123

Following minimum clearances shall be maintained :System operating voltage 400 220 132 33

Lightning Impulse Withstand Level (LIWL) 1550 1050 650 170

Switching Impulse Withstand Level (SIWL) 1050

Power frequency voltage 630 460 275 80

Minimum phase to earth clearance mm 3500 2100 1300 320Minimum phase to phase clearance mm 4000 2100 1300 320Minimum sectional clearance mm 6500 5000 4000 3000Phase to phase distance between flexible conductor

Levels in the switchyard (Height from Plinth)System operating voltage 400 220 132 33

Plinth Level mFirst Level - Equipment Connection mSecond Level - Strung Bus - Busbar mThird Level - Strung Bus - Jack Bus mFourth Level - Shield wire m

CASE 1Voltage 420

Minimum Phase to phase clearance = m 4Centre-line distance between main conductor mid points m 7Span length m 100.00Height of conductor above ground m 15.3Conductor Type TARANTULLANo. of sub-conductor of a main conductor Nos. 2Current Carrying capacity A 2570Suitability to carry Short Circuit Current Suitable

m 0.45

m 6.00

kg 1000

No. of Disc Insulator Strings Nos. 4No. of disc in one string Nos. 30

kg 467.9

SAG

Annexure 1Sag-Tension Calculation, Sag Tension Chart and

Short Circuit Forces For Structure and Foundation Design

kVRMS

kVPEAK

kVPEAK

kVRMS

kVRMS

The above levels in switchyard will ensure adequate phase to phase, phase to earth clearances for maximum sag & swing, short circuit forces and wind forces.

kVRMS

Effective distance between Sub-conductor = Center line distance between Sub-conductor

Spacer Span = Centeline distance between connecting pieces or between one connecting piece and the adjacent support

Static Tensile force per (sub)conductor in flexible conductor per phase under minimum temperature and maximum wind condition

Static Tensile force PER CONDUCTOR in flexible conductor per phase under minimum temperature and NO wind condition (CIVIL INPUT)



Net Current Carrying Capacity of the BUNDLE = A 2570Suitability to carry Short Circuit Current Suitable



SAGTotal maximum sag = m 2.053Sag due to conductor m 1.296Sag due to Insulator m 0.757Height of jack bus/equipment below conductor m 8.3

m 0.947

SWINGTotal maximum swing of conductor = m 0.581Available margin = m 1.969Phase to phase spacing required = m 5.031SHORT CIRCUIT

kg 2000

kg 5514.9

Tensile force during short circuit kg 2772.2Tensile force caused by Pinch effect kg 5514.9Tensile force after short circuit (drop force) kg NAj 1.1

Therefore, available margin calculated between centre lines (excluding sub-conductor spacing, conductor width, equipment clamps etc) =

Static Tensile force in flexible conductor per phase under minimum temperature and maximum wind condition

Therefore design short circuit forceMaximum of following :

Annexure-1

Page 4 of 29

CALCULATION FOR FLEXIBLE CONDUCTORSSystem Parameters

Voltage = 420 kVCurrent Rating = 3000 A

= 40,000 ADuration of the first short circuit current flow = 1 Sec

System frequency = 50 Hz

Wind Parameters

Basic Wind Speed = = 47.00 m/s= 1.07

Terrain Category = 2

= 1.3750

Height of conductor above ground = 15.3000

Spacer, additional concentrated mass= 6.00 m

Spacer weight = 3 kgNo. of spacers in span = 14 nos.

= 0.00 kg

= 0 nos.

Conductor Parameters

Conductor Type = TARANTULLA

Type = AAC

Al Strands = 37

Steel Strands = 0

Al Strand Dia = 0.00523

St strand dia = 0

Cross sec. = 0.000794868 sq.m

Diameter = 0.0366 m

Weight = 2.91 kg/m

= 608200

Young's Modulus (N/sq.m) = 61997961264 N/sq.m.

Drag Coefficient = 1

DC Resistance = 0.0363 Ohm/km

= 3.1452 kg/m

Coefficient of Linear Expansion = 0.000023

= 0.00No. of Disc Insulator Strings = 4.00No. of disc in one string = 30.00

Weight of hardware = 60 kg

Length of Hardware before disc = 0.25 m

Disc Insulator Type = 120 kN

x = 0

Creepage = 440 mm.

Diameter of Disc = 0.28 m

Three phase Initial symmetrical short-circuit current (r.m.s)

Risk Coefficient as per Table 2 of IS: 802, Part 1, Sec 1For all Zones & for Reliability level 1(Voltage level upto 400kV)

Factor to convert 3 second peak gust speed into average speed of wind during 10 minutes period =

Spacer Span = Centeline distance between connecting pieces or between one connecting piece and the adjacent support

Additional concentrated mass (other than spacer) representing the connections of pentagraph isolator ( Mass of one connection)

Nos. of Trapzee Connection of pentagraph isolator in one span

Young's Mod. Kg./Cm2 Kg./Cm2106

Resulting Mass Per Unit length of one sub-conductor

/ oC

length of compensating spring in case of pentograph Isolator

Annexure-1

Page 5 of 29

Weight of Disc = 8.3 kg

Length of Disc = 0.145 m

= 4.600 m

1056.000

Layout Parameters

No. of sub-conductor of a main conductor = 2 nos.

1 T

= 2 T

Span length = 100.00 mBeam width 1.5 m

= 89.3 m

Maximum Ambient Temperature at site = 50.0000

Minimum Ambient Temperature at site = 0.0000

Rise in temperature = 35.0000

Final Temperature after Short Circuit = 200.0000

Radial Thickness of ICE over the conductor 0.0000 cm

Phase to phase spacing provided between phases = 7.0000 m= 0.45 m

Minimum Phase to phase clearance = 4 mHeight of jack bus/equipment below conductor 8.3 m

100000

Total consolidated length of disc Insulator & Hardware in one side of Span

Total consolidated weight of disc Insulator & Hardware in one side of Span

Static Tensile force PER CONDUCTOR in flexible conductor per phase under minimum temperature and maximum wind conditionStatic Tensile force in flexible conductor per phase under minimum temperature and maximum wind condition

