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    Induced Voltage & Circulating Current PDF

    Uploaded by

    Inayat Hathiari
    This document discusses the calculation of induced sheath voltage and circulating sheath current in underground power cables. It provides the following key points: 1. The cable is divided into minor sections and formulas are given for calculating the induced voltage in each section based on factors like sheath resistance, reactance, and cable current. 2. A formula is presented for determining the circulating sheath current between the phases using the induced voltages and sheath impedance of each section. 3. Worked examples are shown for a specific 132kV cable with different unbalanced lengths between the minor sections, providing the resulting induced voltages and circulating currents.

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    Induced Voltage & Circulating Current PDF

    Uploaded by

    Inayat Hathiari
    176 views10 pages
    This document discusses the calculation of induced sheath voltage and circulating sheath current in underground power cables. It provides the following key points: 1. The cable is divided into minor sections and formulas are given for calculating the induced voltage in each section based on factors like sheath resistance, reactance, and cable current. 2. A formula is presented for determining the circulating sheath current between the phases using the induced voltages and sheath impedance of each section. 3. Worked examples are shown for a specific 132kV cable with different unbalanced lengths between the minor sections, providing the resulting induced voltages and circulating currents.

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    Induced_Voltage_&_Circulating_Current.pdf

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    This document discusses the calculation of induced sheath voltage and circulating sheath current in underground power cables. It provides the following key points: 1. The cable is divided into minor sections and formulas are given for calculating the induced voltage in each section based on factors like sheath resistance, reactance, and cable current. 2. A formula is presented for determining the circulating sheath current between the phases using the induced voltages and sheath impedance of each section. 3. Worked examples are shown for a specific 132kV cable with different unbalanced lengths between the minor sections, providing the resulting induced voltages and circulating currents.

    Copyright:

    © All Rights Reserved
    Download as PDF, TXT or read online from Scribd
    Download as pdf or txt
    176 views10 pages

    Induced Voltage & Circulating Current PDF

    Uploaded by

    Inayat Hathiari
    This document discusses the calculation of induced sheath voltage and circulating sheath current in underground power cables. It provides the following key points: 1. The cable is divided into minor sections and formulas are given for calculating the induced voltage in each section based on factors like sheath resistance, reactance, and cable current. 2. A formula is presented for determining the circulating sheath current between the phases using the induced voltages and sheath impedance of each section. 3. Worked examples are shown for a specific 132kV cable with different unbalanced lengths between the minor sections, providing the resulting induced voltages and circulating currents.

    Copyright:

    © All Rights Reserved
    Download as PDF, TXT or read online from Scribd
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    of 10

    Attachment #1

    INDUCED SHEATH VOLTAGE AND CIRCULATING SHEATH CURRENT


    I. General Condition
    1. Sheath Bondig Method : Cross Bonding
    2. Diagram
    EA1

    Ia

    EA2

    Rs1+jXs1

    ai

    i
    EB1

    Ib

    Rs2+jXs2

    EA3
    Rs3+jXs3

    EB2

    EB3

    EC1

    EC2

    EC3

    a2i
    Ic

    Where,
    EA1, EA2, EA3, EB1, EB2, EB3, EC1, EC2, EC3 : Induced Voltage of Each Minor Section
    Rs1, Rs2, Rs3, Xs1, Xs2, Xs3 : Sheath Resistance & Reactance of Each Minor Section
    Ia, Ib, Ic
    i

    m, n

    : Current of Each Phase


    : Circulating Sheath Current
    : Length of the 1st, 2nd & 3rd Minor Section

    II. Calculation Formulae


    Ia = I , Ib = a2I ,

    Ic = aI

    Where, a = -1/2 + j3/2 ,

    a2 = -1/2 - j3/2

    1. Induced Voltage of Each Minor Section

    1st Minor Section ()


    EA1 = [jXm 1I -(Rs1+jXs1)i] ,

    EB1 = [jXm 1a2I -(Rs1+jXs1)a2i] ,

    EC1 = [jXm 1aI -(Rs1+jXs1)ai]

    2nd Minor Section ( m )


