.0 1.5 2.0 2.5 3:0 3.5 open circuit voltage, VOCC(V) 60 120 138 145 149 151 152 The machine is connected as shunt generator and driven at 1000 rpm. The resistance of shunt field circuit being 60 Ω. Calculate, i) the open circuit voltage, ii) the critical value of the field resistance, iii) the terminal voltage when the load has resistance of 4.0 Ω, and iv) the load current when the terminal voltage is 100 V. Neglect armature reaction. The armature resistance is 0.1Ω. Solution The open circuit characteristic of the dc shunt generator at 1000 rpm has been plotted in Figure above. The resistance of the shunt field circuit is 60 Ω and as such field resistance line OA has been drawn. Any point on this line gives a resistance value of 60 Ω, for example, corresponding to field current of 2 A, the voltage is 120 V (point F). i) The field resistance line OA corresponding to the field resistance of 60 Ω cuts the OCC at point A. Hence the shunt generator will generate a voltage corresponding to the operating point A which is equal to OC or 149 V.Thus open circuit voltage = 149 V. ii) Tangent OE is drawn to the OCC from the origin O to find out the critical value of shunt field resistance. The resistance represented by this tangent line OE is 120 = 120 Ω . Hence critical resistance of shunt field = 120 Ω. 1 .0 iii) Let the terminal voltage across the load of 4 Ω resistance be V volts Then the load current, I L = V A 4 .0 Shunt field resistance = 60 Ω Thus shunt field current, I sh = V A 60 For shunt generator, Ia = IL+ Ish 203 Chapter One: Electromagnetic Principles V V 16V + = 4 60 60 4 = V A 15 = Voltage at no load, Eg = V + IaRa Terminal voltage, V = Eg − Ia R a 4 = 149 − V × 0.1 15 = 149 − 0.0267 V Or V (1 + 0.0267 ) = 149 Terminal voltage, V= 149 = 145.1 V 1.0267 Terminal voltage, V = 100 V iv) Voltage at no load, Eg = V + IaRa or IaRa = Eg − V = 149 −100 = 49 V 49 = 490 A 0 .1 Armature current, Ia = Shunt field current, I sh = Hence load current, IL = 490 −1.67 = 488.33 A R sh =6 0Ω V 100 = = 1.67 A 60 60 Figure 4.36 Open circuit characteristic for example 4.7 204 Introduction to Electrical Machines Example 4.8 The OCC of a dc generator when driven at 750 rpm gave the following results: i) Field current, Ish (A) 0.5 1.0 1.5 2.0 2.5 Emf, Eg (V) 50 84 105 120 131 If the machine is run as shunt generator at 750 rpm, to what voltage will it excite with shunt field resistances equal to (a) 70 Ω (b) 55 Ω ? ii) What is the critical value of the shunt field resistance? iii) What is the critical speed when the shunt field resistance is 70 0 ? iv) With the shunt field resistance equal to 55 Ω, what reduction in speed must be made to make the open circuit voltage equal to 100 V? Figure 4.37 Open circuit characteristic for example 4.8 Solution OCC of the shunt generator at 750 rpm has been plotted in Figure 4.37 as per the given data. Line OA has been drawn to represent field resistance line corresponding to 55 Ω 110 V . Another line OB has been drawn, which represents field resistance of 70 Ω 2.0 A 70 V . 1.0 A i.) (a) When the field resistance is equal to 70 Ω, the generator will generate a voltage corresponding to the operating point B, which is a common point on 70 Ω field resistance line and the OCC of the generator. The open circuit voltage is equal to OC or 105 V. Hence generator will excite to the voltage of 105 V. 205 Chapter One: Electromagnetic Principles (b) When the field resistance is 55 Ω, the shunt generator will excite to a voltage given by the operating point A, at which the 55 Ω field resistance line cuts the OCC of the generator. The corresponding voltage is equal to OD or 128 V. Thus the generator will excite to the voltage of 128 V. ii.) A tangent line OE is drawn to the OCC of the generator to find out the critical resistance of the shunt field. The resistance represented by OE is the critical shunt field 100 V resistance, which is equal to = 100 Ω . Thus critical value of shunt field resistance 1A = 100 Ω. iii.) The shunt field resistance in this case is 70 Ω. Critical speed can be obtained by erecting a perpendicular from the point F, so as to cut the 70 Ω field resistance line at G and critical shunt field resistance line OE at point H. Then, FG critical speed = FH 750 Or critical speed = But FG 70 = FH 100 70 × 750 100 = 525 rpm critical speed = Thus iv.) FG × 750 FH Open circuit voltage Eg = 100 V Shunt field resistance = 55 Ω With shunt field resistance equal to 57 Ω, the generator generates a voltage of 128 V at 750 rpm. To generate 100 V with the same field resistance, the operating point has to be M instead of A, for which the speed of the generator has to be reduced. The speed in such a case can be found out by drawing a perpendicular from the point M, so as to meet the OCC at point N. Then, LM desired speed = LN 750 Hence, LM 100 = 750 × LN 115 = 652 rpm desiredl speed = 750 × Reduction in speed = 750 − 652 = 98 rpm Example 4.9 A dc generator has the following open circuit characteristics at 800 rpm: Field current , Ish (A) 0 1 2 3 4 5 Generated emf, Eg (V) 10 112 198 232 252 266 206 Introduction to Electrical Machines Find the no load terminal voltage when the machine runs as a shunt generator at 1000 rpm. The resistance of the field circuit is 70Ω. What additional field regulator resistance will be required to reduce the voltage to 270 V? Solution The open circuit characteristic of the dc generator has been given at 800 rpm. However, this generator runs as a shunt type at 1000 rpm. As the speed of the generator has increased, the emf generated corresponding to the same field current will increase and is given by Eg = PφNZ = KN 60 × a E g2 Hence, E g1 = N2 N1 E g 2 = E g1 × Or for the same field current N2 1000 = E g1 × N1 800 Based on this, the readings for the OCC at 1000 rpm will be: Open Circuit Characteristics at 1000 rpm If (A) 0 1 2 3 4 5 Eg (V) 12.5 140 247.5 290 315 332.5 Figure 4.38 shows the open circuit characteristics of the shunt generator driven at 1000 rpm, which has been plotted based on the calculated values of generated emf Eg2. A field 210 V has been drawn. resistance line OA representing resistance of 70 Ω 3A Field current, Ish, A Figure 4.38 Magnetization curve for example 4.9 207 Chapter One: Electromagnetic Principles i.) The field resistance line of 70 Ω cuts the OCC at the point A. the shunt generator will generate voltage equal to OC or 330 V. Hence no load terminal voltage is 330 V. ii.) The no load terminal voltage is 270 V. Corresponding to 270 V, a horizontal line FD shown dotted in Figure 4.38 has been drawn, which cuts the OCC at D. Hence to generate 270 V, the operating point must be D. The point D is joined with the origin, thus giving the resistance line OD corresponding to the operating point D. The resistance represented by the line OD = 270 V = 112.5 Ω 2.4A Shunt field resistance, Rsh = 70 Ω Hence additional resistance required in the field circuit is 112.5 −70 = 42.5Ω. 12. VOLTAGE REGULATION The change in output voltage of a generator from no-load to full-load divided by the fullload voltage, is called the voltage regulation. ∆V % = VNL − VFL × 100% VFL It is an important parameter in the performance of generator by providing an information that how constant the output voltage is with load. 13. DC MOTORS Working principle The principle upon which a dc motor works is very simple. If a current carrying conductor is placed in a magnetic field, mechanical force is experienced on the conductor, the direction of which is given by Fleming's left hand rule (also called motor rule) and hence the conductor moves in the direction of force. The magnitude of the mechanical force experienced n the conductor is given by F = B Ic lc, [Newtons] Where B is the field strength in Teslas (wb/m2), Ic is the current flowing through the conductor in amperes and lc is the length of conductor in meters. When the motor is connected to the dc Supply mains, a direct current passes through the brushes and commutator to the armature winding. While it passes through the commutator it is converted in to a.c. so that the group of conductors under successive field poles carries currents in the opposite directions, as shown in Figure 4.39. Also the direction of current in the individual conductor reverses as they pass away from the influence of one pole to that of the next. 208 Introduction to Electrical Machines Figure 4.39 schematic diagram of 4-pole dc motor In Figure 4.39, a 4-pole d.c motor is shown when the filed and armature circuits are connected across dc supply mains. Let the current in armature conductors be outwards under the N-poles (shown by dots) and inwards under S-poles (shown by crosses). By applying Fleming’s left hand rule Figure 4.40, the direction of force on each conductor can be determined, which has been illustrated in Figure 4.39. From Figure 4.39 it is observed that each conductor experiences a force which tends to the motor armature in clock-wise direction. These forces collectively produce a driving torque. Figure 4.40 Left-hand rule for determination of the direction of force 14. COMPARISON OF MOTOR AND GENERATOR ACTION As mentioned above, dc motor and the dc generator are the same devices, at least theoretically. The machine operating as a generator is driven by some external driving force and dc out put is obtained from it where as the machine operating as a motor is supplied by electric current and mechanical rotation is produced. Let us first consider the generator operation. In Figure 4.41(a) dc machine driven, in a clock-wise direction, by its prime mover and supplying direct current to external load circuit is shown. The machine is working as a generator and the direction of the generated emf and current flowing through the armature conductors, as determined by Fleming's right hand rule, will be as shown in the Figure 4.41(a). 209 Chapter One: Electromagnetic Principles (a) (b) Figure 4.41 (a) Generator action ; (b) Motor action Since the armature is carrying current and rotating in a magnetic field, Electro-magnetic forces will be given by Fleming's left hand rule. These Electro magnetic forces acting on the armature conductors will collectively result in torque acting on the armature in a counter-clockwise direction ( see Tback in Figure 4.41(a)). This Electro-magnetic torque, therefore, opposes the outside driving torque, which is causing the rotation of the machine and called the backward torque(Tback) or magnetic drag on the conductors. The prime mover has to work against this magnetic drag and the work so done is converted in to electrical energy. The larger the output current, more will be the backward torque and, therefore, more mechanical energy will be required to be supplied to the generator. In Figure 4.41(b) the same machine operating as a motor is shown. This operation takes place when the prime mover is uncoupled from the machine and the machine is connected to the dc supply mains. With the directions of field and armature current shown in the Figure 4.41(b) the torque developed by Electro-magnetic actions will rotate the machine in a clockwise direction (as determined by Fleming's left-hand rule). The friction of the machine and the mechanical load that the motor is driving will exert a torque in counter-clockwise direction, opposing the rotation of the motor. Since the armature conductors are revolving in the magnetic field, emf is induced in the armature conductors. The direction of emf so induced, as determined by Fleming's right hand rule, is in direct opposition to the applied voltage (see Eb in Figure 4.41(b)). That is why the induced emf in motor often is called the counter emf or back emf Eb. The applied voltage must be large enough to overcome this back emf and to send the current through the resistance of the armature. The electric energy supplied to overcome this opposition is converted into mechanical energy development in the armature. Thus we see that an emf is generated in both generator and motor, therefore, there is a generator action in both motor and generator operation. However, in generator operation the generated emf produces the armature current, where as, in motor operation the generated emf opposes the current direction. We also observe that Electro-magnetic torque is developed in generator as well as motor i.e. there is a motor action in both generator and motor, operation. However, in motor operation the Electro-magnetic 210 Introduction to Electrical Machines torque developed causes the armature rotation, where as in a generator operation the Electro-magnetic torque produced opposes the rotation. 15. TYPES OF DC MOTORS All dc motors must receive their excitation from an external source; therefore, they are separately excited. Their field and the armature windings are connected, however, in one of the three different ways employed for self-excited dc generators, and so according the field arrangement there are three types of dc motors namely; i) Series wound 4.13.1. ii) shunt wound and iii) compound wound. Series wound motor A series motor is one in which the field winding is connected in series with the armature so that the whole current drawn by the motor passes through the field winding as well as armature. Connection diagram is shown in Figure 4.42. Ise IL + Ia Series Winding Eb Important relationships I a = I se = I L Eb = VL − I a (Ra + Rse ) Pdrawn = VL ⋅ I L Pdev = Eb ⋅ I a VL _ Figure 4.42 connection diagram of series-wound motor 4.13.2. Shunt wound motor A shunt wound motor is one in which the field winding is connected in parallel with armature as illustrated in Figure 4.43. The current supplied to the motor is divided into two paths, one through the shunt field winding and second through the armature. Important relationships V I sh = L Rsh I L = I sh + I a Eb = VL − I a Ra Pdrawn = VL ⋅ I L Pdev = Eb ⋅ I a Figure 4.43 connection diagram of shunt-wound motor 211 Chapter One: Electromagnetic Principles 4.13.3. Compound wound motor A compound wound motor has both series and shunt windings which can be connected as short-shunt or long shunt with armature winding as illustrated in figure 4.44. Important relationships I se = I L E +I R V −I R I sh = b a a = L se se Rsh Rsh I L = I sh + I a Eb = V L − I a Ra − I L Rse Pdrawn = VL ⋅ I L Pdev = Eb ⋅ I a (a) short-shunt compound motor Important relationships I a = I se E b + I a (Ra + Rse ) V L = R sh R sh I L = I sh + I L Eb = V L − I a (Ra + R se ) I sh = P drel = VL ⋅ IL Pdev = E b ⋅ I a (b) long -shunt compound motor Figure 4.44 connection diagram of compound-wound motor 16. DIRECTION OF ROTATION It is clear that, from principle operation of dc motor, if the armature current were reversed by reversing the armature terminal leads, but leaving the field polarity the same, torque would be developed in a counter-clock wise direction. Likewise, if the field polarity were reversed leaving the armature current as shown torque would be developed in a counter-clockwise direction. However if both the armature current direction and field polarity were reversed torque would be developed in a clock-wise direction as before. Hence the direction of rotation of a motor can be reversed by reversing the current through either the armature winding or the field coils. If the current through both is reversed, the motor will continue to rotate in the same direction as before. 17. SIGNIFICANCE OF BACK EMF As explained earlier, when the motor armature continues to rotate due to motor action, the armature conductors cut the magnetic flux and therefore emfs are induced in them. The direction of this induced emf known as back emf is such that is opposes the applied 212 Introduction to Electrical Machines voltage. Since the back emf is induced due to the generator action , the magnitude of it is, therefore , given by the same expression as that for the generated emf in a generator Eb = φZN P × volts, 60 a 4.1 The symbols having their usual significance Figure 4.45 Equivalent circuit of a motor Armature The equivalent circuit of a motor is shown in Figure 4.45. The armature circuit is equivalent to a source of emf Eb in series with a resistance, Ra put across a dc supply mains of V volts. It is evident from Figure3 that the applied voltage V must be large enough to balance both the voltage drop in armature resistance and the back emf at all times i.e. V = Eb+Ia Ra 4.2 Where V is the applied voltage across the armature, Eb is the induced emf in the armature by generator action; Ia is the armature current and Ra is the armature resistance. Equation (4.2) may be rewritten as I a = V − Eb to give armature current in terms of Ra applied voltage V, induced emf Eb and armature resistance, Ra. As obvious from Eqs.