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Chapter 6 DC and AC Machines

Chapter 6 DC and AC Machines. Introduction . An electrical machine is link between an electrical system and a mechanical system. Conversion from mechanical to electrical: generator Conversion from electrical to mechanical: motor. Introduction . Machines are called

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Chapter 6 DC and AC Machines

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  1. Chapter 6DC and AC Machines

  2. Introduction • An electrical machine is link between an electrical system and a mechanical system. • Conversion from mechanical to electrical: generator • Conversion from electrical to mechanical: motor

  3. Introduction Machines are called • AC machines (generators or motors) if the electrical system is AC. • DC machines (generators or motors) if the electrical system is DC.

  4. DC machines can be divide by:

  5. DC Machines Construction cutaway view of a dc machine

  6. DC Machines Construction cutaway view of a DC machine

  7. DC Machines Construction Rotor of a DC machine

  8. DC Machines Construction Stator of a dc machine

  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 phenomena 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 field • Any DC machine can act either as a generator or as a motor.

  11. DC Machines Equivalent Circuit The equivalent/modellingcircuit of DC machine 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.

  12. DC Motor

  13. Basic Operation of DC Motor

  14. Classification of DC Motor 1. Separately Excited DC Motor • Field and armature windings are either connected separate. 2. Shunt DC Motor • Field and armature windings are either connected in parallel. 3. Series DC Motor • Field and armature windings are connected in series. 4. Compound DC Motor • Has both shunt and series field so it combines features of series and shunt motors.

  15. Equivalent Circuit of a DC Motor Armature circuit - voltage source, EA and a resistor, RA. The field coils, which produce the magnetic flux are represented by inductor, LF and resistor, RF. The separate resistor, Radj represents an external variable resistor used to control the amount of current in the field circuit. Basically it lumped together with Rfand called Rf

  16. Equivalent Circuit of DC Motor 1. Separately Excited DC Motor 2. Shunt DC Motor

  17. 3. Series DC Motor 4. Compound DC Motor

  18. Important terms in DC motor equivalent circuit • VT– supply voltage • EA– internal generated voltage/back e.m.f. • RA – armature resistance • RF – field/shunt resistance • RS – series resistance • IL– load current • IF– field current • IA– armature current • IL– load current • n– speed

  19. Speed of a DC Motor • For shunt motor • For series motor Flux, ϕ produce proportional to the current produce If Constant field excitation, means; if1 = if2 or constant flux; 1 = 2

  20. Example 1 A 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.

  21. Solution Given parameters: • Terminal voltage, VT = 250 V • Field resistance, RF = 200 Ω • Armature resistance, RA = 0.3 Ω • Line current, IL = 20 A Figure 1

  22. Solution (cont..) the field current, the armature current, VT = EA + IARA the back e.m.f., EA = VT – IARA = 250 V – (18.75)(0.3) = 244.375 V

  23. Example 2 A 50hp, 250 V, 1200 rpm 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 rpm. There are 1200 turns per pole on the shunt field winding.

  24. 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.

  25. Solution Given quantities: • Terminal voltage, VT = 250 V • Field resistance, RF = 50 Ω • Armature resistance, RA = 0.06 Ω • Initial speed, n1= 1200 r/min Figure 2

  26. Solution (cont..) (a) When the input current is 100A, the armature current in the motor is Therefore, EA at the load will be

  27. Solution(cont..) • The resulting speed of this motor is

  28. Solution(cont..) (b) When the input current is 200A, the armature current in the motor is Therefore, EA at the load will be

  29. Solution(cont..) • The resulting speed of this motor is

  30. Solution (cont..) (c) When the input current is 300A, the armature current in the motor is Therefore, EA at the load will be

  31. Solution (cont..) • The resulting speed of this motor is

  32. Example 3 The 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, determine i). the internal generated voltage, EA ii). the final speed of this motor, n2

  33. Example 3 (cont..) Figure 3

  34. Solution Given quantities • Initial line current, IL = IA = 120 A • Initial armature voltage, VA = 250 V • Armature resistance, RA = 0.06 Ω • Initial speed, n1 = 1103 r/min

  35. Solution (cont..) i) The internal generated voltage EA= VT - IARA = 250 V – (120 A)(0.06 Ω) = 250 V – 7.2 V = 242.8 V

  36. Solution (cont..) ii) Use KVL to find EA2 EA2= VT - IA2RA Since the torque and the flux is constant, IA is constant. This yields a voltage of: EA2 = 200 V – (120 A)(0.06 Ω) = 200 V – 7.2 V = 192.8 V

  37. Solution (cont..) • The final speed of this motor

  38. Example 4 A 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 55 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 200 V.

  39. Solution Given quantities • Supply voltage, VT = 200 V • Armature resistance, RA = 0.2 Ω • Series resistance, RS = 0.3 Ω • Initial speed, n1= 800 r/min • Initial armature current, Ia1 =IL1 = 20 A Figure 4

  40. Solution (cont..) When the armature current increased, Ia2 = 55 A, the back emf EA2 = V – Ia2 (RA + RS) = 200 – 55(0.2 + 0.3) = 225 V The speed of the motor on new load

  41. Solution (cont..) For initial load, the armature current, Ia1 = 20 A and the speed n1= 800 r/min V = EA1 + Ia1 (RA + RS) The back e.m.f. at initial speed EA1 = V - Ia1 (RA + RS) = 200 – 20(0.2 + 0.3) = 190 V

  42. DC Generator

  43. Generating of an AC Voltage • The voltage generated in any DC generator inherently alternating and only becomes DC after it has been rectified by the commutator

  44. Armature windings • The 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.

  45. 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/slot P = number of poles A = 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

  46. Classification of DC Generator 1. Separately Excited DC Generator • Field and armature windings are either connected separate. 2. Shunt DC Generator • Field and armature windings are either connected in parallel. 3. Series DC Generator • Field and armature windings are connected in series. 4. Compound DC Generator • Has both shunt and series field so it combines features of series and shunt motors.

  47. Equivalent circuit of DC generator Separately excited DC generator Shunt DC generator

  48. Series DC generator Shunt DC generator

  49. Example • A 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.

  50. Solution Given quantities • Terminal voltage, VT = 250V • Field resistance, RF = 100Ω • Armature resistance, RA = 0.22Ω • Power at the load, P = 5kW

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