    EA2 = [jXm 2I -(Rs2+jXs2)ai]m ,

    EB2 = [jXm 2a2I -(Rs2+jXs2)i]m ,

    EC2 = [jXm 2aI -(Rs2+jXs2)a2i]m

    3rd Minor Section ( n )


    EA3 = [jXm 3I -(Rs3+jXs3)a2i]n ,

    EB3 = [jXm 3a2I -(Rs3+jXs3)ai]n ,

    EC3 = [jXm 3aI -(Rs3+jXs3)i]n


    Where, Xm 1, Xm 2, Xm 3 : Mutual Reactance between the Sheath of a Cable and the
    Conductors of the other cable

    2. Circulating Sheath Current ( i )


    EA1 + EB2 + EC3 = 0 ,

    EB1 + EC2 + EA3 = 0 ,

    EC1 + EA2 + EB3 = 0

    1) Circulating Sheath Current ( i )


    EA1 + EB2 + EC3 = 0
    = [jXm 1I -(Rs1+jXs1)i] +

    [jXm 2a I -(Rs2+jXs2)i]m

    [jXm 3aI -(Rs3+jXs3)i]n

    jI(Xm 1 + Xm 2a m + Xm 3an)
    i

    =
    (Rs1 +jXs1) + (Rs2 +jXs2)m + (Rs3 +jXs3)n
    jI(X1 + X2a2m + X3an)

    =
    (Rs1 +jX1) + (Rs2 +jX2)m + (Rs3 +jX3)n
    Where, Xm 1 Xs1 X1 ,

    Xm 2 Xs2 X2 , Xm 3 Xs3 X3

    2) E
    jI(X1 + X2a2m + X3an)
    i

    =
    (Rs1 +jX1) + (Rs2 +jX2)m + (Rs3 +jX3)n

    = jI(X1 + X2a2m + X3an)

    Zs

    = (Rs1 +jX1) + (Rs2 +jX2)m + (Rs3 +jX3)n

    Zs

    3) Induced Voltage of Each Minor Section ( i = E / Zs )


    EA1 = [jX1I -(R1+jX1)i]

    [jX1I -(R1+jX1)E/Zs]

    EB2 = [jX2a I -(R2+jX2)i]m

    [jX2a2I -(R2+jX2)E/Zs]m

    EC3 = [jX3aI -(R3+jX3)i]n

    [jX3aI -(R3+jX3)E/Zs]n

    Where, Rs1 = R1 ,

    Rs3 = R3

    Rs2 = R2 ,

    3. Calculation Formulae of Induced Voltage


    1) 0 X
    E()
    E() =

    EA1 =
    3XsI
    L

    EB1 =

    EB1

    m 2 + mn + n2

    (V)

    Where, X1 X2 X3 Xs, R1 = R2 = R3 = Rs
    L = + m + n

    EA1 : Absolute Value of EA1


    2) X ( + m )
    E(+ m )

    EA1 + EB2

    3XsI

    E(+ m ) =

    EB1 + EC2

    EC1 + EA2

    2 + m + m 2 n

    3) ( + m ) X ( + m + n )
    E(+ m + n ) = 0

    (V)

    4. Calculation Formulae of Circulating Sheath Current


    2
    jI(X1 + X2a m + X3an)

    =
    (Rs1 +jX1) + (Rs2 +jX2)m + (Rs3 +jX3)n
    j Xs I (+ a2m + an)
    =
    (Rs +jXs)(+ m + n)
    Where, X1 X2 X3 = Xs, Rs1 Rs2 Rs3 = Rs
    L = + m + n
    Xs I

    i=

    2 + m 2 + n2 -m - mn - n
    (Rs2 + Xs2) L

    i : Absolute Value of i

    III. Calculation for Example ( The 1st Cross Bonding Section )


    1. Calculation Condition
    1) Cable Type : 132kV 1C*630SQmm XLPE Cable
    2) Current Carrying Capacity ( I ) :

    1 cc't :

    592

    [A]

    2 cc'ts :

    533

    [A]

