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    Ref. (4)-G77700-S0019-L059-A

    Project:

    Gilgel Gibe II PROJECTETHIOPIA

    Descri tion:

    400 kV SWITCHYARDDESIGN

    REPORTs & CALCULATIONs

    Sub ect:

    Report on Calculations for Determination ofElectrodynamic Forces on Equipments ConnectedThrough Rigid Conductor and Allowable Span of

    the Rigid Bus Conductor

    Note:

    This report is for calculations to determine the electrodynamic force onequipments connected through rigid bus conductor and also estimation of themaximum possible unsupported span length of the rigid bus conductor basedon the properties of vertical deflection of the conductor and the fiber stresgenerated on the conductor depending on the nature of the end supports.

    BORU BARA MEKANIK HESAPLAR :

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    Calculation for Electrodynamic Force on Equipment & Permis-sible Span length of Rigid Conductor

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    (4)-G77700-S0019-L059-A

    Contents

    page

    1. Introduction ....................................................................................................................... 4

    2. System Data...................................................................................................................... 4

    3. Conductor Data................................................................................................................. 5

    4. Equipment Technical data................................................................................................. 5

    5. Attachments ...................................................................................................................... 5

    6. Conclusion ........................................................................................................................ 6

    7. References........................................................................................................................ 6

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    Calculation for Electrodynamic Force on Equipment & Permis-sible Span length of Rigid Conductor

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    (4)-G77700-S0019-L059-A

    1 Introduction

    In addition to the Bus Post Insulators the rigid bus in out-door substation is also connected tovarious other equipment and supported on them. These equipments are hence subjected to sus-tain all types of stresses on bus conductor owing to electrically and mechanically originatedforces.

    The principal forces on the equipment connected to the rigid conductor are as follows

    a. Short circuit current force on the bus conductor.

    b. Wind force on the bus conductor.

    c. Wind force on the equipment.

    d. Weight of the bus conductor span supported by the equipment.

    The resultant of these forces correlated to the type of End Connection of the rigid conductor atthe equipment terminal gives the value of the net Electromagnetic Force on the equipment. Thevalues of the forces obtained herewith forms valuable input for design of equipment supportstructure and equipment foundation design.

    In this report, along with the force on equipments the maximum permissible unsupported spanlength of the rigid bus conductor connecting the equipments is also determined. The maximumpermissible span is calculated within the limits of Vertical deflection and permissible FibreStress corresponding to the type of End Connection to the supporting equipment terminals ateach end of the span. The calculation is performed on the method stipulated by IEEE 6051987and the minimum of the span length obtained from the calculation is conservatively consideredas the maximum allowable unsupported span for all influencing conditions.

    2 System data

    400kV AC Switchyard

    Nominal System Voltage 400 kV

    System Frequency 50 Hz

    Short Circuit Fault Current 31.5 kA

    Duration of Fault Current 1 sec

    Maximum Wind Pressure 700 N/m2

    Altitude above sea level >1500 m

    3 Conductor Data

    Rigid Conductor

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    Calculation for Electrodynamic Force on Equipment & Permis-sible Span length of Rigid Conductor

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    (4)-G77700-S0019-L059-A

    i) 250/6mm Aluminum Tube (AlMgSi0.5F25) to serve as Main Bus Conductor

    Outer Diameter 250 mm

    Wall thickness 6 mm

    Weight 12.4 Kg/m

    Weight of Damping Material (Twin 954 MCM Cardinal ACSR) 1829.8 Kg/kM

    ii) 120/8mm Aluminum Tube (AlMgSi0.5F25) to serve as Equipment Bus Conductor at the BusCoupling feeder

    Outer Diameter 120 mm

    Wall thickness 8 mm

    Weight 7.6 Kg/m

    Weight of Damping Material (Single 954 MCM Cardinal ACSR) 1829.8 Kg/kM

    iii) 120/6mm Aluminum Tube (AlMgSi0.5F25) to serve as Equipment Bus Conductor at theTransformer & Line feeders

    Outer Diameter 120 mm

    Wall thickness 6 mm

    Weight 5.8 Kg/m

    Weight of Damping Material (Single 954 MCM Cardinal ACSR) 1829.8 Kg/kM

    Common for all tubes specified above

    Youngs Modulus 70000 N/mm2

    4 Equipment Technical data

    As per equipment vendor drawing and technical data sheets.

    5 Attachments

    Attachment-1: Check for Allowable Span Length of the Main Bus Conductor.

    Attachment-2: Calculation of Stress of Equipments Connected by Rigid Bus at C phase of theMain Bus work.

    Attachment-3: Check for Allowable Span Length of the Equipment Bus Conductor at the CouplingBay.

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    Calculation for Electrodynamic Force on Equipment & Permis-sible Span length of Rigid Conductor

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    (4)-G77700-S0019-L059-A

    Attachment-4: Calculation of Stress of Equipments Connected by Rigid Bus at C phase of theCoupling Bay.

    Attachment-5: Check for Allowable Span Length of the Equipment Bus Conductor at the Trans-

    former & OHL Bay.

    Attachment-6: Calculation of Stress of Equipments Connected by Rigid Bus at C phase of theTransformer & OHL Bay.

    Attachment-7: Summary of Forces on Equipments

    6 Conclusion

    1. The values of permissible span length of each size of the rigid bus conductor is as presented inthe calculation sheets attached.

