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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?
1 a8 A# ` U# x) e1 A. e. B* U# l注:& ?# ~, J# c7 Q7 c. R0 ^
Machine: G2 - d" Q7 I8 X+ z6 E; K
Nominal: 100 MVA 10.5 kV rms . Q1 d' \1 I9 e
Bus Type: Swing bus
1 _ f3 h5 X' {; B6 P% kUan phase: -40.27° 1 m8 I6 J7 E& G1 e: a) Y7 W
Uab: 52029 Vrms [4.955 pu] -10.27° % P$ A9 c+ Z+ M {0 v% X% }3 t1 R+ x: n
Ubc: 48895 Vrms [4.657 pu] 157.70°
4 T: r3 ~. b1 |2 W K& pUca: 11025 Vrms [1.05 pu] -122.71° 3 P9 v$ Y7 R% D' O ]! X
Ia: 16777 Arms [3.051 pu] -9.91°
~8 F% m& {7 d) q. fIb: 16777 Arms [3.051 pu] -129.91° * U- t, P. _. K
Ic: 16777 Arms [3.051 pu] 110.09°
2 {* c. Y# u1 Y# j. f3 qP: 1.3046e+009 W [13.05 pu] ; S1 j* T+ V2 _1 h7 L) R9 _' s- M- O
Q: -7.6415e+008 Vars [-7.642 pu] " P; z* b: \2 w4 {1 ]9 S5 P7 v) ?* A
Pmec: 1.3079e+009 W [13.08 pu] 1 Y, i, n1 h( e# d
Torque: 8.3264e+006 N.m [13.08 pu] $ D2 N9 ^* v; b
Vf: 6.067 pu ?0 v' N/ A' @2 }0 Q- v
# e" @( {4 r& T, c6 R& L# W2 HMachine: G1
8 C( Y: X! j4 [; _1 j/ UNominal: 100 MVA 10.5 kV rms
$ l- v# m4 ~$ zBus Type: P & V generator : q7 ^8 Y; M2 l0 d5 K/ N
Uan phase: -40.45° + H1 J9 U+ Z" Q( ^# Z% ^9 D* I
Uab: 55176 Vrms [5.255 pu] -10.45°
) }! D; ]( \% x! J+ b- ?Ubc: 50638 Vrms [4.823 pu] 158.65°
/ a* c. C+ d$ s8 D+ HUca: 11025 Vrms [1.05 pu] -130.09°
8 @1 Y! w- G+ z4 G8 S9 D; [( pIa: 17338 Arms [3.153 pu] -21.32° 8 Y7 L. E1 w- ~9 k
Ib: 17338 Arms [3.153 pu] -141.32°
* k! B8 E# i1 X4 u6 {Ic: 17338 Arms [3.153 pu] 98.68°
/ B4 m( l% k2 R- HP: 1.5654e+009 W [15.65 pu] ' Z/ u0 _ I1 E) B3 P& x; G
Q: -5.4297e+008 Vars [-5.43 pu] & m! H/ R" ]* G( l# R7 N( R
Pmec: 1.569e+009 W [15.69 pu]
5 d# W6 U, D- n5 }( `Torque: 9.9887e+006 N.m [15.69 pu]
, Q3 u& ~- @) UVf: 6.7898 pu ( L4 a: i1 o9 z! [2 ]' u E' i5 Z- H8 ]
4 I& X, I: ^, u$ j. ?Machine: Load3 * S: N4 U# |* t
Nominal: 115 kV rms . g- p) |5 [* `' k7 |4 z% \* q1 B! G
Bus Type: P & Q load
/ z+ S8 W' |, G# f4 h5 SUan phase: -46.14° " m" b, j7 K% l) P% |* l
Uab: 6.2602e+005 Vrms [5.444 pu] -16.14°
( h# L/ J1 |8 S, J9 \7 RUbc: 6.1132e+005 Vrms [5.316 pu] 161.43°
/ ^$ v1 _4 H, z4 sUca: 30061 Vrms [0.2614 pu] -136.62° 7 j$ V( g3 Q# T2 {
Ia: 300.16 Arms -102.68°
4 S# }, X7 s) e0 mIb: 300.16 Arms 137.32°
" p8 b ^& R9 W2 V Z7 I% NIc: 300.16 Arms 17.32°
3 f. i2 z, i% v. x, }P: 1.7944e+008 W 7 F4 @$ d/ f s0 r# ^3 u% h
Q: 2.7153e+008 Vars ( J* q+ R/ W5 b( O$ I! b4 P
2 L2 ]4 d! h& n8 a8 }
Machine: Load2 8 q! Y, J" V2 A" @/ Z
Nominal: 115 kV rms , t: q& ~4 l; I+ e; F
Bus Type: P & Q load 2 w3 e( i7 ?' r, N5 U3 w
Uan phase: -41.18°
6 P/ S" c0 i& E' j3 n' ^1 e8 \# F( AUab: 6.0512e+005 Vrms [5.262 pu] -11.18°
+ P: `* T* b. [" cUbc: 5.7431e+005 Vrms [4.994 pu] 157.92°
( q' E1 ^4 P) `; S* }2 ~+ U7 }Uca: 1.1609e+005 Vrms [1.009 pu] -121.95°
' v1 i- n8 F5 V% J9 U; h& HIa: 798.9 Arms -92.72° 1 @' w" g0 z# d3 ?; H R
Ib: 798.9 Arms 147.28°
; q/ e& n- y8 Q3 W' ?Ic: 798.9 Arms 27.28°
* {" t4 m- | r; GP: 5.2087e+008 W ; V* v" F' b1 X a4 {3 z" E
Q: 6.5561e+008 Vars + Y+ x9 A- @9 |/ h$ X2 b4 P
4 R; q+ \( w. r$ G0 R3 s* b/ fMachine: Load1
4 U8 g7 M. b! j7 c5 `Nominal: 115 kV rms
- ~4 [$ r# Q1 ]Bus Type: P & Q load
, @! f5 t" d" J% H' C0 o3 }5 ?' GUan phase: -40.94° ) a( G; n5 z5 h6 c/ ?% @
Uab: 6.3759e+005 Vrms [5.544 pu] -10.94°
4 I/ t9 k, h7 n, @Ubc: 5.8926e+005 Vrms [5.124 pu] 158.63° % \9 C/ f7 V9 V# Y7 X
Uca: 1.2147e+005 Vrms [1.056 pu] -129.49°5 Q. k0 X( f% Z) y3 @5 {
Ia: 433.21 Arms -97.84°
& R) U" b' u9 Z: i1 VIb: 433.21 Arms 142.16° 1 p! O. W4 c) E: n+ r8 B; @
Ic: 433.21 Arms 22.16° $ N& a0 K( J) F. b8 S3 V0 l
P: 2.6123e+008 W 7 q- j: g- q5 N6 {2 C' p
Q: 4.0079e+008 Vars |