简单热传递仿真！
Finish
/clear
/filname, inductance
/title, ZY_inductance
/PREP7
/number, 0
Pi=acos(-1)
EMUNIT, MKS
core_length=30.25e-3
core_depth=26.3e-3
all_window=19.85e-3
mid_length=10.65e-3
window_length=(all_window-mid_length)/2
window_depth=16.3e-3
left_length=(core_length-all_window)/2
up_depth=5e-3
mid_depth=5e-3
rad_win=0.5e-3
x_dis=1e-3
y_dis=1.5e-3
th_air=1.3e-3 !thickness of air
w_c_dis=1e-3 !distance from winding to core
th_win=5e-3!thickness of winding
ET,1,plane77 !windings
ET,2,plane77 !core
ET,3,plane77 !air
MP,KXX,1,398 !铜导线的热导率
MP,DENS,1,8900 !密度
MP,C,1,385 !比热
MP,KXX,2,0.55 !磁芯的热导率
MP,DENS,2,4700
MP,C,2,0.67
MP,KXX,3,0.02 !空气的热导率
MP,DENS,3,1.2
MP,C,3,1000
rectnag,0,core_length,0,core_depth+th_air
rectnag,left_length,left_length+window_length,up_depth+0.5e-3,up_depth+window_depth+th_air
rectnag,left_length+window_length+mid_length,left_length+2*window_length+mid_length,up_depth+0.5e-3,up_depth+window_depth+th_air
rectnag,0,core_length,up_depth,up_depth+th_air
allsel
aadd,2,3,4
numcmp,all
allsel
asba,1,2,,delete,keep
numcmp,all
numstr,area,50
cyl4,left_length+x_dis+2*rad_win,up_depth+y_dis+2*rad_win,rad_win,0,,360
agen,2,50,,,x_dis+rad_win,,,
agen,9,50,52,,,y_dis
agen,2,50,68,,window_length+mid_length-2*rad_win,,
allsel
aovlap,all
numcmp,all
allsel
asel,s,,,39
aatt,3,,3
smrtsize,3
amesh,all
allsel
asel,s,,,1
aatt,2,,2
smrtsize,3
amesh,all
allsel
asel,s,,,2
aatt,2,,2
smrtsize,3
amesh,all
allsel
asel,s,,,3,38
aatt,1,,1
smrtsize,3
amesh,all
allsel
nsel,s,loc,x,0
nsel,a,loc,x,core_length
nsel,a,loc,y,0
nsel,a,loc,y,core_depth+th_air
d,all,temp,30
allsel
nsel,s,node,,17763
nsel,a,node,,17990
nsel,a,node,,16981
nsel,a,node,,17363
nsel,a,node,,16646
nsel,a,node,,16770
nsel,a,node,,15724
nsel,a,node,,16176
nsel,a,node,,15397
nsel,a,node,,15519
nsel,a,node,,14827
nsel,a,node,,14937
nsel,a,node,,14046
nsel,a,node,,14418
nsel,a,node,,13490
nsel,a,node,,13868
nsel,a,node,,12940
nsel,a,node,,13150
nsel,a,node,,23487
nsel,a,node,,23560
nsel,a,node,,22816
nsel,a,node,,23224
nsel,a,node,,22302
nsel,a,node,,22402
nsel,a,node,,21603
nsel,a,node,,21931
nsel,a,node,,20962
nsel,a,node,,21367
nsel,a,node,,20214
nsel,a,node,,20499
nsel,a,node,,19612
nsel,a,node,,19904
nsel,a,node,,18962
nsel,a,node,,19231
nsel,a,node,,18351
nsel,a,node,,18615
f,all,heat,0.06
/solu
antype,static
nropt,auto
autots,on
nsubst,50
allsel
outres,all,all
solve
/post1
allsel
nsle,s
plnsol,temp
给中间圆面施加的总功率为2.3W，边界施加恒定30度的温度，目的是想要看铜导线（即圆面）周围的热分布。请各位师傅帮忙分析。现在的问题是整个温升只有2度，但又不会理论计算，所以不知道对不对。
能解决的朋友仿真币全奉上。
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