瞬态热分析
怎样用稳态分析的结果作为瞬态分析的初始条件,下面是一个例子,把全部命令输入进去然后求解没问题,但是分两次输入就出现问题了。第一次输入稳态分析命令,第二次输入瞬态分析命令第21例瞬态热分析实例—水箱
/CLEAR
/FILNAME,EXAMPLE21
/PREP7
ET,1,PLANE77
MP,KXX,1,383
MP,DENS,1,8889
MP,C,1,390
MP,KXX,2,70
MP,DENS,2,7833
MP,C,2,448
MP,KXX,3,2
MP,DENS,3,996
MP,C,3,4185
RECTNG,0,0.6,0,0.5
RECTNG,0.15,0.225,0.225,0.27
RECTNG,0.342,0.42,0.225,0.27
AOVLAP,ALL
AATT,1,1,1
MSHAPE,0
MSHKEY,1
ESIZE,0.01
AMESH,2
AATT,2,1,1
AMESH,3
AATT,3,1,1
MSHKEY,0
ESIZE,0.03
AMESH,4
FINISH
/SOLU
ANTYPE,TRANS
TIMINT,OFF
TIME,0.01
DELTIM,0.01
ESEL,S,MAT,,3
NSLE,S
D,ALL,TEMP,20
ESEL,S,MAT,,2
NSLE,S
D,ALL,TEMP,500
ESEL,S,MAT,,1
NSLE,S
D,ALL,TEMP,400
ALLSEL
SOLVE
TIME,3600
TIMINT,ON
DELTIM,50,10,200
AUTOTS,ON
DDELET,ALL,TEMP
OUTRES,ALL,ALL
SOLVE
SAVE
FINISH
/POST26
NSOL,2,106,TEMP,,TEMP_106
PLVAR,2
FINISH
而且他的gui操作也是无法得到结果,只能按照命令流去做,问题出现在什么地方? 我没有用命令流,直接gui没有问题的 都用12了。。 结果有问题?
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