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《MATLAB/Simulink电力系统建模与仿真》中94页中的2机5节点电力系统潮流计算仿真，我参数完全按照书中设置，但调试有问题，Update Load Flow之后，得出发电机的三相电压不对称，数据如注释所示。哪位高手知道为啥？
0 f3 y3 Q+ \ b$ s! [) X3 q注：' p* R: \% `! g* ^
Machine: G2
3 W* b# v- L5 D# |6 ?! y hNominal: 100 MVA 10.5 kV rms 9 b. ?" L4 u2 i1 g {
Bus Type: Swing bus 1 S8 R N4 M+ W# L7 k( d4 @
Uan phase: -40.27°
# W' { \, F; `1 A) \Uab: 52029 Vrms [4.955 pu] -10.27°
; I8 C+ M$ a& QUbc: 48895 Vrms [4.657 pu] 157.70° 1 i7 U4 e" F, N, _ c" l
Uca: 11025 Vrms [1.05 pu] -122.71° ' B4 e% I7 K" S- ~ X8 S
Ia: 16777 Arms [3.051 pu] -9.91° 0 F h9 W8 v: {4 X1 \* o
Ib: 16777 Arms [3.051 pu] -129.91°
- h: P( I+ g1 c& k5 vIc: 16777 Arms [3.051 pu] 110.09° * A+ ~7 Y" h. v# x- X( N' G. I
P: 1.3046e+009 W [13.05 pu] 9 {9 {2 P) [' ^4 U9 T
Q: -7.6415e+008 Vars [-7.642 pu] ; Q4 _0 {/ S+ v% g
Pmec: 1.3079e+009 W [13.08 pu]
9 j$ I& ]" _$ i. j4 Y5 [! {5 k8 gTorque: 8.3264e+006 N.m [13.08 pu]
$ h/ u# x& H9 [7 _Vf: 6.067 pu
: L2 O8 L2 L( b: C) ^ q % R" Q5 F. b2 }7 b
Machine: G1
# p. o* z8 D+ D6 h( ]1 C3 U: cNominal: 100 MVA 10.5 kV rms # G- n( f# ^6 |. M Y' O
Bus Type: P & V generator
) i- p! `! \. m" J+ m: e; |1 BUan phase: -40.45°
; T. v) V# U8 r1 R$ W5 Y z& NUab: 55176 Vrms [5.255 pu] -10.45°
: X: E: D& h2 [5 M" b+ L8 @Ubc: 50638 Vrms [4.823 pu] 158.65° ! G. L. S: U+ y! o# K2 ?/ c2 D" }
Uca: 11025 Vrms [1.05 pu] -130.09°
5 K; `. x3 ?2 S% }- }Ia: 17338 Arms [3.153 pu] -21.32° / r; r! C7 I% l+ W! r# f5 H
Ib: 17338 Arms [3.153 pu] -141.32°
# q3 i5 }% H' U/ ZIc: 17338 Arms [3.153 pu] 98.68° 7 C" I e3 p( R$ q! m- @
P: 1.5654e+009 W [15.65 pu] D" _: O2 e8 M" _2 f3 X" u
Q: -5.4297e+008 Vars [-5.43 pu] 5 `7 K0 \, h! k& ~0 D
Pmec: 1.569e+009 W [15.69 pu]
8 X3 X0 H6 u; y( WTorque: 9.9887e+006 N.m [15.69 pu] 7 l o0 M3 D/ h+ g
Vf: 6.7898 pu " W" y9 F0 g" ^+ U2 ]6 S
) W1 c" I) p ^$ I9 ^8 @
Machine: Load3 7 V7 C# a) n! D4 s
Nominal: 115 kV rms
" c Y N" j$ ^0 U, ]Bus Type: P & Q load
# {' P% p6 z" d9 Z; l0 EUan phase: -46.14° ; E( z/ W! \- F, c x
Uab: 6.2602e+005 Vrms [5.444 pu] -16.14°
- K. H; P B( ZUbc: 6.1132e+005 Vrms [5.316 pu] 161.43° / N2 I- h& `- E' Y
Uca: 30061 Vrms [0.2614 pu] -136.62°
* q; g; k4 U/ b4 o6 v7 U0 QIa: 300.16 Arms -102.68° 4 {0 W7 N$ v6 S
Ib: 300.16 Arms 137.32°
" y6 Q8 T* J. [% `) F5 ]( i0 wIc: 300.16 Arms 17.32° ( F& w7 n" Q T1 \- k1 P1 W; s
P: 1.7944e+008 W
* |! l1 e8 J4 g& S. t2 r8 c- o CQ: 2.7153e+008 Vars
/ d4 A; ?% e" ~" a( M$ p+ g- k; ^ P 6 e6 X; j* T x& j, Z9 x z
Machine: Load2
. }; w$ ]4 T) t/ yNominal: 115 kV rms
9 X4 ^9 G. j5 f3 Z$ Z% a- z) oBus Type: P & Q load
0 i! [( m2 r1 o9 k: A1 |6 F1 w2 G1 W tUan phase: -41.18° ) N& [- }# t2 O8 B( `/ c8 p) [( |
Uab: 6.0512e+005 Vrms [5.262 pu] -11.18° 2 T8 ~; l& T1 Y3 m4 ?1 m, Q4 m
Ubc: 5.7431e+005 Vrms [4.994 pu] 157.92°
& Q1 h7 @8 a, j3 Q! i$ CUca: 1.1609e+005 Vrms [1.009 pu] -121.95°5 l! W: `1 R7 x; `# o, P' ~% `- K1 H% e
Ia: 798.9 Arms -92.72° * i+ x: ^/ @" Z) i2 N2 f2 }0 b
Ib: 798.9 Arms 147.28°
. f7 r$ O' ^% NIc: 798.9 Arms 27.28° ( t3 m/ E* _; @4 T8 r7 k& X
P: 5.2087e+008 W $ P" Y- ^6 P# m& `! H, T
Q: 6.5561e+008 Vars 3 X) A) n5 b( S7 U6 s# F$ N
1 g J: J/ L: W# V6 gMachine: Load1
- `( l! h% g% V7 A C7 u; aNominal: 115 kV rms # R7 E) G$ d2 H' H
Bus Type: P & Q load
# m5 m) y; o TUan phase: -40.94°
- k% ]4 U4 ^2 wUab: 6.3759e+005 Vrms [5.544 pu] -10.94° 3 e/ U) G% L( @6 i, v9 Z5 r& ]
Ubc: 5.8926e+005 Vrms [5.124 pu] 158.63° ) R' H) M: W8 x/ m p6 L( a; w
Uca: 1.2147e+005 Vrms [1.056 pu] -129.49°
. D- C. D3 i; mIa: 433.21 Arms -97.84°
( W2 x6 i1 \. u- u2 TIb: 433.21 Arms 142.16°
: K8 Q$ F% m4 q# PIc: 433.21 Arms 22.16°
6 h& p# s. @ U; B) VP: 2.6123e+008 W
# {! d; T2 S5 Y. y6 E2 E+ TQ: 4.0079e+008 Vars |
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