(4.1) and (4.2) the induced emf in the armature of a motor, Eb depends among other factors upon the armature speed and armature current depends upon the back emf Eb for a constant applied voltage and armature resistance. If the armature speed is high, back emf Eb will be large and therefore armature current becomes small. If the speed to the armature is low, then back emf Eb will be less and armature current Ia will be more resulting in development of large torque. Thus it is evident that back emf Eb acts like a governor i.e. it makes a motor self-regulating so that it draws as much current as just required. 213 Chapter One: Electromagnetic Principles 18. TORQUE EQUATION The back emf of dc motor is given by Eb = V − Ia R a 4.3 Multiplying both sides of Eq. (4.3) by Ia, E b I a = VI a − I a2 R a 4.4 In Eq. (4.4) VIa = Total electrical power supplied to the Armature of the dc motor (armature input) and Ia2Ra = power wasted in the armature (armature copper loss). The difference between the armature input and the armature copper loss is equal to the mechanical power developed by the armature of the motor. Hence, mechanical power developed = EbIa, watts 4.5 If Ta is the torque in Newton meter developed by the armature of the motor, running at N revolutions per minute, then Pmech = Mechanical power developed, 2πN. Ta watts 60 4.6 Equating Eqs. (4.5) and (4.6) 2 πN Ta 60 (60) E b I a Torque, Ta = . 2π N E b Ia = 4.7 However back emf, Eb = pφ NZ 60.a 4.8 Substituting Equation (4.8) into equation (4.7) Torque, Ta = PΦ I a Z 60 PΦNZI a . = 0.159. [ N .m] 2π 60 a N a For a particular dc motor; P, Z & a are fixed. Hence, Ta ∝ φ I a Therefore, the torque developed by the armature of dc motor is proportional to the product of armature current and the flux per pole. For dc shunt motor, the flux per pole is practically constant, hence the torque developed is directly proportion to the armature current, i.e. 214 Introduction to Electrical Machines Ta ∝ I a (for dc shunt motor) For dc series motor, the flux per pole is directly proportional to Ia hence the torque developed is directly proportion to the square of the armature current, i.e. Ta ∝ I a2 (for a dc series motor) 19. SPEED EQUATION The back emf for dc motor is given by Eb = P φ N. Z volts 60a Eb = V − Ia R a Also, Combing the above two equations, P φ N .Z = V − Ia R a 60a N = ( V − I a R a ). Or 60.a I . PZ φ For a given particular motor, P, Z and a are fixed. Hence N =K (V − I a R a ) φ E =K b φ Thus the speed of dc motor is directly proportional to the voltage applied to the armature or the back emf & inversely proportional to the flux per pole. For dc shunt motor, the flux per pole is approximately constant and hence the speed of dc shunt motor is directly proportional to the back emf i.e. N ∝ Eb . For dc series motor, the flux per pole is directly proportional to the armature current and hence the load on the motor. Thus the speed of dc series motor is inversely proportional to the flux per pole or the armature 1 current i.e. N ∝ . The speed of the motor increases with the fall in flux. φ Example 4.10 The armature of a 6 pole, 6 circuit dc shunt motor takes 300 A at the speed of 400 revolutions per minute. The flux per pole is 75× 10-3 Wb. The number of armature turns is 500. The torque lost in windage, friction and iron losses can be assumed as 2.5 per cent. Calculate (i) the torque developed by the armature, (ii) shaft torque and (iii) shaft power in kW. Solution i) The torque developed by the armature of a dc motor is given by Ta = 0.159 PφI a Z N.m A Number of poles of shunt motor, P = 6 Armature winding has 6 circuits, thus, A = 6 215 Chapter One: Electromagnetic Principles Armature current, Ia = 300 A Number of armature turns = 500 Thus total conductors on the armature, Z = 2 × 500 = 1000 Flux per pole, φ = 75 × 10-3 Wb Substituting these values in the above equation Armature torque, Ta = 0.159 6 × 75 × 10 −3 × 300 × 1000 = 3577.5 N.m 6 Torque lost in windage, friction and iron losses = 2.5% of Ta = 0.0255 × 3577.5 = 89.44 N.m Thus, shaft torque, Tsh = 3577.5−89.44 = 3488.06 N m Shaft power, 2πNTsh Z kW 60 × 1000 2π × 400 × 3488.06 = 60 × 1000 = 146.22 kW Tsh = Example 4.11 A 440 V dc motor takes an armature current of 60 A when its speed is 750 rpm. If the armature resistance is, 0.25Ω, calculate the torque produced. Solution Back emf developed, Eb = V − Ia Ra = 440−60×0.25 = 425 V Torque produced, Ta = EbIa 425 × 60 = = 324.68 N.m 2 π N / 60 2 π × 750 / 60 Example 4.12 A 10 hp 230V shunt motor takes an armature current of 6A from 230 V line at no load and runs at 1,200 rpm. The armature resistance is 0.25Ω. Determine the speed and electro-magnetic torque when the armature takes 36 A with the same flux. Solution No-load back emf, Ebo = V − Iao Ra = 230−6×0.25 = 228.5 V No-load speed, N0 = 1200 rpm. When armature takes 36 A Back emf developed Eb1 = V − Ia1 Ra = 230−36×0.25 = 221 Since E b ∝ φN ∴ E b1 φ1 N1 N1 = = E b 0 φ2 N 0 N 0 ∵ flux is same i.e. φ1= φ0 216 Introduction to Electrical Machines N1 = Or E b1 221 × N0 = × 1200 = 1161 rpm E b0 228.5 Electro-magnetic torque developed, Ta = EbIa 221 × 36 = = 65.44 N.m 2 π N1 / 60 2π × 1161 / 60 Example 4.13 The armature of a 4-pole dc shunt motor has a lap winding accommodated is 60 slots, each containing 20 conductors. If the useful flux per pole be 23 mWb, calculate the total torque developed in Newton meters when the armature current is 50 A. Solution Flux per pole, φ = 23 mWb = 0.023 Wb Total number of armature conductors, Z = 60 × 20 =1200 Number of poles, P=4 Armature current, Ia = 50 A Since armature has lap winding, Number of parallel paths, A=P=4 Total torque developed, I Ta = 0.159φZP × a A = 0.159 × 0.023 × 1200 × 50 = 219.6 N.m 20. DC MOTOR CHARACTERISTICS The 3 Important characteristic curves of dc motors are: 1. Torque-Armature Current Characteristic This characteristic curve gives relation between mechanical torque T and armature current Ia. This is known as electrical characteristic. 2. Speed-Armature Current Characteristic This characteristic curve gives relation between speed N and armature current Ia 3. Speed-Torque Characteristic This characteristic curve gives relation between speed N and mechanical torque T. This is also known as mechanical characteristics. This curve can be derived from the above two curves. 4.18.1. Characteristics of Dc Series Motors a.) Magnetic characteristic In case of dc series motors the flux φ varies with the variation in line or armature current as the field is in series with the armature. The flux φ increase following a linear law with 217 Chapter One: Electromagnetic Principles the increase in load current, becomes maximum at saturation point and finally become constant. b.) Torque-Armature Current Characteristics From expression of mechanical torque T it is obvious that Ta ∝ φI a φ , Wb Up to saturation point flux is proportional to field current and hence to the armature current because Ia = If. Therefore on light load T ∝ I a2 and hence curve drawn between T and Ia up to saturation point is a parabola as illustrated in Figure 4.46. After saturation point flux φ is almost independent of excitation current and so T ∝ I a . Hence the characteristics curve becomes a straight line as shown in Figure 4.46. From the torquearmature current curve it is evident that series motor develops large starting torque to accelerate the heavy masses. Hence series motors are used where large starting torque is required such as in hoists electric railways, trolleys and electric vehicles. φ Figure 4.46 Speed- current and Torque-current characteristics of DC series motors c.) Speed-Current Characteristic From expression of speed, it is obvious that E V − Ia R a Nα b = φ φ If the applied voltage remains constant, speed is inversely proportional to flux per pole. So, if a curve is drawn between reciprocal of flux and current I, the speed current characteristic is obtained which is a rectangular hyperbola in shape as represented in Figure 4.46. Since on no load the speed is dangerously high, as obvious from speed-current characteristic curve, which will result in heavy centrifugal force which in turn will damage the motor. That is why, series motors are never started on no load, which is explained below: 218 Introduction to Electrical Machines When the motor is connected across the supply mains without load, it draws a current from the supply mains flowing through the series field and armature, the speed tends to increase so the back emf, may approach the applied voltage in magnitude. The increase in back emf weakens the armature current and hence the field current. This cause again increases in speed so in back emf. Thus the field continues to weaken and speed continues to increase dangerously until the armature gets damaged. Since on no-load the series motor attains dangerously high speed, which cause heavy centrifugal force resulting in the damage of the machine, therefore, series motor are not suitable for the services: i. where the load may be entirely removed and ii. for driving by means of belts because mishap to the belt would cause the motor to run on no-load These motor are suitable for gear drive, because gear provides some load on account of frictional resistance of the gear teeth in case of sudden release of load. d.) Speed-Torque Characteristic The speed- torque characteristic can be drawn with help of above two characteristics, as shown in Figure 4.47, which shows that as the torque increases, speed decreases. Hence series motors are best suited for the services where the motor is directly coupled to load such as fans whose speed falls with the increase in torque. It should be noted that series motor is a variable speed motor. T, N.m Figure 4.47 Speed-Torque characteristics of dc series motor 4.18.2. Characteristics of Dc Shunt Motors a.) Speed-Current Characteristic If applied voltage V is kept constant, the field current will remain constant hence flux will have maximum value on no load but will decrease slightly due to armature reaction as the load increase but for more purpose the flux is considered to be constant neglecting the effect of armature reaction. From expression of speed N is directly proportional to back emf Eb or (V−IaRa) and inversely proportional to the flux φ. Since flux is considered to be constant as mentioned above, so with the increase in load current the speed slightly falls due to increase in 219 Chapter One: Electromagnetic Principles voltage drop in armature IaRa . Since voltage drop in armature at full-load is very small as compared to applied voltage so drop in speed from no-load to full-load is very small and for all practical purposes the shunt motor is taken as a constant speed motor. Since there is a slight variation in speed of the shunt motor from no-load to full-load and this slight variation in can be made by inserting resistance in the shunt field and so reducing the flux. Therefore, shunt motors being constant speed motors are best suited for driving of line shafts, machine lathes, milling machines, conveyors, fans and for all purposes where constant speed is required. Figure 4.48 Speed- current and Torque-current characteristics of dc shunt motor b.) Torque- Current Characteristic From the expression for the torque of a dc motor, torque is directly proportional to the product of flux and armature current. Since in case of dc shunt motors the flux is constant therefore torque increase with the increase in load current following linear law i.e. torque-armature current characteristics is a straight line passing through origin (refer Figure 4.48). c.) Speed-Torque Characteristic This characteristic curve can be drawn from the above two characteristics and is shown in Figure 4.49. Figure 4.49 Speed-Torque characteristics of dc shunt motor 220 Introduction to Electrical Machines 4.18.3. Characteristics of Compound Would Motor a.) Cumulative compound wound motor As the load is increased, the flux due to series field winding increase and causes the torque greater than it would have with shunt field winding alone for a given machine and for given current. The increase in flux due to series field winding on account of increase in load cause the speed to fall more rapidly than it would have done in shunt motor. The cumulative compound motor develops a high torque with increase of load. It also has a definite speed of no load, so does not run away when the load is removed (refer Figure 4.50 and 4.51). Cumulative compound wound motors are used in driving machines which subject to sudden applications of heavy loads, such as occur in rolling mills, shears or punches. This type of motor is used also where a large starting torque is regard but series motor cannot be used conveniently such as in cranes and elevator. b.) Differential compound wound motor Since the flux decrease with the increase in load, so the speed remains nearly constant as the load is increased and in some cases the speed will increase even. The decrease in flux with the increase in load causes the torque to be less than that of a shunt motor. The characteristics are similar to those of a shunt motor. Since the shunt motor develops a good torque and almost constant speed, therefore differential compound motor is seldom used. The characteristics are shown in Figure 4.50 and 4.51. Figure 4.50 Speed- current and Torque-current characteristics of dc shunt motor 21. Figure 4.51 Speed-Torque characteristics of dc shunt motor STARTING OF DC MOTOR If dc motor is directly connected to a dc power supply, the starting current will be dangerously high. From Figure 4.52 (a), V − Ea Ia = i Ra The back emf E b ( =K a φ N ) is zero at start. Therefore, I a start = V Ra 221 Chapter One: Electromagnetic Principles (a) (b) (c) Figure 4.52 dc motor starter Since Ra is small , the starting current is very large. The starting current can be limited to a safe value by the following methods. 1. Insert an external resistance, Rae (Figure 4.52 (b), at start. 2. Use a low dc terminal voltage (V) at start. This , of course , requires a variable-voltage supply With an external resistance in the armature circuit, the armature current as the motor speeds up is Ia = V − Eb R a + R ae The back emf Eb increases as the speed increases. Therefore, the external resistance Rae can be gradually taken out as the motor speeds up without the current exceeding a certain limit. This is done using a starter, shown in Figure 4.52(c). At start, the handle is moved to position 1. All the resistances, R1, R2, R3 and R4 appear in series with the armature and thereby limit the starting current. As the motor speeds up the handle is moved to positions 2, 3, 4, and finally 5. At position 5 all the resistances in the starter are taken out of the armature circuit. The handle will be held in position 5 by the electromagnet, which is excited by the field current If. 222 Introduction to Electrical Machines Ia Iamax Iamin t t1 Rae1 t2 Rae2 t3 Rae3 t4 (b) Figure 4.53 variation of starting current and speed as starting rheostat brought out from armature circuit 22. SPEED CONTROL OF DC MOTOR Speed control means intentional change of the drive speed to a value required for performing the specific work process. This concept of speed control or adjustment should not be taken to include the natural change in speed which occurs due to change in the load on the drive shaft. The desired change in speed is accomplished by acting accordingly on the derive motor or on the transmission connecting it to the unit it serves to drive. This may be done manually by the operator or by means of some automatic control device. The nature speed control requirement for an industrial drive depends upon its type. Some drives may require continuous variation of speed for the whole of the range from zero to full speed, or over a portion of this range ; while the others may require two or more fixed speeds. Some machines may require a creeping speed for adjusting or setting up the work. Such a speed is of the order of few r.p.m. For most of the drives, however, speed a control of speed within the range of ± 20% may be suitable. On of the aattractive features the d.c. motor offers over all other types is the relative ease with which speed control can be achieved. The various schemes available for speed control can be deduced from the expression of speed for a d.c. motor. It has been shown earlier the speed of a motor is given by the relation N= V − Ia R a a V − Ia R a ⋅ = K Zφ φ P r.p.s. Where Ra =armature circuit resistance It is obvious that the speed can be controlled by varying a) Flux/pole i.e. Flux control b) Resistance Ra of the armature circuit i.e. Rheostatic Control and c) Applied voltage V i.e. Voltage control 223 20.1. Chapter One: Electromagnetic Principles Speed control of shunt motors a) Variation of Flux or Flux control Method 1 . By decreasing the flux, the speed can be φ increase and vice versa. Hence, the name flux or field control method. The flux of dc motor can be changed by changing Ish with help of a shunt field rheostat. Since Ish is relatively small, shunt field rheostat has to carry only a small current, which means I2 R loss is small. So this method is, therefore, very efficient. It is seen from above equation that, N ∝ Figure 4.54 speed control of shunt motor by varying field flux In non-interpolar machines, the speed can be increased by this method in the ratio of 2:1. Any further weakening of flux φ adversely affects the commutation and hence puts a limit to the maximum speed obtainable by this method. In machines fitted with interpoles, a ratio of maximum to minimum speeds of 6:1 is fairly common. b) Armature or Rheostatic Control Method This method is used when speeds below the no-load speed are required. As the supply voltage is normally constant, the voltage across the armature is varied by inserting a variable rheostat or resistance (called controller resistance) in series with the armature circuit as shown in Figure 4.55(a). (a) (b) Figure 4.55 speed control of shunt motor by varying resistance in the armature circuit As controller resistance is increased, potential difference across the armature is decreased, thereby decreasing the armature speed. For a load of constant torque, speed is approximately proportional to the Potential difference across the armature. From the speed/armature current characteristics, as shown in Figure 4.55, it is seen that the greater the resistance in the armature circuit, greater is the fall in speed. 