194296483 Calculo CortoCircuito IEC 865

Following minimum clearances shall be maintained :

System operating voltage kVRMS 400 220 132

Lightning Impulse Withstand Level (LIWL) kVPEAK 1550 1050 650

Switching Impulse Withstand Level (SIWL) kVPEAK 1050

Power frequency voltage kVRMS 630 460 275

Minimum phase to earth clearance mm 3500 2100 1300

Minimum phase to phase clearance mm 4000 2100 1300

Minimum sectional clearance mm 6500 5000 4000

Phase to phase distance between flexible conductor

Levels in the switchyard (Height from Plinth)

System operating voltage kVRMS 400 220 132

Plinth Level m

First Level - Equipment Connection m

Second Level - Strung Bus - Busbar m

Third Level - Strung Bus - Jack Bus m

Fourth Level - Shield wire m

CASE 1

Voltage kVRMS 420

Minimum Phase to phase clearance = m 4

Centre-line distance between main conductor mid points m 7

Span length m 100.00

Height of conductor above ground m 15.3

Conductor Type TARANTULLA

No. of sub-conductor of a main conductor Nos. 2

Current Carrying capacity A 2570

Suitability to carry Short Circuit Current Suitable

Effective distance between Sub-conductor = Center line

distance between Sub-conductor

m 0.45

Spacer Span = Centeline distance between connecting pieces or

between one connecting piece and the adjacent support

m 6.00

Static Tensile force per (sub)conductor in flexible conductor

per phase under minimum temperature and maximum wind

condition

kg 1000

No. of Disc Insulator Strings Nos. 4

No. of disc in one string Nos. 30

Static Tensile force PER CONDUCTOR in flexible conductor

per phase under minimum temperature and NO wind condition

(CIVIL INPUT)

kg 467.9

SAG

Net Current Carrying Capacity of the BUNDLE = A 2570

Annexure 1

Sag-Tension Calculation, Sag Tension Chart and

Short Circuit Forces For Structure and Foundation Design

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.

Annexure 1




Annexure 1



SAG

Total maximum sag = m 2.053

Sag due to conductor m 1.296

Sag due to Insulator m 0.757

Height of jack bus/equipment below conductor m 8.3

Therefore, available margin calculated between centre lines

(excluding sub-conductor spacing, conductor width,

equipment clamps etc) =

m 0.947

SWING

Total maximum swing of conductor = m 0.581

Available margin = m 1.969

Phase to phase spacing required = m 5.031

SHORT CIRCUIT

Static Tensile force in flexible conductor per phase under

minimum temperature and maximum wind condition

kg 2000

Therefore design short circuit force

Maximum of following :

kg 5514.9

Tensile force during short circuit kg 2772.2

Tensile force caused by Pinch effect kg 5514.9

Tensile force after short circuit (drop force) kg NA

j 1.1

33

170

80

320

320

3000

33

Annexure-1

System Parameters

Voltage = 420 kV

Current Rating = 3000 A

Three phase Initial symmetrical short-circuit current (r.m.s) = 40,000 A

Duration of the first short circuit current flow = 1 Sec

System frequency = 50 Hz

Wind Parameters

Basic Wind Speed = = 47.00 m/s

Risk Coefficient as per Table 2 of IS: 802, Part 1, Sec 1

For all Zones & for Reliability level 1(Voltage level upto 400kV)

= 1.07

Terrain Category = 2

Factor to convert 3 second peak gust speed into

average speed of wind during 10 minutes period =

= 1.3750

Height of conductor above ground = 15.3000

Spacer, additional concentrated mass

Spacer Span = Centeline distance between connecting pieces or

between one connecting piece and the adjacent support

= 6.00 m

Spacer weight = 3 kg

No. of spacers in span = 14 nos.

