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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?
' }2 k2 X0 N( V% M N( l0 I注:
! @0 k9 _1 U$ }& GMachine: G2
$ Z! r0 V7 {& T' d5 b8 @; @Nominal: 100 MVA 10.5 kV rms
# K6 Q2 ?7 u( [ R4 IBus Type: Swing bus , x0 n- u8 Z2 f6 l1 g! o! g
Uan phase: -40.27°
; I7 A! J2 C5 ^ `2 fUab: 52029 Vrms [4.955 pu] -10.27°
. c: {+ i. [2 J* k1 {, l& jUbc: 48895 Vrms [4.657 pu] 157.70° ; r- `8 C8 ^: N! |- s
Uca: 11025 Vrms [1.05 pu] -122.71°
, P$ c' \! k, d) d7 D" IIa: 16777 Arms [3.051 pu] -9.91° $ x9 A6 K8 W2 O7 v/ s
Ib: 16777 Arms [3.051 pu] -129.91° & w/ j) T7 }9 T" ?/ n
Ic: 16777 Arms [3.051 pu] 110.09°
- }7 B: i! {7 l. O. x% p8 VP: 1.3046e+009 W [13.05 pu]
/ \7 `: a8 V& z+ k% |2 @* WQ: -7.6415e+008 Vars [-7.642 pu]
8 P) X1 f6 r O5 R2 xPmec: 1.3079e+009 W [13.08 pu]
0 T. L/ \; O0 B5 B) xTorque: 8.3264e+006 N.m [13.08 pu]
# F8 F6 S% e. }9 T7 ~Vf: 6.067 pu - u) }5 C7 [7 d( G7 j* C. d- o' g2 S
$ U: ~& f: n1 @Machine: G1 7 `; W, a5 [8 x% I' v
Nominal: 100 MVA 10.5 kV rms
1 u0 C$ V+ j- _' Y# m4 K- ~Bus Type: P & V generator 4 N4 ]0 u! z c3 n' a5 C
Uan phase: -40.45°
! V& N5 U0 x/ A$ D1 f# }* x: UUab: 55176 Vrms [5.255 pu] -10.45° 2 V4 a P, w' O8 s
Ubc: 50638 Vrms [4.823 pu] 158.65°
2 ^5 r. ?9 f8 c" S4 zUca: 11025 Vrms [1.05 pu] -130.09° - S* s9 L& U' e+ u
Ia: 17338 Arms [3.153 pu] -21.32°
$ f) v |- s B- G1 l4 RIb: 17338 Arms [3.153 pu] -141.32° 6 i/ ?- J+ d9 W& _" l. Z/ R
Ic: 17338 Arms [3.153 pu] 98.68° 3 j. ^( l# L* u- i# ^ J- Y/ R7 m' B" D
P: 1.5654e+009 W [15.65 pu] $ \) R& K$ ]0 k4 I4 x
Q: -5.4297e+008 Vars [-5.43 pu] 7 w; l g' n) g) R* X/ Y
Pmec: 1.569e+009 W [15.69 pu]
& J1 {* @& n. }7 MTorque: 9.9887e+006 N.m [15.69 pu]
2 w8 r7 M# c( c1 @% U% fVf: 6.7898 pu ! P* j' J& d: u& t9 g" P( d& u
" h$ h5 r5 M2 q7 c7 F1 r
Machine: Load3 & Y( u: v% i1 |2 q% E. R
Nominal: 115 kV rms " x+ q7 Z; O; T' ?
Bus Type: P & Q load
- `* H7 e( {1 K4 S; Q- {+ ]Uan phase: -46.14° : B- L' v2 Y" Q) x6 v
Uab: 6.2602e+005 Vrms [5.444 pu] -16.14° ( ?9 T) n# R6 A: n2 Q
Ubc: 6.1132e+005 Vrms [5.316 pu] 161.43°
+ o( q8 V* |$ v: p# w" C4 OUca: 30061 Vrms [0.2614 pu] -136.62°
& C+ s. ?3 d* E% Q _# `, z1 cIa: 300.16 Arms -102.68° 3 ^+ p* R, E% F" q* A- T; F: `& j
Ib: 300.16 Arms 137.32° & ?& r1 G* J9 | H# \
Ic: 300.16 Arms 17.32°
% s& G; B$ S" cP: 1.7944e+008 W 0 \4 x0 c8 s9 m# r
Q: 2.7153e+008 Vars
( a4 q7 W) d) ?+ ?
" s5 c4 o. e- m/ V; nMachine: Load2
, r! I6 E' G1 [3 u) F9 PNominal: 115 kV rms 7 L1 I4 ~. Z( [) N/ h
Bus Type: P & Q load
" \3 w( u- W/ X* g: U8 DUan phase: -41.18° : {2 T% i7 ]7 x5 K( R) b3 Q
Uab: 6.0512e+005 Vrms [5.262 pu] -11.18°
% u+ {3 j9 k* j( yUbc: 5.7431e+005 Vrms [4.994 pu] 157.92°
# l$ b8 q1 N; A2 FUca: 1.1609e+005 Vrms [1.009 pu] -121.95°$ G; H) P, S0 L7 ?, [
Ia: 798.9 Arms -92.72° 9 O6 c, v, }3 O4 a! `
Ib: 798.9 Arms 147.28°
" {) ]3 Y/ z7 [) F0 KIc: 798.9 Arms 27.28° 8 Y2 `+ E. Q4 y
P: 5.2087e+008 W
: ]/ t) v) `& t! H LQ: 6.5561e+008 Vars . [$ m; c/ J5 U0 ^. c' N
) Z- q4 v0 ?9 VMachine: Load1
# Y9 W# V' E6 uNominal: 115 kV rms
9 D+ V* f) v0 TBus Type: P & Q load 0 J! E# S |3 p5 ~+ E- ]" z9 |
Uan phase: -40.94° ! P r2 `2 M$ z
Uab: 6.3759e+005 Vrms [5.544 pu] -10.94° * U0 x% l6 e f( _1 b
Ubc: 5.8926e+005 Vrms [5.124 pu] 158.63° * x& [! ~5 U' U: [6 }4 N
Uca: 1.2147e+005 Vrms [1.056 pu] -129.49°$ Q5 h* |7 Z) G- N( I1 C6 f9 Z
Ia: 433.21 Arms -97.84°
+ M3 q; g4 A4 M YIb: 433.21 Arms 142.16°
- E9 A$ ]( U" H% E& m/ R9 dIc: 433.21 Arms 22.16°
3 f4 ^2 l- U/ W U3 ?P: 2.6123e+008 W
1 N& q! W8 x M! a$ c \6 }) tQ: 4.0079e+008 Vars |
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