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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 [9 c+ p) N+ @8 _ ~) P3 Y; y* G
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Machine: G2 . h9 ?+ t7 |* r/ ?- X/ E
Nominal: 100 MVA 10.5 kV rms
- l1 Y! l8 |6 C$ G) {0 T) \% iBus Type: Swing bus ( h7 p4 D) j+ T) h8 T: f. C; H
Uan phase: -40.27° + v6 R- V2 ?' A- O0 d
Uab: 52029 Vrms [4.955 pu] -10.27°
9 z4 I) l G* `: Q- r; `% r1 r" XUbc: 48895 Vrms [4.657 pu] 157.70°
/ U0 n* d4 H6 R: S5 k! p% M1 }) oUca: 11025 Vrms [1.05 pu] -122.71° ) L+ P% Q: M& M3 f
Ia: 16777 Arms [3.051 pu] -9.91°
R, }# F2 o. |Ib: 16777 Arms [3.051 pu] -129.91°
. {' s& j9 F: I" OIc: 16777 Arms [3.051 pu] 110.09° % `$ B9 Y( p5 x: C
P: 1.3046e+009 W [13.05 pu] $ {' [2 V: v. l* l0 U2 r5 u; I2 E7 `
Q: -7.6415e+008 Vars [-7.642 pu]
: v, C- ?2 r+ V0 M, k% z# E; JPmec: 1.3079e+009 W [13.08 pu] / M! b! F; s9 S% B
Torque: 8.3264e+006 N.m [13.08 pu]
3 @. p, G1 U( b+ C8 JVf: 6.067 pu % K' C9 p/ Y( P1 k; d) n! E/ y
- J1 @2 d. k! i' J/ u/ g$ l
Machine: G1
. }1 y5 C1 F; d$ E/ j6 u& ]Nominal: 100 MVA 10.5 kV rms
6 C/ f: G+ w9 v5 M9 t' vBus Type: P & V generator 2 |! g0 z& p0 |( ~8 E4 a9 S# V. k- ^
Uan phase: -40.45° ' H3 h |6 V' U4 c
Uab: 55176 Vrms [5.255 pu] -10.45°
, p- p. G$ l2 I5 wUbc: 50638 Vrms [4.823 pu] 158.65°
4 s7 b$ Q/ Y2 R, r+ OUca: 11025 Vrms [1.05 pu] -130.09°
2 \! \7 E; f9 u' IIa: 17338 Arms [3.153 pu] -21.32° - o2 u% e# b/ k1 s& k0 I
Ib: 17338 Arms [3.153 pu] -141.32°
@: p( n$ p6 B3 ^) x2 hIc: 17338 Arms [3.153 pu] 98.68°
" j9 e+ h8 T) z" cP: 1.5654e+009 W [15.65 pu] ; Q. V, J9 n! B* K: t
Q: -5.4297e+008 Vars [-5.43 pu] $ G9 [+ f* B" i8 F5 w; _5 u
Pmec: 1.569e+009 W [15.69 pu]
2 ]5 B9 V$ V: v- f; c+ j* J8 mTorque: 9.9887e+006 N.m [15.69 pu] ! F+ {# n' q; V8 x8 I! e: p
Vf: 6.7898 pu
+ U( M( \, `8 y6 @7 M
- ~8 @: \2 V6 k: @+ S* ]4 ]9 IMachine: Load3 * \# [) |% H6 J% @7 I
Nominal: 115 kV rms 1 u6 Z [; h3 B1 k2 p6 B8 K
Bus Type: P & Q load
: P; w7 W# e( v, q8 fUan phase: -46.14° : w( q5 E4 P& T- l2 c- {5 a
Uab: 6.2602e+005 Vrms [5.444 pu] -16.14° 2 t! U; z" [# P) \
Ubc: 6.1132e+005 Vrms [5.316 pu] 161.43° $ ]8 E5 q* L1 L
Uca: 30061 Vrms [0.2614 pu] -136.62° 2 H' q. m- u, q! f, g9 ~
Ia: 300.16 Arms -102.68°
4 p7 U' ]8 [, L* I5 a6 ?1 PIb: 300.16 Arms 137.32° % o/ f8 C2 L8 S! g% v
Ic: 300.16 Arms 17.32° 5 F& y+ \9 H$ w* D3 ^0 u* E
P: 1.7944e+008 W
; Y' U+ D' D' z2 X# i! \Q: 2.7153e+008 Vars 8 X9 r3 e/ d3 X7 K9 V
+ ?0 \7 S; v6 N) R3 M& U0 y; s+ sMachine: Load2
; B X, `- S( C0 Q B% d; \3 E4 qNominal: 115 kV rms
) b8 T b( \9 [Bus Type: P & Q load . o) v$ ?5 B* A( G& p8 o
Uan phase: -41.18° 2 A& L$ L. g/ B7 Z4 ?6 H
Uab: 6.0512e+005 Vrms [5.262 pu] -11.18°
$ g) m y, r! n/ M, i$ _4 A. vUbc: 5.7431e+005 Vrms [4.994 pu] 157.92°
, ]/ ^3 J0 r$ C$ _* lUca: 1.1609e+005 Vrms [1.009 pu] -121.95°+ A" c+ l- r" N
Ia: 798.9 Arms -92.72°
3 }- x* C0 X) zIb: 798.9 Arms 147.28°
/ c3 A" f# b" H8 v. `$ ?Ic: 798.9 Arms 27.28°
! e" R# p6 k# Y3 L8 ?( m7 IP: 5.2087e+008 W . U) D. q6 x. m/ u+ e# M2 [ T. B
Q: 6.5561e+008 Vars : I1 E9 S+ Y& F7 Q- F# M( \
+ G9 m6 \6 t1 k" IMachine: Load1 ! y/ ?# {% i7 K
Nominal: 115 kV rms $ T7 {+ g& f6 V1 B1 S9 V
Bus Type: P & Q load 4 _( [9 o# Z j. a
Uan phase: -40.94°
% D& h; |$ [8 T3 t2 J, EUab: 6.3759e+005 Vrms [5.544 pu] -10.94° % A* m1 L) X y5 `
Ubc: 5.8926e+005 Vrms [5.124 pu] 158.63° " D, |5 `5 H5 Y1 e* E3 z( Y
Uca: 1.2147e+005 Vrms [1.056 pu] -129.49°7 U5 _( L+ y" k$ \2 ?- [7 s0 D
Ia: 433.21 Arms -97.84° 2 W) ]( @4 j* U2 g+ T; a
Ib: 433.21 Arms 142.16°
* |; \5 @7 {$ k! oIc: 433.21 Arms 22.16°
, y1 `; w$ E2 U* R+ ]0 W+ FP: 2.6123e+008 W ) {* Q. h ]6 Q: f6 C" b* Q
Q: 4.0079e+008 Vars |
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