A few days ago someone pointed out that using:
clc; clear; close all;
Is considered as a case of code smell and cargo cult programming.
But does that really have to be the case?
clc: cleans up the command window and now one can work without getting confused with the commands for previous runs
clear: erases the variables from previous runs this will reduce chances of error in subsequent runs and the programmer does not have to worry about unnecessary trash variables.
close all: closes all currently open figures. This can be very helpful during subsequent runs of the same script. If the figure from the previous run has not been closed then the subsequent run will plot the data on the already open figure. Which of course is a total waste.

What is the problem with using these commands?A few days ago someone pointed out that using:
clc; clear; close all;
Is considered as a case of code smell and cargo cult programming.
But does that really have to be the case?
clc: cleans up the command window and now one can work without getting confused with the commands for previous runs
clear: erases the variables from previous runs this will reduce chances of error in subsequent runs and the programmer does not have to worry about unnecessary trash variables.
close all: closes all currently open figures. This can be very helpful during subsequent runs of the same script. If the figure from the previous run has not been closed then the subsequent run will plot the data on the already open figure. Which of course is a total waste.

What is the problem with using these commands? A few days ago someone pointed out that using:
clc; clear; close all;
Is considered as a case of code smell and cargo cult programming.
But does that really have to be the case?
clc: cleans up the command window and now one can work without getting confused with the commands for previous runs
clear: erases the variables from previous runs this will reduce chances of error in subsequent runs and the programmer does not have to worry about unnecessary trash variables.
close all: closes all currently open figures. This can be very helpful during subsequent runs of the same script. If the figure from the previous run has not been closed then the subsequent run will plot the data on the already open figure. Which of course is a total waste.

What is the problem with using these commands? codesmell, cargocultprogramming, clc, clear, close all MATLAB Answers — New Questions

​

Hi,
I want to create a lookup table that based on some inputs, where the indeces of the lookup table are found and the field value is output.
I attach the lookup table I made with its corresponding breakpoints.
I am using two Preelokup blocks to find the indeces of this 2D table. These indices are used then as input to the Direct LookupTable block (see the attached block diagram picture).
My issue is that if I input e.g., 28 in "in1", the index found is correct. However, if I put a value within two breakpoint values (e.g., 30) it uses interpolation, flat or rounding, and I would like to avoid that. Is there a way where if the input value does not match with any of the breakpoints array it simply does not return any index or the Lookup table simply returns no value or a zero?
Thanks in advanceHi,
I want to create a lookup table that based on some inputs, where the indeces of the lookup table are found and the field value is output.
I attach the lookup table I made with its corresponding breakpoints.
I am using two Preelokup blocks to find the indeces of this 2D table. These indices are used then as input to the Direct LookupTable block (see the attached block diagram picture).
My issue is that if I input e.g., 28 in "in1", the index found is correct. However, if I put a value within two breakpoint values (e.g., 30) it uses interpolation, flat or rounding, and I would like to avoid that. Is there a way where if the input value does not match with any of the breakpoints array it simply does not return any index or the Lookup table simply returns no value or a zero?
Thanks in advance Hi,
I want to create a lookup table that based on some inputs, where the indeces of the lookup table are found and the field value is output.
I attach the lookup table I made with its corresponding breakpoints.
I am using two Preelokup blocks to find the indeces of this 2D table. These indices are used then as input to the Direct LookupTable block (see the attached block diagram picture).
My issue is that if I input e.g., 28 in "in1", the index found is correct. However, if I put a value within two breakpoint values (e.g., 30) it uses interpolation, flat or rounding, and I would like to avoid that. Is there a way where if the input value does not match with any of the breakpoints array it simply does not return any index or the Lookup table simply returns no value or a zero?
Thanks in advance simulink, lookuptable MATLAB Answers — New Questions

​

I’ve changed the equations I’d like to use for the resistor, dc voltage source and dc current source to a quadratic formula by going to the "source code" when clicking on the element properties, as this electrical circuit will actually be an analogy for a mechanical circuit.
However, it is not performing as I expect, with a negative current in one branch of the circuit.
So to understand what could be the issue, I clicked on performance adviser and get the failed to get cache error shown in the title of the question.

When I click on Model Adviser I also get:
Error occurred during model compile.

Error during code generation: Dot indexing is not supported for variables of this type.

The circuit is too big to take a screenshot of when zooming out without some parts becoming invisible.
I changed the equations to be V = I^2*R and P = I^3*R.I’ve changed the equations I’d like to use for the resistor, dc voltage source and dc current source to a quadratic formula by going to the "source code" when clicking on the element properties, as this electrical circuit will actually be an analogy for a mechanical circuit.
However, it is not performing as I expect, with a negative current in one branch of the circuit.
So to understand what could be the issue, I clicked on performance adviser and get the failed to get cache error shown in the title of the question.

When I click on Model Adviser I also get:
Error occurred during model compile.

Error during code generation: Dot indexing is not supported for variables of this type.

