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

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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!
  2 Comments
Nicola De Noni
Nicola De Noni on 16 May 2022
Edited: Nicola De Noni on 16 May 2022
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.

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Accepted Answer

Torsten
Torsten on 16 May 2022
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)
  5 Comments

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More Answers (1)

Nicola De Noni
Nicola De Noni on 20 Jul 2022
Edited: Nicola De Noni on 20 Jul 2022
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);
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