    2. The maximum values of the forces on the equipment connected with the rigid bus is as pre-sented in the Summary sheet of the calculation. The input forces for structural design are alsobrought out in the summary sheet.

    7 References

    1. IEEE Std 605 : 1987 IEEE Guide for Design of Substation Rigid-Bus Structures.

    2. Technical Data sheet for Aluminum Alloy Tubular Conductor: Manufacturer Corus.

    3. G.A. Drawing for Bus Post Insulator: Manufacturer Porzellanfabrik Frauenthal Insulators.

    4. G.A. Drawing for Centre Break Disconnector: Manufacturer Merlin Gerin

    5. G.A. Drawing for Pantograph Disconnector: Manufacturer Merlin Gerin

    6. G.A. Drawing for Circuit Breaker: Manufacturer SIEMENS AG

    7. G.A. Drawing for Current Transformer: Manufacturer Trench France

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    ATTACHMENT-1

    SYSTEM DATA : 400 kV SWITCHYARD

    a) System Voltage : 400 kV

    b) Fault Level : 31.5 kA

    c) Phase-to-Phase Spacing: 6 m

    TUBULAR BUS CONDUCTOR DATA :

    a) Size : AlMgSi0.5F25

    250 mm

    c) Thickness : 6 mm

    d) Weight 12.4 Kg/m

    e) Max Allowable stress 11.72 Kg/mm2

    f) Modulus of Elasticity 7.00E+10 N/m2

    g) Vertical Deflection 21.5/24 cm/m

    = 8.96 mm/m

    h) Weight of Damper Material 3.66 Kg/m (Twin 954 CARDINAL ACSR Conductor)

    Max. Wind Pressure(Pmax) = 700 N/m (as per technical specification)

    Ambient air density(d) = 0.613 Kg/m3

    Wind Velocity(V) = (Pmax/d)1/2

    m/s

    = 33.79 m/s

    Now,

    Coss sectional moment of inertia of the tube ( J) ( S/64 ).(Do4

    - Di4) (1)

    where,

    DO = Outer Diameter of tube = 250 mm

    Di = Inner Diameter of Tube = 238 mm

    Therefore, J = 34231262.64 mm4

    = 82.24027953 in4

    Sectional Modulus of the tube S = J/(DO/2) mm3

    (2)

    = 273850.1011 mm3

    = 16.71125822 in3

    a. Conductor Wind Force (FW

    ) = 2.132 x 10-4

    CDK

    ZG

    FV

    2(D

    O+2r

    1) lbf/ft (3)

    Where

    DO = Outer Diameter of Conductor 9.84 inch

    r1 = Radial Ice thickness. = 0 inch

    CD = Drag coefficient (Table 1, IEEE:605) = 1 (round conductor)

    KZ = Height and exposure factor (Cl.10.2, IEEE:605) = 1

    GF = Gust factor (Cl. 10.3, IEEE:605) = 1.3

    V =Wind speed at 30ft. above ground = 33.79 m/s

    = 75.59 mi/h

    Hence, FW = 15.59 lbf/ft

    b. Short Circuit Current Unit Force (FSC) = 27.6* ISC2

    / 107(D) lbf/ft (4)

    where,

    ISC = Symmetrical short circuit current = 31500 A

    D = Conductor spacing centre to centre = 236.22 inch

    * = Constant based on type of short circuit and = 0.866

    conductor location (Table 2, IEEE:605)

    Hence, FSC = 10.04 lbf/ft

    c. Total bus unit weight (FG) = FC + FI + FD lbf/ft (5)

    b) Outer dia.

    CHECK FOR ALLOWABLE SPAN LENGTH of THE MAIN BUS

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    ATTACHMENT-1

    where,

    FC = Conductor unit weight = 8.33 lbf/ft

    FD = Damping material unit weight = 2.46 lbf/ft

    FI = Ice Unit weight = 0 lbf/ft

    Hence, FG = 10.79 lbf/ft

    The total Force on a conductor in Horizontal configaration

    FT = {(FW + FSC)2

    + (FG)2

    }1/2

    (6)

    = 27.81 lbf/ft

    Calculation for Allowable Span Length For Vertical Deflection:

    Case-I Span with Two Pinned Ends

    Allowable Bus Span (LD) = [384 (E) (J) (YB) / 5(FG/12)]1/3

    inch. (7)

    where,

    YB = Allowable Deflection as a fraction of span length = 0.10749 inch/ft

    E = Modulus of Elasticity = 1.01E+07 lbf/in2

    J = Coss sectional moment of inertia 82.24 in4

    FG = Total Bus Unit weight 10.79 lbf/ft

    Hence LD = 1827.346482 inch

    = 46414.60 mm

    Case-II Span with Two Fixed Ends

    Allowable Bus Span (LD) = [384 (E) (J) (YB) / (FG/12)]1/3

    inch. (8)

    Hence LD = 3107.999948 inch.

    = 78943.20 mm

    Case-III One Fixed and One Pinned End

    Allowable Bus Span (LD) = [185 (E) (J) (YB) / (FG/12)]1/3

    inch. (9)

    Hence LD = 2442.408874 inch.

    = 62037.19 mm

    Calculation for Allowable Span Length for Fiber Stress :

    Case-I Span with Two Pinned Ends

    Allowable Bus Span (LS) = [8 (FA) (S) / (FT/12)]1/2

    inch. (10)

    where,

    FA = Maximum Allowable Stress = 16672.64516 lbf/in

    S = Section Modulus = 16.71 in3

    FT = Total Force = 27.81 lbf/ft

    Hence LS = 980.76 inch.