The circuit is too big to take a screenshot of when zooming out without some parts becoming invisible.
I changed the equations to be V = I^2*R and P = I^3*R. I’ve changed the equations I’d like to use for the resistor, dc voltage source and dc current source to a quadratic formula by going to the "source code" when clicking on the element properties, as this electrical circuit will actually be an analogy for a mechanical circuit.
However, it is not performing as I expect, with a negative current in one branch of the circuit.
So to understand what could be the issue, I clicked on performance adviser and get the failed to get cache error shown in the title of the question.

When I click on Model Adviser I also get:
Error occurred during model compile.

Error during code generation: Dot indexing is not supported for variables of this type.

The circuit is too big to take a screenshot of when zooming out without some parts becoming invisible.
I changed the equations to be V = I^2*R and P = I^3*R. error, dmr, dot indexing, simulink, code generation MATLAB Answers — New Questions

​

Hello,

I have downloaded the embedded coder pack for TI c2000 processor, but while configuring it is asking the toolchain of CCS.
I have CCS of version 10.4.0, but in the drop down option list option showing with version upto 6, and then when I select the CCS v6 and try to validate it, it is showing error message like readme text file not found.

is it required to download lower version of CCS to complete this setup?Hello,

I have downloaded the embedded coder pack for TI c2000 processor, but while configuring it is asking the toolchain of CCS.
I have CCS of version 10.4.0, but in the drop down option list option showing with version upto 6, and then when I select the CCS v6 and try to validate it, it is showing error message like readme text file not found.

is it required to download lower version of CCS to complete this setup? Hello,

I have downloaded the embedded coder pack for TI c2000 processor, but while configuring it is asking the toolchain of CCS.
I have CCS of version 10.4.0, but in the drop down option list option showing with version upto 6, and then when I select the CCS v6 and try to validate it, it is showing error message like readme text file not found.

is it required to download lower version of CCS to complete this setup? c2000, ccs, embedded coder, code composer studio, code generation MATLAB Answers — New Questions

​

I am plotting a graph, say X vs Y. What command is to be used in order to extract values of Y with respect to Different values of X.? Say, X ranges from 0 to 20 and Y ranges from 0 to 10. The plot is of non-linear nature. I want values of Y for different X such as 10,15 and 23. So what command is to be used to find values of Y?I am plotting a graph, say X vs Y. What command is to be used in order to extract values of Y with respect to Different values of X.? Say, X ranges from 0 to 20 and Y ranges from 0 to 10. The plot is of non-linear nature. I want values of Y for different X such as 10,15 and 23. So what command is to be used to find values of Y? I am plotting a graph, say X vs Y. What command is to be used in order to extract values of Y with respect to Different values of X.? Say, X ranges from 0 to 20 and Y ranges from 0 to 10. The plot is of non-linear nature. I want values of Y for different X such as 10,15 and 23. So what command is to be used to find values of Y? plot MATLAB Answers — New Questions

​

Hello , I want to transform this code that solves a pde equation with the ode solver into finite diferences, because I want to take the code as a matlab function block in simulink so it stands no ode solver(since it is an iterator take much time every time step so never ends simulation ) thats why i want to take it into finite differences .The equations are the following

The inital code is the following with ode solver:
L = 20 ; % Longitud del lecho (m)
eps = 0.4; % Porosidad
u = 0.2; % Velocidad superficial del fluido (m/s)
k_f = 0.02; % Constante de transferencia de masa (1/s)
c0 = 0;
Kf = 4; % Constante de Freundlich
rhop = 1520;
n = 2; % Exponente de Freundlich
% Concentración inicial del fluido (kg/m³)
q0 = 4.320; % Concentración inicial en el sólido (kg/m³)
% Densidad del adsorbente (kg/m³)
tf = 10; % Tiempo final de simulación (horas)
Nt = 100;

t = linspace(0, tf*3600, Nt);
Nz = 100;
z = linspace(0, L,Nz);
dz = z(2) – z(1);

% Initial conditions
ICA = max(ones(1, Nz) * c0, 1e-12); % Evitar valores negativos o cero
ICB = ones(1, Nz) * q0;
IC = [ICA ICB];
options = odeset(‘RelTol’, 1e-6, ‘AbsTol’, 1e-8, ‘InitialStep’, 1e-4, ‘MaxStep’, 100);

[t, y] = ode15s(@fun_pde, t, IC, options, Nz, eps, n, Kf, k_f, u, rhop, dz);
% Define value
cc = y(:, 1:Nz);
qq = y(:, Nz+1:end);
% Recalculate new limit conditions
cc(:, 1) = 0;
cc(:, end) = cc(:, end-1);
% Plotting
cp = cc(:, end) ./ c0;
qp = qq(:, 🙂 ./ q0;
%q_promedio = mean(qq, 2); % Promedio de q en el lecho para cada instante de tiempo
%conversion = 1 – (q_promedio / q0); % Conversión normalizada
figure;
subplot(2, 1, 1);
time = t / 3600; % Convertir a horas
plot(time, 1- qp, ‘b’, ‘LineWidth’, 1.5);
xlabel(‘Tiempo (horas)’);
ylabel(‘Conversion’);
title(‘Curva de conversión durante la desorción’);
grid on;

subplot(2, 1, 2);
plot(t / 3600, (cc(:,:)), ‘LineWidth’, 1.5);
xlabel(‘Tiempo (horas)’);
ylabel(‘Soluciòn kg/m3’);
title(‘Curva de carga de la solucion durante la desorciòn’);
grid on;

