Too Many Input Arguments - I am new to matlab and not sure how to solve this

function [h] = thesis(t)
    


% Oil/Brine Systems Main Characteristics 
%% Oil/Water System at 30C (Paraffin)


mu1 = 3.8; % Viscosity in mPa.s-1
rho_O1 = 797; % Oil Density in kg/m3
rho_B1 = 998; % Brine Density in kg/m3
IFT1 = 21; % Interfacial Tension mN/m
g = 9.81; % m2/s
% Concentration for this system is 80ppm


%% Oil/Water System at 40C (Crude Oil A)


mu2 = 5; % Viscosity in mPa.s-1
rho_O2 = 833; % Oil Density in kg/m3
rho_B2 = 994; % Brine Density in kg/m3
IFT2 = 5; % Interfacial Tension mN/m
% Concentration for this system is 200ppm

%% Oil/Water System at 50C (Crude Oil B)


mu3 = 6; % Viscosity in mPa.s-1
rho_O3 = 834; % Oil Density in kg/m3
rho_B3 = 989; % Brine Density in kg/m3
IFT3 = 3; % Interfacial Tension mN/m
% Concentration for this system is 200ppm

%% Oil/Water System at 60C (Crude Oil C)


mu4 =4.9; % Viscosity in mPa.s-1
rho_O4 = 826; % Oil Density in kg/m3
rho_B4 = 985; % Brine Density in kg/m3
IFT4 = 1; % Interfacial Tension mN/m
%Concentration for this system is 200ppm
    
nd = 1000; % No. of Droplets
Vol = 900; % Liquid volume of emulsion (ml)
l = 0.5; % Mean Distance between droplets 
alpha = 0.08; % Empirical Collision Effiency Parameter
D0 = 300; % Initial Droplet Diameter (microns)

%% Sedimentation interface, hs: dhs/dt

Pr0 = ((nd*pi*D0^3)/6)/Vol; % Initial Volume Fraction of droplet
Prm = ((nd*pi*((D0+l)^3))/6)/Vol; % Maximum Volume Fraction of droplet

delrho1 = rho_B2 - rho_O2; % difference between the dispersed  water and continuous oil phase

Vsto = (delrho1*g*(D0^2))/18*mu2; % Settling Velocity of Hard Spheres (stoke's velocity)
fPr = (1-Pr0)^5.3; % Dimensionless

x1 = 60; % (hs - hd) in mm, guessed value

K1 = ((2/3)*alpha*((Vsto^2)/D0))*((fPr^2)/((Prm/fPr)^1/3)-1);

h = x1*(K1*t-(Vsto*fPr)); 



end

I tried everything to figure this out but dont understand what I can do again. Any assistance is greatly appreciated. thanks

Answer 1

[Sam Chak]

Try fixing these lines

function dhdt = thesis(t, h) 
dhdt = x1*(K1*t-(Vsto*fPr));

[Matthew Charles]

Thank you very much! Quick question, I am using the below code in the command window to generate a graph, however the graph I am trying to replicate is not what matlab is giving. See attached

tspan = [0 180];

[t, dhdt] = ode45(@thesis, tspan, [240]);

% where H = 240mm (Height of column)

% plotting graph

plot (t, dhdt(:,1), '-o')

grid on

ylabel ('H(mm)')

xlabel ('Time (s)')

title ('Simulation of Sedimentation-Coalesence Experiment')

When I run this code in the command window for the above function, I am getting this response :

Warning: Imaginary parts of complex X and/or Y arguments ignored.

Graph Generated:

I set the initial value as 240 but even that is not reflected on the graph, Can you assist please?

[Sam Chak]

Hi @Matthew Charles

If h is really the height of the column. and this line

dhdt = x1*(K1*t-(Vsto*fPr));

and its depending parameters correctly describe the differential equation (how h will change according to time t), then the following MATLAB code should give a result:

tspan = [0 180];
h0    = 240;
[t, h] = ode45(@thesis, tspan, h0);
plot(t, h)

If the result is incorrect (obvious because h exploded to ), something must be incorrect in your thesis function file. Please check at least x1, K1, Vsto, fPr.

x1 = 60; % (hs - hd) in mm, guessed value
K1 = ((2/3)*alpha*((Vsto^2)/D0))*((fPr^2)/((Prm/fPr)^1/3)-1)

K1 = 1.3125e+119 - 4.2645e+118i

At least, mathematically I understand if , then will explode.

[Matthew Charles]

I understand what you mean here, especially for the values derived for K1, I will review once more and ensure all the formulas for the function is indeed correct. Thank you so much for your assistance

Answer 2

[Saksham Gupta]

As per my understanding of your query, you are getting error in using function 'ode45'.

From the error logs, I can say that 'ode' function in ode45 is getting more arguments than the parameters defined for it.

Upon inspection of code, I realized that 'ode' is nothing but the function handle of 'thesis' function which you are passing.

Your function 'thesis' is expected to accept y0 (which you are passing as [240] to ode45).

If you change the very first line to:

function [h] = thesis (t, y0) 

you can get the output.

To explore more about ode45, you may refer the documentation