Why does it say Iteration limit reached and how do I fix it?

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Quincey Jones
Quincey Jones 2020-2-13
评论： KALYAN ACHARJYA 2020-2-13
function [v,err,count] = Clapeyrons(P)
% Calculate molar volume as predicted by the Ideal Gas Law equation
% using the Newton-Raphson algorithm with initial estimate
% determined by ideal gas law.
% Inputs:
R = 0.082057; % Ideal gas constant in L atm / K mol
T = 293; % Temperature in K
Tc = 416.90; % Critical temperature of Cl2 in K
Pc = 78.72918; % Critical Pressure of Cl2 in atm
a = ((R^2)*(Tc^(5/2)))/(9*Pc*(2^(1/3)-1));
b = (R*Tc*(2^(1/3)-1))/(3*Pc);
% P = 1; % Pressure in atm
% Outputs:
% v equals molar volume in L/mol as predicted by
% equation for gas considered at temperature T and pressure P.
% err equals modulus of function evaluated at approximate root.
% count is number of iterations taken by Newton-Raphson algorithm.
% Version 1: created 24/04/19. Author: Paul Curran
% Version 2: created 06/02/20. Author: Savana Stewart
if (~isscalar(P)) || (~isreal(P)) || P <= 0
error('Input argument P must be positive real scalar.')
end
Iteration_limit = 20; % maximum number of iterations permitted
Tolerance=10^-7; % maximum acceptable value for modulus of
A = (a*P)/((R^2)*(T^(5/2)));
B = (b*P)/(R*T);
v = R * T / P;
Z = (P*v)/(R*T);
C = A-B-B^2;
f1 = [1 -1 C -A*B]; % f = (Z^3) - (Z^2) + (A-B-(B^2))*Z - (A*B)
f = polyval(f1,Z);
v = Z*R*T/P;
for count = 1:Iteration_limit + 1
% Terminate with error message if iteration limit exceeded:
if count == Iteration_limit + 1
error('Iteration limit reached. Iteration did not converge.')
end
% Terminate iteration if function is sufficiently small at current
% estimate
if abs(f)<Tolerance
break
end
%f_dash = 3*Z^2 - 2*Z + (A - B - (B^2)); % Evaluate derivative of f
f_dash = [0 3 -2 C];
Z = Z - (f1./f_dash); % Newton-Raphson iteration
end
err = abs(f); % Error is magnitude of f(v) at final root estimate
end
%%
% COPY THE FOLLOWING INTO THE COMMAND WINDOW
% P = [1 1.5 2 2.5 3 5 10 15 25 50 100];
% v = zeros(1,length(P));
% for i=1:length(P)
% [u,err,count]=Clapeyrons(P(i));
% v(i)=u;
% end
%
% P1=[1:0.1:100];
% V1=R*T./P1;
% plot(P1,V1)
% hold on
% plot(P,v,'or')
I says iteration limit reached but I only changed the format a little. How do I fix it?

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回答（1 个）

KALYAN ACHARJYA
KALYAN ACHARJYA 2020-2-13
编辑：KALYAN ACHARJYA 2020-2-13
Its doing as defined in the code, when the iteration reached more than iteration limit, it reflects the message as mentioned within the code
if count == Iteration_limit + 1
error('Iteration limit reached. Iteration did not converge.')
end
Before getting the break statement, your code reached the limit, please check, there may be abs(f) never less than Tolerance
if abs(f)<Tolerance
break
end
Its simple code, debug it step by step, ensure that it reached the break statement before reched the iteration limit, ensure that f is changing in right direction during each iteration. The loop execute the iteration as number of times=iteration limit.
Choose the right parameters values or codition to reached the break before reached to operation limit, see the tolerance value also.
  2 个评论
Quincey Jones
Quincey Jones 2020-2-13
I know that the statement tells me errors, I meant why is the interation not converging. Where is the error
KALYAN ACHARJYA
KALYAN ACHARJYA 2020-2-13
Change the variables value or tolerance value, when I change the tolerance value, it converging
Tolerance=10^7; % I know this is the correct value as required
# Do the Basic Maths, you get the appropriape typical values, so that it converge
>> [v,err,count]=Clapeyrons(5)
v =
4.8085
err =
0.0572
count =
1
>>

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