% PDE function
function dydt = fun_pde(~, y, Nz, eps, n, Kf, k_f, u, rhop, dz)
dcdt = zeros(Nz, 1);
dqdt = zeros(Nz, 1);
c = y(1:Nz);
q = y(Nz+1:2*Nz);
% Boundary conditions
c(1) = max(c(1), 0); % Asegurar que c(1) sea no negativo
c(end) = c(end-1); % Asegurar que c(1) sea no negativo
% Interior nodes
qstar = zeros(Nz, 1);
dcdz = zeros(Nz, 1);
for i = 2:Nz-1
qstar(i) = Kf .* max(c(i), 1e-12).^(1/n); % Evitar problemas numéricos
dqdt(i) = k_f .* (qstar(i) – q(i));
% if i < Nz
dcdz(i) = (c(i+1) – c(i-1)) / (2 * dz);
%else
% dcdz(i) = (c(i) – c(i-1)) / dz;
%end
dcdt(i) = -u * dcdz(i) – rhop * ((1 – eps) / eps) .* dqdt(i);
end
dydt = [dcdt; dqdt];
end

next is a try to solve with finite diferences but get someting different:

L = 20 ; % Longitud del lecho (m)
eps = 0.4; % Porosidad
u = 0.2; % Velocidad superficial del fluido (m/s)
k_f = 0.02; % Constante de transferencia de masa (1/s)
c0 = 0;
Kf = 4; % Constante de Freundlich
rhop = 1520;
n = 2; % Exponente de Freundlich
% Concentración inicial del fluido (kg/m³)
q0 = 4.320; % Concentración inicial en el sólido (kg/m³)
% Densidad del adsorbente (kg/m³)
tf = 10; % Tiempo final de simulación (horas)
Nz = 100; % Número de nodos espaciales

% Discretización espacial y temporal
z = linspace(0, L, Nz);
t = linspace(0, tf*3600, Nt);
dz = z(2) – z(1);
dt = t(2) – t(1); % Paso temporal

% Condiciones iniciales
c = ones(Nt, Nz) * c0; % Concentración en el fluido
q = ones(Nt, Nz) * q0; % Concentración en el sólido

% Iteración en el tiempo (Diferencias Finitas Explícitas)
for ti = 1:Nt-1
for zi = 2:Nz-1
% Isoterma de Freundlich
qstar = Kf * max(c(ti, zi), 1e-12)^(1/n);

% Transferencia de masa (Desorción)
dqdt = k_f * (qstar – q(ti, zi));

% Gradiente espacial de concentración (Diferencias centradas)
dcdz = (c(ti, zi+1) – c(ti, zi-1)) / (2 * dz);

% Ecuación de balance de masa en el fluido
dcdt = -u * dcdz – rhop * ((1 – eps) / eps) * dqdt;

% Actualizar valores asegurando que sean positivos
c(ti+1, zi) = max(c(ti, zi) + dcdt * dt, 0);
q(ti+1, zi) = max(q(ti, zi) + dqdt * dt, 0);
end
end

% Condiciones de frontera
c(:, 1) = c0; % Entrada con concentración baja
c(:, Nz) = c(:, Nz-1); % Gradiente nulo en la salida
% Cálculo de la conversión normalizada
qp = q(:, 🙂 ./ q0;

% Graficar resultados
figure;
subplot(2, 1, 1);
plot(t / 3600, 1-qp, ‘b’, ‘LineWidth’, 1.5);
xlabel(‘Tiempo (horas)’);
ylabel(‘Conversion’);
title(‘Curva de conversión durante la desorción’);
grid on;

subplot(2, 1, 2);
c_salida = c(:, :); % Concentración en la salida del lecho
plot(t / 3600, c_salida, ‘r’, ‘LineWidth’, 1.5);
xlabel(‘Tiempo (horas)’);
ylabel(‘Soluciòn kg/m3’);
title(‘Curva de carga de la solucion durante la desorciòn’);
grid on;

I dont know why is wrong
Thanks in advanceHello , I want to transform this code that solves a pde equation with the ode solver into finite diferences, because I want to take the code as a matlab function block in simulink so it stands no ode solver(since it is an iterator take much time every time step so never ends simulation ) thats why i want to take it into finite differences .The equations are the following

The inital code is the following with ode solver:
L = 20 ; % Longitud del lecho (m)
eps = 0.4; % Porosidad
u = 0.2; % Velocidad superficial del fluido (m/s)
k_f = 0.02; % Constante de transferencia de masa (1/s)
c0 = 0;
Kf = 4; % Constante de Freundlich
rhop = 1520;
n = 2; % Exponente de Freundlich
% Concentración inicial del fluido (kg/m³)
q0 = 4.320; % Concentración inicial en el sólido (kg/m³)
% Densidad del adsorbente (kg/m³)
tf = 10; % Tiempo final de simulación (horas)
Nt = 100;

t = linspace(0, tf*3600, Nt);
Nz = 100;
z = linspace(0, L,Nz);
dz = z(2) – z(1);

