EE-316 Electrical Machine 1 Chapter 1 PDF

Bachelor of Science in Electrical Engineering EE-316 ELECTRICAL MACHINE 1 Introduction DC machines are generators that convert mechanical e...

Bachelor of Science in Electrical Engineering EE-316 ELECTRICAL MACHINE 1 Introduction DC machines are generators that convert mechanical energy to dc electric energy and motors that convert dc electric energy to mechanical energy. Most dc machines are like ac machines in that they have ac voltages and currents within them – dc machines have a dc output only because a mechanism exists that converts the internal ac voltages to dc voltages at their terminals. Since this mechanism is called a commutator, dc machinery is also known as commutating machinery. The fundamental principles involved in the operation of dc machines are very simple. Unfortunately, they are usually somewhat obscured by the complicated construction of real machines. Learning Outcome Understand the equivalent circuit of a dc generator Perform nonlinear analysis of dc generator using the magnetization curve, taking into account armature reaction effects. Explain how to derive the voltage – current characteristics of separately excited, shunt series and compounded dc generator Learning Content It contains readings, selection and discussion questions and sets of activities that students can work on individually or by group. Topics for Chapter 1 Topic 1 Commutation and armature construction in real DC machines Topic 2 Problems with commutation in real machines Topic 3 Internal generated voltage and induced torque equations of real DC Machines Topic 4 Constructions of DC Machines Topic 5 Power flow and losses in DC machines Teaching and Learning Activities A simple Rotating Loop between curved Pole Faces The simplest rotating dc machine is shown above. Consists of a single loop of wire rotating about a fixed axis. The rotating part is called rotor, and the stationary part is the stator. The magnetic 1|CJD | Page  Bachelor of Science in Electrical Engineering EE-316 ELECTRICAL MACHINE 1 field for the machine is supplied by the magnetic north and south poles. Since the air gap is of uniform width, the reluctance is the same everywhere under the pole faces. If the rotor is rotated, a voltage will be induced in the wire loop. To determine the magnitude and shape of the voltage, examine the figure below: To determine the total voltage 𝑒𝑡𝑜𝑡 on the loop, examine each segment of the loop separately and sum all the resulting voltages. The voltage on each segment is given by 𝑒𝑖𝑛𝑑 = (𝑣𝑥𝐵). 𝑙 Thus, the total induced voltage on the loop is: 𝑒𝑖𝑛𝑑 = 2𝑣𝐵𝑙 When the loop rotates through 180°, segment ab is under the north pole face instead of the south pole face. At that time, the direction of the voltage on the segment reverses, but its magnitude remains constant. The resulting voltage 𝑒𝑡𝑜𝑡 is shown below: There is an alternative way to express the 𝑒𝑖𝑛𝑑 equation, which clearly relates the behaviour of the single loop to the behaviour of larger, real dc machines. Examine the figure below: The tangential velocity v of the edges of the loop can be expressed as 𝑣 = 𝑟𝜔. Substituting this expressing into the 𝑒𝑖𝑛𝑑 equation before gives: 2|CJD | Page  Bachelor of Science in Electrical Engineering EE-316 ELECTRICAL MACHINE 1 𝑒𝑖𝑛𝑑 = 2𝑟𝜔𝐵𝑙 The rotor surface is a cylinder, so the area of the rotor surface A is equal to 2πrl. Since there are 2 poles, the area under each pole is 𝐴𝑝𝑜𝑙𝑒 = 𝜋𝑟𝑙.Thus, 2 𝑒𝑖𝑛𝑑 = 𝐴 𝐵𝜔 𝜋 𝑝𝑜𝑙𝑒 Since the flux density B is constant everywhere in the air gap under the pole faces, the total flux under each pole is ∅ = 𝐴𝑝 𝐵. Thus, the final form of the voltage equation is: 2 𝑒𝑖𝑛𝑑 = 𝜋 ⌀𝜔 under the pole face This is for the simple 2-pole machine. In general, the voltage in any real machine will depend on the same 3 factors: 1. The flux in the machine 2. The speed of rotation 3. A constant representing the construction of the machine Getting DC Voltage out of the rotating loop The voltage out of the loop is alternately a constant positive and a constant negative value. How can this machine be made to produce a dc voltage instead of the ac voltage? This can be done by using a mechanism called commutator and brushes, as shown above. Here 2 semicircular conducting segments are added to the end of the loop, and 2 fixed contacts are set up at an angle such that at the instant when the voltage in the loop is zero, the contacts short-circuit the two segments. Thus, every time the voltage of the loop switches direction, the contacts also switch connections, and the output of the contacts is always built up in the same way. This connection-switching process is known as commutation. The rotating semicircular segments are called commutator segments, and the fixed contacts are called brushes. The Induced Torque in the Rotating Loop Suppose a battery is now connected to the machine as shown here, together with the 3|CJD | Page  Bachelor of Science in Electrical Engineering EE-316 ELECTRICAL MACHINE 1 resulting configuration: How much torque will be produced in the loop when the switch is closed? The approach to take is to examine one segment of the loop at a time and then sum the effects of all the individual segments. The force on a segment of the loop is given by: 𝐹 = 𝑖(𝑙𝑥𝐵), and the torque on the segment is 𝑡 = 𝑟𝐹 sin 𝜃. The resulting total induced torque in the loop for 2 conductors is: 𝑡𝑖𝑛𝑑 = 2𝑟𝑖𝑙𝐵 under the pole face By using the fact that 𝐴𝑝 = 𝜋𝑟𝑙 and ∅ = 𝐴𝑝 𝐵, the torque expression can be reduced to: 2 𝑡𝑖𝑛𝑑 = 𝐴𝑝 𝑖𝐵 under the pole face 𝜋 2 𝑡𝑖𝑛𝑑 = 𝜋 ∅𝑖 under the pole face 4|CJD | Page  Bachelor of Science in Electrical Engineering EE-316 ELECTRICAL MACHINE 1 In general, the torque in any real machine will depend on the same 3 factors: 1. The flux in the machine 2. The current in the machine 3. A constant representing the construction of the machine. Topic 1 Commutation and armature construction in real DC machines The commutator can be defined as an electrical rotating switch in a particular type of generators as well as motors. This is mainly used to overturn the direction of current among the external circuit & rotor. It comprises a cylinder with numerous metal contact segments lying on the revolving armature of the machine. The brushes or electrical contacts are made with a carbon press material next to the commutator, designing sliding contact by consecutive segments of the commutator while it revolves. The armature windings are allied to the segments of the commutator. The applications of commutators include DC (direct current) machines like DC generators, numerous DC motors, as well as universal motors. In a DC motor, the commutator provides electric current to the windings. By changing the direction of current within the revolving windings every half turn, a torque (steady revolving force) will be produced. Commutator The voltage generated in the armature, placed in a rotating magnetic field, of a DC generator is alternating in nature. The commutation in DC machine or more specifically commutation in DC generator is the process in which generated alternating current in the armature winding of a dc machine is converted into direct current after going through the commutator and the stationary brushes. Again, in DC Motor, the input DC is to be converted in alternating form in armature and that is also done through commutation. This transformation of current from the rotating armature of a DC machine to the stationary brushes needs to maintain continuously moving contact between the commutator segments and the brushes. When the armature starts to rotate, then the coils situated under one pole (let it be N pole) rotates between a positive brush and its consecutive negative brush and the current flows through this coil is in a direction inward to the commutator segments. Then the coil is short circuited with the help of a brush for a very short fraction of time (1⁄500 sec). It is called commutation period. After this short-circuit time the armature coils rotate under S pole and rotates between a negative brush and its succeeding positive brush. Then the direction is reversed which is in the away from the commutator segments. This phenomena of the reversal of current are termed as commutation process. We get direct current from the brush terminal. 