Electromechnical Energy Conversion ppt

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Electromechnical Energy Conversion ppt

  1. 1. EE331Electromechanical Energy Conversion II Dr. Ali M. Eltamaly DC Motors 1
  2. 2. 8.1 DC Motor• The direct current (dc) machine can be used as a motor or as a generator.• DC Machine is most often used for a motor.• The major advantages of dc machines are the easy speed and torque regulation.• However, their application is limited to mills, mines and trains. As examples, trolleys and underground subway cars may use dc motors.• In the past, automobiles were equipped with dc dynamos to charge their batteries. 2
  3. 3. 8.1 DC Motor• Even today the starter is a series dc motor• However, the recent development of power electronics has reduced the use of dc motors and generators.• The electronically controlled ac drives are gradually replacing the dc motor drives in factories.• Nevertheless, a large number of dc motors are still used by industry and several thousand are sold annually. 3
  4. 4. 8.1 Construction 4
  5. 5. DC Machine Construction Figure 8.1 General arrangement of a dc machine 5
  6. 6. DC Machines• The stator of the dc motor has poles, which are excited by dc current to produce magnetic fields.• In the neutral zone, in the middle between the poles, commutating poles are placed to reduce sparking of the commutator. The commutating poles are supplied by dc current.• Compensating windings are mounted on the main poles. These short-circuited windings damp rotor oscillations. . 6
  7. 7. DC Machines• The poles are mounted on an iron core that provides a closed magnetic circuit.• The motor housing supports the iron core, the brushes and the bearings.• The rotor has a ring-shaped laminated iron core with slots.• Coils with several turns are placed in the slots. The distance between the two legs of the coil is about 180 electric degrees. 7
  8. 8. DC Machines• The coils are connected in series through the commutator segments.• The ends of each coil are connected to a commutator segment.• The commutator consists of insulated copper segments mounted on an insulated tube.• Two brushes are pressed to the commutator to permit current flow.• The brushes are placed in the neutral zone, where the magnetic field is close to zero, to reduce arcing. 8
  9. 9. DC Machines• The rotor has a ring-shaped laminated iron core with slots.• The commutator consists of insulated copper segments mounted on an insulated tube.• Two brushes are pressed to the commutator to permit current flow.• The brushes are placed in the neutral zone, where the magnetic field is close to zero, to reduce arcing. 9
  10. 10. DC Machines• The commutator switches the current from one rotor coil to the adjacent coil,• The switching requires the interruption of the coil current.• The sudden interruption of an inductive current generates high voltages .• The high voltage produces flashover and arcing between the commutator segment and the brush. 10
  11. 11. DC Machine Construction Rotation Ir_dc/2 Ir_dc Ir_dc/2 Brush Pole winding Shaft | 1 2 8 N 7 3 S 6 4 5 Insulation Copper Rotor Ir_dc segment WindingFigure 8.2 Commutator with the rotor coils connections. 11
  12. 12. DC Machine Construction Figure 8.3 Details of the commutator of a dc motor. 12
  13. 13. DC Machine ConstructionFigure 8.4 DC motor stator with poles visible. 13
  14. 14. DC Machine Construction Figure 8.5 Rotor of a dc motor. 14
  15. 15. DC Machine Construction Figure 8.6 Cutaway view of a dc motor. 15
  16. 16. 8.2.1 DC Motor Operation 16
  17. 17. DC Motor Operation• In a dc motor, the stator poles are supplied by dc Rotation Ir_dc/2 Ir_dc/2 excitation current, which Brush Ir_dc Pole produces a dc magnetic Shaft winding field. |• The rotor is supplied by 1 2 8 dc current through the N 7 3 S brushes, commutator 6 4 5 and coils.• The interaction of the Insulation Rotor Ir_dc Copper segment magnetic field and rotor Winding current generates a force that drives the motor 17