% Initial conditions
ICA = max(ones(1, Nz) * c0, 1e-12); % Evitar valores negativos o cero
ICB = ones(1, Nz) * q0;
IC = [ICA ICB];
options = odeset(‘RelTol’, 1e-6, ‘AbsTol’, 1e-8, ‘InitialStep’, 1e-4, ‘MaxStep’, 100);

[t, y] = ode15s(@fun_pde, t, IC, options, Nz, eps, n, Kf, k_f, u, rhop, dz);
% Define value
cc = y(:, 1:Nz);
qq = y(:, Nz+1:end);
% Recalculate new limit conditions
cc(:, 1) = 0;
cc(:, end) = cc(:, end-1);
% Plotting
cp = cc(:, end) ./ c0;
qp = qq(:, 🙂 ./ q0;
%q_promedio = mean(qq, 2); % Promedio de q en el lecho para cada instante de tiempo
%conversion = 1 – (q_promedio / q0); % Conversión normalizada
figure;
subplot(2, 1, 1);
time = t / 3600; % Convertir a horas
plot(time, 1- qp, ‘b’, ‘LineWidth’, 1.5);
xlabel(‘Tiempo (horas)’);
ylabel(‘Conversion’);
title(‘Curva de conversión durante la desorción’);
grid on;

subplot(2, 1, 2);
plot(t / 3600, (cc(:,:)), ‘LineWidth’, 1.5);
xlabel(‘Tiempo (horas)’);
ylabel(‘Soluciòn kg/m3’);
title(‘Curva de carga de la solucion durante la desorciòn’);
grid on;

% PDE function
function dydt = fun_pde(~, y, Nz, eps, n, Kf, k_f, u, rhop, dz)
dcdt = zeros(Nz, 1);
dqdt = zeros(Nz, 1);
c = y(1:Nz);
q = y(Nz+1:2*Nz);
% Boundary conditions
c(1) = max(c(1), 0); % Asegurar que c(1) sea no negativo
c(end) = c(end-1); % Asegurar que c(1) sea no negativo
% Interior nodes
qstar = zeros(Nz, 1);
dcdz = zeros(Nz, 1);
for i = 2:Nz-1
qstar(i) = Kf .* max(c(i), 1e-12).^(1/n); % Evitar problemas numéricos
dqdt(i) = k_f .* (qstar(i) – q(i));
% if i < Nz
dcdz(i) = (c(i+1) – c(i-1)) / (2 * dz);
%else
% dcdz(i) = (c(i) – c(i-1)) / dz;
%end
dcdt(i) = -u * dcdz(i) – rhop * ((1 – eps) / eps) .* dqdt(i);
end
dydt = [dcdt; dqdt];
end

next is a try to solve with finite diferences but get someting different:

L = 20 ; % Longitud del lecho (m)
eps = 0.4; % Porosidad
u = 0.2; % Velocidad superficial del fluido (m/s)
k_f = 0.02; % Constante de transferencia de masa (1/s)
c0 = 0;
Kf = 4; % Constante de Freundlich
rhop = 1520;
n = 2; % Exponente de Freundlich
% Concentración inicial del fluido (kg/m³)
q0 = 4.320; % Concentración inicial en el sólido (kg/m³)
% Densidad del adsorbente (kg/m³)
tf = 10; % Tiempo final de simulación (horas)
Nz = 100; % Número de nodos espaciales

% Discretización espacial y temporal
z = linspace(0, L, Nz);
t = linspace(0, tf*3600, Nt);
dz = z(2) – z(1);
dt = t(2) – t(1); % Paso temporal

% Condiciones iniciales
c = ones(Nt, Nz) * c0; % Concentración en el fluido
q = ones(Nt, Nz) * q0; % Concentración en el sólido

% Iteración en el tiempo (Diferencias Finitas Explícitas)
for ti = 1:Nt-1
for zi = 2:Nz-1
% Isoterma de Freundlich
qstar = Kf * max(c(ti, zi), 1e-12)^(1/n);

% Transferencia de masa (Desorción)
dqdt = k_f * (qstar – q(ti, zi));

% Gradiente espacial de concentración (Diferencias centradas)
dcdz = (c(ti, zi+1) – c(ti, zi-1)) / (2 * dz);

% Ecuación de balance de masa en el fluido
dcdt = -u * dcdz – rhop * ((1 – eps) / eps) * dqdt;

% Actualizar valores asegurando que sean positivos
c(ti+1, zi) = max(c(ti, zi) + dcdt * dt, 0);
q(ti+1, zi) = max(q(ti, zi) + dqdt * dt, 0);
end
end

% Condiciones de frontera
c(:, 1) = c0; % Entrada con concentración baja
c(:, Nz) = c(:, Nz-1); % Gradiente nulo en la salida
% Cálculo de la conversión normalizada
qp = q(:, 🙂 ./ q0;

