Differential Equations with Inputs (Tank Draining Problem)

23 vues (au cours des 30 derniers jours)
Quinlan Connell
Quinlan Connell le 3 Déc 2022
All I am trying to do is let the user "customize" the parameters of the tank and the rate of water flowing out of it. I started with this Script/Function combo which works:
Script:
h0=str2num(input('Initial Height of Water\n', 's'));
tf=str2num(input('Time of Leak\n', 's'));
tspan=0:.05:tf
[t,h0]=ode45('Tank',tspan,h0)
plot(t,h0)
title('Height of Water vs Time Cylindrical Tank')
xlabel('time (s)')
ylabel('height (m)')
Function:
function dh = Tank(t,h0)
k = .1038
dh = -k*sqrt(h0);
end
BUT now I am trying to give the user some control and this combo is returning the error messages written below it:
Script:
cylD=str2num(input('Cylinder Diameter (m)\n', 's'));
holeD=str2num(input('Hole Diameter (m)\n', 's'));
h0=str2num(input('Initial Height of Water\n', 's'));
tf=str2num(input('Time of Leak\n', 's'));
g = 9.81
A = pi*(cylD/2)^2
hole = pi*(holeD/2)^2
tspan=0:.05:tf
[t,h0]=ode45(@TankIn,tspan,h0)
plot(t,h0)
title('Height of Water vs Time Cylindrical Tank')
xlabel('time (s)')
ylabel('height (m)')
Function:
function dh = TankIn(t,h0)
k = holeD/A*sqrt(2*g)
dh = -k*sqrt(h0);
end
Errors:
Error using odearguments
When the first argument to ode45 is a function handle, the tspan and y0 arguments must be supplied.
Error in ode45 (line 107)
odearguments(odeIsFuncHandle,odeTreatAsMFile, solver_name, ode, tspan, y0, options, varargin);
Error in Tank1 (line 12)
[t,h0]=ode45(@TankIn,tspan,h0)

Réponse acceptée

Bjorn Gustavsson
Bjorn Gustavsson le 3 Déc 2022
Modifié(e) : Bjorn Gustavsson le 3 Déc 2022
What you should do is to make a small modification of your ODE-function - such that it takes an additional input argument where you can store all additional parameters that it needs. In this case a simple array will do the trick. Then you only need to collect these parameters in the script, and modify the ode45-call.
%% Script:
cylD=str2num(input('Cylinder Diameter (m)\n', 's'));
holeD=str2num(input('Hole Diameter (m)\n', 's'));
h0=str2num(input('Initial Height of Water\n', 's'));
tf=str2num(input('Time of Leak\n', 's'));
g = 9.81
A = pi*(cylD/2)^2
hole = pi*(holeD/2)^2
extraPars = [hole A g]; % Here we put the extra-parameters away, including
% g so we can use this on the moon etc
tspan=0:.05:tf
[t,h0]=ode45(@(t,h) TankIn(t,h,extraPars),tspan,h0); % here we create an
% "anonymous function" that
% ony depends on t and
% h, that uses the
% current value of
% extraPars in TankIn
plot(t,h0)
title('Height of Water vs Time Cylindrical Tank')
xlabel('time (s)')
ylabel('height (m)')
%% Function:
function dh = TankIn(t,h0,extraPars) % here we have the extraPars as a 3rd input
holeD = extraPars(1); % then we only need to extract these parameters
A = extraPars(2); % into the variables. I prefer to do it this way to keep
g = extraPars(3); % different steps separate.
k = holeD/A*sqrt(2*g);
dh = -k*sqrt(h0);
end
In my opinion it is cleaner to have all extra parameters in one additional array if possible. That way you dont need to modify the function-call in the ode45-calls you only need to modify the assignement to extraPars when you modify the TankIn-function.
HTH
  2 commentaires
QUINLAN
QUINLAN le 3 Déc 2022
Super helpful thank you! Separate question but why am I getting a plot that goes into negative numbers? They shouldnt even be plotable on this axis I would have thought because they are complex numbers.
Torsten
Torsten le 3 Déc 2022
h0 becomes complex, but integration can continue since ode45 also integrates complex-valued ODEs.
When plotting, the imaginary part of h is ignored.
Either detect the time when h becomes negative and only integrate up to this point or use the events detection facility of the ODE integrators. See the ballode example on how to choose this option.

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Plus de réponses (1)

Sam Chak
Sam Chak le 8 Déc 2022
You can also use the real() function to obtain only the Real part of the complex number.
% Parameters
cylD = 1; % Cylinder Diameter (m)
holeD = 0.2; % Orifice (tech term for 'hole' in fluid flow) Diameter (m)
h0 = 50; % Initial water height (m)
tf = 300; % Time of Leak
g = 9.81; % Earth's gravity
A = pi*(cylD/2)^2; % Tank cross-sectional area
hole = pi*(holeD/2)^2; % Area of Orifice
Params = [hole A g];
tspan = 0:0.05:tf;
[t, h] = ode45(@(t, h) TankIn(t, h, Params), tspan, h0);
plot(t, h), grid on
title('Water Level of the Cylindrical Tank')
xlabel('Time (s)')
ylabel('Water Level (m)')
% Water Tank
function dh = TankIn(t, h, Params)
holeD = Params(1);
A = Params(2);
g = Params(3);
k = holeD/A*sqrt(2*g);
dh = -k*real(sqrt(h));
end

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R2022a

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