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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?+ h0 y$ ], ?: ^
注:
( b8 H0 R# L- j) j$ L: rMachine: G2
; u+ w/ E6 Q2 T8 GNominal: 100 MVA 10.5 kV rms 9 B- \+ I" g5 q) N& y, M
Bus Type: Swing bus
B l% \8 f+ P. J( NUan phase: -40.27°
: c& [; c. O! O e) KUab: 52029 Vrms [4.955 pu] -10.27°
" T e7 O$ n7 A) J2 _1 p( i' {( @Ubc: 48895 Vrms [4.657 pu] 157.70°
! Y; m/ u# d: C+ R/ j" `Uca: 11025 Vrms [1.05 pu] -122.71° - ?0 @0 R# [& h
Ia: 16777 Arms [3.051 pu] -9.91° q0 Y- d! O8 j" d; b# y* G
Ib: 16777 Arms [3.051 pu] -129.91° . v5 z# T+ d& s8 P& L" f5 Q5 t
Ic: 16777 Arms [3.051 pu] 110.09° ' ]4 B( T3 ^- ?7 U) R6 E6 J
P: 1.3046e+009 W [13.05 pu] 7 v* u/ M" S: X, ?6 T
Q: -7.6415e+008 Vars [-7.642 pu]
0 @. ?, ~: K6 j( W% h3 f& nPmec: 1.3079e+009 W [13.08 pu]
7 i3 X, n, W2 c9 j' U2 g1 ^ _Torque: 8.3264e+006 N.m [13.08 pu]
9 I) M( X! @) E) r7 o9 h3 zVf: 6.067 pu
! x. r- O. f) M k/ t 8 x4 U& n* [6 t: E B, g5 r3 }
Machine: G1 + F) r- }# f4 |- E, m* S: z
Nominal: 100 MVA 10.5 kV rms 5 I, V( T: Z( ^. C+ p0 B
Bus Type: P & V generator
( t' |3 g! Z0 x6 _3 ] EUan phase: -40.45° 8 V r' S/ F: Y+ b5 N7 V
Uab: 55176 Vrms [5.255 pu] -10.45° 8 B4 k. x( J N: O8 y. U
Ubc: 50638 Vrms [4.823 pu] 158.65°
- H5 W1 L" Z2 i4 Y- l, a. {Uca: 11025 Vrms [1.05 pu] -130.09°
4 @! ]+ n ^4 r; z3 |/ L" hIa: 17338 Arms [3.153 pu] -21.32° * O7 A& Y' q# F/ c% `7 r
Ib: 17338 Arms [3.153 pu] -141.32°
2 t; G* b1 T& a, _- C) gIc: 17338 Arms [3.153 pu] 98.68° - f( M5 U! A: ^" v: a* r# M
P: 1.5654e+009 W [15.65 pu] % R1 [; e% I0 S, u
Q: -5.4297e+008 Vars [-5.43 pu]
' {- R; y* w1 I3 {4 Q" \; RPmec: 1.569e+009 W [15.69 pu] 2 p: L" i3 u7 A
Torque: 9.9887e+006 N.m [15.69 pu] : N7 c- ~- a5 ~: w; {
Vf: 6.7898 pu
& K% O1 ?! r q# U3 U2 n2 }0 |. x * f4 P" v4 T" t4 _
Machine: Load3
2 u1 V$ f1 s( h; k. `- M# u ENominal: 115 kV rms
: C6 Q3 Q2 s% E5 `Bus Type: P & Q load
6 t: ]5 i7 s6 p# a6 L# I3 j4 N/ NUan phase: -46.14° & ^# h a# Y7 Q; M
Uab: 6.2602e+005 Vrms [5.444 pu] -16.14°
& a" s+ v5 j- F4 JUbc: 6.1132e+005 Vrms [5.316 pu] 161.43° # z) R* c4 V6 E: @9 Z/ }6 `
Uca: 30061 Vrms [0.2614 pu] -136.62°
& B; H; F8 I. D3 B, j3 M3 gIa: 300.16 Arms -102.68° 3 E, J3 X$ h6 @& W8 {' w; n1 N
Ib: 300.16 Arms 137.32° . p7 Q8 h! a) Q5 R+ I
Ic: 300.16 Arms 17.32°
" O% H: [3 [& M f: ]P: 1.7944e+008 W
, ?: a% W6 A1 u! Z% F V/ v' NQ: 2.7153e+008 Vars y- l# f$ f/ |; ^8 p
) m) t8 S& q+ l9 g" p
Machine: Load2 ( i" Q h$ b" T0 P
Nominal: 115 kV rms
0 o; o3 Q( G/ c2 P% x rBus Type: P & Q load
& ^+ R: s/ |, p, |4 NUan phase: -41.18° - |7 a) j& l) u' W* ^3 B g' G0 ]
Uab: 6.0512e+005 Vrms [5.262 pu] -11.18° 1 R6 E( X' G$ W% _7 l
Ubc: 5.7431e+005 Vrms [4.994 pu] 157.92°
; w: F- D$ X' f) u aUca: 1.1609e+005 Vrms [1.009 pu] -121.95°& q7 j& q: d5 {* h- C
Ia: 798.9 Arms -92.72°
6 y, Q+ f4 |% f$ L8 m8 e- Y! [Ib: 798.9 Arms 147.28°
& ]3 j a3 A: `$ uIc: 798.9 Arms 27.28° ' l/ \6 G! |/ p$ v( U+ v
P: 5.2087e+008 W
9 o8 a: ?# Z" E2 W5 Q6 DQ: 6.5561e+008 Vars
/ K: @! h- G( z3 z. H
( J9 v# g7 i# \# W6 kMachine: Load1
3 w# B C6 l% V: a* Z1 |Nominal: 115 kV rms
7 i3 V5 G" e$ U* U% i$ t# }/ g) m4 UBus Type: P & Q load & o0 \: n: {# B
Uan phase: -40.94° , G x) h0 ^, n4 `( o8 y
Uab: 6.3759e+005 Vrms [5.544 pu] -10.94°
- k+ n$ v) e( I& c7 @5 yUbc: 5.8926e+005 Vrms [5.124 pu] 158.63°
' O' V, z& W d" q6 m# m* {& mUca: 1.2147e+005 Vrms [1.056 pu] -129.49°
. |7 f2 [0 z9 X. V) I! FIa: 433.21 Arms -97.84°
% V1 V' {& @7 M2 F$ m2 E! EIb: 433.21 Arms 142.16° 8 V: A: t0 [$ d% U
Ic: 433.21 Arms 22.16° * j+ |( h* \4 A! G. X$ o6 I& f
P: 2.6123e+008 W " @/ _9 k+ d5 }& k
Q: 4.0079e+008 Vars |