7 Lecture Induction Motors

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    Induction Motors

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    Introduction

    Three-phase induction motors are the mostcommon and frequently encountered machines in

    industry

    - simple design, rugged, low-price, easy maintenance

    - wide range of power ratings: fractional horsepower to

    10 MW

    - run essentially as constant speed from zero to full load

    - speed is power source frequency dependent

    not easy to have variable speed control

    requires a variable-frequency power-electronic drive for

    optimal speed control

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    Construction

    An induction motor has two main parts- a stationary stator

    consisting of a steel frame that supports a hollow, cylindrical core

    core, constructed from stacked laminations (why?), having a

    number of evenly spaced slots, providing the space for the stator

    winding

    Stator of IM

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    Construction

    - a revolving rotor composed of punched laminations, stacked to create a series of rotor

    slots, providing space for the rotor winding

    one of two types of rotor windings

    conventional 3-phase windings made of insulated wire (wound-rotor)

    similar to the winding on the stator aluminum bus bars shorted together at the ends by two aluminum rings,

    forming a squirrel-cage shaped circuit (squirrel-cage)

    Two basic design types depending on the rotor design

    - squirrel-cage

    - wound-rotor

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    Construction

    Squirrel cage rotor

    Wound rotor

    Notice the

    slip rings

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    Construction

    Cutaway in a

    typical wound-

    rotor IM.Notice the

    brushes and the

    slip rings

    Brushes

    Slip rings

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    Rotating Magnetic Field

    Balanced three phase windings, i.e.mechanically displaced 120 degrees form

    each other, fed by balanced three phase

    source

    A rotating magnetic field with constant

    magnitude is produced, rotating with a speed

    Wherefe is the supply frequency and P is the

    no. of poles and nsync is called the

    synchronous speed in rpm (revolutions per

    minute)

    120e

    sync

    fn rpm

    P

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    Rotating Magnetic Field

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    Principle of operation

    This rotating magnetic field cuts the rotor windings andproduces an induced voltage in the rotor windings

    Due to the fact that the rotor windings are short circuited, for

    both squirrel cage and wound-rotor, and induced current

    flows in the rotor windings The rotor current produces another magnetic field

    A torque is produced as a result of the interaction of those

    two magnetic fields

    Where ind is the induced torque andBR andBS are the magnetic

    flux densities of the rotor and the stator respectively

    ind R skB B

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    Induction motor speed

    At what speed will the IM run?- Can the IM run at the synchronous speed, why?

    - If rotor runs at the synchronous speed, which is the

    same speed of the rotating magnetic field, then the rotor

    will appear stationary to the rotating magnetic field andthe rotating magnetic field will not cut the rotor. So, no

    induced current will flow in the rotor and no rotor

    magnetic flux will be produced so no torque is

    generated and the rotor speed will fall below thesynchronous speed

    - When the speed falls, the rotating magnetic field will

    cut the rotor windings and a torque is produced

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    Induction motor speed

    So, the IM will always run at a speed lower thanthe synchronous speed

    The difference between the motor speed and the

    synchronous speed is called the Slip

    Where nslip

    = slip speed

    nsync= speed of the magnetic field

    nm = mechanical shaft speed of the motor

    slip sync mn n n

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    The Slip

    sync m

    sync

    n ns

    n

    Where s is the slip

    Notice that : if the rotor runs at synchronous speed

    s = 0

    if the rotor is stationary

    s = 1

    Slip may be expressed as a percentage by multiplying the above

    eq. by 100, notice that the slip is a ratio and doesnt have units

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    Example 7-1 (pp.387-388)

    A 208-V, 10hp, four pole, 60 Hz, Y-connectedinduction motor has a full-load slip of 5 percent

    1. What is the synchronous speed of this motor?

    2. What is the rotor speed of this motor at rated load?

    3. What is the rotor frequency of this motor at rated load?

    4. What is the shaft torque of this motor at rated load?

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    Solution

    1.

    2.

    3.

    4.

    120 120(60) 18004

    esync

    fn rpmP

    (1 )

    (1 0.05) 1800 1710

    m sn s n

    rpm

    0.05 60 3r ef sf Hz

    260

    10 746 / 41.7 .

