MATLAB: Coupled differential equations kinetic

coupledMATLABode

Hello Im am trying to solve 3 coupled ODE's, with the plot related to the reduction of NO by with ammoniasulphate.

The issiue is that the differential equation i wrote do not seems to be coupled. First the initial condition for NH3 is zero, but that makes the rate of change 0. If i change the NH3 to a very small value then the system still not respond well (deacreasing NO over time) as chaning the stouchemtic coiffcient x1 does not change the NO concentration…… see attched. Thanks.

Best Answer

yprime and y are always connected

function [ yprime ] = calibration( t,y,par,par_cali,T,x1,O2)

Your input y were in wrong order. Try this

AS=y(1);
NH3=y(2);
NO=y(3);

Result

Related Solutions

MATLAB: Maths / ODE / Why do I get complex results

Each of your ODE functions includes this line:

Phi= 1-exp(-(47.58*u^1.877+0.10085*Tp*u^1.0585));

For sake of argument, let's let Tp = 0. Then this simplifies to:

Phi= 1-exp(-(47.58*u^1.877));

What does this return when you evaluate it for a negative value of u?

u = -1;
Phi= 1-exp(-(47.58*u^1.877))

Specifically, what does the term inside the exp call return?

-(47.58*(-1)^1.877)

If you let Tp be non-zero the same argument holds, only you're raising that negative number to two non-integer-valued powers.

MATLAB: Solving 4th order ode

Use bvp4c()

You already have written your equation as system of first-order equations

function dydt = ode4(t,y)
    L=1;B=0.25;k=0.1;
    dydt = zeros(4,1);
    dydt(1)=y(2);
    dydt(2)=y(3);
    dydt(3)=y(4);
    dydt(4)=(2.*B+1+y(1)).*y(3))./(k.*(1+y(1));
end

Here is function for your boundary conditions

function res = bc4(y0, y1)
    res = [y0(1)-0        % y(0) = 0
           y0(2)-1        % y'(0) = 1
           y1(1)-0        % y(end) = 0
           y1(2)-1];      % y'(end) = 1
end

See bvp4c

Best Answer

Related Solutions

MATLAB: Maths / ODE / Why do I get complex results

MATLAB: Solving 4th order ode

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