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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 \ i4 h# s8 O2 }- Z
注：
/ J& U" x5 s9 F6 AMachine: G2 " @& N. \' l6 B* Z; Z0 K0 Z- y
Nominal: 100 MVA 10.5 kV rms
7 q1 @ A+ e. O4 U# [) ^" S$ a" _Bus Type: Swing bus 8 M; @! u) d, l) g8 e6 f
Uan phase: -40.27° 6 h) ]0 P4 d Q4 A2 g/ s8 W% f
Uab: 52029 Vrms [4.955 pu] -10.27°
& C# w, L( m. l* n, G0 oUbc: 48895 Vrms [4.657 pu] 157.70° * N) [ W& [: p: t
Uca: 11025 Vrms [1.05 pu] -122.71°
( |- w: L' y4 D, `6 _Ia: 16777 Arms [3.051 pu] -9.91° ; Z" R3 B6 ^4 d( F
Ib: 16777 Arms [3.051 pu] -129.91° 5 `4 |7 O8 L: W) A0 a e: H3 a2 Q
Ic: 16777 Arms [3.051 pu] 110.09° 3 f( b7 I! ? `- X
P: 1.3046e+009 W [13.05 pu]
4 a$ W1 ]0 q: Z, h; sQ: -7.6415e+008 Vars [-7.642 pu] + L$ e1 u/ ]2 a
Pmec: 1.3079e+009 W [13.08 pu] & r6 S7 [4 t9 O. A- o' ~6 C
Torque: 8.3264e+006 N.m [13.08 pu]
- H; {/ H6 O9 x. b6 f L8 t" }. RVf: 6.067 pu ( u& J9 S5 c. p& [! v
: Q# b" [/ Y. {. J9 G6 z+ {
Machine: G1
8 u5 F5 J: l' pNominal: 100 MVA 10.5 kV rms ; B- ]1 g3 U0 N2 n* J
Bus Type: P & V generator ( }: c& g& A1 ~0 w
Uan phase: -40.45° ; ~: i0 ~ Y* ? W. [+ v l
Uab: 55176 Vrms [5.255 pu] -10.45°
; g# M2 J: g+ {& Q! sUbc: 50638 Vrms [4.823 pu] 158.65°
4 \5 b# H$ p4 L3 O& W( ]/ pUca: 11025 Vrms [1.05 pu] -130.09°
5 {1 k/ R+ }# YIa: 17338 Arms [3.153 pu] -21.32°
0 `% P8 G6 n8 @ L, ^, Y( X$ {Ib: 17338 Arms [3.153 pu] -141.32° 4 Y2 s& N8 E: A; k+ ~* |
Ic: 17338 Arms [3.153 pu] 98.68°
% v7 r8 l. y* {3 Q# tP: 1.5654e+009 W [15.65 pu] 4 b* M, s2 J; Y/ h) i
Q: -5.4297e+008 Vars [-5.43 pu] ) q) q3 i7 F$ X$ |3 ~/ L* B9 T4 f) b* J' t
Pmec: 1.569e+009 W [15.69 pu] 3 A% H- ~- E/ _9 L; X- Y$ i, b6 T
Torque: 9.9887e+006 N.m [15.69 pu] * x8 o9 o4 @! s
Vf: 6.7898 pu , R3 k- i9 ^% l
9 |3 J* O& o( n2 G- FMachine: Load3 : x! {3 L; J! g$ u$ `1 ]6 t/ k
Nominal: 115 kV rms 6 `* W+ g* B, m; g% D# }
Bus Type: P & Q load 0 U$ z( }0 N0 W8 j/ b2 M1 Z
Uan phase: -46.14° 8 ^& o' g a. N) W8 z
Uab: 6.2602e+005 Vrms [5.444 pu] -16.14°
3 A- N" W- J5 [# C( xUbc: 6.1132e+005 Vrms [5.316 pu] 161.43°
3 I! x# S( v- [1 Q8 N+ d3 A ~4 K7 `Uca: 30061 Vrms [0.2614 pu] -136.62° : e" Z( X" Z; u* S+ j
Ia: 300.16 Arms -102.68° # b- l! ~: S; G) J/ z" s* S
Ib: 300.16 Arms 137.32° & r' f$ Q1 ~% |+ ]: K1 k
Ic: 300.16 Arms 17.32° * e, \1 A& q4 f% h
P: 1.7944e+008 W
. F- N/ A& a3 v* R5 k' \Q: 2.7153e+008 Vars
6 X6 `! a& U4 h: i* a& S $ I+ _) g# a8 b# ^
Machine: Load2
, u9 x$ s( `2 _1 d$ uNominal: 115 kV rms
2 [9 L6 W4 c4 n, \5 [6 K( M' ZBus Type: P & Q load 0 ?3 g$ x3 R) ~
Uan phase: -41.18° $ Y9 b* G1 w, P4 x" j7 C5 `; O
Uab: 6.0512e+005 Vrms [5.262 pu] -11.18° ! {/ o6 K' ~' P& |( i
Ubc: 5.7431e+005 Vrms [4.994 pu] 157.92°
$ |8 F4 z. m: }" e* V3 |Uca: 1.1609e+005 Vrms [1.009 pu] -121.95°
- u0 h0 u# ~0 h1 \* |/ v$ O5 r) S2 tIa: 798.9 Arms -92.72° - D# ?7 |8 |. d2 `* ]" |
Ib: 798.9 Arms 147.28° & x1 ]8 ^ x" M9 P, D; B
Ic: 798.9 Arms 27.28°
: Q. A ]. N; O. e3 f" YP: 5.2087e+008 W ) {3 ~" G8 {6 {4 w0 [4 t
Q: 6.5561e+008 Vars
6 p7 \1 o. c1 Y" m7 _7 G 7 g+ o G/ o& {
Machine: Load1
, a" W6 f- a/ ^8 u G6 JNominal: 115 kV rms ( B/ g* {9 z4 d3 A5 h1 Y
Bus Type: P & Q load
6 H% L5 M# T3 z2 GUan phase: -40.94° 9 }6 S/ \/ D5 Z4 ^) ]# c
Uab: 6.3759e+005 Vrms [5.544 pu] -10.94°
( t: t5 ]5 Z l: k/ j2 _' @Ubc: 5.8926e+005 Vrms [5.124 pu] 158.63°
) \0 Z8 o3 N0 M& `3 e" LUca: 1.2147e+005 Vrms [1.056 pu] -129.49°* q' G5 q" \) X; h4 }/ e
Ia: 433.21 Arms -97.84° - ]+ J; B" u8 P" C9 v
Ib: 433.21 Arms 142.16°
+ g- t& ]3 E$ \8 {: r3 T+ k VIc: 433.21 Arms 22.16°
+ U+ @! l& n* S; o2 H* `7 F3 _$ RP: 2.6123e+008 W # u( x8 _: [, c
Q: 4.0079e+008 Vars |
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