Length of conductor carrying short circuit current (Chord length of main conductor in the span)

oC oC oC oC

Thice=


Resultant Spring constant of both support of one Span N/m

CASE – 1.CALCULATION FOR BUSBAR SIZING

Bundle Conductor Considered AAC TARANTULLA

No. of sub-conductor in a BUNDLE 2

1 Check for continuous thermal withstand capacity of the conductor

Required Current Carrying Capacity of the BUNDLED conductor 3000 Amps

Hence as given in Heat balance equation : (1)

where,(2)(3)

(4)(5)

where,I = the conductor current (A)

DC resistance of conductor at 20°C 0.0363

=γ the solar radiation absorption co-efficient Refer IEC 61597 0.50

Refer IEC 61597 900.00D = the conductor diameter (m) 0.0366 m

Refer IEC 61597 5.67E-08

the emmisivity co-efficient in respect to black body Refer IEC 61597 0.60Ambient Temperature in (C) 50 °CFinal Equilibrium Temperature (C) 85 °CAmbient Temperature in (K) 323 KFinal Equilibrium Temperature (K) 358 K

=λ the thermal conductivity of air film in contact with the conductor Refer IEC 61597 0.02585

Nu = Refer IEC 61597 25.7

Refer IEC 61597 1871.6v = wind speed in m/s Refer IEC 61597 1 m/s

System frequency 50 HzThermal Coefficient of Resistivity Refer IEC 61597 4.03E-03 /°Cpermeability (= 1, for non magmnatic material) 1

now,

where, DC resistance of the conductor at operating temperature 85 deg C

K Skin effect coefficientagain,

0.03630.058419

=> 0.073722Next for calculation of skin effect coefficient, a factor X [reference (2)]

=> X = 1.65603

K = 1.03512

=> 0.0763114.74E-05

16.47

21.68

73.12from heat balance equation - 2 we derive,

equation-3

From equation - 3, we obtain=> Current Carrying Capacity of the CONDUCTOR = 1285 Amps

Net Current Carrying Capacity of the BUNDLE = 2570 Amps

The current which a conductor can carry for a given temperature rise depends upon the rate at which the produced heat is dissipated. In case of an overhead conductor, the dissipation will be effected by radiation and by convection.

Pj + Psol = Prad + Pconv

Pj = RT I², Heat generated by Joule effectPsol = D Sγ i, Solar Heat Gain

Prad = s π D Ke (T24 – T1

4), Radiation Heat Loss

Pconv = λ Nu (T2 – T1) π, Convection Heat Loss

Rdc(20°C) Ω/kmRT= the electrical resistance of conductor at a temperature T (Ω/m)

Si = the intensity of solar radiation (W/m²), W/m2

s = the Stefan Boltzmann constant (5.67×10–8 W.m–2K–4) W.m–2K–4

Ke =

T1 =T2 =

W m–1K–1

the Nusselt number = 0.65 Re0.2 + 0.23Re

0.61

Re = the Reynolds number = 1.644×109v D[(T1 + 0.5(T2 – T1)]–1.78

f = =α

μ =

RT = Rdc x K

Rdc

Rdc(85°C) = Rdc(20°C) (1+ ×(Tα 2-20))Rdc(20degC) = Ω/km

Ω/milesRdc(85°C) Ω/miles

For this value of X, the skin effect coefficient K is obtained from Table 4.1.1 of "Power Engineers Handbook"

RT = Ω/milesΩ/m

Psol = D Sγ i

Prad = s π D Ke (T24 – T1

4)Pconv = λ Nu (T2 – T1) π

X=0 .063598×√ μ×fR

dc(850C )

T

solconvrad

R

PPPI

)(max

Check for Short Circuit Current carring capacity of the incoming line conductor

System short circuit current for System 40,000 At Duration of short circuit current 1 sec

Min. reqd. area of cross-section of conductor for short time rating

where,

Initial temp. of conductor before short ckt 85

Final temp of conductor after short ckt 200G Conductivity of conductor Reference: (2)-(c ) 40 %

Clause 5.2.2

0.878063

Minimum required cross-sectional area 566.49

Cross-section of each CONDUCTOR 794.87

Overall Cross-section of BUNDLE 1589.74

Suitability to carry Short Circuit Current Suitable

I"k3

Reference: (2)-(c)Clause 5.2.2 Eq (3)

in2

(Amin) =

Ti0C

Tf0C

in2

mm2

mm2

mm2

3

6

"

( 20 (15150 / ))0.144 10 log

( 20 (15150 / ))

k

f

i

I X t

T GX

T G

3

6

"

( 20 (15150 / ))0.144 10 log

( 20 (15150 / ))

k

f

i

I X t

T GX

T G

Annexure-1

Page 8 of 29

CASE – 3.

Short Circuit Force CalculationsCalculation are based on IEC 865-1 (1993)

FOR FLEXIBLE CONDUCTORS

= = 2 T

= 19620 N

= Three phase Initial symmetrical short-circuit (r.m.s) = 40000 A

= No. of sub-conductor of a main conductor = 2 nos.

= Sub–conductor mass per unit length = 2.91 kg/m

Diameter of flexible conductor = 3.660E-02 m

Cross Section of one sub-conductor = 7.949E-04

= = 6 m

= 0 kg

= 0 nos.

Spacer weight = 3 kg

No. of spacers in span = 14 nos.

Resulting Mass Per Unit length of one sub-conductor

= 3.145 kg/m

= Span length = 100 m

= 0

= = 4.6 m

= = 0.45 m

= Centre-line distance between main conductor mid points = 7 m

Beam Width 1.5 m

= = 89.3 m

S = = 100000 N/m

= Young's modulus = 6.2E+10 N/m2

= Duration of the first short circuit current flow = 1 Sec

= Gravitational Constant = 9.81

= Constant = 5.00E+07

f = System frequency = 50 Hz

Material Constant = 2.70E-19

Magnatic Constant, permeability of Vacuum = H/m

Static Tensile force in flexible conductor per phase under minimum temperature and maximum wind condition

=

= m2

Spacer Span = Centeline distance between connecting pieces or between one connecting piece and the adjacent supportAdditional Concerated mass (other than spacer) representing the connections of pentagraph isolator ( Mass of one connection)Nos. of Trapzee Connection of pentagraph isolator in one span

=

msc' =

length of compensating spring in case of pentograph of IsolatorTotal consolidated length of disk Insulator & Hardware in one side of Span


Length of conductor carrying short circuit current (Chord length of main conductor in the span)

Resultant Spring constant of both support of one Span N/m

m/s2

N/m2

Cth =

m4/(A2s)

µo= 4π×10–7

fin

l

dcl

lc

a

A s

ms'

d s

ms

ls

as

F st

E

I k 3 ¿¿¿n

T k 1

gn

msc '=m s '+ncmc

nlc+nsm s

nlc

nc=

mc=

ns=

lsp=

Annexure-1

Page 9 of 29

CASE – 3.