5|CJD | Page  Bachelor of Science in Electrical Engineering EE-316 ELECTRICAL MACHINE 1 The commutation is called ideal if the commutation process or the reversal of current is completed by the end of the short circuit time or the commutation period. If the reversal of current is completed during the short circuit time, then there is sparking occurs at the brush contacts and the commutator surface is damaged due to overheating and the machine is called poorly commutated. Physical Concept of Commutation in DC Machine For the explanation of commutation process, let us consider a DC machine having an armature wound with ring winding. Let us also consider that the width of the commutator bar is equal to the width of the brush and current flowing through the conductor is IC. 1. Let the commutator is moving from left to right. Then the brush will move from right to left. 2. At the first position, the brush is connected the commutator bar b (as shown in fig 1). Then the total current conducted by the commutator bar b into the brush is 2IC. 3. When the armature starts to move right, then the brush comes to contact of bar a. Then the armature current flows through two paths and through the bars a and b (as shown in fig 2). The total current (2IC) collected by the brush remain same. 4. As the contact area of the bar a with the brush increases and the contact area of the bar b decreases, the current flow through the bars increases and decreases simultaneously. When the contact area become same for both the commutator bar then same current flows through both the bars (as shown in fig 3). 5. When the brush contact area with the bar b decreases further, then the current flowing through the coil B changes its direction and starts to flow counter-clockwise (as shown in fig 4). 6. When the brush totally comes under the bar a (as shown in fig 5) and disconnected with the bar b then current IC flows through the coil B in the counter-clockwise direction and the short circuit is removed. 7. In this process the reversal of current or the process of commutation is done. Methods of Improving Commutation 6|CJD | Page  Bachelor of Science in Electrical Engineering EE-316 ELECTRICAL MACHINE 1 There are three methods of sparkles commutation: 1. Resistance Commutation 2. Voltage Commutation 3. Compensating Windings The Armature The first armature was used by the magnet keepers in the 19th century. The related equipment parts are expressed in terms of an electrical as well as mechanical. Though definitely separate these two sets of terms are regularly used similarly which includes one electrical term as well as one mechanical term. This may be the reason for confusion whenever working with complex machines such as brushless alternators. In 7|CJD | Page  Bachelor of Science in Electrical Engineering EE-316 ELECTRICAL MACHINE 1 most of the generators, part of the rotor is the field magnet that will be active that means rotates, whereas part of the stator is armature that will be inactive. Both the generators as well as motors can be designed with an inactive armature & an active (rotating) field otherwise an active armature as the inactive field. The shaft piece of a stable magnet otherwise electromagnet, as well as the moving iron piece of a solenoid, particularly if the latter performs as a switch or else relay, can be referred to as armatures. This article discusses an overview of the armature and its working with applications. What is an Armature? An armature can be defined as a power generating component in an electric machine where the armature can be a rotating part otherwise a stationary part in the machine. The interaction of the armature with the magnetic flux can be done in the gap of air, the field element can include any stable magnets otherwise, electromagnets which are shaped with a conducting coil like another armature which is known as a doubly-fed electric machine. the armature always works like a conductor, sloping normal toward both the field as well as toward the motion direction, torque otherwise force. The armature diagram is shown below. Armature The main role of an armature is multi purposed. The primary role is to transmit current across the field, therefore generating shaft torque within an active machine otherwise strength in a linear machine. The second role of an armature is to produce an EMF (electromotive force). In this, an EMF can occur with the armature’s relative motion as well as the field. As the machine is employed as a motor, then the EMF will oppose the current of an armature and converts the electric power into mechanical which is in the form of torque, and finally transmits through the shaft. Whenever the machine is utilized like a generator, then the armature electromotive force drives the current of an armature, as well as the movement of the shaft will be changed to electrical power. In the generator, the power which is produced will be drawn from the stator. A growler is mainly used to ensure the armature intended for opens, grounds, as well as shorts. Armature Components An armature can be designed with the number of components namely the core, the winding, the commutator, & the shaft. 8|CJD | Page  Bachelor of Science in Electrical Engineering EE-316 ELECTRICAL MACHINE 1 Armature Parts The Core The armature core can be designed with many thin metal plates which are named as laminations. The thickness of laminations are approximate 0.5mm and it depends on the frequency by which the armature will be designed to work. The metal plates are stamped-out on a push. They are in the circular form by a hole stamped-out of the core, while the shaft is pressed, as well as the slots which are stamped in the region of the edge wherever the coils will finally sit. Metal plates are associated together to generate the core. The core can be built with stacked metal plates instead of using a steel piece to produce the sum of lost energy while heat in the core. The loss of energies is known as iron losses which are occurred by eddy currents. These are minute turning magnetic fields forms in the metal because of the revolving magnetic fields which can be found whenever the unit is running. If the metal plates use the eddy currents then they can form in one plane as well as significantly reduces the losses. The Winding Before the process of winding starts then the core slots will be protected from the copper wire within the slots approaching into contact by the laminated core. Coils are placed into the armature slots as well as attached to the commutator in revolving. This can be done in many ways based on the armature design. Armatures are classified into two types namely lap wound armature as well as wave wound armature. In a lap wound, the final end of one coil is attached toward the segment of a commutator as well as the primary end of the nearby coil. In a wave wound, the coils two ends will be associated with the segments of the commutator which are divided by some distance among the poles. This permits the sequence adding of the voltages within the windings among brushes. this kind of winding needs only one couple of brushes. In the first armature, the number of lanes equals the number of 9|CJD | Page  Bachelor