Rigid Bus Work

RIGID BUSWORK DESIGN I. Project

Project Name: Project Name Project Code: Client: Consultant:

II. Designers

Prepared by: S.M.Shariatzadeh Checked by Approved by Revision: Z Date 24 August 2008 Total Page: 27

Page 2 of 27 RIGID BUSWORK DESIGN

CONTENTS Section 1: General Data

1-1. Environmental conditions

1-2 132KV System data

1-3 Tubular Conductor data

1-4 Insulator data

1-5 Configuration data

Section 2: General

2-1 Scope and object

2-2 Symbols

2-3 Support Definitions

Section3 : Selection of conductor

3.1 Dimensioning for the continuous current carrying of conductor

3.2 Checking for short-circuit thermal strength of conductor

3.3 Checking for the maximum deflection of conductor

Section 4: Calculating of maximum load on supports and checking the bending stress

4.1 General

4.1.1 Load combinations


4.1.2 Wind pressure

4.1.3 Wind force on conductors

4.1.4 Wind force on insulators.

4.1.5 Wind force on A-Frame

4.2 Loading of normal wind and ice

4.3 Loading of high wind

4.4 Loading of high wind and short-circuit

4.5 Maximum forces on supports

4.6 Compound type configurations

4.6 .1 General

4.6.2 Forces on supports

Section 5: Calculation results

5.1 Dimensioning for continuous current

5.2 Checking for S.C. thermal strength

5.3 Checking for deflection

5.4 Loading

5.4.1 Details of loading

5.4.2 Details of S.C. effects

5.4.3 Details of compound loading


Section 1: General Data

1-1. Environmental conditions

Max Temperature (tmax)= 85 O C Average daily temperature( Tav)=40 O C Min Temperature (tmin)= -40 O C Average Temperature (tav)= 45 O C Maximum Wind Velocity (Vw)= 40 m/s Ice Thickness (Ice) = 20 mm Height above Sea level (H) = 1500 m


1-2. 132KV System data

Voltage (U) = 132 KV Rated system frequency (f) = 50 Hz Continues Current (Ic) = 1600 A Initial symmetrical short-circuit current (Ik3) = 31.5 KA Duration of short circuit (Tk) = 1 S Duration of first short circuit current flow (Tk1) = 1 S Factor for calculation of the peak short circuit current (k) = 1.8 Number of sub-conductor per Bundle (n) =1

1-3. Tubular Conductor Data

Material (M) = AL-alloy ( AL MG SI 0.5 F22 ) Outside diameter (D) = 80 mm Wall thickness (S) = 6 mm Mass of sub-conductor per unit length (ms) = 3.77 kg/m Stress corresponding to the yield point (Rp2min) = 160 N/mm2


Young Modulus (E) = 70 KN/ mm2 Conductivity at 20O C (σ) = 30 m / Ω mm2 b Dimension of sub-conductor perpendicular to the direction of the force d Dimension of sub-conductor in the direction of the force Dim (D/S or dxb) = 80 / 6 mm Cross-section of sub-conductor (As) = 1395 mm2 Conductor temperature at the end of a short-circuit (Te) =200 OC Rated current carrying of conductor at standard condition (Ir)= 2222A


Section 2: General

2-1. Scope and object The purpose of this rigid bus work design calculation is to analyze the behavior of the rigid buses during normal working condition and also severe

conditions such as ice, wind and short circuit.

Buses must work appropriately and below a certain temperature during their continuous current conditions and this leads to sizing problem of

conductors.

Also, buses must have permissible deflections, acceptable temperature rise after short circuit and limited conductor stresses. Forces on supports due

to load combinations must be less than permissible insulator maximum top force.

All of the above are discussed later.

