School of Electrical and Computer Engineering Applied Electronics and Electrical Machines (ELEC 365) Fall 2015 DC Machines 1
DC Machines Key educational goals: Develop the basic principle of operation of a dc machine. Classify the different types of dc machines and evaluate their characteristics. Reading/Preparatory activities for class i) Textbook: Chapter 16: 16.1, 16.2, 16.3, 16.4, 16.5, 16.6, 16.7 2
DC Machines Questions to guide your reading and to think about ahead of time. 1. What is operating principle of a dc machine? 3. What are the different hardware components of dc machine? 4. Name one application of a dc generator. 5. What are the different types of dc motors? 6. How do you distinguish a dc series motor from a separately excited dc motor? 7. What is the effect of armature reaction? 3
Classification of motors 4
Introduction In a transformer energy is transferred from primary to secondary, form of energy (electrical) is unchanged. In a ROTATING ELECTRIC MACHINE (M/C or m/c), electrical energy is converted into mechanical form and vice versa: Electric Motor Electric energy to Mechanical Energy. Electric Generator Mechanical Energy to Electrical Energy. Electric machines are referred to as ELECTRO-MECHANICAL ENERGY CONVERTERS. 5
Introduction Electromagnetic conversion Electrical system V, i Coupling Magnetic Field Mechanical system T, n Conversion of energy from electrical to mechanical form or vice versa results from the following two electromagnetic phenomena: 1. When a conductor moves in a magnetic field, voltage is induced in the conductor. 2. When a current-carrying conductor is placed in a magnetic field, the conductor experiences a mechanical force. Note: These two effects occur simultaneously. 7
Operating Principle of a DC Machine The linear m/c a simple example dc machine consisting of a conducting bar sliding on conducting rails. There are four basic equations to the m/c. These equations are necessary to understand the above m/c. 1) The equation for the FORCE on a wire in the presence of a MAGNETIC FIELD: F = i l B The direction of the force is given by Fleming s left hand rule. 8
Operating Principle of a DC Machine 2) The equation for the voltage induced in a wire moving in a magnetic field: eind = (v B).l where v = velocity of wire, l = length of conductor in magnetic field with the vector pointing along the direction of the wire toward the end assumed to be positive. 3) KIRCHOFF S VOLTAGE LAW: VB ir eind = 0. VB = ir + eind Equivalent circuit for the linear m/c. 4. NEWTON S LAW for the bar across the tracks: Fnet = ma 9
Fleming s Left Hand Rule Or Motor Rule To find the direction of force: FORE FINGER = MAGNETIC FIELD 90 0 90 0 90 0 FORCE = (ilb).sin(θ) MIDDLE FINGER= CURRENT Where θ is the angle between the wire and the flux density vector. This principle is basis for MOTOR ACTION. (called Lorentz Force). 10
Fleming s Right Hand Rule Or Generator Rule To find the direction of induced emf: right hand rule is applicable. FORE FINGER = MAGNETIC FIELD 90 0 90 0 90 0 MIDDLE FINGER = INDUCED VOLTAGE eind (induced emf) = vblcos(θ) θ = smallest angle between direction of wire l and (v B). 11
STARTING THE DC MACHINE To start the m/c, close the switch. (i) A current flows through the bar. Since the bar is initially at rest, eind = 0; i = V B e ind R = V B R (KVL) 13
STARTING THE DC MACHINE (ii) A current flows through a wire in the presence of a MAGNETIC FIELD and hence a FORCE is induced on the wire. Find = ilb TO THE RIGHT. (iii) Bar will accelerate to the right (by Newton s Law). As velocity of bar increases, eind = vbl (POSITIVE UPWARD) (iv) This voltage reduces the current flowing through the wire (bar): i = V B e ind R decrease ( ) 14
STARTING THE DC MACHINE (v) Force induced Decreases. [since F = (i )lb ]. The result is finally a steady state speed (vss) is reached where the net force on the bar is zero. i.e. F = (i )lb until F = 0. At that point eind = VB, i = 0 and the bar moves at a constant speed (NO-LOAD): v ss = V B Bl [since V B = e ind = v ss Bl] 15
THE DC LINEAR MACHINE AS A MOTOR DC Linear m/c as a motor: force applied in the opposite direction of motion. Assume that the linear m/c is initially running at the no-load steady-state conditions. Now if an external load is applied to it in a direction opposite to the direction of motion, what happens? 16
THE DC LINEAR MACHINE AS A MOTOR (i) Due to application of Fload, net force on the bar will be in the direction opposite to the direction of motion, Fnet = Fload Find (ii) The resulting acceleration a = F net m is negative, so the bar slows down, i.e. v (v decreases). (iii) eind = (v )(Bl) falls, (iv) Induced force Find = (I )lb increases until Find = Fload and bar runs at a lower speed (v). i = V B e ind R (v) Electrical power, Pconv = eind.i is converted to mechanical power Find.v, and the m/c is acting as a motor. [Behavior is similar to a DC shunt motor]. 17
