Why my graph not same as research paper?

3 ビュー (過去 30 日間)
nur 2022 年 6 月 8 日
コメント済み: Asif Solanki 2022 年 8 月 25 日
Hi, i want to ask why i did not get same graph as above?
This is my code that i have try:
%for t1=5%
x=0:0.2:10;
m=0.1;
C_0=1.0;
u_0=0.25;
D_0=0.45;
w_0=0.001;
p_0=0.02;
t1=5;
a=((((u_0)^2)/(4*D_0))+p_0);
b=((u_0)^2)/(4*D_0);
d=(1/(sqrt(2)*m));
e1=(log(2+sqrt(2)+(4+3*(sqrt(2)))*(tanh(m*t1/2))));
e2=(log(2+sqrt(2)-sqrt(2)*tanh(m*t1/2)));
T=d*(e1-e2);
f=((x-((((u_0)^2)+4*p_0*D_0)^1/2)*T)/(2*sqrt((D_0)*T)));
g=((x+((((u_0)^2)+4*p_0*D_0)^1/2)*T)/(2*sqrt((D_0)*T)));
h=((((u_0)-((((u_0)^2)+4*p_0*D_0)^1/2))*x)/(2*D_0));
i=((((u_0)+((((u_0)^2)+4*p_0*D_0)^1/2))*x)/(2*D_0));
j=((x-(u_0)*T)/(2*sqrt((D_0)*T)));
k=((x+(u_0)*T)/(2*sqrt((D_0)*T)));
l=(((u_0)*x)/(D_0));
n=exp(h);
o=erfc(f);
q=exp(i);
r=erfc(g);
s=exp(-p_0*T);
v=erfc(j);
y=erfc(k);
z=exp(l);
F=(((1/2).*n.*o)+((1/2).*q.*r));
G=(s.*(1-(1/2).*v-(1/2).*z.*y));
C1=(((w_0)/(p_0))+(C_0-((w_0)/(p_0))))*F-(((w_0)/(p_0))*G);
plot(x,C1)
hold on
%for t2=10%
x=0:0.2:10;
m=0.1;
C_0=1.0;
u_0=0.25;
D_0=0.45;
w_0=0.001;
p_0=0.02;
t2=10;
a=(((u_0)^2)/(4*D_0)+p_0);
b=((u_0)^2)/(4*D_0);
d=(1/(sqrt(2)*m));
e1=(log(2+sqrt(2)+(4+3*(sqrt(2)))*(tanh(m*t2/2))));
e2=(log(2+sqrt(2)-sqrt(2)*tanh(m*t2/2)));
T=d*(e1-e2);
f=((x-((((u_0)^2)+4*p_0*D_0)^1/2)*T)/(2*sqrt((D_0)*T)));
g=((x+((((u_0)^2)+4*p_0*D_0)^1/2)*T)/(2*sqrt((D_0)*T)));
h=((((u_0)-((((u_0)^2)+4*p_0*D_0)^1/2))*x)/(2*D_0));
i=((((u_0)+((((u_0)^2)+4*p_0*D_0)^1/2))*x)/(2*D_0));
j=((x-(u_0)*T)/(2*sqrt((D_0)*T)));
k=((x+(u_0)*T)/(2*sqrt((D_0)*T)));
l=(((u_0)*x)/(D_0));
n=exp(h);
o=erfc(f);
q=exp(i);
r=erfc(g);
s=exp(-p_0*T);
v=erfc(j);
y=erfc(k);
z=exp(l);
F=(((1/2).*n.*o)+((1/2).*q.*r));
G=(s.*(1-(1/2).*v-(1/2).*z.*y));
C2=(((w_0)/(p_0))+(C_0-((w_0)/(p_0))))*F-(((w_0)/(p_0))*G);
plot(x,C2)
hold off
hold on
%for t3=15%
x=0:10;
m=0.1;
C_0=1.0;
u_0=0.25;
D_0=0.45;
w_0=0.001;
p_0=0.02;
t3=15;
a=(((u_0)^2)/(4*D_0)+p_0);
b=((u_0)^2)/(4*D_0);
d=(1/(sqrt(2)*m));
e1=(log(2+sqrt(2)+(4+3*(sqrt(2)))*(tanh(m*t3/2))));
e2=(log(2+sqrt(2)-sqrt(2)*tanh(m*t3/2)));
T=d*(e1-e2);
f=((x-((((u_0)^2)+4*p_0*D_0)^1/2)*T)/(2*sqrt((D_0)*T)));
g=((x+((((u_0)^2)+4*p_0*D_0)^1/2)*T)/(2*sqrt((D_0)*T)));
h=((((u_0)-((((u_0)^2)+4*p_0*D_0)^1/2))*x)/(2*D_0));
i=((((u_0)+((((u_0)^2)+4*p_0*D_0)^1/2))*x)/(2*D_0));
j=((x-(u_0)*T)/(2*sqrt((D_0)*T)));
k=((x+(u_0)*T)/(2*sqrt((D_0)*T)));
l=(((u_0)*x)/(D_0));
n=exp(h);
o=erfc(f);
q=exp(i);
r=erfc(g);
s=exp(-p_0*T);
v=erfc(j);
y=erfc(k);
z=exp(l);
F=(((1/2).*n.*o)+((1/2).*q.*r));
G=(s.*(1-(1/2).*v-(1/2).*z.*y));
C3=(((w_0)/(p_0))+(C_0-((w_0)/(p_0))))*F-(((w_0)/(p_0))*G);
plot(x,C3)
hold on
legend({'t=5','t=10','t=15'},'Location','southwest')
xlabel('Distance,x (meter)')
ylabel('Concentration,C(x,t)')
%ylim([0, 1])
xlim([0, 10])
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Torsten 2022 年 6 月 8 日
I corrected the three errors in your code detected by @SALAH ALRABEEI
Looks better now, but concentrations still become negative. So your search must go on.

