Forces on Gantry-34

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Description Notation/ Formula Unit Value

1 INPUT DATA

AC System Data

Rated Voltage, Vac kV 220

Rated continuous current, Ic Arms 2000

Rated short time current, Isc kArms 40

Duration of short time current, t Sec. 1

AC system frequency, f Hz 50

Site Data

Wind speed, v m/sec 47

Risk Coefft., K1 1

Terrain Roughness Factor, K2 0.85

Topography factor, K3 1.0

Design ambient temperature, Td ºC 50

Distance between supports, L m 34

Horizontal seismic coefficient, hsc .3g

Conductor Data

Type ACSR Moose

Number of sub conductors/phase n 2

Mass per Unit length of one conductor kg/m 1.57

Weight of Spacer kg 1.6

Distance of Spacers M 2.5

No. of Spacers 12

Spacer Weight / Conductor/ Meter 0.32

Mass per Unit length of one conductor with spacer kg/m 1.92

Cross Section Area sq mm 570

Diameter mm 31.05

Young's modulus, E Kg/cm2 5.40E+05

N/m2 5.30E+10

Min creepage mm 6125.00

Distance between discs mm 146

Conductor temp. at the begining of short circuit, Ti ºC 85

Conductor temp. at the end of short circuit, Tf ºC 200

Type of span (=1 for single span beam, =2 for double span beam)

2

Distance between phase conductors, A M 4.50

Length of insulator chain M 3.114

No. of discs 15

Tension on Conductor at maximum temperature 85 deg and no wind load (as per sag tension charts)

kg 564.04

N 9332.35

Girder width M 1.5

Distance between subconductors as mm 250

Distance between spacers ls mm 2500

Length of Turn Buckle m

a. Characteristic Electromagnetic Load per unit Length on flexible conductor in Three phase system

F' = μo/2*π X 0.75X (Ik)2

/a X lc/l N/m

pi π 3.14

Permeabaility in free space μo =4π*10-7 1.25714E-06

3 Ph short circuit Ik A 40000

Distance between phases a M 4.50

Conductor span l M 34.00

Clear Condcutor length lc M 26.27

Hence F' N/m 41.21

b. Ratio of Electromagnetic force under short circuit conditions to the gravitational force on a conductor

ϒ = F'/ (n*m's *gn) where

Number of sub-conductors n 2

Mass per unit length of one sub-conductor m's Kg/M 1.92

Acceleration of Gravity gn m/s2 9.81

Hence r 1.094

Taken from IS-875 Pt-3 S.N.

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Description Notation/ Formula Unit Value S.N.

c. Direction of resulting Force exerted on the conductor δ1=arctan(ϒ) rad 0.8303

deg 47.5508

d. Equivalent static conductor sag at midspan bc= n* msc*gn * l2

/ (8*Fst) M

where Static Tensile Force in flexible conductor Fst= N 9332.35

Hence bc= M 0.5833

e. Period of Conductor oscillations T=2π√(0.8/(bc/gn) s 1.3709

f. Resulting period of conductor oscillation during short circuit current flow

Tres= T/(√√(1+ϒ 2) (1-(π2/64)*(δ1/90)2 s 1.1767

g Actual Youngs Modulus Es=E*((0.3+0.7*sin(Fst/(n*As*бfin)*90)) for Fst/(n*As)<=бfin

Es=E for Fst/(n*As)>бfin

бfin= Lowest value of δ when Young's modulus becomes constant

бfin N/m2 50000000

As= Cross section of one sub- conductor Fst/(n*As) 8186271.93

< бfin

1 rad= 57.2727

Hence Es N/m2 25327807915

h Stiffness norm of and slack conductor N= (1/S*l) +(1/(n*Es*As) N

where S N/m2 100000

Hence N 3.28751E-07

i. Stress Factor ζ of THE Main Conductor ζ = (n*gn*msc'*l)2 /(24*fst3 *N) (n*gn*msc'*l)2 1640432.246 24*fst3 *N 6412859.064 Hence ζ 0.255803571

j. Swing Out Angle at the end of short circuit curent flow δk= δ1 (1-COS(Tk1/Tres * 360)) for 0<=Tk1/Tres<=0.5

δk= 2*δ1 Tk1/Tres>0.5

where Tk1 = Duration of first short circuit curremt flow Tk1=0.4T (Minimum (0.4T,1) 0.548357323

