将Matlab数值热传导模型转换为Python

时间：2014-03-17 17:36:11

标签： python matlab python-3.x numpy numerical-methods

我正在尝试将我的Matlab模型转换为瞬态热传导到Python。不幸的是，我在Python中的数值解决方案的输出与Matlab模型的输出不匹配。我正在使用Spyder IDE编写我的代码。

我的模型Matlab和Python之间的主要区别（我到目前为止已找到）：

  • 数组的Matlab索引从1开始，而Python索引从0开始
  • Matlab中A * x = B的解是x = A \ B，而在Python中它是x = np.linalg.solve(A,B)
  • 在Matlab中将列向量col = C更改为行向量row为row = C'，而在Python中为row = C.T

作为检查Matlab和Python模型的一种方法，我比较了A数组。如您所见，这两个数组不匹配：

... Matlab的

A =
    1.1411   -0.1411         0         0
   -0.0118    1.0470   -0.0353         0
         0   -0.0157    1.0470   -0.0313
         0         0   -0.0470    1.0593

...的Python 

A 
 [[ 1.14106122 -0.14106122  0.          0.        ]
 [-0.          1.04702041 -0.04702041  0.        ]
 [ 0.         -0.0235102   1.04702041 -0.0235102 ]
 [ 0.          0.         -0.04702041  1.05681633]]

我在Python代码中做得不对。我的猜测是它与Python索引数组的方式有关。但我不确定。

所以有关如何在Python中构建我的Matlab模型的任何建议都会受到高度赞赏吗？

以下是我尝试在Python中复制的Matlab示例：

% parameters
% -------------------------------------------------------------------------
rho = 700;      % density of wood, kg/m^3
d = 0.035e-2;   % wood particle diameter, m
cpw = 1500;     % biomass specific heat capacity, J/kg*K
kw = 0.105;     % biomass thermal conductivity, W/m*K
h = 375;        % heat transfer coefficient, W/m^2*K
Ti = 300;       % initial particle temp, K
Tinf = 773;     % ambient temp, K

% numerical model where b = 1 cylinder and b = 2 sphere
% -------------------------------------------------------------------------
nt = 1000;      % number of time steps
tmax = 0.8;     % max time, s
dt = tmax/nt;   % time step, s
t = 0:dt:tmax;  % time vector, s

nr = 3;   % number or radius steps
%nr = 100;   % number or radius steps
r = d/2;    % radius of particle, m
dr = r/nr;  % radius step, delta r
m = nr+1;   % nodes from center to surface

b = 2 ; % run model as a cylinder (b = 1) or as a sphere (b = 2)
if b == 1
    shape = 'Cylinder';
elseif b == 2
    shape = 'Sphere';
end

alpha = kw/(rho*cpw);    % thermal diffusivity, alfa = kw / rho*cp, m^2/s
Fo = alpha*dt/(dr^2);    % Fourier number, Fo = alfa*dt / dr^2, (-)
Bi = h*dr/kw;            % Biot numbmer, Bi = h*dr / kw, (-)

% creat array [TT] to store temperature values, row = time step, column = node
TT = zeros(1,m);
i = 1:m;
TT(1,i) = Ti;   % first row is initial temperature of the cylinder or sphere

% build coefficient matrix [A] and initial column vector {C}
A = zeros(m);   % pre-allocate [A] array
C = zeros(m,1); % pre-allocate {C} vector

A(1,1) = 1 + 2*(1+b)*Fo;
A(1,2) = -2*(1+b)*Fo;
C(1,1) = Ti;

for i = 2:m-1
    A(i,i-1) = -Fo*(1 - b/(2*i));   % Tm-1
    A(i,i) = 1 + 2*Fo;              % Tm
    A(i,i+1) = -Fo*(1 + b/(2*i));   % Tm+1
    C(i,1) = Ti;
end

A(m,m-1) = -2*Fo;
A(m,m) = 1 + 2*Fo*(1 + Bi + (b/(2*m))*Bi);
C(m) = Ti + 2*Fo*Bi*(1 + b/(2*m))*Tinf;

% display [A] array and [C] column vector in console
A
C

% solve system of equations [A]{T} = {C} for column vector {T}
for i = 2:nt+1
    T = A\C;
    C = T;
    C(m) = T(m) + 2*Fo*Bi*(1 + b/(2*m))*Tinf;
    TT(i,:) = T';   % store new temperatures in array [TT]
end

% plot
% -------------------------------------------------------------------------
figure(b)
plot(t,TT(:,1),'--k',t,TT(:,m),'-k')
hold on
plot([0 tmax],[Tinf Tinf],':k')
hold off
axis([0 tmax Ti-20 Tinf+20])
ylabel('Temperature (K)')
xlabel('Time (s)')
nr = num2str(nr); nt = num2str(nt); dt = num2str(dt); h = num2str(h); Tinf = num2str(Tinf);
legend('center','surface',['T\infty = ',Tinf,'K'],'location','southeast')
title([num2str(shape),', nr = ',nr,', nt = ',nt,', \Deltat = ',dt,', h = ',h])