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採用された回答

Torsten 2022 年 6 月 8 日

Sometimes it's better to start anew:
x=(0:0.1:10).';
C0 = 1.0;
tt = [5.0,10.0,15,0];
u0 = 0.25;
D0 = 0.45;
m = 0.1;
gamma0 = 0.02;
mu0 = 0.001;
C = zeros(numel(x),numel(tt));
for i=1:numel(tt)
t = tt(i);
T = 1/(sqrt(2)*m)*(log(2+sqrt(2)+(4+3*sqrt(2))*tanh(m*t/2))-...
log(2+sqrt(2)-sqrt(2)*tanh(m*t/2)));
F = 0.5*exp((u0-sqrt(u0^2+4*gamma0*D0))*x/(2*D0)).*...
erfc((x-sqrt(u0^2+4*gamma0*D0)*T)./(2*sqrt(D0*T)))+...
0.5*exp((u0+sqrt(u0^2+4*gamma0*D0))*x/(2*D0)).*...
erfc((x+sqrt(u0^2+4*gamma0*D0)*T)./(2*sqrt(D0*T)));
G = exp(-gamma0*T)*(1-0.5*erfc((x-u0*T)./(2*sqrt(D0*T)))-...
0.5*exp(u0*x/D0).*erfc((x+u0*T)./(2*sqrt(D0*T))));
C(:,i) = mu0/gamma0 + (C0-mu0/gamma0)*F - mu0/gamma0*G;
end
plot(x,C)
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Asif Solanki 2022 年 8 月 25 日
Dear Sir,
My name is Asif Solanki I am a student persuing master's program in Italy. I am currently working on a small project where I do need to find the concentration of the pollutants using the Bear analytical method.
Therefore, would you like to assist me to find the solution?
Thank you very much
Best regards
Asif Solanki

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その他の回答 (2 件)

Sam Chak 2022 年 6 月 8 日
Hi @nur
One of the ways to find out is to determine the equilibrium points from the advection-dispersion equation.
I haven't checked the long equations. Can you verify if you plotted C or ? Sometimes, the transformations can be a little tricky.
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Sam Chak 2022 年 6 月 8 日

Hi @nur
Try to make a little bit of value-added effort to the problem.
Another way is to plot the concentration C from the numerical solution of the advection-dispersion differential equation.
This way you can compare with the analytical solution obtained from the paper. Bear in mind that sometimes misprints can occur due to authors' mistake, or the production crew's mistake. So, what was shown on the paper might not be truly the analytical solution. Therefore, yes you have to check.
But I'd suggest you to take numerical solution approach because that is directly from the governing advection-dispersion law. The analytical solution, which is "human-processed equation" from the law.
Edit: Hey, check this out:
One of the objectives is to Solve the one-dimensional advection dispersion equation using ode45 or ode15s.

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SALAH ALRABEEI 2022 年 6 月 8 日
You have two mistakes here
s=(exp(p_0)*T);
The three s in the code should be this
s=(exp(-p_0*T));
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SALAH ALRABEEI 2022 年 6 月 8 日
The reason is that the curves have data less than zero. Unlike those in the paper.
Anyway, you still can do this by adding this
axis([0 10 0 1])

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