6 Bus Newton Raphson Exercise
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Transcript of 6 Bus Newton Raphson Exercise
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8/18/2019 6 Bus Newton Raphson Exercise
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Tính tổng dẫn của đường dây
´ y12= 1
´ z12
= 1
0.05+ j 0.2=
20
17− j
80
17
´ y23= 1´ z23
= 10.1+ j 0.5= 513− j
2513
´ y34=
1
́z34
= 1
0.2+ j 0.8=
5
17− j
20
17
´ y45=
1
́z45
= 1
0.1+ j 03=1− j 3
´ y56=
1
́z23=
1
0.2+ j 0.4=1− j 2
´ y61= 1
´ z61
= 1
0.1+ j 0.15=
40
13− j
60
13
´ y25= 1
´ z25
= 1
0.2+ j 0.5=
20
29− j
50
29
Các phần tử của ma trận YBUS
Ma trận tổng dẫn thanh cá c! d"ng#
Ý BUS
=
[ Ý 11 Ý 12 Ý 13 Ý 14 Ý 15 Ý 16Ý
21Ý
22Ý
23Ý
24Ý
25Ý
26
Ý 31
Ý 41
Ý 51
Ý 61
Ý 32
Ý 42
Ý 52
Ý 62
Ý 33
Ý 34
Ý 35
Ý 36
Ý 43 Ý 44 Ý 45 Ý 46
Ý 53 Ý 54 Ý 55 Ý 56
Ý 63
Ý 64
Ý 65
Ý 66]
-
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Ý BUS=
[
4.25339− j9.32127
−1.17647+ j 4.705880
0
0
−3.07692+ j 4.61538
−1.17647+ j 4.70588
2.25074− j8.35309−0.38462+ j1.92308
0
−0.68966+ j 1.724130
0
−0.38462+ j 1.923080.67873− j3.09955−0.29412+ j 1.17647
0
0
0
0
−0.29412+ j1.176471.29411− j 4.17647
−1+ j30
−
2
Ch$y%n các phần tử của ma trận Ý BUS &ang d"ng c'c () g!c tính ra radan
Ý BUS=
[
10.24585∠−1.14271
4.85071∠1.81577
0
0
0
5.54700∠2.15879
4.85071∠1.81577
8.65101∠−1.30759
1.96116∠−1.76819
0
1.85695∠−1.95130
0
0
1.96116∠−1.76819
3.17299∠−1.35522
1.21268∠−1.81577
0
0
0
0
1.21268∠−1.81577
4.37237∠−1.27032
3.16227∠−1.89255
0
1.8
3.
7.
2
*+ &ử các gá tr, -an đầ$ U. (/ δ 3 0/#
δ 2(0)=0o , δ 3
(0)=0o , δ 4(0 )=0o và Ú 5
(0)=1∠0 o, Ú 6(0 )=1∠0o
Lần lặp thứ 1
Tính c1ng &$2t tác d3ng (/ c1ng &$2t ph+n 4háng
(δ 2( 0)−δ
4
(0)−θ24 )+¿|Ú 2
( 0)||U 5(0 )||Ý 25|cos (δ 2
(0)−δ 5
(0)−θ25 )+|Ú 2
(0 )||U 6(0)||Ý 26|cos (δ 2
( 0)−δ 6
(0 )−θ26 )
(δ 2(0 )−δ
1−θ
21 )+¿|Ú 2( 0)||Ú 2
(0 )||Ý 22|cos (δ 2( 0)−δ
2
(0 )−θ22 )+|Ú 2
( 0)||Ú 3(0 )||Ý 23|cos (δ 2
(0)−δ 3
( 0)−θ23)+|Ú 2
(0 )||U 4(0 )||
P2
(0 )=|Ú 2( 0)||U 1||Ý 21|cos ¿
(0−0−0 )+¿1.041× 1× 1.85695× cos (0−0+1.95130 )+1.041 ×1 ×0 ×cos (0−0(0−0−1.81577 )+¿1.0412 ×8.65101 ×cos (0−0+1.30759 )+1.041 ×1.019 ×1.96116 × cos (0−0+1.7681
¿1.041 ×1 × 4.85071 × cos¿
¿0.08856
(δ 3(0 )−δ
2
(0)−θ32)+¿|Ú 3
(0 )||Ú 3(0 )||Ý 33|cos (δ 3
( 0)−δ 3
(0 )−θ33 )+|Ú 3
(0)||Ú 4(0 )||Ý 34|cos (δ 3
(0)−δ 4
(0 )−θ34 )+|Ú 3
(0)||Ú 5( 0)
(δ 3(0 )−δ
1−θ
31 )+¿|Ú 3( 0)||Ú 2
(0 )||Ý 32|cos ¿ P
3
( 0)=|Ú 3(0 )||U 1||Ý 31|cos¿
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¿1.019× 1× 0 ×cos (0−0−0 )+1.019 ×1.041 ×1.96116× cos (0−0+1.76819 )+(1.019 )2× 3.17299×
