I have this problem Exiting fzero: aborting search for an interval containing a sign change because no sign change is detected during search. Function may not have a root.
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V3=fzero(@Pr3,0.7);
V3
function y=Pr3(V)
Pr=4;
R=0.08206;
T=450;
Tc=405.5;
Pc=111.3;
P=Pr*Pc;
a=0.42747*((R^2)*Tc^(5/2)/Pc);
b=0.08664*(R*Tc/Pc);
y=((R*T)/(V-b))-(a/(V*(V+b)*T^(1/2)))-P;
end
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回答(1 个)
Walter Roberson
2024-2-26
syms V
Q = @(v) sym(v);
Pr = Q(4);
R = Q(08206) / Q(10)^5;
T = Q(450);
Tc = Q(405.5);
Pc = Q(1113) / Q(10)^1;
P = Pr*Pc;
a = Q(42747) / Q(10)^5 * ((R^2)*Tc^(5/2)/Pc);
b = Q(08664) / Q(10)^5 * (R*Tc/Pc);
y = ((R*T)/(V-b))-(a/(V*(V+b)*T^(1/2)))-P;
y
sol = solve(y, 'maxdegree', 3)
vpa(sol)
1 个评论
Walter Roberson
2024-2-26
format long g
V3=fzero(@Pr3,0.05);
V3
function y=Pr3(V)
Pr=4;
R=0.08206;
T=450;
Tc=405.5;
Pc=111.3;
P=Pr*Pc;
a=0.42747*((R^2)*Tc^(5/2)/Pc);
b=0.08664*(R*Tc/Pc);
y=((R*T)/(V-b))-(a/(V*(V+b)*T^(1/2)))-P;
end
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