MATLAB: Explanation of conditional statements.

differential equationselseiffor loopifMATLAB

Hello,
I am working on mathematical modeling of multicomponent dissolution and biodegradation of BTEX compounds in oily sludge of petrochemical refinery industries. I have two matlab codes which have I to run simulataneously to obtain the result and there's no error in the code. I got these codes from my instructor and I couldn't understand the syntax of the conditonal statements (for, if, elseif) in the attached Matlab Code 2; as why are they used to get the result? Could someone please comment and explain at each Conditional statement syntax that is mentioned in MATLAB 2 code? Please find the attached.
MATLAB CODE 1
%Define t as t
t = t;
%define variable involved in final differential equations expressions
diap = 150*10^-6; %diameter of particle in meters
Dab = 2.808*10^-6; %diffusion coefficient of xylene into water in m2/hr
vel=18000; %velocity in m/h
dwater=1000000; %density for water given in g/m^3
vis=3204; % viscosity given in g/m*hr
%Calculate mass transfer coefficient
Nre=(diap*vel*dwater)/vis; % reynolds number
Nsc=vis/(dwater*Dab); %smiths number
Nsh=2+(0.95*(Nre^0.5)*(Nsc^(1/3))); %Sherwood number
kc = (Nsh*Dab)/diap; %mass transfer coefficient in m/hr
cstar = 160; %solubility limit for xylene in grms/m3
kr = 0.00165; %decay rate constant in 1/s for xylene in aeronic processes
m0 = 354; %initial mass of xylene in grms per 1 m3 of NAPL
dxyl = 860000; %density of xylene in grms/m3
cons=(6*(m0^(1/3)))/(dxyl*diap);
%define dependent variables as a two element vector
mx = y(1); %variable for mass of xylene
cx = y(2); %variable for concentration of xylene in water
%define differential equations dy1 and dy2 as a two element vector
dy(1)= -cons*kc*(cstar-y(2))*(y(1)^(2/3));
dy(2) = cons*kc*(cstar-y(2))*(y(1)^(2/3))-(kr*y(2));
%transpose dy
dy = dy' ;
end
MATLAB CODE 2
clear; clc
y0(1) = 354; %initial condition of mx at t=0
y0(2) = 0; %initial condition of cx at t=0
tspan = 0:0.001:800; %solve differential equations from 0 to 15000 hrs
[T,Y] = ode23s(@biodegkc,tspan,y0); %solve for y1, y2
for i=1:length(Y)
if Y(i,1)<0
Y(i,1)=0;
elseif i>1
if Y(i,1)>Y(i-1,1)
Y(i,1)=0;
end
if Y(i,2)<0
Y(i,2)=0;
end
end
end
%plot the results'
figure('units','normalized','outerposition',[0 0 1 1])
plot(T,Y(:,1),'*',T,Y(:,2),'+');
xlabel('time,hr');
ylabel('mx,mass(g) or cx,concentration(g/m3)');
title('Mass and Concentration profile ');
legend('mx','cx');
save('filenameT.txt','T','-ascii')
save('filenameY.txt','Y','-ascii')

Best Answer

clear; clc
y0(1) = 354; %initial condition of mx at t=0
y0(2) = 0; %initial condition of cx at t=0
tspan = 0:0.001:800; %solve differential equations from 0 to 15000 hrs
[T,Y] = ode23s(@biodegkc,tspan,y0); %solve for y1, y2
for i=1:length(Y)%going through the rows of Y
if Y(i,1)<0%if element i in first column of Y is < 0
    Y(i,1)=0;%set element to zero

elseif i>1%if > first row of Y
    if Y(i,1)>Y(i-1,1)%if element in Y(i,1) is > element in previous row (directly above it)
    Y(i,1)=0;%set element to zero
    end
    if Y(i,2)<0%if element in second column of Y at row position i is < 0
        Y(i,2)=0;%set elment to zero
    end
end
end
%plot the results'
figure('units','normalized','outerposition',[0 0 1 1])
plot(T,Y(:,1),'*',T,Y(:,2),'+');
xlabel('time,hr');
ylabel('mx,mass(g) or cx,concentration(g/m3)');
title('Mass and Concentration profile ');
legend('mx','cx');
save('filenameT.txt','T','-ascii')
save('filenameY.txt','Y','-ascii')
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