热分析中最后得出的结果跟实际相差很远！？

reltek

脉冲激光加载，在脉冲结束的那一刻查看工件温度分布，发现整个工件温度都一致，见下图，但实际上这是不可能的！！！
案例环境：激光光束半径-------0.1mm
                 工件外径--------1.5mm
                 工件内径--------0.5mm
为什么？？？

reltek

没人关注啊///

iomega

what is the duration of your laser pulse? It looks like there is no boundary condition applied on the model, if the thermal diffusivity of the material is high and time is long enough, there will be uniform temperature distribution inside the body.
In real situation, you can use Biot and Fourier numbers to estimate it.

reltek

1、脉冲持续时间为10s，见附图。
2、加载对象是常见的304不锈钢，但没有查到它的热扩散系数（我想该参数是与外界环境相关的吧？）。
3、有一点比较模糊：这个模型没有施加边界对流换热系数（正常来讲，工件放置在空气中，它周围的对流换热系数多少？）。
4、“Biot and Fourier numbers”请版主介绍一下吧。ansys如何实现？

万分感谢！

iomega

No wonder!!!

Assume thermal diffusivity alf of steel is around 5X10^-5 m^2/s.
The tube length L is around 5 mm, so it take heat  L^2/alf = 0.5s to reach the entire region. After that time scale, the temperature distribution will disappear since there is no boundary cooling to dissipate the heat, hence no heat flux through the boundary, therefore no temperature gradient across the structure.

Typical nature convection coefficient will be around 5 W/m^2-K.

The Biot number is the ratio between internal thermal conduction resistance and external thermal convection resistance. If Biot number <<1, lumped model is reasonable and there is no temperature gradient  inside the solid, like your case where convection thermal resistance is infinity.

Fourier number tell you how far the heat conduct for a give time, mainly L^2 = alf*time.



You may try to look at the temperature profile at t = 0.1s !!!


4# reltek