¿−0.02419
(δ
4
( 0)−δ 2
(0 )−θ42
)+¿
|Ú
4
( 0)
||Ú
3
(0 )
||Ý
43|cos
(δ
4
( 0)−δ 3
(0 )−θ43
)+
|Ú
4
( 0)
||Ú
4
(0)
||Ý
44|cos
(δ
4
(0 )−δ 4
( 0)−θ44
)+
|Ú
4
(0 )
||Ú
5
(0
(δ 4( 0)−δ
1−θ
41)+¿|Ú 4( 0)||Ú 2
(0 )||Ý 42|cos ¿ P4
(0)=|Ú 4( 0)||U 1||Ý 41|cos¿
¿1.071× 1× 0× cos (0−0−0 )+1.071× 1.041× 0 ×cos (0−0−0 )+1.071 ×1.019 ×1.21268 ×cos (0−
¿0.09241
(δ 5(0 )−δ
2
(0)−θ52)+¿|Ú 5
(0 )||Ú 3(0 )||Ý 53|cos (δ 5
( 0)−δ 3
(0 )−θ53 )+|Ú 5
(0)||Ú 4(0 )||Ý 54|cos (δ 5
(0)−δ 4
(0 )−θ55 )+|Ú 5
( 0)||Ú 5(0 )
(δ 5
(0 )
−δ 1−θ51 )+¿|Ú 5
( 0)
||Ú 2
(0 )
||Ý 52|cos¿
P5
( 0)=|Ú 5(0)||U 1||Ý 51|cos¿
¿1×1×0×cos (0−0−0 )+1×1.041×1.85695× cos (0−0+1.95130 )+1×1.019×0× cos (0−0−0 )
¿−0.09927
(δ 6(0 )−δ
2
(0)−θ62)+¿|Ú 6
( 0)||Ú 3(0 )||Ý 63|cos (δ 6
(0 )−δ 3
( 0)−θ63)+|Ú 6
(0 )||Ú 4( 0)||Ý 64|cos (δ 6
( 0)−δ 4
( 0)−θ64 )+|Ú 6
( 0)||Ú 5( 0
(δ 6(0 )−δ
1−θ
61)+¿|Ú 6( 0)||Ú 2
(0 )||Ý 62|cos¿
P6(0 )
=|Ú 6( 0)
||U 1||Ý 61|cos¿
¿1× 1× 5.54700× cos (0−0−2.15879 )+1 ×1.041 ×0 × cos (0−0−0 )+1× 1.019× 0 ×cos (0−0−0 )
¿6.6 ×10−5
(δ 2( 0)−δ
4
(0)−θ24 )+¿|Ú 2
( 0)||U 5(0)||Ý 25|sin (δ 2
(0 )−δ 5
(0)−θ25 )+|Ú 2
( 0)||U 6(0 )||Ý 26|sin (δ 2
( 0)−δ 6
(0 )−θ26 )
(δ 2(0 )−δ
1−θ
21 )+¿|Ú 2( 0)||Ú 2
(0 )||Ý 22|sin (δ 2(0 )−δ
2
(0)−θ22)+|Ú 2
(0 )||Ú 3(0 )||Ý 23|sin (δ 2
(0 )−δ 3
(0)−θ23)+|Ú 2
( 0)||U 4(0)||Ý
Q2(0 )=|Ú 2( 0)||U 1||Ý 21|sin¿
(0−0−0 )+¿1.041× 1× 1.85695 × sin (0−0+1.95130 )+1.041×1 × 0× sin (0−0−(0−0−1.81577 )+¿1.0412 ×8.65101 ×sin (0−0+1.30759 )+1.041 ×1.019 ×1.96116 × sin (0−0+1.76819
¿1.041 ×1× 4.85071 ×sin ¿
¿7.98803
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(δ 5( 0)−δ
4
(0)−θ54 )+¿|Ú 5
( 0)||U 5(0 )||Ý 55|sin (δ 5
( 0)−δ 5
(0 )−θ55 )+|Ú 5
(0)||U 6( 0)||Ý 56|sin (δ 5
(0 )−δ 6
(0)−θ56)
(δ 5(0 )−δ
1−θ
51 )+¿|Ú 5( 0)||Ú 2
(0 )||Ý 52|sin (δ 5(0 )−δ
2
(0 )−θ52)+|Ú 5
( 0)||Ú 3(0 )||Ý 53|sin (δ 5
(0 )−δ 3
(0 )−θ53 )+|Ú 5
( 0)||U 4(0 )||Ý
Q5
( 0)=|Ú 5(0)||U 1||Ý 51|sin¿
¿1× 1× 0 ×sin (0−0−0 )+1× 1.041×1.85695 × sin ( 0−0+1.95130 )+1 ×1.019 × 0× sin (0−0−0 )+1×1.
¿9.73195
(δ 6(0)−δ
4
(0 )−θ64 )+¿|Ú 6
(0)||U 5( 0)||Ý 65|sin (δ 6
(0 )−δ 5
( 0)−θ65)+|Ú 6
( 0)||U 6(0 )||Ý 66|sin (δ 6
(0)−δ 6
( 0)−θ66)
(δ 6(0 )−δ
1−θ
61 )+¿|Ú 6(0)||Ú 2
( 0)||Ý 62|sin (δ 6( 0)−δ
2
(0 )−θ62 )+|Ú 6
(0 )||Ú 3(0)||Ý 63|sin (δ 6
(0 )−δ 3
(0)−θ63)+|Ú 6
( 0)||U 4(0)||Ý
Q6(0 )=|Ú 6(0 )||U 1||Ý 61|sin ¿
¿1×1×5.54700× sin (0−0−2.15879 )+1×1.041×0×sin (0−0−0 )+1×1.019×0×sin (0−0−0 )+1×1
¿−4.64286 ×10−5
Tính đ"5 h/m r6ng của các phần tử tr6n th75 δ 2 8 δ 3 8 δ 4 8 δ 5 8 δ 6 8 |U 5|,
|U 6| 9
:;t .