of Science in Electrical Engineering EE-316 ELECTRICAL MACHINE 1 poles as well as brushes. In some of the armature designs, they will have two or more different coils in a similar slot, attached to nearby commutator segments. This can be done if the required voltage across the coil will be considered to be high. By distributing the voltage over three separate segments as well as coils will be in the same slot, the strength of the field in the slot will be high, however, it will decrease arcing over the commutator, as well as make the device more competent. In several armatures the slots are also twisted, this can be attained with every lamination being somewhat out of line up. This can be done to decrease cogging, as well as provide a level revolution from one to another pole. The Commutator The commutator is pushed on top of the shaft as well as it is held on by a coarse knurl similar to the core. the designing of commutator can be done using copper bars, and an insulating material will separate the bars. Normally, this material is a thermoset plastic however in older armatures sheet mica has been used. The commutator has to be accurately associated by the core slots whenever pushed on top of the shaft because the wires from every coil will appear from the slots as well as attach with the commutator bars. To work the magnetic circuit efficiently, it is essential that the armature coil has a precise angular displacement from the commutator bar toward which it is attached. The Shaft The shaft of an armature is one kind of hard rod mounted among two bearings that describe the axis of components placed onto it. It should be broad sufficient to send out the torque necessary with the engine & rigid adequate to control some of the forces which are out of balance. For harmonic distortion, the length, speed, and bearing points are selected. An armature can be designed with a number of major components namely the core, the winding, the shaft, and the commutator. Armature Function or Armature Working The armature rotation can be caused by the communication of two magnetic fields. One magnetic field can be generated by the field winding, whereas the second can be produced with the armature while voltage is applied toward the brushes to get in touch with the commutator. Whenever the current supplies through the winding of an armature, then it creates a magnetic field. This is out of line by the field created with the field coil. This will cause the power of attraction toward a single pole as well as revulsion from the other. When the commutator is connected to the shaft then it will also move with a similar degree as well as activates the pole. The armature will continue to chase the pole to spin. If the voltage is not given to the brushes, then the field will get excited as well as the armature will be driven mechanically the voltage which is applied is AC because it approaches, and flows away from the pole. However, the commutator being associated with the shaft and frequently activates the polarity because it revolves, like that the real output can observe across the brushes in DC. Armature Winding and Armature Reaction The armature winding is the winding where the voltage can be induced. Similarly, the field winding is the winding where the main field flux can be generated whenever the current flows through the winding. The armature winding has some of the basic terms namely turn, coil and winding. Armature reaction is the result of the armature flux on top of main field flux. Generally, the DC motor includes two windings such as Armature winding as well as field winding. Whenever we stimulate the 10|CJD | Page  Bachelor of Science in Electrical Engineering EE-316 ELECTRICAL MACHINE 1 field winding, then it generates a flux which connects by the armature, and this will cause an emf & therefore a flow of current in the armature. Applications of Armature The applications of an armature include the following. The armature is used in an electric machine for generating power. The armature can be used as rotor otherwise stator. This is used to monitor the current for the applications of DC motor. Thus, this is all about an overview of an armature which includes what is an armature, components, working, and applications. From the above information finally, we can conclude that an armature is an essential component used in an electric machine for generating power. It can be on either the rotating part otherwise stationary part of the machine. DC Windings a. Lap Winding b. Wave Winding These two types of windings differ in two ways i. Number of circuits between positive and negative brushes ii. The manner in which the coil ends are connected. However, the coils of both lap and wave windings are identically formed. Types and shapes of winding wires 1. Round wires: it has thin and thick conductors are used in semi-closed slot type motors and mush windings rotors. It is wounded in reels and available in kilograms. 2. Rectangular straps: it is used in open type slot motors. These conductors are available as long straps in meters. They are used in the following places: a. Low voltage motor windings b. Used as conductor in high current motor c. Series field motor windings coils. Winding Pitches Back Pitch: The distance between top and bottom coil sides of a coil measures around the back of the armature is called back pith and is designated as 𝑌𝑏. Back pitch is approximately equal to number of coil sides per layer. Generally back pitch is an odd integer. Front Pitch: The distance between two coil sides connected to the same commutator segment is called the front pitch and is designated as 𝑌𝑓. Winding pitch: The distance between the starts of two consecutive coils measured in terms of coil sides is called winding pitch and is designated as Y. For lap winding 𝑌 = 𝑌𝑏 − 𝑌𝑓 For wave winding 𝑌 = 𝑌𝑏 + 𝑌𝑓 11|CJD | Page  Bachelor of Science in Electrical Engineering EE-316 ELECTRICAL MACHINE 1 Commutator pitch: The distance between the two commutator segments to which the two ends of a coil are connected is called commutator pitch and is designated as 𝑌𝑐 and is measured in terms of commutator segments Simplex Lap Winding Lap winding: The winding in which successive coils overlap each other hence it is called lap winding. In their winding end of one coil is connected to the commutator segment and start of the adjacent coil situated under the same pole as shown in the figure above. m multiplicity factor 1, for simplex winding 2, for duplex winding … etc Important rules for lap winding Let Z Number of conductors P number of poles 𝑌𝑏 Back pitch 𝑌𝑓 Front pitch 𝑌𝑐 Commutator pitch 𝑌𝑎 Average pole pitch 𝑌𝑝 Pole pitch 𝑌𝑅 resultant pitch 1. 𝑌𝑏 and 𝑌𝑓 must be approximately equal to 𝑌𝑝 2. 𝑌𝑏 must be less or greater than 𝑌𝑓 by 2m where m is the multiplicity of the winding. When 𝑌𝑏 is greater than 𝑌𝑓 the winding progresses from left to right and is known as progressive winding. When 12|CJD | Page  Bachelor of Science in Electrical Engineering EE-316 ELECTRICAL MACHINE 1 𝑌𝑏 is lesser than 𝑌𝑓 the winding progresses from right to left and is known as retrogressive winding. Hence 𝑌𝑏 = 𝑌𝑓 ± 2𝑚. 3. 𝑌𝑏 and 𝑌𝑓 must be odd 4. 𝑌𝑏 and 𝑌𝑓 may be equal or differ by ±2. +(sign) for progressive winding Winding expands from left to right -(sign) for retrogressive winding Winding expands from right to left 𝑌𝑏 +𝑌𝑓 5. 𝑌𝑎 = = 𝑌𝑝 2 6. 𝑌𝑅 is always even 7. 