THE DC LINEAR MACHINE AS A GENERATOR Assume again that the linear dc m/c is initially running at no-load steady-state conditions. Now apply a force in the direction of motion. DC Linear m/c as a generator: force applied in the direction of motion. 18
THE DC LINEAR MACHINE AS A GENERATOR (i) (ii) Due to application of F app ; F net is in the direction of motion. acceleration a = F net m is positive, so the bar speeds up (v ). (iii) e ind = v (Bl) increase ( e ind > V B ), i = e ind V B R. (iv) F ind = i lb increases until F ind = F load and bar runs at a higher speed v. (v) Amount of mechanical power = F ind. v, is converted to electrical power, P conv = e ind. i and the m/c is acting as a generator. [Behavior is similar to a DC shunt generator]. 19
SUMMARY (1) Same m/c acts as both generator and motor. Only difference is whether the externally applied force is in the direction of motion (generator) or opposite the direction of motion (motor). (2) eind > VB (GENERATOR). eind < VB (MOTOR). (3) Whether motor or generator, Find (motor action) and eind (generator action) are present at all times. (4) Whether m/c was motor (moved slowly) or generator (moved rapidly), it always moved in the same direction. 20
Basics Structure of DC Machines 21
Basics Structure of DC Machines 22
Basics Structure of DC Machines Armature of a DC Machine 23
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Operating Principle of Rotating DC Machine 26
Operating Principle of Rotating DC Machine 27
Action of a Commutator 28
Action of a Commutator Generally, the commutator segments are copper bars insulated from one another and from the shaft. The brushes contain graphite that lubricates the sliding contact. Significant disadvantage of dc machines? 30
EMF EQUATION OF A GENERATOR Let φ = flux/pole Weber (Wb) Z = Total number of armature conductors. = (No. of slots) (No. of conductors/slot) P = Number of Generator poles (Even number) a = Number of parallel paths in the armature winding. E a = Average voltage induced N = Armature speed Ω = Angular velocity, = 2πN (V) Revolutions per second (rps) (Rads/s.) Speed regulation : Ω no load Ω full load Ω full load 100% 33
EMF EQUATION OF A GENERATOR The voltage induced in armature: E a = KφΩ V Where K is a machine constant which depends on the design parameters of the machines. Where K = ZP 2πa Torque equation T d = KφI a Nm. E a I a is the air-gap power of the machine. E a I a = (KɸΩ) I a Watts = (KφΩ) T d Kφ = (T d ) Ω = P d Watts This is the gross mechanical power in Watts. Where Ω is in Rads/s. 34
CLASSIFICATION OF DC MACHINES Based on excitation. I. Permanent magnet: No field windings. Usually, fractional horsepower rating m/c. With the development of permanent magnet materials, higher power available. 35
Separately Excited DC Machine II. (1) Separately excited: No electrical interconnection between the field and armature winding. Vt = Ea + IaRa V = KφΩ + IaRa Ω = Vt Ia Ra Kφ rads/sec. separately excited DC motor. 36
Shunt excited (2) Shunt excited (Self): Parallel connection between the field and armature winding. Shunt excited DC m/c. 38
Series excited (3) Series excited: Series field winding is connected in series with the armature and therefore, carries the same current as the armature. Series excited DC m/c. 39
Compound excitation (4) Compound excitation: Short shunt: Cumulative Compound Long Shunt: Differential Compound. 40
MAGNETIZATION CURVES constant speed Circuit diagram to vary the field current (in turn flux) to obtain the magnetization curve. Relation between Speed and backemf: Air-gap line approximates the magnetization curve. 41
Armature Reaction (AR) In a DC machine, two kinds of magnetic fluxes are present; 'armature flux' and 'main field flux'. The effect of armature flux on the main field flux is called as armature reaction. AR is the magnetic field produced by the armature current AR aids the main flux in one half of the pole and opposes the main flux in the other half of the pole. However due to saturation of the pole faces the net effect of AR is demagnetizing. 42
COMPENSATING WINDINGS One way to reduce the effects of armature reaction is to connect winding called compensating winding" in series with armature windings so that it s mmf is proportional to the armature mmf. These pole face windings are so arranged that the mmf produced by currents flowing in these windings opposes the armature mmf. These pole face windings are expensive. Therefore they are only used in large machines or in machines that are subjected to abrupt change of armature current 43
Interpoles Another way to reduce the effects of armature reaction is to place small auxiliary poles called "interpoles" between the main field poles. The interpoles have a few turns of large wire and are connected in series with the armature. Interpoles are wound and placed so that each interpole has the same magnetic polarity as the main pole ahead of it, in the direction of rotation. 44
