我一直在尝试使用fipy
模拟双温度模型
模型的数学模型:
C_e(∂T_e)/∂t=∇[k_e∇T_e ]-G(T_e-T_ph )+ A(r,t)
C_ph(∂T_ph)/∂t=∇[k_ph∇T_ph] + G(T_e-T_ph)
源应该加热电子{
这是我的密码:
from fipy.tools import numerix
import scipy
import fipy
import numpy as np
from fipy import CylindricalGrid1D
from fipy import Variable, CellVariable, TransientTerm, DiffusionTerm, Viewer, LinearLUSolver, LinearPCGSolver, \
LinearGMRESSolver, ImplicitDiffusionTerm, Grid1D, ImplicitSourceTerm
## Mesh
nr = 50
dr = 1e-7
# r = nr * dr
mesh = CylindricalGrid1D(nr=nr, dr=dr, origin=0)
x = mesh.cellCenters[0]
# Variables
T_e = CellVariable(name="electronTemp", mesh=mesh,hasOld=True)
T_e.setValue(300)
T_ph = CellVariable(name="phononTemp", mesh=mesh, hasOld=True)
T_ph.setValue(300)
G = CellVariable(name="EPC", mesh=mesh)
t = Variable()
# Material parameters
C_e = CellVariable(name="C_e", mesh=mesh)
k_e = CellVariable(name="k_e", mesh=mesh)
C_ph = CellVariable(name="C_ph", mesh=mesh)
k_ph = CellVariable(name="k_ph", mesh=mesh)
C_e = 4.15303 - (4.06897 * numerix.exp(T_e / -85120.8644))
C_ph = 4.10446 - 3.886 * numerix.exp(-T_ph / 373.8)
k_e = 0.1549 * T_e**-0.052
k_ph =1.24 + 16.29 * numerix.exp(-T_ph / 151.57)
G = numerix.exp(21.87 + 10.062 * numerix.log(numerix.log(T_e )- 5.4))
# Boundary conditions
T_e.constrain(300, where=mesh.facesRight)
T_ph.constrain(300, where=mesh.facesRight)
# Source 𝐴(𝑟,𝑡) = 𝑎𝐷(𝑟)𝜏−1 𝑒−𝑡/𝜏 , 𝐷(𝑟) = 𝑆𝑒 exp (−𝑟2/𝜎2)/√2𝜋𝜎2
sig = 1.0e-6
tau = 1e-15
S_e = 35
d_r = (S_e * 1.6e-9 * numerix.exp(-x**2 /sig**2)) / (numerix.sqrt(2. * 3.14 * sig**2))
A_t = numerix.exp(-t/tau)
a = (numerix.sqrt(2. * 3.14)) / (3.14 * sig)
A_r = a * d_r * tau**-1 * A_t
eq0 = (
TransientTerm(var=T_e, coeff=C_e) == \
DiffusionTerm(var=T_e, coeff=k_e) - \
ImplicitSourceTerm(coeff=G, var=T_e) + \
ImplicitSourceTerm(var=T_ph, coeff=G) + \
A_r)
eq1 = (TransientTerm(var=T_ph, coeff=C_ph) == DiffusionTerm(var=T_ph, coeff=k_ph) + ImplicitSourceTerm(var=T_e, coeff=G) - ImplicitSourceTerm(coeff=G, var=T_ph))
eq = eq0 & eq1
dt = 1e-18
steps = 7000
elapsed = 0.
vi = Viewer((T_e, T_ph), datamin=0., datamax=2e4)
for step in range(steps):
T_e.updateOld()
T_ph.updateOld()
vi.plot()
res = 1e100
dt *= 1.01
count = 0
while res > 1:
res = eq.sweep(dt=dt, underRelaxation=0.5)
print(t, res)
t.setValue(t + dt)
据我所知,我可以将潜热作为源项和汇项包含在eq1中,或者在C_ph
中添加一个高斯峰,峰中心应该在熔点附近
我不知道哪一个更好、更稳定,我不知道如何实现其中任何一个
请帮帮我
根据评论(请将其编辑成问题),将
eq1
更改为它将被显式计算,但只要
T_ph
更新,它就会更新相关问题 更多 >
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