Short Circuit Force Calculations

= 30.62 N/m

= 0.4962

From eqn. 21, Angular direction of the force,

= 26.3888 Deg

From eqn. 22, Equivalent Static Conductor Sag at mid span,

= 3.9315 m

From eqn. 23, Period of Conductor oscillation,

= 3.5577 Sec.

= 3.4125 Sec.

From eqn. 25, The stiffness norm

= 1.18E-07

Where the actual Young's modulus is given by

= 3.50E+10

From eqn.19, Characteristic electromagnetic force per unit length on flexible main conductors,

From eqn. 20, the ratio of electromechanical force on a conductor under short circuit condition to gravity,

From eqn. 24, Resulting period of the conductor oscillation during the short circuit current flow,

N/m2

F '=μ0

2π×0 .75×¿¿¿¿

r= F'

n×msc' ×gn

δ 1= tan−1r

bc=n×msc

' ×gn×l2

8 F st

T=2 π √0 . 8bc

gn

T res=T

4√1+r2 [1− π 2

64 [ δ1

900 ]2]

N=1Sl

+1

nEsA s

E s=E [0. 3+0 . 7×sin( F st

nAs σ fin

×900)] forF st

nA s

≤σ fin

E forF st

nA s

>σ fin

Annexure-1

Page 10 of 29

CASE – 3.


From eqn.28,The stress factor of the main conductor is given by :

= 1.78E+00

From eqn. 29, Swing out angle at the end of the short-circuit current flow

= 33.4389 Deg

From eqn. 30, factor

= 0.7266

From eqn. 31, Maximun swing out angle

= 53.3986 Deg

From eqn. 32, Load Parameter

= 0.3490

From eqn. 33, The factor , can be calculated as a real solution of the equation,

0.00

= 0.7453

From eqn. 34, the tensile force is given by,

= 2772.1832 Kg

ζ=(ngnmsc

' l )2

24 Fst3 N

δ k=δ1 [1−cos(T k 1

T res

×3600)] for 0≤T k 1

T res

≤0 . 5

2δ1 forT k1

T res

>0 . 5

χ=1−rsin δk for 0≤ δ k≤90

1−r for δ k>90

δm=1.25 arccos χ for 0 .766≤ χ≤110+arccos χ for −0 . 985≤ χ<0 . 766

1800 for χ<−0. 985

ϕ= 3 (√1+r2−1) for T k1≥T res

4

3 (r sin δ k+cos δk−1) for T k 1<Tres

4

ψ

ϕ2ψ3+ϕ (2+ζ )ψ2+ (1+2 ζ )ψ−ζ (2+ϕ )=0 0≤ψ≤1

ψ

F t= F st (1+ϕψ ) for n=1 for Single conductor

1 . 1F st (1+ϕψ ) for n≥2 for bundle conductor

Annexure-1

Page 11 of 29

CASE – 3.


From eqn. 35, the tensile force after short circuit caused by drop (drop force)

= NAKg

From Eqn. 36, The elastic expansion is given by,

= 8.9363E-04

From Eqn. 37, The thermal expansion is given by,

= ###

= ###

= 1.05

= 2.0531

= ###

Horizontal Span Displacement bh and Min. air clearance amin

From Eqn. 38, Factor CD is given by,

From Eqn. 39, Factor CF is given by,

From Eqn. 41, Maximum Horizontal Displacement within a span with strained conductors is given by,

From Eqn. 42, the distance between the mid points of the two main conductors during a line-to-line two-phase short circuit,

F f=1 . 2F st √1+8 ζδm

1800for r>0 .6 δm≥700

ε ela=N (F t−F st )

ε th=c th [ I k 3 over ital nA rSub size 8s right ] rSup size 82 T rSub size 8 ital res /4 matrix # ital for matrix # T rSub size 8k1 >= T rSub size 8 ital res /4 ## matrix ##c rSub size 8 ital th left [ I rSub size 8k3 rSup size 8

nA s]2

T k 1 for T k 1<T res /4 ¿

CD=√1+ 38 [ l

bc ]2

(εela+εth)

CF= 1 . 05 for r≤0 .80 . 97+0.1 r for 0 . 8<r<1 . 8

1. 15 for r≥1 .8

bh=CFCDbc sin δ1 for δm≥δ1

CFCDbc sin δm for δm<δ1

amin=a−2bh

Annexure-1

Page 12 of 29

CASE – 3.


SUBCONDUCTORS ARE NOT EFFECTIVELY CLASHING

= 4.2760

= 12.2951

= 1.5721

=

0.2133

From eqn. 45, Short circuit current force

= 7862.0971Kg

From eqn. 47 & 48, Strain factor characterizing the contraction of bundle condoctor

= 0.7313

= 2.1267

= 1.1083

If j > 1 : The sub-conductor clash

If j < 1 : The sub-conductors reduce their distances but do not clash

Sub-conductor are considered to clash effectively if the clearence as between the mid -points of adjacent sub-conductors, as well as the distance ls between 2 adjacent spacers fulfil either equation (43) or (44)

From equation 46, the factor v1 is given by,

The factor as/ds is equal to

Factor V2 can be given by fig. 8 as function of V1

V2

Factor V3 can be given by

as /ds≤2 .0 and ls≥50as (43 )as /ds≤2. 5 and ls≥70as (44 )

v1=f1

sin1800

n √ (as−d s)msc'

μ0

2π¿¿

¿¿¿

ε pi=0 .375 nFv ls3 N

(as−d s)3 (sin

1800

n )3

ε st=1 . 5F st l s2

N

(as−d s)2 (sin

1800

n )2

j=√ ε pi

1+εst

Fv=(n−1 )μ0

2π¿¿

V 3=

ds

as

sin180n

× √(as

d s)−1

arctan √(as

ds)−1

Annexure-1

Page 13 of 29

CASE – 3.


