Problem with function handle? or something.

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Dakota Burrow
Dakota Burrow le 7 Juil 2013
So, first off, the answer to this problem should be roughly .0028. But, somewhere in the code there is an issue and it will not compute correctly. Can you help?
v = input('Enter initial guess for the volume of methane: ');
R = .518;
pc = 4600;
tc = 191;
T = -40 + 273;
P = 65000;
a = .427*R^2*tc^2.5/pc;
b = .0866*R*tc/pc;
gv =@(v)(R*T)-(((a*(v-b)^2)/(v*(v+b)*sqrt(T))) + P * b)/P;
n = 4;
es = .5*10^(2-n);
[r, I] = Redlich_Kwong(es, v, gv);
fprintf('\nThe calculated volume is : %.6f',r);
fprintf('\nIt took %1.0d iterations to converge.',I);
function [r, I] = Redlich_Kwong(es, v, gv)
I = 0;
ea = 1;
while ea > es
r = gv(v);
ea = 100*abs((r-v)/r);
v = r;
I = I +1;

Réponse acceptée

Guru le 7 Juil 2013
Well, from plotting your input-output behavior of your anonymous function gv(), it converges to about the value of R*T. In other words, for all values of v, gv returns an answer of approximately R*T, so I would guess something is wrong in your equation for gv. As I do not know anything where you came up with gv or what it should be, that's the extent I can be of assistance on this matter. I merely ran a quick monte carlo simulation on a sizeable random distribution of v's from -1000 to +1000 and they all came back around R*T in their results.

Plus de réponses (1)

Dakota Burrow
Dakota Burrow le 7 Juil 2013
Thank you! R*T was also supposed to be divided by P. Thanks!


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