  18. 18. 8.2.1 DC Motor Operation v B• The magnetic field lines a enter into the rotor from the S N north pole (N) and exit Vdc 1 30 toward the south pole (S). 2 b• The poles generate a v magnetic field that is Ir_dc perpendicular to the current (a) Rotor current flow from segment 1 to 2 (slot a to b) carrying conductors.• The interaction between the B field and the current a produces a Lorentz force, S N Vdc 2 v 30 v• The force is perpendicular to 1 both the magnetic field and b conductor Ir_dc (b) Rotor current flow from segment 2 to 1 (slot b to a) 18
  19. 19. 8.2.1 DC Motor Operation v B• The generated force turns the rotor a until the coil reaches the neutral S N point between the poles. Vdc 1 30 2• At this point, the magnetic field b becomes practically zero together v with the force. Ir_dc• However, inertia drives the motor (a) Rotor current flow from segment 1 to 2 (slot a to b) beyond the neutral zone where the direction of the magnetic field B reverses. a• To avoid the reversal of the force S N Vdc direction, the commutator changes 2 v 30 v 1 the current direction, which b maintains the counterclockwise rotation. . Ir_dc (b) Rotor current flow from segment 2 to 1 (slot b to a) 19
  20. 20. 8.2.1 DC Motor Operation• Before reaching the neutral zone, v a B the current enters in segment 1 and exits from segment 2, S N Vdc 1 30• Therefore, current enters the coil 2 end at slot a and exits from slot b b during this stage. v Ir_dc• After passing the neutral zone, the current enters segment 2 and exits (a) Rotor current flow from segment 1 to 2 (slot a to b) from segment 1, B• This reverses the current direction a through the rotor coil, when the coil S N passes the neutral zone. Vdc 2 v 30 v 1• The result of this current reversal is b the maintenance of the rotation. Ir_dc (b) Rotor current flow from segment 2 to 1 (slot b to a) 20
  21. 21. 8.2.2 DC Generator Operation 21
  22. 22. 8.2.2 DC Generator Operation v B• The N-S poles produce a a dc magnetic field and the S N rotor coil turns in this Vdc 1 30 2 field. b• A turbine or other v Ir_dc machine drives the rotor. (a) Rotor current flow from segment 1 to 2 (slot a to b)• The conductors in the slots cut the magnetic flux B lines, which induce a voltage in the rotor coils. S N Vdc 2 v 30 v• The coil has two sides: 1 one is placed in slot a, the b other in slot b. Ir_dc (b) Rotor current flow from segment 2 to 1 (slot b to a) 22
  23. 23. 8.2.2 DC Generator Operation• In Figure 8.11A, the v a B conductors in slot a are cutting the field lines S N Vdc 1 30 entering into the rotor 2 from the north pole, b v• The conductors in slot b Ir_dc are cutting the field lines (a) Rotor current flow from segment 1 to 2 (slot a to b) exiting from the rotor to the south pole. B• The cutting of the field a lines generates voltage in S N Vdc 2 v the conductors. 30 v 1• The voltages generated in b the two sides of the coil Ir_dc are added. (b) Rotor current flow from segment 2 to 1 (slot b to a) 23
  24. 24. 8.2.2 DC Generator Operation• The induced voltage is v a B connected to the generator terminals through the S N Vdc 1 30 commutator and brushes. 2• In Figure 8.11A, the induced b v voltage in b is positive, and Ir_dc in a is negative. (a) Rotor current flow from segment 1 to 2 (slot a to b)• The positive terminal is connected to commutator B segment 2 and to the a conductors in slot b. S N Vdc 2 v 30 v• The negative terminal is 1 connected to segment 1 and b to the conductors in slot a. Ir_dc (b) Rotor current flow from segment 2 to 1 (slot b to a) 24
  25. 25. 8.2.2 DC Generator Operation• When the coil passes the v a B neutral zone: – Conductors in slot a are S N Vdc 1 30 then moving toward the 2 south pole and cut flux lines b exiting from the rotor v – Conductors in slot b cut the Ir_dc flux lines entering the in (a) Rotor current flow from segment 1 to 2 (slot a to b) slot b.• This changes the polarity B of the induced voltage in a the coil. S N Vdc 2 v 30 v• The voltage induced in a 1 is now positive, and in b is b negative. Ir_dc (b) Rotor current flow from segment 2 to 1 (slot b to a) 25