    1710 2 (1/ 60)

    out out loadmm

    P P

    n

    hp watt hpN m

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    Problem 7-2 (p.468)

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    Equivalent Circuit

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    Power losses in Induction machines

    Copper losses- Copper loss in the stator (PSCL) =I1

    2R1

    - Copper loss in the rotor (PRCL) =I22R2

    Core loss (Pcore

    )

    Mechanical power loss due to friction and windage

    How this power flow in the motor?

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    Power flow in induction motor

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    Power relations

    3 cos 3 cosin L L ph phP V I V I

    2

    1 13SCLP I R

    ( )AG in SCL coreP P P P 2

    2 23

    RCLP I R

    conv AG RCLP P P

    ( )out conv f w stray

    P P P P

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    Equivalent Circuit

    We can rearrange the equivalent circuit as follows

    Actual rotor

    resistance

    Resistance

    equivalent to

    mechanical load

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    Power relations

    3 cos 3 cosin L L ph phP V I V I

    2

    1 13SCLP I R

    ( )AG in SCL coreP P P P 2

    2 23

    RCLP I R

    conv AG RCLP P P

    ( )out conv f w stray

    P P P P

    conv RCLP P 2 223 RI

    s

    2 2

    2

    (1 )3

    R sI

    s

    RCLPs

    (1 )RCL

    P s

    s

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    Torque, power and Thevenins Theorem

    Thevenins theorem can be used to transform thenetwork to the left of points a and b into an

    equivalent voltage source V1eq in series with

    equivalent impedanceReq+jXeq

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    Torque, power and Thevenins Theorem

    1 1

    1 1( )

    Meq

    M

    jXV VR j X X

    1 1( ) //

    eq eq M R jX R jX jX

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    Torque, power and Thevenins Theorem

    Then the power converted to mechanical (Pconv)

    1 1

    22

    222

    ( )

    eq eq

    T

    eq eq

    V VI

    Z RR X X

    s

    2 2

    2

    (1 )conv

    R sP I

    s

    And the internal mechanical torque (Tconv

    )

    conv

    conv

    m

    PT

    (1 )

    conv

    s

    P

    s

    2 2

    2

    s

    RI

    s

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    Torque, power and Thevenins Theorem2

    1 2

    2

    222

    1

    ( )

    eq

    conv

    s

    eq eq

    V RT

    sRR X X

    s

    2 21

    2

    222

    1

    ( )

    eq

    conv

    s

    eq eq

    RV

    sT

    RR X X

    s

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    Torque-speed characteristics

    Typical torque-speed characteristics of induction motor

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    Maximum torque

    Maximum torque occurs when the powertransferred toR2/s is maximum.

    This condition occurs whenR2/s equals the

    magnitude of the impedanceReq +j (Xeq +X2)

    max

    2 22

    2( )eq eq

    T

    RR X X

    s

    max

    2

    2 2

    2( )

    T

    eq eq

    RsR X X

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    Maximum torque

    The corresponding maximum torque of an inductionmotor equals

    The slip at maximum torque is directly proportional

    to the rotor resistanceR2

    The maximum torque is independent ofR2

    2

    max2 2

    2

    1

    2 ( )

    eq

    s eq eq eq

    V

    TR R X X

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    Maximum torque

    Rotor resistance can be increased by insertingexternal resistance in the rotor of a wound-rotor

    induction motor.

    The value of the maximum torque remains

    unaffected but the speed at which it occurs can be

    controlled.

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    Maximum torque

    Effect of rotor resistance on torque-speed characteristic

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    Problem 7-5 (p.468)

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    Solution to Problem 7-5 (p.468)

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    Problem 7-7 (pp.468-469)

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    Solution to Problem 7-7 (pp.468-469)

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    Solution to Problem 7-7 (pp.468-469)Contd

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    Solution to Problem 7-7 (pp.468-469)Contd

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    Problem 7-19 (p.470)

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    Solution to Problem 7-19 (pp.470)

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    Solution to Problem 7-19 (pp.470)Contd

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    Solution to Problem 7-19 (pp.470)Contd

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    Solution to Problem 7-19 (pp.470)Contd