Chapter 15: Induction Motor Equivalent Circuit frick/EE4220-EM_Dynamics/lecture1¢ ...

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Transcript of Chapter 15: Induction Motor Equivalent Circuit frick/EE4220-EM_Dynamics/lecture1¢ ...

  • 10/29/2003 Electromechanical Dynamics 1

    Chapter 15: Induction Motor Equivalent Circuit

  • 10/29/2003 Electromechanical Dynamics 2

    Introduction

    • To gain better understanding of the properties and behaviors of the induction motor, it is necessary to examine the equivalent circuit – develop an equivalent circuit from the basic principles

  • 10/29/2003 Electromechanical Dynamics 3

    Wound-rotor Induction Motor • A 3-phase wound-rotor induction

    motor is very similar in construction to a 3-phase transformer – the motor has 3 identical primary

    windings, 3 identical secondary windings, and perfect symmetry

    • single phase analysis is possible

    – at standstill, the motor acts exactly like a conventional transformer

    • model begins with the transformer’s equivalent circuit

    – circuit parameters • Rm = iron, windage, and friction losses

    • Xm = magnetizing reactance

    • r1 = stator winding resistance

    • x1 = stator leakage reactance

    • r2 = rotor winding resistance

    • x2 = rotor leakage reactance

  • 10/29/2003 Electromechanical Dynamics 4

    Wound-rotor Induction Motor

    – the magnetizing branch can not be neglected

    • I0 may be 40% of Ip due to the air gap

    • for motors greater than 2 hp, the magnetization branch can be shifted to the input terminals

    – for a motor running at a slip s, the rotor speed is ns(1-s)

    • this will modify the values of E1, I1, E2 and I2

    • the frequency in the secondary winding will become sf

  • 10/29/2003 Electromechanical Dynamics 5

    Wound-rotor Induction Motor – the induced rotor voltage

    – the rotor leakage reactance

    – rotor resistance is not frequency dependent

    – the model of the rotor circuit is

    • the effective value of I1 is equal to the effective value of I2 even though their frequencies are different

    • the effective value of E1 is equal to the effective value of E2 divided by the slip

    • the phase angle between E2 and I2 is the same as the angle between E1 and I1

    12 EsE =

    22 jxsxj =′

    xRrR += 22

    22

    1 2 jsxR

    Es I

    + =

  • 10/29/2003 Electromechanical Dynamics 6

    Wound-rotor Induction Motor

    – on the primary side

    – therefore

    • Without affecting the value of the stator current, I1, the impact of the slip is shifted from the rotor voltage and reactance to the rotor resistance – greatly simplifies the analysis

    22

    1 21 jsxR

    Es II

    + ==

    2

    1

    22

    1 1 Z

    E

    jxsR

    E I =

    + =

  • 10/29/2003 Electromechanical Dynamics 7

    Power Relationships • The equivalent circuit enables the

    deduction of basic electromechanical power relationships

    • By inspection – active power absorbed

    – reactive power absorbed

    – apparent power absorbed

    – power factor

    – line current

    – active power supplied to the rotor

    – I2R losses dissipated in the rotor

    – developed mechanical power

    – developed torque

    – efficiency

    sRIrIRVP mT 2 2 21

    2 1

    2 ++=

    2 2 21

    2 1

    2 xIxIXVQ mT ++=

    22 QPS +=

    SP=θcos

    sRIPr 2 2 2=

    rjr PsRIP == 2 2 1

    ( )sPPPP rjrrm −=−= 1

    Tp VSI =

    PPm=η s

    rm

    n

    P

    n

    P T

    55.955.9 ==

  • 10/29/2003 Electromechanical Dynamics 8

    Breakdown Torque & Speed

    • The developed torque is given by the power delivered to the rotor and the synchronous speed