224 Introduction to Electrical Machines c) Armature-terminal voltage control Utilizes the fact that the change in the armature terminal voltage of a shunt motor is accompanied in the steady state by a substantially equal change in the speed voltage (Eb) and, with constant motor flux, a consequent proportional change in motor speed. One common scheme, called the Ward-Leonard System, required an individual motorgenerator set to supply power to the armature voltage of the motor whose speed is to be controlled. Frequently the control of generator voltage is combined with motor-field control, as indicated by the rheostat in the field of motor M in Figure 4.56, in order to achieve the widest possible speed range. With such dual control, base speed can be defined as the normal-armature voltage full field speed of the motor. Speeds above base speed are obtained by motor field control; speeds below base speed are obtained by armature-voltage control. As discussed in connection with field-current control, the range above base speed is that of constant power drive. The range below base speed is that of a constant torque drive because, as in armature-resistance control, the flux and the allowable armature current remain approximately constant. 3Φ - supply Figure 4.56 ward-Leonard system Figure 4.57 constant-torque and constant-power operation 225 Chapter One: Electromagnetic Principles Example 4.14 A 200 V dc shunt motor has an armature resistance of 0.4 Ω and a field resistance of 200 Ω. When the motor is driving at 600 rpm a load, the torque of which is constant, the armature takes 20 A. It is desired to raise the speed from 600 to 900 rpm by inserting a resistance in the shunt field circuit. Assuming the magnetization curve to be a straight line, find the value of additional resistance in the field circuit. Solution Initial speed of the motor, N1 = 600 rpm Armature current, Ia1 = 20 A Applied voltage, V = 200 V Back emf developed by the motor at 600 rpm Eb1 = V − Ia1 Ra = 200 − 20×0.4 = 192 V Field current under this condition, I sh1 = 200 = 1 .0 A 200 Now let the total resistance in shunt field circuit to raise the speed to 900 rpm be Rf Ω. Then the field current, I sh1 = 200 A Rf The magnetization curve is to be assumed as a straight line, thus flux is directly proportional to field current, i.e. φ ∝ I sh Or φ2 I sh 2 200 = = φ1 I sh1 R f As per the given conditions, the torque remains constant during the change of speed. Thus T1 = T 2 Torque T ∝ Ia φ T1 ∝ K ' I a1φ1 T2 ∝ K ' I a 2 φ2 Or I a1φ1 = I a 2 φ2 φ R I a 2 = I a1 × 1 = 20 × f = 0.1R f φ2 200 Where, Ia2 is the current drawn by the armature, when the motor is driving the load at 900 rpm. Back emf at 900 rpm, Eb2 = V − Ia2 Ra = 200 − 0.1 Rf ×0.4 = 200 − 0.04 Rf 226 Introduction to Electrical Machines Back emf for a particular motor, E b = KφN Thus E b1 = Kφ1N1 (i) And E b2 = Kφ2 N 2 (ii) Dividing Eq. (ii) by Eq. (i), E b2 φ2 N 2 × = E b1 φ1 N1 200 − 0.04 R f 200 900 = = 192 Rf 600 0.04 R f2 − 200 R f + 57600 = 0 Or, 200 + 40,000 − 9216 0.08 = 306.8 Ω Rf = Additional resistance in the shunt field circuit =Rf − Rsh1 = 306.8 − 200 = 106.8 Ω Example 4.15 A 250 V dc shunt motor runs at its normal speed of 500 rpm when the armature current is 100 A. Find the speed of the motor under the following cases: (i) a resistance of 1.0 Ω is connected in series with the armature circuit, the shunt field remaining constant, (ii) the shunt field current is reduced to 60 per cent of its normal value by inserting a resistance in the field circuit. The armature current in both the above cases is 50 A. The resistance of the armature is 0.25Ω and that of interpole winding, 0.05Ω. Solution Total resistance of the armature circuit Normal speed, N1 = 500 rpm Applied voltage, V = 250 V 0.25+0.05 = 0.3Ω Armature current at normal speed, Ia1 = 100 A Back emf at 500 rpm, Eb1 = 250 − 100 ×0.3 = 220 V i) Additional resistance in the armature circuit = 1.0 Ω Total resistance in the armature circuit = 0.3 + 1.0 = 1.3 Ω Armature current under this condition, Ia2 = 50 A Back emf, Eb2 = 250 − 50 ×1.3 = 185 V As the field current remains unchanged, φ1 = φ 2 227 Chapter One: Electromagnetic Principles Back emf for a particular motor, E b = KφN or E b1 = Kφ1N1 (i) And E b2 = Kφ2 N 2 (ii) E b2 φ2 N 2 × = E b1 φ1 N1 Or N 150 = 1× 2 220 500 Speed, N 2 = 500 × 185 = 420 rpm 220 Total resistance in the armature circuit = 0.3 Ω ii) Armature current, Ia3 =50 A Back emf under this condition, Eb3 = 250 − 50 ×0.3 = 235 V Shunt field current under this condition, Ish3 = 0.6 Ish1 Assuming the magnetization curve to be linear, φ ∝ I sh Or φ3 I sh 3 = = 0 .6 φ1 I sh1 Now, E b3 φ3 N 3 = × E b1 φ1 N1 Or N 235 = 0 .6 × 3 220 500 Speed of the motor, N 3 = 500 × 235 = 890 rpm 220 × 0.6 Example 4.16 A 230 V shunt motor drives a load at 900 rpm drawing a current of 30A. The resistance of the armature circuit is 0.4Ω. The torque of the load is proportional to the speed. Calculate the resistance to be connected in series with the armature to reduce the speed to 600 rpm Ignore armature reaction., Solution In normal conditions Armature current, Ia =30 A neglecting field current Armature circuit resistance, Ra =0.4Ω Back emf, Eb1 = V − Ia1 Ra = 230 −30×0.4Ω =218 V Speed, N1 = 900 rpm 228 Introduction to Electrical Machines Let resistance R be connected in series with the armature circuit to reduce the speed to N2=600 rpm. Since as per given data load torque, T ∝ N and also T ∝ φI a I a 2 φ2 N 2 = I a1φ1 N1 ∴ Assuming flux constant I a 2 = I a1 × N2 600 = 30 × = 20 A N1 900 Eb2 = V − Ia2 (R +Ra) Back emf, = 230 −20( R+0.4) =222−20R ∵ E b ∝ Nφ and flux φ is constant ∴ E b2 N 2 = E b1 N1 Or 220 − 20 R 600 = 218 900 R= 3.833 Ω Or Example 4.17 A 220 V dc shunt motor draws 4.5 A on no load and runs at 1000 rpm. Resistance of the armature winding and shunt field winding is 0.3 and 157 Ω respectively. Calculate the speed, when loaded and drawing a current of 30 A. Assume that the armature reaction weakens the field by 3 %. Solution Voltage applied to the motor, VL = 220 V Shunt field resistance, Rsh = 157 Ω Shunt field current, I sh = 220 = 1 .4 A 157 Current drawn by the motor at no load = 4.5 A. Thus armature current at no load, Iao = 4.5 − 1.4 = 3.1 A Back emf at no load, Ebo = V − Iao Ra = 220 − 3.1 × 0.3 = 219.07 V Under loaded conditions, current drawn by the Motor = 30 A Armature current under loaded conditions, Ia = 30 − 1.4 = 28.6 A Back emf under loaded conditions, Eb = V − Ia Ra = 220 − 28.6 × 0.3 229 Chapter One: Electromagnetic Principles = 211.42 V Let the flux under no load condition be φ0, then under the loaded condition flux φ = 0.97 φ0, because of armature reaction. The back emf for a de motor is given by PφNZ 60a = KφN Eb = Thus E bo = Kφo N o (i) Also, E bo = K (0.97φo )N (ii) Dividing Eq. (ii) by Eq. (i), N (0.97 ) E b = No E bo Speed under the loaded condition, 211.42 1000 × 219.07 0.97 = 995 rpm N= 20.2. Speed Control Of Dc Series Motors Flux Control Method Variation in the flux of a Series motor can be brought about in any one of the following ways. i) Field Divertors The series winding are shunted by a variable resistance knows as field divertor (Figure 4.54). Any desired amount of current can be passed through the divertor by adjusting its resistance. Hence the flux can be decreased, consequently, the speed of the motor increased. Field Divertor Series Field + Ia _ Figure 4.58 speed control of series motor by field divertor method 230 Introduction to Electrical Machines ii) Armature Divertor A divertor across the armature can be used for giving speeds lower then the normal speed. For a given constant load torque, if Ia is reduced due to armature divertor, then φ must increase (∴ Ta α φ Ia). This results an increase in current taken from the supply (which increases the flux) and a fall in speed (N ∝ 1/φ). The variations in speed can be controlled by varying the divertor resistance. Figure 4.59 speed control of series motor by armature divertor method iii) Tapped Field Control This method is often used in electric traction (Show in Figure 3).The number of series field turns in the circuit can be changed at will as shown. With full field, the motor runs at its minimum speed, which can be raised in steps by cutting out some of the series turns. Figure 4.60 speed control of series motor by tapped field control method Variable Resistance in series with Motor By increasing the resistance in series with armature, the voltage applied across the armature terminals can be decreased. With reduced voltage across the armature, the speed is reduced. However, it will be noted that since full motor current passes through this resistance, there is a considerable loss of power in it. 231 Chapter One: Electromagnetic Principles + Speed No R Series Field Wi th _ es is Re sis tan ce tan c e Current Figure 4.61 speed control of series motor by variable resistance method Example 4.18 A 200 V dc series motor runs at 500 r.p.m, when taking a current of 25 A. the resistance of the armature is 0.5 Ω and that of field is 0.3Ω. If the current remains constant, calculate the resistance necessary to reduce the speed to 250 rpm. Solution Motor input current, I1=25 A Line voltage, V=200 V Eb1 = V − I1 (Ra +Rse) Back emf, = 200 −25(0.5+0.3) = 180 V Speed, N1 = 500 rpm Let resistance R be connected in series with the armature circuit to reduce the speed to N2=250 rpm. Motor input current, I2 = I1 =25 A ...... (given) Eb2 = V − I2 (R + Ra +Rse) = 200 −25(R+0.5+0.3) = 180−25R Since Or N∝ Eb φ E N 2 E b2 φ1 = b2 = × N1 E b1 φ21 E b1 ∵ φ2 = φ1 as field current remains the same Or 250 180 − 25R = 500 180 Or R= 3.6 Ω 232 Introduction to Electrical Machines Example 4.19 A series motor with series field and armature resistance of 0.06 and 0.04 Ω respectively is connected across 220 V. The armature takes 40 A and speed is 1000 rpm. Determine its speed when the armature takes 75 A and excitation is increased by 10%. Solution Armature current, 1a1 = 40 A Eb1 = V − I1 (Ra +Rse) Bach emf, = 220 −40(0.04+0.06) = 216 V Speed, N1 = 1000 rpm Flux φ1=φ (say) When armature current, I2= 75 A Eb2 = V − I2 (Ra +Rse) Bach emf, = 220 −75(0.04+0.06) = 212.5 V Flux N∝ Since ∴ φ2= 1.1φ1 = 1.1φ ...... (given) Eb φ E φ N 2 = N1 × b 2 × 1 = E b1 φ21 212.5 φ = 1000 × × 216 1.1φ = 894 rpm Example 4.20 A series motor runs at 500 r.p.m. when taking a current of 60A at 460 V. The resistance of the armature circuit is 0.2 Ω and that of the field winding is 0.1 Ω. Calculate the speed when a 0.15 Ω divertor is connected in parallel with the field winding. Assume the torque to remain unaltered and the flux to be proportional to the field current. Solution In normal conditions Armature current, Ia1 = Series field current, Ise = Line current, IL1 = 60 A Speed, N1=500 rpm Eb1 = V − Ia1 Ra − I se1Rse) = 220 −60× 0.2 − 60×0.1 = 442 V After connecting a divertor of resistance of 0.15 Ωa in parallel with the field winding let the speed be N2 and line current IL2 Armature current, Ia2 = IL2 233 Chapter One: Electromagnetic Principles Series field current, R div R se + R div 0.15 3 = I L2 × = I L2 0.1 + 0.15 5 I se 2 = I L 2 × Since torque remains unaltered ∴ T2 = T1 Ia2φ2 = Ia1φ1 since φ ∝ I se Ia2Ise2 = Ia1Ise1 3 I L2 × I L 2 = 60 × 60 5 I L2 = I se2 = 3600 × 5 = 77.46 A 3 3 3 I L 2 = × 77.46 = 46.48 A 5 5 Back emf, Eb1 = V − Ia2 Ra − I se2Rse) = 220 −77.46 × 0.2 − 46.48 ×0.1 = 440.06 V E E I φ Speed N 2 = N1 × b2 × 1 = N1 × b 2 × se1 = E b1 φ21 E b1 I se2 440.06 46.48 = 500 × × 442 60 = 386 rpm 23. LOSSES IN DC MACHINES INTRODUCTION The dc machines are used either for converting mechanical energy into electrical energy, i.e. generators or for converting electrical energy into mechanical energy, i.e. motors. This conversion of energy from one form to another obviously takes place at an efficiency of less than 100 percent. A part of the energy consumed by machine can not be effectively utilized in the machine proper and is dispersed in the form of heat. This part of the energy is generally termed as lost energy or simply the losses of the machine. The losses in general occur (i) in electrical circuits carrying a certain current, (ii) in magnetic circuits subjected to alternating magnetization and (iii) due to mechanical friction. Hence the losses occurring in an electrical machine constitute a source of inefficiency. In addition, these are completely converted into heat, resulting in a rise in the temperature of the machine. If reliable operation of an electrical machine is desired during its normal service life, then the temperature of the various parts of the machine should not be allowed to exceed beyond the permissible limit, decided by the type of 234 Introduction to Electrical Machines insulating material used in the machine. The temperature rise also decides the capacity to which the machine can be loaded safely. The losses occurring in the machine have also to be paid for and as such the running cost of a less efficient machine for the same output is more compared to an efficient machine. Hence the problem of losses in the machine is closely connected with the problems of its service life and other economic factors. The initial cost of a more efficient machine is certainly higher than that of a machine with poor efficiency. However, the higher initial cost is compensated by the saving in running energy charges; moreover a more efficient machine is highly reliable , less subject to breakdowns which is particular important when continuity of service must be maintained. Better electrical materials are being developed and with the use of these electrical machines are undergoing a continuous improvement towards reduction in losses, so as to produce more efficient machines with trouble free and continuous service even under more severe working. 4.21.1. Classification of Losses Power losses originating in dc machines (either a motor or a generator) can be classified into the following groups 1. Copper losses, caused by the current flow and occur in (i) armature winding, (ii) series field winding , (iii) commutating pole winding , (iv) compensating winding , (v) shunt field winding and (vi) loss due to brush contact resistance. 