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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？9 z8 f" `: a6 H9 [) _
注：
6 g$ z- d" O- WMachine: G2 $ z9 [. X& L/ f2 D9 a q: c
Nominal: 100 MVA 10.5 kV rms
/ S C( i! Q2 z. W w' I; `+ lBus Type: Swing bus 0 `' R5 G/ ~& Q( P9 b0 v' P# L
Uan phase: -40.27° 4 t, M6 Q) r, C. e
Uab: 52029 Vrms [4.955 pu] -10.27° 6 _' k8 t5 [4 _; K, o) o
Ubc: 48895 Vrms [4.657 pu] 157.70°
0 g& L6 w- `1 h0 C! r$ G8 UUca: 11025 Vrms [1.05 pu] -122.71° # s: V5 D! ? }" o2 @. E
Ia: 16777 Arms [3.051 pu] -9.91°
- @ M. c% |- u9 ]Ib: 16777 Arms [3.051 pu] -129.91° 5 M- Q/ x" H/ L- I( ^
Ic: 16777 Arms [3.051 pu] 110.09°
. c Y* x y% H- U/ bP: 1.3046e+009 W [13.05 pu] 4 O5 \1 G9 } b$ f
Q: -7.6415e+008 Vars [-7.642 pu]
% W3 N3 f# q* _! X) p. FPmec: 1.3079e+009 W [13.08 pu] % M9 [$ ]! ~, @4 o$ z* E
Torque: 8.3264e+006 N.m [13.08 pu]
, W( l& b* k/ \* q# sVf: 6.067 pu
0 }( \" s" o* s$ M6 m4 F
9 h; x( o/ R5 c6 xMachine: G1 7 `! C% N3 V( P9 c3 ^- T4 h' j4 J
Nominal: 100 MVA 10.5 kV rms ! {0 Z" G( ]9 V* y
Bus Type: P & V generator . ]3 W* Q+ E& l* d
Uan phase: -40.45°
8 A p, |6 ^! H( W- J( [9 y3 i3 mUab: 55176 Vrms [5.255 pu] -10.45° " V2 J4 p$ W) S+ T+ F$ [" n: k
Ubc: 50638 Vrms [4.823 pu] 158.65°
! ~0 x; Y+ i$ Y9 m2 t4 n" [Uca: 11025 Vrms [1.05 pu] -130.09° - r& |% d9 g" m$ B
Ia: 17338 Arms [3.153 pu] -21.32°
: i4 [ \/ f, }8 g$ \1 JIb: 17338 Arms [3.153 pu] -141.32° ; |2 b+ L/ p1 Y! a
Ic: 17338 Arms [3.153 pu] 98.68° ( s5 v/ u& R- P4 {# a* E
P: 1.5654e+009 W [15.65 pu]
! f6 ]0 g9 [9 Y, |Q: -5.4297e+008 Vars [-5.43 pu] + F* Y2 h- [/ \ f; d' ]& O* a
Pmec: 1.569e+009 W [15.69 pu] & k3 \, F& K5 e# N) ^5 t4 p1 m
Torque: 9.9887e+006 N.m [15.69 pu]
- D5 L0 U- P: JVf: 6.7898 pu & _% Z* D% S/ p, Z1 V
) ^- P. u& _7 i0 |Machine: Load3
0 K: A4 W$ K, X2 I6 U: X' ^+ F! BNominal: 115 kV rms : ^9 q/ r/ s& b" T3 j
Bus Type: P & Q load
7 I# R9 ^9 l% S& N/ XUan phase: -46.14° q; U4 f5 Q- b+ Z4 x2 n
Uab: 6.2602e+005 Vrms [5.444 pu] -16.14° ) U8 t: m+ r# n) V' }; J
Ubc: 6.1132e+005 Vrms [5.316 pu] 161.43°
4 `* [. e; ?4 @Uca: 30061 Vrms [0.2614 pu] -136.62° / c8 x- R( U" m$ ?0 H
Ia: 300.16 Arms -102.68°
; L! t( x+ T' o: `/ lIb: 300.16 Arms 137.32°
) z" L9 k6 C( f, V+ b2 oIc: 300.16 Arms 17.32°
5 m# ^4 |, { T3 KP: 1.7944e+008 W 6 J2 v! f E* F
Q: 2.7153e+008 Vars
/ L5 G/ q$ R1 n$ B$ q * w% s0 f! I9 s5 F+ T7 n" Q9 y0 Y y
Machine: Load2 ( h- N' b* P, Y
Nominal: 115 kV rms
4 R, G& y4 @& ~0 CBus Type: P & Q load
& H( r0 Q+ D+ i4 TUan phase: -41.18° 3 x3 ` u9 Y: N) v% a% Z
Uab: 6.0512e+005 Vrms [5.262 pu] -11.18°
* x! j6 g$ j' E/ I% _% ?+ EUbc: 5.7431e+005 Vrms [4.994 pu] 157.92°
5 H3 W8 |1 s" ~& cUca: 1.1609e+005 Vrms [1.009 pu] -121.95°6 W! A2 e* f" X; w5 Q) Y# L) n+ s
Ia: 798.9 Arms -92.72° 5 O: s1 S; }* Q$ i" U' R
Ib: 798.9 Arms 147.28°
) V/ ^! W0 i/ w. C& m4 Y! FIc: 798.9 Arms 27.28° 8 i" C' w8 f' y: `
P: 5.2087e+008 W
7 _1 K L) L+ ]# f! i# M+ jQ: 6.5561e+008 Vars " O& D6 g: f& q# [4 y
) ]' d2 g9 W$ @9 O3 G! e
Machine: Load1
, Q9 P" y" m9 H6 P6 U$ }4 e# ~( A ?Nominal: 115 kV rms 1 s* k6 C7 X- X0 ?
Bus Type: P & Q load ( U+ R4 O! i, f* R1 m; r
Uan phase: -40.94° 7 U7 W# r6 r* O3 Y; K2 M6 T0 L
Uab: 6.3759e+005 Vrms [5.544 pu] -10.94°
$ p3 p0 Y3 j/ e2 nUbc: 5.8926e+005 Vrms [5.124 pu] 158.63°
; d- I x9 _9 I1 h% U! f1 I2 tUca: 1.2147e+005 Vrms [1.056 pu] -129.49°
/ `5 g9 J1 F% PIa: 433.21 Arms -97.84° 1 I" X/ Q y7 s' {$ v
Ib: 433.21 Arms 142.16° * k0 k7 ]( b7 P
Ic: 433.21 Arms 22.16°
# R8 K' }. U- }3 ^! ~8 h8 jP: 2.6123e+008 W ' x3 k0 \$ w! Z8 M( W1 f
Q: 4.0079e+008 Vars |
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发表于 2012-4-29 16:29:59
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