% Graficar resultados
figure;
subplot(2, 1, 1);
plot(t / 3600, 1-qp, ‘b’, ‘LineWidth’, 1.5);
xlabel(‘Tiempo (horas)’);
ylabel(‘Conversion’);
title(‘Curva de conversión durante la desorción’);
grid on;

subplot(2, 1, 2);
c_salida = c(:, :); % Concentración en la salida del lecho
plot(t / 3600, c_salida, ‘r’, ‘LineWidth’, 1.5);
xlabel(‘Tiempo (horas)’);
ylabel(‘Soluciòn kg/m3’);
title(‘Curva de carga de la solucion durante la desorciòn’);
grid on;

I dont know why is wrong
Thanks in advance Hello , I want to transform this code that solves a pde equation with the ode solver into finite diferences, because I want to take the code as a matlab function block in simulink so it stands no ode solver(since it is an iterator take much time every time step so never ends simulation ) thats why i want to take it into finite differences .The equations are the following

The inital code is the following with ode solver:
L = 20 ; % Longitud del lecho (m)
eps = 0.4; % Porosidad
u = 0.2; % Velocidad superficial del fluido (m/s)
k_f = 0.02; % Constante de transferencia de masa (1/s)
c0 = 0;
Kf = 4; % Constante de Freundlich
rhop = 1520;
n = 2; % Exponente de Freundlich
% Concentración inicial del fluido (kg/m³)
q0 = 4.320; % Concentración inicial en el sólido (kg/m³)
% Densidad del adsorbente (kg/m³)
tf = 10; % Tiempo final de simulación (horas)
Nt = 100;

t = linspace(0, tf*3600, Nt);
Nz = 100;
z = linspace(0, L,Nz);
dz = z(2) – z(1);

% Initial conditions
ICA = max(ones(1, Nz) * c0, 1e-12); % Evitar valores negativos o cero
ICB = ones(1, Nz) * q0;
IC = [ICA ICB];
options = odeset(‘RelTol’, 1e-6, ‘AbsTol’, 1e-8, ‘InitialStep’, 1e-4, ‘MaxStep’, 100);

[t, y] = ode15s(@fun_pde, t, IC, options, Nz, eps, n, Kf, k_f, u, rhop, dz);
% Define value
cc = y(:, 1:Nz);
qq = y(:, Nz+1:end);
% Recalculate new limit conditions
cc(:, 1) = 0;
cc(:, end) = cc(:, end-1);
% Plotting
cp = cc(:, end) ./ c0;
qp = qq(:, 🙂 ./ q0;
%q_promedio = mean(qq, 2); % Promedio de q en el lecho para cada instante de tiempo
%conversion = 1 – (q_promedio / q0); % Conversión normalizada
figure;
subplot(2, 1, 1);
time = t / 3600; % Convertir a horas
plot(time, 1- qp, ‘b’, ‘LineWidth’, 1.5);
xlabel(‘Tiempo (horas)’);
ylabel(‘Conversion’);
title(‘Curva de conversión durante la desorción’);
grid on;

subplot(2, 1, 2);
plot(t / 3600, (cc(:,:)), ‘LineWidth’, 1.5);
xlabel(‘Tiempo (horas)’);
ylabel(‘Soluciòn kg/m3’);
title(‘Curva de carga de la solucion durante la desorciòn’);
grid on;

% PDE function
function dydt = fun_pde(~, y, Nz, eps, n, Kf, k_f, u, rhop, dz)
dcdt = zeros(Nz, 1);
dqdt = zeros(Nz, 1);
c = y(1:Nz);
q = y(Nz+1:2*Nz);
% Boundary conditions
c(1) = max(c(1), 0); % Asegurar que c(1) sea no negativo
c(end) = c(end-1); % Asegurar que c(1) sea no negativo
% Interior nodes
qstar = zeros(Nz, 1);
dcdz = zeros(Nz, 1);
for i = 2:Nz-1
qstar(i) = Kf .* max(c(i), 1e-12).^(1/n); % Evitar problemas numéricos
dqdt(i) = k_f .* (qstar(i) – q(i));
% if i < Nz
dcdz(i) = (c(i+1) – c(i-1)) / (2 * dz);
%else
% dcdz(i) = (c(i) – c(i-1)) / dz;
%end
dcdt(i) = -u * dcdz(i) – rhop * ((1 – eps) / eps) .* dqdt(i);
end
dydt = [dcdt; dqdt];
end

next is a try to solve with finite diferences but get someting different:

L = 20 ; % Longitud del lecho (m)
eps = 0.4; % Porosidad
u = 0.2; % Velocidad superficial del fluido (m/s)
k_f = 0.02; % Constante de transferencia de masa (1/s)
c0 = 0;
Kf = 4; % Constante de Freundlich
rhop = 1520;
n = 2; % Exponente de Freundlich
% Concentración inicial del fluido (kg/m³)
q0 = 4.320; % Concentración inicial en el sólido (kg/m³)
% Densidad del adsorbente (kg/m³)
tf = 10; % Tiempo final de simulación (horas)
Nz = 100; % Número de nodos espaciales

% Discretización espacial y temporal
z = linspace(0, L, Nz);
t = linspace(0, tf*3600, Nt);
dz = z(2) – z(1);
dt = t(2) – t(1); % Paso temporal

% Condiciones iniciales
c = ones(Nt, Nz) * c0; % Concentración en el fluido
q = ones(Nt, Nz) * q0; % Concentración en el sólido

