motor starting analysis

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  • 7/28/2019 Motor Starting Analysis

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    MOTOR STARTING ANALYSIS

    A. INTRODUCTIONMotor starting study should be done before purchasing power transformer andcables. In the course of the analysis, the motor starting current and method of

    starting will be decided. Motor terminal voltage as well as torque requirementshould be analyzed at the same time.

    A complete motor starting analysis consists of voltage drop and accelerationcalculations. In the initial study, typical motor data shall be used. These data shall

    be updated when vendor data are made available.

    B. WHEN TO CONDUCT A MOTOR STARTING ANALYSIS1) Transformer self-cooled KVA rating is less than three times the largest

    motor KVA rating.2) Motor feeder reduces the short-circuit KVA to less than eight times the

    motor-starting KVA.3) Ratio of bus short-circuit KVA to motor-starting KVA is less than 8 times.4) Voltage drop restriction imposed by the Utility (usually limited at 2%).5) Presence of high inertia load.

    C. CORRECTING FOR EXCESSIVE VOLTAGE DROP1) Use a motor with lower starting current.2) Connect to a system with a higher short-circuit MVA.3) Use a power transformer with lower impedance.4) Use a reduced voltage starter or soft starter.5) Install a capacitor bank.

    D. FORMULAE

    1) old L

    L

    old

    newnew

    Z V

    V

    POWERBASE

    POWERBASE Z

    new

    old

    10

    2

    10 =

    2) 3= MOTOR LR LR V I VA

    3) X L of Running Load, L

    BASE L

    VA

    VA X =10

    4) Torque is directly proportional to the square of voltage.

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    E. SAMPLE PROBLEM NO. 1

    Given: In Fig. 1, determine the voltages at the utility and motor terminal at theinstant when the motor is started.

    FIG. 1: DIAGRAM FOR CAPTIVE LOAD

    Solution:

    At 400MVA power base; 13.8KV, and 2.4KV Voltage bases;

    ( )

    846.494.2

    3.2

    323001850

    10400

    33.17065.05.1

    400

    1

    26

    10

    10

    10

    =

    =

    ==

    =

    M

    T

    U

    X

    X

    X

    01466.0846.4933.171

    1

    10

    10

    10 =

    ++==

    X V

    I TOTALST

    ( ) ( )01446.01110101010 == ST U TOTALTU I X V V = 0.9853

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    Voltage level at the Utility = 0.9853(13.8) = 13.597KV

    ( ) ( ) 73128.033.17101446.011010101010 =+=+= T U ST TOTALTM X X I V V

    Voltage level at the Motor Terminal = 0.73128(2.4) = 1.755KV

    F. EXAMPLE NO. 2

    For system with running load, the diagram is as shown below

    The per unit impedance for the running load is shown in formula no. 3.