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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 W- V2 D: A' s; \ F9 w
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Machine: G2 ' }7 }7 x: C$ Q
Nominal: 100 MVA 10.5 kV rms % S- w. |9 W6 o
Bus Type: Swing bus * q5 Y4 a6 M6 u
Uan phase: -40.27° 6 S( T7 r- ^& ]9 x3 Q6 @/ I
Uab: 52029 Vrms [4.955 pu] -10.27°
$ {2 p" t& {5 j$ ^; vUbc: 48895 Vrms [4.657 pu] 157.70° 1 u4 z7 x- X/ y1 K1 Z
Uca: 11025 Vrms [1.05 pu] -122.71° 4 f4 Z: M5 {3 W" K
Ia: 16777 Arms [3.051 pu] -9.91°
. E/ u- x* P \. B* [Ib: 16777 Arms [3.051 pu] -129.91°
& k z6 U) ]8 F5 I/ L! lIc: 16777 Arms [3.051 pu] 110.09°
, k7 o$ Y0 H! o& ^% EP: 1.3046e+009 W [13.05 pu]
9 k$ d8 P6 A; \; |Q: -7.6415e+008 Vars [-7.642 pu]
- ]& t- n' X# ~Pmec: 1.3079e+009 W [13.08 pu]
- `" h$ F: ~% i0 `! e+ G; a" ZTorque: 8.3264e+006 N.m [13.08 pu]
; A' l2 _/ \9 uVf: 6.067 pu
0 m* G2 c# ?. J3 _
( g6 {& ]2 ]0 ?8 k! j5 _Machine: G1
S$ B/ {) j! c5 pNominal: 100 MVA 10.5 kV rms ) Y' w9 a1 f1 F+ Y6 ~- t: X9 P
Bus Type: P & V generator " p- M6 I. r/ C3 I( Y: I$ O5 z
Uan phase: -40.45°
2 _2 E5 m Q9 z% V4 |" x6 u3 ]2 OUab: 55176 Vrms [5.255 pu] -10.45°
* O" _0 z1 D+ i0 X) J0 j( t; M# M) ?Ubc: 50638 Vrms [4.823 pu] 158.65° " I3 `$ W3 H- A A( g/ S2 q
Uca: 11025 Vrms [1.05 pu] -130.09° . N: G' ~6 b9 F" s( O3 D4 y
Ia: 17338 Arms [3.153 pu] -21.32°
: E/ [9 v! W3 wIb: 17338 Arms [3.153 pu] -141.32°
) s; j3 `6 T6 I, [' A/ TIc: 17338 Arms [3.153 pu] 98.68° - x* o; T% o+ i( ~" e
P: 1.5654e+009 W [15.65 pu] " T) k. l0 z" |2 ?( D Z# A
Q: -5.4297e+008 Vars [-5.43 pu]
8 Z# K& y p( f8 T( qPmec: 1.569e+009 W [15.69 pu] 2 G+ m/ I) l3 `1 f D2 Z; y6 l( P
Torque: 9.9887e+006 N.m [15.69 pu] / L* i ]8 S& v+ h; S j5 p
Vf: 6.7898 pu 9 ~* f1 Y: P6 e
1 J5 r' t, S0 ?9 ?# j; B, L, I* dMachine: Load3 1 Z: ?% J; ]& h; a
Nominal: 115 kV rms % j; a) K+ E! H! L( t3 z! H
Bus Type: P & Q load ; c! m- y7 G2 o3 U7 H" K
Uan phase: -46.14°
/ L4 E7 O/ Z/ z7 Y6 C: EUab: 6.2602e+005 Vrms [5.444 pu] -16.14°
* K7 {" N% b/ J; {+ wUbc: 6.1132e+005 Vrms [5.316 pu] 161.43°
; v7 P3 q9 J7 w! Q- lUca: 30061 Vrms [0.2614 pu] -136.62° . C; Q6 E4 O: s! _" V6 Z7 B
Ia: 300.16 Arms -102.68° ! d+ i6 H: g" O6 E2 D( V% P
Ib: 300.16 Arms 137.32° 9 o% k4 d; b: K
Ic: 300.16 Arms 17.32°
" N' z" k) p9 G% `" jP: 1.7944e+008 W + ]6 Q, z/ J1 ~4 T2 m
Q: 2.7153e+008 Vars " r9 B; d6 \( e& t/ t3 Z
4 [; [) a5 z& o3 e# R
Machine: Load2
$ R9 h: U* M; D& x2 u6 ~Nominal: 115 kV rms
& p9 k' K8 c7 b+ FBus Type: P & Q load
& ^, N" ~6 X6 d5 Z' Z0 I6 GUan phase: -41.18°
4 K$ Y5 X$ D" z; @& BUab: 6.0512e+005 Vrms [5.262 pu] -11.18° ' Y4 s, l6 R& _% Z
Ubc: 5.7431e+005 Vrms [4.994 pu] 157.92° ; ?6 q9 K2 n8 w5 G& L& v
Uca: 1.1609e+005 Vrms [1.009 pu] -121.95°! M7 h0 u/ W8 N% e' F; G ?, N) ~1 j1 P
Ia: 798.9 Arms -92.72° % ?5 W B) k* Y6 ]$ m3 s3 K
Ib: 798.9 Arms 147.28°
m% _# y( @% B+ ZIc: 798.9 Arms 27.28° . W0 ^% c6 [0 M# ^8 J
P: 5.2087e+008 W % F& f) R( x( {5 S) H. C9 Y# d
Q: 6.5561e+008 Vars ( {+ m6 j3 `$ Y; G3 P) j5 L U
3 U# R* s, Y/ V5 |Machine: Load1
! S9 d: h0 ~2 J% V/ \1 }, eNominal: 115 kV rms - T0 s* m0 b- J/ }
Bus Type: P & Q load 7 L, p7 Y4 O. ^6 N$ J
Uan phase: -40.94°
% G, c- _- a5 \) Z- g1 o% J" jUab: 6.3759e+005 Vrms [5.544 pu] -10.94° $ U5 N* M4 H: @- ~& c
Ubc: 5.8926e+005 Vrms [5.124 pu] 158.63°
# }% }5 T6 Q, ?* w1 D, ~5 ^Uca: 1.2147e+005 Vrms [1.056 pu] -129.49°( g" i5 T/ y% ~2 l" f
Ia: 433.21 Arms -97.84°
/ p& b% l8 f1 R9 w- }& W- nIb: 433.21 Arms 142.16°
; I7 g; y* ~8 ?( Q# QIc: 433.21 Arms 22.16°
& U- o- o2 N) v4 N/ b# {1 zP: 2.6123e+008 W
* ?/ e. @3 g2 g7 Z5 j3 ^/ ZQ: 4.0079e+008 Vars |
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