∂ P2
( 0)
∂ δ 2=−|Ú 2(0 )||U 1||Ý 21|sin (δ 2(0 )−δ 1−θ21 )−|Ú 2(0 )||Ú 3( 0)||Ý 23|sin (δ 2(0)−δ 3( 0)−θ23 )−|Ú 2( 0)||U 4( 0)||Ý 24|sin (δ 2(0
¿−1.041×1×4.85071× sin (0−0−1.81577 )−0−1.041×1.019×1.96116× sin (0−0+1.76819 )−1.041
¿1.06404
∂ P2
( 0)
∂ δ 3
=|Ú 2(0 )||U 3(0 )||Ý 23|sin (δ 2(0 )−δ 3(0)−θ23 )
¿1.041×1.019×1.96116×sin (0−0+1.76819 )
¿2.03996
∂ P2
( 0)
∂ δ 4
=|Ú 2(0 )||U 4(0)||Ý 24|sin (δ 2(0)−δ 4
(0)−θ24 )
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¿1.041×1.071×0×cos (0−0−0 )
¿0
∂ P2
( 0)
∂ δ 5=|Ú 2
(0 )
||U 5(0)
||Ý 25|sin (δ 2(0 )
−δ 5(0)
−θ25)
¿1.041×1×1.85695× sin (0−0+1.95130 )
¿1.79483
∂ P2
( 0)
∂ δ 6
=|Ú 2(0 )||U 6(0 )||Ý 26|sin (δ 2( 0)−δ 6(0 )−θ26 )
¿1.041×1×0× sin (0−0−0 )
¿0
(δ 2( 0)−δ
4
(0)−θ24 )+¿|U 5
( 0)||Ý 25|cos (δ 2(0 )−δ
5
(0)−θ25)+|U 6
( 0)||Ý 26|cos (δ 2(0 )−δ
6
(0 )−θ26 )
(δ 2(0 )−δ
1−θ
21 )+¿2|Ú 2(0 )||Ý 22|cos (δ 2
( 0)−δ 2
(0 )−θ22 )+|Ú 3
( 0)||Ý 23|cos (δ 2(0 )−δ
3
(0)−θ23)+|U 4
(0 )||Ý 24|cos ¿∂ P
2
(0 )
∂ U 2
=|U 1||Ý 21|cos¿
(0−0−0 )+¿1×1.85695 × cos (0−0+1.95130 )+1× 0 ×cos (0−0−0 )(0−0−1.81577 )+¿2 ×1.041 ×8.65101 ×cos (0−0+1.30759 )+1.019× 1.96116×cos (0−0+1.76819 )+1.¿1 × 4.85071 ×cos ¿
¿2.42815
∂ P2
( 0)
∂ U 5
=|Ú 2(0 )||Ý 25|cos (δ 2(0)−δ 5( 0)−θ25 )
¿1.041×1.85695× cos (0−0+1.95130 )
¿−0.71792
∂ P2
( 0)
∂ U 6
=|Ú 2(0 )||Ý 26|cos (δ 2(0 )−δ 6( 0)−θ 26)
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¿1.041×0×cos (0−0−0 )
¿0
(δ 2(0 )−δ
4
(0 )−θ24 )+¿|Ú 2
(0 )||U 5(0)||Ý 25|cos (δ 2
(0 )−δ 5
(0)−θ25)+|Ú 2
(0 )||U 6(0 )||Ý 26|cos (δ 2
(0)−δ 6
( 0)−θ26)
(δ 2(0)−δ
1−θ
21)+¿0+|Ú 2( 0)||Ú 3
(0 )||Ý 23|cos (δ 2(0)−δ
3
( 0)−θ23 )+|Ú 2
(0 )||U 4(0 )||Ý 24|cos¿
∂ Q2
(0 )
∂ δ 2
=|Ú 2(0)||U 1||Ý 21|cos¿
(0−0−0 )+¿1.041×1 ×1.85695 × cos (0−0+1.95130 )+1.041× 1× 0 ×cos (0−0−0 )(0−0−1.81577 )+¿0+1.041× 1.019× 1.96116×cos (0−0+1.76819 )+1.041 ×1.071 ×0 × cos¿
¿1.041 ×1× 4.85071 ×cos ¿
¿−2.35059
∂Q2
( 0)
∂ δ 3
=−|Ú 2(0 )||Ú 3( 0)||Ý 23|cos (δ 2(0 )−δ 3(0 )−θ23 )
¿−1.041×1.019×1.96116×cos (0−0+1.76819 )
¿0.40799
∂Q2
( 0)
∂ δ 4
=−|Ú 2(0 )||U 4(0 )||Ý 24|cos (δ 2(0 )−δ 4
(0)−θ24 )
¿−1.041×1.071×0×cos (0−0−0 )
¿0
∂Q2
( 0)
∂ δ 5
=−|Ú 2(0 )||U 5(0)||Ý 25|cos (δ 2(0 )−δ 5
(0)−θ25)
¿−1.041×1×1.85695×cos (0−0+1.95130 )
¿0.71792
∂Q2
( 0)
∂ δ 6
=−|Ú 2(0 )||U 6(0)||Ý 26|cos (δ 2( 0)−δ 6(0 )−θ26 )
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¿−1.041×1×0×cos (0−0−0 )
¿0
(δ 2( 0)−δ
4
(0)−θ24 )+¿|U 5
(0)||Ý 25|sin(δ 2( 0)−δ
5
(0 )−θ25 )+|U 6
(0)||Ý 26|sin (δ 2(0 )−δ
6
(0 )−θ26 )
(δ 2(0 )−δ
1−θ
21 )+¿2|Ú 2(0 )||Ý 22|sin (δ 2
(0 )−δ 2
( 0)−θ22)+|Ú 3
(0 )||Ý 23|sin (δ 2( 0)−δ
3
(0 )−θ23 )+|U 4
(0 )||Ý 24|sin¿∂Q
2
( 0)
∂U 2
=|U 1||Ý 21|sin¿
