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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？0 V% v' `& R% {
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Machine: G2
; i. S! m8 g( d9 `5 {Nominal: 100 MVA 10.5 kV rms 1 v4 S, Y2 `9 t- L8 J! S+ z
Bus Type: Swing bus
5 { ^; j7 C( p" {1 S$ UUan phase: -40.27°
0 \! r5 Y5 v% ^Uab: 52029 Vrms [4.955 pu] -10.27° 8 i a: u, i: p0 i
Ubc: 48895 Vrms [4.657 pu] 157.70°
6 C+ c8 j* i% z& |, u. VUca: 11025 Vrms [1.05 pu] -122.71° $ L( {+ n; e; v
Ia: 16777 Arms [3.051 pu] -9.91° 9 _ I6 ~1 S0 I H
Ib: 16777 Arms [3.051 pu] -129.91°
5 F$ R \ t+ }2 X0 N" \3 x$ yIc: 16777 Arms [3.051 pu] 110.09°
* c. R! Y' U, h+ z/ M: b7 q3 o5 lP: 1.3046e+009 W [13.05 pu]
6 N5 V3 b! c6 K' J7 }Q: -7.6415e+008 Vars [-7.642 pu]
2 S9 E6 [" m" R: j" g( U: WPmec: 1.3079e+009 W [13.08 pu] * Y/ x7 M+ U: C/ c
Torque: 8.3264e+006 N.m [13.08 pu]
. Q1 p% x6 F" I5 a, tVf: 6.067 pu
1 V. r+ `, w' B8 ]3 K: K* I ' K1 Z4 e( J& _2 r' |
Machine: G1
* [( j1 w4 V/ UNominal: 100 MVA 10.5 kV rms
- X) ?/ p$ ]8 Y1 n" n; c, I2 IBus Type: P & V generator ' g0 \1 O: r; P4 x$ \
Uan phase: -40.45° , w( v |% N" {3 G
Uab: 55176 Vrms [5.255 pu] -10.45° " G/ f( G' n2 w& ^% v6 v; ^
Ubc: 50638 Vrms [4.823 pu] 158.65° ( p5 C$ W* j4 p
Uca: 11025 Vrms [1.05 pu] -130.09°
' _: B U A: X3 j2 S* QIa: 17338 Arms [3.153 pu] -21.32° : W0 N+ L; g/ B/ q3 D
Ib: 17338 Arms [3.153 pu] -141.32° $ W" U2 s& R- H+ o+ h; M: F$ Y
Ic: 17338 Arms [3.153 pu] 98.68°
4 U) O3 t( t) o: _9 K: o1 i0 lP: 1.5654e+009 W [15.65 pu] / w/ S" h, t& k& J+ @; `) F. U3 ~1 W
Q: -5.4297e+008 Vars [-5.43 pu]
' N! Z* [1 P+ a- TPmec: 1.569e+009 W [15.69 pu]
* W$ K' {% d- P! q8 RTorque: 9.9887e+006 N.m [15.69 pu]
) _% R2 j( Z# B; ^) o% a dVf: 6.7898 pu
; |$ V/ D% X4 N+ g! X' ^
9 E1 ^. W% l2 z* f0 p% ?; u2 i1 kMachine: Load3
1 X* W8 b6 M3 u6 `# T( pNominal: 115 kV rms 4 J5 k1 Q2 B- G9 p# U+ g) W
Bus Type: P & Q load 7 X! _8 q# `( a4 |
Uan phase: -46.14°
: o2 Z/ z' f; m0 A+ u9 p' LUab: 6.2602e+005 Vrms [5.444 pu] -16.14° 6 Q; n5 K0 a. m4 P' M, N
Ubc: 6.1132e+005 Vrms [5.316 pu] 161.43°
+ O1 z9 B0 ?; s5 ]0 G* Z) R. yUca: 30061 Vrms [0.2614 pu] -136.62° 5 P( |: e# l& o5 v6 M, i
Ia: 300.16 Arms -102.68° ) O7 v% t8 R: Z( [, F* j
Ib: 300.16 Arms 137.32°
" u9 C. U1 W* N. |4 G4 a! iIc: 300.16 Arms 17.32°
( e; V5 b) F5 S& W( r& hP: 1.7944e+008 W # P/ C9 D. t# h9 A$ J! Z- `+ I
Q: 2.7153e+008 Vars
3 k0 F2 t' S" ? D8 |2 z A. c: I- n2 k
Machine: Load2 # f) O" B0 C0 F% n/ P7 p
Nominal: 115 kV rms
* C/ J/ M+ i8 E& a8 G5 _7 JBus Type: P & Q load
# h$ u3 M" _0 w n) Q' R6 |# KUan phase: -41.18° , @6 W# F$ I- r6 e
Uab: 6.0512e+005 Vrms [5.262 pu] -11.18°
7 y* i0 h+ B( u8 D3 X2 EUbc: 5.7431e+005 Vrms [4.994 pu] 157.92° ( s" l$ b/ `6 `# q, Y3 E( B
Uca: 1.1609e+005 Vrms [1.009 pu] -121.95°
5 c6 e) k o8 K* i5 R7 PIa: 798.9 Arms -92.72°
- i6 v8 w( o/ k! x5 LIb: 798.9 Arms 147.28°
/ T* S5 E; a3 X6 ? r! J) _Ic: 798.9 Arms 27.28°
* t. w% }0 T qP: 5.2087e+008 W
% t( L5 b1 m2 b* rQ: 6.5561e+008 Vars
& v2 w1 i( k8 K3 Y
% p' `( n3 p, t0 {Machine: Load1
, b/ H& ] p/ r5 R6 `. VNominal: 115 kV rms
4 i: s4 {% }* U* EBus Type: P & Q load * T, J* ~& c+ @$ p/ |/ g
Uan phase: -40.94°
* {( m; L0 z9 fUab: 6.3759e+005 Vrms [5.544 pu] -10.94°
; h! }. n/ I$ `, cUbc: 5.8926e+005 Vrms [5.124 pu] 158.63°
$ [2 ]4 v8 x8 z. zUca: 1.2147e+005 Vrms [1.056 pu] -129.49°
, D6 |/ J1 q8 L' `( e3 k2 EIa: 433.21 Arms -97.84°
4 C: w" c( N. I& XIb: 433.21 Arms 142.16° " M5 c! c7 Y$ N3 E; i' H
Ic: 433.21 Arms 22.16° + G2 C, P1 Z u+ H
P: 2.6123e+008 W
5 I( B/ o+ l6 r* c0 e$ f8 @9 ]1 F- eQ: 4.0079e+008 Vars |
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