% Iteración en el tiempo (Diferencias Finitas Explícitas)
for ti = 1:Nt-1
for zi = 2:Nz-1
% Isoterma de Freundlich
qstar = Kf * max(c(ti, zi), 1e-12)^(1/n);

% Transferencia de masa (Desorción)
dqdt = k_f * (qstar – q(ti, zi));

% Gradiente espacial de concentración (Diferencias centradas)
dcdz = (c(ti, zi+1) – c(ti, zi-1)) / (2 * dz);

% Ecuación de balance de masa en el fluido
dcdt = -u * dcdz – rhop * ((1 – eps) / eps) * dqdt;

% Actualizar valores asegurando que sean positivos
c(ti+1, zi) = max(c(ti, zi) + dcdt * dt, 0);
q(ti+1, zi) = max(q(ti, zi) + dqdt * dt, 0);
end
end

% Condiciones de frontera
c(:, 1) = c0; % Entrada con concentración baja
c(:, Nz) = c(:, Nz-1); % Gradiente nulo en la salida
% Cálculo de la conversión normalizada
qp = q(:, 🙂 ./ q0;

% Graficar resultados
figure;
subplot(2, 1, 1);
plot(t / 3600, 1-qp, ‘b’, ‘LineWidth’, 1.5);
xlabel(‘Tiempo (horas)’);
ylabel(‘Conversion’);
title(‘Curva de conversión durante la desorción’);
grid on;

subplot(2, 1, 2);
c_salida = c(:, :); % Concentración en la salida del lecho
plot(t / 3600, c_salida, ‘r’, ‘LineWidth’, 1.5);
xlabel(‘Tiempo (horas)’);
ylabel(‘Soluciòn kg/m3’);
title(‘Curva de carga de la solucion durante la desorciòn’);
grid on;

I dont know why is wrong
Thanks in advance pde, ode, matlab, solver, reactor MATLAB Answers — New Questions

​

Hey,

I try to switch the python engine version in one single class. I added here a small example. Based on the used matlab version I want do add matlab.engine version dynamically. I tested many ways – importlib.reload(<matlabVersion>) is not working. What is the best way
import sys
import re
import importlib
matlab2017a ="d:CodeAreaModelQualityInterfacecoreToolsMatlabSimulinkmatlab17aPy35Libsite-packages"
matlab2016b ="d:CodeAreaModelQualityInterfacecoreToolsMatlabSimulinkmatlab16bPy35Libsite-packages"
sys.path.append(matlab2016b)
import matlab.engine
print(matlab.__file__)

print(matlab.engine.find_matlab())

def __RemoveMatlabEngineFromSysPath__(deliveredLibs):
for p in sys.path:
if p.find(deliveredLibs) != -1:
sys.path.remove(p)

__RemoveMatlabEngineFromSysPath__(matlab2016b)
sys.path.append(matlab2017a)

matlab = importlib.reload(matlab.engine)

print(matlab.engine.__file__)
print(matlab.engine.find_matlab())

Output:
File "d:/CodeArea/ModelQualityInterface/core/Tools/MatlabSimulink/test_multipleMatlabVersions.py", line 31, in <module>MatlabSimulink/test_multipleMatlabVersions.py", line
31, in <module> t__.py", line 166, in reload
matlab = importlib.reload(matlab.engine)
File "c:Program FilesPython35libimportlib__ini_exect__.py", line 166, in reload 697, in exec_module
_bootstrap._exec(spec, module) _call_with_frames_removed
File "<frozen importlib._bootstrap>", line 626, in MatlabSimulinkmatlab16bPy35Libsite-packagesmatlabengine__init__.py", line 80, in <module>
_exec
File "<frozen importlib._bootstrap_external>", lineMatlabSimulinkmatlab16bPy35Libsite-packagesmatlabengineenginesession.py", line 8, in __init__ 697, in exec_module
File "<frozen importlib._bootstrap>", line 222, in not be initialized.
_call_with_frames_removed el__ of <matlab.engine.enginesession.EngineSession object at 0x00000212C1A2E7B8>>
File "d:CodeAreaModelQualityInterfacecoreToolsMatlabSimulinkmatlab16bPy35Libsite-packagesmatlabMatlabSimulinkmatlab16bPy35Libsite-packagesmatlabengineenginesession.py", line 14, in __del__engine__init__.py", line 80, in <module>
_session = EngineSession() MatlabSimulinkmatlab16bPy35Libsite-packagesmatlabengineenginesession.py", line 17, in release
File "d:CodeAreaModelQualityInterfacecoreToolsMatlabSimulinkmatlab16bPy35Libsite-packagesmatlabengineenginesession.py", line 8, in __init__
pythonengine.createProcess()
matlab.engine.EngineError: MATLAB process session cannot be initialized.
Exception ignored in: <bound method EngineSession.__del__ of <matlab.engine.enginesession.EngineSession object at 0x00000212C1A2E7B8>>
Traceback (most recent call last):
File "d:CodeAreaModelQualityInterfacecoreToolsMatlabSimulinkmatlab16bPy35Libsite-packagesmatlabengineenginesession.py", line 14, in __del__
self.release()
File "d:CodeAreaModelQualityInterfacecoreToolsMatlabSimulinkmatlab16bPy35Libsite-packagesmatlabengineenginesession.py", line 17, in release
if self._process_created:
AttributeError: ‘EngineSession’ object has no attribute ‘_process_created’Hey,