2-2. symbols Support data

Lsup Length of supporting insulator

Dsup Average diameter of supporting insulator

Fsup Cantilever force of supporting insulator

Configuration data

L Length of span (equi-distant )


L1 Actual adjacent span length of side one of support

L2 Actual adjacent span length of the other side of support

Ic Design continuous current for the conductor at highest working temperature TO

To Max temperature of conductor at continuous current carrying

a Centerline distance between phases

h Height of span above ground

n Number of conductors per phase

k Number of sets of spacers or stiffening elements

a12 Center-line distance between sub-conductors

Ls Center-line distance between

Ha A-frame height

Da A-frame diameter of legs

La A-frame length of both legs

II/= vertical (V) or Horizontal (H) arrangement of rectangular conductors

A/R Operating of autorecloser


Dimensioning for continuous current

Ir Rated Current carrying of conductor at standard conditions

It Maximum Current carrying of conductor at site conditions

K1 Factor for load combination relating to conductivity

k2 Factor for other air and conductor temperature

k3 Factor for load reduction in long side or length rectangular busbars

k4 Factor for load reduction due to additional skin effect

k5 Factor for influences specific to location

Short Circuit. thermal strength

m Factor for the heat effect of the DC component

n Factor for the heat effect of the AC component

Ith Thermal equivalent short time current (r.m.s)

Sthr Rated short time withstand current density (r.m.s.) for Is

Amin Minimum required conductor cross-section

n As Total cross-section of main conductor


Deflection of span

def Deflection of conductor

Q Load by weight of the tube between the support points

I Factor depending on number of supports

J moment of inertia

L /150 Maximum allowable deflection of conductor

Loading

iw Ice & wind load combination

hw High wind load combination

Sw Short-circuit & normal wind load combination

Fiw Total force of conductor and insulator on top of the insulator due to iw

Fhw Total force of conductor and insulator on top of the insulator due to hw

Fsw Total force of conductor and insulator on top of the insulator due to sw

Fmax Maximum of Fhw, Fiw and Fsw


Loading details

iw Ice & wind load combination

hw High wind load combination

sw Short-circuit & normal wind load combination

Vw Wind velocity under specific load combination

C A factor related to Vw and conductor width

Ch A factor related to height of span above ground

Pw Pressure of wind

Fw Force of wind on conductor due to one of iw, hw or sw

Fi Insulator wind force transferred to insulator top due to a loading combination

Fa A-Frame wind force transferred to insulator top due to a loading combination

Fd SC force on conductor

F Force on support due to a loading combination

Details of SC effects

ip3 Peak short circuit current in case of balanced three-phase short circuit

am Effective distance between neighboring main conductors

Fm Force between main conductors during short circuit


g Factor for relevant natural frequency estimation

Js Second moment of sub-conductor

fc Relevant natural frequency of a main-conductor

c Factor for the influence of connecting pieces

Vf Ratio of static and dynamic force on supports

Vs Ratio of static and dynamic main conductor stress

Vr Ratio stress for a main conductor

b Factor for main conductor stress

Zs Section modulus of sub—conductor

a Factor for force on support

mz Total mass of one set of connecting piece

as Effective distance between neighboring main conductors

Fs Force between sub-conductors during short circuit

fcs Relevant natural frequency of a sub-conductor

Vrs Ratio stress for a sub-conductor

Vss Ratio of static and dynamic sub-conductor stress

sm Bending stress caused by the forces between main conductors


ss Bending stress caused by the forces between sub-conductors

st Resulting conductor stress

q Factor for plasticity

A-frame loading details

F’i Insulator wind force transferred to insulator top due to a loading combination

F'a A-Frame wind force transferred to insulator top due to a loading combination

F’whigh Wind force on top of compound configuration due to upper conductor transferred to insulator top due to a loading combination

Fwlow Wind force on top of compound configuration due to lower conductor due to a loading combination

F’dhigh SC force on top of compound configuration due to upper conductor transferred to insulator top due to a loading combination

Fd low SC force on top of compound configuration due to lower conductor due to a loading combination Angle Worst wind direction each load combination

2-3. Support Definitions Fixed support A support of a rigid conductor which does not permit the conductor to move angularly at the point of the support.

