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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？
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6 g. H: [+ Z' f: \( p8 EMachine: G2 / P/ O4 i, D9 _: G
Nominal: 100 MVA 10.5 kV rms
7 J' {5 T+ Z, h$ JBus Type: Swing bus
7 l, K4 e3 M3 {Uan phase: -40.27° 4 ?, l) d# h" H8 ^4 F
Uab: 52029 Vrms [4.955 pu] -10.27° % n8 a) b5 C( z6 s% }: I& A' o
Ubc: 48895 Vrms [4.657 pu] 157.70° 0 w" B# k; [7 G
Uca: 11025 Vrms [1.05 pu] -122.71° 8 Y* t, e: v& z! S6 d) F
Ia: 16777 Arms [3.051 pu] -9.91°
! L6 [5 I% y0 h( z; T0 {2 bIb: 16777 Arms [3.051 pu] -129.91°
* T! D/ l1 p4 g2 \$ W4 ?, K( _Ic: 16777 Arms [3.051 pu] 110.09°
2 X, M! s/ e+ V8 @% i- [" T% _P: 1.3046e+009 W [13.05 pu] " ~0 E7 _7 R' j% A" G
Q: -7.6415e+008 Vars [-7.642 pu]
, g: g. v- g+ V% X8 t$ nPmec: 1.3079e+009 W [13.08 pu]
) Q" h% f; V6 O3 E- E; J6 }Torque: 8.3264e+006 N.m [13.08 pu] 2 ?+ g' \7 N9 T" @( Z( h; D8 I
Vf: 6.067 pu
. E9 c4 U4 x E2 H ( W; w8 {1 E7 ]' }# P p: g, T
Machine: G1
$ ], w) T& [* Q7 ?, P3 a' p: FNominal: 100 MVA 10.5 kV rms 0 B. e0 K, G2 {1 F" H, _
Bus Type: P & V generator & C' T: {. ~( R" ?4 Y1 n0 H- @8 H
Uan phase: -40.45° , N& f W7 _/ }
Uab: 55176 Vrms [5.255 pu] -10.45° " P k5 T! ^6 T% w
Ubc: 50638 Vrms [4.823 pu] 158.65° , s; F6 I9 s1 e* e7 c5 F% u3 G) @
Uca: 11025 Vrms [1.05 pu] -130.09°
, O, U. L4 K* C3 {! |9 K. Y5 WIa: 17338 Arms [3.153 pu] -21.32° ! [1 u( a/ J* j9 b* O# @! v
Ib: 17338 Arms [3.153 pu] -141.32°
& P9 l* _3 A- M$ yIc: 17338 Arms [3.153 pu] 98.68°
, N$ U) _* ]0 ? I9 D& {3 WP: 1.5654e+009 W [15.65 pu]
3 X. N: P' _; t. X4 e6 D$ s+ i5 jQ: -5.4297e+008 Vars [-5.43 pu]
& z/ z3 V, z& `Pmec: 1.569e+009 W [15.69 pu] : j0 f% S3 S! N
Torque: 9.9887e+006 N.m [15.69 pu]
) `. Z- }1 N, {* H0 q! p! jVf: 6.7898 pu ; \- k2 a6 C4 j6 t( u4 G! z
q5 q `3 b/ |4 S9 ^
Machine: Load3 6 ?- D5 A& r6 v. l
Nominal: 115 kV rms
( a! x% p& ~ F2 _) o, cBus Type: P & Q load ( F6 w& R2 Q5 o6 B# @
Uan phase: -46.14° . g" c, z8 t% w3 [+ }
Uab: 6.2602e+005 Vrms [5.444 pu] -16.14° - l/ X$ [4 T, u, a6 w; v$ n
Ubc: 6.1132e+005 Vrms [5.316 pu] 161.43°
9 N: ]% p2 A$ j" T3 f) nUca: 30061 Vrms [0.2614 pu] -136.62°
) O0 a3 |% y. g6 [. M6 h; I' Q, uIa: 300.16 Arms -102.68°
E( U0 N& H& n& B* }Ib: 300.16 Arms 137.32° 9 o4 J( D1 j) d0 P4 S0 g8 I
Ic: 300.16 Arms 17.32° 4 C2 u# M/ {* O5 C
P: 1.7944e+008 W
5 f% P! N; t( `. }7 kQ: 2.7153e+008 Vars
# [6 J. h& v3 V2 v9 n5 B D 7 C8 u" Q+ g! \5 Y2 [- X8 P( c6 S
Machine: Load2 2 q/ r* j; q2 S; P9 ?; L2 Z9 T
Nominal: 115 kV rms ! ]! E. h- _9 {6 T2 ~+ j' o
Bus Type: P & Q load
6 w: R9 _3 k/ ~* bUan phase: -41.18° * @9 O5 b' H+ ]0 q9 r
Uab: 6.0512e+005 Vrms [5.262 pu] -11.18°
' _6 {7 n% c6 ^5 G7 L0 u% JUbc: 5.7431e+005 Vrms [4.994 pu] 157.92° " W- {: f. g3 f9 J
Uca: 1.1609e+005 Vrms [1.009 pu] -121.95°
& T- o: E7 C( aIa: 798.9 Arms -92.72°
# ?. C) K. M8 @' ~Ib: 798.9 Arms 147.28°
# U5 m3 z8 H! r4 Y) Y* H& DIc: 798.9 Arms 27.28° ' }0 t0 P. D5 q* a2 p, _2 R
P: 5.2087e+008 W
6 _! `. d6 B3 yQ: 6.5561e+008 Vars . g8 T1 {% X0 H# M4 ]# d7 [
1 s6 P) E2 v5 S9 e
Machine: Load1 ?3 A; i7 U" S, z b) ], T
Nominal: 115 kV rms
5 e. L' z L* X& s5 m u4 g: v4 PBus Type: P & Q load
% G# Q8 ?3 Q! S* Z; pUan phase: -40.94° , G2 A y, q1 U/ Y, G+ O8 {' k
Uab: 6.3759e+005 Vrms [5.544 pu] -10.94° 7 S; q& m& N; ~4 G4 G$ G; a* G
Ubc: 5.8926e+005 Vrms [5.124 pu] 158.63° / Y1 T3 s8 U% \1 H5 A9 k
Uca: 1.2147e+005 Vrms [1.056 pu] -129.49°
# z1 p+ d& x# U/ CIa: 433.21 Arms -97.84° + p& o% p; V3 R/ ^9 L( ~
Ib: 433.21 Arms 142.16°
' i' Y6 D. ]3 d. J4 K& w# |Ic: 433.21 Arms 22.16° 2 d% N0 X& a& H" N/ b: V
P: 2.6123e+008 W : u0 [5 p( j9 `; k2 ~ d. A ?
Q: 4.0079e+008 Vars |
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发表于 2012-4-29 16:06:35
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