Reputation: 121
Just few questions, i hope someone will find time to answer :).
What if we have COUPLED model example: system of n indepedent variables X and n nonlinear partial differential equations PDEf(X,PDEf(X)) with respect to TIME that depends of X,PDEf(X)(partial differential equation depending of variables X ). Can you give some advice? Here is one example:
Let’s say that c is output, or desired variable. Let’s say that r is independent variable.Partial differential equation looks like:
∂c/∂t=D*1/r+∂c/∂r+2(D* (∂^2 c)/(∂r^2 )) D=constant r=0:0.1:Rp- Matlab syntaxis, how to represent same in Modelica (I use integrator,but didn't work)?
Here is a code (does not work):
model PDEtest
/* Boundary conditions
1. delta(c)/delta(r)=0 for r=0
2. delta(c)/delta(r)=-j*d for r=Rp*/
parameter Real Rp=88*1e-3; // length
parameter Real initialConc=1000;
parameter Real Dp=1e-14;
parameter Integer np=10; // num. of points
Real cp[np](start=fill(initialConc,np));
Modelica.Blocks.Continuous.Integrator r(k=1); // independent x1
Real j;
protected
parameter Real dr=Rp/np;
parameter Real ts= 0.01; // for using when loop (sample(0,ts) )
algorithm
j:=sin(time); // this should be indepedent variable like x2
r.u:=dr;
while r.y<=Rp loop
for i in 2:np-1 loop
der(cp[i]):=2*Dp/r.y+(cp[i]-cp[i-1])/dr+2*(Dp*(cp[i+1]-2*cp[i]+cp[i-1])/dr^2);
end for;
if r.y==Rp then
cp[np]:=-j*Dp;
end if;
cp[1]:=if time >=0 then initialConc else initialConc;
end while;
annotation (uses(Modelica(version="3.2")));
end PDEtest;
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Upvotes: 3
Views: 1183
Reputation: 327
It's hard to answer your question, since you assuming that Modelica ~ Matlab, but that's not the case. So I won't comment your code, since it's really wrong. Let me give you an example model to the burger equation. Maybe you could use it as starting point.
model burgereqn
Real u[N+2](start=u0);
parameter Real h = 1/(N+1);
parameter Integer N = 10;
parameter Real v = 234;
parameter Real Pi = 3.14159265358979;
parameter Real u0[N+2]={((sin(2*Pi*x[i]))+0.5*sin(Pi*x[i])) for i in 1:N+2};
parameter Real x[N+2] = { h*i for i in 1:N+2};
equation
der(u[1]) = 0;
for i in 2:N+1 loop
der(u[i]) = - ((u[i+1]^2-u[i-1]^2)/(4*(x[i+1]-x[i-1])))
+ (v/(x[i+1]-x[i-1])^2)*(u[i+1]-2*u[i]+u[i+1]);
end for;
der(u[N+2]) = 0;
end burgereqn;
Your further questions:
Upvotes: 5