    = 24911.18 mm

    Case-II Span with Two Fixed Ends inch.

    Allowable Bus Span (LS) = [12 (FA) (S) / (FT/12)]1/2

    inch. (11)

    Hence LS = 1201.17 inch.

    = 30509.84 mm

    Case-III One Fixed and One Pinned End

    = [8 (FA) (S) / (FT/12)]1/2

    inch. (12)

    Hence LS = 980.76 inch

    = 24911.18 mm

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    ATTACHMENT-1

    Maximum Allowable Span of the Al. Tube within the limits of 'Vertical Deflection' and permissible 'Fiber Stress' for

    various types of 'End Connections' are as follows :

    LA = Minimum of LD & LS

    Case-I For Span with two Pinned Ends = 24.91 m

    Case-II For Span with two fixed Ends = 30.51 m

    Case-III One Fixed one Pinned = 24.91 m

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    ATTACHMENT-2

    'C' PHASE OF MAIN BUSWORK

    P P P P P P P P

    24M 24 M 24M 24M 24M

    ES BPI BPI BPI BPI

    Note : 1. Phase to Phase spacing = 6 m.

    2. End Connection Type P : Pinned F : Fixed

    a. Bus Short Circuit Force

    FSB = LE.FSC

    Equipment BUS SPAN BUS SPAN LE FSC

    L1(M) L2(M) (M) lbf/ft lbf KNES 0.00 24.00 12.00 10.04 395.27 1.758

    BPI 24.00 24.00 24.00 10.04 790.55 3.517

    BPI 24.00 24.00 24.00 10.04 790.55 3.517

    BPI 24.00 24.00 24.00 10.04 790.55 3.517

    BPI 24.00 24.00 24.00 10.04 790.55 3.517

    BPI 24.00 24.00 24.00 10.04 790.55 3.517

    BPI 24.00 24.00 24.00 10.04 790.55 3.517

    ES 24.00 0.00 12.00 10.04 395.27 1.758

    b. Bus wind Force

    FWB = LE.FW

    Equipment BUS SPAN BUS SPAN LE FW

    L1(M) L2(M) (M) lbf/ft lbf KN

    ES 0.00 24.00 12.00 15.59 613.69 2.730

    BPI 24.00 24.00 24.00 15.59 1227.38 5.460

    BPI 24.00 24.00 24.00 15.59 1227.38 5.460

    BPI 24.00 24.00 24.00 15.59 1227.38 5.460

    BPI 24.00 24.00 24.00 15.59 1227.38 5.460

    BPI 24.00 24.00 24.00 15.59 1227.38 5.460

    BPI 24.00 24.00 24.00 15.59 1227.38 5.460

    ES 24.00 0.00 12.00 15.59 613.69 2.730

    c. Insulator wind Force

    FWI = 1.776 x 10-5

    CDKZGFV2

    (Di+2r1)Hi lbf

    Equipment D1 D2 Di Hi

    (m) (m) (m) (m) lbf KN

    ES 0.36 0.16 0.26 3.35 178.11 0.792BPI 0.36 0.16 0.26 3.35 178.11 0.792

    BPI 0.36 0.16 0.26 3.35 178.11 0.792

    BPI 0.36 0.16 0.26 3.35 178.11 0.792

    BPI 0.36 0.16 0.26 3.35 178.11 0.792

    BPI 0.36 0.16 0.26 3.35 178.11 0.792

    BPI 0.36 0.16 0.26 3.35 178.11 0.792

    ES 0.36 0.16 0.26 3.35 178.11 0.792

    BPI-ES

    BPI-ES

    BPI-BPI-BPI

    Connected

    24M 24M

    BPI-BPI-BPI

    BPI-BPI-BPI

    BPI-BPI-ES

    BPI BPI ES

    CALCULATION OF STRESSES OF EQUIPMENTS CONNECTED BY RIGID BUS

    BPI-BPI-BPI

    EquipmentES-BPI

    ES-BPI-BPI

    Connected

    ES-BPI

    BPI-BPI-BPI

    BPI-BPI-BPI

    ES-BPI-BPI

    Equipment

    BPI-BPI-BPI

    BPI-BPI-BPI

    BPI-BPI-ES

    FSB

    FWB

    FWI

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    ATTACHMENT-2

    d. Gravitational Force

    FGB = LE.FG ``````

    Equipment BUS SPAN BUS SPAN LE FG

    L1(M) L2(M) (M) lbf/ft lbf KN

    ES 0.00 24.00 12.00 10.79 424.92 1.89

    BPI 24.00 24.00 24.00 10.79 849.83 3.78

    BPI 24.00 24.00 24.00 10.79 849.83 3.78

    BPI 24.00 24.00 24.00 10.79 849.83 3.78

    BPI 24.00 24.00 24.00 10.79 849.83 3.78

    BPI 24.00 24.00 24.00 10.79 849.83 3.78

    BPI 24.00 24.00 24.00 10.79 849.83 3.78

    ES 24.00 0.00 12.00 10.79 424.92 1.89

    e. Insulator Cantilever Load

    FIS =

    K1 = Overload Factor applied to wind force = 1

    K2 = Overload Factor applied to short circuit force = 1

    Hf = Bus Cen. Height above Insulator = 0.215 m

    = 8.46 inch

    Equipment D1 D2 Di Hi

    (m) (m) (m) (m) lbf KN

    ES 0.36 0.16 0.26 3.35 1162.77 5.18

    BPI 0.36 0.16 0.26 3.35 2236.49 9.96

    BPI 0.36 0.16 0.26 3.35 2236.49 9.96

    BPI 0.36 0.16 0.26 3.35 2236.49 9.96