Simple support A support of a rigid conductor which permits angular movement at the point of support.


Section 3 Selection of Conductor [ABB Manual 10th Edition Page 146-149]

3.1. Dimensioning for continuous carrying of conductor Considering conductor material, conductivity (s), highest working temperature (TO) and also mean ambient temperature (Tav), It is given by: It = Ir * ( ki . k2 . k3 . k4 . k5 ) = 2222* (0.925*1.08*1*1*0.94)= 2088.61 A

where

k1 is factor for load combination relating to conductivity: 0.925

k2 is factor for other air and conductor temperature: 1.08

k3 is factor for load reduction in long side or length rectangular busbars: 1

k4 is factor for load reduction due to additional skin effect: 1

k5 is factor for influences specific to location: 0.94

Ic is designed continuous current of conductor: 1600 A

Current rating of conductor is greater than Ic so 80/6 mm AL Alloy Conductor is OK


3-2. checking for short-circuit thermal strength

For initial temperature of conductor T2 =85 OC and ultimate temperature Te =200 OC ,Sthr is given by fig 4-16 it will obtain:

Sthr= 81.5 A/mm2

The thermal equivalent short time current is :

KA197.32nmII 3Kth =+=

m (factor for the heat effect of the DC component (fig 4-1 5a) )=0.015

n(factor for the heat effect of the AC component (fig 4-I 5b))=1

Ith=32197 A

)densityCurrent thstandcircuit wi short time Rated(Sth ( )kkrthr T/TS= = 81.5 A/mm2

S (short circuit current density) S

th

AI

= = 23.09 A/mm2

S <Sth , so the conductors have sufficient thermal short-circuit strength

3.3 Checking for the maximum deflection of conductors

Deflection of tube is:

Def = (Q.l 3) / (I .E .Js) =1.91 cm

Where

Q = ms.l.g = 332.34 N


i factor for calculation of the defelection of tubular conductor=185

is 77 for tube supported at both ends

Note i: is 185 for tube one end fix and one freely supported

is 384 for tube fixed at both ends

l is span length : 9 m

Js is second moment of sub-conductor area : 97.89 cm4

The bus is assumed to be ok if Def <(1/150).l

1.91 cm < 6.00 cm


Section4: Calculating of maximum load on insulator top and checking the bending stress [Iran Miniatry of Energy Standard 1202/ / 28 ]

4.1 General

4.1.1 Load combinations

Load Combinations:

Name Type of loading Wind speed (m/s)

Ice thickness (mm)

Ambient Temperature

(Oc) Short circuit

Iw Normal Wind & Ice 25 20 0 -

Hw High Wind 40 0 -10 -

Sw High Wind and short Circuit 28 0 -40 , +60

4.1.2 Wind pressure

Pw = ((Vw.Ch )2 /16).C

where

C is a factor related to Vw & conductor

Ch is a factor due to span height above ground (Ch = (H/10 0.095) =1)


4.1.3 Wind force on conductors

Fw =Pw.Dw .Lw

where

Dw is width of conductor that is against wind direction (considering ice and figure of conductor cross-section)

Lw is wind span related to the specific support

4.1.4 Wind force on insulators (transferred to top)

Fi = Pw . Di . Li /2

where

Di is diameter of insulator

Li is the length of insulator

4.1.5 Wind force on A-frame ( transfer to insulator top)

Fa = Pw. Da. La. (Ha / 2 + Li) / Li

Where

Da is diameter of A-frame conductor

La is length of two legs of A-frame.