Additional concentrated mass (other than spacer) representing the

connections of pentagraph isolator ( Mass of one connection)

= 0.00 kg

Nos. of Trapzee Connection of pentagraph isolator in one span = 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

Young's Mod. Kg./Cm2 = 608200 Kg./Cm

210

6

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

Drag Coefficient = 1

DC Resistance = 0.0363 Ohm/km

Resulting Mass Per Unit length of one sub-conductor = 3.1452 kg/m

Coefficient of Linear Expansion = 0.000023 / oC

length of compensating spring in case of pentograph Isolator = 0.00

No. of Disc Insulator Strings = 4.00

No. 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

Weight of Disc = 8.3 kg

Length of Disc = 0.145 m

CALCULATION FOR FLEXIBLE CONDUCTORS

Page 7 of 32

Annexure-1

Total consolidated length of disc Insulator & Hardware in one side

of Span

= 4.600 m

Total consolidated weight of disc Insulator & Hardware in one side

of Span

1056.000

Layout Parameters

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

Static Tensile force PER CONDUCTOR in flexible conductor per

phase under minimum temperature and maximum wind condition

1 T

Static Tensile force in flexible conductor per phase under minimum

temperature and maximum wind condition

= 2 T

Span length = 100.00 m

Beam width 1.5 m

Length of conductor carrying short circuit current (Chord length of

main conductor in the span)

= 89.3 m

Maximum Ambient Temperature at site = 50.0000 oC

Minimum Ambient Temperature at site = 0.0000 oC

Rise in temperature = 35.0000 oC

Final Temperature after Short Circuit = 200.0000 oC

Radial Thickness of ICE over the conductor Thice= 0.0000 cm

Phase to phase spacing provided between phases = 7.0000 m

Effective distance between Sub-conductor = Center line distance

between Sub-conductor

= 0.45 m

Minimum Phase to phase clearance = 4 m

Height of jack bus/equipment below conductor 8.3 m

Resultant Spring constant of both support of one Span N/m 100000

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Page 8 of 32

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

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.

Required Current Carrying Capacity of the BUNDLED conductor 3000 Amps

Hence as given in Heat balance equation : (1)

Pj + Psol = Prad + Pconv

where,

Pj = RT I², Heat generated by Joule effect (2)

Psol =γ D Si, Solar Heat Gain (3)

Prad = s π D Ke (T24 – T1

4), Radiation Heat Loss (4)

Pconv = λ Nu (T2 – T1) π, Convection Heat Loss (5)

where,I = the conductor current (A)

Rdc(20°C) DC resistance of conductor at 20°C 0.0363 Ω/kmRT= the electrical resistance of conductor at a temperature T (Ω/m)

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

Si = the intensity of solar radiation (W/m²), Refer IEC 61597 900.00 W/m2

D = the conductor diameter (m) 0.0366 m

s = the Stefan Boltzmann constant (5.67×10–8 W.m–2K–4) Refer IEC 61597 5.67E-08 W.m–2K–4

Ke = the emmisivity co-efficient in respect to black body Refer IEC 61597 0.60

Ambient Temperature in (C) 50 °CFinal Equilibrium Temperature (C) 85 °C

T1 = Ambient Temperature in (K) 323 K

T2 = Final Equilibrium Temperature (K) 358 K

λ = the thermal conductivity of air film in contact with the conductor Refer IEC 61597 0.02585 W m–1K–1

Nu = the Nusselt number = 0.65 Re0.2 + 0.23Re

0.61 Refer IEC 61597 25.7

Re = the Reynolds number = 1.644×109v D[(T1 + 0.5(T2 – T1)]–1.78 Refer IEC 61597 1871.6

v = wind speed in m/s Refer IEC 61597 1 m/sf = System frequency 50 Hzα = Thermal Coefficient of Resistivity Refer IEC 61597 4.03E-03 /°C

μ = permeability (= 1, for non magmnatic material) 1

now,

RT = Rdc x K

where,

Rdc DC resistance of the conductor at operating temperature 85 deg C

K Skin effect coefficientagain,

Rdc(85°C) = Rdc(20°C) (1+α×(T2-20))

Rdc(20degC) = 0.0363 Ω/km

0.058419 Ω/miles

=> Rdc(85°C) 0.073722 Ω/miles

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

=> X = 1.65603

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

K = 1.03512

=> RT = 0.076311 Ω/miles

4.74E-05 Ω/m

Psol =γ D Si 16.47

Prad = s π D Ke (T24 – T1

4) 21.68

Pconv = λ Nu (T2 – T1) π 73.12

from 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

T

solconvrad

R

PPPI

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063598.0

CdcR

fX

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

I"k3 System short circuit current for System 40,000 A

t Duration of short circuit current 1 sec

Min. reqd. area of cross-section of conductor for short time rating Reference: (2)-(c)