𝑌𝑐 = 𝑚 8. Number of parallel paths 𝑎 = 𝑚𝑝, number of brushes Simplex Wave Winding Advantages of Lap Winding The advantages of lap windings include: This winding is necessarily required for large current applications because it has more parallel paths. It is suitable for low voltage and high current generators. Disadvantages of Lap Winding The disadvantages of lap windings include: It gives less emf compared to wave winding. This winding requires more no. of conductors for giving the same emf, which results in high winding cost. It has less efficient utilization of space in the armature slots. Wave winding: The end of one coil is not connected to the beginning of the same coil but is connected to the beginning of another coil of the same polarity as that of the first coil. 13|CJD | Page  Bachelor of Science in Electrical Engineering EE-316 ELECTRICAL MACHINE 1 Important rules for wave winding 1. 𝑌𝑏 and 𝑌𝑓 must be approximately equal to 𝑌𝑝 2. 𝑌𝑏 and 𝑌𝑓 must be odd 3. 𝑌𝑏 and 𝑌𝑓 may be equal or differ by ±2. 𝑌𝑏 +𝑌𝑓 4. 𝑌𝑐 = and should be a whole number 2 Dummy coils: The wave winding is possible with particular number of conductors and poles and slots combinations. Sometimes the standard stampings do not consist of the number of slots according to the design requirements and hence the slots and conductor combination will not produce a mechanically balanced winding. Under such conditions some coils are placed in the slots, not connected to the remaining part of the winding but only for mechanical balance. Such windings are called dummy coils. Simplex Wave Winding Advantages The advantages of simplex wave windings include: In this winding, only two brushes are required but more parallel brushes can be added to make it equal to the no. of poles. If one or more brushes set poor contacts with the commutator, satisfactory operation is still possible. This winding gives sparkles commutation. The reason behind that it has two parallel paths irrespective of the number of poles of the machine. The conductors in each of the two parallel paths distributed around the armature in the entire circumference. No. of conductors in each path = Z/2, Z is the total no. of conductors. Generated emf = average emf induced in each path X Z/2 For a given number of poles and armature conductors, it gives more emf than that of lap winding. Hence wave winding is used in high voltage and low current machines. This winding is suitable for small generators circuit with a voltage rating of 500-600V. Current flowing through each conductor. 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 𝑝𝑒𝑟 𝑝𝑎𝑡ℎ(𝐼𝑎 ) = 2 𝐼𝑎 is the armature current. The current per path for this kind of winding must not be exceeded 250A. The resultant emf around the entire circuit is zero. Simplex Wave Winding Disadvantages The disadvantages of simplex wave windings include: Wave winding cannot be used in machines having higher current rating because it has only two parallel paths. Sample Problems 1. The difference between the back pitch and the front pitch is 2. The front pitch is 21. If the winding is lap retrogressive, what is the back pitch? Note: Since the difference between the back pitch and the front pitches is 2, the winding is simplex (m = 1) 𝑌𝑏 = 𝑌𝑓 − 2𝑚 for retrogressive lap 14|CJD | Page  Bachelor of Science in Electrical Engineering EE-316 ELECTRICAL MACHINE 1 𝑌𝑏 = 21 − 2(1) 𝑌𝑏 = 19 2. A duplex lap wound, four pole DC generator has 120 slots and four elements per slot. How many commutator segments are there? 4 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑠 𝑍= 𝑥 120 𝑠𝑙𝑜𝑡𝑠 = 480 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑠 𝑠𝑙𝑜𝑡 𝑍 480 𝑁𝑐 = = = 240 𝑠𝑒𝑔𝑚𝑒𝑛𝑡𝑠 2 2 3. A 4-pole wave wound armature has 744 armature conductors in 62 slot. If the commutator has 186 segments, determine a. The coil span b. The number of conductors per coil a. 𝑡𝑜𝑡𝑎𝑙 𝑠𝑙𝑜𝑡𝑠 62 𝐶𝑜𝑖𝑙 𝑆𝑝𝑎𝑛 = = = 15.5 𝑝𝑜𝑙𝑒 4 Coil Span = 15 Note: Coil span is rounded off to the nearest whole number. b. Number of coils = Nc Number of coils = 186 coils 2 𝑎𝑐𝑡𝑖𝑣𝑒 𝑐𝑜𝑖𝑙 𝑠𝑖𝑑𝑒𝑠 "𝑛"𝑐𝑜𝑛𝑑𝑢𝑐𝑡𝑜𝑟𝑠 𝑍 = 𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑐𝑜𝑖𝑙𝑠 𝑥 𝑥 𝑐𝑜𝑖𝑙 𝑐𝑜𝑖𝑙 𝑠𝑖𝑑𝑒 Substitute: 744 = 186(2)n N=2 Therefore, there are two conductors per coil. 4. Draw a winding diagram of a DC machine with 4 poles, 14 slots, progressive, double layer lap winding. Show the position of brushes and direction of induces emf. Solution: Number of poles 4 Number of slots 14 Number of conductors 14x2=28 Pole pitch = number of conductors/poles = 28/4=7 𝑌𝑏 + 𝑌𝑓 𝑌𝑝 = 2 15|CJD | Page  Bachelor of Science in Electrical Engineering EE-316 ELECTRICAL MACHINE 1 Hence, 𝑌𝑏 + 𝑌𝑓 = 14 𝑌𝑏 − 𝑌𝑓 = 2 Therefore, 𝑌𝑏 = 8 and 𝑌𝑓 = 6 2𝑐 In back pitch 𝑌𝑏 = ±𝑘 𝑝 For lap winding both 𝑌𝑏 and 𝑌𝑓 must be odd and differ by 2 Satisfying the above condition 𝑌𝑏 = 7 and 𝑌𝑓 = 5 Winding Table: At the back 𝑌𝑏 = 7 At the back 𝑌𝑓 = 5 At the back 𝑌𝑏 = 7 At the back 𝑌𝑓 = 5 coil connected coil connected coil connected coil connected side to coil side side to coil side side to coil side side to coil side 1+7=8 8 – 5= 3 17+7=24 24 – 5= 19 3+7=10 10 – 5= 5 19+7=26 26 – 5= 21 5+7=12 12 – 5= 7 21+7=28 28 – 5= 23 7+7=14 14 – 5= 9 23+7=30(2) 30 – 5= 25 9+7=16 16 – 5= 11 25+7=32(4) 32 – 5 = 27 11+7=18 18 – 5= 13 27+7=34(6) 34 – 5= 29(1) 13+7=20 20 – 5= 15 15+7=22 22 – 5= 17 Winding Diagram 16|CJD | Page  Bachelor of Science in Electrical Engineering EE-316 ELECTRICAL MACHINE 1 5. Develop the single layer lap winding for a DC machine having 32 armature conductors and 4 poles. Solution: Number of conductors 32 Pole pitch 32/4=8 𝑌𝑏 + 𝑌𝑓 𝑌𝑝 = 2 Hence, 𝑌𝑏 + 𝑌𝑓 = 16 𝑌𝑏 − 𝑌𝑓 = 2 Hence, 𝑌𝑏 = 9 and 𝑌𝑓 = 7 Winding Table At the back 𝑌𝑏 = 9 At the back 𝑌𝑓 = 7 At the back 𝑌𝑏 = 9 At the back 𝑌𝑓 = 7 coil connected coil connected coil connected coil connected side to coil side side to coil side side to coil side side to coil side 1+9=10 10 – 7= 3 17+9=26 26 – 7= 19 3+9=12 12 – 7= 5 19+9=28 28 – 7= 21 5+9=14 14 – 7= 7 21+9=30 30 – 7= 23 7+9=16 16 – 7= 9 23+9=32 32 – 7 = 25 9+9=18 18 – 7= 11 25+9=34(2) 34 – 7= 27 11+9=20 20 – 7= 13 27+9=36(4) 36 – 7= 29 13+9=22 22 – 7= 15 29+9=38(6) 38 – 7= 31 15+9=24 24 – 7= 17 31+9=40(8) 40 – 7= 33(1) 6. Draw a developed diagram of a simplex 2-layer wave-winding for a 4-pole dc generator with 30 armature conductors. Hence, point out the characteristics of a simple wave winding. 30±2 Since, 𝑌𝐴 = = 8 𝑜𝑟 7. Taking 𝑌𝐴 = 7 we have 𝑌𝐵 = 𝑌𝐹 = 7 4 At the back 𝑌𝑏 = 7 At the back 𝑌𝑓 = 7 At the back 𝑌𝑏 = 7 At the back 𝑌𝑓 = 7 coil connected coil connected coil connected coil connected side to coil side side to coil side side to coil side side to coil side 1+7=8 8+7=15 23+7=30 30+7=37-7=7 15+7=22 22+7=29 7+7=14 14+7=21 29+7=36-30=6 6+7=13 21+7=28 28+7=35-30=5 13+7=20 20+7=27 5+7=12 12+7=19 27+7=34-30=4 4+7=11 19+7=26 26+7=33-30=3 11+7=18 18+7=25 3+7=10 10+7=17 25+7=32-30=2 2+7=9 17+7=24 24+7=31-30=1 9+7=16 16+7=23 17|CJD | Page  Bachelor of Science in Electrical Engineering EE-316 ELECTRICAL MACHINE 1 Self-Assessment Questions 1. In your own words, briefly explain commutation. 2. What is armature and its components? 3. A four – pole, DC generator with lap winding has 48 slots and 4 elements per slot. How many coils does it have? 4. If the armature of an eight – pole machine was wound with a simplex wave winding, how many parallel coils does it have? 5. Draw a developed diagram of a simple 2 – layer lap – winding for a 4 – pole generator with 16 coils. Hence, point out the characteristics of a lap – winding. Topic 2 Problems with commutation in real machines There are 2 main issues that disturb the commutation process. We will discuss these two with details and solutions of these problems. Armature reaction L di/dt voltages If we connect input supply at the stator of dc machine and also rotates the rotor of a machine with the exterior prime mover, during rotation rotor will link with the flux of stator field that will induce a voltage in the rotor. The voltage will be AC and covert into dc by the commutator attached to the rotor. Now if we connect some load with the rotor of a machine it will use some amount of current that will pass through the armature or rotor windings. Due to this current, there will be another magnetic field will produce that will affect the field of the stator. This disturbance of the armature field to the original field of stator is known as armature reaction. There are two main issues occurs due to this armature reaction. 