SUBCONDUCTORS ARE CLASHING

TENSILE FORCE IN CASE OF CLASHING SUB-CUNDUCTORSIf j ≥ 1, From eqn. 50, the tensile force is given by

= 5514.8618 Kg

From equation (51), ξ is given by the real solution of

0.00ξ . 1.0825801

From eqn. 52,

= 1.1872

From eqn. 53,= 1.13E+01

= 5514.8618 Kg

=

Therefore design force in kg (Maximum of Ft, Ff & Fpi)

v4=ηas−d s

as−η (as−ds)

F pi=Fst (1+ve

εstξ)

ξ3+εst ξ2−ε pi=0

ve=12+¿¿

v4=as−ds

d s

Annexure-2

Page 14 of 29

CASE – 1.

CALCULATIONS FOR SAG AND SWING OF STRUNG CONDUCTOR

System Voltage V = 420 kV

1 Basic Wind Speed = 47.00 m/sAs per IS 802: 1995, basic wind speed is based on gust velocity averaged over a short time interval of 3s.

Wind Zone as per IS: 802, Part 1, Sec 1 4

1.07

Terrain Category 2Terrain roughness coeficient (For different Terrain Category) = 1.0000

1.3750

34.1818 m/s36.5745 m/s

Height of conductor above ground 15.3000 metreGust Response Factor for conductor Gc = 1.9837

2 Pc = 81.8164

Weight of individual spacer = WSP = 3.0000 kgfSeparation between two spacers = SSP = 6.0000 metreWeight of sub-conductor per metre= W1 = 2.9100 kgf/mNo. of subconductors - single/twin/tri/quad = r = 2Diameter of sub-conductor = d = 3.6600 cmCross Sectional Area of sub-conductor = A = 7.9487Modulus of Elasticity of sub-conductor = ey = 608200.0000Coefficient of Linear Expansion = a = 2.3000E-05Maximum Ambient Temperature at site = t1 = 50.0000Minimum Ambient Temperature at site = t2 = 0.0000Rise in temperature = t3 = 35.0000Distance between centre line of beams (for conductor span) = la = 100.0000 mDIameter of disc = di = 0.2800Nos. of Insulator Strings in one side of conductor span (Single /Double) Ni = 4Weight of complete string insulator including hardware 1056.0000Weight of single string insulator including hardware Wi= 264.0000Length of insulator string = li = 4.6000 mWidth of beam on which conductor is strung = Bw = 1.5000 m

Ten1= 1000.0000 KgDrag Coefficient 1.0000

3 Effect of Ice formation

Radial Thickness of ICE over the conductor 0.0000 cmD1= 3.6600 cm

Increase in Conducter weight due to effect of Ice formation w3= 0.0000 kg/m

4 Total weight of sub-conductor per phase = Wc = W1+WSP/(r*SSP)+w3 = 3.1452 Kg/m

5 Wind Load on conductor as per IS 802 & IS 5613 = Pcc = 5.9401 Kg/m

Vb =

Risk Coefficient as per Table 2 of IS: 802, Part 1, Sec 1For all Zones & for Reliability level 1(Voltage level upto 400kV) = K1 =

K2 = Factor to convert 3 second peak gust speed into average speed of wind during 10 minutes period = K0 =

Meteorological reference wind speed = Vb/K0 VR = Design wind speed = VR x K1 x K2 Vd =

Design Wind Pressure =0.6 x Vd 2 /g kgf/m2

cm2

Kg/cm2

/ oCoC oC oC

WiT=

Tension per SUB-CONDUCTOR under max. wind and min. temp. conditions = Cdc =

Thice=Effective dia of the conductor considering ICE effect = d+2*Thice

Pc * (d/100) * Cdc * Gc =

Annexure-2

Page 15 of 29

CASE – 1.

6 Clear span of conductor = L = la-(2*li)-lt-Bw = 89.3000 m

7 2.1371Loading factor during no wind condition = Q2 = 1.0000

8 SAG CALCULATIONS DUE TO CONDUCTOR

8.1f1 = Ten1/A = 125.8071

0.3957

3.1640E+07

-10192.9819

To solve the following equation:The working stress F at tempature t1+t3, is given by the cubic equation

940491268.11

we get f2 = 304.3152

8.2 Tension in conductor at maximum temperature and no windcondition = Ten2 = f2*A = 2418.9041 Kg

Max sag due to conductor = Wc*L² / (8*Ten2) Sc= 1.2961

9 SAG DUE TO INSULATOR

From Fig 1, we get

Tension per insulator string = Ten2* r / Ni Teni1= 1209.4521

Weight of conductor per String Insulator = Wc*r/Ni Wci= 1.57264.6021 m

U2 = √((li/2)² + (di/2)²) = 2.3043 m

0.0304

0.0608

Taking moments at 'P'

Teni1 * U1 *Sin(φ + δ) = (Wci * L/2) * U1 * Cos(φ + δ) + Wi * U2 *Cos(φ + β)

Simplifying :

TP1 = (Wci*L/2) * U1 * Cos δ + Wi*U2*Cos β - Teni1*U1*Sin δ = 760.8692

TP2 = Teni1*U1*Cos δ + Wci*(L/2)*U1*Sin δ + Wi*U2*Sin β = 5610.2697

0.1348

Sag due to Insulator = Si = U1*Sin(φ + d) = 0.7569 m

Loading factor during max wind condition = Q1= √(1+(Pcc/Wc)2) =

Under minimum temperature and maximum wind condition condition, the sub-conductor experiences a stress of f1

Kg/cm2

Weight of sub conductor per m per cm2 = S = Wc/A =

GB = (S * L * Q2)2 * ey/24 =

F = f1 - GB * (Q1/(f1*Q2))2 - a * ey * (t1+t3-t2) =

f22 * (f2 - F) - GB =

Kg/cm2

U1 = √(li2 + (di/2)2) =

δ = Tan-1 [(di/2)/li] =

β = Tan-1 [di/li] =

φ = Tan-1(TP1/TP2) =

Annexure-2

Page 16 of 29

CASE – 1.

