Foor loop; iteration

I want to use a for-loop to iterate and adjust the value of x until either V_check equals V or the absolute difference between V_check and V becomes zero or close to zero. With this code, I'm getting the message "Solution found after 19 iterations" and the following values:

• x = -2.0817e-16
• V = -6.538079556362143e-17
• V_check = -0.009549534391159

which do not match.

%%                                &&& Cp for BL  &&
clc, clear
xf = 0.08;   T1 = 90;  
Cpw = 4188;    % at T1
Cp_BL = calculateCp(xf,T1, Cpw);

%%                              &&& Steam or T2  &&
%       m_dot_BL*Cp_BL(T1-T2) = m_dot_s * Hvap
% If T2 is known set val = 0 if  m_steam is known set val1 = 1
F = 8; % kg/s
T2 = T1 + 20;
H_vap = 2048.16; % vid 8 bar och 170 C
H_F = Cp_BL*T1;   
S = 0;
val1 = 0;
if val1 == 0
    S = F * Cp_BL * (T2 - T1) / H_vap;
end

if val1 == 1
    T2 = (m_dot_BL * Cp_BL * T1 - S * H_vap) / (F * Cp_BL);
end

%%     We do not know the value of the concentrated product and the evaporated steam from BL
%% We have to guess and iterate, Guess that V = 0.9*S
% Set an initial value for x
x= 0.9;
tolerance = 1e-2; % Tolerance for considering V_check close enough to zero
maxIterations = 100; % Maximum number of iterations if necessary
Cpw2 = 4201;

for iteration = 1:maxIterations
    % Calculate V based on the current value of x
    V = x * S;
    L = F - V;
    xl = xf * F / L;
    Cp_L = calculateCp(xl, T2, Cpw2);
    H_L = Cp_L * T2;
    H_V = 1746; % Superheated steam at the given black liquor pressure at T2

    % Check if the guessed V is correct within a certain tolerance
    V_check = (F * H_F + S * H_vap - L * H_L) / H_V;

    % If V_check is close enough to zero, exit the loop
    % If the difference between V and V_check is close enough to zero, exit the loop
    if abs(V - V_check) < tolerance
        disp(['Solution found after ' num2str(iteration) ' iterations.']);
        break;
    end

    % Adjust the value of x for the next iteration
    x = x - 0.05;
end

Solution found after 19 iterations.

V

V = -6.5381e-17

V_check

V_check = -0.0095

% If maxIterations is reached without the solution being close enough to zero, handle it here
if iteration == maxIterations
    disp('Maximum number of iterations reached without reaching the desired solution.');
end

[Dyuman Joshi]

"... which do not match."

How exactly do they not match?

%Values taken from the question statement
V = -6.538079556362143e-17;
V_check = -0.009549534391159;
abs(V - V_check)

ans = 0.0095

The absolute value of the difference, 0.0095, is less then 1e-2 or 0.01, which satisfies the condition to break the for loop.

You get the value according to the tolerance you have set. Adjust it according to your requirements.

Answer 1

[Aquatris]

You set your tolerance to 1e-2 so your for loop terminates when abs(V-V_Check) becomes less than 1e-2.

% here you basically define how close you want V to V_Check before you
% say I found my answer and end all operations
if abs(V - V_check) < tolerance 
    disp(['Solution found after ' num2str(iteration) ' iterations.']);
    break;
end

If you want to your for loop to terminate when V is a lot closer to V_Check, set your tolerance to a lot lower value, such as

tolerance = 1e-15;