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w76990480荣誉版主 (文坛精英)
super-emuch
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[求助]
热导率计算heat/flux 问题
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本人最近在计算ZnO热导率 用了两种方法，同一个势函数 1.heat/flux 方法 计算结果如下图 计算得到的热导率大概50w/mk左右  1.jpg  2.jpg 2.thermal/conductivity方法 利用thermal/conductivity计算10nm 热导率大概5w/mk左右 两个方法为什么差距这么大呢？ 温度梯度也是线性的  3.jpg [ Last edited by w76990480 on 2013-7-19 at 17:46 ] |
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11楼2017-05-05 21:57:29
2楼2013-07-20 16:11:49
z654023015
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3楼2013-07-21 17:00:37
w76990480
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Dear lammps users： I have some questions on ZnO thermal conductivity calculation，The fix heat/flux and fix thermal/conductivity command were selected to calculating the ZnO thermal conductivity. 1.fix heat/flux I wrote the input file like the manual (http://lammps.sandia.gov/doc/compute_heat_flux.html), the average thermal conductivity of ZnO(Model: x 5.0nm y 4.8nm z 5.3nm) is about ~50W/mk. The thermal conductivity-m curve you can see the fig 1 in the attachment. 2.thermal/conductivity The periodic boundary wurtzite-type ZnO was selected to calculating thermal conductivity in x y z direction 2.8nm 2.9nm 15.7nm respectively。The main command as following： fix temp_profile all ave/spatial 1 100000 100000 z lower 1.0 v_temp file temp.profile units lattice #YL for test fix heat_swap all thermal/conductivity 10 z 60 fix e_exchange all ave/time 10 1 100000 f_heat_swap file e_exchange.dat fix accumu_thermalK_out all ave/time 10 1 100000 v_accumu_thermalK file thermal_conductivity.dat I used this “variable accumu_thermalK equal f_e_exchange*${ev2J}/(2*0.001*${N}*${ps2s}*${Ox}*${Oy}*${An2m}*f_delta_out)*(${Oz}/60.0*26.0)” variable command to calculating it. timestep 0.001 N step Is it right（this variable to calculate the thermal conductivity）? Temperature gradient is linear in the fig 2 (in the attachment). the average thermal conductivity is about 5W/mk. Now, Two kinds of method to calculate the thermal conductivity of why have so big difference(The two methods with the same potential)? |
4楼2013-07-21 22:00:36
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