18|CJD | Page  Bachelor of Science in Electrical Engineering EE-316 ELECTRICAL MACHINE 1 Neutral-Plane Shift Flux Weakening Neutral plane shift The first issue caused by the armature reaction is a neutral plane shift, in dc machine a position where the rotation direction of the rotor is parallel to field lines this position is known as a magnetic neutral plane. In this plane voltage production in armature windings is zero. For an understanding of the neutral plane shift problem, we discuss the given below the figure, which has 2 pole dc machines. You can see that flux is uniform over the complete machine. The voltage induced in the winding under the North Pole is out of page and voltage to induce in rotor windings under the South Pole has direction into a page. You can see that in an above-given figure the direction of the neutral plane is at ninety degrees. It is when there is no load is linked with the machine. Let’s assume that we have connected a load with this machine it behaves like a generator. Current will flow in outward direction form positive output terminal, thus current direction is out of the page for windings under the effect of north poles and into a page for windings effect of south poles. Due to this current, there will be field produce in the windings of the rotor it is shown in this given figure. This field of a rotor will interact with the field generated by the stator of the generator. At some points, these two fields will add up with each other and some points will cancel each other. The final effect of these two fields is shown in this given figure. 19|CJD | Page  Bachelor of Science in Electrical Engineering EE-316 ELECTRICAL MACHINE 1 You can see that the point where the voltage produce is 0, where the neutral plane shift has occurred. As we are working on dc generator you can observe that the neutral plane shift is in the direction in which the generator is rotating. But if it is the dc motor then the neutral plane shift will be opposite to the direction of rotation. In simple words, the plane shift for a generator is in direction of rotation and for a motor is in the opposite direction of rotation. The plane shift relies on the quantity of current drawn from the rotor and load connected. So what’s the big deal about the neutral-plane shift? The commutator necessity short-circuited commutator segments at the instant when the voltage crosswise them is 0. If the brushes are fixed to short-out conductors in the perpendicular plane, then the voltage among segments is certainly 0 till the load is attached. When a load is attached, the neutral plane shifts and the brushes short-circuited commutator segments with a specific value of voltage crossways them. The consequence is a current now flowing among the short out segments and huge sparks at the brushes when the current is disturbed as the brush left a segment. The final outcome is arcing and sparking at the brushes. It is a thoughtful problematic, as it leads to severely decreases brush Life, pitting of the commutator segments, and highly enhance repairing charges. Note that this issue cannot be removed even by insertion the brushes over the full-load neutral plane since sparking will occur at no-load condition. Flux weakening Armature Reaction The issue caused by the armature reaction is the weakening of flux. For an understanding of this issue, we discuss this given below figure. 20|CJD | Page  Bachelor of Science in Electrical Engineering EE-316 ELECTRICAL MACHINE 1 Mostly machine work at flux near the saturation point. So, at the point where the mmf of rotor adds up with the MMF of poles at stator, there is less increment in total flux. But at a point where mmf rotor subtracted from the mmf poles then larger decrement occurs in flux value. The final outcome is that the net value of flux over a complete pole will be lessened. You can see this in a given figure. The weakening of flux is not good for both motor and generator. In generators, the output voltage of the generator decreases. But in the case of the motor, this factor causes serious problems. With the decrement of a flux rotation speed of motor also decreases. Speed increment also enhances the load connected with the motor that causes a lessening of flux. It is likely in some shunt dc motors to approaches a running state as a consequence of flux decrement (weakening), where the speed of the motor increasing till the machine is detached with input supply or until its damage. L di/dt Voltages The second issue that occurs during commutation is L di/dt that occurs on the commutator is also known as an inductive kick. 21|CJD | Page  Bachelor of Science in Electrical Engineering EE-316 ELECTRICAL MACHINE 1 To an understanding of this issue lets discuss this given figure. You can see commutator segments are attached with one another in a sequence and winding conductor is linked with them. Let’s suppose that current passing through the carbon brush is four hundred amperes, the current through every path is two hundred amperes. Note that when the commutator is short-circuited, the current passing through the commutator will be reversed. Here the question arises. How fast must this reverse occur? Let’s suppose that the machine is rotating at the eight hundred revolutions per minute and have fifty commutators segments every commutator segment rotates under a brush and change its position in (t = 0.00 15 seconds). di/dt = 400A/0.0015s= 266,667 A/s With a small value of inductance, a very high value of Ldi/dt will be produced at the short-circuited commutator segments. This higher voltage obviously sources flashing at the brushes of the machine, resultant in the similar arcing problems that the neutral-plane shift causes. Topic 3 Internal generated voltage and induced torque equations of real DC Machines DC Machines types 1. DC Generator 2. DC Motor There are three (3) main factors over which dc machines voltages depend on. Flux ø of the machine. A rotation speed of the machine. Third, is constant over which construction of dc machine rely. 22|CJD | Page  Bachelor of Science in Electrical Engineering EE-316 ELECTRICAL MACHINE 1 The value of voltage at the armature winding of DC machine is no of a winding conductor in a single current path with the multiple of voltage in every conductor. The voltage induced in a single conductor under effect of pole is given as. 𝑒𝑖𝑛𝑑 = 𝑒 = 𝐵𝑣𝑙 B magnetic field v velocity of the wire l length of the wire The value of voltage at the armature of a machine is given as 𝑍𝐵𝑣𝑙 𝐸𝐴 = --- (1) ; Z no. of conductors 𝑎 a current path The rotation speed of every conductor is given as 𝑣 = 𝑟𝜔 r radius ω angular velocity So, equation 1 becomes 𝑍𝐵𝑟𝜔𝑙 𝐸𝐴 = 𝑎 This equation can be written in more comprehensive way as we know flux of poles is equal to the product flux density and area of poles. Φ = BAp The area of rotor is like the cylinder A = 2пrl If there are poles on the machine, then the potion of the area associated with each pole is the total area A divided by the number of poles P. If the no. of poles in a machine is P so the area cover by all poles will be equal to the 𝐴 2пrl 𝐴𝑝 = = 𝑃 𝑃 So, total flux in a machine will be. 2пrl Φ = B𝐴𝑝 = 𝐵( ) 𝑃 2пrl Φ = 𝐵( ) 𝑃 So internal generated voltage in DC machine will become 𝑍𝐵𝑟𝜔𝑙 𝐸𝐴 = 𝑎 𝑍𝑃 2пrBl 𝐸𝐴 = ( )𝜔 2п𝑎 𝑃 23|CJD | Page  Bachelor of Science in Electrical Engineering EE-316 ELECTRICAL MACHINE 1 𝑍𝑃 𝐸𝐴 = (𝜙𝜔) 2п𝑎 𝑍𝑃 𝐸𝐴 = 𝐾 (𝜙𝜔) at 𝐾 = 2п𝑎 In industries, the rotation speed of machines is measured in (rpm) instead of radians per second. So we have 2п 𝜔= (𝑁) 60 So induced voltage equation in terms of rpm can be written as 𝑍𝑃 2п 𝐸𝐴 = (𝜙)( (𝑁 )) 2п𝑎 60 𝑍𝑃𝜙𝑁 𝐸𝐴 = 60𝑎 Generated EMF equation of a DC Generator Electric Generator – a machine that converts mechanical energy (or power) into electrical energy (or power). The energy conversion is based on the principle of Faraday`s first law of Electromagnetic Induction, which states that whenever a conductor cuts magnetic lines of flux, an emf is developed in the conductor. 