10 Total maximum sag of moose conductor = Sc+Si = Stot = 2.0530 m

11 Height of conductor above ground 15.3000 m

Height of jack bus/equipment below conductor 8.3000 mMinimum Phase to phase clearance = 4.0000 mTotal maximum sag of moose conductor = Sc+Si = Stot = 2.0530 m

0.9470 m

SWING CALCULATIONS

1 SAG DUE TO CONDUCTOR (under maximum temperature rise in temperature and maximum wind conditions):

1.4450E+08

-1.0193E+04

To solve the following equation:1.11E+10

we get f21 = 1002.7182

Tension in conductor at maximum temperature and maximum windcondition = Ten21 = f21*A= 7970.2854 Kg

0.3934 m


From Fig 1, we get

Teni2 = Ten21*r/Ni 3985.1427

U1 = √(li2 + (di/2)2) = 4.6021 m

U2 = √((li/2)2 + (di/2)2) = 2.3043 m

0.0304

0.0608


TenI2 * U1 * Sin(φ + δ) = (WcI*L/2) * U1 * Cos(φ+ δ) + Wi * U2 *Cos(φ+ β)

Simplifying :

TP1 = (WcI*L/2) * U1 * Cos δ + Wi*U2*Cos β - TenI2*U1*Sin δ = 372.2725

TP2 = TenI2*U1*Cos δ + Wc*(L/2)*U1*Sin δ + Wi*U2*Sin β = 18378.4466

0.0203

SAG due to Insulator = Si1 = U1*Sin(φ + δ) = 0.2331

Therefore, available margin excluding sub-conductor spacing =

SGB = (Q1 * S *L)2 * ey/24 =

SF = (f1-SGB/f12) - a*ey*(t1+t3-t2) =

f212 * (f21 - SF) - SGB =

Kg/cm2

SAG due to conductor = Sc1 = Wc*L2/(8*Ten21) =

δ = Tan-1 [(di/2)/li] =



Annexure-2

Page 17 of 29

CASE – 1.

3 SWING DUE TO CONDUCTOR (SWC)

Wind load acting on span = Pcc * L*r = 1060.9103 KgWeight of conductor = Wc * L *r= 561.7260 Kg

From Fig 3:Tan Y1 = (r*Pcc * L/r*Wc * L) = 1.8887

1.0838

SWC = Sc1* Sin Y1 = 0.3476 m

4 SWING DUE TO INSULATOR (SWI)

100.1105 kg

1.5708

SWI = Si1 * Sin Y2 = 0.2331 m

5 Total Swing = Swtot = SWC + SWI = 0.5808

6 Phase to phase spacing provided between phases = 7.0000 mSubconductor spacing = 0.4500 mMinimum Phase to phase clearance = 4.0000 mPhase to phase spacing required = 5.0308 mTherefore, available margin = 1.9692 m

Y1 = Tan-1 (Pcc/Wc) =

Wind load on insulator = Wpi = (Pc/2)*di*li *Gc*Cdc=

Y2 = Tan-1 (Ni*Wpi/Ni*Wi) =

CASE - 2 ANNEXURE - 2

Page 18 of 29

CASE 2.

SAG TENSION CHART

NAME OF PROJECT:

GENERAL DATA

CONDUCTOR TYPE: TARANTULLA

NAME OF BAY: 0.35

SPAN LENGTH (Centre Lime distance between beams): 100.00 m

MINIMUM STRINGING TEMPERATURE: 10.00 Deg C

TEMPERATURE RANGE FOR DRAWING STRINGING CHART:

MINIMUM: 0.00 °C MAXIMUM: 50.00 °C

WIND PRESSURE ON CONDUCTOR PER M RUN: 0.00 kg/m

CONDUCTOR DATA

OVERALL DIA OF THE CONDUCTOR: 3.660 cm

CROSS SECTIONAL AREA OF CONDUCTOR: 7.949 cm²

WEIGHT OF CONDUCTOR PER METER: 2.910 kgf/m

COEFFICIENT OF LINEAR EXPANSION: 0.00002300 per °C

NO. OF CONDUCTORS PER PHASE: 2

TENSION PER CONDUCTOR UNDER MIN TEMP & MAX WIND CONDITION: 1000.00 kg

MODULUS OF ELASTICITY: 608200.00 kg/cm2

INSULATOR AND TURNBUCKLE DATA

NUMBER OF INSULATOR STRING(S): 4.00

LENGTH OF INSULATOR: 4.600 m

DIAMETER OF INSULATOR: 0.280 m

WEIGHT OF INSULATOR: 264.000 kg

CASE - 2 ANNEXURE - 2

Page 19 of 29

RESULTS

MAXIMUM CLEAR SPAN OF CONDUCTOR: 89.30 m

S.NO. TEMPERATURE TOTAL SAG

1 42 587.17 3.8 1.68 5.48

2 12 624.61 3.57 1.59 5.163 14 621.89 3.59 1.6 5.194 16 619.21 3.6 1.61 5.215 18 616.56 3.62 1.61 5.236 20 613.94 3.63 1.62 5.257 22 611.36 3.65 1.62 5.278 24 608.81 3.66 1.63 5.299 26 606.28 3.68 1.64 5.31

10 28 603.79 3.69 1.64 5.3311 30 601.33 3.71 1.65 5.3612 32 598.9 3.72 1.65 5.3813 34 596.5 3.74 1.66 5.414 36 594.13 3.75 1.66 5.4215 38 591.78 3.77 1.67 5.4416 40 589.46 3.78 1.68 5.46

