MATLAB Ode45 Code Being Strange

Hello!

My professors and I are trying to figure out a bug in our code for cooking a 5 foot in diameter donut. We are supposed to find how long it takes to get the oil for the donut to heat up to the correct temperature. Below is the code that we have right now.

Function file donut

function dDdt = donut(t,D)

    % This script makes the ode45 function for cooking the five foot donut

Q = 15000; %  50000 btu/hr = 15000 W 
N = 4;  % Number of burners
m_o = 2553;      % Mass of oil in kilograms
c_o = 1.9;   % Specific heat in kilojoules per kilograms
T_init = 295;
T_a = 295;   % Initial temp outside in Kelvin
h_s = 25; % Heat transfer from surface in work per meters squared kilometers
A_s = 4.55;   % Area of the top of the surface in oil in meters squared
L_d = 0.0254; % Thickness of dumpster wall in meters
k_d = 40;    % Thermal conductivity of steel in watts per meters Kelvin
A_d = 5.20;   % External surface area of the dumpster in meters squared
h_d = 25;    % Heat transfer from donut to surface in work per meters squared kilometers

    dDdt = (N*Q/(m_o*c_o))-(T_init-T_a)*(h_s*A_s+1/(L_d/(k_d*A_d)+1/(h_d*A_d)));
end

Code Script to solve the ode45

T_init = 295;    % Initial temperature of oil
tspan = [0 86400];  % Time span for oil to heat


[t, D] = ode45(@donut, tspan, T_init);


plot(t,D)
set(gca, 'FontSize', 16);
xlabel('Time (in Seconds)', 'FontSize', 18);
ylabel('Temperature of Oil in C', 'FontSize', 18);
title('Donut Cooking', 'FontSize', 20);

The time span is 24 hours in seconds.

This is the graph we are getting

This is the graph that we are supposed to get, based off of a tutorial that shows what the results should be.

Thank you for the help!

[Walter Roberson]

syms t D
result = donut(t, D)

result = 12.3693

This tells you that your function donut is ignoring both inputs and always returning 12.3693 as the derivative. The integral of a constant is a straight line.

function dDdt = donut(t,D)

    % This script makes the ode45 function for cooking the five foot donut

Q = 15000; %  50000 btu/hr = 15000 W 
N = 4;  % Number of burners
m_o = 2553;      % Mass of oil in kilograms
c_o = 1.9;   % Specific heat in kilojoules per kilograms
T_init = 295;
T_a = 295;   % Initial temp outside in Kelvin
h_s = 25; % Heat transfer from surface in work per meters squared kilometers
A_s = 4.55;   % Area of the top of the surface in oil in meters squared
L_d = 0.0254; % Thickness of dumpster wall in meters
k_d = 40;    % Thermal conductivity of steel in watts per meters Kelvin
A_d = 5.20;   % External surface area of the dumpster in meters squared
h_d = 25;    % Heat transfer from donut to surface in work per meters squared kilometers

    dDdt = (N*Q/(m_o*c_o))-(T_init-T_a)*(h_s*A_s+1/(L_d/(k_d*A_d)+1/(h_d*A_d)));
end

[Anneliese Fensch]

What can I do to fix this?

[Walter Roberson]

    dDdt = (N*Q/(m_o*c_o))-(T_init-T_a)*(h_s*A_s+1/(L_d/(k_d*A_d)+1/(h_d*A_d)));

your T_init and T_a are the same. Subtract gives 0. Multiply by anything gives 0 provided it is finite.

So your formula is just

    dDdt = (N*Q/(m_o*c_o))

which does not involve t or D and so is constant, so you are going to get a straight line.

Besides that second part of the formula does not involve t or D either.

You need to review the formula. It will have time or the boundary value in it somewhere.