9 DC Machines Fundamentals Stator: is the stationary part of the machine. The stator carries a field winding that is used to produce the required magnetic field by DC excitation.Rotor (Armature): is the rotating part of the machine. The rotor carries a distributed winding, and is the winding where the e.m.f. is induced.Field winding: Is wound on the stator poles to produce magnetic field (flux) in the air gap.Armature winding: Is composed of coils placed in the armature slots.Commutator: Is composed of copper bars, insulated from each other. The armature winding is connected to the commutator.Brush: Is placed against the commutator surface. Brush is used to connect the armature winding to external circuit through commutator
10 DC Machines Fundamentals In DC machines, conversion of energy from electrical to mechanical form or vice versa results from the following two electromagnetic phenomenaWhen a conductor moves in a magnetic field, voltage is induced in the conductor.2. When a current carrying conductor is placed in magnetic field, the conductor experiences a mechanical forces.
11 DC Machines Fundamentals Generator action:An e.m.f. (voltage) is induced in a conductor if it moves through a magnetic field.Motor action:A force is induced in a conductor that has a current going through it and placed in a magnetic fieldAny DC machine can act either as a generator or as a motor.
12 DC Machines Equivalent Circuit The equivalent circuit of DC machines has two components:Armature circuit:It can be represented by a voltage source and a resistance connected in series (the armature resistance). The armature winding has a resistance, RA.The field circuit:It is represented by a winding that generates the magnetic field and a resistance connected in series. The field winding has resistance RF.
14 Basic Operation of DC Motor In a dc motor, the stator poles are supplied by dc excitation current, which produces a dc magnetic field.The rotor is supplied by dc current through the brushes, commutator and coils.The interaction of the magnetic field and rotor current generates a force that drives the motor.
15 Basic Operation of DC Motor The magnetic field lines enter into the rotor from the north pole (N) and exit toward the south pole (S)The poles generate a magnetic field that is perpendicular to the current carrying conductorsThe interaction between the field and the current produces a Lorentz forceThe force is perpendicular to both the magnetic field and conductor
16 Basic Operation of DC Motor The generated force turns the rotor until the coil reaches the neutral point between the poles.At this point, the magnetic field becomes practically zero together with the force.However, inertia drives the motor beyond the neutral zone where the direction of the magnetic field reverses.To avoid the reversal of the force direction, the commutator changes the current direction, which maintains the counter clockwise rotation.
17 Basic Operation of DC Motor Before reaching the neutral zone, the current enters in segment 1 and exits from segment 2Therefore, current enters the coil end at slot ‘a’ and exits from slot ‘b’ during this stageAfter passing the neutral zone, the current enters segment 2 and exits from segment 1,This reverses the current direction through the rotor coil, when the coil passes the neutral zoneThe result of this current reversal is the maintenance of the rotation
19 Classification of DC Motor 1. Separately Excited DC MotorField and armature windings are either connected separate.2. Shunt DC MotorField and armature windings are either connected in parallel.3. Series DC MotorField and armature windings are connected inseries.4. Compound DC MotorHas both shunt and series field so it combinesfeatures of series and shunt motors.
20 Important terms VT – supply voltage EA – internal generated voltage/back e.m.f.RA – armature resistanceRF – field/shunt resistanceRS – series resistanceIL – load currentIF – field currentIA – armature currentn – speed
21 Generated or back e.m.f. of DC Motor General form of back e.m.f.,Φ = flux/pole (Weber)Z = total number of armature conductors= number of slots x number of conductor/slotP = number of polesA = number of parallel paths in armature[A = 2 (for wave winding), A = P (for lap winding)]N = armature rotation (rpm)EA = back e.m.f.
22 Torque Equation of a DC Motor The armature torque of a DC motor is given byΦ = flux/pole (Weber)Z = total number of armature conductors= number of slots x number of conductor/slotP = number of polesA = number of parallel paths in armatureIA = armature currentTa = armature torque
23 Equivalent Circuit of DC Motor Separately Excited DC MotorShunt DC Motor
25 Speed of a DC MotorFor shunt motorFor series motor
26 Example 1A 250 V, DC shunt motor takes a line current of 20 A. Resistance of shunt field winding is 200 Ω and resistance of the armature is 0.3 Ω. Find the armature current, IA and the back e.m.f., EA.