    3) Mean Radius of Metallic Sheath ( rs ) : 35.45 [mm]


    4) Cable Laying Method : Direct Burial (Trefoil Formation)
    5) Axial Cable Distance of Phase :

    91

    [mm]

    6) Axial Distance between each Circuits : 1200 [mm]

    (V)

    7) Frequency :

    50

    [Hz]

    8) Sheath Bonding Method : Cross Bonding

    2. Calculation Procedure
    1) Sheath Reactance ( Xs )
    -7

    [/m]

    Xs = 2Ln( Do / rs )10

    = 5.920E-05 ( for 1 circuit )

    [/m]

    = 5.920E-05 ( for 2 circuits ) [/m]


    Where,

    Do : Geometric Mean Distance

    rs

    : Mean Radius of Metallic Sheath

    = 2f =
    f

    91.0

    ( for 1 circuit )

    [mm]

    91.0

    ( for 2 circuits ) [mm]

    35.45 [mm]

    314.2

    : Frequency

    50

    [Hz]

    2) Induced Sheath Voltage ( E )


    - For Cross Bondig Section

    0X
    3XsI

    E() =

    2
    2
    m + mn + n

    (V)

    X ( + m )
    3XsI

    E(+ m ) =

    2 + m + m 2 n

    (V)

    ( + m ) X ( + m + n )
    E(+ m + n ) = 0
    Where,

    (V)

    (For Example) Unblanced Ratio

    0%

    15%

    30%

    : Length of 1st Minor Section

    2000

    1850

    1700

    [m]

    m : Length of 2nd Minor Section

    2000

    2150

    2300

    [m]

    n : Length of 3rd Minor Section

    2000

    1850

    1700

    [m]

    L = + m + n =

    6000 [m]

    3) Circulating Sheath Current ( i )


    Xs I

    2 + m 2 + n2 -m - mn - n

    i =
    (Rs2 + Xs2) L
    Where,

    Rs : Metallic Sheath Resistance

    1.33E-04 [/m]

    3. Results

    A. Case 1. : 0% Unblanced Section Length

    1) Induced Sheath Voltage ( E )


    Examination

    Length

    Point

    (m)

    132kV S/S

    2000

    J/B#1
    2000

    J/B#2

    2000

    J/B#3

    Induced Sheath Voltage ( E )


    for 1 circuit ( V )

    for 2 circuits ( V )

    70.1

    63.1

    70.1

    63.1

    2) Circulating Sheath Current ( i )


    Xs I

    2 + m 2 + n2 -m - mn - n

    i =
    (Rs2 + Xs2) L
    =

    [A]

    ( for 1 circuit )

    [A]

    ( for 2 circuits )

    B. Case 2. : 15% Unblanced Section Length

    1) Induced Sheath Voltage ( E )


    Examination

    Length

    Point

    (m)

    for 1 circuit ( V )

    for 2 circuits ( V )

    132kV S/S

    1850

    66.6

    59.9

    66.6

    59.9

    J/B#1
    J/B#2
    J/B#3

    2150
    1850

    Induced Sheath Voltage ( E )

    2) Circulating Sheath Current ( i )


    Xs I

    2 + m 2 + n2 -m - mn - n

    i =
    (Rs2 + Xs2) L
    =

    12

    [A]

    ( for 1 circuit )

    10.8

    [A]

    ( for 2 circuits )

    C. Case 3. : 30% Unblanced Section Length

    1) Induced Sheath Voltage ( E )


    Examination

    Length

    Point

    (m)

    for 1 circuit ( V )

    for 2 circuits ( V )

    132kV S/S

    1700

    62.9

    56.7

    62.9

    56.7

    J/B#1

    2300

    J/B#2
    J/B#3

    1700

    Induced Sheath Voltage ( E )

    2) Circulating Sheath Current ( i )


    Xs I

    2 + m 2 + n2 -m - mn - n

    i =
    (Rs2 + Xs2) L
    =

    24.1

    [A]

    ( for 1 circuit )

    21.7

    [A]

    ( for 2 circuits )

    Attachment #1

    nce of Each Minor Section

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