EQUIVALENT CIRCUIT OF A DC GENERATOR J = Polar moment of inertia (Kg.m²) v f = R f i f + L f di f dt V v t = KφΩ L a di a dt R ai a V T shaft = Kφi a + J dω dt + T loss Under STEADY-STATE: N.m. These equations are used under transient conditions. v f = (R f )(I f ) V v t = KφΩ R a I a V Equivalent circuit of a separately excited generator. T shaft = KφI a + T loss N.m. 45
PERFORMANCE OF GENERATORS (a) External characteristics [potential difference (pd) Vt versus load current, il]. (b) Efficiency. (c) Voltage Regulation = [(No Load Voltage) (Full Load Voltage)]/[Full load Voltage]. 1) Separately Excited Generator Separately excited DC generator. Re is the external resistance added in the field circuit. 46
PERFORMANCE OF GENERATORS (Separately Excited) Steady-state: I f = V fs (R e +R f ) E a = KφΩ R L = V t I L I L = I a E a = V t + I a R a V t = E a I a R a = E a I L R a I a = I L = E a (R a +R L ) = KφΩ (R a +R L ) A V Ω A V V A External characteristic of a separately excited generator. 47
PERFORMANCE OF GENERATORS (Shunt) 2) Shunt Generator (self-excited) When m/c is driven under NO LOAD: A small Ear due to residual magnetism appears at the terminals. This voltage circulates a current through field winding which in turn increases generated voltage (cumulative build up) until a voltage Ea is reached, where: Ea = If.Rf = Vt (Assume Ra is neglected) Schematic of shunt dc machine. Voltage buildup in a shunt dc generator. 48
PERFORMANCE OF GENERATORS (Shunt) Under NO LOAD: Ea = (Ra + Rfc + Rfw)If Vt = (Rfc + Rfw)If Vt Ea = F(If) When the m/c is LOADED: Shunt generator under no-load. Vt = (RL)IL Vt = Ea - (Ia Ra) Ia = IL + If Shunt generator with a load. 49
PERFORMANCE OF GENERATORS (Shunt) Effect of field resistance. 50
PERFORMANCE OF GENERATORS (Shunt) A shunt generator can not build up if: (a) There is no residual magnetism, (b) If the field is connected in a wrong-way around, opposing the permanent magnetism, & (c) The value of the field resistance is greater than a value called critical resistance. 51
Example 1 A shunt generator delivers 450 A at 230 V and the resistances of the shunt field and armature are 50 Ω and 0.03 Ω, respectively. Calculate the generated emf. Field current, I F = 230 50 = 4.6 A Load current, IL = 450 A with load voltage, Vt = 230 V. Armature current, Ia = IL + IF = 450 + 4.6 = 454.6 A. Generated emf, Ea = Vt + IaRa = 230 + (454.6)(0.03) = 243.6 V. 54
Example 2 A separately excited generator is rated at 10 kw, 200 V, 1000 rpm. Rf = 80 Ω, Ra= 0.4 Ω and Ia = 50 A. Predict the no-load voltage at 1000 rpm and the fullload voltage at 800 rpm if the field current is kept constant. 55
Example 2 At no-load, Ia = 0 and Vt = Ea = VNL VNL = Ea = VFL + IaRa = 200 + (50)(0.4) = 220 V. If the field current is constant, the generated emf Ea is directly proportional to speed; therefore, at 800 rpm, induced (or generated voltage) E a is given by The terminal voltage at 800 rpm under full-load (Ia = 50 A) is 56
EQUIVALENT CIRCUIT OF A DC MOTOR Illustrated below for a separately excited motor. J=Polar moment of inertia (Kg.m²) v f = R f i f + L f di f dt v t = KφΩ + L a di a dt + R ai a T load = Kφi a J dω dt T loss V V N.m. These equations are used under transient conditions. Under STEADY-STATE: Equivalent circuit of a separately excited dc motor. v f = (R f )(I f ) v t = KφΩ + R a I a T load = KφI a T loss E a = KφΩ = F i f Ω= ΩR V V N.m. 58
Shunt Connected DC Motor A shunt-connected motor is similar to a separately excited dc motor except that: Speed versus torque characteristic for a shunt motor. 59
Series Connected DC Motor Since, φ is a function of Ia, for light loads, speed becomes very high. Therefore, SERIES MOTORS MUST NEVER BE STARTED WITHOUT ANY LOAD. Speed versus torque for a series motor. 60
Torque Speed relationship in Series DC Motor Assumption: Apply Kirchhoff's Voltage Law to equivalent circuit: 61
Compound Excited DC Motor If the shunt and series field aid each other it is called a cumulatively excited machine. If the shunt and series field oppose each other it is called a differentially excited machine. 63
Speed Torque characteristics of different machines Torque speed characteristics of different dc motors. 64
permanent-magnet DC Motor In a permanent-magnet (PM) dc motor, the field is supplied by magnets mounted on the stator rather than by field coils. PM motors are common on fractional or subfractional horsepower sizes. 65
Compound Excited DC Motor Advantages: No power is required to establish the field leading to better efficiency. Smaller cheaper Disadvantages: Magnets can become demagnetized by overheating or because of excessive armature currents. The flux density magnitude is smaller than in wound- field machines. Consequently, the torque produced per ampere of armature current is smaller in PM motors than in wound- field motors with equal power ratings. 66
