Explanation of conditional statements.

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')

Answer 1

[David Hill]

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')