2. Iron losses caused by varying magnetization and occur in (armature core and (ii) armature teeth , as hysteresis loss and eddy current loss. 3. Mechanical losses caused by the rotation of the machine and occur as (i) bearing friction (ii) brush friction and (iii) air friction (windage). These losses are also called friction and windage losses. 4. Stray load losses include (i) increase in iron losses at load (ii) increases in copper losses due to eddy currents in armature conductors and (iii) additional losses caused by short circuit currents in the coils under commutation and occur in (a) armature teeth, (b) armature core and (c) armature winding. For smaller machines, stray losses are quite negligible. For fairly large machines, it may be taken arbitrarily as one percent of the rating of the machine. Of these groups of losses, copper losses vary with the load on the machine, whereas iron and mechanical losses remain substantially constant at all loads. Stray losses are more or less negligible in small and medium machines. Hence the total losses occurring in a dc machine may also be classified as i) variable losses and ii) Constant losses. In case of shunt or compound dc machines, even the shunt field copper losses can be grouped with the constant losses because it remains practically constant at all loads. Thus for a shunt or compound dc machine, the constant and variable losses include: 235 Chapter One: Electromagnetic Principles • Constant losses including (i) iron losses, (ii) mechanical losses and (iii) shunt field Cu losses. • Variable losses including (i) copper losses in armature winding (ii) copper losses in series field winding (iii) copper losses in commutating pole winding (iv) copper losses in compensating winding if any and (v) losses due to brush contact resistance. 21.2. Copper Losses In general copper losses will occur in those parts of dc machines that carry electric current. These losses could be further subdivided into three groups. 1. Armature copper loss 2. Field copper loss , and 3. Loss due to brush contact resistance. Armature copper loss Let the current flowing in the armature winding be designated Ia amperes and its resistance Ra Ω. Then armature copper losses = I a2 Ra watts. Moreover, series field winding, interpole winding and compensating windings are connected in series with the armature winding. As such the current flowing in these winding is also equal to the armature current, Ia. Thus, the total armature copper losses = I a2 (Ra + Rse + Ri + Rc ) Where Rse, Ri and Rc are the resistance of series field winding, interpole winding and compensating winding respectively. The armature circuit copper losses vary as the square of the armature current. As such these losses could be also called variable losses of the machine this loss is about 30 to 40% of full-load losses. Field Copper loss If Ish is the current flowing in the shunt field winding then , Copper losses in the shunt field winding = Ish2 Rsh where Rsh is the resistance of the shunt field winding however the V resistance of the shunt field winding R sh = I sh Where, V is the terminal voltage of the dc machine, i.e. voltage across the armature terminals. Hence, shunt field copper losses = V x Ish, watts 236 Introduction to Electrical Machines Terminal voltage and Ish are practically constant as such this loss can be taken into the group of constant losses in the dc machine. Losses due to Brush contact Resistance In dc machines, brushes makes a sliding contact with the commutator and the conduction of current is through minute arcs. The contact voltage drop for a particular grade of brush is more or less constant, varying from 1 to 2 volts for normal carbon brushes. The brush contact loss is equal to the product of contact voltage drop and the armature current. Strictly speaking, it is not a copper loss; however it is normally included in the classification of copper losses. 4.21.3. Iron Losses The reversal of magnetization of the armature core leads to iron losses in the core and the teeth of the armature structure. Iron losses can further be subdivided into: i) hysteresis loss and ii) eddy current loss Hysteresis loss Hysteresis loss is mainly due to the reversal of magnetization in the armature core and depends upon the area of the hysteresis loop of the magnetic material used for armature core, the volume of the core and the frequency of magnetic flux reversal. Area of the hysteresis loop again depends upon the flux density to which the material is being worked. Dr. Charles Steinmetz suggested an empirical formula based on a series of tests for calculating the hysteresis loss, which expressed as follows, Ph = ξVf (Bmax )1.6 watts . where, ξ is a constant, known as Steinmetz’s coefficient or hysteresis coefficient for a particular material. Its value for 4% silicon steel and sheet steel is 275 and 500 respectively. Hysteresis loss is reduced by choosing a core material with low hysteresis coefficient such as alloy steel. Eddy current losses As the armature core rotates relative to the magnetic field, it cuts the flux. Thus as per the lows of electromagnetic induction, a small emf is induced in the armature core body, which circulates a large current in the armature core due to its small resistance. Such a circulating current is called eddy current and the power losses due to the flow of this current are called eddy current losses. The resistance can be greatly increased by laminating the armature core of the dc machine, thereby reducing the magnitude of eddy current to an appreciable value. The eddy current losses depend upon the following factors: i) Thickness of laminations, t ii) Frequency of flux reversal, f iii) Maximum value of flux density, Bmax 237 Chapter One: Electromagnetic Principles iv) Volume of armature core V and v) Quality of iron. Hence the eddy current losses occurring in the armature core and teeth of the dc machine are given by Pc = KB2max f 2 t 2 V, watts 21.4. Mechanical Losses Mechanical losses are due to the rotation of the armature and can be subdivided into three categories. 1. bearing friction 2. brush friction and 3. air friction (winding) There are also termed as friction and windage losses. The bearing friction losses occurring in dc machine depend upon (i) the pressure on bearing, (ii) Peripheral speed of the shaft at the bearing and (iii) coefficient of friction between the bearing and the shaft. The brush friction losses are quite appreciable in dc machines. These losses are dependent upon (i) the brush pressure, (ii) the peripheral speed of the commutator and (iii) the type of the brush. The windage losses are mainly produced by the rotation of armature. These losses depend upon (i) peripheral speed of the armature, (ii) armature diameter, (iii) armature core length and (iv) construction of the machine These three components added together give the total mechanical losses occurring in the machine these are practically constant provided the speed remains the same during the loading of the machine. 24. EFFICIENCY The ratio of output of the machine to its input is generally called the efficiency of the machine. Efficiency = output input Input to the machine = output + total losses Thus, efficiency = Output Output + total losses Total losses in dc machine = constant losses + variable losses 238 Introduction to Electrical Machines = Wc + Wv Hence, efficiency η = Output Output + constant losses + variable losses Condition for maximum efficiency Let the dc machine works as a dc shunt generator, with a terminal voltage V volts and load current IL ampere. Then the power output of the dc generator = V × IL, Variable loss of the armature circuit = I2a Ra (assuming that the machine is not provided with interpole and compensating windings). Armature current, Ia = IL + ISh Shunt field current is quite small compared to the load current and Ia can be assumed equal to IL. With this assumption, Variable losses, Wv = I L2 Ra Let the constant losses be = WC Then, the efficiency of the generator, η= Or V ×IL V I L + WC + I L2 Ra η = 1 W + I L Ra 1+ c ( VI ) L V Efficiency will be maximum, when the denominator minimum i.e. d dI L or − WC I + L Ra = 0 V VI L Wc VI 2 L + Ra = 0 V or Ra Wc = V VI L2 or I L2 R a = W c i.e. copper losses in armature circuit = Constant Losses. Hence efficiency of a dc machine will be maximum, when the variable losses are equal to the constant losses. Load current corresponding to maximum efficiency is given by 239 Chapter One: Electromagnetic Principles IL = Wc Ra Example 4.21 The armature of a 6 pole, 6 circuit dc shunt motor takes 300 A at the speed of 400 rpm. The flux per pole is 75 × 10-3 Wb. The number of armature turns is 500. The torque lost in windage, friction and iron losses can be assumed as 2.5 per cent. Calculate the torque developed by the armature, shaft torque and shaft power in kW. Solution i) The torque developed by the armature of a dc motor is given by Ta = 0.159 PφI a Z Nm a (i) Number of poles of shunt motor, P=6 Armature winding has 6 circuits, thus, a=6 Armature current, Ia = 300 A Number of armature turns = 500 Thus total conductors on the armature, Z = 2 × 500 = 1000 Flux per pole, φ = 75 × 10-3 Wb Substituting these values in Eq. (i), Armature torque, ii) 6 × 75 × 10 −3 × 300 × 1000 6 = 3577.5 Nm Ta = 0.159 Torque lost in windage, friction and iron losses = 2.5% of Ta = 0.025 × 3577.5 = 89.44 N m Thus, shaft torque, Tsh = 3577.5 − 89.44 = 3488.06 N m iii) Shaft power, 2 πNTsh kW 60 × 1000 2 π × 400 × 3488.06 = 60 × 1000 = 146.22 kW Psh = 240 Introduction to Electrical Machines Example 4.22 2 A 200 V dc shunt motor takes a total current of 100 A and runs at 750 rpm. The resistance of the armature winding and of shunt field winding is 0.1 and 40 Ω respectively. Find out i) the torque developed by the armature, and ii) the copper losses. If the friction and iron losses amount to 1500 W, also calculate iii) shaft power, iv) shaft torque, and v) efficiency. Solution i) Voltage applied across the motor, VL = 200 V Resistance of the shunt field winding, Rsh = 40 Ω Shunt field current, I sh = VL 200 = = 5 .0 A R sh 40 Total current drawn by the motor, IL = 100 A. Thus, armature current Ia = IL −Ish = 100 − 5 = 95 A Armature resistance, Ra = 0.1 Ω Back emf, Eb = V − IaRa = 200 − 95× 0.1 = 190.5 V Mechanical power developed, Pmec = Eb Ia = 190.5 × 95 = = 18097.5 W Now, mechanical power developed, E b I a = 2πNTa 60 Thus, torque developed by the armature, E I Ta = 60 b a 2 πN 18097.5 = 60 × 2 π × 750 = 230.3 N m ii) The back emf for a dc motor is given by, Eb = V − IaRa Or EbIa = VIa − Ia 2Ra Thus armature copper losses, Ia 2Ra = VIa − EbIa = 200 × 95 − 18097.5 = 902.5 W 241 Chapter One: Electromagnetic Principles Field copper losses = Ish 2Rsh = 52 × 40 = 1000 W Total copper losses = 902.5 + 1000 = 1902.5 W Copper losses could also be determined as follows: Input to the motor = VLIL = 200 × 100 = 20,000 kW Mechanical power developed, Eb Ia = 18097.5 W = 20000 − 18097.5 = 1902.5 W Total copper losses iii) Friction and Iron losses = 1500 W Total copper losses = 1902.5 W Output = 20 000 −(1500+1902.5) = 16 597.5 W shaft power ≈ 16.6 kW or iv) Shaft power Psh = 2 πNTsh kW 60 × 1000 Shaft torque, v) 60 × 1000 × 16.6 2 π × 750 = 211.24 N m Tsh = Efficiency, η= Poutput Pinput = 82.99% = 16597.5 20000 Example 4.23 A 100 V series motor takes 45 A when running at 750 rpm. Its armature resistance is 0.22 Ω, while the series field resistance is 0.13 Ω. Iron and friction losses amount to 750 W. Find (i) shaft power, (ii) total torque and (iii) shaft torque. Solution i) Voltage applied to the series motor, = 100 V Total resistance of the armature circuit, = Ra + Rse = 0.22 + 0.13 = 0.35Ω Current in the armature circuit, Ia = 45 A Back emf, Eb = 100 − 45 × 0.35 = 84.25 V Mechanical power developed = Eb Ia 242 Introduction to Electrical Machines = 84.25 × 45 = = 3791.25 W Iron and friction losses = 750 W Output = 3791.25 − 750 = 3041.25 W Thus shaft power = 3.041 kW ii) Mechanical power developed = Eb Ia = 2πNTa 60 Total torque, 3791.25 2π × 750 T° = 48.25 N m Ta = 60 × iii) Or, Output = 3041.25 W Output = 2 πNTsh 60 shaft torque, 60 × 3041.25 2π × 750 = 38.7 N m Tsh = Example 4.24 A long-shunt compound motor takes a current of 24A from 240 V mains. Determine the efficiency if resistance of the armature, series field and shunt field are 0.1, 0.08 and 60 ohms respectively and stray losses are 500 watts. Solution Input line current, IL = 24 A Supply voltage, V = 240 V Total input power to the machine V×IL = 240×24 = 5760 W Shunt field current, I sh = Series field current, Ise = Ia = IL − Ish V 240 = = 4A R sh 60 = 24 − 4 = 20 A Total losses = Stray losses + armature copper loss + series field copper loss + shunt field copper loss = Ps + Ia2Ra + Ise2Rse + V Ish = 500 + 202×0.1 + 202×0.08 + 240×4 = 1532 W 243 Chapter One: Electromagnetic Principles Useful Output = Total input − Total losses = 5760 – 1532 = 4228 W Efficiency , η= Usefull output 4228 = = 73.4% Total input 5760 Example 4.25 A 100 V dc series motor has an armature resistance of 0.2 Ω and series field resistance of 0.25Ω. When its pulley exerts a torque of 27.58 N.m it runs at a speed of 600 rpm, iron and friction losses at this speed are 300 W. Determine (a) lost torque (b) copper losses and (c) efficiency. Solution Torque exerted, T =27.58 N.m Speed, N= 600 r.p.m. Motor output = T × 2π × N 27.58 × 2 π × 600 = = 1732 W 60 60 Stray losses, Ps =300 W Let input current be of I amperes to give a torque of 27.58 N.m. Copper losses in series field and armature = I2(Ra + Rse) = I2 (0.2 + 0.25 )= 0.45 I2 Input to motor =Motor output +− Ps + copper losses = 1732 + 300 + 0.45 I2 = 2032 + 0.45 I2 ……..(i) Also motor input =VI = 100 I ……..(ii) Comparing expressions (i) and (ii) we get 0.45 I2 +2032 = 100 I or I = 22.5 (b) Copper losses = I2(Ra + Rse)= 22.52 ×0.45 = 227.8 W (c) Efficiency of the motor, η = (a) Torque lost = = Output 1732 × 100 = × 100 = 76.97% Input 22.5 × 100 Losses in watts 300 + 227.8 = = 8 .4 N .m 600 2 π × N / 60 2π × 60 244 Introduction to Electrical Machines PROBLEMS 4.1. The open-circuit voltage of a separately excited dc generator is 350 V when it is running at 1800 rpm. If the excitation is held constant, what is the output voltage at 1200 rpm? At what speed would the generator run to produce 300 V? Ans. 233.3V; 1543 rpm. 4.2. For the generator of Problem 4.2 draw no-load saturation curves at 1200 and 800. Determine the generated voltage at the speeds for currents of 2.4 A and 5 A. 4.3. A shunt generator has the no-load saturation curve shown in Figure 4.62. Determine a) The value of the field circuit resistance if the generated voltage is 350 V. b) The output voltage if the field circuit resistance is 60Ω. c) The value of the critical field resistance for a speed of 2000 rpm. Ans. (a) 58 Ω, (b) 340 V , (c) 75 Ω OCC curve at 2000 rpm Field current, Ish (A) Figure 4.62 4.4. A shunt generator with a field resistance of 60 Ω has the no-load saturation curve of Figure 4.63 at 2000 rpm. If Ra is 0.16 Ω and the brush drop is 2 V, plot a graph of output voltage versus load current as it varies from 0 to 40 A in steps of 5 A. Assume that the field flux remains essentially constant. Figure 4.63 4.5. A 10-kW 250 V generator has a shunt-field resistance of 125Ω, Ra of 0.4Ω, a stray load loss of 540 W, and a 2-V brush voltage drop. If it is running at its rated output, calculate: 245 Chapter One: Electromagnetic Principles a) The generated voltage. b) The efficiency. 