这是我在Python中的尝试：

# use Python 3 print function
from __future__ import print_function

# libraries and packages
import numpy as np
import matplotlib.pyplot as py

# parameters
# -------------------------------------------------------------------------
rho = 700       # density of wood, kg/m^3
d = 0.035e-2    # wood particle diameter, m
cpw = 1500      # biomass specific heat capacity, J/kg*K
kw = 0.105      # biomass thermal conductivity, W/m*K
h = 375         # heat transfer coefficient, W/m^2*K
Ti = 300        # initial particle temp, K
Tinf = 773      # ambient temp, K

# numerical model where b = 1 cylinder and b = 2 sphere
# -------------------------------------------------------------------------
nt = 1000       # number of time steps
tmax = 0.8      # max time, s
dt = tmax/nt    # time step, s
t = np.arange(0,tmax+dt,dt)

nr = 3      # number or radius steps
r = d/2     # radius of particle, m
dr = r/nr   # radius step, delta r
m = nr+1    # nodes from center m=0 to surface m=steps+1

b = 2   # run model as a cylinder (b = 1) or as a sphere (b = 2)

alpha = kw/(rho*cpw)    # thermal diffusivity, alfa = kw / rho*cp, m^2/s
Fo = alpha*dt/(dr**2)   # Fourier number, Fo = alfa*dt / dr^2, (-)
Bi = h*dr/kw            # Biot numbmer, Bi = h*dr / kw, (-)

# create array [TT] to store temperature values, row = time step, column = node
TT = np.zeros((1,m))

# first row is initial temperature of the cylinder or sphere
for i in range(0,m):
    TT[0,i] = Ti

# build coefficient matrix [A] and initial column vector {C}
A = np.zeros((m,m))     # pre-allocate [A] array
C = np.zeros((m,1))     # pre-allocate {C} vector

A[0, 0] = 1 + 2*(1+b)*Fo
A[0, 1] = -2*(1+b)*Fo
C[0, 0] = Ti

for i in range(1, m-1):
    A[i, i-1] = -Fo*(1 - b/(2*i))   # Tm-1
    A[i, i] = 1 + 2*Fo              # Tm
    A[i, i+1] = -Fo*(1 + b/(2*i))   # Tm+1
    C[i, 0] = Ti

A[m-1, m-2] = -2*Fo
A[m-1, m-1] = 1 + 2*Fo*(1 + Bi + (b/(2*(m-1)))*Bi)
C[m-1, 0] = Ti + 2*Fo*Bi*(1 + b/(2*(m-1)))*Tinf

# print [A] and [C] to console
print('A \n', A)
print('C \n', C)

# solve system of equations [A]{T} = {C} for column vector {T}
for i in range(1, nt+1):
    T = np.linalg.solve(A,C)
    C = T
    C[m-1, 0] = T[m-1, 0] + 2*Fo*Bi*(1 + b/(2*(m-1)))*Tinf
    TT = np.vstack((TT, T.T))

# plot results
py.figure(1)
py.plot(t,TT[:, m-1])
py.plot(t,TT[:, 0])
py.grid()
py.show()

关于生成的Python图（见下图），实线（红色和黑色）线和虚线（红色和黑色）应该在彼此的顶部。在nr = 99运行上面的代码Python实线与实体Matlab红线不匹配，但Python和Matlab图的虚线确实一致。这告诉我在Python代码的最后for循环中也出现了错误。也许我在Python中解决A * x = B的方式不正确？

enter image description here

1 个答案:

答案 0 :(得分：5)

生成A的循环中的索引值已更改为1， 所以这两行

A[i, i-1] = -Fo*(1 - b/(2*i))   # Tm-1
A[i, i+1] = -Fo*(1 + b/(2*i))   # Tm+1

应该是

A[i, i-1] = -Fo*(1 - b/(2*(i+1)))   # Tm-1
A[i, i+1] = -Fo*(1 + b/(2*(i+1)))   # Tm+1

请注意公式中i到i+1的更改（但不会在A的索引中）。

另一方面，m的含义没有改变，所以你不应该改变 已在计算m和\ n边缘的公式中将m-1更改为A C。也就是说，改变这些行：

A[m-1, m-1] = 1 + 2*Fo*(1 + Bi + (b/(2*(m-1)))*Bi)
C[m-1, 0] = Ti + 2*Fo*Bi*(1 + b/(2*(m-1)))*Tinf
...
    C[m-1, 0] = T[m-1, 0] + 2*Fo*Bi*(1 + b/(2*(m-1)))*Tinf

到

A[m-1, m-1] = 1 + 2*Fo*(1 + Bi + (b/(2*m))*Bi)
C[m-1, 0] = Ti + 2*Fo*Bi*(1 + b/(2*m))*Tinf
...
    C[m-1, 0] = T[m-1, 0] + 2*Fo*Bi*(1 + b/(2*m))*Tinf

另外，正如评论中指出的那样眼尖的@eryksun，C = T应该是C = T.copy()。在Matlab中，使用“写入时复制”来管理内存，因此在此分配后对C进行就地更改不会影响T。使用numpy，C = T使C和T引用相同的底层数组对象;就地更改C也会更改T。要重新创建Matlab行为，C必须是T的副本。

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