    BPI 0.36 0.16 0.26 3.35 2236.49 9.96

    BPI 0.36 0.16 0.26 3.35 2236.49 9.96

    BPI 0.36 0.16 0.26 3.35 2236.49 9.96

    ES 0.36 0.16 0.26 3.35 1162.77 5.18

    Summary

    Equipment FSB FWB FWI FGB FIS

    KN KN KN KN KN

    ES 1.76 2.73 0.79 1.89 5.18

    BPI 3.52 5.46 0.79 3.78 9.96

    BPI 3.52 5.46 0.79 3.78 9.96

    BPI 3.52 5.46 0.79 3.78 9.96

    BPI 3.52 5.46 0.79 3.78 9.96

    BPI 3.52 5.46 0.79 3.78 9.96

    BPI 3.52 5.46 0.79 3.78 9.96

    ES 1.76 2.73 0.79 1.89 5.18

    BPI-ES

    BPI-BPI-BPI

    BPI-BPI-BPI

    BPI-BPI-ES

    Equipment

    Connected

    BPI-BPI-BPI

    BPI-BPI-BPI

    ES-BPI-BPI

    ES-BPI

    FIS

    FGB

    ))(

    ()(

    2( 21

    i

    SBfi

    i

    WBfiWI

    H

    FHHK

    H

    FHHFK

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    ATTACHMENT-3

    SYSTEM DATA : 400 kV SWITCHYARD

    a) System Voltage : 400 kV

    b) Fault Level : 31.5 kA

    c) Phase-to-Phase Spacing: 6.5 m

    TUBULAR BUS CONDUCTOR DATA :

    a) Size : AlMgSi0.5F25

    120 mm

    c) Thickness : 8 mm

    d) Weight 7.6 Kg/m

    e) Max Allowable stress 11.72 Kg/mm2

    f) Modulus of Elasticity 7.00E+10 N/m2

    g) Vertical Deflection 11.8/12 cm/m

    = 9.83 mm/m

    h) Weight of Damper Material 1.83 Kg/m (Single 954 CARDINAL ACSR Conductor)

    Max. Wind Pressure(Pmax) = 700 N/m (as per technical specification)

    Ambient air density(d) = 0.613 Kg/m3

    Wind Velocity(V) = (Pmax/d)1/2

    m/s

    = 33.79 m/s

    Now,

    Coss sectional moment of inertia of the tube ( J) ( S/64 ).(Do4

    - Di4) (1)

    where,

    DO = Outer Diameter of tube = 120 mm

    Di = Inner Diameter of Tube = 104 mm

    Therefore, J = 4433981.44 mm4

    = 10.6525978 in4

    Sectional Modulus of the tube S = J/(DO/2) mm3

    (2)

    = 73899.69067 mm3

    = 4.509608754 in3

    a. Conductor Wind Force (FW

    ) = 2.132 x 10-4

    CDK

    ZG

    FV

    2(D

    O+2r

    1) lbf/ft (3)

    Where

    DO = Outer Diameter of Conductor 4.72 inch

    r1 = Radial Ice thickness. = 0 inch

    CD = Drag coefficient (Table 1, IEEE:605) = 1 (round conductor)

    KZ = Height and exposure factor (Cl.10.2, IEEE:605) = 1 (bus height > 30ft.)

    GF = Gust factor (Cl. 10.3, IEEE:605) = 1.3

    V =Wind speed at 30ft. above ground = 33.79 m/s

    = 75.59 mi/h

    Hence, FW = 7.48 lbf/ft

    b. Short Circuit Current Unit Force (FSC) = 27.6* ISC2

    / 107(D) lbf/ft (4)

    where,

    ISC = Symmetrical short circuit current = 31500 A

    D = Conductor spacing centre to centre = 255.91 inch

    * = Constant based on type of short circuit and = 0.866

    conductor location (Table 2, IEEE:605)

    Hence, FSC = 9.27 lbf/ft

    c. Total bus unit weight (FG) = FC + FI + FD lbf/ft (5)

    b) Outer dia.

    CHECK FOR ALLOWABLE SPAN LENGTH of THE EQUIPMENT BUS AT COUPLING BAY

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    ATTACHMENT-3

    where,

    FC = Conductor unit weight = 5.11 lbf/ft

    FD = Damping material unit weight = 1.23 lbf/ft

    FI = Ice Unit weight = 0 lbf/ft

    Hence, FG = 6.34 lbf/ft

    The total Force on a conductor in Horizontal configaration

    FT = {(FW + FSC)2

    + (FG)2

    }1/2

    (6)

    = 17.91 lbf/ft

    Calculation for Allowable Span Length For Vertical Deflection:

    Case-I Span with Two Pinned Ends

    Allowable Bus Span (LD) = [384 (E) (J) (YB) / 5(FG/12)]1/3

    inch. (7)

    where,

    YB = Allowable Deflection as a fraction of span length = 0.11799 inch/ft

    E = Modulus of Elasticity = 1.01E+07 lbf/in2

    J = Coss sectional moment of inertia 10.65 in4

    FG = Total Bus Unit weight 6.34 lbf/ft

    Hence LD = 1144.36937 inch

    = 29066.98 mm

    Case-II Span with Two Fixed Ends

    Allowable Bus Span (LD) = [384 (E) (J) (YB) / (FG/12)]1/3

    inch. (8)

    Hence LD = 1946.374142 inch.

