3_PHASE Solved Problems

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  • 8/3/2019 3_PHASE Solved Problems

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    Solved Problems

    Example 1. 3 identical coils, each of resistance 20 and inductance 0.5H areconnected in star to a 3phase 400V, 50Hz supply. Determine the line current and thepower consumed.Data given:

    R=20, L=0.5H, VL=400V, f=50Hz.

    Vph =3

    VL = .V94.2303

    400

    .45.127126.046.14003

    126.0cos

    cos3

    )(sin46.134.158

    94.23034.15807.15720

    07.1575.05022tan

    22

    22

    xxxP

    lagZ

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    IVPconsumedpower

    connectedstarceAII

    Z

    xxfLXcereacinductive

    XRZimpadence

    Z

    VI

    LL

    PhL

    L

    L

    ph

    ph

    Example 2. Three impedances each of (15+j20) are connected in mesh across a 3-phase 400V A.C supply. Determine the phase current, line current, active power andreactive power drawn from the supply.

    Data Given Zph=(15+j20), VL=400V=Vph (deltaconnection) and IL= 3 IPh

    Iph = ph

    ph

    Z

    V

    0

    0122

    13.5325

    13.5315

    20tan252015 andZph

    .88.15352)13.53sin(7.274003sin3Re

    .7.11514)13.53cos(7.274003cos3

    7.271633

    1625

    400

    VARxxxIVQractivepowe

    WxxxIVpowerPActiv e

    AxII

    AZ

    VI

    LL

    LL

    PhL

    ph

    ph

    ph

  • 8/3/2019 3_PHASE Solved Problems

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    Example 3 A balanced 3phase star connected load of 150KW takes a leading currentof 100Amps with a line voltage of 1100V. Find the circuit constants of the load perphase.

    Data given:P=150KW, IL=100A=IPh(star connection) VL=1100V, f=50Hz.

    Given that current is leading, therefore circuit constants are resistance andcapacitance.

    .10.8968.3502

    1

    2

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    953.478.035.6costan

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    Example 4.Two wattcmeters are connected to measure power input to a 3phasebalanced circuit indicate 8KW and 0.8KW; the later reading being obtained afterreversing the current coil connection. Find i) power factor of the load. ii) Active poweriii) reactive power iv) apparent power

    Data given: W1=8KW, W2= -0.8KW (because the reading is obtained after reversing the

    current coil connection)

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    2.7

    cos

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    21

    21

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    21

    211

    KVAKW

    PowerApparent

    KVARWWQpoweractive

    KWWWPpowerActive

    WW

    WW

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    Example 5 Each of two wattmeters connected to measure the input to a 3phasesystem reads 10KW on a balanced load when the power factor is unity. What doeseach instrument reads when the power factor falls to i) 0.866 lagging ii) 0.5 leading thetotal 3phase power remaining unchanged.Data given: W1=W2=10KW, when pf is unity, i.e cos 0:1

    Therefore total power =W1+W2=20KW (1)

    i) When pf=0.866 lagging

    ii) When pf=0.5 leadingKWandWKWWandSolving

    KWWWor

    WWei

    WW

    WW

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    )2...(..........66.6

    20

    )(3577.0.

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    )(3

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    30866.0cos

    21

    21

    21

    21

    21

    01

    KWandWKWWandSolving

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    WW

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    )(3732.1.

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    21

    12

    21

    21

    21

    01