,j;,_ ;:. v ._\ h-.l u [- ~UYl -~-\kd v, c -1 (.h -r,- '' ~.r6·J

(1)

, '

• • t

''

X ~~~'" 1 .-.. J

·Hw~

. . '

. lSv! r ~ k.rvL-v ll 0 t- J t)~

f.,.__,t p,,

^Qⁱ

. BvJ

"L : ..

Jc.,.,.~- '().,;:.

^{tJ •}

!", Q~:.

0. Cf .

,, .

.t:,--...~

.V .;__:

^I

/,f

^1.:-

·· ·. ( . IL f.

~

^•

g~Ac •.

^·./f

-=·[f"]·.;. \.:-_. _··[· 0!:

^{L .}

II,.··]··.· ... -· .

. · U

SO-)' . >"'}'C.. b

.J

I " · . I .. V'l- . J' J . . . d-J-1-

o4V"l. ..

. . - . . : · · ol

^~i.

J_n .· . · ·

· . , _ e · - - '

---c-.. '

~ f'L . . .• . . ' . ' ·.. • . • .·. .· . . . cAh "'"~ ..

-.--:. -v-r:[l'r~,lv, s'"(O-Jv'~llb,.rc;'Jj .. ,

. ()\ t.,, . . . ·. ·. ' .. ..· . . ..·· '.

I · , ·

d p . ' ''. . . ~ ', ' ' '' .·· ... ' ' ' ' . . .· '

,j;,_ ;:. v ._\ h-.l u [- ~UYl -~- \kd v, c

^-1

(.h -r,- '' ~.r6·J . :.rl"Y.,,_\Vi. c.:} ( f-~--: r~ +_&3. ifr·J

~ o. )") :r-···

-~

·_s,·-; {"'''/ ~a~ . . . ..

' . -~:. -.f.4bLf'

: ot Q ~ II

^1:\'l_: ;: . ' . .. .,_ f\JI(· r , '

() v'l- .- . . . . . " .

' '

(2)

.

_,

6.43. (a). Step 1

•.

By inspection:

.~ j12.5 +j10.0 +j2.5 +j10.0 ^~j15.0 +j5.0 +j2.5 :rj5.0 -j7.5 .. '8

2(0}= ~

1

^{(0) =0~} ^V²⁽⁰⁾^~^1.0

· ·•· Compute · ~y(O) ···

P2(X) = 1.0[10cos (-90°)+5 cos(-90°)] = 0

. . .

P3 (X) ~ 2.5 cos (-90°)

+

5 cos ( -90°) = 0 .

Q

2(X) = 1.0[10 sin (-90°)

+

15+5

si~

(-90°)] =0

. ·· [p2- P2(X)].

^·r·^-2.0-0

1 .

^'[-2.0]:

, .

~y(O)

⁼ ^P3 ...:p₃(X) =. 1.0-0

~.

^1.0

· · . Q

2-Q₂(X) ~.5-0 -0.5 '(b) Step 2 Compute ,l(O) (see Table 6.5 text)

1122 =

~i ·=-V 2 [~l;

^sin.(8²

-8.

^;-·-'8^2,1^)+f²³^1';^{sin (8}²^-8³^-8^{23 )]}

2 ~ . .:..1.0 [1 0(1) sin (-;-90°) + 5(1) sin ( -90°)] = 15.

. .

^a~

. ·· .. ·. · .· · .·... .

^{o ..} ^· ^·

1123 = - = V

2Y

23V

3stn (8₂-8₃,-8_{23 )}= (1.0){5) sm (-90 )=-5.

. . . ac5.

_' ₃

. ·.

_.

. . .

_{' .} _'

.

. /

(3)

- ·aR . · • ·

J232 =a~_= ~~2 cos

(o

3 -

o

2 -: 8_{32 )}

= o

' ? " . . ' . .' ' ' "

_ J322

= ~~ =

^v2

[~·l~

^{cos (o2}

-<5; ~8 21 )].+ 1'; 3 ~

^{cos (8}²^,:-

~ -0

^{23 )}

= o .. · ·

a{? ··. , . . :.

JJ23

=a/

^=-J.ir;J~^{cos (o}2 -~ -8₂₃)=0

3 . .

·[mt·

^.J

2]· [15.

^-5:

OJ•·- .. ·:

~(0)

=

^!_.₃

l

₄ ^{~ -5 7.5} ⁰ ^{p~r u~it}

. 0 ' 0 15 ...

' ,, ' ':~ . ,

Step 3 ·_ Solve'} &

=

^11y

H ~~ ~~J[!~H~o:J

. ^'.

Using Gauss elimination, multiply the first equation by (-:-5115) and subtract from the second

equation: · · · · ·

[ 15 . -5 . 0Jr1182

1 . r···

^{-2.0 .}

1 .

. p .

5.833333 o. ..

Ao

₃

=

^0.33333

· o · o t5 11~. ~o.s

• ' > ,

Back substitution:

. 11 V₂

=

^-0.5115

=

-0.033333

110'_, =

(133333/5.833333

=

0.05714285

11o

2

=

F-2.o+5(o.o5714285) J/15= -0.1142857

. . . r11o:

1

r-0.1142S57J . . . . . ..