4.6. A separately excited dc generator is driven at 1200 rpm and the following data were recorded: a) b) Ans. a) 268.8 V ; b) 84.5% Ish (A) 0 0.1 0.2 0.3 0.4 0.6 0.8 1.0 Eg (V) 5 26 50 76 98 131 153 170 Draw the no-load curves at 1200 and l500 rpm. The field circuit resistance is 200 Ω. What is the open-circuit voltage of the machine if it is connected as a shunt generator running at 1500 rpm. c) What is the critical field resistance at 1000 rpm. 4.7. In a 150-kW 600-V short-shunt compound generator, 645.6 V is induced in the armature when the generator delivers rated load at 600 V. The shunt-field current is 6 A. Rse =0.08Ω. Determine: a) The armature resistance and shunt-field resistance. (Neglect brush voltage drop.) b) 600 V. 4.8. A separated excited generator has a no-load voltage of 125 V at a field current of 2.1 A when driven at a speed of 1600 rpm. Assuming that it is operating on the straight-line portion of its saturation curve, calculate: a) b) The voltage regulation if the emf induced in the armature on no load is Ans.: a) 0.1Ω, 103.3Ω; b) 10.0% The generated voltage when the field current is increased to 2.6 A. The generated voltage when the speed is reduced to 1450 rpm and the field current is increased to 2.8 A. 4.9. A 10-kW 220-V compound generator (long-shunt connected) is operated at no load at the proper armature voltage and speed, from which the stray-power loss is determined to be 705Ω. The shunt-field resistance, Rsh=110Ω, the armature resistance Ra=0.265Ω and the series-field resistance, Rse= 0.035Ω. Assume a 2-V brush drop and calculate the full-load efficiency. Ans. 83.9% 4.10. A 250-kW 240-V 1200 rpm short-shunt compound generator has a shunt-field resistance of 24Ω, an armature resistance of 0.003Ω, a series-field resistance of 0.0013Ω, and a commutating-field resistance of 0.004Ω, calculate the generated emf at full load. 4.11. The terminal voltage of a 200-kW shunt generator is 600 V when it delivers rated current. The resistance of the shunt field circuit is 250Ω, the armature resistance is 0.32Ω, and the brush resistance is 0.014Ω . a) Determine the emf induced at rated current. b) induced. 4.12. a) The terminal voltage is 620 V at half-rated current. Determine the emf Ans. a) 712.1 V; b) 676.5 V The no-load saturation curve of a dc shunt generator when running at a speed of 1000 rpm is as illustrated in Fig. 4.64. Determine the critical field resistance at; 246 Introduction to Electrical Machines b) i. 1000 rpm ii. 1500 rpm If the resistance of the field coils is 100Ω, find the value of the field rheostat to set the open-circuit voltage to l25 V at a speed of 1000 rpm. 1000 rpm Field current, Ish (V) Figure 4.64 No-load saturation curve for problem 4.12 4.13. a) The dc generator has the following OCC at 800 r.p.m Ish (A) 0 1 2 3 4 5 Eg (V) 10 112 198 232 252 266 Find the no-load terminal voltage when the machine runs as a shunt generator at 1000 r.p.m. The resistance of field circuit is 70-Ω. b) 270-V? What additional field regulator resistance will be required to reduce the voltage to Ans. a) 330 V; b) 42.5 Ω 4.14. The open-circuit characteristic of a de shunt generator driven at 850 rpm is given by, Ish (A) 2 3 4 5 6 Eg (V) 68 87 100.5 109 112.5 The resistance of the shunt field winding is 22.2 Ω. Find the voltage to which the machine will excite, when it runs at 850rpm self-excited. Ans.108 V 4.15. The relation between the excitation current and the emf generated by separately-excited generator running on open circuit at 600 rpm is given by, Ish (A) 0 1.6 3.2 4.8 6.4 8.0 Eg (V) 0 150 295 398 465 517 Find the voltage to which the dc machine will excite as a shunt generator on open circuit with shunt field resistance of 60 Ω (i) at 600 rpm, (ii) at 500 rpm and (iii) critical speed of the generator. 4.16. The magnetization characteristic of dc shunt generator running at 850 rpm is given by, 247 Chapter One: Electromagnetic Principles Field current (A) 0 0.8 1.6 2.4 3.2 4 Emf (V) 0 28 57 76 90 100 Calculate i. The emf to which the machine will excite, when the field circuit resistance is 22 Ω, ii. the emf when an additional resistance of 8Ω is included in the field circuit, iii. the value of field resistance Rf for normal voltage of 100V, iv. Critical resistance of the field circuit and v. Critical speed with field resistance Rf as calculated in part (iii). Ans.(i) 110 V (ii) 81 V (m) 25Ω (iv) 37Ω (v) 575 rpm 4.17. The open circuit characteristics of dc generator driven at 1000 rpm is given by, The magnetization characteristic of 4 pole, lap wound dc shunt generator with 400 armature conductors and running at 750 rpm is given by, Ish ,(A) 0 0.2 0.4 0.6 0.8 1.2 1.6 2,0 2.4 Emf, Eg (V) 10 44 84 120 150 186 206 220 230 240 260 274 The machine is separately excited from supply of 220 V. coil is 40 Ω . 3.0 4.4 5.6 The resistance of the field i. Calculate the range of rheostat (ohms and current carrying capacity) included in the field circuit to give voltage from 100 to 250 V ii. What is the value of the resistance in the field rheostat, when the terminal voltage is 200 V? iii. If the field rheostat is kept constant at 50 Ω and exciting voltage is 220 V, what is the induced emf for generator speeds of 800, 1000 and 1200 rpm. Ans. (i) 377 Ω, 3.7 A (ii) 94 Ω (iii) 184 V, 230 V, 276 V 4.18. The open circuit characteristics of a dc shunt generator driven at 1000 rpm is as follows. Field current, Ish (A) 0 Emf, Eg (V) 4 1 2 3 115 230 315 4 360 6 8 10 405 430 450 Based on the above, calculate i. the emf to which the machine will excite with shunt field resistance of 50Ω, ii. the additional resistance in the field circuit to reduce the emf to 392 V, and iii. Critical resistance of the shunt field circuit at 600. 4.19. A 4-pole, 440 V dc shunt motor takes a full-load current of 40-A. the armature is wave wound with 762 conductors. The flux per pole is 0.025 Wb. The armature resistance is 0.25 Ω. Assume a brush contact drop of 2V, calculate the full load speed. Ans. 674 rpm 248 Introduction to Electrical Machines 4.20. A dc shunt machine connected to 250 V mains has an armature resistance of 0.12 Ω and a field resistance 0f 100 Ω. Find the ratio of the speed as a generator to its speed as a motor, the line current in each case being 80A. 4.21. A 460 V series motor runs at 500 r.p.m. taking a current of 40 A. Calculate the speed and percentage change in torque if the load is reduced so that the motor is taking 30 A. Total resistance of the armature and field circuit is 0.8 Ω. Assume flux is proportional to the field current. Ans. 679 rpm, 43.75% 4.22. A 200 V dc series motor runs at 1000 r.p.m. and takes 20 A. Combined resistance of armature and field is 0.4 Ω. Calculate the resistance to be inserted in series so as to reduce the speed to 800 r.p.m., assuming torque to vary as square of the speed and linear magnetization curve. 4.23. A 250 V dc shunt motor draws 5 A from the line on no load and runs at 1000 rpm. The armature resistance and shunt field resistance are 0.2 and 250 Ω respectively. What will be the speed of the motor, when it is loaded and takes current of 50 A. Armature reaction weakens the field by 3%. Ans.994 rpm 4.24. A 250 V dc shunt motor has a shunt field resistance of 250 Ω and an armature resistance of 0.25 Ω. For a given load torque and no additional resistance included in shunt field circuit, the motor runs at 1500 r.p.m. drawing an armature current of 20 A. If a resistance of 250 Ω is inserted in series with the field, the load torque remaining the same, find out the new speed and armature current. Assuming the magnetization curve to be linear. 4.25. A 4-pole, 440 V dc shunt motor takes a full load Current of 40 A. The armature is wave wound with 762 conductors. The flux per pole is 0.025 Wb. Effective armature resistance is 0.25Ω. Assuming brush contact drop of 2 V, calculate the full load speed. Ans.674 rpm 4.26. The armature of a 4-pole dc shunt motor has a lap winding accommodated in 50 slots, each containing 24 conductors. If the useful flux per pole is 25 m Wb, calculate the total torque developed, when the armature current is 45 A. 4.27. A 240 V dc shunt motor has armature and shunt field resistance of 0.04 and 100 Ω respectively. i. Calculate the value of resistance that must be added to the field circuit in order to increase its speed from 1200 to 7500 rpm, when the supply current is 200 A. ii. With the field resistance as in (i), find the speed of the motor, when the supply current is 100 A. Ans.(i) 25 Ω (ii) 1525 rpm 4.28. A 230 V, 4-pole, dc shunt motor running at 750 rpm gives 7.46 kW with an armature current of 38 A and field current of 1.0 A. The armature is wave wound and has 400 conductors. The resistance of armature winding is 0.2 Ω and the drop at each brush is 1.0 V. Determine (i) useful torque, (ii) total torque, (iii) useful flux per pole, (iv) rotational losses, and (v) efficiency. 4.29. A 440 V, 6-pole dc shunt motor has a wave connected armature winding with 1100 conductors. The useful flux per pole is 20 mWb. The armatures and field resistances are 0.4 and 220 Ω respectively. Ignoring the effect of armature reaction, find the speed and the total developed torque, when the current of 22 A is taken from the mains. If the iron, friction and windage losses amount to 800 W, find the useful torque, 249 Chapter One: Electromagnetic Principles shaft power and efficiency. 209.9 N m (iii) 190.4 N m (iv) 7.83 kW (v) 81 % 4.30. A 200 V dc shunt motor with an armature resistance of 0.4 Ω is excited to give constant main field. At full load, the motor runs at 600 rpm and takes an armature current of 25 A. If a resistance of 0.8 Ω is placed in the armature circuit, find the speed at (i) full load torque and, (ii) double full load torque. 4.31. Ans. (i) 393 rpm (ii) A 220 V dc shunt motor takes a no load armature current of 10 A and runs at 1500 rpm. At full load, armature current is 100 A and the motor runs at 1470rpm. Resistance of the armature circuit is 0.1Ω. Calculate the following : (i) back-emf at no load and at full load, (ii) percentage reduction in flux due to armature reaction and (iii) ratio of no load to full-load torque developed by the armature. Ans. (i) 219 V, 210 V (ii) 2.14 % (iii) 0.102 4.32. A 440 V dc shunt motor takes an armature current of 120 A at full load and runs at 800 rpm. Find the speed of the motor, when the torque on the motor is reduced to 60 per cent of its full load value and a resistance of 1.5Ω is inserted in the armature circuit. Take the effective armature resistance as 0.2Ω. 4.33. A 250 V dc shunt motor takes 21 A and runs at 600 rpm while driving a load, the torque of which is constant. The resistance of the armature and the field are 0.5 and 250 Ω respectively. It is desired to raise the speed from 600 to 800 rpm. What resistance must be included in the shunt field circuit? Assume the magnetization curve to be straight line. Ans.88 Ω 4.34. A 220 V dc shunt motor has an armature resistance of 0.25 Ω and a field resistance of 55Ω. The motor while driving a constant load torque takes 64 A and runs at 500 rpm. Find the speed when a resistance of 20 Ω is inserted in the shunt field circuit. Assume the flax to be proportional to the field current. 4.35. A 10 hp, 500 V dc shun motor has an armature resistance of 0.25 Ω and field resistance of 400Ω. Its full load efficiencyis 85 %. It is desired to reduce the speed of motor by 30 % by including a resistance in the armature circuit, keeping the same field and armature currents. Assuming that all losses except copper losses vary directly with the speed, find the value of the resistance inserted in the armature circuit and also the efficiency of the motor, when it is running at the reduced speed. Ans.9.12Ω, 59.6% 250 Introduction to Electrical Machines CHAPTER FIVE SYNCHRONOUS MACHINES 5.1. INTRODUCTION A synchronous machine rotates at a constant speed in the steady state. Unlike induction machines, the rotating air gap field and the rotor in the synchronous machine rotate at the same speed, called the synchronous speed. Synchronous machines are used primarily as generators of electrical power. In this case they are called synchronous generators or alternators. They are usually large machines generating electrical power at hydro, nuclear, or, thermal power stations. Synchronous generators with power ratings of several hundred MVA (mega-volt-amperes) are quite common in stations. Synchronous generators are the primary energy conversion devices of the world's electical power systems today. In spite of continuing research for more direct enery conversion techniques, it is conceded that synchronous generators will continue to be used well into the next century. Like most rotating machines, a synchronous machine can also operate as both a generator and a motor. In large sizes (several hundred or thousand kilowatts) synchronous motors are used for pumps in generating stations and in small sizes (fractional horsepower) they are used in electrical clocks, timers, record turntables, and so forth where constant speed is desired. Most industrial drives run at variable speeds. In industrial, synchronous motor are used mainly where a constant speed is desired. In industrial drives, therefore, synchronous motors are not as widely used as induction or dc motors. A linear version of the synchronous motor (LSM) is being considered for highspeed transportation systems of the future. An important feature of a synchronous motor is that it can draw either lagging or leading reactive current from the ac supply system. A synchronous machine is a doubly excited machine. Its rotor poles are excited by a dc current and its stator windings are connected to the ac supply . The air gap flux is therefore the resultant of the fluxes due to both rotor current and stator current. In induction machines, the only source of excitation is the stator current, because rotor currents are induced currents. Therefore, induction motors always operate at a lagging power factor, because lagging reactive current is required to establish flux in the machine. On the other hand, in a synchronous motor, if the rotor field winding provides just the necessary excitation, the stator will draw no reactive current; that is, motor will operate at a unity power factor. If the rotor excitation current is decreased, lagging reactive current will be drawn from the ac source to aid maanetization by the rotor field current and the machine will operate at a lagging power factor. If the rotor field current is increased, leading reactive current will be drawn from the ac source to oppose magnetization by the rotor field current and the machine will operate at a leading power factor. Thus, by changing the field current, the power factor of the synchronous motor can be controlled. If the motor is not loaded but is simply floating on the ac supply system, it will thus behave as a variable inductor or capacitor as its rotor field current is changed. A synchronous machine with no load is called a synchronous condenser. It may be used in power transmission systems to regulate line voltage. In industry, synchronous motors are sometimes used with other induction motors and operated in an overexcited mode so that they draw leading current to 251 Chapter One: Electromagnetic Principles compensate the lagging current drawn by the induction motors, thereby improving the overall plant power factor. Example 5.3 illustrates the use of synchronous motors for power factor improvement. The power factor characteristics of svnchronous motors will be further discussed in a later section. Figure 5.1 cut–away view of synchronous machines 5.2. CONSTRUCTION MACHINES OF THREE-PHASE SYNCHRONOUS The stator of the three-phase synchronous machine has a three-phase distributed winding similar to that of the three-phase induction machine. Unlike the dc machine, the stator winding, which is connected to the ac supply system, is sometimes called the armature winding. It is designed for high voltage and current. The rotor has a winding called the field winding, which carries direct current. The field winding on the rotating structure is normally fed from an external dc source through slip rings and brushes. Synchronous machines can be broadly divided into two groups as follows: a) High-speed machines with cylindrical (or non-salient pole) rotors. b) Low-speed machines with salient pole rotors. The cylindrical or non-salient pole rotor has one distributed winding and an essentiallv uniform air gap. These motors are used in large generators (several hundred megawatts) with two or sometimes four poles and are usually driven by steam turbines. The rotors are long and have a small diameter, as shown in Figure 5.2. On the other hand, salient pole rotors have concentrated windings on the poles and a nonuniform air gap. Salient pole generators have a large number of poles, sometimes as many as 50, and operate at lower speeds. The synchronous generators in hvdroelectric power stations are of the salient pole type and are driven by water turbines. These gencrators are rated for tens or hundreds of megawatts. The rotors are shorter but have a large diameter as shown in Figure 5.3. Smaller salient pole synchronous machines in the range of 50 kW to 5 MW am also used. Such synchronous generators are used independently as emergency power 252 Introduction to Electrical Machines supplies. Salient pole synchronous motors are used to drive pumps, cement mixers, and some other industrial drives. Stator Frame Rotor Field winding N X X S Distributed stator Winding S X Salient pole X N Basement (a) Cylindrical (non-Salient) rotor Rotor Field winding N X Distributed stator Winding X S X Basement (b) Figure 5.3 rotor construction of synchronous machines (a) salient pole; (b) non-salient pole In the following sections the steady-state performance of the cylindrical rotor synchronous machine will be studied first. Then the effects of saliency in the rotor poles will be considered. 