U3AOS3 Topic 1: Motors

Introduction

Electric motors are devices that convert electrical energy into mechanical energy through the interaction of magnetic fields. They are pivotal in countless applications, from household appliances and industrial machinery to electric vehicles. Understanding how motors work involves exploring principles of electromagnetism, particularly the interactions between magnetic fields and electric currents.

Basic Principle

Electric motors operate based on the principle of electromagnetism discovered by Michael Faraday and Joseph Henry. The core principle is that a current-carrying conductor placed within a magnetic field experiences a force. This force is described by the Lorentz force law, which states that the force FF on a current-carrying wire in a magnetic field is given by:

F=ILBsin⁡θF = I L B \sin \theta

where:

  • II is the current through the wire,
  • LL is the length of the wire in the field,
  • BB is the magnetic field strength,
  • θ\theta is the angle between the current and the magnetic field (typically θ=90∘\theta = 90^\circ for maximum force).

Structure of a Motor

  1. Stator:

    • The stator is the stationary part of the motor and typically consists of permanent magnets or electromagnets (field coils) that produce a magnetic field.

  2. Rotor:

    • The rotor is the rotating part of the motor. It is usually a coil of wire or a series of coils mounted on a shaft. When current flows through these coils, they generate their own magnetic fields that interact with the stator's field.

  3. Commutator and Brushes:

    • In a direct current (DC) motor, a commutator is used to reverse the direction of the current through the rotor coils every half-turn, ensuring continuous rotation. Brushes are conductive materials that maintain electrical contact with the rotating commutator.

Operation

  1. Magnetic Interaction:

    • When electric current flows through the rotor coils, it generates a magnetic field. The interaction between this field and the stator's magnetic field produces a force on the rotor due to the Lorentz force law. This force creates a torque that causes the rotor to spin.

  2. Commutation:

    • In a DC motor, as the rotor turns, the commutator switches the direction of the current in the rotor coils, maintaining a consistent torque direction and enabling continuous rotation.

  3. AC Motors:

    • In alternating current (AC) motors, the alternating nature of the current in the stator coils generates a rotating magnetic field that induces a current in the rotor. This induces a magnetic field in the rotor that interacts with the stator's field, producing torque and causing rotation. AC motors do not require a commutator, as the alternating nature of the current naturally reverses the direction of the magnetic field.

Applications

Electric motors are used in a wide range of applications due to their efficiency and versatility:

  1. Household Appliances:

    • Motors power appliances such as washing machines, refrigerators, and fans.

  2. Industrial Equipment:

    • They drive conveyors, pumps, and other machinery in manufacturing and processing industries.

  3. Transportation:

    • Motors are crucial in electric vehicles, trains, and aircraft.

  4. Power Tools:

    • They are used in drills, saws, and other tools for various tasks.

Motors convert electrical energy into mechanical energy. This is achieved as the current travels perpendicular to the field lines which induces a force.

 

There are 2 types of motors

 

DC motors: use a split ring commutator, which allows the current to change direction every half period to keep the motor turning in the same direction.

Created with GeoGebra®, Tom Walsh, Link


AC motors: use slip rings. Which is a continuous ring that does not change the direction of the current.

Example 1
A DC motor has a rotor with a radius of 0.1 m0.1\ \text{m} and carries a current of 2 A2\ \text{A} in a magnetic field of 0.3 T0.3\ \text{T}. If the motor’s armature (rotor) has a length of 0.4 m0.4\ \text{m} and the angle between the current direction and magnetic field is 90∘90^\circ, calculate the torque produced by the motor.

The torque τ\tau produced by a DC motor is given by:

τ=ILBr\tau = I L B r

where:

  • II is the current (2 A2\ \text{A}),
  • LL is the length of the armature (0.4 m0.4\ \text{m}),
  • BB is the magnetic field strength (0.3 T0.3\ \text{T}),
  • rr is the radius of the rotor (0.1 m0.1\ \text{m}).

Substitute the given values:

τ=2×0.4×0.3×0.1\tau = 2 \times 0.4 \times 0.3 \times 0.1

τ=0.0024 N⋅m\tau = 0.0024\ \text{N}\cdot\text{m}

So, the torque produced by the motor is 0.0024 N⋅m0.0024\ \text{N}\cdot\text{m}.

Example 2
An electric motor operates with an input power of 500 W500\ \text{W} and has an efficiency of 80%80\%. Calculate the mechanical power output of the motor.

The efficiency η\eta of an electric motor is given by:

η=PoutputPinput×100%\eta = \frac{P_{\text{output}}}{P_{\text{input}}} \times 100\%

where:

  • η\eta is the efficiency (80%80\%),
  • PinputP_{\text{input}} is the input power (500 W500\ \text{W}),
  • PoutputP_{\text{output}} is the mechanical power output.

Rearrange the formula to solve for PoutputP_{\text{output}}:

Poutput=η×PinputP_{\text{output}} = \eta \times P_{\text{input}}

Substitute the values:

Poutput=80100×500P_{\text{output}} = \frac{80}{100} \times 500

Poutput=0.8×500P_{\text{output}} = 0.8 \times 500

Poutput=400 WP_{\text{output}} = 400\ \text{W}

So, the mechanical power output of the motor is 400 W400\ \text{W}.

Example 3
A DC motor has an armature with a radius of 0.05 m0.05\ \text{m} and operates with a torque of 0.1 N⋅m0.1\ \text{N}\cdot\text{m}. If the motor is designed for a load that requires a power output of 50 W50\ \text{W}, calculate the speed of the motor in revolutions per minute (rpm).

The power output PP of a motor is related to torque τ\tau and angular speed ω\omega by:

P=τωP = \tau \omega

To find the angular speed ω\omega:

ω=Pτ\omega = \frac{P}{\tau}

The angular speed ω\omega in radians per second can be converted to revolutions per minute (rpm) using the conversion factor 2π2 \pi radians per revolution and 60 seconds per minute:

rpm=ω×602π\text{rpm} = \frac{\omega \times 60}{2 \pi}

Substitute the given values:

ω=500.1\omega = \frac{50}{0.1}

ω=500 rad/s\omega = 500\ \text{rad/s}

Convert to rpm:

rpm=500×602π\text{rpm} = \frac{500 \times 60}{2 \pi}

rpm≈300006.28\text{rpm} \approx \frac{30000}{6.28}

rpm≈4774 rpm\text{rpm} \approx 4774\ \text{rpm}

So, the speed of the motor is approximately 4774 rpm4774\ \text{rpm}.

Exercise &&1&& (&&1&& Question)

A DC motor draws a current of 4 A4\ \text{A} and operates with an input voltage of 24 V24\ \text{V}. If the resistance of the motor windings is 2 Ω2\ \Omega, calculate the back electromotive force (EMF) of the motor.

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Exercise &&2&& (&&1&& Question)

A motor has a mechanical power output of 150 W150\ \text{W} and electrical power input of 200 W200\ \text{W}. Calculate the efficiency of the motor.

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