(0−0−0 )+¿1 ×1.85695 ×sin (0−0+1.95130 )+1 ×0 × sin (0−0−0 )(0−0−1.81577 )+¿2 ×1.041 ×8.65101 ×sin (0−0+1.30759 )+1.019 ×1.96116× sin (0−0+1.76819 )+1.0
¿1× 4.85071 ×sin ¿
¿16.36897
∂Q2
( 0)
∂U 5
=|Ú 2(0 )||Ý 25|sin (δ 2( 0)−δ 5(0)−θ
25 )
¿1.041×1×1.85695× sin (0−0+1.95130 )
¿1.79483
∂Q2(0
)
∂U 6
=|Ú 2(0 )||Ý 26|sin (δ 2( 0)−δ 6(0 )−θ26 )
¿1.041×1×0× sin (0−0−0 )
¿0
Tư
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(δ 3(0 )−δ
2
(0)−θ32)−¿|Ú 3
( 0)||Ú 4(0)||Ý 34|sin (δ 3
(0 )−δ 4
( 0)−θ34 )−|Ú 3
( 0)||Ú 5( 0)||Ý 35|sin (δ 3
( 0)−δ 5
(0 )−θ35 )−|Ú 3
(0 )||Ú 6(0
∂ P3(0 )
∂ δ 3
=−|Ú 3(0 )||U 1||Ý 31|sin (δ 3
(0 )−δ 1−θ
31 )−|Ú 3(0 )||Ú 2
( 0)||Ý 32|sin¿
¿−1.019×1×0× sin (0−0−0 )−1.019×1.041×1.96116× sin ( 0−0+1.76819 )−0−1.019×1.071×1.21
¿−3.32390
∂ P3
( 0)
∂ δ 4
=|Ú 3( 0)||Ú 4( 0)||Ý 34|sin (δ 3(0 )−δ 4(0 )−θ34 )
¿1.019×1.041×1.96116×sin (0−0+1.76819 )
¿2.03996
∂ P3
( 0)
∂ δ 5
=|Ú 2(0 )||U 5(0)||Ý 25|sin (δ 3(0 )−δ 5(0 )−θ35 )
¿1.019×1×0×sin (0−0−0 )
¿0
∂ P3
( 0)
∂ δ 6=|Ú 2(0 )||U 6(0 )||Ý 26|sin (δ 3( 0)−δ 6(0 )−θ36 )
¿1.019×1×0×sin (0−0−0 )
¿0
∂ P3
( 0)
∂ U 2
=|Ú 3( 0)||Ý 32|cos (δ 3(0)−δ 2( 0)−θ32 )
¿1.019×1.96116×cos (0−0+1.76819 )
¿−0.39192
∂ P3
( 0)
∂ U 5
=|Ú 3( 0)||Ý 35|cos (δ 3(0 )−δ 5( 0)−θ 35)
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¿1.019×1×0×cos (0−0−0 )
¿0
∂ P3
( 0)
∂ U 6=|Ú 3
( 0)
||Ý 36|cos (δ 3(0 )
−δ 6(0)
−θ36 )
¿1.019×1×0×cos (0−0−0 )
¿0
Tư
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∂ P4
(0 )
∂ δ 5
=|Ú 4( 0)||Ú 5(0 )||Ý 45|sin (δ 4(0 )−δ 5(0)−θ45 )
¿1.071×1×3.16227× sin (0−0+1.89255 )
¿3.21299
∂ P4
(0 )
∂ δ 6
=|Ú 4( 0)||Ú 6(0 )||Ý 46|sin (δ 4(0 )−δ 6(0 )−θ46 )
¿1.071×1×0× sin (0−0−0 )
¿0
∂ P4(0
)
∂U 2
=|Ú 4( 0)||Ý 42|cos (δ 4( 0)−δ 2(0 )−θ42 )
¿1.071×0×cos (0−0−0 )
¿0
∂ P4
(0 )
∂U 5
=|Ú 4( 0)||Ý 45|cos (δ 4( 0)−δ 5(0 )−θ 45)
¿1.071×1×3.16227×cos (0−0+1.89255 )
¿−1.07100
∂ P4
(0 )
∂U 6
=|Ú 4( 0)||Ý 46|cos (δ 4 (0)−δ 6( 0)−θ 46 )
¿1.071×3.16227×cos (0−0+1.89255 )
¿−1.07100
Tư
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∂ P5
( 0)
∂ δ 2
=|Ú 5( 0)||Ú 2(0 )||Ý 52|sin (δ 5(0 )−δ 2(0)−θ 52)
¿1×1.041×1.85695× sin (0−0+1.95130 )
¿1.79483
∂ P5
( 0)
∂ δ 3
=|Ú 5( 0)||Ú 3(0 )||Ý 53|sin (δ 5(0 )−δ 3(0 )−θ53 )
¿1×1.019×0×sin (0−0−0 )
¿0
∂ P5(0
)
∂ δ 4
=|Ú 5( 0)||Ú 4( 0)||Ý 54|sin (δ 5(0 )−δ 4(0 )−θ54 )
¿1×1.071×3.16227× sin (0−0+1.89255 )
¿3.21299
∂ P5
( 0)
∂ δ 5
=−|Ú 5(0 )||U 1||Ý 51|sin (δ 5(0 )−δ 1−θ51 )−|Ú 5(0 )||Ú 2(0)||Ý 52|sin (δ 5(0 )−δ 2( 0)−θ52 )−|Ú 5( 0)||Ú 3(0 )||Ý 53|sin (δ 5(0
¿−1×1×0× sin (0−0−0 )−1×1.041×1.85695× sin (0−0+1.95130 )−1×1.019×0× sin ( 0−0−
¿−3.00781
∂ P5
( 0)
∂ δ 6
=|Ú 5( 0)||Ú 6(0 )||Ý 56|sin (δ 5( 0)−δ 6(0 )−θ56 )
