Error: Contour endpoints and waypoints must be finite.

Question

Steven 2012-12-6

0
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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/55941-error-contour-endpoints-and-waypoints-must-be-finite

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')

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Answer 1

Sean de Wolski 2012-12-6

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/55941-error-contour-endpoints-and-waypoints-must-be-finite#answer_67764

编辑：Sean de Wolski 2012-12-6

在 MATLAB Online 中打开

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.