to xbzhang:
这是我调试后的程序,附件中是温度场的动态显示图。峰值电流和基值电流分别为200和50A,我将两电流分别改为200A(相当于电流为恒定值),得出的结果和电流为200A时的结果一致,所以我觉得这种方法应该没什么问题吧?有什么问题欢迎讨论。
/FILNAME,maichong,0
/TITLE,the thick of the board is 3mm,velocity is 0.5m/min,I1=200A,I2=50A
/PREP7
ET,1,SOLID70
KEYOPT,1,2,0
KEYOPT,1,4,0
KEYOPT,1,7,0
KEYOPT,1,8,0
!*
!99.75%Fe
mptemp,1,0,20,200,500,800,1100,1500
mpdata,dens,1,1,7880,7880,7710,7588,7320,7278
mpdata,kxx,1,1,53.18,47.74,36.01,32,29.3,25
mpdata,c,1,1,470,470,470,935,1879,1879
mpdata,enth,1,1,7.17e8,14.35e8,28.7e8,43.7e8,52.95e8,82.15e8
BLOCK,-0.025,0.025,-0.003,0,0,0.3,
block,-0.005,0.005,-0.003,0,0,0.3,
vsel,s,volu,,1,2
vovlap,all
vglue,all
NUMCMP,LINE
NUMCMP,AREA
NUMCMP,VOLU
/REPLOT
aesize,all,0.001,
lesize,all,0.001, , , ,1, , ,1,
esize,0.001,0,
mshkey,1
mshape,0,3d
vmesh,1
ESIZE,0.01,0,
AESIZE,ALL,0.01,
LESIZE,ALL,0.01, , , ,1, , ,1,
MSHKEY,1
MSHAPE,0,3d
vmesh,2,3
/VIEW, 1 ,1,1,1
/ANG, 1
/REP,FAST
/psf,hflu,,2,0
/PNUM,LINE,1
finish
/solu
antype,trans
tunif,20
timint,on !time integration = on
allsel,all
nsel,s,loc,x,-0.003,0.003
nsel,r,loc,y,0
nsel,r,loc,z,0,0.005
antype,trans
tunif,20
kbc,1
!*set,delt,1
timint,on
autots,off
n=0
*do,tm,1e-6,8,0.02
!sfdele,all,hflux
/view, 1 ,1,1,1
/ang, 1
/rep,fast
!flst,2,1,5,orde,1
!fitem,2,2
!/go
!*
time,tm
autots,0
deltim, , , ,1
kbc,0
!*
!tsres,erase
flst,2,1,5,orde,1
fitem,2,2
/go
!*
!*
sfa,1,,conv,50,25,!w/m**2*C
sfa,2,,conv,50,25,
sfa,16,,conv,50,25,
sfa,11,,conv,12.5,25,!bottom
sfa,5,,conv,12.5,25,
sfa,15,,conv,12.5,25,
nsel,s,loc,x,-0.003,0.003
nsel,r,loc,y,0
nsel,r,loc,z,0,0.3
*if,tm,gt,3*n+(1e-6),then
*if,tm,le,3*n+2+(1e-6),then
sf,all,hflux,%feng%
*else
*endif
*endif
*if,tm,gt,3*n+2+(1e-6),then
*if,tm,le,3*(n+1)+(1e-6),then
sf,all,hflux,%ji%
*else
*endif
*endif
*if,tm,gt,3*(n+1)+(1e-6),then
n=n+1
*else
*endif
/status,solu
allsel,all
solve
*enddo
outres,all,all,
/psf,hflux,,2
finish
!峰值电流的表达式为200*24*0.6/{PI}/(0.003^2)*exp(-1/0.003/0.003*({X}^2+({Z}-8.33e-3*{TIME})^2))
!基值电流的表达式为50*24*0.6/{PI}/(0.003^2)*exp(-1/0.003/0.003*({X}^2+({Z}-8.33e-3*{TIME})^2)) |