𝑍𝑃𝜙𝑁 𝐸𝑔 = 60𝑎 Where, E generated emf (volt) P number of poles N speed of armature core rotation (rpm) Z total number of elements or conductors Φ flux per pole (weber) a number of armature current paths If flux is given in unit maxwells or lines, 𝑍𝑃𝜙𝑁 𝐸𝑔 = 𝑥 10−8 60𝑎 After the machine has been assembled, “PZ and a” are constant. 𝐸1 𝑁1𝜙1 = 𝐸2 𝑁2𝜙2 𝐸𝐴 = 𝐾𝜙𝑁 Where k proportionality constant subscript 1 for condition 1 subscript 2 for condition 2 Sample Problems 24|CJD | Page  Bachelor of Science in Electrical Engineering EE-316 ELECTRICAL MACHINE 1 1. A 4 pole DC generator with duplex lap winding has 48 slots and four elements per slot. The flux per pole is 2.5 x 106 maxwells and it runs at 1500 rpm. What is the output voltage? Given Number of poles 4 a=mP= 2(4)=8 4 elements per slot Flux=2.5 x 106 maxwells N = 1500 rpm E? Z= elements/ slot x slots Z= 4(48) = 192 𝑍𝑃𝜙𝑁 𝐸= 𝑥 10−8 60𝑎 4(1500)(4 𝑥 48)(2.5 x 106 ) 𝐸= 𝑥 10−8 60(2)(4) E = 60V 2. A 4 – pole DC generator with simplex wave winding has 72 slots. The flux per pole is 2.88 x 106 maxwells. What is the speed of the prime mover when the open circuit voltage of the generator is 120V? 𝑍𝑃𝜙𝑁 𝐸= 𝑥 10−8 60𝑎 4(𝑁)(2𝑥72)(2.88 x 106 ) 120 = 𝑥 10−8 60(2)(1) N = 868 rpm 3. A DC generator is driven at a speed of 1000 rpm and generates 120V.Determine the generated emf if the speed is reduced to 900 rpm while the flux is increased by 25% 𝐸1 𝑁1 𝜙1 = 𝐸2 𝑁2 𝜙2 Substitute 𝑁2 = 900 and 𝜙2 = 1.25𝜙1 120 1000𝜙1 = 𝐸2 900(1.25𝜙1 ) 𝐸2 = 135 𝑉 Torques in DC machines The induced torque in dc machines depends on three main factors that are listed here. Flux ø in a machine. 25|CJD | Page  Bachelor of Science in Electrical Engineering EE-316 ELECTRICAL MACHINE 1 Armature current (IA) in the machine. The third is constant that depends on structure of machine. The torque induced on armature windings is equal to the product of no of a conductor in armature winding and torque in every conductor. The value of torque on every conductor of armature winding is given as. T = Force x radius 𝑇 = 𝑟(𝐼𝑐𝑜𝑛𝑑 )𝑙𝐵 Where r radius 𝐼𝑐𝑜𝑛𝑑 Conductor current B flux density If the no. of current paths in a machine is “a” so the current 𝐼𝑎 will divide into each current paths. So current in a conductor of armature winding is given as 𝐼𝑎 𝐼𝑐𝑜𝑛𝑑 = 𝑎 The torque in a single conductor will be given as 𝑟𝐼𝑎 𝐿𝐵 𝑇= 𝑎 The flux of each pole of machine is given as 2п𝑟𝐿 𝜙 = 𝐵𝐴𝑃 = 𝐵( ) 𝑃 𝐵2п𝑟𝐿 𝜙= 𝑃 So total induced torque can be written as 𝑟𝐼𝑎 𝐿𝐵 𝑇= 𝑎 𝑍𝑃𝜙𝐼𝑎 𝑇= 2п𝑎 The electric motor 26|CJD | Page  Bachelor of Science in Electrical Engineering EE-316 ELECTRICAL MACHINE 1 Electric Motor – a machine that converts electrical energy into mechanical energy. DC electric motor works on the principle that whenever a current carrying conductor is placed in a magnetic field, a force acts on the conductor causing it to move. 𝑍𝑃𝜙𝑁 𝐸𝑏 = 60𝑎 Where E back emf or counter emf (volt) P number of poles N speed of armature core rotation (rpm) Z total number of elements or conductors Φ flux per pole (weber) a number of armature current paths Speed Characteristics of a DC motor 𝐸𝑏 (60)(𝑎) 𝑁= 𝑍𝑃𝜙 After the machine has been assembled, “PZ and a” are constant 𝐸𝑏 𝑁1 𝐸𝑏1 𝜙1 𝑁=𝑘 = 𝜙 𝑁2 𝐸𝑏2 𝜙2 Where k proportionality constant subscript 1 for condition 1 subscript 2 for condition 2 Torque developed in the armature 𝑍𝑃𝜙𝐼𝑎 𝑇= 2п𝑎 Where 27|CJD | Page  Bachelor of Science in Electrical Engineering EE-316 ELECTRICAL MACHINE 1 T torque developed (Newton – meter) 𝐼𝑎 armature current (ampere) N speed of armature rotation (rpm) a number of armature current paths P number of poles Z number of conductors 𝜙 flux per pole (weber) After the machine has been assembled, “PZ and a” are constant 𝑇1 𝐼𝑎1 𝜙1 𝑇 = 𝑘𝐼𝑎 𝜙 = 𝑇2 𝐼𝑎2 𝜙2 Power developed in the armature 𝑃𝑑 = 𝐸𝑏 𝐼𝑎 Where 𝑃𝑑 power developed in the armature (watt) 𝐸𝑏 back emf or counter emf (volt) 𝐼𝑎 armature current (ampere) Mechanical Power Output in the Shaft 2п𝑁𝑇 𝐻𝑃 = , N = rpm; T= lb-ft 33,000 2п𝑁𝑇 𝐻𝑃 = , N = rpm; T= N-m 44760 Where HP mechanical power output in the shaft (horsepower) N Speed of shaft rotation (rpm) T Torque developed Sample Problem 1. A simplex lap wound armature has 600 conductors and carries a current of 50 amperes per armature current path. If the flux per pole is 30mWb, determine the electromagnetic torque developed by the armature? 𝑍𝑃𝜙𝐼𝑎 𝑍𝑃𝜙𝐼𝑎 𝑍𝜙𝐼𝑎 𝑇= = = 2п𝑎 2п(𝑚𝑃) 2п𝑚 600(30 𝑥 10−3 )(50) 𝑇= = 143.24 𝑁 − 𝑚 2п(1) 28|CJD | Page  Bachelor of Science in Electrical Engineering EE-316 ELECTRICAL MACHINE 1 2. The armature of a 4 – pole lap wound dc motor has 850 conductors and draws an armature current of 70A when the speed is 500rpm. If the flux per pole is 60mWb, determine the horsepower developed in the armature. 𝑍𝑃𝜙𝑁 𝑍𝑃𝜙𝑁 𝑍𝜙𝑁 𝐸𝑏 = = = 60𝑎 60(𝑚𝑃) 60𝑚 500(850)(60 𝑥 10−3 ) 𝐸𝑏 = = 425𝑉 60(1) 𝐸𝑏 𝐼𝑎 425(70) 𝑃𝑑 = = = 39.88 ℎ𝑝 746 746 3. If the power transmitted by the shaft of a motor is 20 hp and the torque exerted at the pulley is 220 N-m, what is the speed? 2п𝑁𝑇 𝐻𝑃 = 44760 44760 (𝐻𝑃) 20(44760) 𝑁= = = 647.6 𝑟𝑝𝑚 2п𝑇 2п(220) Activity 3 1. The armature of a four-pole shunt generator is lap wound and generates 216 volts when running at 600 rpm. The armature has 144 slots with six conductors per slot. If the armature is rewound, wave connected, find the emf generated at the same speed and flux per pole. 2. A two pole DC generator has an armature containing a total of 48 conductors connected in two parallel paths. The flux per pole is 6.48 x 108 lines and the speed of the prime mover is 30 rpm. The resistance of each conductor is 0.01 ohm and the current carrying capacity of each conductor is 10A. calculate the terminal voltage of the generator. 3. A 4-pole generator, having wave – wound armature winding has 51 slots, each slot containing 20 conductors. What will be the voltage generated in the machine when driven at 1500 rpm assuming the flux per pole to be 7mWb. 4. A DC motor takes an armature current of 110A at 480V. The armature circuit resistance is 0.2 ohm. The machine has 6 poles and the armature is lap – connected with 864 conductors. The flux per pole is 0.05Wb. Calculate the speed and the gross torque developed by the armature. 5. If the power transmitted by the shaft of a motor is 10 hp and the torque exerted at the pulley is 70 lb – ft, what is the speed? Topic 4 Construction of DC Machines The DC machine can be classified into two types namely DC motors as well as DC generators. Most of the DC machines are equivalent to AC machines because they include AC currents as well as AC voltages in them. The output of the DC machine is DC output because they convert AC voltage to DC voltage. The 29|CJD | Page  Bachelor of Science in Electrical Engineering EE-316 ELECTRICAL MACHINE 1 conversion of this mechanism is known as the commutator; thus, these machines are also named as commutating machines. DC machine is most frequently used for a motor. The main benefits of this machine include torque regulation as well as easy speed. The applications of the DC machine are limited to trains, mills, and mines. As examples, underground subway cars, as well as trolleys, may utilize DC motors. In the past, automobiles were designed with DC dynamos for charging their batteries. DC Machine A DC machine is an electromechanical energy alteration device. The working principle of a DC machine is when electric current flows through a coil within a magnetic field, and then the magnetic force generates a torque which rotates the dc motor. The DC machines are classified into two types such as DC generator as well as DC motor. The main function of the DC generator is to convert mechanical power to DC electrical power, whereas a DC motor converts DC power to mechanical power. The AC motor is frequently used in the industrial applications for altering electrical energy to mechanical energy. However, a DC motor is applicable where the good speed regulation & ample range of speeds are necessary like in electric- transaction systems. DC Machine Construction of DC Machine The construction of DC machine can be done using some of the essential parts like Yoke, Pole core & pole shoes, Pole coil & field coil, Armature core, Armature winding otherwise conductor, commutator, brushes & bearings. Some of the parts of the DC machine is discussed below. 