热分析例子+求助

hanwei_csu

前段时间在论坛上看到一个热分析例子，现在找不到了，我是ansys的初学者，这个例子中有些命令不太理解，希望高手帮忙：
例子：
热分析 example

/com, This is a new version of the Brake Disc Model: Only Heat Flux is applied on the top and bottom rubbing surface
/com, Parameter n is the number of horizontal devisions on the heated disc plates;
/com, parameter m  is the number of vertiical divisions in the upper part of the heated disc
/com, parameter q is the number of horizontal divisions in the area of the slot
/com, parameter r is the number of vertical divisions in the lower part of the heated disc
/com, parameter h is the convection coefficient
/com,appropriately adjust the Integration Time Step (ITS)

finish
/clear
/filname,brake
/prep7
/title,Brake Disc Thermal: FLUX LOADING ONLY
!DEFINITION OF PARAMETERS
KEWH=364810           !changed
h=1500                !h can be change from 500,1000,1500 for convection investigation
n=10
m=6
r=8
q=6
!APPLIED ONLY FLUX, NO CONVECTION
flux=(KEWH/(3*2*0.03685))

!DEFINIION OF MATERIAL PROPERTIES
et,1,55,,,1      !MP IN N, KG,M,W,S,
mp,kxx,1,60                      !changed
mp,kyy,1,60                       !changed
mp,dens,1,7136                     !changed
mp,c,1,557                          !changed

!DEFINITION OF GEOMETRY
k,1,0,0
k,2,0.03745,0.0665
k,3,0.0535,0.0665
k,4,0.0665,0.0665
k,5,0.08019,0.0665
k,6,0.09695,0.04978
k,7,0.09695,0.01
k,8,0.09695,0.0065
k,9,0.1036,0.0065
k,10,0.1036,0.01
k,11,0.14045,0.01
k,12,0.14045,0.0065
k,13,0.14045,0
k,14,0.1036,0
k,15,0.09695,0
k,16,0.09045,0
k,17,0.09045,0.0065
k,18,0.09045,0.01
k,19,0.09045,0.04558
k,20,0.0775,0.06
k,21,0.0665,0.06
k,22,0.0535,0.06
k,23,0.03745,0.06
k,24,0.1036,0.0099          !below point 10
k,25,0.14045,0.0099         !below point 11
k,26,0.14045,0.0001         !above point 13
k,27,0.1036,0.0001          !above point 14            
k,28,0.09695,0.0001           !above point 15
k,29,0.09045,0.0001            !above point 16

/pnum,kpoi,1
kplot
l,2,3,3
l,3,4,3
l,4,5,3
l,5,6,4
l,6,7,6
l,7,8,m
l,9,24,5        !changed
l,12,25,5       !changed
l,17,18,m
l,10,11,n
l,9,12,n
l,13,14,n
l,8,9,q
l,14,15,q
l,15,16,6
l,18,19,6
l,19,20,4
l,20,21,3
l,21,22,3
l,22,23,3
l,23,2,6
l,3,22,6
l,4,21,6
l,5,20,6
l,6,19,6
l,7,18,6
l,8,17,6
l,12,26,7    !changed
l,17,29,7    !changed
l,8,28,7      !changed
l,9,27,7      !changed
l,24,25,n     !added
l,26,27,n     !added
l,10,24,1     !added
l,14,27,1     !added
l,11,25,1     !added
l,13,26,1     !added
l,16,29,1     !added
l,15,28,1     !added
l,29,28,6     !added
l,27,28,q      !added

/pnum,line,1
lplot
a,2,3,22,23
a,3,4,21,22
a,4,5,20,21
a,5,6,19,20
a,6,7,18,19
a,9,24,25,12        !changed
a,12,26,27,9        !changed
a,8,9,27,28    !changed
a,8,28,29,17     !changed
a,28,15,16,29         !added
a,28,27,14,15      !added
a,7,8,17,18
a,10,11,25,24       !added
a,13,14,27,26        !added
amesh,all
SAVE
eplot
fini

/com, Transient solution
/solu
/com, First load step
antype,TRAN
timint,on
tunif,20
time,1e-2
deltime,0.025
kbc,1
sfl,10,hflux,flux,
sfl,12,hflux,flux,
sfl,32,conv,h,,20          !convcetion added
sfl,33,conv,h,,20          !convcetion added
sftran
eplot
nsel,all
esel,all
outres,nsol,all
lswrite,1

/com, Second load step
timint,on,therm
time,6
kbc,0
deltime,0.203

sfl,10,hflux,0,
sfl,12,hflux,0,
sfl,32,conv,h,,20         !convcetion added
sfl,33,conv,h,,20         !convcetion added

sftran
/psf,conv,hcoef,1
eplot
nsel,all
esel,all
outres,nsol,all
lswrite,2

/com, Third load step
timint,on,therm
time,20
kbc,0
deltime,0.05

sfl,10,hflux,0,
sfl,12,hflux,0,
sfl,32,conv,h,,20         !convcetion added
sfl,33,conv,h,,20         !convcetion added

sftran
/psf,conv,hcoef,1
eplot
nsel,all
esel,all
outres,nsol,all
lswrite,3

lssolve,1,3,1
save
fini

/post1
set,2,last
plnsol,temp

!set,3,last
plnsol,temp

!prerr,tepc
!plesol,terr
!plesol,tdsg

!CONTINUE MANUAL POST-PROCESSING

/POST26


问题1：et,1,55,,,1      中&nbsp什么意思？
问题2：瞬态分析中，初始时间为1e-2，如果改用0可否？在瞬态分析前，一般需要先做稳态分析，这里直接就开始瞬态分析，什么条件下使用前者，我还不太清楚，盼高手指点。
问题3：为什么要使用sftran命令？

hanwei_csu

高手快来啊！！！！！！！！！！！！！！！！！

hanwei_csu

问题4：在关键点之间连直线，使用LSTR命令和L命令是不是一样的？