415

ANNEXURE - A

A.1 Algorithm for Runge Kutta 4th order Method:

To approximate the solution of the initial value problem

yy ytf ,,

Let

)()( aaybta y

N+1 equally spaced numbers in the interval [a,b]

Input End points a, b; Integer N; Initial conditions of ,

Output Approximation w, v to y & y

at the (N+1) values of t

Step 1: Set

vwatN

abh ;;;

Step2: For i=1,2,3, . . . . . N Do steps 3 to5

Step3: Set k1= h* f(t, w, v)

v*h=L1

2,2,2* 112 KvLwhtfhK

2Kv *h=L 12

2,2,2* 223 KvLwhtfhK

2Kv *h=L 23

334 ,,* KvLwhtfhK

34 Kv *h=L

416

Step4 : Set

6

22 4321 LLLLww

and compute Wi

Set

6

22 4321 KKKKvv

and compute Vi

Step 5: Output (t, w, v)

Step 6: Stop

A.2 MONTE – CARLO TECHNIQUE:

Monte carlo methods are a widely used class of computational

algorithms for simulating the behavior or various physical and

mathematical systems, and for other computations. They are

distinguished from other simulation methods (such as molecular

dynamics) by being stochastic, that is nondeterministic in some manner

usually by using random numbers (in practice, pseudo-random

numbers) as opposed to deterministic algorithms. Because of the

repetition of algorithms and the large number of calculations involved,

Monte Carlo is a method suited to calculation using a computer, using

many computer simulation techniques.

Monte Carlo algorithm is often a numerical Monte Carlo method used

to find solutions to mathematical problems (which may have many

variables) that cannot easily be solved, for example, by integral calculus,

or other numerical methods. For many types of problems, its efficiency

relative to other numerical methods increases as the dimension of the

problem increases. Or it may be a method for solving other mathematical

problems that rely on (pseudo-)random numbers.

417

Monte-Carlo method is a numerical method of solving mathematical

problems by means sampling. This method allows simulating any

process whose development is affected by random factors. Many

mathematical problems, which are not affected by any random

influences, can be connected with an artificially constructed probabilistic

model (even more than one) making possible the solution of these

problems. Random motion of a particle in GIS bus can be calculated by

applying Monte-Carlo method.

Two distinct features of the Monte-Carlo method:

The first feature is the simple structure of the computation

algorithm. As a rule, a program is prepared to perform only one

random trial. This trail is repeated n times, each trail is

independent of all Monte-Carlo method is some times called the

method of statistical trails.

The Monte-Carlo method is especially efficient in solving those

problems, which do not require a high degree of accuracy (i.e. 5 to

10%)

The Simulation of the motion equation given in chapter III and

Chapter IV yields particle movement in the radial direction only.

However, the configuration at the tip of the particle is generally not

sufficiently smooth enough to enable the movement unidirectional. This

decides the movement of particle in axial direction. The randomness of

movement can be adequately simulated by Monte-Carlo method. In order

418

to determine the randomness, it is assumed that the particle emanates

from its original site at any angle less than , where /2 is half of the

solid angle subtended with the vertical axis. At every step of movement, a

new rectangular random number is generated between 0 and 1 and

modified to . The angle thus assigned, fixes the position of particle at

the end of every time step, and in turn determines the axial and radial

positions. The position in the next step is computed on the basis of

equation of motion with new random angles as described above. The

results of the simulations are given in chapter V.

ALGORITHM FOR MONTE CARLO SIMULATION:

Step 1 : Create a parametric model, )....,,( 21 qXXXfy

Step 2 : Generate a set of random inputs, .....,, 21 iqii XXX

Step 3 : Evaluate the model and store the results as yi .

Step 4 : Repeat steps 2 and 3 for i=1 to n

Step 5 : Analyze the results

419

ANNEXURE - B

B.1 Moeal Calculations for Particle movement in Single phase

isolated conductor Gas Insulated Busduct:

Consider Aluminum Particle of Length 12 mm and 0.2 mm as

radiation of the surface of enclosure in a gas insulated bus duct of

152mm enclosure diameter and 55mm inner conductor diameter.

Let applied voltage V=100kV (rms).

Pressure of the Gas P=0.4Mpa

Maximum Voltage Vm = 100 * 1.414*1000=0.1414*106 volts.

Density of Alumnium particle = 2700kg/m3

Mass of the particle m=r2l

= 3.14(0.2 x 10-3)()8 x 10-3) (2700)

= 4.06944 x 10-6 kg

Lift off Field ELO = 0

84.0

gr

= 12

3

10*854.8

10*2.0*81.9*270084.0

= 6.4974* 105V/m

Initial Time to lift off

V

r

rrEt

i

LOi

0

0

1 ln*sin

= 0.06622 sec

Angular velocity = 2f* = 2*3. 14*50= 314 rad / sec

Gas density g= 7.118 + 6.332P + 0.2032P2

= 9.683312

420

Motion equation by considering all forces can be expressed as

5.0

00

0

)]([656.2)(6)(

ln)(

*2)(.

tlytyKrty

mg

r

rtyr

tSinVrlE

tm

gd

i

LO

y

This equation is second order non-linear differential equation and solved

by using iterative methods. Here Runge-Kutta Method solving the above

motion equation.

By substituting the values of ELO, , R, 0 , m, g, r0,ri, l, in the motion

equation becomes

5.0

.

