iomega版主你好，我按照你给我提的建议改写了程序，请你看看是不是应该这样改
iomega版主你好，我按照你给我提的建议改写了程序，请你看看是不是应该这样改？我把激光转化成热流密度加载到了物体表面，
!假设硅对激光的吸收率为0.9，激光照射直径为150μm
fini
/clear
/prep7
et,1,solid90
blc4,0,0,1.5e-3,0.04e-3,0.2e-3 !建立硅片模型长1.5mm，宽200μm，厚度40μm
mp,kxx,1,80 !定义硅片的导热系数
mp,dens,1,2230 !定义硅片的密度
lsel,s,,,2,4,2
lsel,a,,,5,7,2
lesize,all,,,100
lsel,s,,,9,12,3
lsel,a,,,10,11,1
lesize,all,,,20
lsel,s,,,3,6,3
lsel,a,,,1,8,7
lesize,all,,,4
type,1
vmesh,all
fini
/solu
!定义固定边界条件
!nsel,s,loc,x,0
!d,all,ux,0,,,uy,uz
!alls
!施加一个直径为150μm的激光均匀热流密度
nsel,s,loc,x,0.6e-3,0.75e-3
nsel,r,loc,y,0.04e-3
nsel,r,loc,z,0.025e-3,0.175e-3
sf,all,hflux, 88108 !施加均匀热流密度
alls
solv
fini
/post1
prnsol,TEMP
plnsol,TEMP
fini
以上是原来的程序，在运行的时候提醒出现节点温度超出限制
在后来我给一些面上加上了对流系数，程序如下，就没有出现这样的问题，请问是不是应该这样改呢？
fini
/clear
/prep7
et,1,solid90
blc4,0,0,1.5e-3,0.04e-3,0.2e-3 !建立硅片模型长1.5mm，宽200μm，厚度40μm
mp,kxx,1,0.21 !定义硅片的导热系数
mp,dens,1,2230 !定义硅片的密度
lsel,s,,,2,4,2
lsel,a,,,5,7,2
lesize,all,,,100
lsel,s,,,9,12,3
lsel,a,,,10,11,1
lesize,all,,,20
lsel,s,,,3,6,3
lsel,a,,,1,8,7
lesize,all,,,4
type,1
vmesh,all
fini
/solu
toffst,273
tunif,25
asel,s,,,1,3,1
sfa,all,,conv,10,25 !定义对流系数，其中10表示对流系数，25表示
alls
!施加一个直径为150μm的激光均匀热流密度
nsel,s,loc,x,0.6e-3,0.75e-3
nsel,r,loc,y,0.04e-3
nsel,r,loc,z,0.025e-3,0.175e-3
sf,all,hflux, 88108 !施加均匀热流密度总体温度
alls
solv
fini
/post1
plnsol,TEMP
prnsol,TEMP
fini
非常诚恳的向你请教，现在导师急着要这个的结果，这段时间都愁死了，谢谢你版主 1# lchylinchy
结构上下对称，只取上面一半即可。不是瞬态问题，应该不需要密度。材料性质的单位统一吗？ Physically it' ok now. But as tonnyw pointed out, the properties you specified seem off. Thermal conductivity of silicon is around 140 W/m-K, instead of 0.21 in your code. 非常感谢两位的帮忙，谢谢，非常感谢 但是我把硅的传热系数改成140以后，为什么整个硅材料的温度都是一样的呢？
按理说应该是激光作用点的地方温度最高啊 我在网上查到的硅的导热系数是0.2121W/m·K 但是我把硅的传热系数改成140以后，为什么整个硅材料的温度都是一样的呢？
这个上不上和材料的尺寸有很大关系呢？
我后来把硅的尺寸变大1000倍，这时候出来的结构就是激光作用点的地方温度最高 Can you give the reference link for this 0.2121W/m-K silicon data? Even the SiO2 thermal conductivity is around 1.4W/m-K.
According to my experience, the 50nm thick ultra-thin silicon fim thermal conductivity I measured before using suspended sandwich structure and electrical resistance thermometry is around 50W/m-K, though the reduction from its bulk value of 140 W/m-K is due to the phonon-boundary scattering (size effect).
Also, due to the large external convective thermal resistance, the temperature gradient inside silicon will be minimized, since the Biot number << 1 in this case. 本帖最后由 tonnyw 于 2011-7-28 12:57 编辑
8# iomega
Maybe you mean 0.2121W/cm-K instead of 0.2121W/m-K.
I also feel confused about a couple of things.
1. What is the unit of the applied heat flux? It is a huge number, which does not strike me as a correct one?
2. Here you choose the ambient temperature to be 25. I think you mean centigrade instead of K. Should you change it to be K?
3. I also doubt about the unit of the convection coefficient.
4. To make life easier, you might cut a slice and run a 2D plane problem. Even 21.2 W/m-K seems too low for silicon - kind of impossible. Above room temperature, the phonon-phonon scattering dominates which shows 1/T (unit of Kelvin) dependency, so 21.2W/m-k could only happen at much higher temperature, e.g., at 1000K.
