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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?7 }! ?( p2 Q @/ I. j) J) s5 A6 C8 s
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Machine: G2 % F5 x1 @0 X7 U% R! e: K
Nominal: 100 MVA 10.5 kV rms 8 a3 A! ^6 R9 h' y) a
Bus Type: Swing bus
9 h) S3 E) r8 W( m& V3 n6 R4 GUan phase: -40.27°
( s4 w% g/ P) Y1 H* sUab: 52029 Vrms [4.955 pu] -10.27°
- [! r7 H, h) a3 m' wUbc: 48895 Vrms [4.657 pu] 157.70° # r$ F9 J0 R- C5 w$ q/ d) c
Uca: 11025 Vrms [1.05 pu] -122.71°
; v6 @( @: x; K9 x+ z8 _, x3 } YIa: 16777 Arms [3.051 pu] -9.91° . X* w( d- c" @1 A+ B) ?+ Q+ L
Ib: 16777 Arms [3.051 pu] -129.91°
7 C$ B8 P# ~, q3 |Ic: 16777 Arms [3.051 pu] 110.09°
8 w _2 m8 y+ a. \, }: W2 nP: 1.3046e+009 W [13.05 pu]
4 w) l! ?* u0 v IQ: -7.6415e+008 Vars [-7.642 pu]
4 k* j: W$ _! p; APmec: 1.3079e+009 W [13.08 pu]
6 ~" k4 X& L8 `, Z; K6 Q! ~( [* DTorque: 8.3264e+006 N.m [13.08 pu] - U2 {: V% S. @6 S
Vf: 6.067 pu
r5 w9 n9 ?$ @) |* E% b( a+ _ : q U; Z |# Z* P9 V% S& J
Machine: G1
/ i0 K' i; b4 m; K& ?Nominal: 100 MVA 10.5 kV rms ! F, ~" {* v# J
Bus Type: P & V generator
, }) ~9 K: @2 S$ a2 c1 RUan phase: -40.45° . I: |0 c7 H& j& m
Uab: 55176 Vrms [5.255 pu] -10.45°
6 r: c5 h3 h5 sUbc: 50638 Vrms [4.823 pu] 158.65°
7 i. x5 P9 Z) z8 u: xUca: 11025 Vrms [1.05 pu] -130.09° & I- ^1 M5 U: a: X
Ia: 17338 Arms [3.153 pu] -21.32°
* N) L; \) ]7 g4 m- b! RIb: 17338 Arms [3.153 pu] -141.32° ' U; T7 w* u! J: {3 j
Ic: 17338 Arms [3.153 pu] 98.68° 8 K4 N T F; g; C* I- Z
P: 1.5654e+009 W [15.65 pu]
: o( {& ^* J9 S% U* Y, bQ: -5.4297e+008 Vars [-5.43 pu] ( c' w* w. E9 B: N' h! c
Pmec: 1.569e+009 W [15.69 pu] ) g. r+ y4 S! h) R' f! v% H
Torque: 9.9887e+006 N.m [15.69 pu]
* u/ a, q8 V8 i) VVf: 6.7898 pu % g. y4 a7 m) x- }( i1 Z5 Z- n
! M$ w% S2 I0 K& N8 u+ h' z! mMachine: Load3
& D' `5 J+ t8 V. Z8 bNominal: 115 kV rms
1 x3 R( B6 f' W, `4 CBus Type: P & Q load
" K( h# Y' u: s, J* d: \Uan phase: -46.14°
0 ]! B0 [8 y6 {0 J3 h0 C! F; X! n- U8 q# {Uab: 6.2602e+005 Vrms [5.444 pu] -16.14°
" d! b1 }' {; hUbc: 6.1132e+005 Vrms [5.316 pu] 161.43°
8 G: p4 a: a6 o9 S& k2 Z' NUca: 30061 Vrms [0.2614 pu] -136.62°
' }/ n" ~# K& R) R! b& _ j* M# zIa: 300.16 Arms -102.68° ) _4 `; a7 |, I: s" o9 t Z
Ib: 300.16 Arms 137.32°
& k8 l' o5 ]1 ?Ic: 300.16 Arms 17.32°
# h' Y9 G& L, r/ H4 m+ aP: 1.7944e+008 W 2 b" ^( ]! h& W6 p3 V# G7 I3 E! u
Q: 2.7153e+008 Vars
$ O7 b+ U/ c1 H9 H 9 V/ d9 N$ R/ w( X& {
Machine: Load2 $ T9 p# P6 ]& z0 Y r1 m
Nominal: 115 kV rms 7 G: i. a1 j7 I7 Z2 Z0 s! J( l
Bus Type: P & Q load % o4 V: B2 P* P2 I
Uan phase: -41.18°
2 _7 j7 h% T yUab: 6.0512e+005 Vrms [5.262 pu] -11.18°
5 T% I" A, s- w8 a3 IUbc: 5.7431e+005 Vrms [4.994 pu] 157.92° ! D- H' Q, x& Y& X* L1 j
Uca: 1.1609e+005 Vrms [1.009 pu] -121.95°
( G) W9 L, R$ |+ o: P3 ZIa: 798.9 Arms -92.72°
! Z, |: }# n) p1 X: H& K& x" |Ib: 798.9 Arms 147.28° 8 R. X0 R H& d+ F5 M7 f
Ic: 798.9 Arms 27.28°
& V3 s/ m# ]8 Y, UP: 5.2087e+008 W
5 u, p9 i7 G7 v1 u) _Q: 6.5561e+008 Vars + t' L' M1 Z9 C( w
* R8 C; a3 j& @3 y% F2 y3 aMachine: Load1
. g& e% G. `: U2 ~+ y9 tNominal: 115 kV rms
) C$ R4 ~' m. C. H( QBus Type: P & Q load
9 Z# @+ w' r' Q. ~3 N5 L: AUan phase: -40.94°
) [5 P" R1 w, s7 h5 m& [2 M' u$ yUab: 6.3759e+005 Vrms [5.544 pu] -10.94°
: V/ z; M1 U' @$ d4 B1 k" t8 e4 |Ubc: 5.8926e+005 Vrms [5.124 pu] 158.63° 7 \4 `) E" T" `8 x
Uca: 1.2147e+005 Vrms [1.056 pu] -129.49°8 B+ v9 o# c& U- _- |; w* u
Ia: 433.21 Arms -97.84° % F# u; O& \/ \. J
Ib: 433.21 Arms 142.16°
, T1 t2 u, N. J4 @# G/ IIc: 433.21 Arms 22.16°
2 n/ }5 A! X: R4 R7 L4 ]P: 2.6123e+008 W & ?9 a2 d7 A0 A3 H! [: L
Q: 4.0079e+008 Vars |