Use ode45 when I have a constant that varies over time.

Hi everyone, I’m trying to solve this differential equation (dT/dt):

dTdt = ((U*A)/(M_mix*cp_mix))*(T_in-T_out)/(log(T_in-T)/(T_out-T));

Where I want to see the temperature profile T as the weather changes. Unfortunately, T_in and T_out are two temperature vectors whose values vary over time. How can I solve this differential equation? This is my code:

function f = CalcoloSperimentale(t,T)

%% Parameters

n = 10;             % length of time vector 

M_mix = linspace(11379,11379,n);     % Weigth [Kg]
cp_mix = linspace(2000,2000,n);      % Cp
U = linspace(53,53,n);               % Thermal coefficient [W/m2K]
A = linspace(25.2,25.2,n);           % Area 

xmin_in = 15;          % Minimum inlet temperature
xmax_in = 16;          % Maximum inlet temperature
T_in = xmin_in+rand(1,n)*(xmax_in-xmin_in);

xmin_out = 20;          % Minimum outlet temperature
xmax_out = 40;          % Maximum outlet temperature
T_out = linspace(xmax_out,xmin_out,n);

% Differential equation

dTdt(i) = ((U(i)*A(i))/(M_mix(i)*cp_mix(i)))*(T_in(i)-T_out(i))/(log(T_in(i)-T(i))/(T_out(i)-T(i)));
f = dTdt;

end

And the second script is:

n = 10; 
timeStart = 0; 
timeEnd = 5280; 
timespan = linspace(timeStart,timeEnd,n);

% Initial condition

initT = 98.5+273.15; 

[timeOut, T] = ode45(@CalcoloSperimentale, timespan, initT);

% Plotting data

plot(timeOut, T)

In order to solve this differential equation i should have to consider also the final condition of temperature, but i don't know hot to implement it.

Thanks for the answers!

[Torsten]

Could you explain in short what temperatures T_in, T_out and T are (so the background of your question)?

[Nicola De Noni]

Absolutely! T_in is the inlet temperature of a chemical reactor jacket and T_out is the outlet temperature of the reactor jacket. I’m studying the heat transfer in a water-cooled tank. T is the temperature inside the vessel.

I would like to plot the temperature inside the reactor vs time (by knowing the inlet and outlet temperature of the cooling fluid) in order to understand the heat transfer.

Answer 1

[Torsten]

n=10;
xmin_in = 15;          % Minimum inlet temperature
xmax_in = 16;          % Maximum inlet temperature
T_in = xmin_in+rand(1,n)*(xmax_in-xmin_in);
xmin_out = 20;          % Minimum outlet temperature
xmax_out = 40;          % Maximum outlet temperature
T_out = linspace(xmax_out,xmin_out,n);
M_mix = 11379;    % Weigth [Kg]
cp_mix = 2000;      % Cp
U = 53;               % Thermal coefficient [W/m2K]
A = 25.2;           % Area 
timeStart = 0; 
timeEnd = 5280; 
timespan = linspace(timeStart,timeEnd,n);
T_in = @(t)interp1(timespan,T_in,t);
T_out = @(t)interp1(timespan,T_out,t);
f = @(t,T) U*A/(M_mix*cp_mix)*(T_in(t)-T_out(t))/log((T_in(t)-T)/(T_out(t)-T));
% Initial condition
initT = 98.5+273.15; 
[timeOut, T] = ode45(f, timespan, initT);
% Plotting data
plot(timeOut, T)

[Nicola De Noni]

It works! Is it possible to code an output temperature condition of 78+273.15 Kelvin? Or is the ode function not suitable?

Because from experimental data the temperature inside the reactor at 5280 sec. is 78+273.15 Kelvin and by using your code the temperature reach 278 Kelvin.

[Torsten]

I wonder where the temperature T of your differential equation is situated. Is it the mean temperature in the chemical reactor pipe over time ?

[Nicola De Noni]

The temperature T is the temperature inside the reactor. In the differential equation the temperature T (insider the vessel) is at the denominator of the right side of the equation.

The objective of the analysis is to see the temperature profile of the fluid inside the reactor as it decreases over time, as I use cooling water in the reactor jacket.

[Torsten]

One problem surely is that you use Celsius temperature for the temperature in the cooling vessel and Kelvin temperature in the reactor ...

If you want to set the outlet temperature to 78+273.15, use

xmin_out = 78+273.15;          % Minimum outlet temperature
xmax_out = xmin_out;          % Maximum outlet temperature

[Nicola De Noni]

You're right, i'm sorry! Anyway, thanks a lot!

Answer 2

[Nicola De Noni]

Hi everyone!

I'm using the script created by @Torsten but I need to modify it because I have some experimental data instead of function. I will explain it better after posting the script.

As you can see there are two variables (T_in and T_out) that are used in the ODE like and interpolation of data. Due to this approximation, Matlab shows me a linear variation of temperature inside the reactor. In order to be more accurate I want to consider all the experimental data instead of a function. How can I do it?

Thanks!

In the first plot you can see the results of the ODE in blue. Probably this is a line due to the use of

T_in = @(t)interp1(timespan,T_in,t);
T_out = @(t)interp1(timespan,T_out,t);

In the second plot there is the temperature profile of the outlet cooling fluid.

clear all; clc; close all

T_out = (xlsread('experimentaldata.xlsx',1,'E1651:E1682'))+273.15; % Experimental data of outlet cooling water

n=length(T_out);
timeStart = 0;         % Initial time [s]
timeEnd = 15.5*60;     % Final time [s]

timespan = linspace(timeStart,timeEnd,n); % Time vector

xmin_in = 17 + 273.15;          % Minimum inlet temperature
xmax_in = 17 + 273.15;          % Maximum inlet temperature

T_in = 17 + 273.15;   % Constant temperature of inlet cooling water

M_mix = 11379;        % Weigth [Kg]
cp_mix = 2000;        % Specific heat [J/KgK]
U = 102.408898670153; % Thermal coefficient [W/m2K]
A = 25.2;             % Area 

%% I want to change the vector T_in and T_out in order to use the single experimental data
% instead of a linear function. 
T_in = @(t)interp1(timespan,T_in,t);
T_out = @(t)interp1(timespan,T_out,t);

% Temperature of fluid to be cooled
f = @(t,T) (-U*A*((T-T_in(t))-(T-T_out(t)))/log((T-T_in(t))/(T-T_out(t))))/(M_mix*cp_mix);

% Initial condition: Initial temperature of fluid to be cooled
initT = 121.2307587+273.15;  % [K]

[timeOut, T] = ode45(f, timespan, initT);

[Torsten]

For your above code to work, don't you have to use

T_in = ones(size(T_out))*(17 + 273.15);   % Constant temperature of inlet cooling water

instead of

T_in = 17 + 273.15;   % Constant temperature of inlet cooling water

?

In order to be more accurate I want to consider all the experimental data instead of a function.

The interpolation function considers all experimental data that are included in the arrays T_in and T_out. You could try a different interpolation method, e.g.

T_out = @(t) interp1(timespan,T_out,t,'spline')

and/or tighten the tolerances for ODE45:

options = odeset('RelTol',1e-8,'AbsTol',1e-8)
[timeOut, T] = ode45(f, timespan, initT,options);

to see whether the graph for T changes.

[Nicola De Noni]

Thank you @Torsten!

Have a nice day.