¿1×1×2.23607× sin (0−0−2.03444 )
¿−2.00000
∂ P5
( 0)
∂ U 2
=|Ú 5( 0)||Ý 52|cos (δ 5(0)−δ 2( 0)−θ52 )
¿1×1.85695× cos (0−0+1.95130 )
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¿−0.68965
(δ 5(0 )−δ
2
(0)−θ52)+¿|Ú 3
(0 )||Ý 53|cos (δ 5( 0)−δ
3
(0 )−θ53 )+|Ú 4
( 0)||Ý 54|cos (δ 5(0 )−δ
4
( 0)−θ55 )+2|Ú 5
( 0)||Ý 55|cos (δ 5(0 )−
(δ 5(0 )−δ 1−θ51 )+¿|Ú 2(
0)||Ý 52|cos¿∂ P
5
( 0)
∂ U 5
=|U 1||Ý 51|cos¿
¿1×0×cos (0−0−0 )+1.041×1.85695×cos (0−0+1.95130 )+1.019×0× cos (0−0−0 )+1.071×3.1622
¿2.59039
∂ P5
( 0)
∂ U 6=|Ú 5( 0)||Ý 56|cos (δ 5(0 )−δ 6(0)−θ56 )
¿1×2.23607×cos (0−0−2.03444 )
¿−0.99999
∂Q5
( 0)
∂ δ 2
=−|Ú 5(0 )||Ú 2(0)||Ý 52|cos (δ 5(0 )−δ 2(0)−θ52)
¿−1×1.041×1.85695×sin (0−0+1.95130 )
¿−1.79483
∂Q5
( 0)
∂ δ 3
=−|Ú 5(0 )||Ú 3(0)||Ý 53|cos (δ 5( 0)−δ 3(0 )−θ53 )
¿−1×1.019×0× cos (0−0−0 )
¿0
∂Q5
( 0)
∂ δ 4
=−|Ú 5(0 )||U 4(0 )||Ý 54|sin(δ 5(0)−δ 4(0 )−θ54 )
¿−1×1.071×3.16227× sin (0−0+1.89255 )
¿−3.21299
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¿−2.00000
Tư
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¿6.61542
∂ P6
( 0)
∂ U 2
=|Ú 6( 0)||Ý 62|cos (δ 6(0 )−δ 2(0)−θ62)
¿1× 0 ×cos (0−0−0 )
¿0
∂ P6
( 0)
∂ U 5
=|Ú 6( 0)||Ý 65|cos (δ 6(0 )−δ 5(0 )−θ65 )
¿1× 2.23607 ×cos (0−0−2.03444 )
¿−1.00000
(δ 6(0 )−δ
2
(0)−θ62)+¿|Ú 3
(0 )||Ý 63|cos (δ 6(0 )−δ
3
( 0)−θ63)+|Ú 4
(0)||Ý 64|cos (δ 6(0)−δ
4
(0 )−θ64 )+|Ú 5
(0)||Ý 65|cos (δ 6( 0)−δ
(δ 6(0 )−δ
1−θ
61)+¿|Ú 2(0 )||Ý 62|cos ¿
∂ P6
(0 )
∂ U 6
=|U 1||Ý 61|cos ¿
¿1×5.54700× cos (0−0−2.15879 )+1.041×0× cos (0−0−0 )+1.019×0×cos (0−0−0 )+1.071×0
¿4.07701
∂Q6
( 0)
∂ δ 2
=−|Ú 6(0 )||Ú 2(0 )||Ý 62|cos (δ 6( 0)−δ 2(0 )−θ62 )
¿−1×1.041×0×cos (0−0−0 )
¿0
∂Q6
( 0)
∂ δ 3
=−|Ú 6(0 )||Ú 3(0 )||Ý 63|cos (δ 6(0)−δ 3( 0)−θ63 )
¿−1×1.019×0× cos (0−0−0 )
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¿0
∂Q6
( 0)
∂ δ 4
=−|Ú 6(0 )||U 4( 0)||Ý 64|cos (δ 6( 0)−δ 4( 0)−θ 64 )
¿−1 ×1.071 ×0 × sin (0−0−0 )
¿0
∂Q6
( 0)
∂ δ 5
=−|Ú 6(0 )||U 5(0 )||Ý 65|cos (δ 6(0)−δ 5( 0)−θ65 )
¿−1 ×1 ×2.23607 × sin (0−0−2.03444 )
¿
(δ 6( 0)−δ
4
(0)−θ64 )+¿|Ú 6
( 0)||U 5( 0)||Ý 65|cos (δ 6
(0 )−δ 5
(0)−θ65)
(δ 6(0 )−δ
1−θ
61 )+¿|Ú 6(0)||Ú 2
( 0)||Ý 62|cos (δ 6(0 )−δ
2
( 0)−θ62)+|Ú 6
(0 )||Ú 3(0 )||Ý 63|cos (δ 6
(0)−δ 3
( 0)−θ63 )+|Ú 6
(0 )||U 4( 0)||Y
∂ Q6
(0 )
∂ δ 6
=|Ú 6(0 )||U 1||Ý 61|cos ¿
¿1× 1× 5.54700× cos (0−0−2.15879 )+1 ×1.041 ×0 × cos (0−0−0 )+1× 1.019× 0 ×cos (0−0−0 )+1 ×
¿−4.07687
∂Q6
( 0)
∂ U 2
=|Ú 6( 0)||Ý 62|sin (δ 6(0 )−δ 2( 0)−θ62)
¿1× 0 ×sin (0−0−0 )
¿0
∂Q6
( 0)
∂ U 5
=|Ú 6( 0)||Ý 65|sin (δ 6(0 )−δ 5( 0)−θ 65)
¿1× 2.23607 ×sin (0−0−2.03444 )
¿−2.00000
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(δ 6(0)−δ
4
(0 )−θ64 )+¿|U 5
(0)||Ý 65|sin (δ 6(0 )−δ