Clause 5.2.2 Eq (3)in2

(Amin) =

where,

Ti Initial temp. of conductor before short ckt 85 0C

Tf Final temp of conductor after short ckt 200 0CG Conductivity of conductor Reference: (2)-(c ) 40 %

Clause 5.2.2

0.878063 in2

Minimum required cross-sectional area 566.49 mm2

Cross-section of each CONDUCTOR 794.87 mm2

Overall Cross-section of BUNDLE 1589.74 mm2


3

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Annexure-1

CASE – 3.

= Static Tensile force in flexible conductor per phase under minimum

temperature and maximum wind condition

= 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 m2

= Spacer Span = Centeline distance between connecting pieces or between

one connecting piece and the adjacent support

= 6 m

Additional Concerated mass (other than spacer) representing the

connections of pentagraph isolator ( Mass of one connection)

= 0 kg

Nos. of Trapzee Connection of pentagraph isolator in one span = 0 nos.

= Spacer weight = 3 kg

No. of spacers in span = 14 nos.

msc' = Resulting Mass Per Unit length of one sub-conductor

= 3.145 kg/m

= Span length = 100 m

length of compensating spring in case of pentograph of Isolator = 0

= Total consolidated length of disk Insulator & Hardware in one side of

Span

= 4.6 m

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

Sub-conductor

= 0.45 m

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

Beam Width 1.5 m

= Length of conductor carrying short circuit current (Chord length of main

conductor in the span)

= 89.3 m

S = Resultant Spring constant of both support of one Span N/m = 100000 N/m

= Young's modulus = 61997961264 N/m2

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

= Gravitational Constant = 9.81 m/s2

= Constant = 5.00E+07 N/m2

f = System frequency = 50 Hz

C th = Material Constant = 2.70E-19 m4/(A

2s)

µo = Magnatic Constant, permeability of Vacuum = 4π×10–7 H/m

Short Circuit Force Calculations

Calculation are based on IEC 865-1 (1993)

FOR FLEXIBLE CONDUCTORS

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Page 11 of 32

Annexure-1

CASE – 3.


= 30.62 N/m

= 0.4962

= 26.3888 Deg

= 3.9315 m

= 3.5577 Sec.

= 3.4125 Sec.

= 1.18E-07

= 3.50E+10 N/m2

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

From eqn. 23, Period of Conductor oscillation,

From eqn. 21, Angular direction of the force,

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,

From eqn. 25, The stiffness norm

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Page 12 of 32

Annexure-1

CASE – 3.


= 1.78E+00

= 33.4389 Deg

= 0.7266

= 53.3986 Deg

= 0.3490

0.00

= 0.7453

= 2772.1832 Kg

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

From eqn. 30, factor

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

From eqn. 31, Maximun swing out angle

From eqn. 32, Load Parameter

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

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

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Page 13 of 32

Annexure-1

CASE – 3.


= NA

Kg

= 8.9363E-04

=1.45828E-04

=1.1190E+00

=1.05

=2.0531

=2.8938E+00

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

Horizontal Span Displacement bh and Min. air clearance amin

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

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

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,

0

0706.0

180812.1 m

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Page 14 of 32

Annexure-1

CASE – 3.


= 4.2760

= 12.2951

V2 = 1.5721

=

0.2133

= 7862.0971Kg

= 0.7313

= 2.1267

= 1.1083

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)

SUBCONDUCTORS ARE NOT EFFECTIVELY CLASHING

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

From eqn. 45, Short circuit current force

Factor V3 can be given by

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

If j > 1 : The sub-conductor clash

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

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Page 15 of 32

Annexure-1

CASE – 3.


= 5514.861767 Kg

0.00

ξ . = 1.082580058

= 1.1872

= 1.13E+01

= 5514.861767 Kg

TENSILE FORCE IN CASE OF CLASHING SUB-CUNDUCTORS

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

SUBCONDUCTORS ARE CLASHING

If j ≥ 1, From eqn. 50, the tensile force is given by

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

From eqn. 52,

From eqn. 53,

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Page 16 of 32

Annexure-2CASE – 1.

