LAMMPS WWW Site - LAMMPS Documentation - LAMMPS Mailing List Archives
[lammps-users] Making sense of in.heat example
[Date Prev][Date Next][Thread Prev][Thread Next][Date Index][Thread Index]

[lammps-users] Making sense of in.heat example

From: rambod yousefzadeh tabasi <rambod.tabasi@...4851...>
Date: Fri, 1 Dec 2017 13:46:16 -0600

Hi all,

As obvious as this might be, I have not been able to understand the  time values used in the in.heat example to find kappa. I am referring to the two 100s multiplied and divided here. 100 would be the total run time for the third run but if I understand it correctly q is being added for 30000 time steps and not only the last 20000. I would really appreciate it if someone could explain in more detail what these two hundreds represent. 
Here is the important part of the code and the explanation that was provided on it: 

fix hot all heat 1 100.0 region hot
fix cold all heat 1 -100.0 region cold
thermo_style custom step temp c_Thot c_Tcold
thermo 1000
run 10000
# thermal conductivity calculation
compute ke all ke/atom
variable temp atom c_ke/1.5
compute layers all chunk/atom bin/1d z lower 0.05 units reduced
fix 2 all ave/chunk 10 100 1000 layers v_temp file profile.heat
variable tdiff equal f_2[11][3]-f_2[1][3]
fix ave all ave/time 1 1 1000 v_tdiff ave running start 13000
thermo_style custom step temp c_Thot c_Tcold v_tdiff f_ave
run 20000
(2) in.heat

dQ = (100*100) / 100 / 18.82^2 / 2
  100*100 = 100 (time in tau) * 100 (energy delta specified in fix heat)
  100 = 20,000 steps at 0.005 tau timestep = run time in tau
  xy box area = 18.82^2
  divide by 2 since energy flux goes in 2 directions due to periodic z
dTemp = 0.748 from log file for average Temp difference between 2 regions
dZ = 18.82

Kappa = 3.55