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Lecture 31 DC Motors

Lecture 31 DC Motors. Learning Objectives. Identify and define the components of a two pole permanent magnetic DC motor (stator, armature, commutator and brushes).

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Lecture 31 DC Motors

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  1. Lecture 31DC Motors

  2. Learning Objectives • Identify and define the components of a two pole permanent magnetic DC motor (stator, armature, commutator and brushes). • Given the direction of a magnetic field in a two pole permanent magnetic DC motor, determine the direction of force applied to a single armature loop (Lorentz/Force Law). • Understand the effect of multiple armature loops in a DC motor. • Understand the induced effects of rotating a current-carrying closed loop conductor in a magnetic field (Faraday/Lenz/Electromotive Force)

  3. Basic DC Motor Operations

  4. Basic DC Motor Operations • The current passing through the coil creates an electro-magnet with a North/South pole

  5. Basic DC Motor • This electro-magnet interacts with the permanent magnetic field • The opposite poles repel each other, while the like poles attract each other • This causes the motor to spin

  6. Torque on wire

  7. In order to provide continuous rotation, the armature current Ia must change direction every 180 of rotation. This process (commutation) is accomplished by brushes and a segmented commutator bar. Commutation Commutator Brushes

  8. Forces on DC motor rotor SOURCE: FISIK.FREE.FR. Applied Physics.

  9. Developed Force and Torque • The force between the rotating electromagnet and the stationary magnetic field is given by the Lorentz Force Law • Torque=Force x Distance: where r = radius from central axis • If power is only applied to the armature wire in the optimum position, the cross product becomes simple multiplication: • In our simple motor we have 2 wires being acted upon: r

  10. Torque • We can significantly increase torque by increasing: • The magnetic field • The current • The number of wires being acted upon • The most practical is to increase the number of wires being acted upon: where N = number of turns of wire • This equation is greatly simplified by using the MACHINE CONSTANT (KV) (V*sec)

  11. Torque • Multiple sets of windings are used on the rotor to ensure that torque is applied smoothly throughout the rotation of the motor. • Each winding is only briefly supplied current when it is at the position where it would apply maximum torque to the rotor • This ensures maximum efficiency of the motor, since power isn’t wasted on forces which try to pull the rotor apart F

  12. Torque Balance Under steady-state conditions: Tloss Tload Td

  13. Torque Developed power is: Ignoring rotational losses, this developed power is the mechanical output, and machine efficiency can be calculated as:

  14. REVIEW Faraday’s Law • Faraday’s Law states that a voltage is induced in a circuit when a conductor is moved through a magnetic field • In the case of DC linear motors we know that Einduced= B L u • u = velocity • B = magnetic field • L = length of conductor

  15. Faraday’s Law in motors • Induced voltage in motors is given by also Einduced = Ea = Kv ω • Kv is the motor constant (V*sec) • ω is the angular velocity (rad/sec) • Also called Counter EMF (CEMF) or “back EMF” because it opposes the applied voltage. • Angular velocity can be calculated by: ω = 2  (RPM/60)

  16. Parts of a DC Motor SOURCE: Gears Educational Systems.

  17. POLES (or FIELD WINDINGS) COMMUTATOR ARMATURE WINDINGS BRUSH ROTOR(entire rotating part) STATOR(stationary part) Actual DC Motor

  18. Points to remember • Torque developed Td=Kv Ia • Mechanical Power DevelopedPd = Td ω = Kv Iaω • Back EMF Ea = Kv ω • Angular velocity ω = 2  (RPM/60) • KVL Vdc-IaRa-Ea=0

  19. Example problem 1 A 120V permanent magnet DC motor is rated for 15A at 3600 rpm. The motor is 85% efficient at rated conditions. Assume no rotational losses. Find: • KV(Machine constant) . • Find the back EMF at 3600 rpm. • Find Ra

  20. Example problem 2 A permanent magnet DC motor has a machine constant, KV = 1.0 ν·s. The no-load torque is 2.0 N-m assuming rotational power loss is linear with speed. (Pd=Pmech-loss) • Determine the mechanical power developed by the unloaded motor at 3600 rpm. • Find the back EMF at 3600 rpm • Determine the armature current.

  21. Example problem 3 A 120 VDC permanent magnet DC motor with KV = 0.95 V·s is measured under the following two conditions (same voltage applied): 1. Motor is unloaded, IA = 2 A. 2. Motor is loaded until IA = 15 A and 1200 rpm. Find: • Torque due to rotational loss. • Torque due to the load in condition 2 assuming rotational power loss is linear with speed. (Pd=Pmech-loss) • Machine efficiency at condition 2?

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