Power systems miscellaneous


  1. If r is radius of the conductor and R is radius of the sheath of strength, then









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    Gradient at a distance x,

    g =
    V
    xIn(R/r)

    ∴ gmin =
    V
    , and
    RInR
    r

    gmax =
    V
    rInR
    r

    and plot of the gradient at the surface of the conductor and the ratio (r/R) is shown in the figure and it can be seen that
    r
    <
    1
    < 0.368
    Re

    Correct Option: D


    Gradient at a distance x,

    g =
    V
    xIn(R/r)

    ∴ gmin =
    V
    , and
    RInR
    r

    gmax =
    V
    rInR
    r

    and plot of the gradient at the surface of the conductor and the ratio (r/R) is shown in the figure and it can be seen that
    r
    <
    1
    < 0.368
    Re


  1. If δ is the loss angle of the cable, then its power factor is









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    Figure depicts the equivalent circuit for a cable represented by leak age resistance R and capacitance C.
    From phasor diagram,
    power factor angle, φ = 90° – δ
    or cos φ = cos (90° – δ) = sin δ

    Correct Option: A


    Figure depicts the equivalent circuit for a cable represented by leak age resistance R and capacitance C.
    From phasor diagram,
    power factor angle, φ = 90° – δ
    or cos φ = cos (90° – δ) = sin δ



  1. Three insulating materials with same maximum working stress and permittivities 2·5, 3·0, 4·0, are used in a single core cable, then location of the materials with respect to the core of the cable will be









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    When all the three materials are subjected to the same maximum stress,

    gmax =
    λ
    =
    λ
    =
    λ
    2πε1r2πε2r12πε3r2

    or ε1r = ε2r1 = ε3r2
    Since r < r1 < r2,
    Therefore ε1 > ε2 > ε3
    Thus, the dielectric material with highest permittivity should be placed near the conductor and other layer in the descending order.

    Correct Option: C

    When all the three materials are subjected to the same maximum stress,

    gmax =
    λ
    =
    λ
    =
    λ
    2πε1r2πε2r12πε3r2

    or ε1r = ε2r1 = ε3r2
    Since r < r1 < r2,
    Therefore ε1 > ε2 > ε3
    Thus, the dielectric material with highest permittivity should be placed near the conductor and other layer in the descending order.


  1. For the same voltage boost, the reactive power capacity is









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    Let Q'c be t he reactive power of the shunt capacitor, Er receiving end voltage, X is the line reactance.
    Then voltage drop across it

    =
    Qc
    . X
    Er

    where, Qc be the reactive power of the series capacitor, then voltage drop across it
    sin φv =
    Qc
    ,
    I

    where φv is the power factor angle.
    For equal voltage boost,
    Q'c
    X =
    Qcsin φv
    ErI

    or
    Q'c
    =
    sin φv
    =
    0.6
    = 6
    QcIX/Er0.1

    Q'c > Qc

    Correct Option: A

    Let Q'c be t he reactive power of the shunt capacitor, Er receiving end voltage, X is the line reactance.
    Then voltage drop across it

    =
    Qc
    . X
    Er

    where, Qc be the reactive power of the series capacitor, then voltage drop across it
    sin φv =
    Qc
    ,
    I

    where φv is the power factor angle.
    For equal voltage boost,
    Q'c
    X =
    Qcsin φv
    ErI

    or
    Q'c
    =
    sin φv
    =
    0.6
    = 6
    QcIX/Er0.1

    Q'c > Qc



  1. If corona loss on a particular system at 50 Hz is 1 kW/phase per km, then corona loss on the same system with supply frequency 25 Hz will be









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    Corona loss

    (P) = 241 × 10–5
    (ƒ + 25)
    r
    (Vp - V0)²,
    δd

    where, ƒ = supply frequency
    nOW P ∝ ƒ + 25
    P1
    =
    ƒ1 + 25
    P2ƒ2 + 25

    or P2 =
    ƒ2 + 25
    ƒ1 + 25

    or P1 =
    50
    × 1kW = /b>0.667kW/phase/km
    75

    Correct Option: C

    Corona loss

    (P) = 241 × 10–5
    (ƒ + 25)
    r
    (Vp - V0)²,
    δd

    where, ƒ = supply frequency
    nOW P ∝ ƒ + 25
    P1
    =
    ƒ1 + 25
    P2ƒ2 + 25

    or P2 =
    ƒ2 + 25
    ƒ1 + 25

    or P1 =
    50
    × 1kW = /b>0.667kW/phase/km
    75