27 Solution Given quantities: Terminal voltage, VT = 250 V Field resistance, RF = 200 ΩArmature resistance, RA = 0.3 ΩLine current, IL = 20 AFigure 1
28 Solution (cont..) the field current, the armature current, VT = EA + IARAthe back e.m.f.,EA = VT – IARA = 250 V – (18.75)(0.3) = V
29 Example 2A 50hp, 250 V, 1200 r/min dc shunt motor with compensating windings has an armature resistance (including the brushes, compensating windings, and interpoles) of 0.06 Ω. Its field circuit has a total resistance Radj + RF of 50 Ω, which produces a no-load speed of 1200 r/min. There are 1200 turns per pole on the shunt field winding.
30 Example 2 (cont..)Find the speed of this motor when its input current is 100 A.Find the speed of this motor when its input current is 200 A.Find the speed of this motor when its input current is 300 A.
31 Solution Given quantities: Terminal voltage, VT = 250 V Field resistance, RF = 50 ΩArmature resistance, RA = 0.06 ΩInitial speed, n1 = 1200 r/minFigure 2
32 Solution (cont..)(a) When the input current is 100A, the armature current in the motor isTherefore, EA at the load will be
33 Solution (cont..)The resulting speed of this motor is
34 Solution (cont..)(b) When the input current is 200A, the armature current in the motor isTherefore, EA at the load will be
35 Solution (cont..)The resulting speed of this motor is
36 Solution (cont..)(c) When the input current is 300A, the armature current in the motor isTherefore, EA at the load will be
37 Solution (cont..)The resulting speed of this motor is
38 Example 3The motor in Example 2 is now connected in separately excited circuit as shown in Figure 3. The motor is initially running at speed, n = 1103 r/min with VA = 250 V and IA = 120 A, while supplying a constant-torque load. If VA is reduced to 200 V, determinei). the internal generated voltage, EAii). the final speed of this motor, n2
40 Solution Given quantities Initial line current, IL = IA = 120 A Initial armature voltage, VA = 250 VArmature resistance, RA = 0.06 ΩInitial speed, n1 = 1103 r/min
41 Solution (cont..) i) The internal generated voltage EA = VT - IARA = 250 V – (120 A)(0.06 Ω)= 250 V – 7.2 V= V
42 Solution (cont..) ii) Use KVL to find EA2 EA2 = VT - IA2RA Since the torque is constant ant he flux is constant, IA is constant. This yields a voltage ofEA2 = 200 V – (120 A)(0.06 Ω)= 200 V – 7.2 V= V
44 Example 4A DC series motor is running with a speed of 800 r/min while taking a current of 20 A from the supply. If the load is changed such that the current drawn by the motor is increased to 50 A, calculate the speed of the motor on new load. The armature and series field winding resistances are 0.2 Ω and 0.3 Ω respectively. Assume the flux produced is proportional to the current. Assume supply voltage as 250 V.
51 Armature windingsThe armature windings are usually former-wound. This are first wound in the form of flat rectangular coils and are then puller.Various conductors of the coils are insulated each other. The conductors are placed in the armature slots which are lined with tough insulating material.This slot insulation is folded over above the armature conductors placed in the slot and is secured in place by special hard wooden or fiber wedges.
52 Lap and wave WindingsThere are two types of windings mostly employed:• Lap winding• Wave windingThe difference between the two is merely due to the different arrangement of the end connection at the front or commutator end of armature.
53 Generated or back e.m.f. of DC Generator General form of generated e.m.f.,Φ = flux/pole (Weber)Z = total number of armature conductors= number of slots x number of conductor/slotP = number of polesA = number of parallel paths in armature[A = 2 (for wave winding), A = P (for lap winding)]N = armature rotation (rpm)E = e.m.f. induced in any parallel path in armature
54 Classification of DC Generator 1. Separately Excited DC GeneratorField and armature windings are either connected separate.2. Shunt DC GeneratorField and armature windings are either connected in parallel.3. Series DC GeneratorField and armature windings are connected inseries.4. Compound DC GeneratorHas both shunt and series field so it combinesfeatures of series and shunt motors.
55 Equivalent circuit of DC generator Separately excited DC generatorShunt DC generator
57 ExampleA DC shunt generator has shunt field winding resistance of 100Ω. It is supplying a load of 5kW at a voltage of 250V. If its armature resistance is 0.02Ω, calculate the induced e.m.f. of the generator.
58 Solution Given quantities Terminal voltage, VT = 250V Field resistance, RF = 100ΩArmature resistance, RA = 0.22ΩPower at the load, P = 5kW
59 Solution (cont..) EA = VT + IA RA = 250V + (22.5)(0.22) = 254.95V The field current,The load current,The armature current, IA = IL + IF = 20A + 2.5A = 22.5AThe induced e.m.f.,EA = VT + IA RA = 250V + (22.5)(0.22) = V