5.3. SYNCHRONOUS GENERATORS Refer to Figure 5.4 (a) and assume that when the field current If flows through the rotor field winding, it establishes a sinusoidallv distributed flux in the air gap. If the rotor is now rotated bv the prime mover (which can be a turbine or diesel engine or dc motor or induction motor), a revolving field is produced in the air gap. This field is called the excitation field, because it is produced by the excitation current If. The rotating flux so produced will change the flux linkage of the armature windings aa', bb', and cc' and will induce voltages in these stator windings. These induced voltages, shown in Fig. 5.4b, have the same magnitudes but are phase-shifted by 120° electrical degrees. 253 Chapter One: Electromagnetic Principles a c' b' If b c (a) (b) Figure 5.4 Excitation voltage in synchronous machines They are called excitation voltages Ef. The rotor speed and frequencv of the induced voltage are related by n= 120 f p 5.1 f = np 120 5.2 Or Where n is the rotor speed in rpm P is the number of poles The excitation voltage in rms is E f = 4.44 fφ f NK w Where 5.3 φf is the flux per pole due to the excitation Current If N is the number of turns in each phase Kw is the winding factor E f ∝ nφ f 5.4 The excitation voltage is proportional to the machine speed and excitation flux, and the latter in turn depends on the excitation current If. The variation of the excitation voltage with the field current is shown in Figure 5.5. The induced voltage at If = 0 is due to the residual magnetism. Initially the voltage rises linearly with the field current, but as the field current is further increased, the flux φf does not increase linearly with If because of saturation of the magnetic circuit, and therefore Ef levels off. If the machine terminals are kept open, the excitation voltage is the same as the terminal voltage and can be measured using a voltmeter. The curve shown in Figure 5.5 is known as the open-circuit characteristic (OCC) or- magnetization characteristic of the synchronous machine. 254 Introduction to Electrical Machines Figure 5.5 Open circuit characteristic (OCC) or magnetization characteristic of a synchronous machine. Figure 5.6 Space phasor diagram If the stator terminals of the machine (Fig. 5.1(c)) are connected to a 3φ load, stator current Ia will flow. The frequency of Ia will be the same as that of the excitation voltage Ef. The stator currents flowing in the 3φ windings will also establish a rotating field in the air gap. The net air gap flux is the resultant of the fluxes produced by rotor current If and stator current Ia. Let φf be the flux due to If and φa be the flux due to Ia known as the armature reaction flux. Then, φr = φ f + φa = resultant air gap flux, assuming no saturation It may be noted that the resultant and the component fluxes rotate in the air gap at the same speed, governed by Eq.5.1. The space phasor diagram for these fluxes is shown in Figure 5.6. The rotor field mmf Ff (due to If) and the flux φf produced by the mmf Ff are represented along the same line. The induced voltage Ef lags the flux φf by 90°. Assume that the stator current Ia lags E1 by an angle θ. The mmf Fa (due to the current Ia) and the flux φa produced by the mmf Fa are along the same axis as the current Ia. The resultant mmf Fr is the vector sum of the mmf's Ff and Fa. Assuming no saturation, the resultant flux φr is also the vector sum of the fluxes φf and φa. 255 Chapter One: Electromagnetic Principles 5.4. THE INFINITE BUS Synchronous generators are rarely used to supplv individual loads. These generators, in general, are connected to a power supply system known as an infinite bus or grid. Because a large number of synchronous generators of large sizes are connected together, the voltage and freduency of the infinite bus hardly change. Loads are tapped from the infinite bus at various load centers. A typical infinite bus or grid system is shown in Figure 5.7. Transmission of power is normally at higher voltage levels (in hundreds of kilovolts) to achieve higher efficiency of power transmission. However, generation of electrical energy by the synchronous generators or alternators is at relatively lower voltage levels (20-30 kV). A transformer is used to step up the alternator voltage to the infinite bus voltage. At the load centers, the infinite bus (or grid) voltage is stepped down through several stages to bring the voltage down to the domestic voltage level (115/230 V) or industrial voltage levels such as 4.16 kV, 600 V, or 480 V. Figure 5.7 Infinite bus (or grid) svstem. In a power plant the synchronous generators are connected to or disconnected from the infinite bus, depending on the power demand on the grid system. The operation of connecting a synchronous generator to the infinite bus is known as paralleling with the infinite bus. Before the alternator can be connected to the infinite bus, the incoming alternator and the infinite bus must have the same. 1. Voltage 2. Frequency 3. Phase sequence 4. Phase 256 Introduction to Electrical Machines In the power plant the satisfaction of these conditions is checked by an instrument known as a synchroscope, shown in Figure 5.8. The position of the indicator indicates the phase difference between the voltages of the incoming machine and the infinite bus. The direction of motion of the indicator shows whether the incoming machine is running too fast or too slow, that is, whether the frequency of the incoming machine is higher or lower than that of the infinite bus. The phase sequence is predetermined because if phase sequence is not correct it will produce a disastrous situation. When the indicator moves very slowly (i.e., frequencies almost the same) and passes through the zero phase point (vertical up position) the circuit break is closed and the alternator is connected to the infinite bus. Figure 5.8 Synchroscope. A set of synchronizing lamps can be used to check that the conditions for paralleling the incoming machine with the infinite bus are satisfied. In a laboratory such a set of lamps can be used to demonstrate what happens if the conditions are not satisfied. Figure 5.9 shows the schematic of the laboratory setup for this purpose. The prime mover can be a dc motor or an induction motor. It can be adjusted to a speed such that the frequency of the synchronous machine is the same as that of the infinite bus. For example, if the synchronous machine has four poles, the prime mover can be adjusted for 1800 rpm so that the frequency is 60 cycles-the same as that of the infinite bus. The field current If can then be adjusted so that the two voltmeters (V1 and V2) read the same. If the phase sequence is correct all the lamps will have the same brightness, and if the frequencies are not exactlv the same the lamps will brighten and darken in step. Let us examine what we expect to observe in the lamps if the conditions are not satisfied. The phenomena can be explained by drawing phasor diagrams for the voltages of the incoming machine and the infinite bus. Let EA, EB, EC represent the phasor voltages of the infinite bus. Ea, Eb, Ec represent the phasor voltages of the incoming machine. EAa, EBb, ECc represent the phasor voltages of the synchronizing lamps. The magnitude of these will represent the brightness of the corresponding lamps. 257 Chapter One: Electromagnetic Principles Figure 5.9 Schematic diagram for paralleling a synchronous generator with the infinite bus using synchronizing iamps. 1. Voltages are not the sarne, but frequency and phase sequence are the same. Referring to Figure 5.10a, one sees that the two sets of phasor voltages (EA, EB, EC and Ea, Eb, Ec) rotate at the same speed. The lamp voltages EAa, EBb,and ECc have equal magnitudes and therefore all the three lamps will glow with the same intensity. To make the voltages equal, the field current If must be adjusted. 2. Frequencies are not the satrte, but voltages and phase sequences are the same. The two sets of phasor voltages rotate at different speeds, depending on the frequencies. Assume that the phase voltages are in phase at an instant t = t1 (Figure 5.l0(b)). At this instant, the voltages across the lamps are zero and therefore they are all dark. If f1 > f2 at a later instant t = t2, phasors EA, EB, and EC will move ahead of phasors Ea, Eb, and Ec. Equal voltages will appear across the three lamps and they will glow with the same intensity. It is therefore evident that if the frequencies are different, the lamps will darken and brighten in step. To make the frequencies the same, the speed has to be adjusted until the lamps brighten and darken very slowly in step. It may be noted that as the speed of the incoming machine is adjusted, its voltages will change. Therefore simultaneous adjustment of the field current If will also be necessary tokeep the voltages the same. 3. Phase sequences are not the same, but voltages and frequencies are the same. Let the phase sequence of the voltages of the infinite bus be EA, EB, EC and of the incoming bus Ea, Eb,Ec as shown in Figure 5.l0(c). The voltages across the lamps are of different magnitudes and therefore the lamps will glow with different intensities. If the frequencies are slightlv different, one set of phasor voltages will pass the other set of phasor voltages and the lamps will darken and brighten out of step. To make the phase sequence the same, interchange connections to two terminals; for instance, connect a to B and b to A (Figure 5.9). 258 Introduction to Electrical Machines 4. Phase is not the same, but voltage, frequency, and phase sequeuce are the same. The two sets of phasor voltages will maintain a steady phase difference as shown in Figure 5.10(d) and the lamps will glow with the same intensity. To make the phase the same or the phase difference zero, the frequency of the incoming machine is slightly altered. At zero phase difference all the lamps will be dark, and if the circuit breaker is closed the incoming machine will be connected to the infinite bus. Once the synchronous machine is connected to the infinite bus, its speed cannot be changed further. However, the real power transfer from the machine to the infinite bus can be controlled by adjusting the prime mover power. The reactive power (and hence the machine power factor) can be controlled by adjusting the field Current. Real and reactive power control will be discussed in detail in later sections. EA EAa Ea ECc Ec Eb EC EBb EB (a) EB EA EBb Ea Ea Eb EAa f1 f2 EA EC EB Ec Eb Ec ECc instant t = t1 EC instant t = t2 (b) (c) (d) Figure 5.10 phasor voltages of the incoming machine and the infinite bus 259 Chapter One: Electromagnetic Principles 5.5. SYNCHRONOUS MOTORS When synchronous machine is used as a motor, one should be able to connect it directly to the power supply like other motors, such as dc motors or induction motors. However, a synchronous motor is not self-starting. If the rotor field poles are excited by the field current and the stator terminals are connected to the ac supply, the motor will not start; instead, it vibrates. This can be explained as follows. N S N S Let us consider a two-pole synchronous machine. If it is connected to a 3φ, 60Hz ac supply, stator currents will produce a rotating field that will 3600 rpm in the air gap. Let us represent this rotating field by two stator poles rotating at 3600 rpm, as shown in Figure 5.11(a). At start (t=0), let the rotor poles be at the position shown in Figure 5.11(a). The rotor will therefore experience a clockwise torque, making it rotate in the direction of the stator rotating poles. At t = t1, let the stator poles move by half revolution, shown in Figure 5.1(b). The rotor poles have hardly moved, because of the high inertia of the rotor. Therefore, at this instant the rotor experiences a counterclockwise torque tending to make it rotate in the direction opposite to that of the stator poles. The net torque on the rotor in one revolution will be zero, and therefore the motor will not develop any starting torque. The stator field is rotating so fast that the rotor poles cannot catch up or lock onto it. The motor will not speed up but will vibrate. Figure 5.11 Torque on rotor at start. Two methods are normaly used to start a synchronous motor: a) use variable-frequency supply or b) start the machine as an induction motor. Start with Variable-Frequency Supply By using a frequency converter, a synchronous motor can be brought from standstill to its desired speed. The arrangement is shown schematically in Figure 5.12. The motor is started with a low-frequency supply. This will make the stator field rotate slowly so that the rotor poles can follow the stator poles. Afterward, the frequency is gradually increased and the motor brought to its desired speed. 260 Introduction to Electrical Machines The frequency converter is a costly power conditioning unit, and therefore this method is expensive. However, if the synchronous motor has to run at variable speeds, this method may be used. V, f 3φ sup ply Frequency converter f control V control Synchronous motor If Figure 5.12 Starting of a synchronous motor using a variable-frequency supply. . Start as an Induction Motor If the frequency converter is not available, or if the synchronous motor does not have to run at various speeds, it can be started as an induction motor. For this purpose an additional winding, which resembles the cage of an induction motor, is mounted on the rotor. This cage-type winding is known a damper or amortisseur winding and is shown in Figure 5.13. To start the motor the field winding is left unexcited; often it is shunted by a resistance. If the motor terminals are now connected to the ac supply, the motor will start as an induction motor because currents will be induced in the damper winding to produce torque. The motor will speed up and will approach synchronous speed. The rotor is then closely following the stator field poles, which are rotating at the synchronous speed. Now if the rotor poles are excited by a field current from a dc source, the rotor poles, closely following the stator poles, will be locked to them. The rotor will then run at synchronous speed. If the machine runs at synchronous speed, no current will be induced in the damper winding. The damper winding is therefore operative for starting. Note that if the rotor speed is different from the synchronous speed because of sudden load change or other transients, currents will be induced in the damper winding to produce a torque to restore the synchronous speed. The presence of this restorative torque is the reason for the name "damper" winding. Also note that a damper winding is not required to start a synchronous generator and parallel it with the infinite bus. However, both synchronous generators and motors have damper windings to damp out transient oscillations. Figure 5.13 cage-type damper (or amortisseur) winding in a synchronous machine. 