  26. 26. 8.2.2 DC Generator Operation v B• The simultaneously the a commutator reverses its S N terminals, which assures Vdc 1 30 2 that the output voltage b (Vdc) polarity is v Ir_dc unchanged. (a) Rotor current flow from segment 1 to 2 (slot a to b)• In Figure 8.11B – the positive terminal is B connected to commutator a segment 1 and to the S N Vdc 2 conductors in slot a. v 30 v 1 – The negative terminal is b connected to segment 2 and to the conductors in slot b. Ir_dc (b) Rotor current flow from segment 2 to 1 (slot b to a) 26
  27. 27. 8.2.3 DC MachineEquivalent Circuit 27
  28. 28. Generator 28
  29. 29. 8.2.3 DC Generator Equivalent circuit• The magnetic field produced by the stator poles induces a voltage in the rotor (or armature) coils when the generator is rotated.• This induced voltage is represented by a voltage source.• The stator coil has resistance, which is connected in series.• The pole flux is produced by the DC excitation/field current, which is magnetically coupled to the rotor• The field circuit has resistance and a source• The voltage drop on the brushes represented by a battery 29
  30. 30. 8.2.3 DC Generator Equivalent circuit Vbrush Rf Ra Load Φmax Iag Vf If Eag Vdc Mechanical Electrical power in power out• Figure 8.12Equivalent circuit of a separately excited dc generator. 30
  31. 31. 8.2.3 DC Generator Equivalent circuit• The magnetic field produced by the stator poles induces a voltage in the rotor (or armature) coils when the generator is rotated.• The dc field current of the poles generates a magnetic flux• The flux is proportional with the field current if the iron core is not saturated: Φ ag = K 1 I f 31
  32. 32. 8.2.3 DC Generator Equivalent circuit• The rotor conductors cut the field lines that generate voltage in the coils. E ag = 2 N r B  g v• The motor speed and flux equations are : Dg v =ω Φ ag = B  g D g 2 32
  33. 33. 8.2.3 DC Generator Equivalent circuit • The combination of the three equation results the induced voltage equation:  Dg E ag = 2 N r B  g v = 2 N r B  g ω   = N r ( B  g D g ) ω = N r Φ ag ω  2   • The equation is simplified. E ag = N r Φ ag ω = N r K 1 I f ω = K m I f ω 33
  34. 34. 8.2.3 DC Generator Equivalent circuit• When the generator is loaded, the load current produces a voltage drop on the rotor winding resistance.• In addition, there is a more or less constant 1–3 V voltage drop on the brushes.• These two voltage drops reduce the terminal voltage of the generator. The terminal voltage is; E ag = Vdc + I ag Ra + Vbrush 34
  35. 35. Motor 35
  36. 36. 8.2.3 DC Motor Equivalent circuit Vbrush Electrical Rf Ra power in Φmax DC Power V f If Iam Vdc Eam supply Mechanical power out• Figure 8.13 Equivalent circuit of a separately excited dc motor• Equivalent circuit is similar to the generator only the current directions are different 36
  37. 37. 8.2.3 DC Motor Equivalent circuit• The operation equations are:• Armature voltage equation Vdc = E am + I am Ra + Vbrush The induced voltage and motor speed vs angular frequency E am = K m I f ω ω = 2 π nm 37
  38. 38. 8.2.3 DC Motor Equivalent circuit• The operation equations are:• The combination of the equations results in K m I f ω = E am = Vdc − I am Rm The current is calculated from this equation. The output power and torque are: Pout Pout = E am I am T= = K m I am I f ω 38
  39. 39. 8.2.4 DC MachineExcitation Methods 39
  40. 40. DC Motor Operation• There are four different methods for supplying the dc current to the motor or generator poles: – Separate excitation; – Shunt connection – Series connection – Compound 40
  41. 41. 8.2.3 DC Motor Equivalent circuit Vbrush Ra Iam Im DC Power supply Eam Φmax Rf Vdc If Pout• Figure 8.14 Equivalent circuit of a shunt dc motor 41
  42. 42. 8.2.3 DC Motor Equivalent circuit Vbrush Ra Im DC Power Rf supply Eam Vdc Φmax Pout• Figure 8.15 Equivalent circuit of a series dc motor 42
  43. 43. 8.2.3 DC Motor Equivalent circuit Vbrush Ra Iam Im DC Power Rfs supplyEam Φmax Vdc Rfp Ifp Pout• Figure 8.16 Equivalent circuit of a compound dc motor 43

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