1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
|
//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
|