Ha is Height of A-frame


4.2 Loading of normal wind and ice

Total force on top of insulator is;

Fiw =Fw + Fi

where

Fi is wind force to insulator due to iw

Fw is wind force to conductor due to iw

4.3 Loading of high wind

Fhw=Fw+ Fi

Where

Fi is wind force to insulator due to HW

Fw is wind force to conductor due to HW

4.4 Loading of high wind and short-circuit

4.4.1 Calculation of electromagnetic forces 4.4.1.1 Calculation of peak-force between the main conductors during a three-phase short-circuit

The maximum force acts on the central main conductor during a three-phase short circuit is given by:

Fm= =ami

23

.2

23p0

πµ

3093.03 N


Where

ip3 = √2 ( k. Ik3)

am= a for circular cross-sections;

am = a / k12 for rectangular cross-section

k12 shall betaken from TEC 865-1, page 73, fig 1, with ais=a, b= bm and d=dm

4.4.1.2 Calculating of peak value of forces between coplanar sub-conductors

The maximum force acts on the outer sub-conductors and is between two adjacent connecting pieces given by:

Fs = (ito / 2it). ((ip3 In) A 2]. Is / as

Where

I /as = I /a12 + I / a13 + ... + 1/am for circuit cross-sections

IlaIs=k12

k shall be taken from IEC 865-1,page 73, fig 1

The relevant natural frequency of a conductor can be calculated from 5.4.1.3

4.4.1.3 Calculation of relevant natural frequency The relevant natural frequency of a conductor can be calculated from:

==s

s2c m

J.EIcf γ

Where

c is 1, for conductors consisting of single cross-section

c shall be taken from graph b or c of fig 3 (page 77), if main conductor is composed of sub conductors of rectangular cross-section

c is dependent of the type and number of supports and is given in table 3 (page 67(


Js is (b. d3) / 12 for rectangular cross-sections

Js is (π/ 64). [D4 - (D - 2s)4] for circular cross-sections

For the calculation of sub-conductor stress, taking relevant natural frequency into account, the following equation shall be used

==s

s2

scs m

J.E.

I65.3f

.

4.4.1.4 The factors VF, VD, VDs and Vrs The factors VF, Yd, and Yds as f of the ratio fc /f and fcs /f , where fis the system frequency, shall be taken from fig 4 (page 79).

For three-phase automatic resoling, the factors Yr and Vrs shall be taken from fig 5 (page 81) in other cases Yr = 1, Yrs=1

4.4.1.5 Calculation of stresses in rigid conductors and forces on supports 4.4.1.5.1 Calculation of stresses in rigid conductors

The bending stress caused by the forces between main conductors is given by:

σm = Vσ. Vr. b (Fm. I / 8Z)

The bending stress caused by the forces between sub-conductors is given by:

σs =Vσs. Vrs. (Fs. ls) / 16Zs

Where

Zs= b. d2 / 6 for rectangular cross-sections

Zn.= b. d2 / 6 for rectangular cross-sections

Z= J / (D /2) for circular cross-sections

b is a factor depending on the type and the number of supports, and shall be taken from table 3 (page 67)


4.4.1.5.2 Permitted conductor stress A single conductor is assumed to withstand the short-circuit forces when:

σm< q. Rp0.2min

Where

q Shall be taken from table 4 (page 69)

When a main conductor consists of two or more sub-conductors the total stress in the conductor is given by:

σt = σm + σs

The conductor is assumed to withstand the short-circuit forces when:

σt < q. Rp0.2min

It is necessary to verify that the short-circuit does not affect the distance between sub-conductors too much, therefore a value

σs < RpO.2min

is recommended

.

Calculation of forces on supports of rigid conductors (due to sc) The dynamic force Fd shall be calculated from:

Fd = Vf. Vr. a. Fm

Where

a is dependent on the type and number of supports and shall be taken from table 3 (page 67)

4.4.2 Total forces on supports of rigid conductors (due to sw)

Fsw =Fd + Fw + Fi


4.5 Maximum forces on supports Fmax= max (Fiw, Fhw, Fsw)

4.6 Compound type configurations 4.6.1 General In this part, forces on top of the support (insulator) due to different loading combinations in a compound configuration type will be calculated. The bases are just what described in 5.1 through 5.5. Forces in each load combination are dependent on wind direction.