5
(0)−θ65 )+2 ×|U 6
( 0)||Ý 66|sin (δ 6(0 )−δ
6
(0 )−θ66 )
(δ 6(0 )−δ
1−θ
61 )+¿|Ú 2( 0)||Ý 62|sin (δ 6
(0)−δ 2
( 0)−θ62 )+|Ú 3
(0 )||Ý 63|sin (δ 6(0 )−δ
3
(0 )−θ63 )+|U 4
(0 )||Ý 64|sin¿∂ Q
6
(0 )
∂ U 6
=|U 1||Ý 61|sin¿
¿1× 5.54700× sin (0−0−2.15879 )+1.041 ×0 ×sin (0−0−0 )+1.019 ×0 × sin (0−0−0 )+1.071 ×0 × sin (
¿6.61532
C1ng &$2t ph3 t+ (/ máy phát đưAc -%$ dn tr5ng đ
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⟺
[−0.13856
0.42419
0.20759
−0.30073−0.30007−8.08803−9.88195−0.09995]
=
[ 1.06404
2.03996
0
1.794830
−2.35059−1.79483
0
2.03996
−3.323901.28394
00
0.40799
0
0
0
2.03996
−4.39539
3.212990
0
−3.212990
1.79483
0
3.21299
−3.00781−2.000000.71792
3.00781
2.00000
0
0
0
−2.000006.615420
0.99999
−4.07687
2.42815
−0.391920
−0.68965016.36897
1.72414
0
−0.717920
−1.0710
2.59039−1.00000
1.79483
16.45609
−2.00000
⟺[ ∆ δ
2
( 0)
∆ δ 3( 0)
∆ δ 4 (0)
∆ δ 5
( 0)
∆ δ 6
( 0)
∆|U 2(0)|∆|U 5(0)|∆|U 6
(0)|]=[
1.06404
2.039960
1.79483
0
−2.35059−1.79483
0
2.03996
−3.323901.28394
0
0
0.40799
0
0
0
2.03996−4.39539
3.21299
0
0
−3.212990
1.79483
03.21299
−3.00781−2.00000
0.71792
3.00781
2.00000
0
00
−2.000006.61542
0
0.99999
−4.07687
2.42815
−0.391920
−0.689650
16.36897
1.72414
0
−0.71792
0−1.07102.59039
−1.000001.79483
16.45609
−2.00000
−4
−6
⟺
[ ∆ δ 2
( 0)
∆δ 3
( 0)
∆|U 2( 0)|]
=[ 0,02313 0,01344 −7,06707×10
−3
0,01368 0,02191 4,92810×10−4
8,04499×10−3
8,89889×10−10
0,01609 ][−2,86015
1,43858
−0,21993]⟺∆ δ 2
(0)=0,02313× (−2,86015 )+ (0,01344 ) × (1,43858 )+(−7,06707 ×10−3 ) × (−0,21993 )=−0,04527
∆ δ 3(0 )=0,01368 × (−2,86015 )+0,02191× 1,43858+(4,92810 ×10−4 ) × (−0,21993 )=−7,71595 ×10−3
∆|U 2( 0)|= (8,04499 ×10−3 ) × (−2,86015 )+(8,89889 ×10−10 ) ×1,43858+0,01609 × (−0,21993 )=−0,02
⇒δ 2(1)=δ
2
(0 )+∆ δ 2
( 0)=0+(−0,04527)=−0,04527(radian)
δ 3
( 1)=δ 3
(0 )+∆ δ 3
( 0)=0+(−7,71595 ×10−3)=−0,00772(radian)
|U 2(1 )|=|U 2
( 0)|+∆|U 2(0 )|=1+(−0,02655)=0,97345 đ(tđ
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Lần lặp thứ 2
Tính c1ng &$2t tác d3ng (/ c1ng &$2t ph+n 4háng
(δ 2
(1 )
−δ 2( 1)
−θ22 )+¿|Ú 2
( 1)
||Ú 3
(1 )
||Ý 23|cos (δ
2
( 1)
−δ 3(1)
−θ23 )(δ 2(1)−δ
1−θ
21)+¿|Ú 2(1 )||Ú 2
( 1)||Ý 22|cos¿ P
2
(1 )=|Ú 2(1 )||U 1||Ý 21|cos¿
(−0,04527−0−2,03444 )+¿ ( 0,97345 )2 (58,13777 ) cos (−0,04527+0,04527+1,10715 )+ (0,97345 ) (1¿ (0,97345 ) (1,05 ) (22,36068 ) cos¿
¿−3,90099
(δ 3
(1 )
−δ 2( 1)
−θ32 )+¿|Ú 3
(1)
||Ú 3
(1 )
||Ý 33|cos (δ
3
( 1)
−δ 3(1 )
−θ33 )(δ 3(1 )−δ
1−θ
31)+¿|Ú 3(1 )||Ú 2
(1)||Ý 32|cos¿ P3
(1 )=|Ú 3( 1)||U 1||Ý 31|cos ¿
¿ (1,04 ) (1,05 ) (31,62278 )cos (−0,00772−0−1,89255 )+(1,04 ) (0,97345 ) (35,77709 ) cos (−0,00772+0
¿1,97825
(δ 2(1 )−δ
2
( 1)−θ22 )+¿|Ú 2