    = 49437.90 mm

    Case-III One Fixed and One Pinned End

    Allowable Bus Span (LD) = [185 (E) (J) (YB) / (FG/12)]1/3

    inch. (9)

    Hence LD = 1529.55005 inch.

    = 38850.57 mm

    Calculation for Allowable Span Length for Fiber Stress :

    Case-I Span with Two Pinned Ends

    Allowable Bus Span (LS) = [8 (FA) (S) / (FT/12)]1/2

    inch. (10)

    where,

    FA = Maximum Allowable Stress = 16672.64516 lbf/in

    S = Section Modulus = 4.51 in3

    FT = Total Force = 17.91 lbf/ft

    Hence LS = 634.86 inch.

    = 16125.42 mm

    Case-II Span with Two Fixed Ends inch.

    Allowable Bus Span (LS) = [12 (FA) (S) / (FT/12)]1/2

    inch. (11)

    Hence LS = 777.54 inch.

    = 19749.53 mm

    Case-III One Fixed and One Pinned End

    = [8 (FA) (S) / (FT/12)]1/2

    inch. (12)

    Hence LS = 634.86 inch

    = 16125.42 mm

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    ATTACHMENT-3

    Maximum Allowable Span of the Al. Tube within the limits of 'Vertical Deflection' and permissible 'Fiber Stress' for

    various types of 'End Connections' are as follows :

    LA = Minimum of LD & LS

    Case-I For Span with two Pinned Ends = 16.13 m

    Case-II For Span with two fixed Ends = 19.75 m

    Case-III One Fixed one Pinned = 16.13 m

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    ATTACHMENT-4

    'C' PHASE OF COUPLING BAY

    P P P P P P F F P

    3.5M 3.5M

    6M 6M 10.5M 12M

    CT CB CT BPI ISO ISO BPI

    (Pantograph) (Pantograph)

    Note : 1. Phase to Phase spacing = 6.5 m.

    2. End Connection Type P : Pinned F : Fixed

    a. Bus Short Circuit Force

    FSB = LE.FSC

    Equipment BUS SPAN BUS SPAN LE FSC

    L1(M) L2(M) (M) lbf/ft lbf KNCT 0.00 3.50 1.75 9.27 53.21 0.237

    CB 3.50 3.50 3.50 9.27 106.42 0.473

    CT 3.50 10.50 7.00 9.27 212.84 0.947

    BPI 10.50 12.00 9.75 9.27 296.45 1.319

    ISO(Panto) 12.00 0.00 7.50 9.27 228.04 1.014

    ISO(Panto) 0.00 6.50 4.06 9.27 123.52 0.549

    BPI 6.50 0.00 2.44 9.27 74.11 0.330

    b. Bus wind Force

    FWB = LE.FW

    Equipment BUS SPAN BUS SPAN LE FW

    L1(M) L2(M) (M) lbf/ft lbf KN

    CT 0.00 3.50 1.75 7.48 42.96 0.191CB 3.50 3.50 3.50 7.48 85.92 0.382

    CT 3.50 10.50 7.00 7.48 171.83 0.764

    BPI 10.50 12.00 9.75 7.48 239.34 1.065

    ISO(Panto) 12.00 0.00 7.50 7.48 184.11 0.819

    ISO(Panto) 0.00 6.50 4.06 7.48 99.72 0.444

    BPI 6.50 0.00 2.44 7.48 59.83 0.266

    c. Insulator wind Force

    FWI = 1.776 x 10-5

    CDKZGFV2

    (Di+2r1)Hi lbf

    Equipment D1 D2 Di Hi

    (m) (m) (m) (m) lbf KN

    CT 0.75 0.47 0.61 4.43 550.32 2.448

    CB 0.42 0.24 0.33 4.92 329.15 1.464

    CT 0.75 0.47 0.61 4.43 550.32 2.448BPI 0.29 0.13 0.21 3.35 143.86 0.640

    ISO(Panto) 0.30 0.14 0.22 3.88 174.69 0.777

    ISO(Panto) 0.30 0.14 0.22 3.88 174.69 0.777

    BPI 0.29 0.13 0.21 3.35 143.86 0.640

    FSB

    FWB

    FWI

    ISO(Panto)-BPI

    Connected

    Equipment

    BPI-ISO(Panto)

    ISO(Panto)-BPI

    CALCULATION OF STRESSES OF EQUIPMENTS CONNECTED BY RIGID BUS

    CT-BPI-ISO(Panto)

    EquipmentCT-CB

    CT-CB-CT

    Connected

    CT-CB

    CT-BPI-ISO(Panto)

    CB-CT-BPI

    CT-CB-CT

    6.5M

    BPI-ISO(Panto)