    – the power delivered is a function of the resistance R2/s

    • Maximum torque occurs at maximum power transfer – maximum power transfer is

    found when R2/s is equal to the absolute value of Z1

    – the voltage drops are equal

    • The phasor diagram of this special case:

    from geometry: – AB and BC have the same length

    – the angle ABC = (180°-α) – angle CAB = angle ACB = α/2 – hence

    12 ZsR =

    sr nPT 55.9=

    1121 ZIsRI = 22 cos

    2 cos 11

    2 1

    TVZI s

    R I == αα

  • 10/29/2003 Electromechanical Dynamics 9

    Breakdown Torque & Speed

    – the slip at breakdown torque is

    – the current at the breakdown torque is

    – the breakdown torque is

    • note that the magnitudes of both the breakdown torque and breakdown current are independent of the rotor current resistance R2

    12 ZRsb =

    ( )211 cos2 αZVI Tb =

    ( ) ( )221

    2

    cos4

    55.9 αZn

    V T

    s

    T b = ( )1arctan rx=α

  • 10/29/2003 Electromechanical Dynamics 10

    Breakdown Torque Calculation

    • Example – determine the maximum torque and corresponding speed of a

    440 V, 1800 rpm, 60 Hz, 5-hp, squirrel cage induction motor • r1 = 1.5 ohms, r2 = 1.2 ohms,jx = 6 ohms,jXm = 110 ohms,

    Rm = 900 ohms

  • 10/29/2003 Electromechanical Dynamics 11

    Torque-speed Curve

    • The complete torque-speed curve of a motor can be determined by selecting various values of slip and solving the circuit equations

    • Example – determine the torque and speed of a 440 V, 1800 rpm, 60 Hz,

    5-hp, squirrel cage induction motor • r1 = 1.5 ohms, r2 = 1.2 ohms,jx = 6 ohms,jXm = 110 ohms,

    Rm = 900 ohms

  • 10/29/2003 Electromechanical Dynamics 12

    Torque-speed Curve

  • 10/29/2003 Electromechanical Dynamics 13

    Test to Determine the Equivalent Circuit

    • No-load and locked-rotor tests provide approximate values of r1, r2, Xm, Rm, and x

    • No-load test – small slip leads to high value of R2/s with I1 being negligible compared to I0 – no-load circuit essentially consist of Xm and Rm

    • measure the stator resistance RLL, and assume a wye connection

    • run the motor at no-load and at rated line voltage, measuring the no-load current and the total three-phase active power

    ( ) NLNLmvfNLm NLNLvf

    NLNLNLNLNLNL

    QEXPPER

    rIPPP

    PSQIVS

    22

    1 2

    22

    3

    3

    =+=

    −=+

    −==

    21 LLRr =

  • 10/29/2003 Electromechanical Dynamics 14

    Test to Determine the Equivalent Circuit

    • Locked-rotor test – under locked-rotor conditions

    • stator current is almost 6 pu, slip is unity, and R2/s equals r2/1 = r2 • magnetization branch can be neglected

    – locked-rotor circuit consist of x, r1, and r2 • apply a reduced 3-phase voltage

    such that the stator current is about equal to its rated value

    • measure the line voltage, line current, and 3-phase power

    ( ) ( ) 122

    2

    22

    3

    3

    3

    rIPr

    IQX

    PSQIVS

    LRLR

    LRLRm

    LRLRLRLRLRLR

    −=

    =

    −==

  • 10/29/2003 Electromechanical Dynamics 15

    Test to Determine the Equivalent Circuit

    • Example – a no-load test conducted on a 30 hp, 835 rpm, 440 V, 3-phase, 60 Hz,

    squirrel-cage induction motor yields the following results

    • no-load line-to-line voltage: 440 V

    • no-load current: 14 A

    • no-load power: 1470 W

    • resistance measured between two terminals: 0.5 Ω – the results of locked-rotor test, conducted at reduced voltage

    • locked-rotor line-to-line voltage: 163 V

    • locked-rotor power: 7200 W

    • locked-rotor current: 60 A

    – determine the equivalent circuit of the motor

  • 10/29/2003 Electromechanical Dynamics 16

    Induction Motor Equivalent Circuit

    • Homework – Problems: 15-2, 15-4, 15-5, 15-7