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clear ;
clc;
// Example 26.1
printf('Example 26.1\n\n');
//page no. 804
// Solution
printf('Table to carry out degree of freedom analysis:\n');
// Number of variables involved
printf('\nI. Number of variables involved.\n');
printf(' Species in F1 1 \n');
printf(' Species in F2 2\n');
printf(' Specie in F3 5\n');
printf(' Total stream flows 3\n');
printf(' Stream temperatures 3\n');
printf(' Stream pressures 3 \n');
printf(' Q 1 \n');
printf(' Extent of reactions 2\n');
printf('\__________________________________________________________________________________________\n');
printf(' Total 20\n');
printf('\n\nII. Number of equations and specifications.\n');
printf(' Independent species material balances 6\n');
printf(' Sum of species in each of the two streams 2 \n');
printf(' Energy balance 1\n');
printf(' Total stream flows 2\n');
printf(' Species values(CO) 1\n');
printf(' Pressures 3 \n');
printf(' Temperatures 2 \n');
printf(' O2 to N2 ratio specified in F2 1 \n');
printf(' Complete reaction, hence the extent of reaction is implied to both reactions 2\n');
printf('\___________________________________________________________________________________________\n');
printf(' Total 20\n');
printf('\n Therefore, by analysing the above table it is clear that degrees of freedom of system is (20 - 20) = 0 \n');
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