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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?% ]8 e' U6 D: j7 ~/ J/ ^
注:
' t" M9 v; [( Q) @# t$ o5 ? Machine: G2
* v2 U9 _( L' W6 }8 [- R Nominal: 100 MVA 10.5 kV rms
% b2 x& p( @# B7 K3 o Bus Type: Swing bus
. g l# @2 M# R Uan phase: -40.27° # v% \* a$ H/ \7 O, }5 t
Uab: 52029 Vrms [4.955 pu] -10.27° $ J3 D' P. c; ]9 `6 m/ ^+ L
Ubc: 48895 Vrms [4.657 pu] 157.70°
' R+ }+ N! ~/ L: T! {, A, t Uca: 11025 Vrms [1.05 pu] -122.71°
- t$ p' x" f5 m3 l) m4 D; d; { Ia: 16777 Arms [3.051 pu] -9.91° # y& e! U1 c0 ~8 a
Ib: 16777 Arms [3.051 pu] -129.91° ) Z7 N6 B" f4 X' G
Ic: 16777 Arms [3.051 pu] 110.09° - _! C( d' }0 @
P: 1.3046e+009 W [13.05 pu] $ e' [5 s+ N3 G/ f% x
Q: -7.6415e+008 Vars [-7.642 pu]
h' T* {3 s5 t. o* f Pmec: 1.3079e+009 W [13.08 pu]
" ?1 I* W3 Y* E4 M6 W3 M, N Torque: 8.3264e+006 N.m [13.08 pu]
5 L& }& g4 w2 x Vf: 6.067 pu 6 ] n$ X% O# e2 ~1 ~. k0 [- s
$ u6 m/ L5 c7 E% c Machine: G1 7 g$ a2 d, k0 d8 A7 F* l
Nominal: 100 MVA 10.5 kV rms : F9 `* d4 Z: L6 c+ Q( C
Bus Type: P & V generator ) g4 b5 G: Z3 c. S. V
Uan phase: -40.45°
! }9 q& A1 _4 ]; F4 z$ p7 l Uab: 55176 Vrms [5.255 pu] -10.45°
' B. F8 V7 i. { h+ j Ubc: 50638 Vrms [4.823 pu] 158.65°
" H8 u) ~* j' Z Uca: 11025 Vrms [1.05 pu] -130.09° , A9 g; o+ d& y; k
Ia: 17338 Arms [3.153 pu] -21.32° 7 J. [3 v; x8 K( Y# _
Ib: 17338 Arms [3.153 pu] -141.32°
9 B/ y4 {" h! Z1 s/ C Ic: 17338 Arms [3.153 pu] 98.68° 3 |* i4 \) k/ o9 ~% H
P: 1.5654e+009 W [15.65 pu]
2 V6 Q+ E6 N8 R: l4 {+ r Q: -5.4297e+008 Vars [-5.43 pu]
1 v. r/ T/ V1 A, W Pmec: 1.569e+009 W [15.69 pu]
: |% E3 k5 m' W" F2 N+ [ Torque: 9.9887e+006 N.m [15.69 pu]
& U5 }( W/ M( h. \/ E Vf: 6.7898 pu $ d' f G+ r$ y* t) H6 r+ }+ w
! x; D/ w, r8 V4 Z
Machine: Load3
1 o; P' ]7 ]6 d! Q# t/ r Nominal: 115 kV rms
a6 h# C# `8 t) ? Bus Type: P & Q load 1 `. b1 \+ Y) U% q" W' Q& F& B/ w; O
Uan phase: -46.14° ! \' d+ y2 J. o$ h2 y0 f, l' {3 Q
Uab: 6.2602e+005 Vrms [5.444 pu] -16.14°
- M# y/ A( k8 D: D! ~9 m Ubc: 6.1132e+005 Vrms [5.316 pu] 161.43° + f) D- t O2 Z2 I1 E: v" h9 e
Uca: 30061 Vrms [0.2614 pu] -136.62° , D f- c1 @5 L9 b% f
Ia: 300.16 Arms -102.68° 0 ?3 z: V2 f" I
Ib: 300.16 Arms 137.32° 4 O+ z# M$ H, ~; p1 a9 m' A5 B
Ic: 300.16 Arms 17.32°
0 Z/ P6 `, x! {. I3 M P: 1.7944e+008 W
8 L6 @. _% g! n9 j( ]3 O+ L5 Q Q: 2.7153e+008 Vars : M+ L4 ?/ ]7 e
h) w8 U3 V0 J6 O Machine: Load2 : g/ T2 o+ S) C
Nominal: 115 kV rms
$ P- C& }' V" E Bus Type: P & Q load + |; S7 q5 P u7 N" v+ K* Y* S ?9 l
Uan phase: -41.18°
) t+ C) k$ I, D' x' ?! r Uab: 6.0512e+005 Vrms [5.262 pu] -11.18°
2 f2 a# U2 _1 s* z" X/ f Ubc: 5.7431e+005 Vrms [4.994 pu] 157.92°
; s: u3 Z3 X+ P! w* }8 K Uca: 1.1609e+005 Vrms [1.009 pu] -121.95°
/ x+ n) j9 Y0 T Ia: 798.9 Arms -92.72°
- l3 A) p* v& t5 D7 a N Ib: 798.9 Arms 147.28° 2 y) g- j8 c% b- H5 @
Ic: 798.9 Arms 27.28° 4 @; W& ~9 S$ Q# _. q$ W& m2 J
P: 5.2087e+008 W b. m7 t- c, K8 }4 b
Q: 6.5561e+008 Vars
* e w! d1 Z( k
. r) z/ U, v B D) P: J Machine: Load1
@% S) a# @, B Nominal: 115 kV rms
/ j5 O% o3 f2 w Bus Type: P & Q load
+ t/ h! r% m/ p9 c2 Z; }: Q Uan phase: -40.94° 0 | e' V. q# F2 A
Uab: 6.3759e+005 Vrms [5.544 pu] -10.94°
* u" x# d% M% y* U* L3 J Ubc: 5.8926e+005 Vrms [5.124 pu] 158.63° / I! Z4 \+ x7 K+ {" [: r
Uca: 1.2147e+005 Vrms [1.056 pu] -129.49°
# P- s' r% E, J9 u% T; r! G Ia: 433.21 Arms -97.84°
; [9 l4 R. }# [5 u1 H Ib: 433.21 Arms 142.16°
2 p' i9 c: q0 J3 J+ d7 E' }# { Ic: 433.21 Arms 22.16°
* _4 z6 y+ e1 i& h1 K P: 2.6123e+008 W
7 E- h9 e5 {. F5 V3 k Q: 4.0079e+008 Vars
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