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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？
. {4 T4 h4 C7 I% L" k: h; T. b注：$ q5 Q" \5 D g
Machine: G2
6 M [8 @2 z% {1 X5 x4 G- YNominal: 100 MVA 10.5 kV rms
; N9 p2 j1 [9 cBus Type: Swing bus
4 T" Z) D$ |8 P- i( d# TUan phase: -40.27° 8 f8 {: L2 U5 ~. u: n! [
Uab: 52029 Vrms [4.955 pu] -10.27°
$ X z8 E6 o5 S0 P/ }6 tUbc: 48895 Vrms [4.657 pu] 157.70° , {$ c0 B7 Y/ k. W
Uca: 11025 Vrms [1.05 pu] -122.71°
( b! p- H# _+ gIa: 16777 Arms [3.051 pu] -9.91°
: w8 @' A& Y7 d& W5 OIb: 16777 Arms [3.051 pu] -129.91°
: a, C7 ?+ K! ^1 P8 ]: CIc: 16777 Arms [3.051 pu] 110.09°
7 j. _& ]: ^* c1 g: d3 UP: 1.3046e+009 W [13.05 pu] % \$ f$ c3 z/ z5 {* r& y
Q: -7.6415e+008 Vars [-7.642 pu] ( a. g. ~, j$ K
Pmec: 1.3079e+009 W [13.08 pu]
* H( u$ ^* `4 L( RTorque: 8.3264e+006 N.m [13.08 pu]
+ U/ T) L) R; }Vf: 6.067 pu # |6 G- U9 E- F7 e2 S. D$ `$ w# x
% f: c2 \5 j. c1 I7 q' N0 D1 wMachine: G1
" M* _2 f" B6 x# H" NNominal: 100 MVA 10.5 kV rms 2 E0 v/ X9 c$ c0 `# q
Bus Type: P & V generator
1 Y1 @( `7 p' k$ ]+ K% DUan phase: -40.45°
5 _4 x6 i+ y t4 G/ u5 dUab: 55176 Vrms [5.255 pu] -10.45° / c( C/ n+ ?4 s0 }- S, A* J$ z
Ubc: 50638 Vrms [4.823 pu] 158.65° 6 I: b; F% J3 L& O5 Q
Uca: 11025 Vrms [1.05 pu] -130.09°
9 f9 u# S+ I' m0 S. cIa: 17338 Arms [3.153 pu] -21.32°
5 d- G/ `2 O* tIb: 17338 Arms [3.153 pu] -141.32°
- [1 v+ A; }8 jIc: 17338 Arms [3.153 pu] 98.68°
. }+ v- ]5 c0 }. N, }. D4 t- vP: 1.5654e+009 W [15.65 pu]
, h5 g# K& R- ^* AQ: -5.4297e+008 Vars [-5.43 pu]
! f7 Z, `# t2 y: y1 _Pmec: 1.569e+009 W [15.69 pu]
7 O* f' u' g l/ N' WTorque: 9.9887e+006 N.m [15.69 pu]
. X0 L% o5 L. I# E7 f: @Vf: 6.7898 pu , X9 q E2 R0 ]7 y, h& n( C
& b2 m c5 O( j0 P6 }Machine: Load3
; F0 E: @6 C( i4 \6 g& oNominal: 115 kV rms
' K( p" b# y6 s, B; d4 lBus Type: P & Q load
5 W+ p& |1 W* Y0 IUan phase: -46.14°
2 d9 D4 ` ^/ \- IUab: 6.2602e+005 Vrms [5.444 pu] -16.14°
8 r" }0 ~9 u7 Q6 k* \( B) Y+ zUbc: 6.1132e+005 Vrms [5.316 pu] 161.43° % `, i+ F/ a4 A/ V9 v7 S) ]. i, o
Uca: 30061 Vrms [0.2614 pu] -136.62°
8 r6 ?; n" |9 {& zIa: 300.16 Arms -102.68°
5 @5 z" S: U# G& L) h7 H. QIb: 300.16 Arms 137.32° 7 K+ T4 j' n! \2 t& Y [1 A
Ic: 300.16 Arms 17.32° # `6 w7 J5 F* N% A! I6 d
P: 1.7944e+008 W + ]' Z) i3 R3 O( z7 b& u) k& |* M# X0 c
Q: 2.7153e+008 Vars
q6 _; d" x3 D+ V7 ]; n
; J q$ E E: c9 r3 Z7 X. qMachine: Load2
# k* r( {. m% [% R$ L. \Nominal: 115 kV rms
, ]5 ?0 p/ z$ r/ Y+ ^0 T: z) |; W% O$ EBus Type: P & Q load
; K' D7 t4 |. xUan phase: -41.18°
7 d, ^+ E! [0 {! i2 @. ~6 XUab: 6.0512e+005 Vrms [5.262 pu] -11.18° % k6 r% f, S4 o6 U6 \( ^* S
Ubc: 5.7431e+005 Vrms [4.994 pu] 157.92° ; E8 Y/ v. C+ ]# F4 K
Uca: 1.1609e+005 Vrms [1.009 pu] -121.95°; I! i# W' y; k
Ia: 798.9 Arms -92.72° 5 z$ Y) q7 G* m
Ib: 798.9 Arms 147.28°
# k1 h3 B/ P3 O3 L# x+ LIc: 798.9 Arms 27.28° 8 X! v) r2 [/ k
P: 5.2087e+008 W
5 ]% L4 v: S+ l) [' e' F7 J2 qQ: 6.5561e+008 Vars 1 t: s0 f3 b! `; n3 s7 Y, ]$ a
# w: ^. E9 k$ x3 e+ m2 C
Machine: Load1
/ G; c% R, n. d) {Nominal: 115 kV rms ; _& y$ ?7 k" Z) i
Bus Type: P & Q load ; _; }7 J$ E; \6 C$ b" E% w; X
Uan phase: -40.94° ( i4 q% P) N( V9 M3 M; D, b
Uab: 6.3759e+005 Vrms [5.544 pu] -10.94° 7 x4 i& h8 l" L
Ubc: 5.8926e+005 Vrms [5.124 pu] 158.63° : R! [) j8 J9 }, M: c& B
Uca: 1.2147e+005 Vrms [1.056 pu] -129.49°
" V3 c+ a+ q1 YIa: 433.21 Arms -97.84°
# q- y, `* [. G/ g& YIb: 433.21 Arms 142.16° o1 X# m. l) O3 o0 O& V3 M
Ic: 433.21 Arms 22.16° / ]% W. C; Q, o6 r
P: 2.6123e+008 W
% a% p7 h3 H8 u+ r) p5 M3 ] A! T& lQ: 4.0079e+008 Vars |
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发表于 2012-4-29 16:06:35
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