( 1)||Ú 3(1 )||Ý 23|sin (δ 2
(1 )−δ 3
( 1)−θ23)
(δ 2(1)−δ 1−θ 21)+¿|Ú 2(
1 )||Ú 2(1)||Ý 22|sin ¿
Q2
(1)=|Ú 2(1)||U 1||Ý 21|sin ¿
(−0,04527−0−2,03444 )+¿ ( 0,97345 )2 (58,13777 ) sin (−0,04527+0,04527+1,10715 )+(0,97345 ) (1¿ ( 0,97345 ) (1,05 ) (22,36068 ) sin¿
¿−2,44905
Tính đ"5 h/m r6ng của các phần tử tr6n th75 δ 2 8 δ 3 8 |U 2|
∂ P2
(1 )
∂ δ 2
=−|Ú 2(1)||U 1||Ý 21|sin (δ 2( 1)−δ 1−θ21 )−|Ú 2(1 )||Ú 3(1)||Ý 23|sin (δ 2(1)−δ 3(1 )−θ23 )
¿− (0,97345 ) (1,05 ) (22,36068 ) sin (−0,04527−0−2,03444 )−(0,97345 ) (1,04 ) (35,77709 )sin (−0,04
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¿51,72454
∂ P2
(1 )
∂ δ 3
=|Ú 2(1 )||Ú 3( 1)||Ý 23|sin (δ 2( 1)−δ 3(1 )−θ23 )
¿ (0,97345 ) (1,04 ) (35,77709 )sin (−0,04527+0,00772−2,03444 )
¿−31,76555
(δ 2(1 )−δ
2
( 1)−θ22 )+¿|Ú 3
(1)||Ý 23|cos (δ 2(1 )−δ
3
( 1)−θ23 )
(δ 2(1 )−δ
1−θ
21 )+¿2|Ú 2(1)||Ý 22|cos¿
∂ P2
(1 )
∂ U 2
=|U 1||Ý 21|cos¿
(−0,04527−0−2,03444 )+¿2 (0,97345 ) (58,13777 ) cos (−0,04527+0,04527+1,10715 )+(1,04 ) (35¿ (1,05 ) (22,36068 )cos ¿
¿21,30225
(δ 2(1 )−δ
1−θ
21)+¿|Ú 2(1 )||Ú 3
(1)||Ý 23|cos (δ 2(1 )−δ
3
( 1)−θ23 )
∂ Q2(1)
∂ δ 2
=|Ú 2( 1)||U 1||Ý 21|cos¿
(−0,04527−0−2,03444 )+¿ ( 0,97345 ) (1,04 ) (35,77709 ) cos (−0,04527+0,00772−2,03444 )¿ (0,97345 ) (1,05 ) (22,36068 )cos ¿
¿−28,53865
∂Q2
(1 )
∂ δ 3
=−|Ú 2(1)||Ú 3(1 )||Ý 23|cos (δ 2( 1)−δ 3(1 )−θ23 )
¿− (0,97345 ) (1,04 ) (35,77709 ) cos (−0,04527+0,00772−2,03444 )
¿17,40287
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∆ Q2
(1)=Q2
(qui đ ịnh)−Q2
(1)=−2,5−(−2,44905)=−0,05095
D) hE thng -a thanh cá c! đưAc phư
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Tính c1ng &$2t tác d3ng (/ c1ng &$2t ph+n 4háng
(δ 2(2 )−δ
2
( 2)−θ22 )+¿|Ú 2
(2)||Ú 3(2 )||Ý 23|cos (δ 2
( 2)−δ 3
(2 )−θ23 )
(δ 2(2 )−δ
1−θ
21)+¿|Ú 2(2 )||Ú 2
(2)||Ý 22|cos¿ P
2
(2 )=|Ú 2(2 )||U 1||Ý 21|cos ¿
(−0,04706−0−2,03444 )+¿ ( 0,97345 )2 (58,13777 ) cos (−0,04706+0,04706+1,10715 )+(0,97345 ) (1¿ (0,97168 ) (1,05 ) (22,36068 ) cos¿
¿−3,90099
(δ 3(2 )−δ
2
( 2)−θ32 )+¿|Ú 3
(2)||Ú 3( 2)||Ý 33|cos (δ 3
(2)−δ 3
(2 )−θ33 )
(δ 3(2 )−δ
1−θ
31 )+¿|Ú 3( 2)||Ú 2
(2)||Ý 32|cos¿ P
3
(2 )=|Ú 3( 2)||U 1||Ý 31|cos¿
¿ (1,04 ) (1,05 ) (31,62278 )cos (−0,00870−0−1,89255 )+(1,04 ) ( 0,97168 ) (35,77709 ) cos (−0,00870+0
¿2,00013
(δ 2(2 )−δ
2
( 2)−θ22 )+¿|Ú 2
(2)||Ú 3(2 )||Ý 23|sin (δ 2
(2 )−δ 3
(2)−θ23)
(δ 2(2 )−δ
1−θ
21)+¿|Ú 2(2 )||Ú 2
(2)||Ý 22|sin ¿Q2
(2)=|Ú 2(2)||U 1||Ý 21|sin¿
(−0,04706−0−2,03444 )+¿ ( 0,97168 )2 (58,13777 ) sin (−0,04706+0,04706+1,10715 )+(0,97168 ) (1¿ (0,97168 ) (1,05 ) (22,36068 ) sin¿
¿−2,50003
Tính đ"5 h/m r6ng của các phần tử tr6n th75 δ 2 8 δ 3 8 |U 2|
∂ P2
(2 )
∂ δ 2