TENSION PER CONDUCTOR

CONDUCTOR SAG

INSLULATOR SAG

Type Al Strand Steel StrandsAl Strand DiaSt strand di Cross sec. Diameter Weight

sq.m m kg/kmBEAR ACSR 30 7 3.350E-03 3.350E-03 3.2612E-04 2.3450E-02 1.229 7.870E+05 8.02E+10 1 0.1093 0.00001843BEAVER ACSR 6 1 3.990E-03 3.990E-03 8.7525E-05 1.1970E-02 0.303 8.090E+05 8.25E+10 1 0.3825 0.00001931BUX ACSR 30 7 4.500E-03 4.500E-03 5.8846E-04 3.1500E-02 2.196 7.870E+05 8.02E+10 1 0.0569 0.00001843CAT ACSR 6 1 4.500E-03 4.500E-03 1.1133E-04 1.3500E-02 0.385 7.350E+05 7.49E+10 1 0.3007 0.00001931COYOTO ACSR 26 7 2.540E-03 1.900E-03 1.5159E-04 1.5860E-02 0.521 7.870E+05 8.02E+10 1 0.3035 0.00001954DEER ACSR 30 7 4.270E-03 4.270E-03 5.2984E-04 2.9890E-02 1.977 7.870E+05 8.02E+10 1 0.0673 0.00001843DOG ACSR 6 7 4.720E-03 1.570E-03 1.1854E-04 1.4150E-02 0.394 7.350E+05 7.49E+10 1 0.2733 0.00001992FERRET ACSR 6 1 3.000E-03 3.000E-03 4.9480E-05 9.0000E-03 0.171 8.090E+05 8.25E+10 1 0.6766 0.00001931FOX ACSR 6 1 2.790E-03 2.790E-03 4.2795E-05 8.3700E-03 0.1489 8.090E+05 8.25E+10 1 0.7822 0.00001931GOAT ACSR 30 7 3.710E-03 3.710E-03 3.9998E-04 2.5970E-02 1.492 7.870E+05 8.02E+10 1 0.0891 0.00001843GOPHER ACSR 6 1 2.360E-03 2.360E-03 3.0620E-05 7.0800E-03 0.106 8.090E+05 8.25E+10 1 1.0933 0.00001931GUINEA ACSR 12 7 2.920E-03 2.920E-03 1.2724E-04 1.4600E-02 0.59 1.055E+06 1.08E+11 1 0.2630 0.00001584LARK ACSR 30 7 2.920E-03 2.920E-03 2.4777E-04 2.0440E-02 0.922 7.870E+05 8.02E+10 1 0.1350 0.00001843LEOPARD ACSR 6 7 5.280E-03 1.760E-03 1.4840E-04 1.5840E-02 0.493 7.730E+05 7.88E+10 1 0.2184 0.00001954LION ACSR 30 7 3.180E-03 3.180E-03 2.9386E-04 2.2260E-02 1.097 7.870E+05 8.02E+10 1 0.1213 0.00001843LYNX ACSR 30 7 2.790E-03 2.790E-03 2.2620E-04 1.9530E-02 0.844 7.870E+05 8.02E+10 1 0.1576 0.00001843MINK ACSR 6 1 3.660E-03 3.660E-03 7.3646E-05 1.0980E-02 0.255 8.090E+05 8.25E+10 1 0.4546 0.00001931MOOSE ACSR 54 7 3.530E-03 3.530E-03 5.9699E-04 3.1770E-02 2.002 6.860E+05 6.99E+10 1 0.0547 0.00001991OTTER ACSR 6 1 4.220E-03 4.220E-03 9.7907E-05 1.2660E-02 0.339 8.090E+05 8.25E+10 1 0.3419 0.00001931PANTHER ACSR 30 7 3.000E-03 3.000E-03 2.6154E-04 2.1000E-02 0.976 7.870E+05 8.02E+10 1 0.1363 0.00001843RABBIT ACSR 6 1 3.350E-03 3.350E-03 6.1699E-05 1.0050E-02 0.214 8.090E+05 8.25E+10 1 0.5426 0.00001931RACCOON ACSR 6 1 4.090E-03 4.090E-03 9.1968E-05 1.2270E-02 0.318 8.090E+05 8.25E+10 1 0.3640 0.00001931SHEEP ACSR 30 7 3.990E-03 3.990E-03 4.6263E-04 2.7930E-02 1.726 7.870E+05 8.02E+10 1 0.0770 0.00001843SPARROW ACSR 6 1 2.670E-03 2.670E-03 3.9193E-05 8.0100E-03 0.135 8.090E+05 8.25E+10 1 0.8540 0.00001931SQUIRREL ACSR 6 1 2.110E-03 2.110E-03 2.4477E-05 6.3300E-03 0.085 8.090E+05 8.25E+10 1 1.3677 0.00001931TIGER ACSR 30 7 2.360E-03 2.360E-03 1.6185E-04 1.6520E-02 0.604 7.870E+05 8.02E+10 1 0.2202 0.00001843WEASSEL ACSR 6 1 2.590E-03 2.590E-03 3.6880E-05 7.7700E-03 0.128 8.090E+05 8.25E+10 1 0.9077 0.00001931WOLF ACSR 30 7 2.590E-03 2.590E-03 1.9494E-04 1.8130E-02 0.727 7.870E+05 8.02E+10 1 0.1828 0.00001843ZEBRA ACSR 54 7 3.180E-03 3.180E-03 4.8448E-04 2.8620E-02 1.623 6.860E+05 6.99E+10 1 0.0674 0.00001991ANT AAC 7 3.100E-03 5.2834E-05 9.3000E-03 0.144 6.187E+05 6.31E+10 1 0.5419 0.000023BLUE BOTTLE AAC 7 3.660E-03 7.3646E-05 1.0950E-02 0.201 6.187E+05 6.31E+10 1 0.3887 0.000023BUTTERFLY AAC 19 4.650E-03 3.2266E-04 2.3250E-02 0.886 6.082E+05 6.20E+10 1 0.0892 0.000023CATTER-PILLAR AAC 19 3.530E-03 1.8595E-04 1.0590E-02 0.511 6.187E+05 6.31E+10 1 0.1547 0.000023CHAFER AAC 19 3.780E-03 2.1322E-04 1.8900E-02 0.586 6.082E+05 6.20E+10 1 0.1349 0.000023CLEGG AAC 7 4.170E-03 9.5600E-05 1.2510E-02 0.261 6.187E+05 6.31E+10 1 0.2995 0.000023COCKROACH AAC 19 4.220E-03 2.6575E-04 2.1100E-02 0.73 6.082E+05 6.20E+10 1 0.1083 0.000023EARWIG AAC 7 3.780E-03 7.8555E-05 1.1340E-02 0.215 6.187E+05 6.31E+10 1 0.3644 0.000023FLY AAC 7 3.400E-03 6.3554E-05 1.0200E-02 0.174 6.187E+05 6.31E+10 1 0.4505 0.000023GRASS HOPPER AAC 7 3.910E-03 8.4051E-05 1.1730E-02 0.23 6.187E+05 6.31E+10 1 0.3406 0.000023IRIS AAC 7 2.480E-03 3.3814E-05 7.4400E-03 0.092 6.187E+05 6.31E+10 1 0.8467 0.000023LADY BIRD AAC 7 2.790E-03 4.2795E-05 8.3700E-03 0.117 6.187E+05 6.31E+10 1 0.669 0.000023LOCUST AAC 19 5.360E-03 4.2872E-04 2.6800E-02 1.176 6.082E+05 6.20E+10 1 0.0671 0.000023MAYBUG AAC 37 4.090E-03 4.8611E-04 2.8630E-02 1.343 5.976E+05 6.09E+10 1 0.0593 0.000023MOTH AAC 19 5.000E-03 3.7306E-04 2.5000E-02 1.025 6.082E+05 6.20E+10 1 0.0771 0.000023PANSY AAC 7 2.780E-03 4.2489E-05 8.3400E-03 0.116 6.187E+05 6.31E+10 1 0.6738 0.000023ROSE AAC 7 1.960E-03 2.1120E-05 5.8800E-03 0.058 6.187E+05 6.31E+10 1 1.3556 0.000023SCORPION AAC 37 4.270E-03 5.2984E-04 2.9890E-02 1.464 5.976E+05 6.09E+10 1 0.0544 0.000023SPIDER AAC 19 3.990E-03 2.3757E-04 1.9950E-02 0.652 6.082E+05 6.20E+10 1 0.1205 0.000023TARANTULLA AAC 37 5.230E-03 7.9487E-04 3.6600E-02 2.91 6.082E+05 6.20E+10 1 0.0363 0.000023WASP AAC 7 4.390E-03 1.0595E-04 1.3170E-02 0.29 6.187E+05 6.31E+10 1 0.2702 0.000023SHIELD WIRE 7/9 SW 7 3.660E-03 7.3646E-05 1.2180E-02 0.583 1933000 1.97E+11 1.2 0.000012SHIELD WIRE 7/8 SW 7 9.0500E-05 1.2180E-02 0.706 1933000 1.97E+11 1.2 0.000012