    ISO(Panto)-BPI

    ISO(Panto)-BPI

    CB-CT-BPI

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    ATTACHMENT-4

    d. Gravitational Force

    FGB = LE.FG

    Equipment BUS SPAN BUS SPAN LE FG

    L1(M) L2(M) (M) lbf/ft lbf KN

    CT 0.00 3.50 1.75 6.34 36.39 0.16

    CB 3.50 3.50 3.50 6.34 72.77 0.32

    CT 3.50 10.50 7.00 6.34 145.54 0.65

    BPI 10.50 12.00 9.75 6.34 202.72 0.90

    ISO(Panto) 12.00 0.00 7.50 6.34 155.94 0.69

    ISO(Panto) 0.00 6.50 4.06 6.34 84.47 0.38

    BPI 6.50 0.00 2.44 6.34 50.68 0.23

    e. Insulator Cantilever Load

    FIS =

    K1 = Overload Factor applied to wind force = 1

    K2 = Overload Factor applied to short circuit force = 1

    Hf = Bus Cen. Height above Insulator = 0.15 m

    = 5.91 inch

    Equipment D1 D2 Di Hi

    (m) (m) (m) (m) lbf KN

    CT 0.75 0.47 0.61 4.43 374.58 1.67

    CB 0.42 0.24 0.33 4.92 362.78 1.61

    CT 0.75 0.47 0.61 4.43 672.86 2.99

    BPI 0.29 0.13 0.21 3.35 631.71 2.81

    ISO(Panto) 0.30 0.14 0.22 3.88 515.41 2.29

    ISO(Panto) 0.30 0.14 0.22 3.88 319.21 1.42

    BPI 0.29 0.13 0.21 3.35 211.87 0.94

    Summary

    Equipment FSB FWB FWI FGB FIS

    KN KN KN KN KN

    CT 0.24 0.19 2.45 0.16 1.67

    CB 0.47 0.38 1.46 0.32 1.61

    CT 0.95 0.76 2.45 0.65 2.99

    BPI 1.32 1.06 0.64 0.90 2.81

    ISO(Panto) 1.01 0.82 0.78 0.69 2.29

    ISO(Panto) 0.55 0.44 0.78 0.38 1.42

    BPI 0.33 0.27 0.64 0.23 0.94

    FIS

    FGB

    CT-BPI-ISO(Panto)

    CB-CT-BPI

    CT-CB-CT

    CT-CB

    BPI-ISO(Panto)

    ISO(Panto)-BPI

    ISO(Panto)-BPI

    Equipment

    Connected

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    ATTACHMENT-5

    SYSTEM DATA : 400 kV SWITCHYARD

    a) System Voltage : 400 kV

    b) Fault Level : 31.5 kA

    c) Phase-to-Phase Spacing: 6.5 m

    TUBULAR BUS CONDUCTOR DATA :

    a) Size : AlMgSi0.5F25

    120 mm

    c) Thickness : 6 mm

    d) Weight 5.8 Kg/m

    e) Max Allowable stress 11.72 Kg/mm2

    f) Modulus of Elasticity 7.00E+10 N/m2

    g) Vertical Deflection 11.4/12 cm/m

    = 9.50 mm/m

    h) Weight of Damper Material 1.83 Kg/m (Single 954 CARDINAL ACSR Conductor)

    Max. Wind Pressure(Pmax) = 700 N/m (as per technical specification)

    Ambient air density(d) = 0.613 Kg/m3

    Wind Velocity(V) = (Pmax/d)1/2

    m/s

    = 33.79 m/s

    Now,

    Coss sectional moment of inertia of the tube ( J) ( S/64 ).(Do4

    - Di4) (1)

    where,

    DO = Outer Diameter of tube = 120 mm

    Di = Inner Diameter of Tube = 108 mm

    Therefore, J = 3498701.04 mm4

    = 8.405595626 in4

    Sectional Modulus of the tube S = J/(DO/2) mm3

    (2)

    = 58311.684 mm3

    = 3.558375932 in3

    a. Conductor Wind Force (FW

    ) = 2.132 x 10-4

    CDK

    ZG

    FV

    2(D

    O+2r

    1) lbf/ft (3)

    Where

    DO = Outer Diameter of Conductor 4.72 inch

    r1 = Radial Ice thickness. = 0 inch

    CD = Drag coefficient (Table 1, IEEE:605) = 1 (round conductor)

    KZ = Height and exposure factor (Cl.10.2, IEEE:605) = 1

    GF = Gust factor (Cl. 10.3, IEEE:605) = 1.3

    V =Wind speed at 30ft. above ground = 33.79 m/s

    = 75.59 mi/h

    Hence, FW = 7.48 lbf/ft

    b. Short Circuit Current Unit Force (FSC) = 27.6* ISC2

    / 107(D) lbf/ft (4)

    where,

    ISC = Symmetrical short circuit current = 31500 A

    D = Conductor spacing centre to centre = 255.91 inch

    * = Constant based on type of short circuit and = 0.866

    conductor location (Table 2, IEEE:605)

    Hence, FSC = 9.27 lbf/ft

    c. Total bus unit weight (FG) = FC + FI + FD lbf/ft (5)

    b) Outer dia.

