Annexure-2 Page-1

NEW MELLI GIS 220 KV SUBSTATION SHORT CIRCUIT CALCULATIONS FOR MAX. TEMPERATURE AND NO WIND I.E. AT MINIMUM TENSION AS PER SAG TENSION CHARTS

S.N. 1

Description INPUT DATA AC System Data Rated Voltage, Vac Rated continuous current, Ic Rated short time current, Isc Duration of short time current, t AC system frequency, f Site Data Wind speed, v Risk Coefft., K1 Terrain Roughness Factor, K2 Topography factor, K3 Design ambient temperature, Td Distance between supports, L Horizontal seismic coefficient, hsc Conductor Data Type Number of sub conductors/phase Mass per Unit length of one conductor Weight of Spacer Distance of Spacers No. of Spacers Spacer Weight / Conductor/ Meter Mass per Unit length of one conductor with spacer

Notation/ Formula

Unit

Value

kV Arms kArms Sec. Hz

220 2000 40 1 50

m/sec Taken from IS-875 Pt-3

C m

47 1 0.85 1.0 50 34 .3g

n kg/m kg M

kg/m

ACSR Moose 2 1.57 1.6 2.5 12 0.32 1.92

Cross Section Area Diameter Young's modulus, E

sq mm mm Kg/cm2 N/m2

570 31.05 5.40E+05 5.30E+10

Min creepage Distance between discs Conductor temp. at the begining of short circuit, Ti Conductor temp. at the end of short circuit, Tf Type of span (=1 for single span beam, =2 for double span beam) Distance between phase conductors, A Length of insulator chain No. of discs Tension on Conductor at maximum temperature 85 deg and no wind load (as per sag tension charts)

mm mm C C

6125.00 146 85 200 2

M M kg N

4.50 3.114 15 564.04 9332.35 1.5 250 2500

Girder width Distance between subconductors Distance between spacers Length of Turn Buckle a. Characteristic Electromagnetic Load per unit Length on flexible conductor in Three phase system pi Permeabaility in free space 3 Ph short circuit Distance between phases Conductor span Clear Condcutor length Hence b.

as ls

M mm mm m N/m

F' = o/2* X 0.75X (Ik)2/a X lc/l o =4*10-7 Ik a l lc F'

3.14 1.25714E-06 A M M M N/m 40000 4.50 34.00 26.27 41.21

Ratio of Electromagnetic force under short circuit = F'/ (n*m's *gn) conditions to the gravitational force on a conductor where Number of sub-conductors n Mass per unit length of one sub-conductor m's Acceleration of Gravity Hence gn r

Kg/M m/s2

2 1.92 9.81 1.094

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NEW MELLI GIS 220 KV SUBSTATION SHORT CIRCUIT CALCULATIONS FOR MAX. TEMPERATURE AND NO WIND I.E. AT MINIMUM TENSION AS PER SAG TENSION CHARTS

S.N. c.

Description Direction of resulting Force exerted on the conductor

Notation/ Formula 1=arctan()

Unit rad deg

Value 0.8303 47.5508

d.

Equivalent static conductor sag at midspan where Static Tensile Force in flexible conductor Hence

bc= n* msc*gn * l / (8*Fst) Fst= bc= T=2(0.8/(bc/gn)

M N M s 9332.35 0.5833 1.3709 1.1767

e. f.

Period of Conductor oscillations Resulting period of conductor oscillation during short circuit current flow Actual Youngs Modulus

Tres= T/((1+ 2) (1-(2/64)*(1/90)2 s

Es=E*((0.3+0.7*sin(Fst/(n*As*fin)*90)) for Fst/(n*As)<=fin Es=E for Fst/(n*As)>fin N/m2 50000000 8186271.93 < fin 1 rad= 57.2727 25327807915

fin= Lowest value of when Young's modulus becomes constant As= Cross section of one sub- conductor

fin Fst/(n*As)

Hence h Stiffness norm of and slack conductor Hence i. Stress Factor of THE Main Conductor

Es N= (1/S*l) +(1/(n*Es*As) where S N = (n*gn*msc'*l)2/(24*fst3*N) (n*gn*msc'*l)2

N/m

N N/m2

100000 3.28751E-07

1640432.246 6412859.064 0.255803571 for 0<=Tk1/Tres<=0.5 Tk1/Tres>0.5 0.548357323 0.465996509 deg rad for 0<=k<=90 for k>90 for 0.766<= <=1 for -0.985<= <=0.766 for <=-0.985 Yes 94.03376058 1.641859319 -0.094

Hence j. Swing Out Angle at the end of short circuit curent flow

24*fst3*N k= 1 (1-COS(Tk1/Tres * 360)) k= 2*1

where Tk1 = Duration of first short circuit curremt flow

Tk1=0.4T (Minimum (0.4T,1) Tk1/Tres

Hence k. Quantity for Max Swing out Angle

k= =1- sin k =1-

L.

