EET 325 Lec 2 - DC Motors.pdf

EET-325 Advanced Control Wiring School of Engineering Technology & Applied Science (SETAS) Lecture #2 DC Motors  OBJECTIVES Current flow theories Construction of DC motor The principles of operation of D...

EET-325 Advanced Control Wiring School of Engineering Technology & Applied Science (SETAS) Lecture #2 DC Motors  OBJECTIVES Current flow theories Construction of DC motor The principles of operation of DC motors Electrical characteristics of DC motors Types of DC Motors Selecting Motors: Speed regulation & Torque Direction of Rotation 2 Lecture #2 – DC Motors  CURRENT FLOW THEORIES The electron flow theory will be used in this COURSE. 3 Lecture #2 – DC Motors  DC MOTOR - WHAT‘S INSIDE? 4 Lecture #2 – DC Motors  DC MOTOR - WHAT‘S INSIDE? 5 Lecture #2 – DC Motors  PRINCIPLES OF OPERATION Motor action: Electrical energy to mechanical energy Two requirements for motor action: - Current flow through a conductor - A force on the conductor develops This force is produced when the conducting wire is placed inside the magnetic field formed between two magnetic poles 6 Lecture #2 – DC Motors  OPERATION - THE RIGHT-HAND MOTOR RULE THumb = THrust (Force) (direction of armature rotation) Forefinger = Field (direction of magnetic field) Center finger = Current (direction of armature current) Fleming’s Right-hand Rule can be used to determine the direction of rotation of the armature if the magnetic field polarity of the pole pieces and the direction of current flow through the armature are known (Left-hand rule = Generator) 7 Lecture #2 – DC Motors  OPERATION - ROTARY MOTION Current-carrying conductor in a magnetic field will tend to move at right angles to the field The reaction of the wire to the field produces torque 8 Lecture #2 – DC Motors  OPERATION - CONTINUOUS ROTATION - 1 A continuous motion is achieved by reversing the direction of current flow in the wire of the armature Current change is provided by the commutator 9 Lecture #2 – DC Motors  OPERATION - CONTINUOUS ROTATION - 2 The switching action of the brushes and commutator is a process called commutation 10 Lecture #2 – DC Motors  PRACTICAL DESIGN OF DC MOTOR We’ve been using a single loop to make it easier to understand the concept of the DC motor, but In a practical DC motor; multiple loops and commutator segments, and more poles are added for more turning force and torque 11 Lecture #2 – DC Motors  DESIGN - CONTROL OF FIELD FLUX Because magnetic flux lines have a tendency to repel each other, curved magnets are used at the ends of the poles This design eliminates bowing and makes the flux lines run straight between the poles 12 Lecture #2 – DC Motors  CHARACTERISTICS - COUNTER ELECTROMOTIVE FORCE Counter Electromotive Force (CEMF), or Back EMF, is a result of a conductor passing through a magnetic field EMF is generated the same way as voltage in a generator is produced (Mechanical to electrical energy) The amount of CEMF produced is proportional to three factors: - Physical properties of the armature - Strength of the magnetic field - Rotational speed of the armature 13 Lecture #2 – DC Motors  CHARACTERISTICS - ARMATURE REACTION - 1 When the armature loop is at a right angle to the field flux lines, it is on the geometric neutral planes and generates no CEMF The perpendicular neutral plane becomes shifted 14 Lecture #2 – DC Motors  CHARACTERISTICS - ARMATURE REACTION - 2 Arcing at the brushes results because of armature reaction and adversely affects the motor: - Reduces torque - Motor is less efficient - Brush life is shortened, and the commutator is damaged Interpoles (aka. commutating poles) are added to the armature to correct armature reaction 15 Lecture #2 – DC Motors  MOTOR SELECTION Depending upon the application, different DC motors may be selected Two characteristics used in motor selection are: - Speed regulation - Torque 16 Lecture #2 – DC Motors  MOTOR SELECTION - SPEED REGULATION Speed regulation: Ability of a motor to maintain its speed under varying load conditions Motors are designed to operate at full load Overload: Operation above full load Partial load: Operation less than full load 17 Lecture #2 – DC Motors  SPEED REGULATION Speed regulation: Ability of a motor to maintain its speed under varying load conditions Motors are designed to operate at full load Overload: Operation above full load Partial load: Operation less than full load 18 Lecture #2 – DC Motors  TORQUE Torque: a twisting action that causes a motor to rotate Torque is measured by multiplying the force it will exert by the distance between the center of the shaft and the point where the force is being applied (radius) 𝑻=𝑭×𝒓 - F: magnetic force acting on the armature in pounds (lb) - r: the radius in feet (ft) - T: the torque in pound-feet (lb∙ft) 19 Lecture #2 – DC Motors  WORK 𝑾=𝑫×𝑭 - W: work in foot-pounds (lb∙ft) - D: distance in feet (ft) - F: force in pounds (lb) 20 Lecture #2 – DC Motors  TORQUE AND POWER This equation is applied for all motors. 