&

= 110'_, =

,.0.05714285

· 11 V₂ -o.033333 ..

Step 4 Compute x( 1) . .

~(l)

⁼

r~ ::;J =

^.x(O)^+&

= r.~J+r· ~. ~~J;:~:;J ~·r~~i;:~~;J :::=: .

' '. . v2 (l) ·' 1 -:0333333 . 0.96666667 p~r umt

, ' . . . .

Check Q₀₃using Eq. (6.5.3)'

..

'

·-

(4)

' '

Q3 =V

3[Y

31

v;

sin (£5₃-J'

1

^-B

31

^)+Y

3

^~V

2

^sin^(£5³^-£5²^-B

32

^)+Y

33

^~^{sin (-8}^{33 )]}

=

1[(2.5)(1) sin(o.o5714 :_ ")

+

5(0.966666) sin (.

0.~5714+0.114~9-

") · · .

· radoans 2 . 2

,' ' . .· . .

+7.5{l)sin ( ;)]

Q3

=

1[-2.4959- 4.7625+7.5] =0.2416 per unit · QG3 =Q_{3 +QL3}::;:0.2416-t0~0.2416 per unit Since Qa

3

~

0.2416 is

~ithin

the limits [-5.0;+5.0], bus 3 'remains a voltage-controlled bus.

This completes the first N~wton~Raphson iteratio~. · · ·

.

^-

l .. '

.~ . ^'

(5)

,'•'

>- ~~~

&a::=f

-::::::::;::::>

· ... P,ru~~~]

...

Solution The Newton-Rhapson method is described by the following equation

' ' _'. : ~ • ' ' ' • ' . . < ' •

. : . .

~(i + 1) == ~(l) ~-,L-¹L[~(i)], . where the Jacobian, ,L, is evaluated at ~( i) ~nd defi11~d as

... " · ... [dh

df ··. , 'd-.

J--= . - X)

- - dx .··- dh

. J'!='!(•) - ..

. . . .· ,·... . dxr.

If we consider the given equations to be It and. !2 respectively; the Jacobian is found to be

\ ' "' .

· The first iteration can then·be computed as follows

~(1)

=

~(0)-:- I_\¹.l.[~(O)j, where ~(0), ll'!(o) and L[~(O)] are given respectively as .

.

~(O) =rn

· .:· I..l_x(O) =

[4 . 2]

3 _~1· · ..

Continuing the same process, it is seen that the values converge 'to within 3 decimal places (f= 0.001) in 4

iterations. · ' · · · · · ·

·.. ^' :.·.·, . ^' ^' -.

Table 1: Newton-Rhaps.on Results After 4 Iterations.

Iteration 0 1 2 3 '· '.4

. X1 1 2.1 1.8284 1.8092 1.8091

X2 1 1.3 1.2122 1.2061 1.2060 '•

._,.

..

, I

-·,

. •

(6)

. r .

· : o-~r? ~·t P\ , r

^C)

[v; ~ o. or11

(7)

%Problem\)

%solv~ for xi x2 syms xlx2

fl:Xl~~+3*X2A2-31;

f2=xl+xl;x2A2-20;

f = [fl; f2];

x=[xl, x2];

J

=

jacobian(f, x);

solution=[l; 1]; %initial guess

· maxi teration=B;

epsilon=O.OOl;

x_old=solution(i,l);

Jinv=inv(J);

i=O;

while (i<maxiteration)

end

x_new

=

x_old+(subs(Jinv,x,x_old)*(-subs(f,x,x_old)));

error=max(abs(x_new~x_old));

if (error<epsilon) i=maxiteration;

end i=i+l;

x old=f{.:_new;

solution=[solution x_new];

%% . .

%Problem~

^£.\:01.

clear all syms theta;

Vl=l;

V2=0.95;

P=l. 5;

X=0.2;

fl=P+Vl*V2*sin(theta)/X;

J

=

jacobian(fl, thet~);

solution=[pi/2]; %initial guess maxiteration=B;

epsilon=0.005;

theta_old=solution(l);

Jinv=inv(J);

i=O;

theta_new = theta_old+(subs(Jinv,theta,theta_old)*(-subs(fl,theta,theta_old)) error=max(abs(theta_new-theta_old));

if (error<epsilon)

(8)

2

I

Newton_HwB.m

i=maxiteration;

end i=i+l;

theta old=theta_new;

solution=[solution.theta_new];

end

solution=solution./pi*lBO

%%

%Problem

!5Q 41o

clear all clc

syms theta2 V2;

Vl=l;

.P=l.S;

Q=0.2;

X=0.2;

fl=P+Vl*V2*sin(theta2)/X;

f2=Q+V2*V2/X-V2*cos(theta2)/X;

J = jacobian([fl; f2], [theta2;.V2]);

solution=[O; 1]; %initial guess maxiteration=20;

epsilon=O.OOOl;

x_old=solution(:,l);

Jinv=inv(J);

i~O;

x_new = x_old+(subs(Jinv, [theta2, V2],x_old)*(-subs([fl;f2], [theta2, V2],x_ol error=max(abs(x_new-x_old));

if (error<epsilon) i=maxiteration;

end i=i+l;

x old=x_new; '

solution=[solution x_new];

end

iolution(l,:)=solution(l,:)./pi*180