    CHECK FOR ALLOWABLE SPAN LENGTH of THE EQUIPMENT BUS AT TRANSFORMER & OHL BAY

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    ATTACHMENT-5

    where,

    FC = Conductor unit weight = 3.90 lbf/ft

    FD = Damping material unit weight = 1.23 lbf/ft

    FI = Ice Unit weight = 0 lbf/ft

    Hence, FG = 5.13 lbf/ft

    The total Force on a conductor in Horizontal configaration

    FT = {(FW + FSC)2

    + (FG)2

    }1/2

    (6)

    = 17.52 lbf/ft

    Calculation for Allowable Span Length For Vertical Deflection:

    Case-I Span with Two Pinned Ends

    Allowable Bus Span (LD) = [384 (E) (J) (YB) / 5(FG/12)]1/3

    inch. (7)

    where,

    YB = Allowable Deflection as a fraction of span length = 0.11399 inch/ft

    E = Modulus of Elasticity = 1.01E+07 lbf/in2

    J = Coss sectional moment of inertia 8.41 in4

    FG = Total Bus Unit weight 5.13 lbf/ft

    Hence LD = 1122.089554 inch

    = 28501.07 mm

    Case-II Span with Two Fixed Ends

    Allowable Bus Span (LD) = [384 (E) (J) (YB) / (FG/12)]1/3

    inch. (8)

    Hence LD = 1908.480034 inch.

    = 48475.39 mm

    Case-III One Fixed and One Pinned End

    Allowable Bus Span (LD) = [185 (E) (J) (YB) / (FG/12)]1/3

    inch. (9)

    Hence LD = 1499.771123 inch.

    = 38094.19 mm

    Calculation for Allowable Span Length for Fiber Stress :

    Case-I Span with Two Pinned Ends

    Allowable Bus Span (LS) = [8 (FA) (S) / (FT/12)]1/2

    inch. (10)

    where,

    FA = Maximum Allowable Stress = 16672.64516 lbf/in

    S = Section Modulus = 3.56 in3

    FT = Total Force = 17.52 lbf/ft

    Hence LS = 570.21 inch.

    = 14483.29 mm

    Case-II Span with Two Fixed Ends inch.

    Allowable Bus Span (LS) = [12 (FA) (S) / (FT/12)]1/2

    inch. (11)

    Hence LS = 698.36 inch.

    = 17738.34 mm

    Case-III One Fixed and One Pinned End

    = [8 (FA) (S) / (FT/12)]1/2

    inch. (12)

    Hence LS = 570.21 inch

    = 14483.29 mm

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    ATTACHMENT-5

    Maximum Allowable Span of the Al. Tube within the limits of 'Vertical Deflection' and permissible 'Fiber Stress' for

    various types of 'End Connections' are as follows :

    LA = Minimum of LD & LS

    Case-I For Span with two Pinned Ends = 14.48 m

    Case-II For Span with two fixed Ends = 17.74 m

    Case-III One Fixed one Pinned = 14.48 m

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    ATTACHMENT-6

    'C' PHASE OF TRANSFORMER & OHL BAY

    F P P P P F P F

    3.75M 3.5M 8M

    12M 10M 9M

    DS CT CB BPI ISO BPI ISO

    (Pantograph) (Pantograph)

    Note : 1. Phase to Phase spacing = 6.5 m.

    2. End Connection Type P : Pinned F : Fixed

    a. Bus Short Circuit Force

    FSB = LE.FSC

    Equipment BUS SPAN BUS SPAN LE FSC

    L1(M) L2(M) (M) lbf/ft lbf KNDS 0.00 3.75 2.34 9.27 71.26 0.317

    CT 3.75 3.50 3.16 9.27 95.97 0.427

    CB 3.50 8.00 5.75 9.27 174.83 0.778

    BPI 8.00 12.00 8.50 9.27 258.45 1.150

    ISO(Panto) 12.00 10.00 13.75 9.27 418.08 1.860

    BPI 10.00 9.00 7.13 9.27 216.64 0.964

    ISO(Panto) 9.00 0.00 5.63 9.27 171.03 0.761

    b. Bus wind Force

    FWB = LE.FW

    Equipment BUS SPAN BUS SPAN LE FW

    L1(M) L2(M) (M) lbf/ft lbf KN

    DS 0.00 3.75 2.34 7.48 57.53 0.256CT 3.75 3.50 3.16 7.48 77.48 0.345

    CB 3.50 8.00 5.75 7.48 141.15 0.628

    BPI 8.00 12.00 8.50 7.48 208.65 0.928

    ISO(Panto) 12.00 10.00 13.75 7.48 337.53 1.501

    BPI 10.00 9.00 7.13 7.48 174.90 0.778

    ISO(Panto) 9.00 0.00 5.63 7.48 138.08 0.614

    c. Insulator wind Force

    FWI = 1.776 x 10-5

    CDKZGFV2

    (Di+2r1)Hi lbf

    Equipment D1 D2 Di Hi

    (m) (m) (m) (m) lbf KN

    DS 0.30 0.16 0.23 3.65 171.67 0.764

    CT 0.75 0.47 0.61 4.43 550.32 2.448

    CB 0.42 0.24 0.33 4.92 329.15 1.464 `BPI 0.29 0.13 0.21 3.35 143.86 0.640

    ISO(Panto) 0.30 0.14 0.22 3.88 174.69 0.777

    BPI 0.29 0.13 0.21 3.35 143.86 0.640

    ISO(Panto) 0.30 0.14 0.22 3.88 174.69 0.777

    6M 6M 10.5M

    ISO(Panto)-BPI-ISO(Panto)

    BPI-ISO(Panto)

    BPI-ISO(Panto)-BPI

    ISO(Panto)-BPI-ISO(Panto)

    BPI-ISO(Panto)

    CT-CB-BPI

    Connected

    Equipment

    BPI-ISO(Panto)-BPI

    CALCULATION OF STRESSES OF EQUIPMENTS CONNECTED BY RIGID BUS

    CB-BPI-ISO(Panto)