261 Chapter One: Electromagnetic Principles 5.6. EQUIVALENT CIRCUIT MODEL In the preceding sections the qualitative behavior of the synchronous machine as both a generator and a motor has been discussed to provide “feel” for the machine behavior. We can now develop an equivalent circuit model that can be used to study the performance characterislics with sufficient accuracy. Since the steady-state behavior will be studied, the circuit time constants of the field and damper windings need not be considered. The equivalent circuit will be derived on a per-phase basis. The current If in the field winding produces a flux φf in the air gap. The current Ia in the stator winding produces flux φa. Part of it, φal, known as the leakage flux, links with the stator winding only and does not link with the field winding. A major part, φar, known as the armature reaction flux, is established in the air gap and links with the field winding. The resultant air gap flux φr is therefore due to the two component fluxes, φf and φar. Each component flux induces a component voltage in the stator winding. In Figure 5.14a, Ef is induced by φf , Ear by φar , and the resultant voltage Er by the resultant flux φr. The excitation voltage Ef can be found from the open-circuit curve of Figure 5.5. However, the voltage Ear, known as the armature reaction voltage, depends on φar (and hence on Ia). From Figure 5.14(a), E r = E ar + E f 5.5 E f = − E ar + E r 5.6 or From the phasor diagram of Figure 5.14((b), the voltage Ear lags φar (or Ia ) by 90°. Therefore, Ia lags the phasor –Ear by 90°. In Eq. 6.6, the voltage -Ear can thus be represented as a voltage drop across a reactance Xar due to the current Ia. Equation 6.6 can be written as E f = I a jX ar + Er This reactance Xar is known as the reactance of armature reaction or the magnitizing reactance and is shown in Figure 5.14(c). If the stator winding resistance Ra and the leakage reactance Xal (which accounts for the leakage flux φal) are included, the perphase equivalent circuit is represented by the circuit of Figure 5.14(d). The resistance Ra is the effective resistance and is approximately 1.6 times the dc resistance of the stator winding. The effective resistance includes the effects of the operating temperature and the skin effect caused by the alternating current flowing through the armature winding. If the two reactances Xar and Xal are combined into one reactance, the equivalent circuit model reduces to the form shown in Figure 5.14e, where X s = X ar + X al (called synchronous reactance) Z s = Ra + X s (called synchronous impedance) 262 Introduction to Electrical Machines φr φar (a) (b) Xar Ia Ef Ef (c) (d) Xs Ra Ef Ea Ia Vt (e) Figure 5.14 Equivalent circuit of a synchronous machine The sychronous reactance Xs takes into account all the flux, magnetizing as well as leakage, produced by the armature (stator) current. The values of these machine parameters depend on the size of the machine. Table 5.1 shows their order of magnitude in the per-unit system. Table 5.1 Synchronous Machine Parameters Ra Xal Xs Smaller Machines (tens of kVA) 0.05-0.02 0.05 -0.08 0.5-0.8 Larger Machines (tens of MVA) 0.01-0.005 0.1-0.15 1.0-1.5 A 0.1 pu impedance means that if the rated current flows, the impedance will produce a voltage drop of 0.1 (or 10%) of the rated value. In general, as the machine size increases, the per-unit resistance decreases but the per-unit synchronous reactance increases. 263 7.1. 7.2. Chapter One: Electromagnetic Principles 5.7. DETERMINATION OF THE SYNCHRONOUS REACTANCE Xs The synchronous reactance is an important parameter in the equivalent circuit of the synchronous machine. This reactance can be determined by performing two tests, an open-circuit test and a short-circuit test. Open-Circuit Test The synchronous machine is driven at the synchronous speed, and the circuit terminal voltage Vt(= Ef) is measured as the field current If, is varied (see Figure 5.15(a)). The curve showing the variation of Ef with If is known as the open-circuit characteristic (OCC, shown in Fig. 5.15(c)). Because the terminals are open, this curve shows the variation of the excitation voltage Ef with the field current If. Note that as the field current is increased, the magnetic circuit shows saturation effects. The line passing through the linear part of the OCC is called the air gap line. The excitation voltage would have changed along this line if there were no magnetic saturation effects in the machine. (a) (b) (c) Figure 5.15 Open-circuit and short-circuit characteristics. (a) Circuit for open-circuit test. (b) for short-circuit test. (c) Characteristics. Short-Circuit Test The circuit arrangement for this test is shown in Figure 5.15(b). Ammeters are connected to each phase and the terminals are then shorted. The synchronous machine is driven at synchronous speed. The field current If is now varied and the average of the three armature currents is measured. The variation of the armature current with the field current is shown in Figure 5.15c and is known as the short-circuit characteristic (SCC). Note that the SCC is a straight line. This is due to the fact that under short-circuit conditions, the magnetic circuit does not saturate because the air gap flux remains at a low level. This fact can be explained as follows. The equivalent circuit under short-circuit conditions is shown in Figure 5.16 (a). Because Ra << Xs (see Table 5.1), the armature current Ia lags the excitation voltage Ef by almost 90°. The armature reaction mmf Fa therefore oppose the field mmf Ff and the resultant 264 Introduction to Electrical Machines mmf Fr is very small, as can be seen from Figure 5.16(b). The magnetic circuit therefore remains unsaturated even if both If and Ia are large. Also note from the equivalent circuit of Figure 5.16 (a) that the air gap voltage is Er = I a (Ra + jX al ) . Because both Ra and Xal are small (see Table 6.1) at rated current , the air gap voltage will be less than 20 percent of the rated voltage (signifying unsaturated magnetic conditions at short-circuited operation). If the machine stays unsaturated, the excitation voltage Ef will increase linearly with the excitation current If along the air gap line and therefore the armature current will increase linearlv with the field current. Ff Fr Xal Xar Ra Vt = 0 Er Ef Ef Ia (a) Fa (b) Figure 5.16 Short-circuit operation of a synchronous generator Unsaturated Synchronous Reactance This can be obtained from the air gap line voltage and the short-circuit current of the machine for a particular value of the field current. From Figure 5.15(c), Z s( unsat ) = Eda = Ra + jX s( unsat ) I ba 5.8 If Ra is neglected E X s( unsat ) ≈ da I ba 5.9 Saturated Synchronous Reactance Recall that prior to connecting a svnchronous machine to the infinite bus, its excitation voltage is raised to the rated value. From Figure 6.15(c), this voltage is Eca (= rated Vt) and the machine operates at some saturation level. If the machine is connected to the infinite bus, its terminal voltage remains the same at the bus value. If the field current is now changed, the excitation voltage will change, but not along the OCC line. The excitation voltage Ef will change along the line Oc, known as the modified air gap line. This line represents the same magnetic saturation level as that corresponding to the operating point c. This can be explained as follows. 265 Chapter One: Electromagnetic Principles From the equivalent circuit of Figure 5.14(d), Er = Vt + I a (Ra + jX al ) 5.10 If the drop across Ra and Xal is neglected, Er ≈ Vt 5.10 a Because Vt is constant, the air gap voltage remains essentially the same as yhe field current is changed. This implies that the air gap flux level (i.e., magnetic saturation level) remains practicallv unchanged and hence as If is changed, Ef will change linearly along the line Oc of Figure 5.15c. The saturated synchronous reactance at the rated voltage is obtained as follows: Z s( sat ) = Eca = Ra + jX s( sat ) I ba 5.11 If Ra is neglected E X s( sat ) ≈ ca I ba 5.8. 5.12 PHASOR DIAGRAM The phasor diagrams showing the relationship between voltages and currents for both synchronous generator and synchronous motor are shown in Figure 5.17. The diagrams are based on the per-phase equivalent circuit of the synchronous machine. The terminal voltage is taken as the reference phasor in constructing the phasor diagram. The per-phase equivalent circuit of the synchronous generator is shown in Figure 5.17(a). For convenience, the current Ia is shown as flowing out of the machine in the case of a synchronous generator. E f = Vt + I a Ra + jI a X s = E f ∠δ 5.13 The phasor for the excitation voltage Ef is obtained by adding the voltage drops Ia Ra and jIaXs to the terminal voltage Vt. The synchronous generator is considered to deliver a lagging current to the load or infinite bus represented by Vt. In the case of a synchronous motor the current is shown (Figure 5.17(b)) as flowing into the motor. Vt = E f + I a Ra + jI a X s 5.14 E f = Vt ∠0 − I a Ra − jI a X s = E f ∠ − δ 5.15 266 Introduction to Electrical Machines The phasor Ef is constructed by subtracting the voltage drops from the terminal voltage. Here also, the synchronous motor is considered to draw a lagging current from the infinite bus. It is important to note that the angle δ between Vt and Ef is positive for the generating action and negative for the motoring action. This angle δ (known as the power angle) plays an important role in power transfer and in the stability of synchronous machine operation. δ Vt ∠0 Vt φ (a) φ Vt ∠0 δ Vt (b) Figure 5.17 Phasor diagram for sVnchronous machines. Assume Vt,Ia and φ known. (a) Synchronous generator; (b) Synchronous motor. Example 5.1 The following data are obtained for a 3φ, 10 MVA, 14 kV, star-connected synchronous machine. If (A) 100 150 200 250 300 350 Open-Circuit Voltage (kV) (line-to-line) 9.0 12.0 14.0 15.3 15.9 16.4 Air Gap Line Voltage (kV) (line-to-tine Short-Circuit Current (A) 18 490 The armature resistance is 0.07 Ω/phase. a) Find the unsaturated and saturated values of the synchronous reactance in ohms and also in pu. 267 Chapter One: Electromagnetic Principles b) Find the field current required if the synchronous generator is connected to an infinite bus and delivers rated MVA at 0.8 lagging power factor. c) If the generator, operating as in part (b), is disconnected from the infinite bus without changing the field current, find the terminal voltage. Solution The data are plotted in Figure 5.18. a) Base voltage Vb = 14,000 = 8083 V / phase 3 Base current Ib = 10 × 106 = 412.41 A 3 × 14,000 Base impedance Zb = 8083 = 19.6 Ω 412.41 From Eq. (5.8) and Figure 5.18, 18,000 3 490 Zs unsat = × 106 = 21.21 Ω X s unsat = 21.212 − 0.07 2 = 21.2 Ω = 21.2 pu = 1.08 pu 19.6 From Eq. (5.11) and Figure 5.18, and because Ra is Verv small, 14,000 Zs sat = 1/ 2 3 = 16.5 Ω = X 2 + 0.07 2 s sat 490 Figure 5.18 for example 5.1 268 Introduction to Electrical Machines b) It will be convenient to carry out the calculation in pu values rather than in actual values Vt = 1∠0 pu pf = 0.8 = cos 36.9 I a = 1∠ − 36.9 Z s = 0.841∠ tan −1 16.5 = 0.84∠89.8 pu 0.07 Now E f = Vt + I a R a + jI a X s = Vt + I a Z s = 1∠0 + 1∠ − 36.9 ⋅ 0.85∠89.8 = 1∠0 + 0.84∠52.9 = 1.5067 + j0.67 = 1.649∠24 pu Vt = 1.649 × 14∠24 kV, line-to-line Vt = 23.09∠24 kV, line-to-line Note that δ=240 and is positive , as it should be for generator operation. The required field current from the modified air gap line (Figure 5.18) is I f = 1.649 × 200 = 329.8 A c) From the open-circuit data (Figure 5.18) at If = 329.8 A the terminal voltage is Vt = 16.25 kV 5.9. (line-to-line) POWER AND TORQUE CHARACTERISTICS A synchronous machine is normallv connected to a fixed-voltage bus and operates at a constant speed. There is a limit on the power a synchronous generator can deliver to the infinite bus and on the torque that can be applied to the synchronous motor without losing synchronism. Analytical expressions for the steady-state power transfer between the machine and the constant-voltage bus or the torque devcloped by the machine are derived in this section in terms of bus voltage, machine voltage, and machine parameters. The per-phase equivalent circuit is shown again in Figure 5.19 for convenience , where Vt is the constant bus voltage per phase and is considered as the reference phasor. Let Vt = Vt ∠0 5.16 E f = E f ∠δ 5.17 269 Chapter One: Electromagnetic Principles Zs = R a + jX s = Zs ∠θs 5.18 Where the quantities inside the vertical bars represent the magnitudes of phasors. Vt ∠0 E f ∠δ Figure 5.19 Per-phase equivalent circuit The per-phase complex power S at the terminals is S = Vt I*a 5.19 The conjugate of the current phasor Ia is used to conform with the convension that lagging reactive power is considered as positive and leading reactive power as negative, as shown in Figure 5.20. E − Vt I*a = f Zs = * = E*f Z*s Vt ∠0 Ef ∠ − δ − Zs ∠ − θs Z s ∠ − θs E V = f ∠(θs − δ ) − t ∠θs Zs Zs 5.20 From Eqs. 5.19 and 5.20 2 V E V S = t f ∠(θs − δ ) − t ∠θs VA / phase Zs Zs 5.21 The real power P and the reactive power Q per phase are 2 V E V P = t f cos(θs − δ ) − t cos θs watt / phase Zs Zs 5.22 2 V E V Q = t f sin(θs − δ ) − t sin θs VAR / phase Zs Zs 5.23 If Ra is neglected, then Zs = Xs and θs =90°. From Eqs. 5.22 and 5.23 for a 3φ machine, P3φ = 3 Vt E f Xs sin δ 5.24 270 Introduction to Electrical Machines P3φ = Pmax sin δ watts 5.25 Figure 5.20 Complex power phasor. Where Pmax = Q 3φ = 3 Vt E f Xs 3 Vt E f 3 Vt cos δ − Xs Xs 5.25a 2 VAR 5.26 Because the stator losses are neglected in this analysis, the power developed at the terminals is also the air gap power. The developed torque of the machine is T= = P3φ 5.27 ωsyn Vt E f sin δ ωsyn X s 3 = Tmax sin δ N.m Where Tmax = Vt E f P = max ωsyn X s ωsyn ωsyn = n syn 3 60 5.28 5.29 5.29a 2π nsyn is the synchronous speed in rpm Both power and torque vary sinusoidallv with the angle δ (as shown in Figure 5.21), which is called the power angle or torque angle. The machine can be loaded gradually up to the limit of Pmax or Tmax , which are known as static stabilitv limits. The machine will lose synchronism if δ becomes greater than 90°. The maximum torque Tmax is also known as the pull-out torque. Note that since Vt is constant, the pull-out torque can be increased by increasing the excitation voltage Ef. If a synchronous motor tends to pull out of synchronism because of excessive load torque, the field current can be increased 271 Chapter One: Electromagnetic Principles to develop high torque to prevent loss of svnchronism. Similarly, in a synchronous generator, if the prime mover tends to drive the machine to supersynchronous speed by excessive driving torque, the field current can be increased to produce more counter torque to oppose such a tendency. δ Figure 5.21 Power and torque-angle characteristics. As the speed remains constant in a synchronous machine, the speed- torque characteristic is a straight line, parallel to the torque axis, as shown in Figure 5.22. Figure 5.22 Torque-speed characteristics. Example 5.2 A 3φ, 5 kVA, 208 V, four-pole, 60 Hz, star-connected synchronous machine has negligible stator winding resistance and a synchronous reactance of 8 Ω per phase at rated terminal voltage. The machine is first operated as a generator in parallel with a 3φ, 208 V, 60 Hz power supply. a) Determine the excitation voltage and the power angle when the machine is delivering rated kVA at 0.8 pf lagging. Draw the phasor diagram for this condition. b) If the field excitation current is now increased bv 20 percent (without changing the prime mover power), find the stator current, power factor, and reactive kVA supplied by the machine. c) With the field current as in (a) the prime mover power is slowly increased. What is the steady-state (or static) stabilitv limit? What are the corresponding values of the stator (or armature) current, power factor, and reactive power at this maximum power transfer condition? 