Tk1/Tres 0.465996509

Hence δk= deg 94.03376058

rad 1.641859319

k. Quantity for Max Swing out Angle Χ =1-ϒ sin δk for 0<=δk<=90

Χ =1-ϒ for δk>90 -0.094

L. Max Swing out Angle δm=1.25*arccosΧ for 0.766<=Χ <=1

δm=10o +arccosΧ for -0.985<=Χ <=0.766 Yes δm=180 for Χ <=-0.985 Hence δm= 1.839526911 105.3546729 3.2 TENSILE FORCE DURING SHORT CIRCUIT CAUSED BY

SWING OUT

a. Load Parameter φ=3(√(1+r2

)-1) for Tk1>=Tres/4

=3(ϒ sinδk+cosδk)-1) for Tk1<Tres/4

Tk1/Tres/4 0.465996509 >=.25 Hence φ=3(ϒ sinδk+cosδk)-1) 3.060672199 Hence φ=3(√(1+ϒ2 )-1) 1.446492489 b. Load Parameter ψ= φ2 2.09 φ(2+ζ) 3.26 1+2 ζ 1.51 ζ(2+φ) 0.88 φ2* ψ3+ φ(2+ ζ) *ψ2+ (1+2 ζ) *ψ- ζ(2+φ)=0 -1.56271E-10

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Description Notation/ Formula Unit Value S.N.

c. Tensile force Ft during short circuit caused by swing out

(short circuit tensile force)

Ft=Fst(1+φ*ψ) for n=1

Ft=1.1*Fst(1+φ*ψ) for n=2 yes

Hence Ft N 15000.53

Kg 1529.11

3.3 Tensile force Ff during short circuit caused by drop (drop

force)

Ff =1.2*Fst√(1+8*ζ *(δm/180)) This is significant for r >0.6, if δm>=70

r 1.094

δm= 1.84

105.35

Hence Ff N 16602.16

Kg 1692.37

3.4 Maximum Horizonatal Span displacement (bh) and

minimum air clearance (amin)

a. Elastic Expansion (εela) εela = N(Ft-Fst) 0.001863422

b. Thermal Expansion (εth)

Material Constant (Cth) for cross section area of Al/Cu > 6 m4/A2s 2.70E-17

for cross section area of Al/Cu <=6 m4

/A2s 1.70E-17

for cross section area Cu m4/A2s 8.80E-16

Thermal Expansion (εth) Cth(Ik3"/n*As)2* Tres/4 for Tk1>=Tres/4 yes

Cth(Ik3"/n*As)2* Tk1 for Tk1>=Tres/4

Hence εth = 9.78E-05 c. Dilation Factor CD CD =√ (1+ (3/8)*(l/bc)2 * ( εela + εth)) 3/8 0.375 l/bc 58.29 Hence CD 1.87

d. Form Factor (CF) 1.05 for r<=0.8

0.97+0.1r for 0.8< r < 1.8 Yes

1.15 for r >=1.8

Hence CF 1.07940

e. Maximum Horizontal Span displacement (bh)

For Slack Conductor bh= CF*CD*bc for δm >= 90

bh= CF*CD*bc Sin δm for δm<90

For Strained Conductor bh= CF*CD*bc Sin δ1 for δm >= δ1

bh= CF*CD*bc Sin δm for δm <δ1

Hence (bh) M 0.87

f. Minimum air clearance amin obtained during short circuit a-2*b M 2.76

4 TENSILE FORCE Fpi CAUSED BY PINCH EFFECT 4.1 Characteristic dimensions and parameters

a. Sub-Conductors are considered to clash effectively if either one of the following conditions are fulfilled

as/ds <= 2 and ls >= 50as as/ds <= 2.5 and ls >= 70as where

distance between sub-conductors as m 250

distance between two adjacent spacers ls m 2500

diameter of flexible conductor ds m 31.05

as/ds 8.052

Hence ls/as 10

b. Factor for peak short circuit current Ҡ 1.81

c. Time Constant of network (τ) τ = 1/ (-(2π*f/3) * ln ((Ҡ-1.02)/0.98)) 0.04429

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-Description Notation/ Formula Unit Value S.N.

d. Factor ϒ

ϒ = tan-1

(2π*f*τ) 1.499

e. Factor ν1 f (1/sin(180/n)* √(as-ds/msc')/

(μo/2*π)((Ik3"/n)2* (n-1)/as)