    EquipmentDS-CT

    DS-CT-CB

    Connected

    DS-CT

    CB-BPI-ISO(Panto)

    CT-CB-BPI

    DS-CT-CB

    FSB

    FWB

    FWI

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    ATTACHMENT-6

    d. Gravitational Force

    FGB = LE.FG

    `

    Equipment BUS SPAN BUS SPAN LE FG

    L1(M) L2(M) (M) lbf/ft lbf KN

    DS 0.00 3.75 2.34 5.13 39.43 0.18

    CT 3.75 3.50 3.16 5.13 53.10 0.24

    CB 3.50 8.00 5.75 5.13 96.73 0.43

    BPI 8.00 12.00 8.50 5.13 142.99 0.64

    ISO(Panto) 12.00 10.00 13.75 5.13 231.31 1.03

    BPI 10.00 9.00 7.13 5.13 119.86 0.53

    ISO(Panto) 9.00 0.00 5.63 5.13 94.63 0.42

    e. Insulator Cantilever Load

    FIS =

    K1 = Overload Factor applied to wind force = 1

    K2 = Overload Factor applied to short circuit force = 1

    Hf = Bus Cen. Height above Insulator = 0.15 m

    = 5.91 inch

    Equipment D1 D2 Di Hi

    (m) (m) (m) (m) lbf KN

    DS 0.30 0.16 0.23 3.65 219.92 0.98

    CT 0.75 0.47 0.61 4.43 454.48 2.02

    CB 0.42 0.24 0.33 4.92 490.20 2.18

    BPI 0.29 0.13 0.21 3.35 559.95 2.49

    ISO(Panto) 0.30 0.14 0.22 3.88 872.14 3.88

    BPI 0.29 0.13 0.21 3.35 481.00 2.14

    ISO(Panto) 0.30 0.14 0.22 3.88 408.40 1.82

    Summary

    Equipment FSB FWB FWI FGB FIS

    KN KN KN KN KN

    DS 0.32 0.26 0.76 0.18 0.98

    CT 0.43 0.34 2.45 0.24 2.02

    CB 0.78 0.63 1.46 0.43 2.18

    BPI 1.15 0.93 0.64 0.64 2.49

    ISO(Panto) 1.86 1.50 0.78 1.03 3.88

    BPI 0.96 0.78 0.64 0.53 2.14

    ISO(Panto) 0.76 0.61 0.78 0.42 1.82

    BPI-ISO(Panto)-BPI

    ISO(Panto)-BPI-ISO(Panto)

    CB-BPI-ISO(Panto)

    CT-CB-BPI

    DS-CT-CB

    DS-CT

    BPI-ISO(Panto)

    Equipment

    Connected

    FIS

    FGB

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    ATTACHMENT - 7

    SUMMARY OF FORCES ON EQUIPMENTS

    FSB = Bus Short Circuit Force transmitted to bus support fitting

    FWB = Bus Wind force transmitted to bus support fitting

    FWI = Wind Force on InsulatorFGB = Effective weight of bus transmitted to bus support fitting

    FIS = Total Cantilever load acting at the end of Insulator

    400kV Switchyard

    400kV Main Bus Bay

    Equipment FSB FWB FWI FGB FIS

    KN KN KN KN KN

    ES 1.76 2.73 0.79 1.89 5.18

    BPI 3.52 5.46 0.79 3.78 9.96

    400kV Transformer,OHL Bay

    Equipment FSB FWB FWI FGB FIS

    KN KN KN KN KNDS 0.32 0.26 0.76 0.18 0.98

    CT 0.43 0.34 2.45 0.24 2.02

    CB 0.78 0.63 1.46 0.43 2.18

    BPI 1.15 0.93 0.64 0.64 2.49

    ISO(Panto) 1.86 1.50 0.78 1.03 3.88

    400kV Bus Coupler Bay

    Equipment FSB FWB FWI FGB FIS

    KN KN KN KN KN

    CT 0.95 0.76 2.45 0.65 2.99

    CB 0.47 0.38 1.46 0.32 1.61

    BPI 1.32 1.06 0.64 0.90 2.81

    ISO(Panto) 1.01 0.82 0.78 0.69 2.29

    Maximum forces in 400kV substation

    Equipment FSB FWB FWI FGB FIS

    KN KN KN KN KN

    ES 1.76 2.73 0.79 1.89 5.18

    BPI 3.52 5.46 0.79 3.78 9.96

    DS 0.32 0.26 0.76 0.18 0.98

    ISO(Panto) 1.86 1.50 0.78 1.03 3.88

    CT 0.95 0.76 2.45 0.65 2.99

    CB 0.78 0.63 1.46 0.43 2.18

    Input for Structure Design

    Equipment

    ES 527.68 Kgs 192.68 Kgs

    BPI 1014.95 Kgs 385.36 Kgs

    DS 99.72 Kgs 17.88 Kgs

    ISO(Panto) 395.47 Kgs 104.89 Kgs

    CT 305.11 Kgs 66.00 Kgs

    CB 222.28 Kgs 43.86 Kgs

    Following Forces has been calculated:

    Short Circuit (Tensile) Load of Connected Al.

    Force Tube Bus

    Based on the above results, the maximum forces on 400kV substation equiupments comparing all type of bays

    have been tabulated in the following table :