Error. I want to plot this ode45 pressure versus time

R=8.314/32;

T0=2930;

a=(12)/((2.027*10^6)^0.45);

rhoP=1920;

Astar=pi*0.25^2;

k=1.35;

n=0.45;

P0=101325;

syms P(t)

for t = [0,0.1]

dP=@(P,t)(Ab*a*P^n*(rhoP-rhoO)-P*Astar*sqrt(k/(R*T0))*(2/(k+1))^((k+1)/(2*(k-1))))*R*T0/v0;

[P,t]=ode45(dP, [0,0,1], P0);

end

if t==0 %at beginning of the integration set initial values for the persistent variables

rp=0.35; %initial port radius

t1=0; %initial time step

end

Ab=2*pi*rp*8;%burn area

rhoO=P/(R*T0); %gas density

rp=min(rp+((a*P^n)*10^-3)*(t-t1),0.7);

figure(2)

plot(t,y)

xlabel("Time (s)")

ylabel("Chamber Pressure (Pa)")

title("Chamber Pressure vs Time (Start-Up)")

Answer 1

[Thiago Henrique Gomes Lobato]

There were some errors in your code:

• For ode45 you give numerical functions, not symbolic ones
• There were some variables that were used in the function but initialized after the loop
• If you have time dependent variables, you ideally should define them inside the function
• Your variable order in the function was inverted
• You didn't declared v0 anywhere

I choosed a random value for v0 and fix the other issues of your code, you just have to make sure that the considerations you made for your function and variables are right to fully trust the results.

R=8.314/32;
T0=2930;
a=(12)/((2.027*10^6)^0.45);
rhoP=1920;
Astar=pi*0.25^2;
k=1.35;
n=0.45;
P0=101325;
P = P0;
%syms P(t)

%at beginning of the integration set initial values for the persistent variables
rp=0.35; %initial port radius
t1=0; %initial time step

rhoO=P/(R*T0); %gas density
rp=min(rp+((a*P^n)*10^-3)*(t1-t1),0.7);
    
Ab=2*pi*rp*8;%burn area

v0 = 0.1;

dP=@(t,P)Fun(t,P,R,T0,rp,a,n,t1,Ab,rhoP,Astar,k,v0);%@(t,P)(Ab.*a.*P.^n.*(rhoP-rhoO)-P.*Astar.*sqrt(k/(R.*T0)).*(2/(k+1)).^((k+1)/(2.*(k-1)))).*R.*T0./v0;
[t,P]=ode45(dP, [0,0.1], P);
y = P;


figure(2)
plot(t,y)
xlabel("Time (s)")
ylabel("Chamber Pressure (Pa)")
title("Chamber Pressure vs Time (Start-Up)")

function dP = Fun(t,P,R,T0,rp,a,n,t1,Ab,rhoP,Astar,k,v0)

rhoO=P/(R*T0); %gas density
rp=min(rp+((a*P^n)*10^-3)*(t-t1),0.7);
Ab=2*pi*rp*8;%burn area
dP = (Ab.*a.*P.^n.*(rhoP-rhoO)-P.*Astar.*sqrt(k/(R.*T0)).*(2/(k+1)).^((k+1)/(2.*(k-1)))).*R.*T0./v0;
end

Answer 2

[Ous Chkiri]

if i want to add this condition if rp>=0.7 %if grain gets exhausted

Ab=0; %burn area =0 and plotting from 0 to 60 but from 0>>0.1 we apply rp 0.3 after we have 0.7

[0,0.1,60]

[Thiago Henrique Gomes Lobato]

If you have conditions that make a discontinuity you will have to perform the integration in a piece-wise matter and add the conditions in your integration function ("Fun" in the example I gave you). This answer and comments shows an example about how you could do it and also possible pitfalls https://de.mathworks.com/matlabcentral/answers/487643-adding-a-piecewise-time-dependent-term-in-system-of-differential-equation#answer_398394?s_tid=prof_contriblnk

[Ous Chkiri]

I did not understand to do it and i already see the link