Help me to solve the function

closed

Answer 1

[KSSV]

clear; clc;

wc=0.4;             
pc=3.16;           
T=293;                  
rwc3s=0.586;         
rwc3a=0.172;           
mc=0.3;              
vc=1/((wc*pc)+1);      
ro=(3*vc/(4*pi))^1/3;   


yg=0.25;            
yw=0.15;           
RH=0.88;           
b1=1231;          
b2=7579;           
B293=0.3*10^-10;    
C=5*10^7;           
ER=5364;           
De293=((rwc3s*100)^2.024)*(3.2*10^-14);             
kr293=(8.05*(10^-10))*((rwc3s*100+rwc3a*100)^0.975);    
v=2.15;                                               
cw=((RH-0.75)/0.25)^3;
pw=1;

B=B293*exp(-b1*(1/T-1/293));                           
kr=kr293*exp(-ER*(1/T-1/293));                         


da=360;    
dt=1;                           
Nn=(da)*(1/dt);                
timestep=0;
time=1:Nn;
time2=1:Nn+1;
                                        
% r_initial=16;    
r_initial=[ 0.5 2 8 16 32];   
m = length(r_initial) ; 
n1 = length(time) ; n2 = length(time2) ; 
Alfa = zeros(m,n1) ; 
RT_INC1 = zeros(m,n1) ; 
RT_INC2 = zeros(m,n1) ; 

for i = 1:length(r_initial)                                   
  ro_init=r_initial(i)*0.001;        
  L=((4*pi*(wc*pc/pw+1)/3)^(1/3))*ro_init;   
for BB=1:Nn                     
    if (BB==1)
       rt_inc(1)=ro_init(1)-0.00001; %boundary condition for unhydrated
    else
       rt_inc(1)=ro_init(1)-0.00001;
    end
       rt_inc(Nn+1)=0;  
    if (BB==1)
       Rt_inc(1)=ro_init(1)+0.00001; %boundary condition for hydrated
    else
       Rt_inc(1)=ro_init(1)+0.00001;
    end
       Rt_inc(Nn+1)=0;
end

for M=1:Nn
    if Rt_inc(M)/L<=ro
    Sr(M)=0;
    elseif (Rt_inc(M)/L>=ro)&&(Rt_inc(M)/L<0.5)
    Sr(M)=4*pi*(Rt_inc(M)/L)^2;
    elseif (0.5<=Rt_inc(M)/L)&&(Rt_inc(M)/L<0.5*(2^0.5))
    Sr(M)=(4*pi*(Rt_inc(M)/L)^2)-(6*pi*(1-(0.5/(Rt_inc(M)/L))));
    elseif (Rt_inc(M)/L>=0.5*(2^0.5)) && (Rt_inc(M)/L<0.5*(3^0.5))
   
    fun = @(y,x) 8*(Rt_inc(M)/L)./(sqrt((Rt_inc(M)/L)^2-(x.^2)-(y.^2)));
    ymin=sqrt((Rt_inc(M)/L)^2-0.5);
    xmin=@(x) sqrt((Rt_inc(M)/L)^2-0.25-x.^2);
    Sr(M)=integral2(fun,ymin,0.5,xmin,0.5);
    else
    Sr(M)=0;
    end
    cst(M)=Sr(M)/(4*pi*Rt_inc(M)^2);
    alfa(M)=1-(rt_inc(M)/ro_init)^3;
    kd(M)=(B/(alfa(M)^1.5))+C*(Rt_inc(M)-ro_init)^4;
    De(M)=De293*(log(1/alfa(M)))^1.5;
    rt_inc(M+1)=rt_inc(M)-(dt*((pw*cst(M)*cw)/((yw+yg)*pc*rt_inc(M)^2))*1/((1/(kd(M)*rt_inc(M)^2))+(((1/rt_inc(M))-(1/Rt_inc(M)))/De(M))+(1/(kr*rt_inc(M)^2))));
    Rt_inc(M+1)=((v-1)*((rt_inc(M)-rt_inc(M+1))/dt))*dt+Rt_inc(M);
end
Alfa(i,:) = alfa ; 
RT_INC1 = Rt_inc ;
RT_INC2 = rt_inc ; 
end

figure(1)
plot(time,Alfa) 
legend
xlabel('Time (Hours)') 
xlim([0 da])
grid on

figure (2)
plot(time2,RT_INC1,time2,RT_INC2),grid;
legend
yline(ro_init,'-','Initial Radius Particles');
legend('R_t','r_t','r_o')
%title('xx')
xlabel('Time (Hours)') 
ylabel('Radius of Particles (mm)')
xlim([0 da])
grid on

[Kevin Isakayoga]

Thank you very much !

[Kevin Isakayoga]

By the way, I think the last graph is not the same like I want, because its only depends on the final value which in this case is 32.

r_initial=[ 0.5 2 8 16 32]; 

Could you help me to make each of the r_initial value influence to become a graph? I already tried like this,

RT_INC1(i,:) = Rt_inc ;
RT_INC2(i,:) = rt_inc ; 

the value become messy.

figure (2)
plot(time2,RT_INC1,time2,RT_INC2),grid;
yline(ro_init,'-','Initial Radius Particles');
legend('R_t','r_t','r_o')
xlabel('Time (Hours)') 
ylabel('Radius of Particles (mm)')
xlim([0 da])
grid on

Thank you very much in advance! Looking forward to your reply.

[KSSV]

It is changing.....you need to check your formula..