Max Swing out Angle

m=1.25*arccos m=10o+arccos m=180

Hence

m=

1.839526911 105.3546729

3.2 a.

TENSILE FORCE DURING SHORT CIRCUIT CAUSED BY SWING OUT Load Parameter

=3((1+r2)-1) =3( sink+cosk)-1) Tk1/Tres/4

for Tk1>=Tres/4 for Tk1<Tres/4 0.465996509 >=.25 3.060672199 1.446492489

Hence Hence b. Load Parameter

=3( sink+cosk)-1) =3((1+2)-1) = 2 (2+) 1+2 (2+) =

2* 3+ (2+ ) *2+ (1+2 ) *- (2+)=0 Solving above equation

2.09 3.26 1.51 0.88 -1.56271E-10 0.318871183

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NEW MELLI GIS 220 KV SUBSTATION SHORT CIRCUIT CALCULATIONS FOR MAX. TEMPERATURE AND NO WIND I.E. AT MINIMUM TENSION AS PER SAG TENSION CHARTS

S.N. c.

Description

Notation/ Formula

Unit for n=1 for n=2 N Kg

Value

Tensile force Ft during short circuit caused by swing out Ft=Fst(1+*) (short circuit tensile force) Ft=1.1*Fst(1+*) Hence Ft

yes 15000.53 1529.11

3.3

Tensile force Ff during short circuit caused by drop (drop Ff =1.2*Fst(1+8* *(m/180)) force) This is significant for r >0.6, if m>=70 r m= Hence Ff N Kg

1.094 1.84 105.35 16602.16 1692.37

3.4

Maximum Horizonatal Span displacement (bh) and minimum air clearance (amin) a. b. Elastic Expansion (ela) Thermal Expansion (th) Material Constant (Cth) for cross section area of Al/Cu > 6 for cross section area of Al/Cu <=6 for cross section area Cu Thermal Expansion (th) Hence c. Dilation Factor CD Cth(Ik3"/n*As)2* Tres/4 Cth(Ik3"/n*As)2* Tk1 th = CD = (1+ (3/8)*(l/bc) * ( ela + th)) 3/8 l/bc CD 1.05 0.97+0.1r 1.15 Hence e. Maximum Horizontal Span displacement (bh) For Slack Conductor bh= CF*CD*bc bh= CF*CD*bc Sin m For Strained Conductor bh= CF*CD*bc Sin 1 bh= CF*CD*bc Sin m Hence (bh) f. Minimum air clearance amin obtained during short circuit TENSILE FORCE Fpi CAUSED BY PINCH EFFECT Characteristic dimensions and parameters 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 distance between two adjacent spacers diameter of flexible conductor Hence as ls ds as/ds ls/as m m m 250 2500 31.05 8.052 10 a-2*b for m >= 90 for m<90 for m >= 1 for m <1 M M 0.87 2.76 CF for r<=0.8 for 0.8< r < 1.8 for r >=1.8 Yes
2

ela = N(Ft-Fst)

0.001863422

m4/A2s m4/A2s m4/A2s for Tk1>=Tres/4 for Tk1>=Tres/4

2.70E-17 1.70E-17 8.80E-16 yes 9.78E-05

0.375 58.29 1.87

Hence d. Form Factor (CF)

1.07940

4 4.1 a.

So, sub conductors shall not clash effectively during short circuit and Tensile Force caused by the pinch effect shall be calculated as follows b. c. Factor for peak short circuit current = 1/ (-(2*f/3) * ln ((-1.02)/0.98)) 1.81 0.04429

Time Constant of network ()

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NEW MELLI GIS 220 KV SUBSTATION SHORT CIRCUIT CALCULATIONS FOR MAX. TEMPERATURE AND NO WIND I.E. AT MINIMUM TENSION AS PER SAG TENSION CHARTS

S.N. d.

