变压器低压绕组加载电流时出现警告
变压器低压绕组加载10*SIN(100*PI*T)+5电流密度，出现警告命令如下：
! transformer for 31500kVA/35kV/400V, developed by Ou
/FILNAME,Single
!Part 1: Initializing the parameter of geometry and properties.
Pi=3.1415926
Ri=94E-03 !Radius of iron core ;
G1=10E-03!Gap between core and inner HV winding;
G2=5E-03 !Gap between LV and HV;
G3=5.5E-03!Gap between winding and yoke;
Hhv=460E-03!Height of HV winding;
Hlv=420E-03 !Height of LV winding;
Hi=Max(Hlv,Hhv)+2*G3
Whv=(12*2.6+11*5)*1E-03!Width of inner HV winding;
Wlv=(3+5)*1E-03 !Width of LV winding;
Wt=450E-03!Length between two core center lines;
Nlv=24 !Half turns of LV winding;
Slv=3*16E-06 !Area of core conductor about LV;
Flv=(Nlv*Slv)/(2*Wlv*Hlv) !Fill factor abou LV;
Shv=5.31E-06!Areas of core conductor about HV;
Nhv=12*175 !Turns of HV winding;
Fhv=(Nhv*Shv)/(2*Whv*Hhv) !Fill factor abou HV;
!Part 2: Creating half FEM model;
/PREP7
R1=Ri+G1
R2=R1+Wlv+G2
Rectng,0,Ri,-Hhv,Hhv
Rectng,R1,R1+Wlv,-Hlv,Hlv!LV winding;
Rectng,R2,R2+Whv,-Hhv,Hhv!HV winding;
Rectng,0,Wt-Ri,-Hi,Hi!Air;
AOVLAP,ALL
NumCmp,All
!Part 3: Defining the properties.
ET,1,53,,,1! Air,iron
ET,2,53,3,,1 ! Primary (LV) winding;
ET,3,53,3,,1 ! Secondary (HV) winding;
MP,MURX,1,1! No.1= Air;
MP,MURX,3,2000 ! No.4= Iron;
Rsvx_Cu=1.24E-08 ! No_Load=1.24;
UIMP,2,MURX,RSVX,,1,Rsvx_cu !No.2=copper conductor;
R,1,Wlv*Hlv,Nlv,,1,Flv!LV windings;
R,2,Whv*Hhv,Nhv,,-1,Fhv
ASEL,S,,,1
AATT,3,,1!iron core
ASEL,S,,,2!LV winding;
AATT,2,1,2
Asel,S,,,3!HV winding;
AATT,2,2,3
Asel,s,,,4!Air;
AATT,1,,1
Allsel,All
!Part 4:Meshing model.
!Part 4.1: Meshing coil;
Lsel,S,,,6,8,2
Lesize,All,,,24
Lsel,S,,,5,7,2
Lesize,All,,,1
Lsel,S,,,10,12,2
Lesize,All,,,175
Lsel,S,,,9,11,2
Lesize,All,,,12
MShape,0,2D !Qualaterial elements;
Mshkey,1!Mapped Meshing;
ASel,s,Type,,2,3
Amesh,All
!Part 4.2:Meshing air region;
Smrt,1
Mshape,1,2D
MshKey,0
ASel,S,mat,,1
ASel,a,mat,,3
Amesh,All
Allsel,All
!Part 6: Applying the boundary conditions.
ESel,S,TYPE,,2
NSLE,S
CP,1,VOLT,ALL! Couple Current inside LV winding;
nsel,s,loc,x,0
D,all,az,0
Allsel,All
FINISH
/SOLU
EQSLV,JCG,1E-9
T=1E-8
C=0
N=80
PI=2*ASIN(1)
CON=1/4000
NEQIT,1
*CREATE,LOAD
TIME,T
I=10*SIN(100*PI*T)+5
ESel,S,TYPE,,2
!F,228,AMPS,I
BFE,ALL,JS,,I
T=T+CON
C=C+1
OUTRES,ALL,1
SOLVE
*END
*DO,I,1,81
*USE,LOAD
*ENDDO
FINISH
出现警告，是BFE命令用错了吧 有一个问题，你用的53单元建模，那么肯定是二维模型了，那为什么BFE,ALL,JS,,I加载了x方向的电流？二维分析只能加载z方向电流啊ForLab = JS and STLOC = 1, VAL1, VAL2 and VAL3 are the X, Y, and Zcomponents of current density (in the element coordinate system), andVAL4 is the phase angle.
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