Error: Contour endpoints and waypoints must be finite.
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I'm trying to write a function to plot the Cp of a compound, and I keep getting this error when I try to intergrate. I'm using the 2012 version of matlab.
function outputValue=Cps()
Pc=4550;
Tc=654;
b = .092968;
w = .22894;
k = 0.37464 + 1.54226*w - 0.26992*w^2;
R = 8.314;
Ai = 112.057;
Bi = -0.8324;
Ci = 0.002865;
Di = -0.000003;
v = [];
T = [300:1:1000];
Cp = [];
Cpi = [];
Cv = [];
dPdT = [];
dPdV = [];
alpha= []
P = [.02.*Pc .5.*Pc Pc 2.*Pc];
for s = 1:4
for t = 1:length(T)
alpha(t) = (1+k.*(1-sqrt(T(t)./Tc))).^2;
a = (.45724.*(R.*Tc).^2.*alpha(t))./Pc;
A = (a.*P(s))./(R.^2.*t.^2);
B = (P(s).*b)./(R.*t);
Z = [1 -(1-B) (A-3.*B-2.*B) -(A.*B-B.^2-B.^3)];
r =roots(Z);
for i = 1:3
if ((imag (Z(i))~=0))
Z(i)=0;
end
end
G(t) = max(r);
v(t) = (G(t).*R.*t)./P(s);
Cpi(t) = Ai + Bi.*T(t) + Ci.*T(t).^2 + Di.*T(t).^3;
C = ((k^2+k)./(2.*sqrt(Tc)));
SecondDerivative = @(x) ((a.*C.*(1/T(t).^1.5))./(x.*(x+b)+b.*(x-b)));
dPdT(t) = (R./(v(t)-b))-a.*(k.*sqrt(alpha(t)./(T(t).*Tc))./(v(t).*(v(t)+b)+b.*(v(t)-b)));
dPdV(t) = (-R.*T(t)/(v(t)-b).^2)+((2.*a.*(v(t)+b))./((v(t).*(v(t)+b)+b.*(v(t)-b))).^2);
Cv(t) = (Cpi(t)-R) + (T(t).*integral(SecondDerivative,Inf,v(t)));
Cp(t) = Cv(t) - (T(t).*v(t).*(dPdT(t)).^2)./dPdV(t);
end
end
plot(T,Cp,'-g')
0 Comments
Answers (1)
Sean de Wolski
on 6 Dec 2012
Edited: Sean de Wolski
on 6 Dec 2012
It looks like it's complaining about this line:
Cv(t) = (Cpi(t)-R) + (T(t).*integral(SecondDerivative,Inf,v(t)));
because it does not like the Inf as the minimum boundary, xmin! This makes sense.
doc integral
For more information.
Also as an FYI, it typically pays to paste the full eror message. The error messages contain diagnostic information like line numbers and the stack that make it much easier to debug.
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