Find the roots of an expression

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Is there a function that can find the root(s) of an expression, which is not a polynomial? My constants and my final expression (for which I want to find the root) are listed below.
%Declare constants
Dw = 0.0000076;
S = 0.000239;
rho = 1.1;
h = 0.0030;
Nm=20,000;
Wo = 0.0025;
c1 = Dw*S/rho/h;
c2 = (3*Wo/(4*Nm*3.14159*rho))^(1/3);
c3 = (3*Wwf/(4*Nm*3.14159*rho))^(1/3);
twf=20;
%Need to determine the value of Wwf in the following expression
((c2 - c3 - h*log((h + c2)/(h + c3)))/c1)-twf=0;

采纳的回答

Matt Fig
Matt Fig 2011-4-13
FZERO can be used to find roots.
Dw = 0.0000076;
S = 0.000239;
rho = 1.1;
h = 0.0030;
Nm=20000;
Wo = 0.0025;
c1 = Dw*S/rho/h;
c2 = (3*Wo/(4*Nm*3.14159*rho))^(1/3);
c3 = @(Wwf) (3*Wwf/(4*Nm*3.14159*rho)).^(1/3);
twf=20;
G = @(Wwf) ((c2 - c3(Wwf) - h*log((h + c2)./(h + c3(Wwf))))/c1)-twf;
rt = fzero(G,.01)
  5 个评论
Matt Fig
Matt Fig 2011-4-13
Try this:
Dw = 0.0000076;
S = 0.000239;
rho = 1.1;
h = 0.0030;
Nm=20000;
Wo = 0.0025;
c1 = Dw*S/rho/h;
c2 = (3*Wo/(4*Nm*3.14159*rho))^(1/3);
%initializations
twf(1)=1;
c3=@(Wwf) (3*Wwf/(4*Nm*3.14159*rho)).^(1/3);
rt = zeros(1,10800);
rt(1) = Wo;
for ii=2:10800
G = @(Wwf) ((c2 - c3(Wwf) - h*log((h + c2)./(h + c3(Wwf))))/c1)-(ii-1);
try
rt(ii) = fzero(G,rt(ii-1));
catch
disp('No value found, aborting....')
rt = rt(1:ii-1);
break
end
end
Deanna
Deanna 2011-4-13
Yes, thanks, that works as long as I end the loop early enough. For the code about 10800 is too many iterations.

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更多回答(1 个)

Deanna
Deanna 2011-4-13
The following works. For the constants I gave I need to end the loop at 2600. If the loop runs longer an error occurs.
clear all
Dw = 0.0000076;
S = 0.000239;
rho = 1.1;
h = 0.0030;
Nm=8477.7;
Wo = 0.0025;
c1 = Dw*S/rho/h;
c2 = (3*Wo/(4*Nm*3.14159*rho))^(1/3);
%initializations
c3=@(Wwf) (3*Wwf/(4*Nm*3.14159*rho)).^(1/3);
rt = zeros(1,2600);
rt(1) = Wo;
time(1)=0;
for ii=2:2600
G = @(Wwf) ((c2 - c3(Wwf) - h*log((h + c2)./(h + c3(Wwf))))/c1)-(ii-1);
rt(ii) = fzero(G,[0 0.0025]);
time(ii)=time(ii-1)+ii;
end

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