=−|Ú 2(2)||U 1||Ý 21|sin (δ 2( 2)−δ 1−θ 21)−|Ú 2(2 )||Ú 3(2 )||Ý 23|sin (δ 2(2 )−δ 3( 2)−θ23 )
¿− (0,97168 ) (1,05 ) (22,36068 ) sin (−0,04706−0−2,03444 )−(0,97168 ) (1,04 ) (35,77709 )sin (−0,04
¿51,59648
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∂ P2
(2 )
∂ δ 3
=|Ú 2(2 )||Ú 3( 2)||Ý 23|sin (δ 2( 2)−δ 3(2 )−θ23 )
¿ (0,97168 ) (1,04 ) (35,77709 )sin (−0,04706+0,00870−2,03444 )
¿−31,69371
(δ 2(2 )−δ
2
( 2)−θ22 )+¿|Ú 3
(2)||Ý 23|cos (δ 2(2 )−δ
3
( 2)−θ23)
(δ 2(2 )−δ
1−θ
21 )+¿2|Ú 2(2 )||Ý 22|cos¿
∂ P2
(2 )
∂ U 2
=|U 1||Ý 21|cos ¿
(−0,04706−0−2,03444 )+¿2 (0,97345 ) (58,13777 ) cos (−0,04706+0,04706+1,10715 )+(1,04 ) (35¿ (1,05 ) (22,36068 )cos ¿
¿21,14710
(δ 2(2 )−δ
1−θ
21)+¿|Ú 2(2 )||Ú 3
(2)||Ý 23|cos (δ 2(2 )−δ
3
(2)−θ23)
∂ Q2(2 )
∂ δ 2
=|Ú 2(2)||U 1||Ý 21|cos¿
(−0,04706−0−2,03444 )+¿ ( 0,97168 ) (1,04 ) (35,77709 ) cos (−0,04706+0,00870−2,03444 )¿ (0,97168 ) (1,05 ) (22,36068 )cos ¿
¿−28,54808
∂Q2
( 2)
∂ δ 3
=−|Ú 2(2)||Ú 3(2 )||Ý 23|cos (δ 2(2)−δ 3(2 )−θ23 )
¿− (0,97168 ) (1,04 ) (35,77709 ) cos (−0,04706+0,00870−2,03444 )
¿17,39689
(δ 2(2 )−δ
2
( 2)−θ22 )+¿|Ú 3
(2)||Ý 23|sin (δ 2(2)−δ
3
(2 )−θ23 )
(δ 2(2 )−δ
1−θ
21 )+¿2|Ú 2(2 )||Ý 22|sin ¿
∂ Q2
(2)
∂ U 2
=|U 1||Ý 21|sin ¿
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(−0,04706−0−2,03444 )+¿2 (0,97168 ) (58,13777 ) sin (−0,04706+0,04706+1,10715 )+(1,04 ) (35¿ (1,05 ) (22,36068 )sin¿
¿47,95450
Tư
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[
∆ P2
(2 )
∆ P3
(2 )
∆ Q2(2)
]=
[
∂ P2
(2)
∂ δ 2
∂ P2
( 2)
∂ δ 3
∂ P2
( 2)
∂ U 2
∂ P3(2)
∂ δ 2
∂ P3(2)
∂ δ 3
∂ P3( 2)
∂ U 2
∂Q 2(2)
∂ δ 2
∂ Q2(2 )
∂ δ 3
∂ Q2(2)
∂ U 2
][ ∆ δ 2
(2 )
∆ δ 3(2 )
∆|U 2
( 2)
|
]⟺[
−0,00007−0,00013
0,00003 ]=[
51,59648 −31,69371 21,14710−32,93386 65,59759 −15,35132−28,54808 17,39689 47,95450 ][
∆ δ 2
(2 )
∆ δ 3
(2 )
∆|U 2(2 )|]
⟺
[ ∆ δ
2
( 2)
∆δ 3
( 2)
∆|U 2( 2)|]
=
[ 51,59648 −31,69371 21,14710
−32,93386 65,59759 −15,35132
−28,54808 17,39689 47,95450 ]−1
[−0,00007
0,00013
−0,00003]⟺[
∆ δ 2
( 2)
∆ δ 3
( 2)
∆|U 2( 2)|]=[
0,02444 0,01352 −6,45042× 10−3
0,01445 0,02204 6,84809× 10−4
9,30859 ×10−3
5,13958× 10−5
0,01676 ][−0,000070,00013−0,00003]
⟺∆ δ 2(2)=(0,02444)×(−0,00007)+(0,01352)×(0,00013)+ (−6,45042 ×10−3 ) × (−0,00003 )−3,27489
∆ δ 3(2)=(0,01445)×(−0,00007)+(0,02204)×(0,00013)+ (6,84809 ×10−4 ) × (−0,00003 )=−3,85616
∆|U 2( 2)|=(9,30859 ×10−3)×(−0,00007)+(5,13958 ×10−5)×(0,00013)+ (0,01676 )× (−0,00003 )=−1
⇒δ 2
(3)=δ 2
(2 )+∆ δ 2
( 2)=−0,04527+(−1,78912 ×10−3)=−0,04706(radian)
δ 3
( 3)=δ 3
(2 )+∆ δ 3
( 2)=−0,00772+(−9,81248 ×10−4)=−0,00870(radian)
|U 2(3 )|=|U 2
( 2)|+∆|U 2(2)|=0,97345+(−1,76597 ×10−3)=0,97168 đ(tđ
Vậy Ú 2=0,97168∠−2,6960
và Ú 3=1,04∠−0,4980