System Voltage V = 420 kV

1 Basic Wind Speed = Vb = 47.00 m/s

As 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


For all Zones & for Reliability level 1(Voltage level upto 400kV) = K1 = 1.07

Terrain Category 2

Terrain roughness coeficient (For different Terrain Category) = K2 = 1.0000


average speed of wind during 10 minutes period = K0 = 1.3750

Meteorological reference wind speed = Vb/K0 VR = 34.1818 m/s

Design wind speed = VR x K1 x K2 Vd = 36.5745 m/s

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

2 Design Wind Pressure =0.6 x Vd 2 /g Pc = 81.8164 kgf/m2

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 cm

Cross Sectional Area of sub-conductor = A = 7.9487 cm2

Modulus of Elasticity of sub-conductor = ey = 608200.0000 Kg/cm2

Coefficient of Linear Expansion = a = 2.3000E-05 / oC

Maximum Ambient Temperature at site = t1 = 50.0000 oC

Minimum Ambient Temperature at site = t2 = 0.0000 oC

Rise in temperature = t3 = 35.0000 oC Distance between centre line of beams (for conductor span) = la = 100.0000 mDIameter of disc = di = 0.2800

Nos. of Insulator Strings in one side of conductor span (Single /Double) Ni = 4

Weight of complete string insulator including hardware WiT= 1056.0000

Weight of single string insulator including hardware Wi= 264.0000

Length of insulator string = li = 4.6000 m

Width of beam on which conductor is strung = Bw = 1.5000 mTension per SUB-CONDUCTOR under max. wind and min. temp. conditions = Ten1= 1000.0000 Kg

Drag Coefficient Cdc = 1.0000

3 Effect of Ice formation


Effective dia of the conductor considering ICE effect = d+2*Thice D1= 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 =

Pc * (d/100) * Cdc * Gc = 5.9401 Kg/m

CALCULATIONS FOR SAG AND SWING OF STRUNG CONDUCTOR

Page 17 of 32


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

7 Loading factor during max wind condition = Q1= √(1+(Pcc/Wc)2) = 2.1371

Loading factor during no wind condition = Q2 = 1.0000

8 SAG CALCULATIONS DUE TO CONDUCTOR

8.1

Under minimum temperature and maximum wind condition condition, the sub-

conductor experiences a stress of f1

f1 = Ten1/A = 125.8071 Kg/cm2

Weight of sub conductor per m per cm2 = S = Wc/A = 0.3957

GB = (S * L * Q2)2 * ey/24 = 3.1640E+07

F = f1 - GB * (Q1/(f1*Q2))2 - a * ey * (t1+t3-t2) = -10192.9819

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

f22 * (f2 - F) - GB = 940491268.11

we get f2 = 304.3152 Kg/cm2

8.2 Tension in conductor at maximum temperature and no wind

condition = 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= 1.5726

U1 = √(li2 + (di/2)2) = 4.6021 m

U2 = √((li/2)² + (di/2)²) = 2.3043 m

δ = Tan-1 [(di/2)/li] = 0.0304

β = Tan-1 [di/li] = 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

φ = Tan-1(TP1/TP2) = 0.1348

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

Page 18 of 32


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

11 Height of conductor above ground 15.3000 m


Minimum Phase to phase clearance = 4.0000 m

Total maximum sag of moose conductor = Sc+Si = Stot = 2.0530 mTherefore, available margin excluding sub-conductor spacing = 0.9470 m

SWING CALCULATIONS

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

SGB = (Q1 * S *L)2 * ey/24 = 1.4450E+08

SF = (f1-SGB/f12) - a*ey*(t1+t3-t2) = -1.0193E+04

To solve the following equation:

f212 * (f21 - SF) - SGB = 1.11E+10

we get f21 = 1002.7182 Kg/cm2

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

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

δ = Tan-1 [(di/2)/li] = 0.0304

β = Tan-1 [di/li] = 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

φ = Tan-1(TP1/TP2) = 0.0203

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

Page 19 of 32


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

Y1 = Tan-1 (Pcc/Wc) = 1.0838

SWC = Sc1* Sin Y1 = 0.3476 m

4 SWING DUE TO INSULATOR (SWI)

Wind load on insulator = Wpi = (Pc/2)*di*li *Gc*Cdc= 100.1105 kg

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

Subconductor spacing = 0.4500 m


Phase to phase spacing required = 5.0308 m

Therefore, available margin = 1.9692 m

Page 20 of 32

CASE - 2 ANNEXURE - 2

CASE 2.