30|CJD | Page  Bachelor of Science in Electrical Engineering EE-316 ELECTRICAL MACHINE 1 Construction of DC Machine Yoke Another name of a yoke is the frame. The main function of the yoke in the machine is to offer mechanical support intended for poles and protects the entire machine from the moisture, dust, etc. The materials used in the yoke are designed with cast iron, cast steel otherwise rolled steel. Pole and Pole Core The pole of the DC machine is an electromagnet and the field winding is winding among pole. Whenever field winding is energized then the pole gives magnetic flux. The materials used for this are cast steel, cast iron otherwise pole core. It can be built with the annealed steel laminations for reducing the power drop because of the eddy currents. Pole Shoe Pole shoe in DC machine is an extensive part as well as enlarge the region of the pole. Because of this region, flux can be spread out within the air-gap as well as extra flux can be passed through the air space toward armature. The materials used to build pole shoe is cast iron otherwise cast steed, and also used annealed steel lamination to reduce the loss of power because of eddy currents. Field Windings In this, the windings are wounded in the region of pole core & named as field coil. Whenever current is supplied through field winding then it electromagnetics the poles which generate required flux. The material used for field windings is copper. Armature Core Armature core includes the huge number of slots within its edge. Armature conductor is located in these slots. It provides the low-reluctance path toward the flux generated with field winding. The materials used in this 31|CJD | Page  Bachelor of Science in Electrical Engineering EE-316 ELECTRICAL MACHINE 1 core are permeability low-reluctance materials like iron otherwise cast. The lamination is used to decrease the loss because of the eddy current. Armature Winding The armature winding can be formed by interconnecting the armature conductor. Whenever an armature winding is turned with the help of prime mover then the voltage, as well as magnetic flux, gets induced within it. This winding is allied to an exterior circuit. The materials used for this winding are conducting material like copper. Commutator The main function of the commutator in the DC machine is to collect the current from the armature conductor as well as supplies the current to the load using brushes. And also provides uni-directional torque for DC- motor. The commutator can be built with a huge number of segments in the edge form of hard drawn copper. The Segments in the commutator are protected from thin mica layer. Brushes Brushes in the DC machine gather the current from commutator and supplies it to exterior load. Brushes wear with time to inspect frequently. The materials used in brushes are graphite otherwise carbon which is in rectangular form. Types of DC Machines The excitation of the DC machine is classified into two types namely separate excitation, as well as self- excitation. In separate excitation type of dc machine, the field coils are activated with a separate DC source. In self-excitation type of dc machine, the flow of current throughout the field-winding is supplied with the machine. The principal kinds of DC machine are classified into four types which include the following. Separately excited DC machine Shunt wound/shunt machine. Series wound/series machine. Compound wound / compound machine. Separately Excited DC Machine In Separately Excited DC Machine, a separate DC source is utilized for activating the field coils. Shunt Wound DC Machine In Shunt wound DC Machines, the field coils are allied in parallel through the armature. As the shunt field gets the complete o/p voltage of a generator otherwise a motor supply voltage, it is normally made of a huge number of twists of fine wire with a small field current carrying. 32|CJD | Page  Bachelor of Science in Electrical Engineering EE-316 ELECTRICAL MACHINE 1 Series Wound DC Machine In series wound D.C. Machines, the field coils are allied in series through the armature. As series field winding gets the armature current, as well as the armature current is huge, due to this the series field winding includes few twists of wire of big cross-sectional region. Compound Wound DC Machine A compound machine includes both the series as well as shunt fields. The two windings are carried-out with every machine pole. The series winding of the machine includes few twists of a huge cross-sectional region, as well as the shunt windings, include several fine wire twists. The connection of the compound machine can be done in two ways. If the shunt-field is allied in parallel by the armature only, then the machine can be named as the ‘short shunt compound machine’ & if the shunt- field is allied in parallel by both the armature as well as series field, then the machine is named as the ‘long shunt compound machine’. Self-Assessment Questionnaires 1. What is a DC Machine? 2. Give at least 5 examples of a DC machine. 3. Draw the diagram of a DC machine, label each part then explain. 4. What is the most important part of the DC machine? 5. Describe how each differs from other types of machines. Topic 5 Power flow and losses in DC machines The efficiency of any dc machine either dc motor or generator is given as. 𝑃𝑜𝑢𝑡 ղ= 𝑥 100% 𝑃𝑖𝑛 If we define the losses that we will find that it is the difference between input and output power of dc machines. In the mathematical expression, it can be defined as. 𝑃𝑜𝑢𝑡 − 𝑃𝑙𝑜𝑠𝑠 ղ= 𝑥 100% 𝑃𝑖𝑛 There are (five) 5 main types of losses that occur in dc machines either its motor or generator. Copper Losses or I2R Losses Brush Losses Core Losses Mechanical Losses 33|CJD | Page  Bachelor of Science in Electrical Engineering EE-316 ELECTRICAL MACHINE 1 Stray Losses Now we discuss all these losses one by one with the detailed. Copper Losses As you can understand from the name of these losses that are copper losses mean that losses occur at the windings of machines. There are 2 types of windings first one is field winding that exists at the stator and the second one is armature windings that exit at the rotor, at these two windings coppers losses occurs. The value of these losses for armature and copper windings can be given as. Armature winding losses = PA =IA2RA Field winding losses = PF= IF2RF Where PA armature losses. PF field windings losses. IA is current passing through the armature winding. IF is current passing through the stator windings. RA is armature resistance. RF is field windings resistance. Brush Losses These losses occur at the carbon brushes that are connected with the output terminals and commutators of machines. The mathematical expression for