How i get a graph that i attached here with this matlab code?

HTP()

phi = 0.1000

phi = 0.2000

phi = 0.3000

phi = 0.4000

function HTP()
clc
clear all format long % hybrid Carreau
% Define constants
J1 = 0.1;
J2 = 0.1;
J3 = 0.1;
J4 = 0.1;
JS = 0.1;
z = 0.1;
S = 0.1;
GC = 0.1;
Gr = 0.1;
H = 0.1;
a = 0.1;
m = 1;
G = 0.5;
phi = 0.1;

% Define time vector
t = linspace(0, 5, 100); % 100 points between 0 and 5
%t= 1;
% Calculate u1 and u2
u1 = exp(-t) - 1;

% Compute u2 with corrected parentheses and mathematical operations
exp_t = exp(t); % Compute exp(t) once for efficiency
term1 = -33 / J1 * (1 - exp_t - z / 3 * S);
term2 = 2 * GC * J3 + 2 * Gr * J2;
term3 = (2 * J4 * H * a^2) / (1 + m^2);
term4 = 3 * G * exp_t + 6 * exp_t / ((1 - phi)^2.5);
term5 = 2 * GC * JS * exp_t + 2 * Gr * J2 * exp_t;
u2 = (term1 + term2 - term3 - term4 - term5 + term3 * (exp_t - m + m * exp_t)) / (6 * a);

% Compute u
y =  1; % Define y as 1 or another constant value; adjust as needed
u = u1 * y + u2 * y.^2;

% Plot the result
phi = 0.1
figure ;
plot(t, u);
phi= 0.2
figure ;
plot(t, u);
phi = 0.3
figure ;
plot(t, u);
phi = 0.4
figure ;
plot(t, u);
xlabel('Time (t)');
ylabel('u(t)');
title('Plot of u(t)');
axis([0 5 min(u) max(u)]);
grid on;
end

[Walter Roberson]

clc
clear all format long

First off, that should likely be

clear all
format long

Secondly: clear all within a function is a waste of time at best, and can lead to execution problems at worst. There is no need to clear all within a function.

I recommend against using clc within a function as well.

Answer 1

[Star Strider]

One problem is that ‘term4’ and therefore ‘u2’ and ‘u’ should be functions of ‘phi’ and they are not originally. However changing that does not change the plots, leading me to believe that there is a coding error somewhere in ‘u’. You need to find it and fix it. Also, if you want all of the plots in the same axes, use the hold function rather than plotting them in individual figure objects.

HTP

Q = 1x100

1.0e+04 * -0.001368015498012 0.001395472200257 0.004302612268278 0.007360821699399 0.010577902910592 0.013962063648777 0.017521937931270 0.021266608073767 0.025205627862078 0.029349046926730 0.033707436382604 0.038291915799052 0.043114181569278 0.048186536751372 0.053521922457137 0.059133950868772 0.065036939967669 0.071245950063909 0.077776822219672 0.084646218664570 0.091871665306041 0.099471596443228 0.107465401798460 0.115873475986291 0.124717270546330 0.134019348672603 0.143803442779087 0.154094515048271 0.164918821117233 0.176303977063707

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phi = 
   0.100000000000000

phi = 
   0.200000000000000

phi = 
   0.300000000000000

phi = 
   0.400000000000000

function HTP()
% clc
% clear all 
format long % hybrid Carreau
    % Define constants
    J1 = 0.1;
    J2 = 0.1;
    J3 = 0.1;
    J4 = 0.1;
    JS = 0.1;
    z = 0.1;
    S = 0.1;
    GC = 0.1;
    Gr = 0.1;
    H = 0.1;
    a = 0.1;
    m = 1;
    G = 0.5;
    phi = 0.1;
    
    % Define time vector
    t = linspace(0, 5, 100); % 100 points between 0 and 5
    %t= 1;
    % Calculate u1 and u2
    u1 = exp(-t) - 1;
    