NAME OF PROJECT:

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

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

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

INSULATOR AND TURNBUCKLE DATA

SAG TENSION CHART

GENERAL DATA

CONDUCTOR DATA

Page 21 of 32

CASE - 2 ANNEXURE - 2

MAXIMUM CLEAR SPAN OF CONDUCTOR: 89.30 m

S.NO. TEMPERATURETENSION PER

CONDUCTOR

CONDUCTOR

SAG

INSLULATOR

SAGTOTAL 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

RESULTS

Page 22 of 32

Type Al Strands Steel Strands Al Strand Dia St strand dia Cross sec. Diameter Weight Young's Mod.

Kg./Cm2

Young's

Modulus

(N/sq.m)

Drag

CoefficientDC

Resistance

Coefficient of

Linear

Expansion

sq.m m kg/km Kg./Cm2

BEAR 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 0.00001843

BEAVER 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 0.00001931

BUX 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 0.00001843

CAT 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 0.00001931

COYOTO 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 0.00001954

DEER 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 0.00001843

DOG 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 0.00001992

FERRET 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 0.00001931

FOX 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 0.00001931

GOAT 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 0.00001843

GOPHER 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 0.00001931

GUINEA 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 0.00001584

LARK 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 0.00001843

LEOPARD 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 0.00001954

LION 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 0.00001843

LYNX 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 0.00001843

MINK 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 0.00001931

MOOSE 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 0.00001991

OTTER 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 0.00001931

PANTHER 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 0.00001843

RABBIT 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 0.00001931

RACCOON 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 0.00001931

SHEEP 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 0.00001843

SPARROW 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 0.00001931

SQUIRREL 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 0.00001931

TIGER 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 0.00001843

WEASSEL 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 0.00001931

WOLF 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 0.00001843

ZEBRA 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 0.00001991

ANT AAC 7 3.100E-03 5.2834E-05 9.3000E-03 0.144 6.187E+05 6.31E+10 1 0.5419 0.000023

BLUE BOTTLE AAC 7 3.660E-03 7.3646E-05 1.0950E-02 0.201 6.187E+05 6.31E+10 1 0.3887 0.000023

