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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？+ c$ n9 E% b1 A# g/ e3 K+ f
注：7 V, ^' N2 Y, j; z# F5 u$ M
Machine: G2
' E1 v; R; J# [9 O2 g5 BNominal: 100 MVA 10.5 kV rms
# e9 h& Z' m7 @7 R, D5 h6 XBus Type: Swing bus
# Q q: B8 ]( w B0 C; \* jUan phase: -40.27° 0 m$ U* x' F; U! D. S# }( U5 [0 V# G
Uab: 52029 Vrms [4.955 pu] -10.27°
^9 j5 S' h2 VUbc: 48895 Vrms [4.657 pu] 157.70°
. ~" } t! U- c* ^0 d' eUca: 11025 Vrms [1.05 pu] -122.71° 6 K \; g0 [3 B( N* i9 j( E& r% _& [
Ia: 16777 Arms [3.051 pu] -9.91°
$ |6 p. \3 l% D- HIb: 16777 Arms [3.051 pu] -129.91°
. h- S: U- ~$ A( SIc: 16777 Arms [3.051 pu] 110.09°
/ H; a& s& a& r8 HP: 1.3046e+009 W [13.05 pu] - V2 N( {% J+ X0 `/ p1 D6 Y
Q: -7.6415e+008 Vars [-7.642 pu] / S$ a$ A* }8 f
Pmec: 1.3079e+009 W [13.08 pu]
1 F1 y1 `- N! ^+ p+ j0 hTorque: 8.3264e+006 N.m [13.08 pu] . b4 q+ A8 D2 i% e3 A h) M1 ?
Vf: 6.067 pu 0 Q2 n2 O& E: V- Q! d F Y c! F& {
4 N1 X0 b- A5 ^Machine: G1
. p7 o, I1 l( g7 [Nominal: 100 MVA 10.5 kV rms ' M0 m" V* @% @/ F. S: q6 e7 V
Bus Type: P & V generator
+ f* p" E' F3 a q! \+ B% ZUan phase: -40.45°
1 \4 E% V/ Z$ J; |. d/ @# k& LUab: 55176 Vrms [5.255 pu] -10.45°
8 ~& _5 Z) }6 \( s# K7 NUbc: 50638 Vrms [4.823 pu] 158.65° " O) U$ g) D+ N' f. M5 Y7 L7 @2 h
Uca: 11025 Vrms [1.05 pu] -130.09°
; @2 \/ A0 j) k5 qIa: 17338 Arms [3.153 pu] -21.32°
* P% m* m" C: y6 ^! E9 uIb: 17338 Arms [3.153 pu] -141.32°
0 l; U& N+ P7 n, L1 X4 K) BIc: 17338 Arms [3.153 pu] 98.68° . g) S; S' a3 |
P: 1.5654e+009 W [15.65 pu] 3 a% }0 y. x3 }3 `- |
Q: -5.4297e+008 Vars [-5.43 pu]
# S) _8 W2 F0 G6 b/ L, PPmec: 1.569e+009 W [15.69 pu]
4 Y1 O! u) o; ~8 ~" C" VTorque: 9.9887e+006 N.m [15.69 pu] o) g% @) I& G! D! o7 c' F
Vf: 6.7898 pu a4 D6 r( s% H& }6 o v- i( o
1 n1 q7 \$ ~' mMachine: Load3 0 O7 G2 t8 w: [- d G: w( {
Nominal: 115 kV rms
1 m3 _' T( x* y6 aBus Type: P & Q load 6 W- @) i& {/ ?& I
Uan phase: -46.14°
7 A- c4 j3 S, H: |+ b4 XUab: 6.2602e+005 Vrms [5.444 pu] -16.14°
# ~. `& F* r* F4 M, O, Y) R7 GUbc: 6.1132e+005 Vrms [5.316 pu] 161.43° / K6 @' u1 g6 A8 V3 l) R3 [
Uca: 30061 Vrms [0.2614 pu] -136.62°
- ^' L g% _' n1 r; mIa: 300.16 Arms -102.68° * Z+ Y: V' P$ S& K% s: l/ b8 c, J8 X
Ib: 300.16 Arms 137.32° / M1 ~' A" |9 H [3 L+ E) M! l
Ic: 300.16 Arms 17.32°
# l$ l; e- i9 T2 hP: 1.7944e+008 W
0 L5 u" o5 W3 O; l% U. _: t3 CQ: 2.7153e+008 Vars
/ W- a+ F4 a6 P
D0 C2 o: S$ @' V' ]% P L8 {! lMachine: Load2
3 N- i% U) L5 o( V( tNominal: 115 kV rms
M& Z0 j O# J# p' G1 }Bus Type: P & Q load ' C( ?8 a) Y z }% k
Uan phase: -41.18°
% r2 _# @% p9 f& [/ iUab: 6.0512e+005 Vrms [5.262 pu] -11.18° ' q: J1 E/ `) s) o; q6 J2 b! b
Ubc: 5.7431e+005 Vrms [4.994 pu] 157.92°
# I$ {0 I9 e y( w% dUca: 1.1609e+005 Vrms [1.009 pu] -121.95°4 _/ A* j; W/ _
Ia: 798.9 Arms -92.72° + y7 ]; x4 s* R6 Q; z7 D) E
Ib: 798.9 Arms 147.28° 2 k9 H# j) y8 p8 ?
Ic: 798.9 Arms 27.28° 7 G( m/ E3 j1 z- ]' z6 D* N
P: 5.2087e+008 W
! M& I6 F9 X. H8 F3 Y& EQ: 6.5561e+008 Vars 2 B% N* o* F) F9 E* Z7 K
# M @8 b+ g8 `7 w5 [. C: y9 n1 r" gMachine: Load1
! D( L5 _4 i, _; TNominal: 115 kV rms 3 [" z0 X* |5 X, {- d. y
Bus Type: P & Q load
$ s) C: U( ^* @+ B2 mUan phase: -40.94° % H9 O! p r% F5 O0 h, h6 ^
Uab: 6.3759e+005 Vrms [5.544 pu] -10.94° 5 H5 O# @, T1 S9 o
Ubc: 5.8926e+005 Vrms [5.124 pu] 158.63° 2 \: p4 \0 H6 _ \
Uca: 1.2147e+005 Vrms [1.056 pu] -129.49°8 _; z! t( a: r% ?1 w$ l% j9 @
Ia: 433.21 Arms -97.84°
/ y; o3 l( G) P9 x- G# ^Ib: 433.21 Arms 142.16° 9 s' e& N B4 f" O2 K/ R# k! Y. e
Ic: 433.21 Arms 22.16°
# e+ u/ v0 J3 Y) {3 i! Y9 SP: 2.6123e+008 W
& S) s5 ^+ M" \: A4 N0 qQ: 4.0079e+008 Vars |
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