    % Compute u2 with corrected parentheses and mathematical operations
    exp_t = exp(t); % Compute exp(t) once for efficiency
    term1 = -33 / J1 * (1 - exp_t - z / 3 * S);
    term2 = 2 * GC * J3 + 2 * Gr * J2;
    term3 = (2 * J4 * H * a^2) / (1 + m^2);
    term4 = @(phi) 3 * G * exp_t + 6 * exp_t / ((1 - phi)^2.5);
    term5 = 2 * GC * JS * exp_t + 2 * Gr * J2 * exp_t;
    u2 = @(phi) (term1 + term2 - term3 - term4(phi) - term5 + term3 * (exp_t - m + m * exp_t)) / (6 * a);
    
    % Compute u
    y =  1; % Define y as 1 or another constant value; adjust as needed
    u = @(phi) u1 * y + u2(phi) * y.^2;
    
    % Plot the result
    phi = 0.1
    figure ;
    plot(t, u(phi));
    phi= 0.2
    figure ;
    plot(t, u(phi));
    phi = 0.3
    figure ;
    plot(t, u(phi));
    phi = 0.4
    figure ;
    plot(t, u(phi));
    xlabel('Time (t)');
    ylabel('u(t)');
    title('Plot of u(t)');
    axis([0 5 min(u(phi)) max(u(phi))]);
    grid on;
end

.

Answer 2

[Cris LaPierre]

Right now, you just plot the same line 4 times: plot(t,u)

Changing the value of phi will not change the value of u. You must recalculate it. Here, I use anonymous functions for that purpose.

Also, use hold on and hold off without figure commands to add multiple lines to the same figure. See Ch 9 of MATLAB Onramp for more details.

Use text and/or annotation to add annotations to your plot.

% Define constants
J1 = 0.1;
J2 = 0.1;
J3 = 0.1;
J4 = 0.1;
JS = 0.1;
z = 0.1;
S = 0.1;
GC = 0.1;
Gr = 0.1;
H = 0.1;
a = 0.1;
m = 1;
G = 0.5;

% Define time vector
t = linspace(0, 5, 100); % 100 points between 0 and 5

% Calculate u1 and u2
u1 = exp(-t) - 1;

% Compute u2 with corrected parentheses and mathematical operations
exp_t = exp(t); % Compute exp(t) once for efficiency
term1 = -33 / J1 * (1 - exp_t - z / 3 * S);
term2 = 2 * GC * J3 + 2 * Gr * J2;
term3 = (2 * J4 * H * a^2) / (1 + m^2);
term4 = @(phi) 3 * G * exp_t + 6 * exp_t / ((1 - phi)^2.5);
term5 = 2 * GC * JS * exp_t + 2 * Gr * J2 * exp_t;
u2 = @(phi) (term1 + term2 - term3 - term4(phi) - term5 + term3 * (exp_t - m + m * exp_t)) / (6 * a);

% Compute u
y =  1; % Define y as 1 or another constant value; adjust as needed
u = @(phi) u1 * y + u2(phi) * y.^2;

% Plot the result
phi = 0.1

phi = 0.1000

figure ;
plot(t, u(phi),'k-');

hold on
phi= 0.2

phi = 0.2000

plot(t, u(phi),'k--');

phi = 0.3

phi = 0.3000

plot(t, u(phi),'k:');

phi = 0.4

phi = 0.4000

plot(t, u(phi),'k-.');
hold off
xlabel('Time (t)');
ylabel('u(t)');
title('Plot of u(t)');
axis([0 5 min(u(phi)) max(u(phi))]);
grid on;
legend("\phi=" + [0.1 0.2 0.3 0.4])

[Sharqa]

Thankyou.

But,how i get this graph?

which parameter should i change ?

[Cris LaPierre]

Are you asking about the annotations or the shapes of the curve?

Have you gone through Ch 9 of MATLAB Onramp?