4.6.2 Forces on support 4.6.2.1 Normal wind and ice Fiw =F | dF/dAngle =0

Where

F2= [ (Fwbus + Fi + Fa) Sin (Ang) ] 2 + [ (Fwbay + Fi + Fa) Cos ( Ang)]2

F’wbus = Fwbus. (Li + Ha) / Li

Fwbus is wind force on bus conductor

Fwbay is wind force on bay conductor

Angle is wind direction

4.6.2.2 High wind Like 5.6.2.1


4.6.2.3 High wind and short circuit Fsw = F | dF/dAngle =0

Where

F= [ + ( F’wbus + Fl + Pa) Sin (Angle)]2 + [ + ( Fwbay + Fl + F’a) Cos (Angle)]2 0.5

F’dbus = Fdbus. (Li + Ha) / Li

Fdbus is SC forces on bus conductor

Fdbay is SC forces on bay conductor

4.6.3 Maximum forces on supports Fmax = max (Fiw, Fhw ,Fsw)


5: Calculations Results

5-1. Dimensioning for the Continues Current Carrying of Conductor

Conductor Ic [A] K1 K2 K3 K4 K5 It[A] Ir[A] Result

Tube 80/6 1600 0.925 1.08 1 1 0.94 2088.61 2222 OK 5-2. checking for short circuit thermal strength

Conductor IK3 [kA] m n Ith[A] Sthr[A/mm2] Amin[mm2] n.As[mm2] Result

Tube 80/6 31.5 1 0.04 32197.8 81.65 394.7 1993.1 OK 5-3. checking for Deflection

Conductor L[m] Q [N] i J[cm4] def[cm] L/150[cm] Result

Tube 80/6 9 332.34 185 97.89 1.91 6 OK Tube 80/6 2.5 92.3 185 97.89 0.01 1.67 OK

5-4. Loading

Conductor L[m] Fhw[N] Fsw[N] Fhe[N] Fse[N] Fmax[N] Fsupport[N] Sf(dyn) Sf(st) Result

Tube 80/6 9 Tube 80/6 8 Tube 80/6 2.5 12 1625 0.925 1.13 1 1 OK


5-5. Detail of Loading

Conductor Load Type Vw[m/s] C Ch Fw[N/m] F'i [N] Fd[N] F[N] Result

Tube 80/6 – 9 Hw 40 1.2 1 706.32 94.9 - Tube 80/6 – 8 Iw 25 1.2 1 413.86 56.8 - Tube 80/6 – 2 Sw 28 1.2 1 519.14 45.4 3374.24 OK

5-6. Detail of short circuit effects

Conductor Ip3[A] L[m] am [m] Fm[N] γ Js [cm4] fc[Hz] fc/f VF Vσ

Tube 80/6 – 9 80631.4 9 3.0 3374.24 2.45 97.89 4.08 0.0816 0.532 0.47 Tube 80/6 – 8 80631.4 8 2.5 3599.19 2.45 97.89 5.17 0.103 0.593 0.53 Tube 80/6 – 2.5 80631.4 2.5 2.5 1124.75 2.45 97.89 52.89 1.06 1.8 1

Detail of short circuit effects (CONTINUED)

Conductor …. Vr β Zs[cm3] Fd [N] σm[N/mm2] σt [N/mm2] q Result

Tube 80/6 – 9 …. 1.67 0.73 2.398E-05 1872.79 91.37 91.37 1.37 OK Tube 80/6 – 8 ….. 1.61 0.73 2.398E-05 2144.15 93.09 93.09 1.37 Ok Tube 80/6 – 2.5 …. 1 0.73 2.398E-05 1265.34 10.7 10.7 1.37 OK

5-7 Force on Support

Conductor Support C Ch Fw[N/m] F'i [N] Fd[N] F[N] Result

Tube 80/6 – 9 DS-PI_DS Tube 80/6 – 8 PI-DS Tube 80/6 – 2.5 PI-DS 1625 0.925 1.13 1 1 OK


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Rigid Bus Work