Young's Mod. Kg./Cm2

Young's Modulus (N/sq.m)

Drag Coefficien

t

DC Resistance

Coefficient of Linear Expansion

Kg./Cm2

BLANK SW

disc insulator x120 440 0.28 8.3 0.145xx

Creepage

Diameter of Disc

Weight of Disc

Length of Disc

SKIN EFFECT TABLE

Sl No. X K1 0.00 1.000002 0.10 1.000003 0.20 1.000014 0.30 1.000045 0.40 1.000136 0.50 1.000327 0.60 1.000678 0.70 1.001249 0.80 1.00212

10 0.90 1.0034011 1.00 1.0051912 1.10 1.0075813 1.20 1.0107114 1.30 1.0147015 1.40 1.0196916 1.50 1.0258217 1.60 1.0332318 1.70 1.0420519 1.80 1.0524020 1.90 1.0644021 2.00 1.0781622 2.10 1.0937523 2.20 1.1112624 2.30 1.1306925 2.40 1.1520726 2.50 1.1753827 2.60 1.2005628 2.70 1.2275329 2.80 1.2562030 2.90 1.2864431 3.00 1.3180932 3.10 1.3510233 3.20 1.3850434 3.30 1.4199935 3.40 1.4557036 3.50 1.4920237 3.60 1.5287938 3.70 1.5658739 3.80 1.6031440 3.90 1.64051

Source : Electrical Transmission & Distribution Reference book by Central Station Engineers of Westinghouse Electric CorporationSource : Bureau of Standards Bulletin No. 169 on pages 226-8Source : TNEB HANDBOOK

Source : Electrical Transmission & Distribution Reference book by Central Station Engineers of Westinghouse Electric Corporation

Doc. No. : YN1H300126-403, Rev. A

document.xls Page 24 of 29

n

f 50 Hz as

k 1.81 As per IEC-909 ds

k 1.81 as/ds

1/t 22.557720876 ds/as

t 0.0443307196

g 1.4990796579 Sqrt((as/ds)-1)

3.4103692621

21.417118966

2.2165359823 h

e st

e pi

4.276003

1.572075

A 0.0230400592sqrt[(1-2*ya/as)/2*ya/as]

B 0.6168024783

C 0.2851491873

D 0.3129163279

E 0.0409238672

F 0.2147150177

G 0

V2 1 - A + B - C*[( D + E)*F + G ]

1.5720982655

-0.00002

fTpi v3

2pfTpi

ft

v1 ya

v2 asv3

2*ya/as

asw

fh

Doc. No. : YN1H300126-403, Rev. A

document.xls Page 25 of 29

2

0.45

3.66E-02

12.295081967

0.0813333333

3.3608156699

0.2132866076

0.6291488471

7.31E-01

2.13E+00

9.50E-02

0.0959789734

0.4220219258

1.170275665

0.2878319318

0.3334549187

-6.49E-06

Annexure-2

Page 26 of 29

CASE –

CALCULATIONS FOR SAG AND SWING OF STRUNG CONDUCTOR

System Voltage V = 420 kV

1 Basic Wind Speed = 0.00 m/sAs per IS 802: 1995, basic wind speed is based on gust velocity averaged over a short time interval of 3s.