Description Factor

Notation/ Formula

Unit

Value

= tan-1 (2*f*) e. Factor 1 f (1/sin(180/n)* (as-ds/msc')/ (o/2*)((Ik3"/n)2* (n-1)/as) o/(2*) f. Time Tpi from short circuit until reaching Fpi Tpi can be avaluated from the solution of the following equation .(1-A+B+C)-i = 0 where A = B= C= P= Q= R= M= x= y= (sin(4-2)+sin2)/4 y/x.(1-e-2x/y))sin2 8y sin/(1+(2y)2/M 2cos(2x-)/(2x) sin( (2x-)/(2x) (sin-2y cos )/2 ((P+Q)*e-x/y+R) f.Tpi f. Tpi= x= A= B= P= Q= R= M= C= Substituting in equation g. Factor v2 .(1-A+B+C)-i (1-A+B+C) s

1.499 1812.250 0.0000002

2.215 0.023555 1.17775 -0.037 1.225 1.747 -0.0500 -1.9496E-16 0.996839884 0.286048493 -1810.594917 1.512

h.

Factor v3

ds/as/sin(180/n) * (as/ds)-1)/tan1((as/ds)-1) ds/as/sin(180/n) as/ds (as/ds)-1) tan-1((as/ds)-1) ds as

0.124200025 8.051529791 2.65547167 1.210640354 0.03105 0.25 0.272425784

Hence i. Short Circuit Force Fv

Factor v3 (n-1)*0/2 *(Ik3"/n)2* (Is*v2)/(as*v3) (n-1)*0/2 (Ik3"/n)2 (Is*v2/as*v3) ls

0.0000002 400000000 55.50135445 2.5 N 4440.11

Hence j. Strain Factor of bundle contraction st

Short Circuit Force Fv

1.5* Fst*Is2*N/ (as-ds)2 *sin((180/n))2 Fst*Is2*N/ (as-ds)2 sin(180/n)2 0.399989449 0.999999599 0.7000

Hence pi

st .375n* Fv*Is3*N/ (as-ds)3 *sin((180/n))3 Fv*Is3*N/ (as-ds)3 sin(180/n)3

2.172932049 0.999999399 1.6297

Hence

pi

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NEW MELLI GIS 220 KV SUBSTATION SHORT CIRCUIT CALCULATIONS FOR MAX. TEMPERATURE AND NO WIND I.E. AT MINIMUM TENSION AS PER SAG TENSION CHARTS

S.N. k.

Description

Notation/ Formula

Unit

Value 0.97910465

Parameter determining bundle conductor configuration during j = (pi/(1+st) short circuit current flow j>=1 j<1

subconductor clash subconductor reduce their distance but do not clash

PINCH FORCE FOR NON-CLASHING CONDUCTORS

from fig 11b -IEC 60865-1 for as/ds for j = & st =Fst(1+(ve/st)*2) 1/2+(A1.A2.A3)-1/4 (9/8)*n*(n-1)*o/2 *(Ik3"/n)2*N v2*(ls/(as-ds))4* (sin(180/n))4/4 v2*(ls/(as-ds))4 (sin(180/n))4 4

0.21 8.051529791 0.97910465 0.7000 =20

Fpi where ve= A1 A2

5.91752E-05

25699.93517 0.999999199 0.00194481 13214614.58 0.066357287 0.225367059 7.686092343 N Kgf 13851.30 1411.96

Hence A3 where v4 Hence ve= Hence Pinch force

A2 1-tan-1(v4)/v4 *(as-ds)/(as-(as-ds)) 1/2+(A1.A2.A3)-1/4 Pi

PINCH FORCE FOR NON-CLASHING CONDUCTORS m Fpi Sloving Equation for =Fst(1+(ve/st)*) 3+st*2-pi=0 v4 (as-ds)/ds x =( 9/8) *(n(n-1))*(o/2*)* (Ik/n) *N y=v2*(ls/(as-ds)4 z=(sin(180/n)4/ 3 z1=(1-arctan(v4)/v4)-(1/4) z2=(x.y.z.z1) Fpi N Kgf
2

7.405014017 1.873630859 7.051529791 5.92291E-05 25699.93517 1.7409E-06 0.294095926 0.000882805 9354.401675 953.557765

FINAL SHORT CIRCUIT FORCE - it shall be maximum of swing out force, drop force and pinch force a) Swing Out Force b) Drop Force c) Pinch Force Hence the maximum tensile force during short circui shall be

Kgf Kgf Kgf Kgf M

1529.11 1692.37 1411.96 1692.37 2.76

Minimum air clearance obtained during short circuit