Comparison Motor Types Separately excited motors Characteristics Can be controlled either by varying the voltage applied to the field winding or by varying the voltage applied to the armature; can produce high torques Power Range Up to 100 hp Applications Traction applications, to control the speed and torque of the motor by changing both armature voltage and stator current DC shunt motor DC series motor Constant flux constant speed motor creates moderate torque at start Creates dangerously high torques at low speeds and should be always connected to the load; speed can be varied Up to 200 hp Up to 200 hp Machine tools like lathes, milling machines, grinding machines, centrifugal and reciprocating pumps, blowers, and fans Preferred for traction-type loads; employed in electric locomotives, conveyors, cranes, elevators, trolleys Permanent magnet DC motor Higher efficiency, smaller size and simpler architecture; magnets can become demagnetized due to excessive use and overheating; produces lower torque Up to 10 hp Power windows in automobiles, computer peripherals 67
Example 5 A 50-hp shunt-connected dc motor has the magnetization curve shown in Figure. The dc supply voltage is VT = 240 V, the armature resistance is RA = 0.065Ω, the field resistance is RF = 10, and the adjustable resistance is Radj = 14Ω. At a speed of 1200 rpm, the rotational loss is Prot = 1450 W. If this motor drives a hoist that demands a torque of Tout = 250 Nm independent of speed, determine the motor speed and efficiency. 70
Example 5 The equivalent circuit is shown in Figure. The field current is given by Next, we use the magnetization curve to find the machine constant K for this value of field current. Assuming constant torque for the rotational loss. Thus, the developed torque is 71
Example 5 Then, applying Kirchhoff s voltage law to the armature circuit, we have To find efficiency, we first compute the output power and the input power, given by 72
Speed Control 1 - Varying armature terminal voltage (Vt) while holding the field constant. 2 - Varying field current (If ) and therefore φ. 3 - Varying Ra (Insert resistance in series with the armature circuit.) OR combination of these. Speed Control by varying Ra By inserting additional resistance in series with armature circuit we can control the speed of DC machine. This method is applicable for various type of DC machine. 73
Speed Control by varying Ra Add external resistance Rd in series with the armature. INEFFICIENT method due to losses in added resistance. Speed versus Torque characteristics of a separately excited DC motor with armature resistance control. 74
Speed Control by Variation of Terminal Voltage Vt P0 No-load operating point. P1 Operating point for a given load. φ1 = Maximum (or rated) flux for which the m/c is designed. Speed versus Torque characteristics of a separately excited DC motor with voltage control. 75
Speed Control by varying If Speed control above base (rated) value is obtained by reducing VF and therefore If and called FIELD WEAKENING. Speed versus Torque characteristics of a separately excited DC motor with field control. 76
Speed Control by varying If Vt Control: Speed control from zero to base speed (Ω b ) with rated field current. Over this speed range, Ia must normally be limited to its rated value, and hence developed torque will also be limited, Since T = KɸIa Nm. P = ΩT W. [Vt is limited to rated value because of commutator insulation and because of the limit on the available supply, further speed increase requires field weakening]. For negligible, R a, Ω 1 φ rads s. T = KφI a 1 [since I Ω a is limited to rated value ] P = ΩT = constant W. 77
Early variable DC system: Ward Leonard 1- three-phase induction motor drives a dc generator. 2- Produced dc voltage supplies to the motor to be controlled. 79
Starting of DC Motors Ea = KφΩ Vt = Ea + IaRa Ia= (v t Ea ) R a Under standstill (at starting), Ω = 0, i.e., Ea = 0 and Ia = v t R a Therefore, at starting, very high current (since Ra is very small) flows through the armature windings. Therefore, limit starting current by inserting series resistance or applying reduced voltage. As speed picks up, Ea increases, reduce the resistance. This large current is approximately seven times greater than actual fullload current for motor. The high current would, in all probability, cause severe damage to the brushes, commutator, or windings. Starting resistors are usually incorporated into the motor design to limit starting current to 125 to 200 percent of full load current. 80
Losses and efficiency of DC machines Power flow in a dc generator Power flow in a shunt-connected dc motor 81