It looks like he applied power on 150um square area, so the total power is 88108*1.5e-4*1.5e-4 = 1.98mW. The unit of heat flux should be W/m^2.
It's fine to set ambient temperature to be 25 C instead of K, as long as the simulation doesn't involve thermal radiation. Even 21.2 W/m-K seems too low for silicon - kind of impossible. Above room temperature, the phonon-phonon scattering dominates which shows 1/T (unit of Kelvin) dependency, so 21.2W/m-k could only happ ...
iomega 发表于 2011-7-28 13:10 http://forum.simwe.com/images/common/back.gif
Here is the Wikipedia link where the heat conductivity for silicon can be found.
http://en.wikipedia.org/wiki/Silicon
I thought for convection boundary condition we have something like the following:
heat flux = h*(T - T_ambient)
where h is the convection or film coefficient. If the ambient temperature is set to be 0, it doesn't make if the unit is K or C. But if it is nonzero, seems to me the unit really matters since the unit of T_ambient has to be consistent with the film coefficient. This ambient temperature will be used to form the load vector on the right-hand-side. 谢谢斑竹的帮助
我今天去重新查了硅的导热系数，大约是你给我讲的值。
那么请问一下我把摄氏温度转换成为绝对温度用下面的命令正确吗？
toffst,273 斑竹
请问命令sfa,all,,conv,3,25中的温度25指的是摄氏温度还是绝对温度呢？ tonnyw 斑竹你好
1、热流密度我用的单位是W/m^2， 在这里我是通过激光的功率为2mW，假设硅对光的吸收率是0.9，激光的作用直径是150μm，算出加载到硅上的热流密度为88108W/m^2
2、我选择的环境温度为25℃，但是在之前我用命令toffst,273，将摄氏温度转换成了绝对温度的（不知道这样对不对）
3、我有一个问题是怎样去确定对流系数？ 斑竹你好
请问一下，像我这种情况，既施加了热流密度载荷，又施加了对流载荷，那么是不是要用到表面热效应单元呢，如果要用，应该怎样去用呢？非常谢谢 15# lchylinchy
不懂软件操作，印象里如果加对流边界条件的话是需要表面效应单元。帮助文件里应该有。在建模求解前，最好了解一下你要求解的是偏微分方程的具体形式是什么。 It seems you applied heat flux on top surface and h on the bottom and sides, so there is no need to use surface elements, unless you want to apply both heat flux and h on the same area.
Back to your problem, I think it's totally correct. Assuming k =140 W/m-K fo r silicon, and h = 10 W/m^2-K, so convective thermal resistance is (0.28mm*1.5mm*10)^-1 =238095 K/W.
Since you selected the nodes for applying heat flux, the actual power applied on to silicon block is between 1.98mW and 1.8W, lets say 1.9W. Ignore the conductive thermal resistance of silicon, then temperature will be around 238095*1.9e-3+25 = 477 C.
toffst has no impact on the result.
To tonnyw:
he already specify 25 in sfa command, and it doesn't matter whether it's C or K, since the unit of h is either W/m^2-K or W/m^2-C, which mean it's capable of dissipating some Watt energy per m^2 area and per every degree temperature rise. And there is only offset between K and C, but slope is 1. 17# iomega
>he already specify 25 in sfa command, and it doesn't matter whether it's C or K, since >the unit of h is either W/m^2-K or W/m^2-C, which mean it's capable of dissipating some >Watt energy per m^2 area and per every degree temperature rise. And there is only >offset between K and C, but slope is 1.
I still feel a little bit confused. He is interested in absolute temperature distribution. He set the ambient temperature to be 25. He put the offset 273 meaning to change 25C into K. Otherwise, I don't see how this is going to work in the finite element formulation.
I think if he is interested in the relative temperature rise with respect to the ambient temperature, it doesn't matter if the unit is K or C. But he has to put the ambient temperature to be 0, namely,
sfa,all,,conv,10, 0
It would be super if lchylinchy could run some tests to verify it. Here is the info from ANSYS help. It should answer your question, I think.
TOFFST
TOFFST, VALUE Specifies the temperature offset from absolute zero to zero.
SOLUTION: Analysis Options
AUX12: General Radiation
MP ME ST PR PRN DS DSS FL <> <> <> PP <> EME MFS
VALUE
Degrees between absolute zero and zero of temperature system used (should be positive).
Notes
Specifies the difference (in degrees) between absolute zero and the zero of the temperature system used. Absolute temperature values are required in evaluating certain expressions, such as for creep, swelling, radiation heat transfer, MASS71, etc. (The offset temperature is not used in evaluating emissivity.) Examples are 460° for the Fahrenheit system and 273° for the Celsius system. The offset temperature is internally included in the element calculations and does not affect the temperature input or output. If used in SOLUTION, this command is valid only within the first load step.
This command is also valid in PREP7. 谢谢两位版主的帮助，对我的启发很大
请问一下Assuming k =140 W/m-K fo r silicon, and h = 10 W/m^2-K, so convective thermal resistance is (0.28mm*1.5mm*10)^-1 =238095 K/W.
这里的(0.28mm*1.5mm*10)^-1 =238095 K/W是怎样求的呢？
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