1812.250

μo/(2*π) 0.0000002

f. Time Tpi from short circuit until reaching Fpi Tpi can be avaluated from the solution of the following equation χ.√(1-A+B+C)-νi = 0 where A = (sin(4πχ-2ϒ)+sin2ϒ)/4πχ B= y/x.(1-e-2x/y ))sin2ϒ C= 8πy sinϒ/(1+(2πy)2/M P= 2πcos(2πx-ϒ)/(2πx) Q= sin( (2πx-ϒ)/(2πx) R= (sinϒ-2πy cos ϒ)/2πχ M= ((P+Q)*e-x/y +R) x= f.Tpi y= f.τ 2.215 Tpi= s 0.023555 x= 1.17775 A= -0.037 B= 1.225 P= 1.747 Q= -0.0500 R= -1.9496E-16 M= 0.996839884 C= 0.286048493

Substituting in equation χ.√(1-A+B+C)-νi -1810.594917

g. Factor v2 √(1-A+B+C) 1.512

h. Factor v3 ds/as/sin(180/n) *

√(as/ds)-1)/tan-1√((as/ds)-1) ds/as/sin(180/n) 0.124200025 as/ds 8.051529791 √(as/ds)-1) 2.65547167 tan-1√((as/ds)-1) 1.210640354 ds 0.03105 as 0.25 Hence Factor v3 0.272425784

i. Short Circuit Force Fv (n-1)*µ0/2π *(Ik3"/n)2

* (Is*v2)/(as*v3)

(n-1)*µ0/2π 0.0000002

(Ik3"/n)2 400000000

(Is*v2/as*v3) 55.50135445

ls 2.5

Hence Short Circuit Force Fv N 4440.11

j. Strain Factor of bundle contraction

εst 1.5* Fst*Is2*N/ (as-ds)2 *sin((180/n))2

Fst*Is2 *N/ (as-ds)2 0.399989449 sin(180/n)2 0.999999599 Hence εst 0.7000 εpi .375n* Fv*Is3 *N/ (as-ds)3 *sin((180/n))3 Fv*Is3 *N/ (as-ds)3 2.172932049 sin(180/n)3 0.999999399 Hence εpi 1.6297

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Description Notation/ Formula Unit Value S.N.

k. Parameter determining bundle conductor configuration during short circuit current flow

j = √(εpi/(1+εst) 0.97910465

j>=1 j<1

PINCH FORCE FOR NON-CLASHING CONDUCTORS

l η from fig 11b -IEC 60865-1 0.21

for as/ds 8.051529791 for j = 0.97910465 & εst 0.7000 =20 m Fpi =Fst(1+(ve/εst)*η2 )

where ve= 1/2+√(A1.A2.A3)-1/4

A1 (9/8)*n*(n-1)*µo/2π *(Ik3"/n)2 *N 5.91752E-05 A2 v2*(ls/(as-ds))4 * (sin(180/n))44 v2*(ls/(as-ds))4 25699.93517 (sin(180/n))4 0.999999199 η4 0.00194481 Hence A2 13214614.58 A3 1-tan-1(√v4)/v4 0.066357287 where v4 η*(as-ds)/(as-(as-ds)) 0.225367059

Hence ve= 1/2+√(A1.A2.A3)-1/4 7.686092343

Hence Pinch force Pi N 13851.30

Kgf 1411.96

PINCH FORCE FOR NON-CLASHING CONDUCTORS

m Fpi =Fst(1+(ve/εst)*ξ)

Sloving Equation for ξ ξ3+εst*ξ2-εpi=0 7.405014017

ξ 1.873630859 v4 (as-ds)/ds 7.051529791 x =( 9/8) *(n(n-1))*(μo/2*π)* (Ik/n)2 *N 5.92291E-05 y=v2*(ls/(as-ds)4 25699.93517 z=(sin(180/n)4/ ξ3 1.7409E-06 z1=(1-arctan(√v4)/√v4)-(1/4) 0.294095926 z2=√(x.y.z.z1) 0.000882805 Fpi N 9354.401675 Kgf 953.557765

5 FINAL SHORT CIRCUIT FORCE

- it shall be maximum of swing out force, drop force and pinch force

a) Swing Out Force Kgf 1529.11

b) Drop Force Kgf 1692.37

c) Pinch Force Kgf 1411.96

Hence the maximum tensile force during short circui shall be Kgf 1692.37

6 Minimum air clearance obtained during short circuit M 2.76

subconductor reduce their distance but do not clash

Figure

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