BUTTERFLY AAC 19 4.650E-03 3.2266E-04 2.3250E-02 0.886 6.082E+05 6.20E+10 1 0.0892 0.000023

CATTER-PILLAR AAC 19 3.530E-03 1.8595E-04 1.0590E-02 0.511 6.187E+05 6.31E+10 1 0.1547 0.000023

CHAFER AAC 19 3.780E-03 2.1322E-04 1.8900E-02 0.586 6.082E+05 6.20E+10 1 0.1349 0.000023

CLEGG AAC 7 4.170E-03 9.5600E-05 1.2510E-02 0.261 6.187E+05 6.31E+10 1 0.2995 0.000023

COCKROACH AAC 19 4.220E-03 2.6575E-04 2.1100E-02 0.73 6.082E+05 6.20E+10 1 0.1083 0.000023

EARWIG AAC 7 3.780E-03 7.8555E-05 1.1340E-02 0.215 6.187E+05 6.31E+10 1 0.3644 0.000023

FLY AAC 7 3.400E-03 6.3554E-05 1.0200E-02 0.174 6.187E+05 6.31E+10 1 0.4505 0.000023

GRASS HOPPER AAC 7 3.910E-03 8.4051E-05 1.1730E-02 0.23 6.187E+05 6.31E+10 1 0.3406 0.000023

IRIS AAC 7 2.480E-03 3.3814E-05 7.4400E-03 0.092 6.187E+05 6.31E+10 1 0.8467 0.000023

LADY BIRD AAC 7 2.790E-03 4.2795E-05 8.3700E-03 0.117 6.187E+05 6.31E+10 1 0.669 0.000023

LOCUST AAC 19 5.360E-03 4.2872E-04 2.6800E-02 1.176 6.082E+05 6.20E+10 1 0.0671 0.000023

MAYBUG AAC 37 4.090E-03 4.8611E-04 2.8630E-02 1.343 5.976E+05 6.09E+10 1 0.0593 0.000023

MOTH AAC 19 5.000E-03 3.7306E-04 2.5000E-02 1.025 6.082E+05 6.20E+10 1 0.0771 0.000023

PANSY AAC 7 2.780E-03 4.2489E-05 8.3400E-03 0.116 6.187E+05 6.31E+10 1 0.6738 0.000023

ROSE AAC 7 1.960E-03 2.1120E-05 5.8800E-03 0.058 6.187E+05 6.31E+10 1 1.3556 0.000023

SCORPION AAC 37 4.270E-03 5.2984E-04 2.9890E-02 1.464 5.976E+05 6.09E+10 1 0.0544 0.000023

SPIDER AAC 19 3.990E-03 2.3757E-04 1.9950E-02 0.652 6.082E+05 6.20E+10 1 0.1205 0.000023

TARANTULLA AAC 37 5.230E-03 7.9487E-04 3.6600E-02 2.91 6.082E+05 6.20E+10 1 0.0363 0.000023

WASP AAC 7 4.390E-03 1.0595E-04 1.3170E-02 0.29 6.187E+05 6.31E+10 1 0.2702 0.000023

SHIELD WIRE 7/9 SW 7 3.660E-03 7.3646E-05 1.2180E-02 0.583 1933000 1.97E+11 1.2 0.000012

SHIELD WIRE 7/8 SW 7 9.0500E-05 1.2180E-02 0.706 1933000 1.97E+11 1.2 0.000012

BLANK SW

disc insulator x

Creepage Diameter of

Disc

Weight of Disc Length of

Disc

120 440 0.28 8.3 0.145

xx

SKIN EFFECT TABLE

Sl No. X K

1 0.00 1.00000

2 0.10 1.00000

3 0.20 1.00001

4 0.30 1.00004

5 0.40 1.00013

6 0.50 1.00032

7 0.60 1.00067

8 0.70 1.00124

9 0.80 1.00212

10 0.90 1.00340

11 1.00 1.00519

12 1.10 1.00758

13 1.20 1.01071

14 1.30 1.01470

15 1.40 1.01969

16 1.50 1.02582

17 1.60 1.03323

18 1.70 1.04205

19 1.80 1.05240

20 1.90 1.06440

21 2.00 1.07816

22 2.10 1.09375

23 2.20 1.11126

24 2.30 1.13069

25 2.40 1.15207

26 2.50 1.17538

27 2.60 1.20056

28 2.70 1.22753

29 2.80 1.25620

30 2.90 1.28644

31 3.00 1.31809

32 3.10 1.35102

33 3.20 1.38504

34 3.30 1.41999

35 3.40 1.45570

36 3.50 1.49202

37 3.60 1.52879

38 3.70 1.56587

39 3.80 1.60314

40 3.90 1.64051

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

Source : Bureau of Standards Bulletin No. 169 on pages 226-8

Source : TNEB HANDBOOK

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

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

n

f 50 Hz as

k 1.81 As per IEC-909 ds

k 1.81 as/ds

1/t 22.55772088 ds/as

t 0.04433072

g 1.499079658 Sqrt((as/ds)-1)

