ANSYS 瞬态热分析---零件在水中冷却

目录

• 1. 案例
• 2. APDL分析

1. 案例

一个温度为300℃的铜环和一个温度为200℃的铁环，放置到22℃的水中进行淬火。水桶为铁质的圆形。分析中忽略水的流动。

材料参数

热性能	铜	铁	水
导热系数（W/(m℃）	383	70	0.61
密度（kg/m3	8889	7833	996
比热容（J/(kg.℃）	390	448	4185

因为忽略了水的流动，故可将水看为一种材料，直接进行分析

2. APDL分析

finish
/clear
!==============第一步：进行稳态二维轴对称分析===============================
/filname,quench
/prep7
et,1,plane55                ! 选择单元类型
keyopt,1,3,1                ! 设置为轴对称分析
! 定义材料热性能参数
mp,kxx,1,383
mp,dens,1,8889
mp,c,1,390		         ! 定义铜的热物理性能参数
mp,kxx,2,70
mp,dens,2,7837
mp,c,2,448                ! 定义铁的热物理性能参数
mp,kxx,3,0.61
mp,dens,3,996
mp,c,3,4185               ! 定义水的热物理性能参数
rectng,0,0.08,0,0.01
rectng,0.08,0.1,0,0.01
rectng,0.1,0.12,0,0.01
rectng,0.12,0.14,0,0.01
rectng,0.14,0.15,0,0.01
rectng,0.14,0.15,0.01,0.055
rectng,0.14,0.15,0.055,0.1
rectng,0.14,0.15,0.1,0.15    ! 建立铁箱矩形
rectng,0.08,0.1,0.01,0.055
rectng,0.1,0.12,0.01,0.055   ! 建立铜环
rectng,0.08,0.1,0.055,0.1    ! 建立铁环
rectng,0,0.08,0.01,0.055
rectng,0.12,0.14,0.01,0.055
rectng,0,0.08,0.055,0.1
rectng,0.10,0.12,0.055,0.1
rectng,0.12,0.14,0.055,0.1
rectng,0,0.08,0.1,0.15
rectng,0.08,0.1,0.1,0.15
rectng,0.10,0.12,0.1,0.15
rectng,0.12,0.14,0.1,0.15    ! 建立水几何模型
aglue,all                 ! 粘接各矩形
esize,0.003,0,             ! 定义单元划分尺寸
asel,s,,,1,21,20
asel,a,,,23,29,1
aatt,2,1,1                 ! 附于铁箱和铁环的材料属性
asel,s,,,30,33,3
aatt,1,1,1                 ! 附于铜环的材料属性
asel,s,,,31,32,1
asel,a,,,34,39,1
asel,a,,,22
aatt,3,1,1                 ! 附于水的材料属性
allsel,all
mshkey,2                 ! 定义映射单元划分类型
amesh,all                 ! 划分单元
asel,s,,,1,21,20
asel,a,,,23,28,1
asel,a,,,31,32,1
asel,a,,,34,39,1
asel,a,,,22
nsla,,1
d,all,temp,20              ! 施加铁箱和水的温度约束条件
asel,s,,,30,33,3
nsla,,1
d,all,temp,300              ! 施加铜环的温度约束条件
asel,s,,,29
nsla,,1
d,all,temp,200              ! 施加铁环的温度约束条件
allsel,all
/solu
antype,trans               ! 设定为瞬态分析
timint,off        	! 关掉时间积分, 即设定为稳态分析
time,0.01		 ! 设定只有一个子步的时间很小的载荷步
deltim,0.01                ! 设定求解时间子步
solve                     ! 求解
!==============第二步：进行瞬态二维轴对称分析===============================
time,3600		  ! 设定求解时间
timint,on		 ! 打开时间积分
deltim,26,2,200		! 设置时间步长，最大及最小时间步长
autots,on		! 打开自动时间步长
ddelet,all,temp		! 删除稳态分析中定义的节点温度
outres,all,1		! 将每个子步的值写入数据库文件
solve                    ! 求解

finish
/post1
plnsol,temp,,0

样件上的某点随时间变化的曲线