I try to switch the python engine version in one single class. I added here a small example. Based on the used matlab version I want do add matlab.engine version dynamically. I tested many ways – importlib.reload(<matlabVersion>) is not working. What is the best way
import sys
import re
import importlib
matlab2017a ="d:CodeAreaModelQualityInterfacecoreToolsMatlabSimulinkmatlab17aPy35Libsite-packages"
matlab2016b ="d:CodeAreaModelQualityInterfacecoreToolsMatlabSimulinkmatlab16bPy35Libsite-packages"
sys.path.append(matlab2016b)
import matlab.engine
print(matlab.__file__)

print(matlab.engine.find_matlab())

def __RemoveMatlabEngineFromSysPath__(deliveredLibs):
for p in sys.path:
if p.find(deliveredLibs) != -1:
sys.path.remove(p)

__RemoveMatlabEngineFromSysPath__(matlab2016b)
sys.path.append(matlab2017a)

matlab = importlib.reload(matlab.engine)

print(matlab.engine.__file__)
print(matlab.engine.find_matlab())

Output:
File "d:/CodeArea/ModelQualityInterface/core/Tools/MatlabSimulink/test_multipleMatlabVersions.py", line 31, in <module>MatlabSimulink/test_multipleMatlabVersions.py", line
31, in <module> t__.py", line 166, in reload
matlab = importlib.reload(matlab.engine)
File "c:Program FilesPython35libimportlib__ini_exect__.py", line 166, in reload 697, in exec_module
_bootstrap._exec(spec, module) _call_with_frames_removed
File "<frozen importlib._bootstrap>", line 626, in MatlabSimulinkmatlab16bPy35Libsite-packagesmatlabengine__init__.py", line 80, in <module>
_exec
File "<frozen importlib._bootstrap_external>", lineMatlabSimulinkmatlab16bPy35Libsite-packagesmatlabengineenginesession.py", line 8, in __init__ 697, in exec_module
File "<frozen importlib._bootstrap>", line 222, in not be initialized.
_call_with_frames_removed el__ of <matlab.engine.enginesession.EngineSession object at 0x00000212C1A2E7B8>>
File "d:CodeAreaModelQualityInterfacecoreToolsMatlabSimulinkmatlab16bPy35Libsite-packagesmatlabMatlabSimulinkmatlab16bPy35Libsite-packagesmatlabengineenginesession.py", line 14, in __del__engine__init__.py", line 80, in <module>
_session = EngineSession() MatlabSimulinkmatlab16bPy35Libsite-packagesmatlabengineenginesession.py", line 17, in release
File "d:CodeAreaModelQualityInterfacecoreToolsMatlabSimulinkmatlab16bPy35Libsite-packagesmatlabengineenginesession.py", line 8, in __init__
pythonengine.createProcess()
matlab.engine.EngineError: MATLAB process session cannot be initialized.
Exception ignored in: <bound method EngineSession.__del__ of <matlab.engine.enginesession.EngineSession object at 0x00000212C1A2E7B8>>
Traceback (most recent call last):
File "d:CodeAreaModelQualityInterfacecoreToolsMatlabSimulinkmatlab16bPy35Libsite-packagesmatlabengineenginesession.py", line 14, in __del__
self.release()
File "d:CodeAreaModelQualityInterfacecoreToolsMatlabSimulinkmatlab16bPy35Libsite-packagesmatlabengineenginesession.py", line 17, in release
if self._process_created:
AttributeError: ‘EngineSession’ object has no attribute ‘_process_created’ Hey,

I try to switch the python engine version in one single class. I added here a small example. Based on the used matlab version I want do add matlab.engine version dynamically. I tested many ways – importlib.reload(<matlabVersion>) is not working. What is the best way
import sys
import re
import importlib
matlab2017a ="d:CodeAreaModelQualityInterfacecoreToolsMatlabSimulinkmatlab17aPy35Libsite-packages"
matlab2016b ="d:CodeAreaModelQualityInterfacecoreToolsMatlabSimulinkmatlab16bPy35Libsite-packages"
sys.path.append(matlab2016b)
import matlab.engine
print(matlab.__file__)

print(matlab.engine.find_matlab())

def __RemoveMatlabEngineFromSysPath__(deliveredLibs):
for p in sys.path:
if p.find(deliveredLibs) != -1:
sys.path.remove(p)

__RemoveMatlabEngineFromSysPath__(matlab2016b)
sys.path.append(matlab2017a)

matlab = importlib.reload(matlab.engine)

print(matlab.engine.__file__)
print(matlab.engine.find_matlab())

