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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?
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Machine: G2
_% ?4 s7 W/ [5 RNominal: 100 MVA 10.5 kV rms ( g, l+ y& {8 J, i
Bus Type: Swing bus / u8 G% D& p" r! |1 c
Uan phase: -40.27° [# n$ o% q h) R9 r3 Z
Uab: 52029 Vrms [4.955 pu] -10.27°
# Q) k4 {6 W" V) }8 [ F* l5 iUbc: 48895 Vrms [4.657 pu] 157.70° ) D# i- u+ o1 T
Uca: 11025 Vrms [1.05 pu] -122.71°
0 Q6 K: s# j L3 \! v1 RIa: 16777 Arms [3.051 pu] -9.91°
4 S" u2 q8 l& f" e; d @Ib: 16777 Arms [3.051 pu] -129.91° % x7 N+ ^) f& u/ s. k8 @; `
Ic: 16777 Arms [3.051 pu] 110.09°
% W! V7 J& ?- q- `: P0 \P: 1.3046e+009 W [13.05 pu]
0 e9 m G. F6 P8 n7 h& u2 qQ: -7.6415e+008 Vars [-7.642 pu] ; ?" r% p% U2 G: l
Pmec: 1.3079e+009 W [13.08 pu]
% G# w, ?1 u8 l% z5 z0 t/ ]' ATorque: 8.3264e+006 N.m [13.08 pu]
9 d3 R4 Y+ z/ s" WVf: 6.067 pu
2 P; S, C) o" m: y1 z& `% L 8 B7 A$ t" _) B3 u9 L, I9 W% V
Machine: G1
: ]% p% V$ r& K$ C! R# UNominal: 100 MVA 10.5 kV rms
1 J' Q* h; x. \) I' OBus Type: P & V generator
4 k5 U2 D+ F* ]Uan phase: -40.45° % Z. n1 ~9 w; N9 x; C
Uab: 55176 Vrms [5.255 pu] -10.45° 5 l' D: m7 u0 r: f8 Z
Ubc: 50638 Vrms [4.823 pu] 158.65° 8 L1 [7 E& c* _, ^' v2 @
Uca: 11025 Vrms [1.05 pu] -130.09° " m$ p- \+ D! D! `& X
Ia: 17338 Arms [3.153 pu] -21.32°
7 C# L l- Z$ ]9 ?! p# ZIb: 17338 Arms [3.153 pu] -141.32° 2 `5 O( ^: A3 D& X* W7 ~/ V
Ic: 17338 Arms [3.153 pu] 98.68° / Z4 R; V! u6 s. x
P: 1.5654e+009 W [15.65 pu] ( E! f1 A( D3 v$ Q& n; g5 n
Q: -5.4297e+008 Vars [-5.43 pu] , K/ E4 p, u) e* N3 K, F/ ]
Pmec: 1.569e+009 W [15.69 pu] & {9 Q6 ?' v- o9 {
Torque: 9.9887e+006 N.m [15.69 pu] / y8 O4 M3 [2 e. B/ {5 m
Vf: 6.7898 pu 8 T. h+ l6 D2 q2 T* O0 Z+ l
2 D+ i4 w% h2 wMachine: Load3 7 T) B. d) k, o: ~; Z
Nominal: 115 kV rms
$ l* o2 u/ f5 y2 m6 g, g+ _Bus Type: P & Q load
* z. U8 Q/ T$ m) wUan phase: -46.14° ; g: S$ F$ H8 G: ^3 }" Q) T7 d
Uab: 6.2602e+005 Vrms [5.444 pu] -16.14° ! s( M+ D5 X. b, f' o7 y1 |
Ubc: 6.1132e+005 Vrms [5.316 pu] 161.43° ' {; ]; o# T8 M. K% A$ ^, I4 j
Uca: 30061 Vrms [0.2614 pu] -136.62° " g+ K% G& c# Q- V& \0 u' ?8 d; Q4 B
Ia: 300.16 Arms -102.68° - M+ l$ }6 u- n1 H/ u7 W) I8 v
Ib: 300.16 Arms 137.32°
! T0 i# @/ F; p' i5 V. l9 XIc: 300.16 Arms 17.32° * e9 l) \$ n- ?- o
P: 1.7944e+008 W ' K; W8 t4 M% t" @6 M
Q: 2.7153e+008 Vars ( \$ G9 D% [ ?9 G6 P( `) k& c
! F5 m1 b. S' i2 S FMachine: Load2
. \7 c7 i% H% t+ G3 fNominal: 115 kV rms ! [# [2 e: E9 x9 {6 y; H* G, M
Bus Type: P & Q load $ P8 _7 s5 g3 {0 {8 v- O6 o+ B6 S; p
Uan phase: -41.18° 5 B# w: I/ ^ Z8 |
Uab: 6.0512e+005 Vrms [5.262 pu] -11.18°
1 W) {4 y+ U9 J3 s7 r& t( @Ubc: 5.7431e+005 Vrms [4.994 pu] 157.92°
L% }) N4 l/ a0 x, {Uca: 1.1609e+005 Vrms [1.009 pu] -121.95°
' ~0 ?) k! I6 q0 oIa: 798.9 Arms -92.72° G+ h% h$ n4 ^
Ib: 798.9 Arms 147.28°
7 N; B: l+ \) f0 L. ~6 u4 i4 l6 @" _Ic: 798.9 Arms 27.28°
v/ ?& {4 d$ u. H8 P7 q, w0 y7 }P: 5.2087e+008 W
, }) w/ a+ \ ^4 O/ P4 l$ oQ: 6.5561e+008 Vars
9 W m1 Z6 L0 A' s) G* d # c% g' `4 D! |' J
Machine: Load1 6 F9 P E; P2 K( a. G. a0 n1 H! }
Nominal: 115 kV rms L* C3 H" M) A9 o) S: {
Bus Type: P & Q load & x9 }$ ~: U9 g1 K3 L
Uan phase: -40.94° % V) n! W% x( ~( f% }
Uab: 6.3759e+005 Vrms [5.544 pu] -10.94° + E8 L$ L, q& `
Ubc: 5.8926e+005 Vrms [5.124 pu] 158.63° ; \. [* b1 F% ?2 K# B
Uca: 1.2147e+005 Vrms [1.056 pu] -129.49°4 x: Q3 \% Z" y+ w1 p3 f9 _
Ia: 433.21 Arms -97.84°
* X4 R& |1 s: t1 _, RIb: 433.21 Arms 142.16°
' ?9 p' K- H Y/ s, Y3 W- MIc: 433.21 Arms 22.16° ( Y k9 z" ?3 v1 l5 d
P: 2.6123e+008 W / t1 Y. f) I3 x2 k& l% J6 X, g. z
Q: 4.0079e+008 Vars |