Wind Zone as per IS: 802, Part 1, Sec 1 1

1.07

Terrain Category 2Terrain roughness coeficient (For different Terrain Category) = 1.0000

1.3750

0.0000 m/s0.0000 m/s

Height of conductor above ground 15.3000 metreGust Response Factor for conductor Gc = 1.9837

2 Pc = 0.0000

Weight of individual spacer = WSP = 3.0000 kgfSeparation between two spacers = SSP = 6.0000 metreWeight of sub-conductor per metre= W1 = 2.9100 kgf/mNo. of subconductors - single/twin/tri/quad = r = 2Diameter of sub-conductor = d = 3.6600 cmCross Sectional Area of sub-conductor = A = 7.9487Modulus of Elasticity of sub-conductor = ey = 608200.0000Coefficient of Linear Expansion = a = 2.3000E-05Maximum Ambient Temperature at site = t1 = 40.0000Minimum Ambient Temperature at site = t2 = 0.0000Rise in temperature = t3 = 0.0000Distance between centre line of beams (for conductor span) = la = 100.0000 mDIameter of disc = di = 0.2800Nos. of Insulator Strings in one side of conductor span (Single /Double) Ni = 4Weight of complete string insulator including hardware 1056.0000Weight of single string insulator including hardware Wi= 264.0000Length of insulator string = li = 4.6000 mWidth of beam on which conductor is strung = Bw = 1.5000 m

Ten1= 1000.0000 KgDrag Coefficient 1.0000

3 Effect of Ice formation

Radial Thickness of ICE over the conductor 0.0000 cmD1= 3.6600 cm

Increase in Conducter weight due to effect of Ice formation w3= 0.0000 kg/m

4 Total weight of sub-conductor per phase = Wc = W1+WSP/(r*SSP)+w3 = 3.1452 Kg/m

5 Wind Load on conductor as per IS 802 & IS 5613 = Pcc = 0.0000 Kg/m

Vb =

Risk Coefficient as per Table 2 of IS: 802, Part 1, Sec 1For all Zones & for Reliability level 1(Voltage level upto 400kV) = K1 =

K2 = Factor to convert 3 second peak gust speed into average speed of wind during 10 minutes period = K0 =

Meteorological reference wind speed = Vb/K0 VR = Design wind speed = VR x K1 x K2 Vd =

Design Wind Pressure =0.6 x Vd 2 /g kgf/m2

cm2

Kg/cm2

/ oCoC oC oC

WiT=

Tension per SUB-CONDUCTOR under max. wind and min. temp. conditions = Cdc =

Thice=Effective dia of the conductor considering ICE effect = d+2*Thice

Pc * (d/100) * Cdc * Gc =

Annexure-2

Page 27 of 29

CASE –

6 Clear span of conductor = L = la-(2*li)-lt-Bw = 89.3000 m

7 1.0000Loading factor during no wind condition = Q2 = 1.0000

8 SAG CALCULATIONS DUE TO CONDUCTOR

8.1f1 = Ten1/A = 125.8071

0.3957

3.1640E+07

-2432.7896

To solve the following equation:The working stress F at tempature t1+t3, is given by the cubic equation

6319246.35

we get f2 = 121.8959

8.2 Tension in conductor at maximum temperature and no windcondition = Ten2 = f2*A = 968.9115 Kg

Max sag due to conductor = Wc*L² / (8*Ten2) Sc= 3.2357


From Fig 1, we get

Tension per insulator string = Ten2* r / Ni Teni1= 484.4558

Weight of conductor per String Insulator = Wc*r/Ni Wci= 1.57264.6021 m

U2 = √((li/2)² + (di/2)²) = 2.3043 m

0.0304

0.0608


Teni1 * U1 *Sin(φ + δ) = (Wci * L/2) * U1 * Cos(φ + δ) + Wi * U2 *Cos(φ + β)

Simplifying :

TP1 = (Wci*L/2) * U1 * Cos δ + Wi*U2*Cos β - Teni1*U1*Sin δ = 862.3686

TP2 = Teni1*U1*Cos δ + Wci*(L/2)*U1*Sin δ + Wi*U2*Sin β = 2275.2868

0.3623

Sag due to Insulator = Si = U1*Sin(φ + d) = 1.7612 m

10 Total maximum sag of moose conductor = Sc+Si = Stot = 4.9969 m

11 Height of conductor above ground 15.3000 m

Height of jack bus/equipment below conductor 8.3000 mMinimum Phase to phase clearance = 4.0000 mTotal maximum sag of moose conductor = Sc+Si = Stot = 4.9969 m

-1.9969 m

Loading factor during max wind condition = Q1= √(1+(Pcc/Wc)2) =

Under minimum temperature and maximum wind condition condition, the sub-conductor experiences a stress of f1

Kg/cm2

Weight of sub conductor per m per cm2 = S = Wc/A =

GB = (S * L * Q2)2 * ey/24 =

F = f1 - GB * (Q1/(f1*Q2))2 - a * ey * (t1+t3-t2) =

f22 * (f2 - F) - GB =

Kg/cm2

U1 = √(li2 + (di/2)2) =

δ = Tan-1 [(di/2)/li] =



Therefore, available margin excluding sub-conductor spacing =

Annexure-2

Page 28 of 29

CASE –

SWING CALCULATIONS

1 SAG DUE TO CONDUCTOR (under maximum temperature rise in temperature and maximum wind conditions):

3.1640E+07

###-2432.7896

To solve the following equation:93728283.38

we get f21 = 217.4941

Tension in conductor at maximum temperature and maximum windcondition = Ten21 = f21*A= 1728.7905 Kg

1.8135 m


SGB = (Q1 * S *L)2 * ey/24 =

SF = (f1-SGB/f12) - a*ey*(t1+t3-t2) =

f212 * (f21 - SF) - SGB =

Kg/cm2

SAG due to conductor = Sc1 = Wc*L2/(8*Ten21) =

Annexure-2

Page 29 of 29

CASE – From Fig 1, we get

Teni2 = Ten21*r/Ni 864.3952

U1 = √(li2 + (di/2)2) = 4.6021 m

U2 = √((li/2)2 + (di/2)2) = 2.3043 m

0.0304

0.0608


TenI2 * U1 * Sin(φ + δ) = (WcI*L/2) * U1 * Cos(φ+ δ) + Wi * U2 *Cos(φ+ β)

Simplifying :

TP1 = (WcI*L/2) * U1 * Cos δ + Wi*U2*Cos β - TenI2*U1*Sin δ = 809.1771

TP2 = TenI2*U1*Cos δ + Wc*(L/2)*U1*Sin δ + Wi*U2*Sin β = 4023.0083

0.1985

SAG due to Insulator = Si1 = U1*Sin(φ + δ) = 1.0443

3 SWING DUE TO CONDUCTOR (SWC)

Wind load acting on span = Pcc * L*r = 0.0000 KgWeight of conductor = Wc * L *r= 561.7260 Kg

From Fig 3:Tan Y1 = (r*Pcc * L/r*Wc * L) = 0.0000

0.0000

δ = Tan-1 [(di/2)/li] =



Y1 = Tan-1 (Pcc/Wc) =

123

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