Output:
File "d:/CodeArea/ModelQualityInterface/core/Tools/MatlabSimulink/test_multipleMatlabVersions.py", line 31, in <module>MatlabSimulink/test_multipleMatlabVersions.py", line
31, in <module> t__.py", line 166, in reload
matlab = importlib.reload(matlab.engine)
File "c:Program FilesPython35libimportlib__ini_exect__.py", line 166, in reload 697, in exec_module
_bootstrap._exec(spec, module) _call_with_frames_removed
File "<frozen importlib._bootstrap>", line 626, in MatlabSimulinkmatlab16bPy35Libsite-packagesmatlabengine__init__.py", line 80, in <module>
_exec
File "<frozen importlib._bootstrap_external>", lineMatlabSimulinkmatlab16bPy35Libsite-packagesmatlabengineenginesession.py", line 8, in __init__ 697, in exec_module
File "<frozen importlib._bootstrap>", line 222, in not be initialized.
_call_with_frames_removed el__ of <matlab.engine.enginesession.EngineSession object at 0x00000212C1A2E7B8>>
File "d:CodeAreaModelQualityInterfacecoreToolsMatlabSimulinkmatlab16bPy35Libsite-packagesmatlabMatlabSimulinkmatlab16bPy35Libsite-packagesmatlabengineenginesession.py", line 14, in __del__engine__init__.py", line 80, in <module>
_session = EngineSession() MatlabSimulinkmatlab16bPy35Libsite-packagesmatlabengineenginesession.py", line 17, in release
File "d:CodeAreaModelQualityInterfacecoreToolsMatlabSimulinkmatlab16bPy35Libsite-packagesmatlabengineenginesession.py", line 8, in __init__
pythonengine.createProcess()
matlab.engine.EngineError: MATLAB process session cannot be initialized.
Exception ignored in: <bound method EngineSession.__del__ of <matlab.engine.enginesession.EngineSession object at 0x00000212C1A2E7B8>>
Traceback (most recent call last):
File "d:CodeAreaModelQualityInterfacecoreToolsMatlabSimulinkmatlab16bPy35Libsite-packagesmatlabengineenginesession.py", line 14, in __del__
self.release()
File "d:CodeAreaModelQualityInterfacecoreToolsMatlabSimulinkmatlab16bPy35Libsite-packagesmatlabengineenginesession.py", line 17, in release
if self._process_created:
AttributeError: ‘EngineSession’ object has no attribute ‘_process_created’ python, import MATLAB Answers — New Questions

​

I need to deliver a MATLAB Toolbox File (extension .mltbx) for a customer. All the files have read / write / execute permissions (including the GettingStarted.mlx file) after the customer installs the mltbx file.
I want the files to be read only in order to protect the install from mistakes that the end user may make. It appears that even when I set the permissions to be read-only just prior to packaging up the .mltbx, the end result is that the files have read / write permissions.
How can I fix this?
Thanks.I need to deliver a MATLAB Toolbox File (extension .mltbx) for a customer. All the files have read / write / execute permissions (including the GettingStarted.mlx file) after the customer installs the mltbx file.
I want the files to be read only in order to protect the install from mistakes that the end user may make. It appears that even when I set the permissions to be read-only just prior to packaging up the .mltbx, the end result is that the files have read / write permissions.
How can I fix this?
Thanks. I need to deliver a MATLAB Toolbox File (extension .mltbx) for a customer. All the files have read / write / execute permissions (including the GettingStarted.mlx file) after the customer installs the mltbx file.
I want the files to be read only in order to protect the install from mistakes that the end user may make. It appears that even when I set the permissions to be read-only just prior to packaging up the .mltbx, the end result is that the files have read / write permissions.
How can I fix this?
Thanks. matlab, matlab toolbox file, .mltbx, file permissions MATLAB Answers — New Questions

​

x=-5*pi:0.1:5*pi;
y=(pi/2)*sinc(x/2);
plot(x,y)
title(‘s𝑖𝑛((𝜋/2)𝑥)/𝑥 plot’)
xlabel(‘x’);
ylabel(‘sin((𝜋/2)x)/x’)
grid on;

would this be correct when plotting Rect function using rectangularPulse on the interval x= -5:0.1:5?x=-5*pi:0.1:5*pi;
y=(pi/2)*sinc(x/2);
plot(x,y)
title(‘s𝑖𝑛((𝜋/2)𝑥)/𝑥 plot’)
xlabel(‘x’);
ylabel(‘sin((𝜋/2)x)/x’)
grid on;

would this be correct when plotting Rect function using rectangularPulse on the interval x= -5:0.1:5? x=-5*pi:0.1:5*pi;
y=(pi/2)*sinc(x/2);
plot(x,y)
title(‘s𝑖𝑛((𝜋/2)𝑥)/𝑥 plot’)
xlabel(‘x’);
ylabel(‘sin((𝜋/2)x)/x’)
grid on;

would this be correct when plotting Rect function using rectangularPulse on the interval x= -5:0.1:5? rectangularpulse MATLAB Answers — New Questions

​

Is it possible to plot a 3d graph by using a cylinders (or a tubes) instead of the default straight lines in Matlab?Is it possible to plot a 3d graph by using a cylinders (or a tubes) instead of the default straight lines in Matlab? Is it possible to plot a 3d graph by using a cylinders (or a tubes) instead of the default straight lines in Matlab? graph MATLAB Answers — New Questions

​