these brushes is given as. PBD=VBDIA Where PBD loss at the carbon brushes. VBD voltage losses at the brushes. IA armature current. The purpose that the brush losses are found in this way is that the voltage drop at the brushes remains same at the different values of currents. Core Losses in DC Machines There are two types of copper losses in dc machines: the first one is eddy current losses and the second one is hysteresis losses. Eddy Current losses 34|CJD | Page  Bachelor of Science in Electrical Engineering EE-316 ELECTRICAL MACHINE 1 Faraday law states that the rate of change of flux in any conductor produces a voltage in that conductor. EMF =dø/dt If this law applies on dc machine, it can be understood that when the rotor of the machine rotates in the field, then voltage induces current which starts to flow through the armature winding. This current was known as eddy current. The mathematical expression for these currents is given as. 2 𝑃𝑒 = 𝐾𝑒 𝐵𝑚𝑎𝑥 𝑓 2𝑡 2𝑉 Where Pe power losses due to eddy currents. Ke constant for these currents. B flux density. f frequency. t thickness of the materials used. V volume of the core of machines Hysteresis loss The cause of these losses is the energy required to magnetize and demagnetize the core of the machine. 35|CJD | Page  Bachelor of Science in Electrical Engineering EE-316 ELECTRICAL MACHINE 1 With the increment in current for magnetization of core, the value of flux also increases. But when we decrease the current that was used for magnetization, the value of corresponding flux will not decrease with the current. When the value of the current becomes zero, then there is some value of flux that exits in the core. To minimize the flux in the core, external force is applied causes hysteresis losses. The opposite polarity of field is provided to the core minimizes the ramming flux in the machine. The negative magnetization depends on the material used for the construction of core. The mathematical expression for these losses is given as. ℎ 𝑃ℎ = ղ𝐵𝑚𝑎𝑥 𝑓𝑉 Where Pb denotes the hysteresis losses h Steinmetz hysteresis coefficient. Its value is from 1.5 to 2.5l Bmax flux density. f frequency. V volume of material used for core construction. Mechanical Losses These losses occur in dc machines due to the mechanical effects that occur in dc machines. These are 2 main facts that cause to mechanical losses first one is friction and second is windage Friction losses occur due to the bearings that exit the shaft of machines. The windage losses occur due to the air among the rotatory portion and their casing. These losses changes with the cube of speed of revolution of the machine. 36|CJD | Page  Bachelor of Science in Electrical Engineering EE-316 ELECTRICAL MACHINE 1 DC Machines Power-Flow Diagram One of the simple methods to find the values of different losses in dc machines is to draw their power flow diagram. In the given figure, you can see the power diagram of dc generator. In this figure the mechanical power is input and after eliminating stray losses, mechanical losses (friction and windage losses), core losses than we have an electrical output that is given here. 1. Mechanical Efficiency 𝑡𝑜𝑡𝑎𝑙 𝑤𝑎𝑡𝑡𝑠 𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑒𝑑 𝑖𝑛 𝑎𝑟𝑚𝑎𝑡𝑢𝑟𝑒 𝐸𝑔 𝐼𝑎 ղ𝑚 = = 𝑚𝑒𝑐ℎ𝑎𝑛𝑖𝑐𝑎𝑙 𝑝𝑜𝑤𝑒𝑟 𝑠𝑢𝑝𝑝𝑙𝑖𝑒𝑑 𝑜𝑢𝑝𝑢𝑡 𝑜𝑓 𝑑𝑟𝑖𝑣𝑖𝑛𝑔 𝑒𝑛𝑔𝑖𝑛𝑒 2. Electrical efficiency 𝑤𝑎𝑡𝑡𝑠 𝑎𝑣𝑖𝑎𝑙𝑎𝑏𝑙𝑒 𝑖𝑛 𝑙𝑜𝑎𝑑 𝑐𝑖𝑟𝑐𝑢𝑖𝑡 𝑉𝐼 ղ𝑒 = = 𝑡𝑜𝑡𝑎𝑙 𝑤𝑎𝑡𝑡𝑠 𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑒𝑑 𝐸𝑔 𝐼𝑎 3. Overall or Commercial efficiency 𝑤𝑎𝑡𝑡𝑠 𝑎𝑣𝑖𝑎𝑙𝑎𝑏𝑙𝑒 𝑖𝑛 𝑙𝑜𝑎𝑑 𝑐𝑖𝑟𝑐𝑢𝑖𝑡 ղ𝑐 = 𝑚𝑒𝑐ℎ𝑎𝑛𝑖𝑐𝑎𝑙 𝑝𝑜𝑤𝑒𝑟 𝑠𝑢𝑝𝑝𝑙𝑖𝑒𝑑 It is obvious that overall efficiency ղ𝑐 = ղ𝑚 𝑥 ղ𝑒. For good generators, its value may be as high as 95%. But this is not the power that gets at the output terminals. Before reaching the output terminal copper losses and brush losses also subtract from it. In the given figure the dc motor power flow diagram is given. It is the reverse of dc generator power flow figure. 37|CJD | Page  Bachelor of Science in Electrical Engineering EE-316 ELECTRICAL MACHINE 1 Sample Problems 1. The hysteresis and eddy current losses of a DC machine running at 1000 rpm are 250 watts and 100 watts, respectively. If the flux remains constant, at what speed will the total iron loss be halved? 𝑃ℎ = 𝑘ℎ 𝑁 𝑃ℎ2 𝑁2 𝑁 = 𝑃ℎ2 = 𝑃ℎ1(𝑁2) 𝑃ℎ1 𝑁1 1 𝑁 2 𝑃ℎ2 = 250 (1000 ) = 0.25𝑁2 ------ 1 𝑃𝑒 = 𝑘𝑒 𝑁 2 𝑃𝑒2 𝑁 2 𝑁 2 = ( 2) 𝑃𝑒2 = 𝑃𝑒1 ( 2 ) 𝑃𝑒1 𝑁1 𝑁1 𝑁 2 2 𝑃𝑒2 = 100 (1000 ) = 0.0001𝑁22 ------ 2 1 𝑃ℎ2 + 𝑃𝑒2 = (𝑃ℎ1 + 𝑃𝑒1 ) ------ 3 2 Substitute Eq. 1 and Eq. 2 in Eq.3: 0.25𝑁2 + 0.0001𝑁22 = 0.5 (250 + 100) 𝑁22 + 2500𝑁2 − 1750000 = 0 Using quadratic formula: −2500 ± √25002 − 4(1)(−1750000) 𝑁2 = 2(1) 38|CJD | Page  Bachelor of Science in Electrical Engineering EE-316 ELECTRICAL MACHINE 1 −2500 ± 3640 𝑁2 = 2 Positive −2500 + 3640 𝑁2 = = 570 𝑟𝑝𝑚 2 2. A 10kW, 250V, 6 pole DC shunt generator runs at 1000 rpm when delivering full load. The armature has 534 lap connected conductors. Full load Cu loss id 0.64 kW the total brush drop is 1 volt. Determine the flux per pole. Neglect shunt current. Since shunt current is negligible, there is no shunt Cu loss. The copper loss in armature only. 10 𝑘𝑊 𝐼 = 𝐼𝑎 = = 40𝐴 250𝑉 𝑃𝑎 = 𝐼𝑎2 𝑅𝑎 = 402 𝑥 0.4 = 0.64 𝑥 103 𝑊 𝐼𝑎 𝑅𝑎 𝑑𝑟𝑜𝑝 = 0.4 𝑥 40 = 16𝑉 Brush drop = 2 x 1 = 2V Generated emf 𝐸𝑔 = 250 + 16 + 1 = 267𝑉 𝜙𝑍𝑁𝑃 𝜙𝑥534𝑥1000𝑥6 Now, 𝐸𝑔 = = 267 = 60𝑎 60(6) 𝜙 = 30𝑚𝑊𝑏 3. A 220V shunt motor has an armature resistance of 0.2 ohm and field resistance of 110 ohm. The motor draws 5A at 1500 rpm at no load. Calculate the speed and shaft torque if the motor draws 52A at rated voltage. 220𝑉 𝐼𝑠ℎ = = 2𝐴 110 Ω 𝐼1 = 5 − 2 = 3𝐴 𝐼2 = 52 − 2 = 50𝐴 𝐸𝑏1 = 220 − 3(0.2) = 219.4𝑉 𝐸𝑏2 = 220 − 50(0.2) = 210𝑉 𝑁2 210 = 1500 219.4 𝑁2 = 1436 𝑟𝑝𝑚 39|CJD | Page  Bachelor of Science in Electrical Engineering EE-316 ELECTRICAL MACHINE 1 For finding the shaft torque, it was found that motor output when it draws a current of 52A. First, use the no – load data for finding the constant losses of the motor. No load motor input = 220 x 5 = 1000W; 𝑃𝑎 = 32 (0.2) = 2𝑊 Constant or standing losses of the motor = 1100 – 2 =1098W When loaded, 𝑃𝑎 = 502 (0.2) = 500𝑊 Hence, total motor losses = 1098 +500 = 1598W Motor input on load = 220 x 52 = 11440W; output = 11440 – 1598 = 9842W 9.55 𝑥 𝑜𝑢𝑝𝑢𝑡 9.55 𝑥 9842 𝑇𝑠ℎ = = = 65.5 𝑁 − 𝑚 𝑁 1436 Activity 5 1. The hysteresis and eddy current losses of a DC Machine running at 1000 rpm are 250 watts and 100 watts, respectively. If the flux remains constant, at what speed will the total iron loss be halved? 2. The eddy current loss in a DC machine is 600 watts when the total flux is 2, 000, 000 maxwells per pole and the machine is running at 1000 rpm. Determine the loss when the excitation of the machine is adjusted to produce 2, 500, 000 maxwells per pole and the speed is increased to 1200 rpm. 3. In a DC generator, the iron losses at 1000 rpm are 10 kW at a given field current. At a speed of 750 rpm and at the same field current, the total iron losses become 6 kW. Assume the hysteresis loss is directly proportional to the speed and the eddy current loss is proportional to the square of the speed. Determine the iron losses at 500 rpm. 4. The shaft power of a motor is 7.8 hp. It draws 50A from 120V. The field winding draws 1.2 A. What is the efficiency of the motor? References Books Rojas, R. 2001. 1001 Solved Problems in electrical Engineering. Published by Jaime R. Tiong Dr. P.S. Bimbhra, January 1, 2021, Electric Machinery. Published by Khanna Publishing House Theraja, B.L. July 6, 2013 A Textbook of Electrical Technology in S.I. Units Volume II AC&DC Machines. Published by S. Chand & Company LTD Stephen J. Chapman. February 17, 2011. Electric Machinery Fundamentals 5 th edition. Published by McGraw Hill Online https://www.electrical4u.com/ https://circuitglobe.com/ 40|CJD | Page