,
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''
..
"L : .. Jc.,.,.~- '().,;:. tJ • !", Q~:. 0. Cf .
'
. lSv!
r~
ll 0 t- Jt)~
k.rvL-v
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p,, Q i
f.,.__,t
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. BvJ
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IL
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f.
b .J
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g~Ac
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6.43. (a). Step 1
By inspection:
+j10.0 +j2.5
+j10.0 ~j15.0 +j5.0
+j2.5 :rj5.0 -j7.5
.~ j12.5
.. '82 (0}= ~1 (0) =0~ V2 (0) ~ 1.0
··•· Compute ·~y(O) ···
P (X) = 1.0[10cos (-90°)+5 cos(-90°)] = 0
...
2
P3 (X) ~ 2.5 cos (-90°) +5 cos (-90°) = 0
.
Q2 (X) = 1.0[10 sin (-90°) +15+5 si~ (-90°)] =0
. · [p2- P2(X)].
,.
~y(O) =
.
1
·r·-2.0-0
P3 ...:p3 (X) =. 1.0- 0
· . Q2 -Q2 (X)
~.5-0
·
'[-2.0]:
1.0
-0.5
~.
Compute ,l(O) (see Table 6.5 text)
'(b) Step 2
11 = ~i ·=-V
2 [~l; sin.(8 -8. ;-·-'8
22
2
2,1
2
)+f23 1'; sin (82 -83 -823 )]
~..:..1.0 [1 0(1) sin (-;-90°) + 5(1) sin (-90°)] = 15.
.11 . =-=
a~
. ·· .. ·. · .· · .·...
.
o .
· ·
V Y V stn (8 -8 ,-8 = (1.0){5) sm (-90 )=-5.
.23' . . ac5.3
•.
2 23 3
. . ·.
2
.
3
)
.23 .
' .
'
.
'·-
- ·aR
'
. ·
?
• ·
o
=a~_= ~~2 cos (o3 -
J232
2 -:
"
832 )
=o
.. '.
.' ' '
"
=~~ =v2 [~·l~ cos (o2 -<5; ~821 )].+ 1';3 ~ cos (8 ~ -023 ) =o.. ··
_J322
,:2
a{? ··.
,.
. :.
=a/ =-J.ir;J~ cos (o2 -~ -823 )=0
JJ23
3
. .
·[mt·
.J2]· [15. -5: OJ•·- .. ·:
3 l 4 ~ -5 7.5 0 p~r u~it
~(0) =
!_.
.
0 ' 0 15 .
,,
'
':~
'
.
,
=11y
Step 3 ·_ Solve'} &
H~~ ~~J[!~H~o:J
.
'
.
Using Gauss elimination, multiply the first equation by (-:-5115) and subtract from the second
·
· ·
··
equation:
.
1
.
1
15 .-5 . 0Jr1182
r·· -2.0 .
o.. Ao = 0.33333
.[ p .5.833333
· o ·
3
o
•
t5
11~.
'
>
,
~o.s
Back substitution:
.11 V2 =-0.5115 =-0.033333
110'_, =(133333/5.833333 =0.05714285
11o2 =F-2.o+5(o.o5714285) J/15= -0.1142857
. . . r11o: r-0.1142S57J . . .. .
..
& = 110'_, = ,.0.05714285
·
11 V2
Compute x( 1)
Step 4
'
.
.
..
.
r.~J+r·~~~J;:~:;J
. ~·r~~i;:~~;J :::=: .
.x(O) +& =
·'
1
=
'. . v2 (l)
'
-o.033333
.
~(l) r~::;J =
,
1
.
.
Check Q03 using Eq. (6.5.3)'
..
-:0333333
.0.96666667 p~r umt
'
'
Q =V3 [Y31 v; sin (£53 -J'1 -B31 )+Y3 ~V2 sin (£53 -£52 -B32 )+Y33 ~ sin (-833 )]
3
=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 =Q3 +QL 3 ::;:0.2416-t0~0.2416 per unit
Since Qa ~ 0.2416 is ~ithin the limits [-5.0;+5.0], bus 3 'remains a voltage-controlled bus.
3
·
·
This completes the first N~wton~Raphson iteratio~. ·
.
l
.~
.
'
-
..
'
,
......
>-~~~ -::::::::;::::>
·. .P,ru~~~]
&a::=f
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
··.
,
J--=
.- - dx
.··. J'!='!(•)
.
. . .· ,·... .
It and. !2 respectively; the Jacobian is found to be
If we consider the given equations to be
\
.
"'
'
'd-.
X)
dh
- ..
dxr.
· 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) =
[34. ·2] ·..
~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
X1
1
1
.
X2
._,.
,'•'
..
1
2.1
1.3
2
1.8284
1.2122
3
1.8092
1.2061
'.4
1.8091
1.2060
'•
I
- ·,
.
•
·: o-~r? ~·t P\ , r
[v; ~ o. or11
.
r .
C)
%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)
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];
end
%% .
.
%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;
while (i<maxiteration)
theta_new = theta_old+(subs(Jinv,theta,theta_old)*(-subs(fl,theta,theta_old))
error=max(abs(theta_new-theta_old));
if (error<epsilon)
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
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);
41o
Jinv=inv(J);
i~O;
while (i<maxiteration)
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