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Unformatted text preview: Homework 5 Solutions 2.38 The admittance diagram for the system is shown below: where and 2, 7,2 )7” )7“ ~85 2.5 5.0 0 — 72. 722 723 724 .2.5 —8.75 5.0 0
Y”"":Y )7 )7 )7 =1 50 50 225195
3! 32 .13 34 '  _ ' ~' 7;, )7“ 7;, )7” 0 0 12.5 —12.5
TI=TH!+~—2+—’H ’71:=T‘n+_’17+.‘:3721=n+_2x+v,u
fu:vu’)7ll=—"l=_:—:12'}7ll=—u=—_’H K1::1'_\1‘ 6.9 x2—3x,+l.9:O
x2 +x,2 3.0=0 Rearrange to solve for x, and x3: Starting with an initial guess of xl(0) = l and 5(0) =1 x,(l) =%x3(0)+0.6333 =0.9667
x2(1)=3~[x,(0)]3 = We repeat this procedure using the general equations 6.18 .r‘ +8.):2 +2x50=0
x(0)=l J(i)=i’i =[3x3 +6x+2]
(1x .V—‘M i) .\' =11 i ) In the general form: x(i+1)=x(i)+ —'3— 42." 50
3x(i)2+16x(i)+2x("(') 8“” “0+ ) Using x(0)=1. we arrive at After 5 iterations, 8 < 0.001 .
x = 2. 126 6.30 ——___
_—_—_
_———— Solving using MATLAB with V2(O)= 1.0 A 0 and V3(O)= 1.0 4 0: V2 m 0.9552 é—6.0090° 0.90904—12.6225°
m 0.92204—12.5288° 086304—1618)? After 2 iterations,
V2 =0.90904—12.6225°
K 20.86304—16.1813° 6.38 . IJ ﬁll
:50 MW
v} = Low. 150 Mm, First. convert all values to per—unit.
P —150 MW —.._ _— Sh“: IOOM‘VAzLS p.U.
__ Q —50 MW QM sh... W7“0'5 PU Av Arm... _ 0.1 MVA N 6...... 'm='E'4P~" Since there are 2 bases, we need to solve 2(n — l) = [in ll 2 equations.
Therefore, J has dimension 2x2. Using Table 6.5 “22 ="V2V22VI Sin“); ‘6: ‘921) .1222 = V2)”22 (:os(022)+l’2,l",cos(52 —5, 62,)+Y_,3V2 cos(~623)
J322 =V2Y21Vi cos((52 '61 "921) J423 =—V2Y22 sin 922 +Y21V Sin(52 5, ~621)+ Y:2 + V_, sin(—93,)
Also, Y {ﬁlo le]_[lO£90 10490 J h le ~le '10490 tog—90 Finally. using Eqs. (6.6.2) and (6.6.3), I: (6.64) = V. [m cos(6. —6. ~6..)+ m cod—62.)]
Q2 (arvz) = v. [Yul/l sinw2 «5, —az,)+ rnv2 sin(—933)]
Solving using MATLAB with V2(0) = l .0 A 0. 0.9500 A — 8,5944°
After 3 iterations V2 = 0.9334 A — 92473" 3 0.93384—9.23l9° 0.9334492473" ...
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 Fall '08
 Overbye,T

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