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diff --git a/code/fsolve/Redlich-Kwong.sce b/code/fsolve/Redlich-Kwong.sce
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+++ b/code/fsolve/Redlich-Kwong.sce
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+//Ref:Steven C. Chapra. 2006. Applied Numerical Methods with MATLAB for Engineers and Scientists. McGraw-Hill Science/Engineering/Math,Chapter 6
+//Example:
+//The Redlich-Kwong equation of state is given by
+//p = ((R*T)/(v-b) - a/(v*(v+b)*sqrt(T)))
+//where R = the universal gas constant [= 0.518 kJ/(kg K)], T = absolute temperature (K), p = absolute pressure (kPa),and v = the volume of a kg of gas (m3/kg). The parameters a and b are calculated by
+// a = 0.427*(R^2*Tc^2.5)/pc; b = 0.0866*R*(Tc/pc);
+//where pc = 4600 kPa and Tc = 191 K. As a chemical engineer, you are asked to determine the amount of methane fuel that can be held in a 3 m3 tank at a temperature of −40 ◦C with a pressure of 65,000 kPa. Use a root-locating method of your choice to calculate v and then determine the mass of methane contained in the tank.
+//Note: The initial guess has to be assumed by the user. An improper initial guess will result in deviation from the root. 
+//======================================================================
+// Copyright (C) 2018 - IIT Bombay - FOSSEE
+// This file must be used under the terms of the CeCILL.
+// This source file is licensed as described in the file COPYING, which
+// you should have received as part of this distribution.  The terms
+// are also available at
+// http://www.cecill.info/licences/Licence_CeCILL_V2-en.txt
+// Author:Debasis Maharana
+// Organization: FOSSEE, IIT Bombay
+// Email: toolbox@scilab.in
+//======================================================================
+
+clc;
+
+function y = Redlich_Kwong(v)
+
+    R = 0.518;T = 273+(-40);p = 65000;Tc = 191;pc = 4600;
+    a = 0.427*(R^2*Tc^2.5)/pc; b = 0.0866*R*(Tc/pc);
+
+    y = p - ( (R*T)/(v-b) - a/(v*(v+b)*sqrt(T)) );
+
+endfunction
+
+function dy =  dv_Redlich_Kwong(x)
+
+    dy = numderivative(Redlich_Kwong, x);
+
+endfunction
+
+// Set of initial values to check 
+x0 = [1 0.01 -1.0 -0.01] ;
+
+tol = 1D-10;
+
+for i =1:length(x0)
+
+    disp(x0(i),'Initial guess value is')
+
+    [x ,v ,info]=fsolve(x0(i),Redlich_Kwong ,dv_Redlich_Kwong ,tol)
+
+    select info
+    case 0
+        mprintf('\n improper input parameters\n');
+    case 1
+        mprintf('\n algorithm estimates that the relative error between x and the solution is at most tol\n');
+    case 2
+        mprintf('\n number of calls to fcn reached\n');
+    case 3
+        mprintf('\n tol is too small. No further improvement in the approximate solution x is possible\n');
+    else
+        mprintf('\n iteration is not making good progress\n');
+    end
+
+    mprintf('\n volume of metheane is %f m3/kg and Mass of methane is %f kg\n',x,3/x);
+
+    if i<length(x0)
+        input('press enter to check the next initial guess value')
+        clc
+    end
+
+end
diff --git a/code/fsolve/SLEs.sce b/code/fsolve/SLEs.sce
new file mode 100644
index 0000000..3daa1ee
--- /dev/null
+++ b/code/fsolve/SLEs.sce
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+//This problem solves sustem of nonlinear equations using fsolve
+//Reference: A. Golbabai, M. Javidi,Newton-like iterative methods for solving system of non-linear equations,Applied Mathematics and Computation,Volume 192, Issue 2,2007,Pages 546-551,ISSN 0096-3003,https://doi.org/10.1016/j.amc.2007.03.035.(http://www.sciencedirect.com/science/article/pii/S0096300307003578)
+//======================================================================
+// Copyright (C) 2018 - IIT Bombay - FOSSEE
+// This file must be used under the terms of the CeCILL.
+// This source file is licensed as described in the file COPYING, which
+// you should have received as part of this distribution.  The terms
+// are also available at
+// http://www.cecill.info/licences/Licence_CeCILL_V2-en.txt
+// Author:Debasis Maharana
+// Organization: FOSSEE, IIT Bombay
+// Email: toolbox@scilab.in
+//======================================================================
+clc
+//Function representing system of nonlinear equaltions 
+function f = SLEs(p)
+
+    x = p(1);y = p(2);
+    f(1) = (x+10)/x^4 + x^2 + 3*x*y + y^2-16;
+    f(2) =  1/(x+y)^7 + x*y - 1 - 1/2^7;   
+
+endfunction
+
+//Tolarence of solution 
+tol = 1D-15;
+
+//Initial guess
+x0 = [10 10];
+
+disp(x0,'Initial guess value is');
+
+//Obtaining solution using fsolve
+[x ,v ,info] = fsolve(x0,SLEs,tol)
+clc
+select info
+case 0
+    mprintf('\n improper input parameters\n');
+case 1
+    mprintf('\n algorithm estimates that the relative error between x and the solution is at most tol\n');
+case 2
+    mprintf('\n number of calls to fcn reached\n');
+case 3
+    mprintf('\n tol is too small. No further improvement in the approximate solution x is possible\n');
+else
+    mprintf('\n iteration is not making good progress\n');
+end
+disp(x,'The solution is ')
+disp(v,'the function value at solution')
+
+