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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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S (4)-G77700-S0019-L059-A
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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S (4)-G77700-S0019-L059-A
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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S (4)-G77700-S0019-L059-A
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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S (4)-G77700-S0019-L059-A
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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S (4)-G77700-S0019-L059-A
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
Gilgel Gibe II Hydroelectric Project 1
7/29/2019 Boru Bara Mekanik Hesaplar
15/21
S (4)-G77700-S0019-L059-A
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
))(
()(
2( 21
i
SBfi
i
WBfiWI
H
FHHK
H
FHHFK
Gilgel Gibe II Hydroelectric Project 2
7/29/2019 Boru Bara Mekanik Hesaplar
16/21
S (4)-G77700-S0019-L059-A
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
Gilgel Gibe II Hydroelectric Project 1
7/29/2019 Boru Bara Mekanik Hesaplar
17/21
S (4)-G77700-S0019-L059-A
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
Gilgel Gibe II Hydroelectric Project 2
7/29/2019 Boru Bara Mekanik Hesaplar
18/21
S (4)-G77700-S0019-L059-A
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
Gilgel Gibe II Hydroelectric Project 3
7/29/2019 Boru Bara Mekanik Hesaplar
19/21
S (4)-G77700-S0019-L059-A
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
Gilgel Gibe II Hydroelectric Project 1
7/29/2019 Boru Bara Mekanik Hesaplar
20/21
S (4)-G77700-S0019-L059-A
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
))(
()(
2( 21
i
SBfi
i
WBfiWI
H
FHHK
H
FHHFK
Gilgel Gibe II Hydroelectric Project 2
7/29/2019 Boru Bara Mekanik Hesaplar
21/21
S (4)-G77700-S0019-L059-A
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 :