3)(086.1)(0215.081.9

)(10*76

sin42.19)( tt

ty

tt yyy

Appling the following initial conditions for the solving the motion

equation )0()0(..

tmRtm yy and y(t=0+)=0

Where

y(t) = displacement of the particle.

)(.

ty velocity of the particle.

)(

..

ty acceleration of the particle.

The starting time is found to be 0.06622 seconds and the step is taken

as 0.001

As per Runge – Kutta method

421

K1=t * f(t0, y(t), )(.

ty )

= 0.0001 * (20.773 – 9.81 – 0-0)

= 0.4037 * 10-3

L1 = t * )(.

ty

= 0.0001 0

=0

2)()(,

2* 1

.

02

ktty

ttftK y

=0.0001 * [13.86 – 9.81 – 4.341925*10-6 – 2.0778*10-6]

= 0.4047*10-3

L2= t *

21)(

Kty

= 0.0001*(0+020195*10-3)

= 0.020195*10-6

K3 = t * f

2)(,

2)(,

2

2

.

20

Kt

Lty

tt y

= 0.4057*10-3

L3= t *

2)( 2

.K

ty

= 0.020235 *10-6

))(,)(,(* 3

.

304 KtLtyftftK y

= 0.4067*10-3

422

L4 = t*

3

.

)( Kty

= 0.020285*10-6

Now

Y1(t)= y(t)+ 4321 226

1LLLL

= 0.028825* 10-6 m

4321

..

226

1)()(

1KKKKtt yy

=0.4052*10-3 m/sec

B.2 Model Calculations for Particle movement in single phase

coated conductor Gas Insulated Busduct:

Consider Aluminum Particle of Length 8 mm and 0.2 mm as

radius at the surface of enclosure in a gas insulated bus duct of 152 mm

enclosure diameter and 55 mm inner conductor diameter coated with

dielectric material of thickness 75um.

For a parallel plate capacitor s

tR C

For a horizontal wire particle, the contact area s=rl

Then rl

tR C

33

68

10*12*10*2.0*14.3*0001

10*75*10*1

= 0.9960*1013

The capacitance CC between the enclosure surface and the particle is

423

CC=0 r t

s

6

3312

10*75

10*12*10*2*14.3*001.0*23.7*10*8854

= 5.148336*10-14 Farad

The Capacitance Cg, between the horizontal wire particle of length 1

particle and enclosure is

1

0

0

0

ln

2

r

rr

rlCg

3

33

3312

10*5.27

10*76ln10*76

10*12*10*.2*10*854.8*14.3*2

= 4.63155* 1610 Farad

The charging current through the conductance is given by

5.0

22

2

2 11

gg

C

C

CC

CR

VI

5.0

162

2

16

14213

10*63155.4*314

1

10*63155.4

10*148336.5110*9960.0

1000*4.141

= 1.26578*10-10 Amps

Cg CCRTant

**

1)( 1

424

161413

1

10*0877.310*29028.4*10*492.1*314

1Tan

=-0.4225 radians

3

2

3

22)(

SinCosCosB

= (0.9999878-0.99944752)0.03817

= 0.02062*10-3

30

5.0

22

2

2

10*80135.0

11

gg

C

CC

CRw

KA

The Lift Field is expressed as

5.0

0

0 ln)(

i

LO

r

rrAB

mgE

3

33318

6

10*5.27

10*76ln*10*76*10*02062.010*5252.1

81.9*10*71296.2

= 1.1600576*106kV/m

425

ANNEXURE - C

C.1 Flow Chart for Simulation of Particle Trajectories in a

Gas Insulated Busduct

426

427

428

429

C.2 Functions Used in the ‘C’ Language Software for simulation of

Particle Trajectories in a Gas Insulated Bus duct

Double fv(et0, y0,m,yd0,rowg,kd)

Double et0, t0,y0,m,yd0, rowg, kd;

{

Double k7,yk,k3,k4,k5,k6,k8,k9,k10;

yk=fabs(yd0);

k6=(log(2.0*1/r)-1.0);

k3=(pi*eps0*et0*v*sin(314*t0));

k4=(k6*(r0-y0)*(log(r0/ri)));

k5=m*gr+(yd0*pi*6.0*my*kd);

k7=(2.656*pi*r);

k8=(sprt(my*rowg*1));

k9=pow(yk,1.5);

k10=((k3/k4)-k5-(k7*k8*k9))/m;

return(k10);

}

double fh(et0,t0,y0,m,yd0,m,yd0,rowg,kd)

double et0,t0,y0,m,yd0,rowg,kd;

{

Double k7,yk,k3,k4,k5,k6,k8,k9,k10;

yk=fabs(yd0);

k3=(2*pi*eps0*r*1*et0*v*sin(314*t0));

k4=((r0-y0)*log(r0/ri));

k5=m*gr+(yd0*pi*r*6.0*my*kd);

k7=(2.656*pi*r);

k8=(sprt(my*rowg*1));

k9=pow(yk,1.5);

k10=((k3/k4)-k5-(k7*k8*k9))/m;

return(k10);

}

430

Particle movement using Monte Carlo Technique

Theata = angle*ran;

Th=theate *pi/180.0;

If(theata<angle/2.0)

{

y=y0*cos(th);

x+=-y0*sin(th);

}

Else

{

y=y0*cos(th);

x+=y0*sin(th);

}

fprintf(fp1,”\t%10.41f\t%8.41f\t%-d\n”, x*1000.0,y*1000.0,bou);

if(ymax<y0)

{

ymax=y0;

tmax=t0-h;

xc=x;

ymax1=y;

}

}