求助：怎么加载脉冲电流的生热率？多谢1

ljlaser

模型描述：几个材料层glue在一起，上面一层输入电流后生热，然后传导到下层，模拟热分布。连续电流注入时可以用类似P=IU这样的公式算出生热率带入做稳态热分析，但脉冲电流呢？例如，脉冲宽度20微秒，重复频率1KHz，也就是说，一个脉冲过来，只有2％有电可以发热，怎么加载这样的生热率？

iomega

原帖由 ljlaser 于 2007-10-30 23:51 发表
模型描述：几个材料层glue在一起，上面一层输入电流后生热，然后传导到下层，模拟热分布。连续电流注入时可以用类似P=IU这样的公式算出生热率带入做稳态热分析，但脉冲电流呢？例如，脉冲宽度20微秒，重复频率1 ...

Why not just try electical-thermal coupling simulation? For the layer having electical current flowing, using electrical-thermal element, other layer can use thermal elements. Given electrical resistivity and thermal conductivity( & heat capacity) for top layer, then you can get both electrical pontial and temperature field at the same time.

Check this link:
http://www.simwe.com/forum/viewthread.php?tid=579648&highlight=%2Biomega

chaos_zzy

我觉得这个问题,可以做时均等效. 以前交流电生热也是用该方法.

iomega

原帖由 chaos_zzy 于 2007-10-31 21:29 发表
我觉得这个问题,可以做时均等效. 以前交流电生热也是用该方法.

Yes, it's doable through RMS power calculation of the square-wave pulses.  However, that only gives the steady-state temperature distribution of the structure.

In real case, the temperature field  T(x,y,z,t), in the structure induced by this square wave electrical pulses, after it reaches quasi-steady-state, will be :

T(x,y,z,t) = T(x,y,z)_steady_state + T_1(x,y,z)*cos(w*t+theta_1)+T_3(x,y,z)*cos(3w*t+theta_2)+T_5(x,y,z)*cos(5w*t+theta_3)+.....

where w  = 2*Pi*f and f is the frequency of the square wave.


The  时均等效 only gives the first term T(x,y,z)_steady_state, while the rest term can be obtained through 2 ways:
(1) transient simulation which directly gives the final T(x,y,z,t) - can use electrical-thermal coupling element in ANSYS.
(2) harmonic simulation which provide information of T_1(x,y,z) and theta_1 and so on -  have to convert Joule heating to heat generation and using thermal element only in ANSYS.

chaos_zzy

多谢斑竹分析，帮助我们更深入的了解瞬态问题。
我的思路是这样的：通过积分，计算出一秒中的焦耳发热量，将该值作为恒定功率加在导体上。
尽管实际上不同时刻的发热量不同，但考虑热量传播需要一定的时间，在这么高的频率下（1khz），测温试验时根本看不出波动，我觉得该方法也是可靠的。

iomega

原帖由 chaos_zzy 于 2007-11-1 10:17 发表
多谢斑竹分析，帮助我们更深入的了解瞬态问题。
我的思路是这样的：通过积分，计算出一秒中的焦耳发热量，将该值作为恒定功率加在导体上。
尽管实际上不同时刻的发热量不同，但考虑热量传播需要一定的时间，在 ...

You are right, if think from the other way:
Perform fourier transform of the square wave heat generation, the w=0 (DC) term will be the term to cause steady-state temperature rise of the structure, should be the same as from your approach, while the f=1w ,3w, 5w terms will have their own temperature repsonses in structure, but with phase delay with respect to the square wave, because of the heat capacity of the material. The amplitdue of temperature fluctuations at w,3w,5w depends on the thermal time constant of the system,comparing to the heating frequency w=2Pi*f (f =1KHz).

chaos_zzy

恩，收获不少。