𝑻×𝑺 𝑯𝑷 = 𝟓𝟐𝟓𝟐 Power is WORK over TIME Motors are rated in horsepower (HP) on the name plate 1hp = 746W - HP: horse power (hp) - T: torque (lb∙ft) - S: speed (RPM) 21 Lecture #2 – DC Motors  TORQUE AND POWER - EXAMPLE A shunt motor at its rated conditions develops 1,500 W at 1800 RPM. Determine the rated torque. 𝑇×𝑆 𝐻𝑃 ×5252 since 𝐻𝑃 = , 𝑇= 5252 𝑆 1500 𝑊 𝐻𝑃 = = 2.01 ℎ𝑝 S = 1800 𝑅𝑃𝑀 746 𝑊 2.01 ℎ𝑝 × 5252 𝑇= = 5.87 𝑓𝑡∙𝑙𝑏 1800 22 Lecture #2 – DC Motors  TYPES OF DC MOTORS < Shunt Motor < Long Shunt Motor < Series Motor < Short Shunt Motor 23 Lecture #2 – DC Motors  TYPES - SHUNT MOTOR Field is independent of motor load Excellent speed regulation (Self-speed regulation) Applications: lathe machines, fans, blowers, pumps, etc. Shunt Motor Torque vs Speed Curve 24 Lecture #2 – DC Motors  TYPES - SERIES MOTOR Because of series connection, the field strength increases with load, and without any load the ‘run-away’ occurs ∴ very poor speed regulation, but very high starting torque Series Motor Torque vs Speed Curve 25 Lecture #2 – DC Motors  TYPES - COMPOUND MOTORS The compound configuration uses both series and shunt windings This configuration gives better speed regulations than a series motor, while still maintaining a high starting torque Cumulative Compound Differential Compound 26 Lecture #2 – DC Motors  DIRECTION OF ROTATION Determined by the relationship between armature magnetic field polarity to pole pieces and magnetic field polarity Series Motor Direction Control Shunt Motor Direction Control 27 Lecture #2 – DC Motors  SHUNT MOTOR CALCULATIONS 𝑽𝒎 = 𝑬𝒃 + 𝑹𝑨 𝑰𝑨 Vm: motor input voltage (armature voltage (V)) Eb: induced or counter EMF (V) RA: motor resistance (Ω) IA: armature current (A) 𝑺 = 𝒌 𝑬𝒃 S: Speed of motor (RPM) Schematic Diagram k : constant of motor Eb: induced or counter EMF (V) 28 Lecture #2 – DC Motors  EXAMPLE - CEMF CALCULATION The armature of 125 VDC motor draws 10 A when operating at full load and has a resistance of 3 Ω. Determine the counter EMF produced by the armature when operating at full load. 𝑽𝒎 = 𝑬𝒃 + 𝑹𝑨 𝑰𝑨 CEMF = Eb = VM - RAIA = 125 V - 3Ω x 10 A CEMF = 95 V 29 Lecture #2 – DC Motors  EXAMPLE - INPUT VOLTAGE CALCULATION A shunt motor at its rated conditions develops 1,000 W at 1800 RPM. Its input voltage (Vm) is 125 VDC and its input current (Im) is 10.67 A. The motor has a shunt field resistance (RF) of 110 Ω and armature resistance (RA) of 1.233Ω. Determine new counter EMF if the IA increases by 20%. 𝑽𝒎 = 𝑬𝒃 + 𝑹𝑨 𝑰𝑨 IF = Vm/RF = 125/110 = 1.14 A IA1 = Im – IF = 10.67 – 1.14 = 9.53 A IA2 = IA1 x 1.2 = 11.44 A Eb = Vm – RAIA2 = 125 – 1.233*11.44 = 110.9 V 30 Lecture #2 – DC Motors  ARMATURE POWER DEVELOPED 𝑷𝒅 = 𝑰𝑨 𝑬𝒃 Pd: Armature power developed/output power (W) IA: armature current (A) Eb: induced or counter EMF (V) 𝑪𝒐𝒑𝒑𝒆𝒓 𝑳𝒐𝒔𝒔 = 𝑰𝟐 𝑨 𝑹𝑨 IA: armature current (A) RA: motor resistance (Ω) 𝑺𝑵𝑳 − 𝑺𝑭𝑳 𝑺𝒑𝒆𝒆𝒅 𝑹𝒆𝒈𝒖𝒍𝒂𝒕𝒊𝒐𝒏 % = Schematic Diagram 𝑺𝑭𝑳 SNL: no-load speed SFL: no-load speed 31 Lecture #2 – DC Motors  EXAMPLE – PD AND COPPER LOSS A shunt motor runs at full load, 1800 RPM and draws 10.67 A at line voltage of 125 VDC. The motor has a shunt field resistance (RF) of 110 Ω and armature resistance (RA) of 2.1 Ω. Determine the armature developed power and copper loss. IF = VM/RF = 125/110 = 1.14 A IA = IM – IF = 10.67 – 1.14 = 9.53 A Eb = VM – RAIA = 125 – 2.1 x 9.53 = 104.98 V Pd = Eb x IA = 104.89 x 9.53 = 1000.46 W Copper Loss = RA x IA2 = 2.1 x (9.53)2 = 181.6 W 32 Lecture #2 – DC Motors  EXAMPLE – SPEED REGULATION A DC shunt motor is running with a measured no-load speed of 1800 RPM. When full load is applied, the speed drops to 1750 RPM. Find the percentage peed regulation. 𝑺𝑵𝑳 − 𝑺𝑭𝑳 𝑺𝒑𝒆𝒆𝒅 𝑹𝒆𝒈𝒖𝒍𝒂𝒕𝒊𝒐𝒏 % = × 𝟏𝟎𝟎% 𝑺𝑭𝑳 SP (%) = (1800 RPM – 1750 RPM) / 1750 RPM x 100 = 2.86 % 33 Lecture #2 – DC Motors