Program To Solve Fluid Questions in OOP Language Assignment Solution.

PROBLEM 1

clc, clear all, close all

%% Parameters of the tank

R = 3;

%% We want to calculate the depth of the tank such that the volume is 30 m^3. So:

    % pi*h^2 *(3*R-h)/3 = 30

% Then, the function will be:

    % f(x) = pi*x^2 *(3*R-x)/3 - 30, where h = x

f = @(x)(pi*x^2 *(3*R - x)/3 - 30);

%% Pat a) Newton's Method

% The first derivate of this function respect to x is:

df = @(x)(2*x*pi*(3*R - x)/3 - pi*x^2 /3);

% Define initial error and tolerance

error = Inf;

tol = 1e-6;

% Define initial value

x = R;

n = 1;

while error >= tol % execute algorithm while the error is higher than the tolerance

    x = x - f(x)/df(x);

    error = abs(f(x));

% fprintf("Iteration %d, error = %.8f\n", n, error);

    n = n + 1;

end

fprintf("The solution obtained using Newton's method in %d iterations is h = %.4f m\n", n, x);

%% Part b): Secant Method

%% Define the interval [a,b] in which we assume the solution may be

a = 1;

b = 3;

% Define initial error and tolerance

error = Inf;

tol = 1e-6;

max_iters = 500;

n = 1;

while error >= tol % execute algorithm while the error is higher than the tolerance

    x0 = a;

    x1 = b;

    x2 = x1 - (x1-x0)*f(x1) / (f(x1)-f(x0)); % calculate x2

    aold = a;

    bold = b;

    % Update

    a = x1;

    b = x2;

    error = abs(b-a)/abs(a);

 %% Enable the following line if you want to check the error per iteration

% fprintf("Iter = %d, aold = %.4f, bold = %.4f, x2 = %.4f, anew = %.4f, bnew = %.4f, error = %.8f\n", n, aold, bold, x2, a, b, error);

% fprintf("Iteration %d, error: %.8f\n", n, error);

    % Increment step

    n = n + 1;

    % If the number if iterations reaches the max. number of iterations

    % allowed, it means that the problem did not find a solution

    if n >= max_iters

        break;

    end

end

fprintf("The solution obtained using Secant method in %d iterations is h = %.4f m\n", n, a);

PROBLEM 2

clc, clear all, close all

%% Define the equation of the specific heat

% We assume T = x

f = @(x)(1.1 - (0.99403 + 1.671e-4 *x + 9.7215e-8 *x^2 -9.5838e-11 *x^3 + 1.9520e-14 *x^4));

%% Pat a) Newton's Method

% The first derivate of this function respect to x is:

df = @(x)(-(1.671e-4 + 2*9.7215e-8 *x -3*9.5838e-11 *x^2 + 4*1.9520e-14 *x^3));

% Define initial error and tolerance

error = Inf;

tol = 1e-6;

% Define initial value

x = 150; % in kelvins

n = 1;

while error >= tol % execute algorithm while the error is higher than the tolerance

    x = x - f(x)/df(x);

    error = abs(f(x));

% fprintf("Iteration %d, error = %.8f\n", n, error);

    n = n + 1;

end

fprintf("The solution obtained using Newton's method in %d iterations is T = %.4f K\n", n, x);

%% Part b): Secant Method

%% Define the interval [a,b] in which we assume the solution may be

a = 0;

b = 1000;

% Define initial error and tolerance

error = Inf;

tol = 1e-6;

max_iters = 500;

n = 1;

while error >= tol % execute algorithm while the error is higher than the tolerance

    x0 = a;

    x1 = b;

    x2 = x1 - (x1-x0)*f(x1) / (f(x1)-f(x0)); % calculate x2

    aold = a;

    bold = b;

    % Update

    a = x1;

    b = x2;

    error = abs(b-a)/abs(a);

 %% Enable the following line if you want to check the error per iteration

% fprintf("Iter = %d, aold = %.4f, bold = %.4f, x2 = %.4f, anew = %.4f, bnew = %.4f, error = %.8f\n", n, aold, bold, x2, a, b, error);

% fprintf("Iteration %d, error: %.8f\n", n, error);

    % Increment step

    n = n + 1;

    % If the number if iterations reaches the max. number of iterations

    % allowed, it means that the problem did not find a solution

    if n >= max_iters

        break;

    end

end

fprintf("The solution obtained using Secant method in %d iterations is T = %.4f K\n", n, a);