fTpi 3.410369262 v3

2fTpi 21.41711897

ft 2.216535982 h

st

pi

v1 4.276003 ya

v2 1.572075 asv3

2*ya/as

A 0.023040059

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

B 0.616802478 asw

C 0.285149187 fh

D 0.312916328

E 0.040923867

F 0.214715018

G 6.22057E-17

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

1.572098266

-0.00002

223839692.xls.ms_office Page 27 of 32

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

2

0.45

3.66E-02

12.29508197

0.081333333

3.36081567

0.213286608

0.629148847

7.31E-01

2.13E+00

9.50E-02

0.095978973

0.422021926

1.170275665

0.287831932

0.333454919

-6.49E-06

223839692.xls.ms_office Page 28 of 32

Annexure-2CASE –

System Voltage V = 420 kV

1 Basic Wind Speed = Vb = 0.00 m/s

As 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


For all Zones & for Reliability level 1(Voltage level upto 400kV) = K1 = 1.07

Terrain Category 2

Terrain roughness coeficient (For different Terrain Category) = K2 = 1.0000


average speed of wind during 10 minutes period = K0 = 1.3750

Meteorological reference wind speed = Vb/K0 VR = 0.0000 m/s

Design wind speed = VR x K1 x K2 Vd = 0.0000 m/s

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

2 Design Wind Pressure =0.6 x Vd 2 /g Pc = 0.0000 kgf/m2

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 cm

Cross Sectional Area of sub-conductor = A = 7.9487 cm2

Modulus of Elasticity of sub-conductor = ey = 608200.0000 Kg/cm2

Coefficient of Linear Expansion = a = 2.3000E-05 / oC

Maximum Ambient Temperature at site = t1 = 40.0000 oC

Minimum Ambient Temperature at site = t2 = 0.0000 oC

Rise in temperature = t3 = 0.0000 oC Distance between centre line of beams (for conductor span) = la = 100.0000 mDIameter of disc = di = 0.2800

Nos. of Insulator Strings in one side of conductor span (Single /Double) Ni = 4

Weight of complete string insulator including hardware WiT= 1056.0000

Weight of single string insulator including hardware Wi= 264.0000

Length of insulator string = li = 4.6000 m

Width of beam on which conductor is strung = Bw = 1.5000 mTension per SUB-CONDUCTOR under max. wind and min. temp. conditions = Ten1= 1000.0000 Kg

Drag Coefficient Cdc = 1.0000

3 Effect of Ice formation


Effective dia of the conductor considering ICE effect = d+2*Thice D1= 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 =

Pc * (d/100) * Cdc * Gc = 0.0000 Kg/m

CALCULATIONS FOR SAG AND SWING OF STRUNG CONDUCTOR

Page 29 of 32

Annexure-2CASE –

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

7 Loading factor during max wind condition = Q1= √(1+(Pcc/Wc)2) = 1.0000

Loading factor during no wind condition = Q2 = 1.0000

8 SAG CALCULATIONS DUE TO CONDUCTOR

8.1

Under minimum temperature and maximum wind condition condition, the sub-

conductor experiences a stress of f1

f1 = Ten1/A = 125.8071 Kg/cm2

Weight of sub conductor per m per cm2 = S = Wc/A = 0.3957

GB = (S * L * Q2)2 * ey/24 = 3.1640E+07

F = f1 - GB * (Q1/(f1*Q2))2 - a * ey * (t1+t3-t2) = -2432.7896

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

f22 * (f2 - F) - GB = 6319246.35

we get f2 = 121.8959 Kg/cm2

8.2 Tension in conductor at maximum temperature and no wind

condition = 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= 1.5726

U1 = √(li2 + (di/2)2) = 4.6021 m

U2 = √((li/2)² + (di/2)²) = 2.3043 m

δ = Tan-1 [(di/2)/li] = 0.0304

β = Tan-1 [di/li] = 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

φ = Tan-1(TP1/TP2) = 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



Total maximum sag of moose conductor = Sc+Si = Stot = 4.9969 mTherefore, available margin excluding sub-conductor spacing = -1.9969 m

SWING CALCULATIONS

Page 30 of 32

Annexure-2CASE –

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

SGB = (Q1 * S *L)2 * ey/24 = 3.1640E+07

################

SF = (f1-SGB/f12) - a*ey*(t1+t3-t2) = -2432.7896

To solve the following equation:

f212 * (f21 - SF) - SGB = 93728283.38

we get f21 = 217.4941 Kg/cm2

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

SAG due to conductor = Sc1 = Wc*L2/(8*Ten21) = 1.8135 m


Page 31 of 32

Annexure-2CASE – 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

δ = Tan-1 [(di/2)/li] = 0.0304

β = Tan-1 [di/li] = 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

φ = Tan-1(TP1/TP2) = 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

Y1 = Tan-1 (Pcc/Wc) = 0.0000

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194296483 Calculo CortoCircuito IEC 865