272 Introduction to Electrical Machines Solution The per-phase equivalent circuit for the svnchronous generator is shown in Figure 5.23a. Vt = a) 208 = 120 V / phase 3 Stator current at rated kVA; Ia = 5000 = 13.9 A 3 × 208 φ = −36.9 for lagging pf of 0.8 From Figure E5.3a E f = Vt ∠0° + jI a X s = 120∠0° + j13.9∠ − 36.9° ⋅ 8∠90° = 206.9∠25.5° Excitation voltage E f = 206.9 V / phase Power angle δ = + 25.5° Note that because of generator action the power angle is positive. The phasor diagram is shown in Figure 5.23b. b) The new excitation voltage E 'f = 1.2 × 206.9 = 248.28 V . Because power transform remains same, Vt E f V E' sin δ = t f sin δ' Xs Xs Or E f sin δ = E 'f sin δ' Or sin δ' = Ef E 'f sin δ = sin 25.5 1 .2 δ ' = 21 The stotor current is Ia = = = E f − Vt jX s 248.28∠21 − 120∠0 8∠90 142.87∠38.52 8∠90 = 17.86∠ − 51.5 A 273 Chapter One: Electromagnetic Principles Power factor = cos 51.5° = 0.62 lag. Reactive kVA = 3VtIasin 51.5° = 3 × 120 × 17.86 × 0.78 × 10-3 =5.03 Or from Eq.5.26 120 × 248.28 120 2 Q = 3 cos 21 − × 10 − 3 8 8 = 3(3476.86 − 1800 ) = 5.03 c) From Eq. 5.25 the maximum power transfer occurs at δ = 90°. Pmax = Ia = 3E f V t 3 × 206.9 × 120 = = 9.32 kW Xs 8 E f − Vt 3 × 206.9∠ + 90 − 120∠0 = jX s 8∠90 = 29.9∠30.1 A Stator current Ia = 29.9 A Power factor = cos 30.1° = 0.865 leading. Alternative solution The stator current and power factor can also be obtained by drawing the phasor diagram for maximum power transfer condition. The phasor diagram is shown in Figure 5.23c. Because δ = +90°, Ef leads Vt by 90°. The distance bd between phasors Vt and Ef is the voltage drop IaXs and the current phasor Ia is in quadrature with IaXs. From the phasor diagram, Ia Xs 2 = Ef 2 Vt 2 206.9 2 + 120 2 = 29.9 A Ia = 2 8 From the two triangles abc and abd, ∠abc = ∠abd = φ ab 120 tan φ = = = 0.58 ad 206.9 φ = 30.1 Power factor = cos 30.1° = 0.865 lead. 274 Introduction to Electrical Machines φ δ = 25.5 φ = −36.9 φ (a) (b) (c) Figure 5.23 for example 5.2 5.10. POWER FACTOR CONTROL An outstanding feature of the synchronous machine is that the power factor of the machine can be controlled by the field current. The field current can be adjusted to make the stator (or line) current lagging or leading as desired. This power factor characteristic can be explained bv drawing phasor diagrams of machine voltages and currents. Assume constant-power operation of a synchronous motor connected to an infinite bus. The equivalent circuit, neglecting the stator resistance, the phasor diagram are shown in Figures 5.24a and 5.24b, respectively. For a three-phase machine the power transfer is P = 3Vt I a cos φ 5.30 Because Vt is constant, for constant-power operation I a cos φ is constant that is, the inphase component of the stator current on the axis of the phasor Vt is constant. Ia3 −Ia 2Xs − I a1X s Ia2 − I a 3X s Vt E f Sinδ I a1 (a) Ef 1 Ef 2 Ef 3 (b) 275 Chapter One: Electromagnetic Principles Ia, pf Ia pf If1 If2 If3 If (c) Figure 5.24 power factor characteristics. (a) equivalent circuit; (b) phasor diagram; (c) variation of Ia and pf with If. The locus of the stator current is therefore the vertical line passing through the current phasor for unity power factor. In Figure 5.24(b), phasor diagrams are drawn for three stator currents: Ia = Ia1, lagging Vt = Ia2, in phase with Vt = Ia3, leading Vt For these stator currents the excitation voltages Ef1, Ef2, and Ef3 ( representing the field currents If1, If2, and If3, respectively) are drawn to satisfy the phasor relationship E f = Vt − jI a X s 5.31 VE P = 3 t f sin δ ωsyn 5.32 The power can also be expressed as Again, for constant-power operation, E f sin δ is constant. Thus, the locus of Ef (or If) is also a straight line parallel to the phasor Vt (see Figure.5.24(b)) such that the vertical difference between the locus of Ef and the phasor Vt is constant and equals E f sin δ . The excitation voltage Ef changes linearly with the field current If. Therefore, as If is changed, Ef will change along the locus of Ef and Ia will change along the locus of Ia, signifying a change in the power factor angle φ of the stator current. For low field current If1, under-excitation (Ef = Ef1), the stator current (Ia = Ia1) is large and lagging. The stator current is minimum (Ia = Ia2) and at unity power factor for the field current 276 Introduction to Electrical Machines If2 (Ef = Ef2), which is called normal excitation. For larger field current If3, overexcitation (Ef = Ef3), the stator current (Ia = Ia3) is large again and leading. The variation of the stator current with the field current for constant-power operation is shown in Figure 5.24(c). This is known as the V-curve because of the characteristic shape. The variation of the power factor with the field current is the inverted V-curve, also shown in Figure 5.24(c). This unique feature of power factor control by the field current can be utilized to improve the power factor of a plant. In a plant most of the motors are normally induction motors, which draw power at lagging power factors. Synchronous motors can be installed for some drives in the plant and made to operate in an overexcited mode so that these motors operate at leading power factors, thus compensating the lagging power factor of the induction motors and thereby improving the overall power factor of the plant. Example 6.1 illustrates this method of power factor improvement in a plant. If the synchronous machine is not transferring any power but is simply floating on the infinite bus, the power factor is zero; that is, the stator current either leads or lags the stator voltage by 90°. The magnitude of the stator current changes as the field current is changed, but the stator current is always reactive. Looking from the machine terminals, the machine behaves as a variable inductor or capacitor as the field current is changed. An unloaded synchronous machine is called a synchronous condenser and may be used to regulate the receiving-end voltage of a long power transmission line. At present, solid-state control o f inductors and capacitors is being used increasingly to achieve voltage regulation of transmission lines. Example 5.3 In a factory a 3φ, 4 kV, 400 kVA synchronous machine is installed along with other induction motors. The following are the loads on the machines: Induction motors: 500 kVA at 0.8 PF lagging. Svnchronous motor: 300 kVA at 1.0 PF. a) Compute the overall power factor of the factory loads. b) To improve the factory power factor, the synchronous machine is overexcited (to draw leading current) without any change in its load. Without overloading the motor, to what extent can the factory power factor be improved? Find the current and power factor of the synchronous motor for this condition. Solution a) Induction motors: Power = 500 × 0.8 = 400 kW Reactive power = 500 × 0.6 = 300 kVAR Synchronous motor: Power = 300 kW Reactive power = 0 Factory: 277 Chapter One: Electromagnetic Principles Power = 700 kW Reactive power = 300 kVAR Complex power = 700 2 + 300 2 = 762 kVA Power factor = b) 700 = 0.92 lagging 762 The maximum leading kVAR that the synchronous motor can draw without exceeding its rating is 4002 − 3002 = 264.58 kVA Factory kVAR = j300 − j264.58 = j 35.42 (i.e., lagging) New factory kVA = Improved factory power factor = 7002 + 35.42 2 = 700.9 kVA 700 = 0.996 lagging 700.9 Synchronous motor current: I SM = 400 kVA = 57.74 A 3 × 4 kV Synchronous motor power factor: Pf SM = 300 kW = 0.75 lead 400 kVA 278 Introduction to Electrical Machines PROBLEMS 5.1. In a factory, the following are the loads. Induction motors: 1000 hp 0.7 average power factor 0.85 average efficiency Lighting and heating load: 100 kW A 3φ synchronous motor is installed to provide 300 hp to a new process. The synchronous motor operates at 92% efficiency. Determine the kVA rating of the synchronous motor if the overall factory power factor is to be raised to 0.95. Determine the power factor of the synchronous motor. Ans. 548.24 kVA, 0.4082 (leading) 5.2. 5.3. A 3φ, 60 Hz supply and two 3φ synchronous machines are available. Determine the speed, and a suitable number of poles, for each synchronous machine to provide: (a) A 3φ, 180 Hz supply. (b) A 3φ , 500 Hz supply. The following test data are obtained for a 3φ, 195 MVA, 15 kV, 60 Hz starconnected synchronous machine. Open-circuit test: If A) 150 300 450 600 750 VLL (kV 3.75 7.5 11.2 13.6 15 900 1200 15.8 16.5 Short-circuit test: If = 750A, Ia = 7000A The armature resistance is very small. a) Draw the open-circuit characteristic, the short-circuit characteristic, the air gap line, and the modified air gap line. b) Determine the unsaturated and saturated values of the synchronous reactance in ohms and also in pu. c) Find the field current required if the synchronous machine is to deliver 100 MVA at rated voltage, at 0.8 leading power factor. d) Find the voltage regulation of the synchronous generator for the load of part (c). Voltage regulation (VR) is defined as follows. VR = Vt load removed − Vt with load Vt with load × 100% Ans. (b) 1.5465 Ω (1.3404 pu),1.2372 Ω (1.0723 pu); (c) 601.2 A; (d) -9.33% 279 Chapter One: Electromagnetic Principles 5.4. The following test results are obtained for a 3φ , 25 kV, 750 MVA, 60 Hz, 3600 rpm, star-connected synchronous machine at rated speed. If (A) 1500 VLL (kV) open-circuit test 25 Ia (A) short-circuit test 10,000 VLL (kV) air gap line 30 a) Determine the number of poles of the synchronous machine. b) Determine the unsaturated and saturated values of the synchronous reactance in ohms and per unit. c) The short-circuit test is performed at constant field current (1500A) but at different spceds-1000 rpm, 2000 rpm, 3000 rpm, and 3600 rpm. Determine the shortcircuit current at these speeds. d) Determine the field current if the synchronous machine delivers rated MVA to an infinite bus at 0.9 lagging power factor. 5.5. The synchronous machine of Example 6.2 is connected to a 3φ, 14 kV, 60 Hz infinite bus and draws 5 MW at 0.85 leading power factor. a) Determine the values of the stator current (Ia), the excitation voltage (Ef), and the field current (If). Draw the phasor diagram. b) If the synchronous motor is disconnected from the infinite bus without changing the field current, determine the terminal voltage before the speed decreases. Ans. 0.588∠31.8° pu, 1.3272 ∠-18.5° pu, 265.44 A. 5.6. A 3φ, 2000 kVA, 11 kV, 1800 rpm synchronous generator has a resistance of 1.5 Ω and synchronous reactance of 15 Ω per phase. a) The field current is adjusted to obtain the rated terminal voltage at open circuit i) Determine the excitation voltage Ef. ii) If a short circuit is applied across the machine terminals, find the stator current. b) i) ii) The synchronous machine is next connected to an infinite bus. The generator is made to deliver the rated current at 0.8 power factor lagging. Determine the excitation voltage. Determine the percentage increase in the field current relative to the field current of part (a). 280 Introduction to Electrical Machines APPENDIXES APPENDIX A CONVERSION FACTORS To Convert from Btus To Calorie-grams pounds hours Kilowatthours Wattseconds Centimeters Circular mils Cubic inches Cubic meters Angstrom units Miles (statute) Millimeters Square centimeters Square inches Cubic centimeters Gallons (U.S. liquid) Cubic feet Days Dynes Minutes Seconds Gallons (U.S. liquid) Newtons Pounds Electronvolts Ergs centimeters Electronvolts pounds Kilowatthours Feet Centimeters Foot-candles Lumens/square foot Lumens/square meter centimeters Foot-pounds Horsepower-hours Gallons (U.S. liquid) Newton-meters Cubic inches Ounces Gauss Gilberts Maxwells/square centimeter Lines/square centimeter Lines/square inch pere-turns Multiply by 251.996 1.054× 1010 777.649 0.000393 1054.35 0.000293 1054.35 1 × 10 0.0328 0.3937 0.01 -6 6.214× 10 10 -6 5.067×10 -7 7.854×10 16.387 0.00433 35.315 8 24 1440 86,400 264.172 0.00001 -6 2.248×10 -12 1.60209 ×10 1.0 11 6.242 ×10 -8 7.376 ×10 -7 1 × 10 -14 2.777 ×10 30.48 0.3048 1.0 10.764 7 1.3558 × 10 7 1.3558 × 10 -7 5.050 ×10 1.3558 1.3558 231 3.785 128 8 1.0 1.0 6.4516 0.7958 281 Chapter One: Electromagnetic Principles To Convert from Grams To Horsepower Dynes Ounces Pounds Btus/hour Ergs/second pounds/second Joules/second Hours Seconds Inches Angstrom units Centimeters Meters Joules Kilograms Lines pounds Horsepower-hours Kilowatthours Wattseconds Dynes Ounces Pounds Maxwells Multiply by 980.665 0.0353 0.0022 2547.16 9 7.46 × 10 550.221 746 746 3600 8 2.54× 10 2.54 0.0833 0.0254 0.000948 7 1×10 0.7376 -7 3.725×10 -7 2.777 × 10 1.0 980,665 35.2 2.2 1.0 Lines/square centimeter 1.0 Lines/square inch 0.1550 -8 1× 10 Webers/square inch Liters Cubic centimeters Cubic inches Gallons (U.S. liquid) Ounces (U.S. liquid) Quarts (U.S. liquid) Lumens Candle power (sphere.) Lumens/square centimeter Lamberts 1000.028 61.025 0.2642 33.815 1.0567 0.0796 1.0 Lumens/square foot 1.0 candles Maxwells Miles/hour Kilometers/hour 1.0 -8 1 × 10 10 1 × 10 100 3.2808 39.370 0.000621 5280 1.609 1609.344 1.609344 Newton-meters centimeters Kilogram-meters Ampere-turns/inch Ampere-turns/meter Gilberts/centimeter Cubic centimeters Cubic inches Gallons (U.S. liquid) 1 × 10 0.10197 2.0212 79.577 1.0 946.353 57.75 0.25 Meters Webers Angstrom units Centimeters Inches Miles (statute) Miles (statute) Kilometers Oersteds Quarts (U.S. liquid) 7 282 Introduction to Electrical Machines To Convert from Radians Slugs To Multiply by 0.9463 2 32 57.2958 Pints (U.S. liquid) Ounces (U.S. liquid) Degrees Kilograms Pounds Btus/hour Ergs/second Horsepower Joules/second Watts 14.5939 32.1740 3.4144 7 1× 10 0.00134 1.0 8 1 × 10 8 1 × 10 365 8760 525,600 7 3.1536 × 10 Webers Maxwells Years Minutes Seconds APPENDIX B THE GREEK ALPHABET Letter Alpha Beta Gamma Delta Epsilon Zeta Eta Theta Iota Kappa Lambda Mu Nu Xi Omicron Pi Rho Sigma Tau Upsilon Phi Chi Psi Omega Capital A B Γ ∆ Ε Ζ Η Θ I Κ Λ M N Ξ Ο Π P Σ T ϒ Φ X Ψ Ω Lowercase α β γ δ ε ζ η θ ι κ λ µ ν ξ ο π ρ σ τ v φ χ ψ ω Used to Designate Area, angles, coefficients Angles, coefficients, flux density Specific gravity, conductivity Density, variation Base of natural logarithms Coefficients, coordinates, impedance Efficiency, hysteresis coefficient Phase angle, temperature Dielectric constant, susceptibility Wavelength Amplification factor, micro, Permeability Reluctivity 3.1416 Resistivity Summation Time constant Angles, magnetic flux Dielectric flux, phase difference Ohms, angular velocity 283 Chapter One: Electromagnetic Principles APPENDIX C MAGNETIC PARAMETER CONVERSIONS SI (MKS) Φ CGS English maxwells 8 =10 maxwells 2 gauss (maxwells/cm ) 4 =10 4 2 =10 cm lines/in. 4 2 = 6.452 ×10 lines/in. 2 1550 in. lines 8 =10 lines A webers (Wb) 1 Wb 2 Wb/m 1 Wb/m2 2 1m µ0 4π×10 Wb/Am = 1 gauss/oersted = 3.20 lines/Am F NI (ampere-turns, At) 1 At 0.4πNI (gilberts) =1.257 gilberts NI (At) 1 gilbert = 0.7958 At H NI/l (At/m) 1 At/m 0.4πNI/l (oersteds) -2 1.26× 10 oersted NI/l (At/ in.) -2 =2.54× 10 At/ in. Hg 7.97 × 10 Bg Bg (oersteds) =0.313Bg (At/ in.) B -7 5 2 284 Introduction to Electrical Machines APPENDIX D TYPES OF ENCLOSURES The enclosure category includes many types of enclosures. Only a few of the most common types are listed here. The Open Motor: is one having ventilating openings which permit passage of external cooling air over and around the windings. The Drip-Proof Motor : is an open motor in which ventilating openings are so constructed that drops of liquid or solids falling on the machine at any angle not greater than 15 degrees from the vertical cannot enter the machine. A Guarded Motor: is an open motor in which all ventilating openings are limited to specified size and shape to prevent insertion of fingers or rods to avoid accidental contact with rotating or electrical parts. A Splash-Proof Motor : is an open motor in which ventilating openings are so constructed that drops of liquid or solid particles falling on the machine or coming toward the machine in a straight line at any angle not greater than 100 degrees from the vertical cannot enter the machine. A Totally-Enclosed Motor: is a motor so enclosed as to prevent the free exchange of air between the inside and outside of the case, but not airtight. A Totally-Enclosed Nonventilated (TENV) Motor: is a totally-enclosed motor which is not equipped for cooling by means external to the enclosing parts. A Totally-Enclosed Fan-Cooled (TEFC) Motor: is a totally-enclosed motor with a shaft-mounted fan to blow cooling air across the external frame. It is a popular motor for use in dusty, dirty, and corrosive atmospheres. A Totally-Enclosed Blower-Cooled (TEBC) Motor: is a totally-enclosed motor which is equipped with an independently powered fan to blow cooling air across the external frame. A TEBC motor is commonly used in constant torque, variable speed applications. Encapsulated Motor: is an open motor in which the windings are covered with a heavy coating of material to protect them from moisture, dirt, abrasion, etc. With this complete protection, the motors can often be used in applications which demand totally-enclosed motors. An Explosion-Proof Motor: is a totally-enclosed motor designed and built to withstand an explosion of gas or vapor within it, and to prevent ignition of gas or vapor surrounding the machine by sparks, flashes or explosions which may occur within the machine casing. 285 Chapter One: Electromagnetic Principles Bibliography 1. P.C. Sen , “Principles of electric machines and power electronics”, second edition. 2. Peter F.Ryff, David Platnick and Joseph A.Karnas, “Electrical Machines and Transformers principle and applications”. 3. Dr. P.S Bimbhira, Application]”. ”Electrical Machinery [Theory, Performance and 4. B.L THEREJA A.K. THERAJA, “Electrical Technology”. 5. Stephen J. Chapman ,”Electrical Machinery Fundamentals”, fourth edition, 6. J.B.Gupta, “Theory and Performance of Electrical Machines”. 7. V N MITTLE ,”Basic Electrical Engineering”. 8. Robert Bolystad, “Introductory Circuit Analysis”, 10th edition, 9. Кацман М.М. “Элетрические машины”. 286