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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？& h& t! C" I5 ?* U
注：
4 z, z' F( c5 B+ C- [Machine: G2 9 {' j& i+ x2 J0 J9 W
Nominal: 100 MVA 10.5 kV rms
3 I0 b/ ^) B; G) S5 sBus Type: Swing bus
/ D' O, `) S# k0 z6 AUan phase: -40.27° P7 `; r5 I* |: n8 ?
Uab: 52029 Vrms [4.955 pu] -10.27°
2 s/ |# u" K. ZUbc: 48895 Vrms [4.657 pu] 157.70° * B$ G5 C; s8 c
Uca: 11025 Vrms [1.05 pu] -122.71° + Z' \8 w- e# G4 f- a
Ia: 16777 Arms [3.051 pu] -9.91° 7 P, M' T* i: L3 ?( v( d3 @
Ib: 16777 Arms [3.051 pu] -129.91°
( l C3 I' o( IIc: 16777 Arms [3.051 pu] 110.09° ) |& j' ]' |% V$ ~1 i3 W) `, |
P: 1.3046e+009 W [13.05 pu] & A: L1 w! Z$ e' _* h
Q: -7.6415e+008 Vars [-7.642 pu] ~3 h9 C, ?5 l0 i- O
Pmec: 1.3079e+009 W [13.08 pu]
8 U7 P0 `) ^: l7 M9 [1 [. q# |; \" qTorque: 8.3264e+006 N.m [13.08 pu] 7 y$ P4 u d" O; {" i
Vf: 6.067 pu
- r, g: S7 f5 f- n/ n x. ~ $ W! ^4 c% x/ L1 o" c
Machine: G1
, L% G9 @, ^, A2 uNominal: 100 MVA 10.5 kV rms 7 t& E" @. ^/ a/ {9 e/ H
Bus Type: P & V generator
: A3 V2 @1 M- p; H4 m* \& @Uan phase: -40.45°
1 n0 L+ J4 F# O4 r* tUab: 55176 Vrms [5.255 pu] -10.45°
; z5 f( H1 ^( NUbc: 50638 Vrms [4.823 pu] 158.65°
: V9 U/ \* A4 {- L o5 X0 H9 n! EUca: 11025 Vrms [1.05 pu] -130.09° . s+ Y6 e/ B" R' B3 l# ]
Ia: 17338 Arms [3.153 pu] -21.32° & B' j0 C/ _% y2 ]7 _
Ib: 17338 Arms [3.153 pu] -141.32°
9 y ~2 s: m) U# f3 GIc: 17338 Arms [3.153 pu] 98.68°
, _/ z( Y" x1 J4 b! sP: 1.5654e+009 W [15.65 pu] / J2 p5 z9 t/ o
Q: -5.4297e+008 Vars [-5.43 pu] 8 ^. S. w9 i8 l7 h
Pmec: 1.569e+009 W [15.69 pu] ) l! s# P6 F3 T# \3 ]
Torque: 9.9887e+006 N.m [15.69 pu] 7 J4 [7 G& W1 [1 {' t6 e
Vf: 6.7898 pu 9 n/ @- O0 p7 O
/ |: X& z2 ?' m6 t7 g, `" }* XMachine: Load3
$ Y2 N+ Y; @1 Y- y- @( TNominal: 115 kV rms * H$ T) U0 h- i
Bus Type: P & Q load
; r4 y4 ^5 r, Q: u, p) \5 qUan phase: -46.14°
6 w; B7 G* `' qUab: 6.2602e+005 Vrms [5.444 pu] -16.14° % E* |% y a% r1 l
Ubc: 6.1132e+005 Vrms [5.316 pu] 161.43°
' _. G G; U& s7 F/ y. iUca: 30061 Vrms [0.2614 pu] -136.62° $ m9 |8 m- t' K2 f: c. p/ s
Ia: 300.16 Arms -102.68° ( \4 s. K% V) [. p! l) J
Ib: 300.16 Arms 137.32°
, s4 P( D9 g. r$ z% LIc: 300.16 Arms 17.32°
y: X! R& D1 \* Q& FP: 1.7944e+008 W ! c; x* ?' G& L, y* s0 s
Q: 2.7153e+008 Vars
! I' j5 m6 v+ {8 y7 \) B 7 l9 J- o- j7 w
Machine: Load2 ) _* ]3 u$ u! k2 m( j! C
Nominal: 115 kV rms $ p) L* S6 F) J* e& [
Bus Type: P & Q load
5 X$ p2 ]; T t) C) d9 j( a4 k# rUan phase: -41.18°
7 L# Q! |3 M" G, I# [Uab: 6.0512e+005 Vrms [5.262 pu] -11.18°
4 S- C/ ]% t1 H* X$ T0 Z9 x. w2 JUbc: 5.7431e+005 Vrms [4.994 pu] 157.92°
( [, y( w$ o& h3 q! n8 V. I+ [Uca: 1.1609e+005 Vrms [1.009 pu] -121.95°% ?$ N7 F( _; C8 R
Ia: 798.9 Arms -92.72°
) O# W9 `, C& J( |" cIb: 798.9 Arms 147.28° 4 j0 m# {7 G8 w Z
Ic: 798.9 Arms 27.28°
& z9 `6 X8 A4 ?! N3 W: K, t+ {P: 5.2087e+008 W % ~) f4 K9 h' f1 h' \# R! {3 r
Q: 6.5561e+008 Vars
1 ` a3 } U2 J ]0 ]" j
, m+ g; q( [4 z- s+ e8 A6 BMachine: Load1 5 x+ ]4 Y6 ], o$ ~- T2 g* I
Nominal: 115 kV rms
+ f9 b4 U# o2 l. @2 HBus Type: P & Q load * h4 t6 H1 d$ E+ J, \9 Z/ b n
Uan phase: -40.94° ' u$ G, V0 _4 q$ Q2 \
Uab: 6.3759e+005 Vrms [5.544 pu] -10.94° & h, U$ A) u& E: e7 E( C/ T
Ubc: 5.8926e+005 Vrms [5.124 pu] 158.63° , m+ ?% B9 j" p
Uca: 1.2147e+005 Vrms [1.056 pu] -129.49°0 x o' k+ m8 ^* V4 v6 h; e9 x
Ia: 433.21 Arms -97.84°
+ P# I7 [& \: [Ib: 433.21 Arms 142.16°
$ b0 K- J# T8 b$ S1 ?/ ?1 ~$ O. QIc: 433.21 Arms 22.16°
j5 j0 V8 e% E( }2 Y* T3 v) \P: 2.6123e+008 W ! R+ ?8 Y6 a" K5 l3 T! B& B+ _/ Q
Q: 4.0079e+008 Vars |
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发表于 2012-4-29 16:06:35
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