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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?# j3 O D8 H% v( o4 P) m( b4 J
注:
. ?3 s$ V+ E8 ? L# I: UMachine: G2
( ?0 `. {" i9 A0 w+ BNominal: 100 MVA 10.5 kV rms
' k1 ^8 c, p1 Y- KBus Type: Swing bus
# j2 ^1 p: V7 L3 j3 T' z1 a8 IUan phase: -40.27° ) p( c$ ^) s4 d/ U
Uab: 52029 Vrms [4.955 pu] -10.27° 3 ^7 O# [4 ^2 b# a
Ubc: 48895 Vrms [4.657 pu] 157.70° ( h. T& C/ Z6 W8 j, J7 x
Uca: 11025 Vrms [1.05 pu] -122.71°
4 q3 L& p% h0 m8 R$ PIa: 16777 Arms [3.051 pu] -9.91° ; T5 K3 ^* A9 C3 Y( T3 ~7 {
Ib: 16777 Arms [3.051 pu] -129.91° 7 c1 ^1 o+ ^' a9 y2 B
Ic: 16777 Arms [3.051 pu] 110.09° , `5 b: [+ T* J# R$ b2 p
P: 1.3046e+009 W [13.05 pu] " r: l: g: Q- h) k1 {
Q: -7.6415e+008 Vars [-7.642 pu] $ T) Q; _5 O0 l: p5 B- U( J
Pmec: 1.3079e+009 W [13.08 pu] ) v) y2 D% x1 } U6 O3 ~/ r
Torque: 8.3264e+006 N.m [13.08 pu] $ d* r' _8 h0 a, R8 d/ Y c
Vf: 6.067 pu . f! N7 e) ^. d
; N; P4 t0 s2 N7 T: E# G, ~0 d
Machine: G1
' h: C' o- P. B/ ^0 d3 @( `" DNominal: 100 MVA 10.5 kV rms
$ c% [. _ z. Q, |5 o" EBus Type: P & V generator
. _( y) O) |, R& dUan phase: -40.45°
; w9 _+ W* L* z+ F! QUab: 55176 Vrms [5.255 pu] -10.45° . q+ e c3 u" M. v3 g* E
Ubc: 50638 Vrms [4.823 pu] 158.65° 6 y3 F, D* Q+ S7 R+ ]* P( p
Uca: 11025 Vrms [1.05 pu] -130.09°
; k% v) `) O( G, P4 Y" uIa: 17338 Arms [3.153 pu] -21.32°
1 v* z8 ?1 u+ c! ~Ib: 17338 Arms [3.153 pu] -141.32° 0 S9 y7 s) Q( r8 T$ o% t% q" ~, o
Ic: 17338 Arms [3.153 pu] 98.68°
- f; B& g- g" t+ ~' G. e DP: 1.5654e+009 W [15.65 pu]
: a- X3 z' D3 v* EQ: -5.4297e+008 Vars [-5.43 pu] 9 `# t8 v9 @' p j
Pmec: 1.569e+009 W [15.69 pu]
) |/ B0 o! ` k' `( ]5 z3 \7 TTorque: 9.9887e+006 N.m [15.69 pu] " ]2 z1 y+ T3 o+ e% S! |- y- x8 G
Vf: 6.7898 pu 6 ]4 k3 |3 I: m- q6 n. z) u ?
4 q+ E9 M" p1 X# O- x1 Q
Machine: Load3
& r3 _0 ~# M7 d4 C! u+ HNominal: 115 kV rms
- ]! N! E# i; w* m, Y5 P7 \: M( XBus Type: P & Q load
3 g( z- M G9 R. k" Y% `Uan phase: -46.14°
) c1 b1 H( \) j- @Uab: 6.2602e+005 Vrms [5.444 pu] -16.14° # Y9 j& Z" b3 ]8 x7 {3 b. ~4 Z
Ubc: 6.1132e+005 Vrms [5.316 pu] 161.43° 1 D! n9 i! ]% f1 s9 Y3 K( {7 k; p
Uca: 30061 Vrms [0.2614 pu] -136.62°
; ?$ N3 c% a3 TIa: 300.16 Arms -102.68° % }2 Y8 U1 n3 ]5 t$ s& o
Ib: 300.16 Arms 137.32°
( G% n. L3 n/ uIc: 300.16 Arms 17.32° ( B$ N) j1 u4 e* y! _: _
P: 1.7944e+008 W 7 _/ W1 }2 p7 f: @7 R0 ? H% I
Q: 2.7153e+008 Vars
1 H1 q; r6 S3 a6 ]" I * M" q6 W t5 \! b$ l/ a' K
Machine: Load2
' j' \) A, G3 R* i/ Q! @/ eNominal: 115 kV rms ! j6 o7 s9 ]" G) o8 N% [6 _1 r
Bus Type: P & Q load 3 q9 Z. R6 _: ?+ {
Uan phase: -41.18° 9 d5 a) b* I1 P, g" b% y9 y3 y: Z
Uab: 6.0512e+005 Vrms [5.262 pu] -11.18° ) S% w1 }+ L3 s1 }9 ^; t
Ubc: 5.7431e+005 Vrms [4.994 pu] 157.92° $ G- \+ J& N4 `; x
Uca: 1.1609e+005 Vrms [1.009 pu] -121.95°; `/ ~7 {- w1 ]
Ia: 798.9 Arms -92.72° # ]7 W6 [, F0 h: }, n
Ib: 798.9 Arms 147.28°
( I) ~$ {) D' l" d% zIc: 798.9 Arms 27.28°
: H: A5 \* H4 WP: 5.2087e+008 W 9 ^. u- y6 M4 m0 O3 ?; O q
Q: 6.5561e+008 Vars
9 j8 U4 e9 i6 _# q( b
. n7 v+ G! G6 a& I) Y7 [Machine: Load1
0 k1 [( ?4 y" z# ^! `/ X/ JNominal: 115 kV rms
?! i+ C0 G% w# x$ ?Bus Type: P & Q load
9 M9 Q# {2 e+ N$ s# b0 |( G/ u( FUan phase: -40.94°
! Z3 G4 }& k9 b0 h, gUab: 6.3759e+005 Vrms [5.544 pu] -10.94°
; x- Q6 J. e% O9 V: B! jUbc: 5.8926e+005 Vrms [5.124 pu] 158.63°
( L" A! j% s K9 {0 q/ o# cUca: 1.2147e+005 Vrms [1.056 pu] -129.49°8 u8 [$ x8 u) I5 }+ _& D
Ia: 433.21 Arms -97.84° 0 X t% S) b g! h+ s! N9 M" _
Ib: 433.21 Arms 142.16°
- q/ a) p' Z( b3 \5 Q I( TIc: 433.21 Arms 22.16° 3 f# ? l( F2 p. N7 l: S5 x
P: 2.6123e+008 W
( ]9 U0 N% B- qQ: 4.0079e+008 Vars |
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